Transcript for:
AP Calculus AB and BC Free Response Insights

hi there I'd like to tell you a few things about using and navigating this video though if you're in a rush you can look at the video chapters and Skip around as you please this video consists of 55 fully solved AP Cal ab and BC free response questions each solution has been checked with the College Board scoring guidelines to ensure that they would all earn full marks the problems have been sorted by topic and within each topic have been sorted by year naturally each free response question meaningfully deals with multiple topics in the AP Calculus curriculum so in the pin comment you can find timestamps just like the video chapters but with a little bit more detail to better help you in finding questions relevant to your needs that pinned comment will also contain notes and Corrections regarding any errors found in the video after its release though there should not be any meaningful errors because I have watched the entire video to check each topic is also preceded by a brief intermission where I'll try to give you some helpful advice for the upcoming section there's a link in the description to download a free pdf that contains all problems we've solved in this video in their proper order if you're looking for more General calculus content there are links in the description to my Calculus 1 course and Calculus 1 exercises playlists which together contain hundreds of additional videos if you have any questions about the solutions presented in this video please don't hesitate to leave a comment but please do include a Tim stamp lastly I've spent hundreds of just on hard drive storage while creating this video so if you find it helpful I hope you'll consider supporting What I Do by joining Wrath of maath as a channel member for access to exclusive and early videos you can also donate on patreon or PayPal there are links in the description and every bit helps good luck studying good luck on the exam and thanks for watching first up is the graph analysis section I encourage you to be very aware of what graph you're actually looking at is it the graph of the most important function in the problem or is it the graph of that function's derivative or is it the graph of some other function entirely also recall that the slope of the graph at any given point is the value of the function's derivative at that point and that the area underneath the curve in the graph is the value of an integral this is problem five from the 2010 AP Cal AB exam this was part of the no calculator section we got a nice picture here let's take a look at what the question is the function G is defined and differentiable on the closed interval from -7 to 5 and satisfies G of 0 equals 5 the graph of G Prime of X the derivative of G consists of a semicircle and three line segments as we can see here so make sure you notice this is the graph of G Prime not the graph of G part A is to find G of 3 and G of -2 let's start off with G of 3 this is pretty simple to find if you remember how integrals and derivatives work we have a given value of G which is G of0 we know G of 0 is five so we can find G of three if we take five and then add the sum of the change of G from0 to three by that I mean we have to add the integral of G Prime of X DX from zero because it's at zero that we know G is five and that's our only known value so we have to start there and then we accumulate these changes these G primes up to three the desired value if this isn't clear here at a glance you might be more comfortable noticing that the integral of G Prime of x from 0 to 3 just this part here is clearly going to be G of three we're undoing the derivative because of the integral and then plugging in the upper bound so G of 3 minus G of 0 right and then well G of 0 we know is five so if we have this addition of five on the left that's going to cancel out with the G of0 and all that's going to be left is G of 3 clearly the equation is true then G of 3 equals g of 3 exactly what we want now let's go ahead with evaluating this we're going to have 5 plus this integral now to evaluate this integral we can look at the given graph of the derivative the integral of the derivative gives us the area under the derivative so we can calculate this integral by calculating some area in particular we need the area under the curve the area under the derivative from X = 0 up to x = 3 which is right there so we need all of this shaded area and we can see that this sh uh this shaded area is made up of a quarter Circle half a semicircle and this right triangle the area of this quarter circle is 1/4 the area of the whole circle the area of a whole circle is pi r 2 and R the radius in this case is 2 so this is going to be < * 2^ 2 or < * 4 but we only want the area of a quarter of this circle so we have to multiply it by a fourth so that's this area here now we just need the area of this right triangle that of course is 1 12 base time height the base from here to here is just one and the height we see is three because that's the y-coordinate of that upper point so the base of one and the height of three and then we just have to simplify this five is the same as 10 Hales I'm writing it in terms of halves because here at the end we have three halves so let's add three halves over there and then the pi term will'll leave at the end 1/4 piun * 4 is just a whole Pi because the 4th and the four cancel out finally 10 + 3es is 13 Hales so our final answer is 13 Hales + Pi again we start at the known value we know that g of 0 is five and then we have to accumulate G's changes by integrating its derivative from zero where we started to 3 if we do that we get G of 3 it's 13 plus pi we're not done with a yet though because we also have to find G of -2 we find G of -2 in the same way take the point we already know five right G of 0 is 5 and then we need to accumulate backwards to -2 so this is the integral from zero where we are starting to -2 we're adding up those rates of change so we're integrating G Prime of X with respect to X this is just going to be five and then the integral of G Prime again is the area under the curve but because we're going backwards from 0 to -2 which I'll shade in blue that's this area here from 0 to x = -2 because we're going backwards we're just going to have to hit this area with a negative sign this area we can see is again just the area of the same sort of quarter Circle we already calculated that area as Pi so this is just NE Pi again it's negative because we're going backwards from 0 to -2 so this is going to be 5 + piun basically what's happening is that we're starting at G of 0 which we know to be five and then by integrating backwards we're undoing the change that occurred from -2 to 0 and thus that gives us G of -2 which we find is just 5 minus pi and now we can move on to Part B find the x coordinate of each point of inflection of the graph of yal G ofx on the interval from -7 to 5 and explain our reasoning in order to find the points of inflection of G of X we're going to have to look at the graph of the derivative a point of inflection is where the derivative goes from increasing to decreasing or decreasing to increasing inreasing so by looking at the graph we can see that those occur at xal 0 2 and 3 at xal 0 the derivative goes from increasing to decreasing at x = 2 the derivative goes from decreasing to increasing and at xal 3 the derivative goes from increasing to decreasing if we look at another Point like x = -2 the derivative is increasing on both sides of -2 so that would not be a point of inflection just for a counter example and there it is written out again we find the points of inflection by looking at where the derivative switches from decreasing to increasing or from increasing to decreasing so those are our points of inflection at xal 0 2 and three moving on then to part C the function H is defined by h of x = g of x - a half x^2 we want to find the x coordinate of each critical point of H where X is between -7 and 5 and classify each critical point as the location of a relative minimum relative maximum or neither and explain our reasoning let's begin by taking the derivative of H so that we can find the critical point since h ofx = g ofx - half x^2 H Prime the derivative is going to be G Prime of X and then apply the power rule for this part that's going to be Min - x Okay g Prime of x - x let's write that down here for part c h Prime of X is G Prime of x - x now to find critical points we need to see where this derivative either doesn't exist or where it's equal to zero we're only interested in the interval between -7 and 5 and we were told that g is differentiable on that entire interval and of course there's no problem subtracting X on that interval either so there's nowhere on the interval where the derivative doesn't exist so the only critical points are going to be where this derivative is equal to zero now if this equals z g Prime of x - x equals 0 then of course we have that g Prime of X would equal x so where does that occur it may help to think about this geometrically G Prime of x = x wherever the derivative G Prime intersects the line Y = X so let's come back to our graph of G Prime and graph the line Y = X the line Y = X passes through 0 0 it passes through 1 one it passes through 33 so the line should look something like like this the points we're looking for are where this derivative the graph in Black intersects this line yal X that would be where G Prime of xal X we can see clearly this occurs at the point 33 that's a point where G Prime of x equals x because they're both three and it also appears to occur at one point on this semicircle and nowhere else now to figure out exactly where it occurs on this sem Circle it will help to write an equation for it so you have to know how to write the equation of a circle centered at the origin with a radius of two the equation of a circle centered at the origin with radius 2 is right here we're only interested in the semicircle that's above the x axis we can find that by just solving for y and that's going to be the square < TK of 4 - x^2 so the Der G Prime of X between X being -2 and 2 the derivative is this the square TK of 4 - x^2 that's an equation for this semicircle here again this is between X being ne-2 and pos2 now with this equation we can figure out at exactly what x coordinate it intersects the line Y = X so we'll just write x = the square < TK of 4 - x^2 and solve this to do that we'll square both sides so x^2 = 4 - x^2 add x^2 to both sides so 2x^2 = 4 divide both sides by 2 so x^2 = 2 and then we find that x = the square TK of 2 it's not the case that x equal < tk2 even though that would satisfy this equation we can see just from our picture that the point we're looking for has a positive x coordinate thus where does G Prime of x equal x where are these critical points that would be at x = 3 and as we just found X = < tk2 now we just have to classify these critical points in order to classify the critical points we need to think about the behavior of this derivative around these critical points if the derivative switches from positive to negative for example that means that the function H switches from increasing to decreasing and a point like that would be a maximum so let's take a look back at our picture and focus on these critical points x = < tk2 and x = 3 if we look at X = < tk2 we can see that to the left of this so over here to the left of this the derivative G Prime is greater than x right it is the upper function so if we were to look at G Prime of x minus X which is H Prime right H Prime is G Prime - x clearly to the left of < tk2 that's going to be positive because G Prime we see is bigger than x okay so it's positive there now what about to the right well to the right of < tk2 X is above G Prime so to the right of < tk2 it's negative and it stays that way afterwards we see that y = x remains above G Prime so at < tk2 the derivative H Prime switches from positive because this is bigger than this to negative because now this G Prime is smaller than this and here that reasoning is written out we could clearly see that between 0 and < tk2 G Prime of X is greater than x and thus H Prime of X is positive on the other hand between < tk2 and 5 we found that H Prime of X is less than or equal to zero it is nonpositive because G Prime of X was less than or equal to X we see G Prime is underneath y = x okay so at < tk2 it switches from positive to nonpositive thus H has a relative maximum at xal < tk2 it's positive derivative so it's increasing and then it's either staying the same or decreasing based on our evidence and neither a minimum nor a Maxum um at x = 3 that's because at x = 3 that's in this interval and we see that the derivative doesn't switch sign in that interval throughout that entire interval which contains x = 3 the derivative H Prime is less than or equal to zero so there's no switching of signs in fact the only time the derivative is zero on this interval from < tk2 to 5 is at that critical point xal 3 everywhere else in this interval H Prime is negative so certainly that would not be xal 3 the location of a Min or a Max and that completes our solution to free response question five from the 2010 apal AB exam this is free response question three from the 2016 AP Cal AB exam no graphing calculator allowed we're not going to need one let's take a look at the problem the figure above shows the graph of the piece wise linear function f so each separate piece here is a line for X between -4 and 12 the function G is defined as an integral of the function f from T to X part A asks does G have a relative minimum a relative maximum or neither at xal 10 and we need to justify our answer in order to determine if there is a relative minimum or a relative maximum here at x = 10 remember we're talking about the function G so we need to consider G's derivative and since G is defined as the integral of f ofx the derivative of G is actually just F ofx and so what we have here is a graph of the derivative of G so this is a lot of information so here for part A we know that g Prime of X is actually F ofx because G is defined as an integral of f of T so when we look at this graph we're looking at G Prime and we're asking does it have a minimum or a maximum or neither here xal 10 well for it to have a minimum or a maximum G's derivative would have to go from positive to negative meaning it's increasing and then decreasing or it would have to go from negative to positive meaning it's decreasing and then increasing and if we look at xal 10 here on this graph the function just is negative the whole time it doesn't switch signs and again this is the derivative of G because it's F ofx and so in fact G does not have an extreme value at xal 10 because its derivative which is what we have a graph of does not change sign at xal 10 and there is our answer written out the function G has neither A Min nor a Max at xal 10 because the derivative of G which is f ofx is less than or equal to zero between here and here so when it passes 10 it doesn't switch signs it's just negative that whole time so no minimum no maximum on to Part B does the graph of G have a point of inflection at xal 4 and we need to justify our answer a point of inflection occurs when the second derivative switches sign which means the first derivative would have to switch from increasing to decreasing or from decreasing to increasing now we have a graph of the derivative here so let's look at xals 4 that's right here does the derivative switch from increasing to decreasing or vice versa yes it does because this graph is the derivative we see that it's increasing here and then right at xal 4 it switches to decreasing so in fact there is a point of inflection in the graph of g at xal 4 there is our solution the function G does have a point of inflection at xal 4 because G Prime of X is f ofx and we see that F ofx switches from increasing to decreasing at xal 4 so yes there is a point of inflection on to part C part C asks us to find the absolute minimum value and the absolute maximum value of G on the interval from4 to 12 and to justify our answers now to find the absolute Max and Min we want to make sure we check all candidates so let's first note where the relative mins and Maxes are that would be where the derivative switches sign and we see that happens at xal -2 it switches sign at xal POS 6 it switches sign and that's it but we also have to check the end points on this interval -4 and 12 so we'll check the value of g at all of these candidates the end points and where the derivative switches sign and we'll see where the Max and where the Min occur here is a table to organize our information for part C we need to check the value of g at these four candidates beginning with xal -4 by definition G of -4 is the integral from pos2 to -4 of f of T DT so this is the area underneath the curve from 2 to -4 which looks roughly like that now this is moving in a backwards direction from 2 to -4 so what we'll do is just throw a negative in front and then we can calculate this as if we're moving in a positive direction from -4 to pos2 First going from -4 to -2 we need the area of that triangle there which is2 multipli the base of two multiplied by the height of --4 and then we have to add the area of this triangle from -2 to pos2 that's going to be 1/2 multiplied by the base which is four multiplied by the height which is four and then just do the arithmetic this is going to be -4 so G of -4 is -4 let's move on then to G of -2 by definition G of -2 is going to be the area under F from pos2 to -2 so we just need to find the area of that triangle and then multiply it by1 because again we're moving in a backwards Direction all right so that's just going to be 12 multiplied by the base the base is -4 because we're going from the positive to the negative so -4 and then multiply by the height which is four like I said we need to multiply the area of the triangle by1 which is what we're doing right there we're treating That Base as a Nega all right so this turns out to be then 8 so G of -2 is8 now G of 6 if we look at the picture that's pretty easy G of 6 is just going to be the area underneath the curve from 2 to 6 which is just the area of that triangle there which is just 1/2 multiplied by the base of four multiplied by the height of four and that is equal to 8 so G of 6 is 8 lastly we have to evaluate G of 12 12 which is going to be the area underneath the curve from pos2 all the way up to pos2 so it will look something like that now these two triangles are actually just going to cancel out so we can just focus on this area here between 10 and 12 that's going to be 1 12 multiplied by the base of 2 multiplied by the height of -4 and so this turns out to be -4 so G of 12 is-4 thus by looking at the table we checked all the possible candidates we can see that the absolute maximum is 8 which occurs at xal 6 so the absolute Max is 8 which occurs at x = 6 and the absolute minimum on this interval occurs at -2 where we have a minimum of8 absolute minimum is8 at X x = -2 finally Part D for X between -4 and 12 so this whole interval we want to find all intervals for which G of X is less than or equal to zero now remember G of X is the area under the curve but starting at two so for example this area of the triangle appears that it should be positive because it's above the x-axis however because we're moving in this direction in the negative way because we're starting at two so we're moving in the negative Direction This positive area is actually negative so all of this space is a place where G of X would be negative now some of that negative area would start to get cancelled out because this area that actually looks negative is going to be counted as positive because we're moving in the backwards Direction however even though this cancels out some of this it doesn't cancel out all of it so all the way from pos2 to -4 G's going to be negative that whole time we can think through a similar thing going the other direction starting at two if we go this way G is positive this whole time because the area is above the x-axis and we're moving in the positive direction but that positive area gets exactly canceled out by this rectangle here so by the time we get to xal 10 G is zero and then afterward it's going to be negative because this area here will be negative and we're traversing it in the positive direction so in total G is going to be negative from 2 all the way over to -4 and it's going to be negative from 10 to 12 so for Part D we can say that g of X is less than or equal to zero is that what the question asked yes less than or equal to zero g of X is less than or equal to zero when X is like we said between 2 and -4 so we'll say when X is less than or equal to positive2 greater than or equal to -4 and it is also less than or equal to zero when X is greater than or equal to 10 and less than or equal to 12 so we'll say it's also less than or equal to Zer when X is greater than or equal to 10 and less than or equal to 12 and there's one more look at our solution that completes free response question three from the 2016 AP Cal AB exam this is free response question six from the 2017 AP Cal AB exam this is part of the no calculator section and there's quite a bit going on here it says let F be the function defined by F ofx = cosine of 2x+ e to the sin x let G be a differential iable function and we have a table right here giving values of G and its derivative G Prime at a few values of X finally let H be the function whose graph consisting of five line segments is shown in this figure here so we have one function given algebraically one function in a table and one function in a graph part A find the slope of the line tangent to the graph of F at xal pi to find the slope of a tangent line we just need to take the derivative so we'll find F Prime and then we'll plug in x = Pi so here for part A the slope M of the tangent line is f Prime the derivative of our function evaluated at PI right we're looking for the slope of the tangent line at xal Pi so let's take the derivative of F and plug in pi the derivative of cine 2x is negative sine of 2x but by the chain rule we also have to multiply by the derivative of the inside function the derivative of 2x is 2 so we multiply by two out front so -2 sin of 2x that's the derivative of cosine of 2x we also have to add the derivative of e to the sinx the derivative of e to the power of a function is just e to the power of that function multiplied by that function's derivative the derivative of sinx is cosin X so this is our derivative F Prime but we want to evaluate it at Pi so let's go ahead and replace these X's with pies which of course is going to work very nicely with these trig functions all right after plugging in pies what do we get well s of 2 pi is zero so this whole term goes away sine of Pi up here is also zero so this is e to the power of 0 which is one so all that's left is cosine pi and surely you can remember your unit circle cosine Pi is over there where the x coordinate is -1 so that's just going to be then -1 that is the slope of the tangent line at xal Pi the tangent to this function f all right Part B let K be the function defined by K of xal H of f ofx f ofx is the function we were just working with h remember is the function shown in this graph the question is to find K Prime of Pi so let's go ahead and write Part B here we are looking for K Prime of Pi so let's just find K Prime of X and then we'll plug Pi in for x k is a composite function it's h of f ofx so to find the derivative K Prime we need to use the chain rule the chain rule begins with the derivative of the outside function the outside function is H so that's H Prime of leave the inside function unchanged then you just have to multiply by the derivative of the inside function in this case that's F Prime of X all right but we of course want to evaluate this derivative at Pi so K Prime of Pi is going to be H Prime of f of pi multiplied by F Prime of Pi but we just evaluated frime of Pi in part A we know that's 1 so let's just replace frime of Pi with1 all right now let's see what is f of Pi well looking up at our function what F ofx actually is we can see that F of Pi would be cosine of 2 pi plus e to the sin Pi now cosine of 2 pi is the same as cosine of 0 which is 1 and e to the sin Pi is e to the 0 which is also 1 so this is 1 + 1 or 2 so we can rewrite this as H Prime of f of Pi is 2 so this is H Prime of 2 multiplied by -1 so we'll still put that ne1 there now what is h Prime of 2 well we'll have to look at our graph to figure that out since we want to know the derivative of H at 2 we're looking right here so here's x = 2 and we're focusing on this line segment the derivative of H is the slope of H which is easy to figure out when H just behaves like a line what is the slope of this line well we see that when it goes down one it goes over 1 2 3 rise over run so the slope is- 1/3 down one over 3 okay so H Prime of 2 the slope of the function H at xal 2 is just the slope of this line which is - 1/3 so this is -1 which I'll move to the front1 * - 1/3 which of course is just positive 1/3 all right let's move on to part C let M be the function defined by MX = G of -2X * H ofx we want to find M Prime of 2 so this is a lot like the previous questions except now we're getting the G function involved which is in this table and we're going to have to use the product rule because of course we're now dealing with a product of functions we're dealing with G * H so let's put problem C down here and we're trying to evaluate M Prime of 2 let's begin by just finding what M Prime of X is and then we'll plug in xal 2 we of course need to use the product rule U Prime V plus v Prime U so looking at this product let's start off with the derivative of G the derivative of G with that -2X inside of it is going to be G Prime of -2X and then multiply by the derivative of the inside function the derivative of -2X is -2 so that would be the U Prime in the product rule U Prime V so we also have to multiply by the second function here h of X U Prime V plus v Prime u v Prime is the Der of the second function so H Prime of x times the first function unchanged G of -2X all right that's M Prime of X and now we just need to plug in x = 2 what is M Prime of 2 well it's going to be -2 multiplied G Prime of -4 because that's -2 * 2 multiplied by h of 2 plus h Prime of 2 uh but h Prime of 2 we already figured out was - 1/3 that showed up in the previous problem H Prime of 2 is 1/3 so let's just replace that with 13 get a little head start there and then we have G of -4 all right hopefully we can figure out what all of these values are let's consult the table for the values of G and G Prime G Prime of -4 we see is -1 and G of -4 is pos5 so we can rewrite this expression it's going to be -2 * G Prime of -4 which we know is -1 so -2 * -1 multiplied by h of 2 + - 13 * G of -4 which we see is pos5 so this is times pos5 now we can simplify and also figure out what H of2 is for starters we have -2 * -1 which is just pos2 and then looking at the graph what is h of 2 xal 2 is right there and so h of 2 is the Y value right there now how do we know what the Y value here is well we know that this line segment starts at the origin 0 0 and we already said that its slope is - 1/3 so if we move over to units we must go down 2/3 so the y coordinate is -2/3 that is H of 2 so we'll replace that right there we're replacing that with -2/3 and then over here on the right we have - 1/3 * 5 so that's - 5/3 finally this is -4/3 - 5/3 so that's 9/3 which is just -3 that completes part C so let's finish with part D is there a number c in the closed interval from -5 to -3 such that g Prime of C is equal to-4 and justify our answer for this part D we must use the mean value theorem we know that g is differentiable and thus it's also continuous on the closed interval between -5 and -3 we are told as much G is a differentiable function and we have some values of G and its derivative that's what's going to let us use the mean value theorem since G is differentiable and continuous from -5 to3 the mean value theorem tells us whatever G's average rate of change over this interval is there must be some point in the interval where its instantaneous rate of change its derivative equals the average rate of change so let's calculate G's average rate of change from X5 to x = -3 to do this we'll set up a fraction the calculation is as follows G of -3 the ending value minus the starting value G of5 this tells us how much G changed over the interval and then we have to divide by the interval width which is just the end point -3 minus the beginning Point -5 all right so this is how much the function changed divided by the interval width G of -3 we see from the table is pos2 and G of5 is pos1 so this is 2 - 10 and then the denominator -3 -5 is -3 + 5 which is just 2 we see then that this is 2 - 10 which is8 / pos2 so this is4 since the average rate of change over the interval is -4 We Know by the mean value theorem that yes there does have to be some point C in this interval such that the derivative of g at this point C is -4 the instantaneous rate of change has to equal the average rate of change at some point on this interval and there is that conclusion written out and that completes our solution to free response question six from the 2017 AP Cal AB exam next up is the continuity section for this section just recall the definition of continuity it's important you know how to use it we say that a function is continuous at a point x C if the limit of the function as X approaches C from the left equals the limit of the function as X approaches C from the right and both of those limits equal the function's value this is free response question six from the 203 AP Cal AB exam let F be the function defined by this piecewise expression F ofx = the < TK of x + 1 for X values between 0 and 3 inclusive and for X greater than 3 but less than or equal to 5 FX = 5 - X part A asks is f continuous at xal 3 explain why or why not by definition of continuity in order for f ofx to be continuous at xal 3 the limit of f ofx from the left must equal the limit of f ofx from the right and they both must equ equal the function's value at x = 3 let's begin by evaluating the limit of f ofx as X approaches three from the left as X approaches three from the left the behavior of f ofx is this the sare < TK of x + 1 because X approaching 3 from the left means that X is less than 3 now the < TK of x + 1 is continuous at xal 3 so to evaluate this limit we can just plug in x = 3 so the limit of f ofx as X approaches 3 from the left this is the limit as X approaches 3 from the left of the square < TK of x + 1 which is just the square root of 3 + 1 which is the < TK of 4 which is of course 2 it's also worth pointing out that this is the function's actual value at three because this is the function's behavior when X is less than or equal to 3 from the left the function's limit is 2 which is also the function's value at three now we just have to check the limit of the function as X approaches 3 from the right when X approaches 3 from the right X is greater than 3 and so the behavior of the function is 5 - x clearly then the limit of the function as X approaches three from the right is 5 - 3 which is 2 so yes indeed this is a continuous this function at x = 3 Because as X approaches 3 from the left the function approaches 2 the function's actual value at three is two and the limit of the function as X approaches 3 from the right is two so yes it is continuous therefore I'll use the three dots for therefore f ofx is continuous at x = 3 moving on to Part B find the average value Val of f ofx on the closed interval from 0 to 5 to find the average value of a function over an interval we just accumulate the function over that interval which means we integrate it and then divide by the interval length it's like adding up the values in a data set and then dividing by the number of values so what we need to do is integrate our function f of x on the closed interval from 0 to 5 so we integrate from 0 to 5 and then of three halves and then divide by the new Power which means we multiply by its reciprocal and we'll have to evaluate that from 0 to 3 and then we have this other integral the integral of 5 - x that's just going to be 5x - 12x^2 again we're just using the power rule and we are evaluating that from 3 to 5 all right so let's begin to evaluate this we have 1/5 out front and then plugging three into this expression gives 2/3 multiplied 4 to the 3 4 to the 3es you can think about as applying the square root first so the square root of 4 which is 2 and then cubing it which is 8 so 2/3 * 8 so 16/3 and then if we plug in zero we are just going to get 2/3 so 16/3 - 2/3 then we have our other integral plugging five into to this expression is going to give us 25 - 25 over 2 and that is just 25 over two so we will just write 25 over two right if you have 25 and you take away half of 25 you get half of 25 and then if we plug in three we will get 15 minus 9 / 2 15 - 9 / 2 15 is the same as 30 / 2 so so this is 30/2 - 9 /2 which is 21/2 and because this resulted from plugging in the lower bound we need to subtract it so minus 21 / 2 now we can finish this up we still have 1/5 out front 16/3 - 2/3 is 14/3 and then 25es - 21es is 4 Hales so plus 4 Hales now 4 over 2 is the same as 2 and two is the same as 6 over3 so that we have common denominators now this then is 1/5 multiplied 20 over3 and then we can cancel a factor of five out of the numerator and denominator giving us a final answer of 4/3 and that's the average value of the function over the closed interval from 0 to five moving on to part C suppose the function G is defined by and here we have a peie wise function similar to F ofx but a little bit different and K and M here are constants we're told that G's differentiable at xal 3 so using that information we must find the values of K and M now since G is differentiable at xal 3 firstly we know that means that g is continuous at xal 3 so the limit from the left must equal the limit from the right which means these two expressions have to equally each other at xal 3 but also G is differentiable at xal 3 so if we take the derivative of the top and the bottom those derivatives must equal each other at xal 3 there can't be a sudden change in the slopes at xal 3 since G is differentiable so let's put all this stuff into equations and get to solving beginning with continuity like we said since G's differentiable at xal 3 that implies G must be continuous at xal three so the limit from the left must equal the limit from the right now the limit from the left is found by looking at this expression and the limit from the right concerns that expression so let's plug in xal 3 to both of these expressions and set them equal to each other the first one becomes K multiplied the square < TK of 3 + 1 so K multiplied the < TK of 4 which is just 2K and because G's continuous here it must equal this guy with x = 3 plugged in which is just 3 m + 2 we will use this equation but before we do let's go back to differentiability not only must the limit of G of X as X approaches 3 from the left equal the limit of G of X as X approaches 3 from the right but this must also be true about the derivatives G Prime because G is differentiable at xal 3 so let's take the derivative and see what equation that gets us taking the derivative of this part of G produces k / 2 multiplied x + 1 to the2 we're just using the power rule and keep in mind we're differentiating with respect to X not with respect to K or M K and M are both constants the derivative of MX + 2 of course is just m so a x = 3 it must be the case that these two things are equal now plugging in x = 3 makes x + 1 become 3 + 1 which is just 4 k/ 2 * 4 to the - 12 is K over 2 * theare < TK of 4 and so that's just 2 * 2 so K over 4 because of course 4 to the 1 half that's just 1 over 4 in a square root so that's why you have that rooot 4 in the denominator so m equals K over4 because G is differentiable at xal 3 now we can plug that equation into this one which resulted from continuity and then we'll be able to solve for everything so we have that 2K is equal to 3 * m but in terms of K that's 3 K over 4 and then plus 2 then we can subtract 34s K from both sides 2K is 84s K so in total we have 54s k equal 2 multiply both sides by 4 over 5 and we get that k equal 85s plugging this into our equation for M gives us that m equal 8 over 20 which is the same as 2 over 5 and so that's our value for K that's our value for M and that completes our solution to free response question 6 from the 2003 apal AB exam this is free response question two from the 2011 AP Cal AB formb exam this is part of the graphing calculator section A 12,000 L tank of water is filled to capacity at time T equals z water begins to drain out of the tank at a rate modeled by R of T which is measured in L hour and R is given by this piece-wise defined function so the way in which water is draining out of the tank changes at t equal 5 you can see that Sudden Change there part A asks is R continuous at tal 5 and show the work that leads to our answer in order to determine if R is continuous at T equals 5 we need to see if the limit from the left which is this Behavior is equal to the Limit from the right which has this Behavior here these limits are written out the limit of R of t as T approaches five from the left based on the definition of R of t r of T is equal to this when T approaches five from the left because when T approaches five from the left T is less than or equal to five on the other hand when T approaches five from the right the behavior of R of T is this because when approaches 5 from the right T is greater than 5 now we have to evaluate and compare both of these limits for this one we can just substitute 5 into the function 600 * 5 is 3,000 so this is just 3,000 divided by 8 8 goes into 225 times which means it goes into 1,125 Times which means it goes into 3,300 75 times now on the bottom we can also just substitute five into this function and we'll bust out a calculator for that this is about 367.8 if you plug this into a calculator so we can see as we approach five from the left we get 375 375 is also the value of the function at t equal 5 you can see this is T less than or equal to 5 we have that behavior so so this is also R of five but that's not equal to the limit of the function as T approaches 5 from the right as T approaches five from the right the limit is about 367.8 they're not equal hence R of T is not continuous at tal 5 part B says find the average rate at which water is draining from the tank between tal 0 and T = 8 the rate at which water is drain in from the tank is this pie wise function so to find the average rate we just need to integrate this function over the interval in question from 0 to 8 and divide by the interval length which is 8 so for Part B we need to integrate our function R of T that's the rate of change of water in the tank we need to integrate this from 0 to 8 and then divide it by 8 dividing it by 8 is the same as multiplying by 1/8 so let's go ahead ahead and do this this integral though because the function is piecewise we're going to have to split it up at the point where Its Behavior changes at tal 5 so we will write the integral from 0 to 5 and from 0 to 5 that's the behavior of the function 600t over t + 3 and then separately we will add the integral from 5 to 8 and from from 5 to 8 the behavior of the function is this 1 e to the .2t and we will consult the graphing calculator for this I'll save the multiplication by 1/8 for the end I'm going to press math and then option nine the integration function we are integrating from 0 to 5 for this first integral what we're integrating is 600 * T but it's more convenient to use X on the calculator so 600x / x + 3 using a lot of parentheses here this integral of course is with respect to X and then we're adding the second integral so again I'll press the math button and option nine for the integration function integrating from 5 to 8 1,000 multiplied by E whoops let me rewrite that e to the power of .2x instead of T into grading with respect to X we get about 264 and then I'll just divide this by 8 so then our answer is about 28.5 3 and remember this is an average rate of change of water so this is in uh lers per hour right that was the volume and time measurement we're using so liters hour all right let's move on to part C part C says find our Prime of three using correct units explain the meaning of that value in the context of this problem so we can take the derivative of r at tal 3 the behavior of r at tal 3 is given here so we could take the derivative of that using the quotient rule but we might as well avoid the trouble and just use the calculator since we can as far as the interpretation which we will write out what the meaning of this is this is the rate of change of the rate at which water is draining out of the tank water is draining out of the tank but not at a constant rate that rate is changing and R Prime is how much the rate is changing it's the rate of change of the rate of change so coming back over to the calculator in order to calculate RP Prime of 3 again I'll press the math button and then option 8 for the derivative function we're taking the derivative with respect to X we're going to use x here in the calculator and because T equals 3 we're looking for the derivative at tal 3 this is the expression we're using so 600x / x + 3 and we're evaluating this at 3 and we get about 50 and because this is the rate of change of the rate at which water is draining out of the tank this is lers per hour per hour so lers per hour squared and here is our written interpretation of this derivative value at time t equal 3 hours the rate at which water is draining out of the tank which is R oft that rate is increasing because the derivative is positive it's increasing by 50 L per hour squared more water is draining out of the tank more quickly finally Part D write but do not solve an equation involving an integral to find the time a when the amount of water in the tank is 9,000 L let's do Part D up here where all the information is conveniently available to us we know the amount of water in the tank began at 12,000 L because it said it was filled to capacity so it must be that 12,000 minus the accumulation of the rate at which water is draining out of the tank as in the integral the integral of that rate at which water is leaving from sometime zero or I should say the time zero the time at which water begins draining to sometime a that's the unknown this must equal 9,000 right if water drains from T equals 0 to this unknown time a then eventually there will be 9,000 L of water left in the tank and we do not have to solve this so we are done there just explaining this again the initial amount of water in the tank is 12,000 L we subtract the accumulation of the rate at which water is leaving from zero to the unknown time a and the time a is some time that satisfies this equation that if we were moving water for this long we would have 9,000 L left and that completes our solution to free response question 2 from the 2011 AP Cal AB formb exam this is free response question 4 from the 2012 AP Cal AB exam let's read it the function f is defined by F ofx = the S < TK of 25 - x^2 which you may recognize as the equation of the top half of a circle centered at the origin with radius 5 and X is between -5 and 5 part A is very straightforward we're just looking for the derivative F Prime of x so let's go ahead and find the derivative of f ofx in order to do that it may help to rewrite F ofx first note that a square root is the same as a power of 1/2 so the square root of 25 - x^2 is the same as 25 - x^2 all to the power of 1 12 and this makes it more clear that to find the derivative F Prime we simply need to use the power rule and the chain rule we have to use the chain rule of course because f ofx is a composite function the outside function is this power of 1/2 how do you take the derivative of something to the 1/2 well you have to bring that power of 1/2 down as a factor leave the inside function unchanged 25 - x^2 and then reduce the power by one so now the power is half that's just the power rule but then by the chain rule because we have a function inside of this 25 - x^2 we also need to multiply by the derivative of that function and the derivative of 25 - x^2 is -2X and this is frime of X although there is a little bit of simplification we can do we can bring the negative to the front and of course the 1/2 and the 2 cancel out so there is our answer frime of x = --x x * 25 - x^2 the2 again we wrote the negative and the X out front and the half and the 2 cancel out so that's our answer for a Part B write an equation for the line tangent to the graph of F at xal -3 okay for a tangent line we need two things a point and a slope for the slope we will be able to use the derivative that we just found and the point is even easier than that the point point we know has an x coordinate of -3 to find the y-coordinate we just need to plug -3 into our function because we know the tangent line has to meet the function at x = -3 so plug -3 into the function to see what the y-coordinate has to be here's our function let's plug -3 into it f of -3 is going to be the square root of 25 - -3 2 is 9 so this is the square < TK of 25 - 9 which is the < TK of 16 which is 4 so the point that our line is tangent 2 is -34 that's where our tangent line is intersecting the curve we can find the slope of the tangent line at this point by plugging -3 into the derivative which we just found so let's plug -3 into this derivative that's going to be -3 remember there's that out front and then in parentheses we have 25 minus -3 2 which is just 9 and this is to the power of2 now what is this well the negative and the negative in the front cancel out so we just have a three and let's go ahead and write this as a fraction this has a negative power 25- 9 so I'm going to give it a positive power by putting it in the denominator then we would have this three in the numerator and in the denominator we would have 25 - 9 to the2 25 - 9 though is 16 so that would be 16 to the 1/2 but a power of 1/2 is the same as a square root so this is the same as 3 over the square root of 16 the square root of 16 of course is Just 4 so this is 3 over 4 that is the slope of our tangent line then we can use point slope form to write the equation of the tangent line Y minus the y coordinate which we found to be 4 equals the slope which we found to be 3 over 4 multiplied x minus the x coordinate which of course is -3 but x - -3 is the same as x + 3 so let's just write it that way because that looks nicer and that's it for Part B moving on to part C let G be the function defined by g ofx equals a pie wise function it is equal to F ofx when X is between -5 and -3 inclusive and it's equal to x + 7 when X is greater than -3 and less than or equal to pos5 the question is is G continuous at xal -3 which is where its definition switches atg -3 it switches from F ofx to to x + 7 so is it continuous there to answer this we must use the definition of continuity the definition of a function being continuous at a point let's say -3 in this case is that the limit of G ofx as X approaches -3 from the left should equal the limit of G of X as X approaches -3 from the right and those one-sided limits both have to equal the function's value at -3 that's what it would mean for G to be continuous at -3 now when X approaches -3 from the left G of X behavior is f of x because X approaching -3 from the left means that X is less than -3 so the behavior of G ofx would just be F ofx now what is it approaching then from the left well we can find that by just plugging -3 into F ofx we've already done that and we know that it would equal 4 so the limit of G of X as X approaches -3 from the left is 4 now what about from the right if x is approaching -3 from the right then the behavior of G is here because it is here where X is greater than -3 so it can approach -3 from the right the behavior of G of X is that it's x + 7 so what's the limit from the right well plug in -3 -3 + 7 is pos4 which was the same as the limit from the left finally if we actually plug -3 into the function G what does it actually equal at -3 well it is equal to F of -3 because remember it's on this piece of the function where X is allowed to equal -3 so G of -3 we see is just F of -3 which is 4 so yes it is continuous and let me write those details SS out here's the argument we just went through G is continuous at xal -3 and we justify this using the definition that's how we were asked to do it uh we know that the limit of G of X as X approaches -3 from the left is the limit as X approaches -3 from the left of f ofx because that's G of x's behavior when X is less than -3 it behaves like f ofx and this limit we find by just plugging in it turns out to be four which is also the value of g at -3 it is 4 now if we approach -3 from the right we found that it also equal 4 when it's behaving like this line x + 7 it's also approaching four as we go tog -3 from the right so the function's value equals the limit from the left equals the limit from the right indeed G is continuous at xal -3 that is part C so let's go ahead and finish with Part D find the value of the integral from 0 to 5 of x * 25 - x^2 DX of course you can see f ofx is showing up in this integral which is why this is another part of this problem all right I've just copied this integral down here is Part D you should notice at a glance that this is a pretty easy U substitution problem if we let U equal 25 - x^2 so I'll write that U = 25 - x^2 well then d U is just -2X DX which practically matches the remainder of the integrand X DX and we can make them match exactly if we multiply both sides of this equation by2 if we do that then on the left we have2 du and on the right we just have X DX so you can see how we'll be able to make that replacement and cleanly write this integral in terms of U one other thing I'll point out notice if this x was not here this x that enables us to use U substitution then you would have to notice that this is the equation of the top half of a circle like I mentioned earlier and then finding the area underneath it is pretty straightforward since its radius is five and we're integrating from0 to 5 so you'd actually be looking for the area of a quarter Circle in that case actually just be a very straightforward geometry problem if you recognize that but that's not the problem we've got we've got this x here so let's proceed with the U substitution so this is the integral we can rewrite the bounds so that they are appropriate for U if x is 0 then U is 25 - 0^ 2ar which is just 25 so in terms of U the lower bound is 25 now if x is equal to 5 then U is 25 - 5^ 2 which is Z so the upper Bound for U is zero now what are we integrating we're integrating the square root of U because 25 - x^2 is U and then the X DX is getting replaced by half du with of course a negative as well let's just bring that negative out front and you know what let's put the negative half out front as a whole and then we just have the DU there okay let's proceed with this integral this is pretty straight C again you should recognize that the square root of U is the same as U to the2 so this is a simple reverse power rule integral you might say we will have a - one2 out front and then to do the reverse power rule we increase the power by 1 1 12 + 1 is 12 + 22 which is three Hales so U to the 3es then divide by this this new power dividing by this fractional power is the same as multiplying by its reciprocal 2/3 and we are evaluating this from the lower bound 25 to the upper bound of zero all right we'll proceed with this we still have that negative half out front if we plug the upper bound zero into this what we're going to get is zero and then subtract plugging in the lower bound so we're subtracting 2/3 multiplied by 25 to the^ of 3 finally this is2 the zero of course doesn't matter 25 to the 3es is the square OT of 25 cubed the square root of 25 is 5 and 5 cubed is 125 so what we have in Brackets which I'll just use parentheses now is 2/3 multiplied by 125 again we got that by first applying the power of 1 12 which is just a square root square root of 25 is 5 and then we apply that power of three cubing the five to get 125 all right we're nearly there the negatives will cancel out in the numerator we would have 2 * 125 which is just 250 and in the denominator we have 2 * 3 which is just 6 now we just have to reduce this fraction we can reduce both things by a factor of two the numerator divided 2 is 125 and the denominator divided by two is 3 we of course could have just canceled the twos up here to begin with and hey look that's our answer and that completes our solution to free response question 4 from the 2012 AP Cal AB exam next up are a few ivt and mvt problems for these problems recall that the intermediate value theorem requires continu it while the mean value theorem requires continuity and differentiability although of course differentiability implies continuity the intermediate value theorem says how if a function is continuous it can't pass from a value F of a to a value F of B without also hitting all of the intermediate values whereas the mean value theorem says that whatever a function's average rate of change is over an interval it must be that its derivative is equal to that average rate of change at at some point in the interval of course provided that the function is differentiable good luck this is free response question six from the 2006 AP Cal AB formb exam a car travels on a straight track during the time interval from 0 to 60 where T is measured in seconds the car's velocity V measured in feet per second and acceleration a measured in feet per second per second are continuous functions the table above shows selected values of these functions you can see time is in the top row then velocity then acceleration part A says using appropriate units explain the meaning of the integral of the absolute value of the Velocity function from 30 to 60 in terms of the car's motion then approximate this integral using a trapezoidal approximation with three sub intervals as determined by the table let's start with the first part of the question the meaning of the integral of the absolute value of the Velocity function the absolute value of velocity is speed and if we integrate speed we get the total distance traveled if we were to just integrate velocity then positive and negative velocities would cancel out in the end you would just get the displacement of the vehicle but since we're integrating the absolute value which is the speed we're going to get the total distance traveled whether it's negative velocity or positive velocity it's all counted the same because of the absolute value so for part A this integral is the distance in feet that the car travels from T = 30 to T = 60 it's not the displacement it's just the total accumulated distance traveled all right the other part of part A is to actually approximate this integral using a trapezoidal approximation with three sub intervals as determined by the table going from 30 to 60 you can see in the table there's only three sub intervals that we can use going from 30 to 35 35 to 50 and 50 to 60 for a trapezoidal approximation it may help to do a quick sketch let's say that tal 30 is right there so our first trapezoid is going to go from 30 to 35 and at 30 the velocity is -14 but we are integrating the absolute value of the Velocity so this is actually going to be positive 14 so maybe positive 14 is right there and this is going up to that and then at 35 the velocity is -10 but again we've got absolute value so that's getting changed to positive 10 and maybe that's about that high and so there is our first trapezoid now the next trapezoid Which I guess I'll do in a different color goes from 35 to 50 so maybe over here is 50 equal spacing here doesn't really matter we're just doing a quick sketch so that it's clear how we're going to calculate the area the velocity at 50 is zero so in fact our second trapezoid is actually just a triangle and then lastly we go from 50 to 60 the velocity at 60 is 10 and so that's going right back up here to 10 and that gives us that triangle there and I should label this is at tal 60 so to approximate this integral from 30 to 60 the integral of the absolute value of the Velocity function approximating this using a trapezoidal sum let's see what are the areas first we have the area of this trapezoid Which is the average of the bases one2 multiplied by this side plus that side the length of this side which is a base of the trapezoid is 14 plus the length of this side which is the other base of the trapezoid Which was 10 so 1/2 the sum of the bases multiplied by the height the height of the trapezoid is this distance here which is five so that's the first area we can just go ahead and cut and paste that down here to our answer and now let's move on to these two triangles the area of this triangle is 12 time its base which is 15 multiplied by its height which is just this distance here which is 10 and then the second triangle right here has an area of 1/2 time the base which is 10 multiplied by the height which is also 10 so we just need to add all of these areas together to get our approximation doing the math here 14 + 10 is 24 * 12 is 12 * 5 is 60 and then 12 * 15 * 10 is 75 and then finally 12 * 10 * 10 is 50 so the approximation is 185 and the unit because this is just distance traveled in feet this is 185 ft all right let's go ahead and just shrink this approximation down so we have a little bit more room to work with and move on to Part B using appropriate units explain the meaning of the integral of acceleration from 0 to 30 in terms of the car's motion and find the exact value of this integral now now acceleration is change in velocity if you accumulate or integrate change in something you get the net change in that thing so if we accumulate these changes in velocity we just get the net change in velocity now acceleration is the rate of change of velocity and if we accumulate or integrate a rate of change what we get is net change so this integral is just the change in velocity of the car from tal 0 to T = 30 and there that answer is written out this integral is the car's change in velocity we want to make sure we use units that's velocity in feet per second from tal 0 to T = 30 all right the other part of the question was asking for the exact value of this integral now we can just use the fundamental theorem of calculus here the integral of acceleration from 0 to 30 is just going to be the velocity evaluated at 30 minus the velocity evaluated at zero because velocity is the anti-derivative of acceleration all right and now we can just refer to the table V of 30 is -14 V of 0 is -20 so -14 minus -20 that's -14 + 20 which is 6 and our units are feet second so the change in velocity was 6 ft per second from t = 0 to T = 30 on two part C for T between 0 and 60 must there be a time T when the velocity is equal to -5 and of course justify our answer so this is an intermediate value theorem problem must there be a place on this interval where the velocity is equal to -5 well the velocity as we're told told is a continuous function and the velocity does pass -5 at some point it has to because we see right here from 35 to 50 the velocity goes from -10 to 0 so yes it must have passed -5 at some point so this is our answer for part C because V of 35 which is -10 is less than -5 Which is less than zero which is V of 50 -5 is between the velocity at 35 and the velocity at 50 and the velocity is continuous so by the intermediate value theorem there must be some value of T in this interval between 35 and 50 which is also in the interval between 0 and 60 such that the velocity is equal to5 moving on to Part D for T between 0 and 60 must there be a time T when the acceleration equals zero and justify our anwers we could use the intermediate value theorem here on the acceleration function or we could use the mean value theorem on the velocity function if we come up to the table we do not see the acceleration function ever pass zero we never see it go from positive to negative but we only have a selection of the values so just because the table of acceleration values doesn't make it clear that it hit zero doesn't mean that it never hit zero let's consider using the mean value theorem applied to the velocity function is there any interval over which the average rate of change of the Velocity is zero the answer is yes because we see it t equals 0 the velocity is -20 and at tal 25 the velocity is also -20 over this interval the average rate of change of the Velocity was Zero because it went from -20 to -20 and since that's the case there must have been some point where the instantaneous rate of change of the Velocity thus the acceleration was zero that's the mean value theorem there's our answer for Part D because V of 0 equals V of 25 they were both -20 the average rate of change is zero over that interval and so the mean value theorem guarantees a t in that interval from 0 to 25 so that the derivative of the Velocity function the instantaneous rate of change which is the acceleration is equal to zero I'll also point out that according to the College Board scoring guidelines for this question you could mention the hypotheses of these theorems or you could mention the theorems you don't have to do both I could say by the mean value theorem all of this is true or I could say because the functions are continuous and differentiable on the interval all of this stuff is true that completes our solution to free response question six from the 2006 AP Cal AB formb exam this is free response question one from the 2011 AP Cal AB exam this is part of the graphing calculator section we will use a calculator throughout this problem a cylindrical can of radius 10 mm is used to measure rainfall in Stormville the can is initially empty and Rain enters the can during a 60-day period the height of water in the can is modeled by the function s where s of T is measured in millimet and T is measured in days from 0 to 60 the rate at which the height of the water is rising in the can is given by S Prime of t equal 2 sin of 03t + 1.5 all right part A according to the model what is the height of the water in the can at the end of the 60 day period we know the Can was initially empty so at the start the zero day T equals 0 the height of water in the can is zero then all we have to do is accumulate the rate of change of the height of the water in the can so integrate S Prime from 0 to 60 that's just like accumulating all of the rainfall so the answer to our question s of 60 the height of the water in the can on the end of the 60th day is going to be the integral from 0 to 60 of S Prime S Prime of course is 2 multiplied s of 0.03 T and plus 1.5 and we are integrating with respect to T and we'll just go ahead and calculate this on a calculator on the calculator I'll press the math button and then option nine to access the integration function we are integrating from 0 to 60 the function we're integrating is 2 sin of 03t but we'll use the variable X on the calculator plus 1.5 and we're integrating with respect to X and it's about 171.8cm in the height of the water at any moment that is S Prime of T So to find the average rate of change in the height of water in the can over the entire 60-day period all we have to do is add up the rates of change over the entire interval which is actually this this integral that we just calculated and then divide by the amount of time that passed divide by the interval WID so the average value is going to be the integral from 0 to 60 S Prime of T DT just like in the previous part and then we need to divide this by the interval width divide by 60 which is the same as multiplying by 1 over 60 so in fact all we have to do is divide 1 71.8 mm by 60 days and this will be about 2.86 3 millime per day the average rate of change in the height of the water is about 2.86 3 mm per day moving on to part C assuming no evaporation occurs at what rate is the volume of water in the can changing at time t equal 7 and indicate units of measure to answer this question let's start off by writing a function for the volume of water in the can the volume of water in the can since it's a cylinder is going to be the area of the circular base base multiplied by the height the area of the circular base isk r² and R we know is 10 mm the radius is 10 mm so < * r^ 2 is < * 100 that's the area of the circular base and then for volume we need to multiply by the height which is the S of T function so this is a function for volume it is of course a function of T and for part C we are asked at what right the volume of water in the can is changing at time t equal 7 so all we have to do is take the derivative of this function and plug in 7 V Prime of 7 is going to be well it's just going to be 100 pi multiplied by S Prime of 7 and we can just consult the calculator to calculate this so on the calculator I'll type 100 pi and then we have to multiply by S Prime of 7 now we don't have to use the calculator's derivative function because S Prime is actually what the question gives us the question gives us S Prime so that's going to be multiplied by 2 * s of .3 * 7 because the question is about tal 7 and we want to make sure that we put this whole derivative in parenthesis so let me put a parthy in front there and at the end we have plus 1.5 and this is about 62.2 28 since the height is measured in millimet the volume is measured in millim cubed and since T is measured in days this rate of change of the volume is cubic millimet per day so on the seventh day the volume of water in the can is increasing by about 62. 218 cubic millim per day finally Part D during the same 60-day period we've been discussing rain on Monsoon Mountain accumulates in a can identical to the one in Stormville the height of the water in the can on Monsoon mountain is modeled by the function M where m equals this the height M of T is measured in millimeters and T is measured in days between 0 and 60 let D of t equal m Prime of T minus S Prime of T where S Prime of course is the S Prime that we've been considering this whole time we want to apply the intermediate value theorem to the function D on the interval from 0 to 60 to justify that there exists a time from 0 to 60 at which the heights of the water in the two cans are changing at the same rate now all of our functions are continuous it's clear that S Prime is continuous and M is a polinomial so it's continuous and so is its derivative M Prime so our difference function D of T is also continuous so it's possible that we could apply the intermediate value theorem the way that we'll do that is plug in zero to D of T and then also plug in 60 to D of T hopefully they have opposite signs and in order for that to be the case since D of T is continuous it must pass through zero which means there must be some time at which the difference in these rates of change is zero so for Part D we will look at D of 0 plugging zero into our difference function which is M Prime of 0 0 minus S Prime of 0 and then we will also plug in 60 the final time in our interval D of 60 of course is equal to M Prime of 60 - S Prime of 60 now a little bit of this will be easier to do without the calculator I don't really feel like typing all of this into the calculator it's fairly easy to just find the derivative and then we'll just plug 60 into that on the calculator let's write the derivative right up here where we have a little bit of room M Prime of t that's going to be 1 400 multipli just using the power rule 9 t^2 minus 60t + 330 all right so we will just plug 60 and 0 into this for our calculator although plugging in zero is really easy because this goes away and this goes away and so if we plug in zero all we're going to get is 330 over 400 which is just 30 3 3 over 40 so when we plug in 0 we're going to have 33 over 40 That's M Prime of 0 and then going to look at S Prime plugging 0 into S Prime will produce 2 * s of 0 which is 0 plus 1.5 so that's actually just 1.5 or three Hales so minus 3 haves this is clearly negative because 33 over 40 is less than 1 and we're subtracting ing three halves which is greater than one all right now let's consider the second part where we plug in 60 and hopefully this will be positive thus the intermediate value theorem will apply to show us that D the function D must have passed zero at some point since it went from negative to positive let's do the calculation and see here on the calculator I'll start off with M Prime of 60 I already know what M Prime is it is up here so now I'll just plug 60 into this I'll save the division by 400 for the end so we have 9 * 60 2 - 60 * 60 + 330 and then divide this whole thing by 400 this turns out to be 72.85 then we will want to find S Prime of 60 S Prime we know is 2 * s of blah blahy blah let's plug 60 into that 2 * s of 03 * 60 and then + 1.5 this is about 3447 and this is clearly positive so yes we can use the ivt since D of T is continuous and D of0 is negative while D of 60 is positive the intermediate value theorem guarantees that there must be some time in the interval where D of TT takes on that intermediate value of zero it starts off less than zero it finishes above zero so it must have passed zero at some time in the interval and that completes our solution to free response question one from the 2011 AP Cal AB exam this is problem three from the 2013 apal AB exam hot water is dripping through a coffee maker filling a large cup with coffee the amount of coffee in the cup at time te from 0 to 6 is given by differentiable function C where T is measured in minutes selected values of c of T measured in ounces are given in the table above which we see here part A asks us to use the data in the table to approximate C Prime of 3.5 and show the computations that lead to our answer and indicate units of measure since the function C is the ounces of coffee that are in the cup C Prime is going to be the change in ounces per unit of time which is minutes so ounces per minute now of course we don't know what c of T is explicitly so we can't find a function for the derivative but we can approximate the value of the derivative at 3.5 by looking at what the average rate of change of the function C of T is from 3 to 4 because 3.5 is in the middle of that interval so if we look at the average rate of change from 3 to 4 we should get a decent approximation of C Prime of 3.5 so for part A we are trying TR to find C Prime of 3.5 and we know that it should be approximately the change in the ounces of coffee from time T = 3 to time T = 4 so C of 4 minus C of 3 divided by the length of that interval which is just 4 minus 3 now let's refer to the table C of 4 is 12.8 and C of 3 is 11.2 so their difference is 1.6 the denominator 4 - 3 is obviously 1 and so this is just 1.6 and like we said this is going to be in ounces per minute moving on then to Part B is there a time T from 2 to 4 at which C Prime of T is equal to 2 and we must justify our answer for this we are using the mean value theorem C of T we know is a differentiable and thus continuous function so we can use the value theorem as long as we find the pieces necessary in the table in order to use the mean value theorem to conclude that the derivative is equal to two at some time we need to find some interval in the table such that the average rate of change over that interval is equal to two so for some examples if we look at the interval of time from 2 to three the change in ounces looks to be 2.4 but the change change in minutes is just one so the average rate of change would be 2.4 that's not equal to two so that's not going to work we are not restricted to looking at intervals of length only one however if we look at an interval like this with length two the average rate of change over this interval is 12.8 - 8.8 that's 4 divided 4 - 2 that's 4 / 2 which is two which is exactly what we're looking for we have an interval over which the Aver average rate of change is two thus at some point in that interval the mean value theorem would guarantee that the derivative is equal to two the interval we were just looking at was from 2 to 4 so C of 4 minus C of 2 / 4us 2 we found that this actually equals 4 over 2 which equals 2 so we have an interval over which the average rate of change is two we can apply the mean value theorem since we were told that c is differentiable and thus continuous we can use the mean value theorem to conclude there must have been a time T from 2 to 4 so between these two values for which C Prime of T was equal to that average rate of change of two moving on to part C use a midpoint sum with three sub intervals of equal length indicated by the data in the table to approximate the average number of ounces of coffee in the cup that's what this is it's adding up the ounces of coffee in the cup C of T and and then dividing by the interval length so this is over that whole time from 0 to 6 using correct units explain the meaning of this integral which is what I just said it is the average amount of coffee in the cup because it's the sum of all the amounts of coffee in the cup divided by the time right divided by six 6 minutes in this case now let's use the table to help organize our information that we'll use for the midpoint sum the interval length is six that's the length of this entire higher interval and we're splitting it up into three sub intervals so each sub interval should have a length of two thus one rectangle will go from 0 to two with the height as determined by the midpoint which is the height at tal 1 which is 5.3 the next rectangle will go from 2 to 4 and have the height as determined by the C value at the midpoint which is 11.2 and then the next rectangle will go from 4 to 6 with its height again determined by the C value at the midpoint which is five and the height is 13.8 so our approximation for this integral is going to be 1 16 multiplied by the sum of the areas of these rectangles their areas are just base times height and each one has a base of two so let's suppose we just pull that factor of two out so it's 1 16 * 2 multiplied by now we just have to worry about the heights which we've Jed down 5.3 11.2 and 13.8 this is equal to 1 16 * 2 is 1/3 and then adding up everything in Brackets that's going to be let's see 16 29 and then 30 and then 30.3 so multiplied by 30.3 and 13 of 30.3 is 10.1 and again this is in ounces it's the average number of ounces of coffee in the cup over the time interval and of course the problem also asked us to explain the meaning of this integral and so there we go our approximation for its value was 10.1 o using the indicated midpoint sum and then this is the explanation of what it means the average amount of coffee in the cup and ounces over the time interval finally Part D the amount of coffee in the cup in ounces is modeled by this function using this model we are asked to find the rate at which the amount of coffee in the cup is changing when T equals 5 this is an easy part to finish with I'll try to squeeze it in right here Part D we just have to take the derivative and then plug in tal 5 so let's see what is B Prime of T well the derivative of 16 will just be zero and then taking the derivative of -16 e to .4 T we will have just -16 * e4t because the exponential function is its own derivative but then we have to multiply by the derivative of the inside function the derivative of .4 T is just .4 so all that's left for us to do now is simplify this derivative a little bit and then plug in t = 5 -16 multiplied .4 is just positive 6.4 so this is 6.4 * e .4 T and I'd say that's a good point to plug in tal 5 b Prime of 5 is going to be 6.4 multiplied by E to the4 * 5 that's just e to -2 and so we might want to finish by just rewriting this with positive exponents so 6.4 over e^ 2 and that's at the perfect time because it sounds like my laundry is done let's just make sure that we get our units in here this is ounces per minute because it is a derivative of this function which was giving us the number of ounces of coffee in the cup as a function of the minutes so if we take its derivative we'll get the rate of change of ounces as the minutes pass so ounces per minute and that will complete our solution to free response question three from the 2013 AP Cal a exam next up are the linear motion problems for these problems recall that we are talking about motion on a line so a negative velocity indicates leftwards motion on the line whereas positive velocity indicates rightwards motion also recall that in order to determine if the speed of a particle is increasing at a point in time you have to check if the velocity and acceleration have the same sign I think think of acceleration as pulling on velocity so if they have the same sign the acceleration is pulling in the same direction that the velocity is going and so the particle is speeding up that doesn't make much sense hopefully it'll make sense when you see the problems good luck this is free response question one from the 2011 AP Cal AB exam this is part of the calculator section we will have to bust out our calculator a few times for this question let's read it for T between 0 and 6 a particle is moving along the x axis the particle's position X of T is not explicitly given the velocity of the particle is given by V of t equal 2 sin of e the T 4 + 1 the acceleration is a of t = e the T 4 cine e to the T over4 and we also know that X of0 the initial position of the particle is is two all right that's a whole lot of information part A is the speed of the particle increasing or decreasing at time tal 5.5 and give a reason for your answer remember that speed is the absolute value of velocity so if the velocity is getting more extreme either more positive or more negative then the speed is increasing to figure out if that's happening we need to assess both the Vel velocity and the acceleration if they have the same sign then yes the speed is increasing if the velocity is positive and the acceleration is positive that would mean the particle is moving to the right and it's doing so at a faster Pace because the acceleration is positive and the velocity is positive if the velocity is negative that would mean the particle is moving to the left and if the acceleration is also negative that would mean that the particle is speeding up with that leftward motion so if the velocity and acceleration have the same sign then the particle is speeding up now I've opened up our graphing calculator again for part A we're going to look at the velocity at the time of Interest which is 5.5 and the acceleration at 5.5 if they have the same sign then indeed the particle is speeding up so let's start with the velocity here is our velocity function function and we need to plug in 5.5 so let's type that into our calculator very carefully it's going to be 2 * s of e to the T over 4 but we're plugging in 5.5 for T So 5.5 over 4 and then + one we see that's about .4 53 before we write that down let's just also check the acceleration that's right here here okay so we'll plug this in we have 12 so 1/ 2 and maybe we'll put that in parentheses just to make sure we're being careful with the calculator 12 multiplied by e to the T over 4 again T is 5.5 so we're plugging in 5.5 multiplied by cosine of e to the 5.5 over 4 and that's it press enter okay so we see both the velocity which is about. 453 and the acceleration which is about 1359 both of these are negative so indeed at tal 5.5 the speed of the particle is increasing it is moving to the left because the velocity is negative and it's doing so at a faster rate as time goes on because the acceleration is also negative and there is our reasoning written out moving on to Part B find the average velocity of the particle for the time period between 0 and six the average value of a function is found by just integrating the function across the interval in question so to find average velocity we'll integrate velocity from 0 to 6 and then just divide by the interval length so we will divide it by six it's kind of like adding up all the values in data set and then dividing by the total number of values in that data set all right so we'll write this down the average velocity whoops don't need two B's there the average velocity is what well it's 16 because again we have to divide by the interval length which is 6 1 16 times the integral from 0 to 6 of velocity the velocity function of course is given we'll go ahead and plug this into the calculator all right again we'll plug this in very carefully we have 1 / 6 multiplied by I'll press the math button to access all my functions and choose option nine the integration function we are integrating from 0 to 6 and now I have to put in the velocity function which I know is 2 sin of e t 4 so 2 multiplied by S of e to the I'm just going to use x as the variable divided 4 and also + one and we're integrating with respect to X again I'm just replacing t with X doesn't really matter we did also multiply by the one six at the front we'll just press enter and there it is I just changed the mode of my calculator to decimals so I could get a decimal answer exactly how you would do that on your calculator if it was reporting fractions may vary a lot of calculators will have a secondary approximately op option on their Enter key so you could press second enter and you would get an approximation instead of an exact value but it depends on your calculator so for me I've got this decimal approximation it is about 1. n49 that is the average velocity of the particle on the interval from 0 to 6 again all we did was integrate the velocity over that interval and divide by the interval length moving on to part C find the total distance traveled by the particle from time tal 0 to T = 6 this is pretty similar to the previous problem we will again need an integral that we will use our calculator for in order to calculate distance now how do you calculate the distance traveled when you have the functions that we've got in this situation well you don't want to integrate velocity because some velocities may be positive some may be negative and that could cancel some of the dist dist out remember we are trying to calculate distance so we don't care if the particle goes 100t one way and then 100 ft back and its displacement would be zero we would want to count that distance as 200 now in order to represent that in an integral we actually have to integrate the speed which has no regard for sign because it's the absolute value of the Velocity function so what is the distance well we have to integrate from 0 to six again the speed and the speed like I said is the absolute value of the velocity of course having a calculator for this makes it a breeze so coming back to the calculator I'll press the math button to access my functions and press number nine the integration function again we're integrating from 0 to 6 and what we're integrating is the absolute value so I will press math and then go to the Numb tab and then press the absolute value function this is the absolute value of velocity so now I have to type in the velocity function which is 2 s of e to the T we're just going to replace t with X ID 4 and then plus 1 I have to correct this accidentally put times 4 this is divided 4 and then plus one outside of the sign function but still in the absolute value bars and we're integrating this with respect to X so this is the integral of the speed which is the absolute value of velocity from 0 to 6 we see it's about 1257 so there we go the distance traveled over this interval is about 12573 we don't integrate velocity because some of the velocities could cancel each other out since they could be positive or negative we just want to count all the distance traveled no matter the direction so we integrate speed all right moving on to Part D for T between 0 and 6 the particle changes Direction exactly once we need to find the position of the particle at that time all right there are a couple steps to this problem if we want to find the position of the particle at the time when it changes direction we first need to figure out when it changes Direction now a change in direction will be marked by a change in the sign of velocity if we go from positive to negative velocity or vice versa that means that the direction has changed from right to left or from left to right so let's open up our graphing calculator and see where the velocity on this interval from 0 to 6 passes zero where the velocity changes sign going back to my calculator I'll press y equals so I can enter in the velocity function and then look at its graph the velocity function is 2 sin of e to the x / 4 and + 1 now I can press the graph button to get a look at the graph and we want to look at the interval between 0 and 6 and see where this velocity function passes zero and changes sign and we see it appears to be right there the velocity goes from positive to negative at T = 5158 it also passes zero and changes sign to the right over here around 7 but that's outside of the interval we're inter rested in so where does it hit zero well in the interval of interest that we have the Velocity hits zero at about 51955 and with that information we can find the position of the particle at the time when it changes Direction it changes Direction at tal 51955 so what is the position at that time well it's going to be the initial position plus the accumulation or integral of the velocities from 0 to the time in question this is like taking that starting position which we know that's why we're starting there and then adding up all the change from that position we know from T equals 0 to the time we want tal 51955 now of course the position at time equals 0 is 2 so this is 2 X of 0 is 2 plus this integral of velocity from 0 to 51955 when it comes to finding position we do want to integrate velocity because whether the velocity is positive or negative is very important to the position and this final thing of course we will deal with on our calculator we need 2 plus 2 plus the integral so I'll press the math button and then function n the integration function the integral from 0 to 5 .1 1955 that time at which the particle changes Direction and then we're integrating the velocity function which we've entered a million times by now it's 2 sin of e to x / 4 + 1 outside of the sign function and we're integrating with respect to X press enter thankfully we're in decimal mode so we get our decimal approximation this is about 14135 so at the time when the particle changes Direction which is at T = 51955 the position of the particle is about 14135 we're able to figure out quite a bit of information even though we don't know the position function explicitly and that completes our solution to free response question 1 from the 2011 AP Cal AB exam this is free response question two from the 2013 AP Cal AB exam a particle moves along a straight line for t between 0 and 5 the velocity of the particle is given by this equation and the position of the particle is given by S of T it's known that the initial position s of 0 is 10 part A asks us to find all values of T in the interval from 2 to 4 for which the speed of the particle is two the speed of the particle is the absolute value of the Velocity function so using a calculator we can just find the intersections of the absal of the Velocity function and Y equal 2 heading on to the calculator I'll press the math button and then go to the number suction to access the absolute value function now I'll enter in the velocity which is -2 Plus in parentheses t^2 but we're using the variable X on the calculator plus 3 x to the power of 6 over 5 6 over 5 is just 1.2 and then - x cubed I'll also graph y = 2 and then I want to see where these things intersect now I'll go ahead and press the graph button and find the intersections on the interval in question which is between 2 and four how you find the intersections of course will depend on your calculator I'm just going to zoom in and locate them so the intersections between 2 and 4 are these two intersections over here one of them is at T = 3.1 27 so when T = 3127 and the other intersection up here is when t equal 3473 moving on to Part B write an expression involving an integral that gives the position s of T use this expression to find the position of the particle at time T = 5 now we'll be able to find the position by just integrating the velocity and using what we know about the initial position so what position does a particle have at any given time T well it starts off with a position of 10 so we would have 10 and then plus the integral from 0 to T of the rate of change of its position which is the velocity now we're integrating the velocity but velocity is given as a function of t t is in the bounds of our integral so we'll have to use a different variable often we would pick something like s but we don't want to use S because s is reserved for the position function so let's just say V of u v of U du this is an integral expression for the position function s of T then we can use the calculator and our function to find the position at time tal 5 so what is s of five well we will consult with the calculator it's going to to be 10 the initial position plus the integral of the Velocity from 0 to the time in question which is five so I'll press the math button and then option n for the integration function we are integrating from 0 to 5 and the function we are integrating is this big velocity function I've entered it into the integral and of course I'm using X rather than t on the calculator and we get about 9.27 9.20 7 that's the position of the particle at time five moving on to part C find all * T in the interval from 0 to 5 at which the particle changes Direction in order for the particle to change direction the velocity must pass zero changing from positive to negative it's not sufficient to just hit zero the velocity could come down to zero but then go back up still the whole time it's not negative we need the velocity to go from positive to negative like this positive to negative or negative to positive that would mean that the particle has changed Direction so here on the calculator I've entered in the velocity function no absolute value bars this time and we can the graph remember we're focused on the interval from 0 to 5 so let's see where does the velocity pass from positive to negative or negative to positive well it would appear right here here at T = 536 it goes from negative to positive so yes that is a change in direction from moving to the left to moving to the right because it's negative to positive and then here when T is about 3318 the sign of the Velocity goes from positive to negative so it was moving to the right but then it started moving to the left and here that answer is written out we see when the velocity equals zero at these two times in the interval and at both of these times there is a change in sign and so it is at those two times that our particle changes Direction finally Part D is the speed of the particle increasing or decreasing at time tal 4 and we need to give a reason for our answer the speed again is the absolute value of the Velocity that's going to be increasing if the acceleration has the same sign as the velocity if the velocity were positive but the acceleration were negative that would mean our particle is moving to the right but the negative acceleration means that it's doing so more slowly so it would be slowing down but if our velocity is positive and our acceleration is positive then not only is the particle moving to the right but it's also getting faster with that rightward motion so in order to determine if the speed of the particle is increasing or decreasing at time t equal 4 we need need to see if the velocity and acceleration have the same sign or opposite signs back on our calculator we can just look at our graph and see that when T equals 4 it's clear the velocity is negative V of 4 is less than Zer now what about the acceleration well we could try to figure that out from the graph since that is the derivative of velocity but we might as well just use the calculators built-in derivative function to get a precise answer so I'll press the math button option 8 for the derivative function we are taking the derivative with respect to X and then just plug in the velocity function but in terms of X so -2 plus parentheses x^2 + 3x to the power of 6 fths which is 1.2 - x to the^ of 3 and we are evaluating this at x = 4 and we see that this the acceleration is also negative so V of 4 is negative and the acceleration a of 4 is about 22.3 which is also negative since the velocity and acceleration have the same sign the particle is speeding up when T equals 4 and there that answer is written out not only is the particle moving to the left at this time but it's doing so more quickly because of that negative acceleration and that will complete our solution to to fre response question two from the 2013 AP Calculus AB exam this is fre response question two from the 2021 AP Cal AB exam this was part of the calculator section we will use our calculator a good amount throughout this problem a particle p is moving along the x axis the velocity of particle P at time T is given by this sign function for T between 0 and Pi at time T equals 0 the initial position of particle p is x = 5 a second particle Q also moves along the x-axis the velocity of particle Q at time T is given by this function here for T between 0 and Pi its initial position at t equal 0 is x = 10 part A is to find the positions of particles p and Q at time T = 1 so for part A we're going to have to take the initial position of each particle and then integrate or add up the rates of change or the velocities from T equals 0 to tal 1 so first let's find the position of particle P at time tal 1 we'll denote that XP of one and again this is going to be the initial position of particle P which was given to us as five and then we need to add the integral from t = 0 to T = 1 and and what we have to integrate is the velocity if we add up all of those rates of change we'll get the total change in position from tal 0 to tal 1 and when we add that to the initial position we'll get the actual position at tal 1 now what was the velocity function which we need to integrate well for particle P the velocity function is this s of t to the 1.5 so we are integrating sine of t to the 1.5 and this integral is taken with respect to T and we'll open up the calculator to figure this out I also set up the integral here for the position of particle Q at time tal 1 same idea the initial position which was given to us is 10 plus the integral of the Velocity function from tal 0 to tal 1 so coming over to the calculator I'm going to press the math button and then press number nine the integration function starting with the position of particle P we are integrating from 0 to 1 the function we're integrating is sine of we'll use x instead of t for convenience this is X the^ of 1.5 and this integral is taken with respect to X this is about 3707 let's not forget to add five to it we have to get that five that initial position and okay so we see this is about 53706 I'll round that to 53707 all right as for particle Q we have to do 10 the initial position plus I press the math button number nine the integration function we're integrating from 0 to 1 this function is x - 1.8 again we're using X instead of t for convenience in our calculator multiplied by 1.25 to the power of X and this integral is taken with respect to X this gives us an answer of about 8564 or 4 and there's our conclusion at time tal 1 P has this position and particle Q has this position all right moving on to Part B are particles p and Q moving toward each other or away from each other at time tal 1 and we need to explain our reasoning now looking back at our answer for part A we see that particle Q is to the right of particle P because particle p is at about x = 5 particle Q is at about x = 8.5 so Q is to the right of P if we sketch that out real quick maybe particle p is right here and particle Q is right here the important part is that Q is to the right of P so if they're moving towards each other then particle P should have a positive velocity and particle Q should have a negative velocity it's moving to the left for them to be moving away from each other particle P should have a negative velocity it's moving to the left and particle Q should have a positive velocity it's moving to the right it is also possible that they could be moving the same direction and yet they could be getting closer together or further apart but let's first see what their velocities are and we can talk about that situation if necessary we'll begin with the velocity of particle p and the time we're interested at again is time t equal 1 so what's the velocity of particle P at time tal 1 well we know what the function for the velocity of particle p is that is s of T the 1.5 so this would be S of 1 to the 1.5 if you plug this into your calculator you'll get about 0.841 so the velocity of particle p is positive so maybe I'll sketch that on our number line it's looking something like that as for the velocity of particle Q at time T = 1 remember that the velocity of particle Q is t - 1.8 * 1.25 to the T So at T = 1 this would be 1 - 1.8 multiplied by 1.25 to the 1 which of course is just 1.25 and this is just -1 so we see the velocity of particle p is positive whereas the velocity of particle Q is negative it's less than zero so indeed these particles are getting closer to each other they're moving towards each other and there's our logic written out since the position of particle P at time tal 1 is to the left of particle q and the velocity of p is positive meaning it's moving to the right whereas the velocity of Q is negative meaning it's moving to the left p and Q are moving toward each other at the time t equal 1 let me add that this is at the time tal 1 all right let's move on then to part C find the acceleration of particle Q at time tal 1 is the speed of particle Q increasing or decreasing at time T equals 1 and explain our reasoning to determine if the speed of the particle is increasing or decreasing we need to compare the sign of the acceleration to the sign of the Velocity since we're talking about a one-dimensional situation we have a particle moving on a number line its direction is indicated by the sign of the Velocity so its speed will be increasing if its acceleration has the same sign as the velocity if the velocity and the acceleration are both negative the particle is moving to the left and it's speeding up with that leftward motion if the velocity and acceleration are both positive then the particle is moving to the right and it's speeding up in that rightward motion so let's see what the acceleration is and then we'll compare the sign of that to the signs of the veloc time T = 1 and we see it's positive it's about 1.02 685 so the acceleration at this time is positive remember we already assessed the velocity at time T equals 1 we already know the velocity of particle Q at time tal 1 is -1 which is obviously less than zero so since the sign of the acceleration is opposite the sign of the Velocity the particle is actually slowing down the velocity is negative so it's moving to the left but the acceleration is positive which means it's actually moving to the left but it's slowing down and remember that's because for a negative velocity a positive acceleration means that the velocity is getting closer to zero closer to zero means the speed is going down so the speed of particle Q is decreasing at time T equals one because the velocity and acceleration have opposite signs all right moving on to part D find the total distance traveled by particle p over the time interval from 0 to Pi to find the total distance traveled all we have to do is integrate the speed function from tal 0 to tal Pi you don't want to integrate velocity because some of the velocities may be positive and some may be negative which would cancel out if you integrate velocity you get displacement which is useful but it's not useful for this if you just want to find how much distance was traveled you don't want to think about what whether it's left motion or right motion you're just trying to add up all the distance traveled so you need to integrate the speed function speed is the absolute value of velocity so here for Part D we are going to integrate from time tal 0 to T = Pi the absolute value of the Velocity function of particle P this integral of course is taken with respect to time and we can do this on a calculator press the math button we'll press number nine for the integration function we're integrating from 0 to Pi and what we're integrating is the speed which is the absolute value of the Velocity so again I press the math button I go to the num Tab and function one is the absolute value function inside of this I put the velocity which is sine of t or in our case x to the 1.5 all right we've got the velocity function in the absolute value we're taking the integral with respect to X and there we go the total distance traveled from tal 0 to tal Pi is about 1. n31 finally there is our conclusion over the time interval from 0 to Pi the total distance traveled by particle p is about 1. n31 and that completes our solution to free response question 2 from the 2021 AP Cal AB exam this is free response question 6 from the 2022 AP Cal AB exam this is part of the no calculator section particle P moves along the x axis such that for time T greater than zero its position is given by this function x p of T particle Q moves along the Y AIS such that for time T greater than zero its velocity is given by this function VQ of T at time tal 1 the position of particle Q is y q of 1 = 2 remember it's moving along the Y AIS so for the time tal 1 the position on the y- AIS of particle Q is 2 part A asks us to find VP of T the velocity of particle P at time T that's pretty easy because we are given the position function of particle P this is the position function so to find the velocity function all we're going to have to do is take the derivative of the position function so we will write this VP of T the velocity of particle P at time T is equal to to x p Prime of T that's just the derivative of the position function here's XP of T and we just need to take its derivative the derivative of 6 is zero and the derivative of -4 e to the t is positive4 e to the T the e to the T doesn't change but you have to multiply by the derivative of the exponent the derivative of the exponent is ne1 which is why the negative in front of the four gets canceled out and that is part a moving on to Part B Part B asks us to find AQ of T the acceleration of particle Q at time T and then find all * t for T greater than Z when the speed of particle Q is decreasing and justify our answer all right let's start off by finding the acceleration to do that we just need to take the derivative of the Velocity function so a q of T we will write equals V Q Prime of T just take the derivative of the Velocity function the velocity function is given to us here 1 / T ^2 which is the same as T -2 so taking the derivative of that is just applying the power rule -2t to the -3 that is the acceleration of particle Q at time T now the other part of the problem asks us to find when the speed of the particle is decreasing the speed is the absolute value of the Velocity so the speed is going to be decreasing when the sign of the velocity and acceleration are opposites so if this is particle Q moving along the Y AIS if its velocity is positive then it's moving in the positive direction but if its acceleration is negative then it's moving in the positive direction increasingly slowly because that negative acceleration is kind of pulling it back on the other hand if the velocity were negative it would be moving down the y- AIS but if its acceleration is positive then its speed would be decreasing because the acceleration being positive would mean that Q is kind of being pulled back up even though it is moving down so again we need the velocity and acceleration to have negative signs in order for the speed to be decreasing so the speed of particle Q is decreasing when the velocity and acceleration have opposite signs it's pretty easy to see that this is when T is positive because the acceleration will clearly be negative t to the -3 is just positive but it's getting hit with that -2 so acceleration will be negative but the velocity up here is 1/ t^2 that's always going to be positive so when T is positive the acceleration and velocity will have opposite signs and thus the speed will be decreasing and we were only looking for posit POS times when this is happening so this would be our answer T greater than Z the velocity of particle Q is negative which means that it is moving down the Y AIS but the acceleration is positive which means that negative velocity is actually getting smaller in magnitude on to part C we are asked to find y q of T the position of particle Q at time T now we are given the velocity of particle Q so in order to find the position of particle Q we're going to have to integrate the velocity function so let's start this process for part C we are looking for yq of T the position function of particle Q that's going to be the integral what we can do is say we're integrating from 1 to T and then the position of particle Q at time one we can add that at the start so y q of 1 we know what that is plus and then we can integrate the velocity from 1 to T the velocity we know is given by this function here 1 / T ^2 and I'll just write that as T -2 although we can't use T because T is in the bound of the integral so let's use S S -2 DS all right y q of 1 was given in the problem as two that was the position at time tal 1 so this is going to be 2 plus the integral of s -2 evaluated from 1 to T the integral of s to the -2 is s the1 that's just using the reverse power rule this is from 1 to T So finally finishing up the evaluation here we have 2 plus plugging in t gives us negative t to the particle P so we are taking the limit as T goes to Infinity of the position of particle P the position of particle p is given to us as 6 - 4 e - t so 6 - 4 e to the- T now as T goes to Infinity this exponent T is going to get very very negative so this term is actually going to be going to zero it'll be 4 e to some massive negative power which is like four / e to a massive power so that thing is going to zero thus this limit is going to be six so in the long run particle p is going to be about 6 units away from the origin on the X AIS on the other hand let's look at particle Q so for particle Q we're taking the limit as T goes to Infinity of the position function of particle Q which we just found 3 minus T1 and a very similar thing will be happening here T the1 it might be more helpful in this case to write that as 1 / T clearly as T goes to Infinity then this expression will just be approaching three because 1/t will be approaching zero so in the long run particle Q is going to be about three units away from the origin on the Y AIS clearly then since 6 is greater than three particle P will be further away from the origin and there that conclusion is and this completes our solution to free response question 6 from the 2022 AP Cal AB exam next up are a few implicit differentiation problems for these problems just recall that you have to keep your eye out for derivative rules that might sneak under your nose if we have something like x * s of Y = 5 then you have to recall that for x * s of Y the product rule is necessary because we have a product of two functions but also the chain rule is necessary because Y is taken to implicitly be a function of X so the derivative of s of Y for example is cosine of y * Dy DX good luck this is free response question 6 from the 1999 AP Cal AB exam in the figure above line L is tangent to the graph of y = 1x^2 at Point p with coordinates W 1/ w^2 where W is greater than Zer so p is just some point of tangency with a positive x coordinate point Q has coordinates w0 this is just the point of tangency projected down to the xaxis and we'll say the tangent line L crosses the x-axis at a point we'll call R which has coordinates k0 part A asks us to find the value of K when w = 3 so for the line tangent to this curve at an x coordinate of three where does that line intersect the x axis to figure that out we'll write an equation of the tangent line and then see what x value will cause y to be zero to write the equation of the tangent line we will use point slope form which begins with Y minus the y coordinate now if W equal 3 well the y coordinate will be 1/ 3 2 so 1/ 9 so y - 1 9 this equals the slope which we'll put there in parentheses multiplied by x minus the x coordinate the x coordinate is 3 so All That Remains is to find the slope the slope of the tangent line to find that we need to take the derivative y = 1X sared then y Prime is -2 /x cub that is just the power rule so the slope of this tangent line at x = 3 will have a slope of -2 over 3 cubed so -2 over 27 and thus we have our complete tangent line equation at x = 3 then we can replace y with 0er and solve for x although rather than writing it as X we were told that the line intersects the x axis at a point we're going to say has coordinates K 0 so the x coordinate we're calling K so let's just replace x with K and we will solve for that all right on the left we have 0 - 1 9th which of course is just - 1 9th on the right we have -2 over 27 K we are Distributing this and then -2 27 * -3 which is + 6 over 27 now 6 over 27 we can reduce that fraction to 2 over 9 and now we can subtract 2 9th from both sides of this equation thus on the left we have -3 99 which is the same as - 13 and this equals -2 over 27 K to finish solving for K we just multiply both sides by -27 / 2 so on the left we will have positive 27 the two negatives cancel out and in the denominator we'll have 3 * 2 which is 6 and this equals K now we can just finish up by reducing the fraction and we have that k equal 9 over 2 3 goes into 27 nine times and it goes into 6 twice thus k equal 9es moving on then to Part B for all W greater than zero find K in terms of W so do what we just did but instead of plugging in W = 3 like we did here to find the value of the derivative and whatnot we just need to keep W arbitrary so again we will begin with our tangent line Y minus the y coordinate if we're keeping W arbitrary then the y-coordinate will be 1 over w^2 this is the y coordinate of the point of tangency so y - 1 w^2 and then we need a slope which is the derivative which as we said before is -2 over W cubed and then this needs to get multiplied by x minus the x coordinate we're letting W be the x coordinate of the point of tangency now again we will replace y with zero and we will replace x with K so 0 - 1/ w^2 = -2 over W Cub multiplied by if Y is 0 we're saying that the x coordinate is K minus W and now we just have to solve for K in terms of w to do that we will Begin by again Distributing on the right side of the equation so -2 over W cubed * K that's going to give us -2 K Over W cubed and now we just simplify W cubed over W ^2 is just going to leave W and then -3 over -2 is just positive 3es so Final Answer k equal 3 W so if you look at a line tangent to this curve 1 /x^2 if the x coordinate of the point of tangency is positive and we call it w then that tangent line would intersect the x axis at x = 3 W moving on then to part C suppose that W is increasing at the constant rate of 7 units per second so our tangent line which has an x coordinate of w is moving forward on the curve so w is increasing at the constant rate of 7 units per second when w equals 5 what is the rate of change of K with respect to time so this is a related rates problem and we're going to have to use the equation we just found K = 3 W in the problem we are given the rate of change of w it is a constant 7 so we could write DW let me write that W again DW DT we know to equal 7 we are looking for dkdt we want to know what that is when w = 5 five right that's what the problem asked us it asks when w equals 5 what's the rate of change of K with respect to time so what we'll do is take the equation we found in the previous problem K = 3 W and then differentiate the left and the right sides with respect to time on the left that's going to give us DK DT and on the right that's going to give us three Hales DW DT but DW DT we know to be constant of 7 so this in fact is equal to 3 * 7 which is 21 /2 and so that is our answer when w equals 5 the rate at which K is changing is 21/2 its rate of change as we see actually does not depend on W so 21/2 that's our answer all right moving on to Part D suppose that W is increasing at the constant rate of 7 units per second when w equals 5 what's the rate of change of the area of triangle pqr I'm not sure what this symbol is that's a a rendering error that's supposed to be triangle pqr with respect to time determine whether the area is increasing or decreasing at this instant so let's look at our picture we're talking right now about this triangle triangle pqr and we're wondering how its area is changing based on the rate of change of w which we know to be seven so let's do Part D up here by our handy dandy picture we'll Begin by writing an equation for the area since we need to figure out how the area is changing we're going to need to begin with an equation for area the area of a triangle is 1 12 base time height so in this case that's going to be 1/2 the height is very easy because the height is just that length there which is the y-coordinate of 1 over w^ 2ar now the base length of the triangle this part here that's going to be the change in x coordinates from Q to R we know that Q has coordinates w0 it's just the projection of Point P onto the X AIS and point R we previously said had coordinates k0 it's just wherever the tangent line intersects the x axis but earlier in the problem we figured out that K is actually just 3 W so then the base length is 3 wus W that's the change in X and that is just 12 w or w/ 2 thus the equation for area we can simplify to 1 over 4 W now that we have our equation for area we can differentiate the left and right sides on the left we have the derivative of the area with respect to time that's going to be-1 over 5^ 2ar or 25 and then D wdt we know to equal 7 and then this is just -7 over 100 so -7 hundreds or 0.07 that's how the area is changing thus since it's negative the area is decreasing and that completes our solution to free response question six from the 1999 AP Cal AB EX exam this is problem five from the 2ab exam let's read it consider the curve given by xy^2 - x Cub y = 6 the first part of the question is to show that dydx equals this fraction so we're just going to have to do some implicit differentiation and make sure that our result matches the answer given there this is a no calculator question so we won't be graphing this or anything like that uh but I went ahead and graphed it anyway and I'll show you what that looks like just because it's interesting we will of course not appeal to this graph during our solution since that's not allowed but that's what it looks like in case you were curious pretty cool all right so let's go ahead with our implicit differentiation I will just rewrite the equation here we have xy^ 2 - x Cub y = 6 and what we're going to do is is take the derivative with respect to X on the left side of the equation and on the right side of the equation on the right side of the equation of course the derivative I'll put that this is part A as well the derivative of a constant the derivative of six is just zero on the left we'll have a little bit more going on we're going to need to apply the product rule in fact because this first term is x * y^2 and this second term is X cub time y so let's start with x y^ 2 if we call X our F and we call Y squar our G then the product rule of course here is frime G Plus G Prime F FR Prime is just the derivative of x which is 1 so frime G is just y^2 and then we need to add G Prime f g Prime the derivative of y^2 is going to be 2 y that's like the derivative of the outside function to Y but then Y is itself a function so we need to also use the chain rule here and that's where the signature dydx of implicit differentiation comes from it's like the derivative of an inside function so that is our G Prime let's not forget that we have to multiply that by F which in this case is X let's move this equal Z over to the right because we're going to need a little bit more room now we have to take the derivative of x cubed Y and this whole thing is getting subtracted so we will subtract this derivative now again we're applying the product rule f is I should probably put this a little bit differently so you can see the exponent there F is X cubed right so F Prime is going to be 3x^2 and so FR Prime G is 3x^2 * y remember the dydx terms only show up when we take the derivative of y so it's going to show up in this next part where we have G Prime f g is y so G Prime is literally just dydx and then we have to multiply this by F which is X cubed and close the parentheses remember we're subtracting this whole thing and this is equal to zero now we just have to solve for dydx solving for dydx is super easy you can do it step by step if you need to uh but I think it's really easy to just do this in one Fell Swoop I'll show you how I think about it so I'm going to write dydx equals what is it equal well the first thing I would do is move all the non dydx terms to the right side of the equation for example y^ SAR needs to get subtracted from both sides so there's going to be a minus y^2 over here what else do we have without a dydx the only other thing is minus 3x^2 y so we would have to add add 3x^2 y to both sides after we move everything without a dydx term over we have to divide by everything that's with a dydx term really what we do is gather those terms together factor out the dydx and then divide by everything else so in this case that's going to be a 2yx that we see with a dydx that would get divided so 2 YX and then we also have a Min - x cub getting multiplied by dydx so we'd also have- X cubed in the denominator and that's how we solve for Dy DX now we just need to make sure that this matches the answer that was part of the question and if we scroll up we see we're supposed to have 3x^2 y yes - y^ 2 yes / 2x y yes we have that - x Cub so there we go I will take one last step of just rewriting this exactly as it appeared in the question all right let's move on to Part B we are supposed to find all points on the curve whose x coordinate is one and write an equation for the tangent line at each of these points so we'll take the original equation defining the curve and replace x with one and then find all the possible y values those will tell us the full sets of points where the x coordinate is one on this curve I'm switching to purple here just for kicks so let's rewrite this this equation but replacing the X's with ones so that's going to give us y^2 - y = 6 and now we just have to solve this quadratic for y let's subtract six from both sides so y^2 - y - 6 = 0 and then we can Factor this into y - 3 * y + 2 = 0 we know this factorization works because -3 * + 2 is - 6 and -3 + 2 is -1 so we see this is a correct factorization and this gives us the possibilities that y = 3 or y = -2 that of course is just the zero product property so now let's come over here and write our tangent lines we will do the tangent to the 13 remember this came from X equaling 1 so that's the x coordinate and then we'll also find the tangent line at the point 1 -2 now at the point 13 using point slope form we of course would have y - 3 equals the slope which we'll have to find multiplied x -1 now to find this slope we're just going to plug this point into the derivative dydx that we found in part A and I'll put that in red right down here so what we're doing is taking dydx and we're plugging in - -2 which is just y + 2 this has to equal the slope which we'll find in a minute multiplied by x -1 to find the slope of course we'll just plug this point 1 -2 into the derivative just like we did before so we are now plugging in x = 1 and y = -2 what is this going to equal well the derivative says we have 3x^2 y first X is 1 so x^2 is 1 Y is -2 so 3 Y is -6 and then subtract y^2 so in this case that's -4 in the denominator we have 2X Y which is just going to be -4 and then subtract X cubed thus we'll subtract 1 -6 - 4 is -10 and the denominator is -5 so this is -10 / -5 and so the slope is pos2 and so we can replace our box with the slope turns out it is 2 and so this is the tangent line y + 2 = 2 * x -1 and we can leave it in point slope form all right it is on to part C find the x coordinate of each point on the curve where the tangent line is vertical that means the slope is infinite so to speak so to find where that happens where the tangent line is vertical we're going to need to see where the denominator of the derivative is equal to zero that's what will make it infinite if the denominator is zero then we just have to check that where the denominator is zero the numerator is not also zero as long as the numerator is anything else we'll be fine so let's begin by setting the denominator equal to zero the denominator of our derivative is 2x y - x Cub so we have that 2x y - x Cub is equal to 0 and thus 2x y = x Cub we can actually just divide both sides of the equation here by X because if we come up to the definition of our curve it's given by this equation and if x was equal to Z this equation would never be true because 0- 0 does not equal 6 so X cannot be equal to 0 thus it's totally fine to divide both sides of this equation by zero thus we have 2 y = = x^2 again this is the denominator of our derivative which only applies on the curve and on the curve X is not equal to zero from here we can divide both sides by two and find that y = 12 x² so where is the tangent line vertical well it's vertical on the curve where y = 1x^2 so let's go back to our original equation and replace y with 12x x^ 2 and see what x values come out of that one more time what we did was set the denominator of the derivative equal to 0 that led us to this y = 12x^2 but we're only interested in points that are actually on the curve so we're going to plug this into the curve and see what x values will satisfy this this is the original equation let's go ahead and replace the Y's with 12x^2 so first we're going to have x times 12 x^ 2^ 2ar which is4 x 4 and then we're going to subtract X cubed multiplied by 12 x^2 I'll put that in parentheses this of course equals 6 on the right side of the equation x * 4 x 4th is a 4 x to 5th and then - x Cub * half x^2 is - 24s x 5 I'm just rewriting 1/ 12 as 24s because I want common denominators and this equals six on the right side of the equation now we can go ahead and combine these on the left to get minus 1/4 x ^ of 5 = 6 multiply both sides by -4 and we have that X to 5 equal -24 and thus X is the fifth root of -4 we can't take the square root of a negative but we can take an odd root of a negative so this is defined this is a real number so it appears we have a vertical tangent where x equal the 5th root of NE -4 however we do need to verify that the numerator of our derivative is non zero here to confirm that we actually have a vertical tangent we need to make sure the numerator of the derivative is non zero we know that X is the 5th root of -24 and we can plug that into this equation for y in order to find the appropriate y-coordinate remember we assumed this equation for y was true and plugged it into the original equation for the curve to get this x value so to get the yv value just plug this x into this equation for y then we'll plug the X and Y into the numerator of the derivative the numerator of the derivative for your reference is 3x^2 y - y^2 when we plug this value for x into this equation for y this is what we get 12 x^ 2ar so the fifth root of 24 becomes 24 to the 25s the negative goes away because of the squaring when we plug in for y so you square that negative goes away and it's just 24 25ths with that 1/2 out front then we plug this value for x and this value for Y into the numerator of the derivative that gives us 3 * x^2 squaring X again gets rid of the negative so we just have this multiplied by Y which we see here and then we subtract y^ 2ar which we see here squaring y turns the half to a 4 and the 24 25ths becomes 24 to the 4ths now this is clearly nonzero because it's clearly a positive number because what we have on the left is 3es 24 to 45s if you were to combine those 3 Hales 24 to 45s and we're subtracting a smaller number 1/4 24 to the 45s so we've got 3 Hales 24 4 fths minus the smaller number a 4th 24 to 4 fths so clearly this is positive and more importantly that means it's non zero so indeed X = the 5th root of -24 is the x coordinate of a vertical tangent and that completes our solution to problem 5 of the 2000 AP Cal AB exam this is free response question six from the 2001 AP Cal AB exam this is part of the no calculator section the function f is differentiable for all real numbers that's nice the point 31/4 is on the graph of y = f ofx and the slope at each point Point XY on the graph is given by this equation part A asks us to find the second derivative of y with respect to X and evaluate it at the point 3 1/4 so we will start with part a right down here we'll be pretty straight forward we just need to take the derivative of both sides of this equation so on the left we'll have the second derivative of y with respect to X and on the right we're going to have to deal with a little bit of implicit differentiation because we have that y there so the second derivative of y with respect to X is the derivative of dydx which is given to us is y^ 2 * 6 - 2x now to take the derivative of this we are going to have to be careful and make sure we use the product rule we have this function y^2 which we could call F and this function here which we could call G so in the context of the product rule we've got to do fime G Plus G Prime f beginning with F Prime we're going to have 2y don't forget the Dy DX this is implicit differentiation and then add or sorry multiply by G so frime g g is 6 Min - 2x then we add G Prime F the derivative of G is just -2 so this is plus -2 multiplied F which is just y^ 2 so -2 y^ 2 now we could simplify this a little bit because we know what dydx is so we could replace it with this expression but we're just asked to evaluate the second derivative at this point 31/4 so let's just do the evaluation there's no need to mess with this expression a bunch since dydx is part of this expression though let's start by evaluating what Dy DX is at the point in question 34 so we're plugging in x = 3 and Y = 1/4 into this expression if we just focus on the 6 - 2x for a second if x = 3 6 - 2x is 0 and so in fact dydx is actually 0er at this point in question and so this whole term in the second derivative would get canceled out because the dydx factor is zero so all we have to evaluate is actually that right there so we are looking for the second derivative of y with respect to X evaluated at the point 314 this first part of the second derivative is just going to be zero and then we have to plug in y = 1/4 to -2 y^ 2 so that's going to be -2 * a 4^ squar a 4 squar is 1/16th and so our final answer for part A is -8 moving on to Part B Part B asks us to find y = F ofx by solving the given differential equation with this given initial condition in order to do this we're going to use separation of variables so we'll divide both sides of this equation by y^2 so the Y's are all together on the left side and then we'll multiply both sides by DX so the X's are together on the right side then we can integrate both sides of the equation and use the given initial condition to solve for the arbitrary constant so we'll put Part B down here and I will just rewrite the given differential equation dydx = y^2 * 6 - 2x like I said we're going to divide both sides by y^2 so on the left we're going to have y to -2 and we're going to multiply both sides by DX so there's no DX on the left anymore and on the right we have 6 - 2x DX now we can integrate both sides of this equation equation because we've separated the variables on the left the integral of y -2 Dy is going to be Y the1 in theory we also have the addition of an arbitrary constant but let's imagine that we collect all of our arbitrary constants on the right side of this equation so we won't worry about it on the left side on the right the integral of 6 - 2x is 6x - x^2 and then we'll put the arbitrary constant over here now so we're going to have y1 equals let's put the x^2 first which is now positive and then - 6X and then minus C because we're multiplying by -1 but C is -13 so minus -3 which is the same as+ 13 finally we can invert both sides of this equation to finish getting y by itself thus on the left we'll have y and on the right we're just going to have 1 over x^2 - 6X + 13 that's the answer and that completes our solution to free response question 6 from the 2001 AP Cal AB exam up next are some related rates problems when solving related rates problems make sure not to plug in any known quantities for things that change before taking the derivative when you plug in a known quantity for something that changes over time you are fixing yourself at a moment in time and thus taking the derivative with respect to time would not make any sense so don't be too hasty when plugging in known values good luck this is free response question six from the 2002 AP Cal AB formb exam got a very nice picture for this one ship a seen here is traveling du West toward Lighthouse rock at a speed of 15 km per hour ship B seen here is traveling due north away from Lighthouse rock at a speed of 10 km per hour note that means X is decreasing because ship a is traveling this way whereas Y is increasing because ship B is traveling this way all right let X be the distance between ship a and Lighthouse rock at time T and let y be the distance between ship B and Lighthouse rock at time t as shown in the figure above very nice part A find the distance in kilometers between between ship a and ship B when x = 4 and y equal 3 so X is equal to 4 Y is equal to 3 we just need to find this segment here that's just Pythagorean theorem and if you know your Pythagorean triples you already know the answer is five if we were to call the distance between them let's not use D because we're going to take some derivatives later let's use R we would have that R2 = 3^2 + 4 squar by the Pythagorean theorem 3 S is 9 4 S is 16 so that would be 25 so R would equal the square < TK of 25 of course that is where we get our answer of five and the units here are kilom moving on to Part B find the rate of change in kilm per hour of the distance between the two ships when x = 4 and Y = 3 so for this we are again going to use the Pythagorean theorem but we can't plug in three and four immediately because Y is only equal to 3 and x equal to 4 for a moment generally speaking these quantities are changing so the relationship we need is r^ 2 = x^2 + y^2 now we can take the derivative of both sides of this equation in order to relate the rates on the left we apply the power rule to get 2 R but then remember we also have implicit differentiation going on because R changed with respect to time so we have drdt and then same thing on the right 2x multili DX DT plus 2 y multiplied 2 uh multiplied Dy DT now we can start to plug in known information we're focused on the moment when X is 4 and Y is 3 so we can replace x with 4 and Y with three but we also know when X is 4 and Y is 3 we figured out that R has to be five that was part a so on the left we have 2 * 5 which is 10 multiplied Dr DT on the right we have 2 * X but X we know is four so that's 8 dxdt we also know because it says in the problem how ship a is moving it says ship a is traveling toward Lighthouse rock at a speed of 15 km/ hour so dxdt is -5 this segment this distance is being reduced by 15 km hour so5 and then we have plus 2 y but we know that Y is 3 so 2 Y is 6 and dydt is going to be 10 because B is moving away from the Rock at 10 km/ hour so 6 multiplied 10 all right everything's in place we just have to solve for drdt on the right side we have 8 * -15 which is -10 plus 60 so -60 is the sum and then divide by 10 so -6 and this is a rate of change of a distance it is kilm per hour finally part C let Theta be the angle shown in the figure so that's this angle here and we want to find the rate of change of theta in radians per hour again when X is 4 and Y is 3 so we want to relate this angle Theta to these sides X and Y we can do that not using S because s of theta would be y over the hypotenuse not using cosine which would be X over the hypotenuse we need to relate these sides using tangent tangent of theta is opposite y over adjacent X so this is just like Part B but we are using a different equation because we're concerned with a different quantity this time we're concerned with the angle tangent of theta like we said is opposite y over adjacent which is X now we can take the derivative of both sides of this equation and eventually solve for D Theta DT on the left the derivative of tangent is secant SAR Theta and then we are multiplying by D Theta DT because the angle changes as time passes and then on the right we are unfortunately going to need the quotient rule first we'll have the derivative of the numerator which is dy DT and I suppose we're going to have to have a little bit more room to work with here so let me move this down so first the derivative of the numerator so Dy DT multiplied by the denominator which is X and then minus the derivative of the denominator dxdt multiplied by the numerator which is y and then all of this gets divided by the denominator squared now we can plug in the known pieces secant sare of theta we can figure out by looking at the triangle we know that X is 4 and Y is 3 remember that secant is the reciprocal of cosine and cosine is adjacent over hypotenuse so if cosine of theta is equal to adjacent X over hypotenuse which we know to be five in this case cosine of theta is 4 over 5 so secant is 5 over 4 so secant squared is 25 over 16 so we have 25 over 16 D Theta DT now let's worry about the right side on the right it's very much like the previous problem dydt we know is pos1 X was given to us as 4 dxdt we know is -15 so - -15 y we know is 3 so -5 * 3 and then again X is 4 so the denominator will be 4^ SAR or 16 now let's just simplify this side a little bit 10 * 4 is 40 and then 15 * 3 is 45 So This is 40 + 45 which is 85 and then this is getting divided by 16 now we'll multiply both sides by Sorry by 16 over 25 to get D Theta DT by itself so on the left we'll have D Theta DT and now we're multiplying 85 over 16 by 16 over 25 this of course produces 85 over 25 which we can simplify five goes into 85 17 times and five goes into 25 five times so final answer is three 3.4 and let's not forget our units this is an angle and so the rate of change is radians per hour it is the angles rate of change and that is our solution to free response question 6 from the 2002 AP Cal AB formb exam this is problem five from the 2003 AP Cal AB exam let's read it a coffee pot has the shape of a cylinder with radius 5 in which we see here let H be the depth of the coffee in the pot which is also measured in inches H is a function of time measured in seconds the volume v of coffee in the pot is changing at the rate of -5 Pi root H cubic in per second so the volume of coffee in the pot is decreasing coffee is leaving the pot and thus the height of the coffee is going to be decreasing as well part A is to show that the height of the coffee is changing with respect to time in this way < TK over 5 so we just need to find dhdt we can do that using the volume of a cylinder equation which is given to us v = < r^2 h the cylinder we're thinking about is this cylinder of coffee inside the cylindrical coffee pot if we take the derivative of the left side of the the volume equation with respect to T we're going to get dvdt which is going to work out just great because dvdt was given to us we know that dvdt equals -5 piun < TK so let's go ahead and write that DV DT we know is -5 piun all right now I'm going to rewrite the volume equation volume = PK r^2 H and we can go ahead and take the derivative of both sides remember that on the right the only function is H Pi is a constant and R is a constant as well the radius of the body of coffee is going to be 5 in the radius of the containing cylinder even as coffee leaves the pot that radius is not going to change all right so taking the derivative on the left with respect to time we get DV d t on the right R 2 is 5^ SAR because the radius is fixed at 5 so this is just going to be 25 pi multiplied by taking the derivative DH DT now we're trying to solve for dhdt we know what dvdt is so let's replace dvdt with its given value which is -5 piun h now we'll just divide both sides of this equation by 25 Pi in order to solve for dhdt and I'm going to write dhdt on the left so this will be equal to this ID 25 pi and when we divide by 25 Pi 5 over 25 just leaves a factor of five in the denominator the pies cancel out and of course we still have that negative and thus we see that dhdt equals < TK over 5 just as desired so we can move on to Part B given that H equal 17 at time tal 0 we want to solve this differential equation for H as a function of T So This is actually a separation of variables differential equation question we need to separate the H's we'll move the H's to the left and separate the t's get those to the right then integrate both sides of that equation and use this initial condition in order to solve for the arbitrary constant so let's start setting this up and I'll write it in red we're going to take this equation divide both sides by root H and multiply both sides by DT thus we're going to have 1 overun H DH on the left and on the right we're going to have 15 DT Now 1 /un H is the same as H to the2 so I'm just going to write it like that because that's easier to integrate now we integrate the left side and we integrate the right side on the left using the reverse power rule to integrate this function we increase the power of2 by 1 thus it becomes a power of positive half and then we divide by this new power dividing by a half is the same as multiplying by two in theory we also have to add an arbitrary constant but that's going to happen on the right side as well so let's just gather the arbitrary constants on the right side this is going to equal the integral of -5 which with respect to T which is just -5 T and then we have all our arbitrary constants we'll just call that A+ C so how do we find H as a function of t well first let's use our initial condition in order to solve for C the arbitrary constant the initial condition is that when T equals 0 H is 17 so let's plug that in we have 2 multiplied 17 to the 12 which is the square otk of 17 equals -5 * T but we're plugging in zero for T so that's just zero and this equals c so there we go that's the arbitrary constant it's 2 * the < TK of 17 so I guess we'll just go ahead and replace that c right now let's write this as+ 2 * theare < TK of 17 and then I'll move this equation off to the side we don't really need that anymore now all that's left for us to do is to solve for H to do that we'll divide both sides of this equation by two and then Square both sides of the equation dividing both sides by two turns -15 T into -110 T and 2 < tk7 becomes just < TK 17 and then we would Square both sides to get rid of that exponent of 1/2 and thus we have our function for H in terms of t h = -110 t + the < TK of 17 all squared and we're in luck because we can very easily use our solution to Part B to answer part C part C asks at what time T is the coffee pot empty well the coffee pot would be empty when the height reaches zero that would mean that the coffee is all the way at the bottom there is no coffee when the height is zero that's when there's no coffee so we need to come to this equation and plug in zero for H and then solve for T now if we plug in zero for H I'll write that this is part C if we plug in zero for H we have that 0 equals this thing squared but if 0 equals something squared then 0 must just equal that something so 0 = -110 t plus the square of 17 again we need this to be zero because we need the height to be zero because that represents there being no coffee in the pot and then we just solve this equation for T so we'll subtract < tk7 from both sides so we have < tk7 = -110 t and then multiply both sides by -10 to finish getting T by itself thus we have that t equal 10 * the sare < TK of 17 and that is when the coffee pot is empty after about 10 * the square < TK of 17 seconds and that completes our solution to problem five from the 2003 AP Cal AB exam this is free response question five from the 2005 AP Cal AB Form B exam consider the curve given by y^2 = 2 + XY part A asks us to show that the derivative of y with respect to X is y over 2 y - x so this is just a simple implicit differentiation problem for part A we are going to have to take this equation and we can differentiate the left side and the right side and then solve for Dy DX if we take the derivative of the left side we get 2 y but then we have to apply the chain rule this is the key step of implicit differentiation we got to multiply by Dy DX because y also changes with respect to X this equals taking the derivative on the right side the derivative of two is zero so we don't have to write that for the derivative of XY we are going to need the product rule if we call XF and Y G then the product rule is fime G Plus G Prime f f Prime would just be one and G would be Y and then we'd have to add G Prime f g Prime is dy DX and F is just X and again that's just applying the product rule all right All That Remains is to solve for Dy DX let's subtract X dydx from both sides so that the dydx terms are together on the left then we will have 2y Dy DX Min - x Dy DX equal y and you can see how this is going to lead to the expected result on the left we'll factor dydx out of both terms so we have dydx multiplied 2 y - x = y and then finally divide both sides by 2 y - x to finish solving for Dy DX and we get exactly what we expected Dy DX equals y / 2 y - x and going back to part A you can see that is the answer we were supposed to get we don't even have to do any weird simplification all right Part B find All Points XY on the curve where the line tangent to the curve has slope 12 we know the slope is the derivative which is given by this expression here y over 2 y - x so we just need this expression the derivative to equal 1/2 so here that is I've set the derivative equal to 1/2 let's multiply both sides by two and multiply both sides by 2 y - x to get rid of the fractions then on the left we're going to have 2 Y and on the right we multiplied both sides by two so the 1/2 is gone and it's just going to be 2 y - x on the right side then we can subtract 2 y y from both sides and we find that 0 = --x which of course is the same as saying x = 0 if x equals 0 then what is y well we can refer back to the equation of the curve y^2 = 2 + XY if x is zero this XY term just goes away and so we have y^2 = 2 and if y^2 = 2 then y must be plus or minus < tk2 thus there are two points where the slope of the tangent to the curve is 1 12 there is the point 0 < tk2 and the 0 < tk2 all right let's go ahead and move on to part C show that there are no points XY on the curve where the line tangent to the curve is horizontal a horizontal tangent means that the slope or the derivative at that point is zero so this is like Part B except we need to take that derivative and set it equal to zero now the only way that this derivative is going to be zero is if the numerator which is y is equal to zero however that just can't ever be the case because if we look at the equation of the curve if Y is zero then 0^2 = 2 + x * 0 clearly this has no Solutions so if the derivative is zero which we need for a horizontal tangent that means Y Must Be zero because Y is the numerator of the derivative and So based on the equation of the curve we would need 0^2 to equal 2 + x * 0 which would mean we need 0 to equal 2 that obviously is not possible thus there are no points on the curve where the tangent has a slope of zero we see that assuming there is such a point results in a contradiction moving on to Part D let X and Y be functions of time T that are related by this equation we've been working with at time t equal 5 the value of y is three and dydt is 6 we want to find the value of dxdt at time t equal 5 so this is like a related rates problem now this is our expression for Dy DX but now we're thinking about derivatives with respect to time so what is dy DT well it's still the case that y changes with respect to X but now they're also both changing with respect to time so dydt would be Dy DX because again y does change change as X changes but because X changes with respect to time it's the chain rule here we also have to multiply by DX DT so dydt is just dydx but multiplied by the change in X with respect to time now we can use the known information to figure out what dxdt is at tal 5 we were told that at Tal 5 Y is equal to 3 and dydt is equal to 6 if y equals 3 we can use the given equation of the curve to solve for x the equation of the curve was y s = 2 + XY so plugging in y = 3 we have 9 = 2 + 3x subtracting two from both sides gives us seven = 3x and so X = 7 over 3 now let's start to plug in our known information at tal 5 dydt is 6 so on the left we have 6 and then on the right y in the numerator is 3 in the denominator we have 2 * 3 - 7 over3 multiplied by dxdt which is what we are trying to solve for now 6 is the same as 18/3 so this is 18/3 - 7/3 but that's the same as 11/3 so let's just write that as 11/3 but then 3 over 11/3 is the same as 3 * 3 over 11 which is just 9 over 11 so now we can multiply both sides by 11 over9 in order to solve for dxdt so dxdt must equal 6 * 11 / 9 which is 66 over 9 which reducing by a factor of three is 22 over 3 and perhaps we should specify with notation that this is the value of dxdt when T equals 5 and that is our answer 22/3 and that completes our solution to free response question five from the 2005 apal AB Form B exam next up are problems on Extreme values and concavity for these problems recall that it often comes down to finding critical points which is where the derivative equals zero or doesn't exist but a critical point is not necessarily the location of an extreme value if xal C is a critical point we have to check if the derivative changes sign at that point in order to determine if it's an extreme value or or not if a derivative goes from positive to negative the function goes from increasing to decreasing and thus that critical point would be a maximum on the other hand if the derivative goes from negative to positive the function goes from decreasing to increasing and thus that point is a minimum you can also use a second derivative and assess concavity to classify critical points good luck this is free response question two from the 1998 AP Cal AB exam let's read it let F be the function given by FX = 2x * e 2x part A asks us to find the limit of f ofx as X goes to negative infinity and the limit as X goes to positive Infinity so let's put part A down here and we will start by evaluating the limit of our function as X goes to NE Infinity if x is going to negative Infinity then 2x is also going to negative Infinity which means this 2x in the exponent of e is going to negative Infinity so we are in fact dividing this 2x by increasingly large powers of e because e has a negative exponent so really it's in the denominator and we could write it that way perhaps to make this a little more clear this is the same as the limit as X goes to negative Infinity of 2x in the numerator divided by e to the absolute value of 2x in the denominator again the power of E is negative because X is going to negative Infinity but we can make that power positive by moving the e to the denominator at this point it's pretty clear that the limit equals zero sure the numerator and denominator are both going to infinity or negative Infinity it's the numerator going to negative infinity and the denominator going to positive Infinity but hopefully you know exponential growth e 2x in the denominator far outpaces the linear growth of 2x in the numerator you could use Lae Tal's rule to formalize this but regardless the denominator is going to Infinity far quicker than the numerator is and so this limit is zero for Part B the limit is X goes to positive infinity or excuse me this isn't Part B this is the second part of part a for X going to positive Infinity this is quite obviously going to positive Infinity you just have 2X which is getting really big and E 2x which is getting really big so you could say this limit is infinity or it doesn't exist remember that a limit diverging to Infinity is just a particular type of nonexistence let's move on then to the actual Part B of this problem find the absolute minimum value of the function f and justify that our answer is an absolute minimum to investigate extreme values we of course need to find critical points so for Part B we will Begin by taking the derivative finding F Prime and then setting that equal to zero now f is a product of functions you might call 2x U and e to the 2x V then to take the derivative we need to use the product rule U Prime V plus v Prime u u Prime would just be two so U Prime V is 2 e 2x if V is e 2x then V Prime is e 2x * 2 because 2 is the derivative of the inside function 2x so V Prime is e 2x * 2 which I'll write is 2 * e 2X and then you have to multiply that by U which is 2x finally we'll set this equal to zero and solve for x to find critical points as we go ahead and set this equal to Z I'm going to factor 2 e 2x out of both terms so we have 2 e 2x multipli by 1 + 2x and we're setting this equal to zero note also that this derivative clearly exists everywhere so the only critical points will be where the derivative equals zero now 2 e 2x is never 0 so the only time the derivative equals 0 is when 1 + 2x = 0 which only happens when X is -2 thus we have our critical point now whether or not an absolute minimum occurs at this critical point is going to depend on the behavior of the derivative to the left and to the right of the critical point for xal -2 to be the location of an absolute minimum it must be the case that the function is decreasing to the left of - one2 and then increasing afterwards so decreas and then increasing to see if that's the case we can take a sample point from the left of one2 and plug it into the derivative and see if it's positive or negative and then take a sample point from the right of negative one2 and plug it in the derivative and see if it's positive or negative so to the left of negative - one2 let's choose a simple Point like sayga 1 what is f Prime of1 plugging -1 into the derivative gives us 2 2 e to the -2 I'm plugging it in here and then 2 * 2 * -1 is -4 so -4 * e to -2 and clearly this is less than zero because we have 2 e the -2 which is positive but then we're subtracting a bigger positive number we're subtracting 4 e to the -2 this clearly is negative which means to the left of -2 F Prime must be less than zero to the right of -2 let's plug in a sample Point like zero zero is a very easy option so F Prime of zero what is that well that's going to be 2 * e to the 0 and then this second term in the derivative is just zeroed out and so this is frime of 0 which is clearly positive because it's just equal to 2 * 1 so to the right of -2 frime is positive thus x = -12 is the location of an absolute minimum but remember Part B asked us to find the absolute minimum not just the location of it the x coordinate is the location but we need to actually plug the x coordinate into to our function in order to find what the absolute minimum actually is so let's plug -2 into our function f now f of x is 2x e to the 2x so we're going to have 2 x which is -2 time e to 2x so e to 2 * -2 what is this well 2 * -2 is -1 and then that's getting multiplied by e to the 2 * -2 so that's e to the1 this clear CLE L is1 over e thus we have our conclusion because the derivative is negative for X less than2 and the derivative is positive for X greater than2 the function is decreasing and then increasing it switches right at5 and so F of - one2 which is 1 over e must be the absolute minimum of the function on to part C what is the range of F Well we just figured out that the minimum value F takes on is-1 over e and we know from part A that the function goes to positive Infinity as X goes to positive Infinity so there is no upper bound on this function we clearly know the lower bound is -1/ e and so its range is everything from -1 / e to positive Infinity so for part C we can say that the range is everything from -1 over e which is included the function does actually attain that minimum value and then there is no upper bound so everything up to positive Infinity not including positive Infinity because positive Infinity is not a number finally let's take care of Part D part D says consider the family of functions defined by y = BX e to BX X where B is a nonzero constant we've been working with the particular function where B is 2 2x e 2x so considering this more general family we want to show that the absolute minimum value of BX e to the BX is the same for all nonzero values of B so the absolute minimum should not depend on B the absolute minimum should be the same we just found the absolute minimum of 2x e 2x was 1/ e so then it should be the case that the absolute minimum of any function BX e to BX is 1 E provided that b is non zero so let's try to prove that we're considering y = BX e to the BX to determine anything about the absolute minimum of such a function we'll need to take the derivative and find the critical points y Prime is well again it's just product rule we have B multiplied e to the BX that's like U Prime V and then we're going to have b^ 2 x e to the BX that's like V Prime U now this derivative is defined everywhere so in investigating critical points All That Remains is to set it equal to zero as we do that we'll Factor b e to the BX out of both terms so we have b e to the BX multiplied 1 + BX and we're setting this equal to zero assuming that b is non z b e to BX is also non Z so the only critical point occurs when 1 + BX equal 0 1 + b x only equals 0 when X is -1/ B just like in the previous part of this problem we had xal -1 /2 so if this function has an absolute minimum it must occur at X = -1 / B so let's try evaluating the function at x = -1 / B at x = -1/ b we have that y would equal B multiplied by -1 / B multiplied by e to the B multiplied by -1/ B simplifying B * -1 over B is -1 and e to the B * -1/ B is e to the -1 this as expected is1 over e now how do we actually know that each one of these functions y = BX e to the BX has a minimum we know that if it has a minimum then it occurs at -1 overb and this is what the minimum is but how do we know they're all minimums how do we know that there's not one of these that's actually a maximum for certain value of B well to proceed with this investigation we can assume that b is positive if you consider a positive case like y = 2x e 2x what's the difference if we make that negative like Y = -2x e to the -2X it's going to be the same except the Y values are going to occur at the opposite X values it's as if we didn't change 2 to -2 and instead change the X's tox and when you make a change like this what amounts to just changing X tox that just causes a reflection across the y AIS something like this to something like this it's just a reflection across the y- AIS which does not change the minimum value so we can assume that b is greater than zero the minimum values will be the same for the negative B's in that case we can again consider a sign chart here's -1/ B and since B is positive this is a negative number so to the right of -1 over B is the point0 and F Prime of 0 is clearly positive if we plug zero in here this term goes away and this is just B * 1 again we're assuming B is positive so in the case where X is zero the derivative is positive and so the function is increasing to the right of-1 overb on the other hand to the left of-1 overb would be a number like -2 overb and if we plug -2 overb into F Prime this is what we get to make the notation more clear let me specify that right now I'm considering F ofx to be the general function BX e to the BX so fime just looks like this and if we plug in -2 over B this is what we get which is clearly negative because it's a positive number b e -2 minus a bigger positive number minus 2B e to -2 so clearly that guy is negative so the function will go from decreasing to increasing in all cases and so we can State our conclusion for Part D that any function like this any function y will have an absolute minimum of -1/ e and this applies for all nonzero values of B and that completes our solution to free response question two from the 1998 AP Cal AB exam this is problem four from the 1999 AB exam let's read through it suppose that the function f has a continuous second derivative for all X and that F of 0 = 2 frime of 0 is -3 F Prime of 0 is zero we'll let G be a function whose derivative is given by the following equation involving F and F Prime part A is to write an equation of the line that's tangent to the graph of F at the point where x equal 0 so we'll use point slope form for this remember the point slope form of a line is y minus the y coordinate what's our y-coordinate well we want the line to be tangent to F where x is 0 and we know that when x equal 0 f is 2 so the y-coordinate is that output of the function which is 2 so we'll have y - 2 equals the slope what's the slope well frime of 0 is given to us it's -3 that's the slope of a tangent line at x = -3 on this function so the slope is-3 and then X of course minus the x coordinate which is zero and we don't even have to write that so if we want to put this into slope intercept form which we don't have to do but this is a really simple line so perhaps we might as well we get y = -3x + 2 so that is our tangent line let's move on to Part B is there sufficient information to determine whether or not the graph of f has a point of inflection when x equals 0 we have to recall that a point of inflection is a point where the graph changes concavity where the second derivative goes from negative to positive or from positive to negative in this case there is not sufficient information the only thing we know about the second derivative is that it is continuous and when x equals 0 the second derivative is also zero but we don't know the values of the second derivative to the left or to the right of zero so we can't say if it switches sign and thus we can't say if in fact xal 0 is a point of inflection moving on then to part C given that g of 0al 4 write an equation of the line tangent to the graph of g at the point where x equal 0 so this will be a lot like part A we'll use the point slope form of a line in order to get our tangent line point slope form is Yus the y coordinate we want this line to be tangent to the graph of G so the y coordinate is the output of G when X is zero and we're given that is four so we have Yus 4 this has to equal the slope of the tangent line at xal 0 multiplied by X of course minus the x coordinate which is zero we don't have to write minus 0 so now let's just focus on this slope this slope is G Prime of 0 so we just need to figure out what that is we're given that g Prime is this big long expression so let's just plug xal 0 into that g Prime of 0 is going to be e to -2 * 0 multiplied by 3 * F of 0 + 2 * frime of 0 and now we can plug in the known pieces of information here e to the -2 * 0 is just e to 0 which is one so this part doesn't really matter 3 * F of0 what is that well looking at the question F of 0 remember is 2 so 3 * F of 0 is 6 then we must add 2 * frime of 0 fime of 0 is -3 so 2 * frime of 0 is -6 thus the derivative of g at x = 0 is 6 + -6 which is zero and so we can plug zero into that box thus the right side of this equation is actually just zero and so we can add four to both sides and find that the equation of the tangent line is just y = 4 a horizontal line finally let's move on to Part D part D says show that g Prime of X the second derivative of g equals this big long expression and then we want to say does G have a local maximum at x equals 0 or not and we need to justify our answer okay so this works out pretty nicely because once we take the second derivative we'll find that it equals this and then we can just plug xal 0 into that second derivative that will give us the concavity if the concavity is downwards so the second derivative takes on a negative value well that will tell us that in fact we have a maximum and if when we plug xal 0 into the second derivative we get a positive number that tells us it's concave up and we have a minimum so let's go ahead and take the derivative of the derivative of G remember that g Prime was given to us as this expression e -2X * all this stuff so I'm going to go ahead and Shrink our work for parts a through C just so I can scroll up a little a little bit and we can see the whole problem while we work through part D so to find G Prime of X we'll just go ahead and take the derivative of both sides of this equation that of course gives us GP Prime on the left and on the right we'll need some derivative rules first notice that it's a product of functions an exponential function times this sum of functions so the product rule says the derivative of this will be U Prime V plus v Prime U the derivative of the first function times the second function so we'll start with that the derivative of the first function is -2 e to -2X using a little bit of chain rule there and then we have to multiply by the second function which is just 3 F ofx + 2 fime of X so that is our U Prime V from the product rule now we need to add V Prime U I'll do that on a second line so that I don't run out of room here so let's think about the derivative of 3 F ofx + 2 frime of X that's our second function which we need the derivative of here and that's pretty straightforward that's just going to be 3 F Prime of X and then plus 2 fpre of X and this needs to get multiplied by the first function which is e to -2X now we can factor out and e to -2X from both of these terms we can also then distribute the -2 through these parentheses and do some simplification so here we are after factoring out the E to -2X we have e to -2X that's getting multiplied by everything here in parentheses we have -2 * 3 F ofx and -2 * 2 fime of X and we have 3 fime of X and we have 2 fpre of X so now let's just go ahead and combine like terms we can combine these fimes and it looks like that's it -4 fpre of x + 3 fime of X is -1 frime of X and so now we can see that indeed this second derivative of G matches what we were supposed to get in the question e to the -2X * -6 FX - frime of x + 2 fpre of X now to determine if G has a local maximum at xal 0 we will determine its concavity by plugging xal 0 into this second derivative what is g Prime of 0 well when we plug Z into the exponential that's just going to give us one so let's not worry about that F of0 we know is equal to 2 so -6 * F of 0 is going to be -12 then we have to subtract fime of 0 fime of 0 is -3 so if we subtract that we are adding three finally we have to add to fou Prime of 0 FP Prime of 0 is 0 so 2fp Prime of 0 is just 0 so this is -12 + 3 which is -9 thus we have a concave down function here at xals 0 and so indeed we have a maximum and there is my written out conclusion and that completes our solution to problem four from the 1999 exam this is problem six from the 2008 AP Cal AB exam so this would have been the final question part of the no calculator section let's read it let F be the function given by F ofx equal Ln X overx for all X greater than Z it's nice enough to just give us the derivative here so we don't even have to use the quotient rule to find that and part A is this write an equation for the line tangent to the graph of F at x = e squared so pretty straightforward to write the equation of a tangent line easiest way to do it is usually point slope form and we do that like this y minus the Y coordinate the y-coordinate is found by just plugging the x coordinate into the function so we'll just plug e^2 because remember that's the x coordinate we'll plug that into the function and this equals the slope which we'll find by plugging e s into the derivative multiplied by xus the x coordinate all right let's just evaluate these pieces and we will be done F of E2 what is that well we just plug e s into our function so it's Ln of e^2 / E2 natural log of E2 is just two because Ln and E cancel out so this is just 2 over e^2 as for the derivative F Prime of e^2 which is going to give us our slope we've got to plug e^2 into this now same thing Ln of e^2 is just 2 so this is going to be 1 - 2 in the numer Ator which is -1 / e^ 2^ 2 which is e 4 in the denominator so this is -1 over e 4 all right now having evaluated the pieces our tangent line equation in point slope form is Yus the y coordinate which we found to be 2 over e^2 equals the slope which we found to be -1 over e to 4 multiplied by xus the x coordinate the x coordinate of course is e^2 and that is part a moving on then to Part B find the x coordinate of the critical point of f so the wording suggests there is one critical point and we have to find its x coordinate determine whether this point is a relative minimum a relative maximum or neither for the function and justify our answer we can determine if the critical point is the location of a maximum or minimum by seeing if the derivative switches sign and if so how it switches sign so let's get into this this is part B the first part of the question is to find the x coordinate of the critical point of the function the critical point of a function is where the derivative either doesn't exist or is equal to zero at this point we would use the quotient rule to find the derivative if it hadn't already been given to us the question already told told us the derivative is 1 - Ln X /x^2 now you may think x = 0 here would be a critical point because the derivative would not exist at xals 0 but in fact that's not a critical point because xals 0 isn't even in the domain of the function the function isn't defined at x equals 0 so this would not be a critical point so then what is a critical point of this function well we've got to see where the derivative equals 0 the derivative is going to equal 0 when 1 - Ln X is equal to 0 this of course means that 1 equal Ln X and the natural log of what is equal to 1 well that of course is just e this means the E must equal x then the question is is this the location this critical point is this the location of a maximum or a minimum or neither we have to think about what happens to our derivative as we pass from X being less than e to being greater than e well let's think about this when X is less than e we know that the natural log of x has to be less than one the natural log is going to be less than one because you need to raise e to fractional powers to get something that's less than e and if Ln X is less than one in this case then 1 minus lnx is positive so we could say to the left of e the derivative is positive now what happens to the right of E when X is greater than E when X is greater than e of course Ln X is greater than one because e needs to be raised to Powers bigger than one to get something bigger than e and if Ln X is greater than 1 then 1 minus Ln X is going to be less than zero I should point out we're only discussing the numerator because x^2 is always positive so we'll have no effect on the sign of the derivative so yes the derivative does switch signs when x equals e how does it switch signs well to the left of e the derivative is positive so our function is increasing and then to the right of e the derivative is negative it is decreasing thus at xal e we have a maximum the derivative goes from positive so an increasing function to negative so the function is decreasing sorry I just hit the mic and there our reasoning is written out F ofx has a relative maximum at xal e the critical point because frime of X the derivative switches from positive to negative at that point let's move on then to part C the graph of the function f has exactly one point of inflection find the x coordinate of this point a point of inflection you should recall is a point where function changes concavity which means it second derivative switches signs so for part C we are going to have to find the second derivative and perhaps this is why they gave us the first derivative there's no need to test our knowledge of the quotient rule twice we'll have to use the quotient rule once in order to get the derivative of the derivative so we can find the second derivative and see if there are any points of inflection or in fact we know there is a point of inflection we just have to find it because we're looking for where this second derivative switches signs and in order for it to switch signs it's going to have to pass zero so this is a lot like finding critical points and using that first derivative okay since the first derivative we know is 1 - lnx /x^2 it's pretty straightforward to apply the quotient rule to find the second derivative you have to start with U Prime V the derivative of the numerator times the denominator the derivative of the numerator is just -1 /x so we'll have - 1 /x that's the derivative of the numerator and then we have to multiply by the denominator so multiply by x^2 and then subtract V Prime U the derivative of the denominator multiplied by the numerator the derivative of the denominator is 2X and the numerator is 1 - lnx all right so that is the numerator based on the quotient Rule and of course all of this is over v^ 2 the denominator squared and x^2 squar is x 4 so that is our second derivative and then we're just doing some simplification -2X * Ln X is positive 2x Ln X so move that to the front then we have Min -2X right here and over here a factor of X cancels out this is actually just Min - x so between the -2X and the - x we have - 3x and of course x 4 is still in the denominator now everything here has a factor of X so we can actually cancel an X out of everything and so this simplifies to 2 Ln x - 3 / X cubed and that's not so bad necessarily at a point of inflection ction the second derivative has to change signs and thus it's got to hit zero so we'll set the numerator equal to 0 because that's the only way the second derivative can equal zero is if the numerator equals zero so 2 Ln x - 3 equal 0 and then we just have to solve this add three to both sides so 2 Ln x = 3 divide both sides by 2 so lnx = 3es and then exponentiate both sides right put it all a power of E E and Ln cancel out and we just have xal e to the 3es and yes indeed this is a point of inflection because when X is less than e to the 3es lnx is going to be less than three Hales and thus 2 lnx is going to be less than three so this will be negative but once X passes e to the 3es this is going to be positive because this term is going to be bigger than 3 of course the X cubed here doesn't affect anything because when X is close to e to the three Hales X is positive and so X cubed is also positive this is the location of our one point of inflection where the concavity switches finally onto Part D we are asked to find the limit of f ofx the limit of this function as X approaches zero from the right we of course can approach Zero from the left because the natural log is not defined for negative values we'll also not be able to just plug zero in because log is not defined at xals 0 and we can't divide by zero either and maybe we can sneak Part D in right here here's Part D the limit as X approaches zero from the right of our function and our function is Ln X overx this is actually pretty straightforward let's just think about what happens here if we approach Zero from the right on the natural log of X the natural log if you could sketch it real quick it veers off to negative Infinity as X approaches zero from the right and X of course is just a line it's looking like that as we approach Zero from the right so we're dividing by these small positive numbers and the numerator is massive negative numbers so this is definitely negative Infinity again that's because as we approach Zero from the right on the natural log we get really big negative numbers you need to raise e to really big negative powers in order to get really small positive numbers those numbers that are close to zero Meanwhile we're dividing these massive negative numbers by very small positive numbers in the denominator which only makes them bigger but doesn't change their sign so the limit is negative infinity and according to the AP college board scoring guidelines for this question you could also simply say that the limit does not exist because going to negative Infinity is just a special type of not existing and that completes our solution to the final question from the8 AP calab exam this is free response question five from the 2008 AP calab formb exam let's read it let G be a continuous function with G of 2 equal to 5 the graph of the pie wise linear function G Prime the derivative of G is shown here from x = -3 up to x = 7 part A says find the x coordinate of all points of inflection of the graph of G of X for X between -3 and 7 and justify our answer a point of inflection is somewhere the second derivative switches signs which means it's somewhere the first derivative goes from increasing to decreasing looking at the graph we see the first derivative remember that's what the graph is it's G Prime the graph goes from increasing to decreasing right here at xal 1 and it goes from decreasing to increasing right here at xal 4 at both of those places the second derivative must change sign because the first derivative has a change in increasing decreasing Behavior so there is our answer for part A G Prime changes from increasing to decreasing at xal 1 and from decreasing to increasing at xal 4 so those are the two points of inflection moving on to Part B Part B says find the absolute maximum value of G on the interval from -3 to 7 and justify our answer to have an absolute maximum the function must go from increasing to decreasing which means the derivative must go from positive to negative we see the derivative goes from positive to Nega ative at the point x = 2 and that is it however looking for extreme values we do also need to check the end points at x = -3 and at x = 7 now this is just the graph of the derivative so how are we going to use this to find values of the function to see which one of these candidates actually is the location of the absolute Max well we'll be able to do that by using integrals we can integrate the derivative and we also have this known point for the function G we know that g of 2 is five so we can integrate from there and find more values of G so here are our candidates the end points of the interval -3 and 7 and that one point where G Prime changes from positive to negative at xal 2 so let's assess the values of g at these points beginning with G of -3 we'll be able to see which one's the maximum once we check all of these values to find G of -3 we'll take the value of G we know that's G of 2 and then just accumulate G's rates of change so the derivative G Prime of x from that known point of two backwards to the desired point of -3 we can compute this integral of course using area calculations and the provided graph G of 2 to begin with we know is five so this is going to be five plus let's let's go ahead and calculate this integral the integral of G Prime from 2 to -3 is the area under the curve from here to here so that's all of this area and also we need all of this area so this is just a couple triangle area calculations now remember we're moving backwards right we're going from 2 to -3 so this area that looks positive because it's above the x-axis we actually need to count as negative because we're moving backwards and this area that looks negative because it's below the x-axis we actually need to count as a positive again because we're moving backwards all right so what are the areas of these triangles well it's just 1/2 base times height so beginning with this one which will make negative it's - one2 the base goes from -1 to two so that is a base of three and the height well it just goes up to one so the height is one and then we need to add we're going to make this area positive that's going to be 12 times the base the base goes from -1 to -3 so it's a base of two and then multiply by the height which is -4 but again we want this area to be positive so we're going to multiply by a height of pos4 another way to think about this is that the base is actually -2 because we're moving back two and the height is actually -4 because it's down four and so you can see those negatives cancel out and it just remains positive so here's the value of our integral with those areas of the triangles we just calculated let's go ahead and simplify this this is five which is the same as 10es and then here we have Min - 3es and then here we have + 8 Hales so in total this is 15 hves or 7.5 all right now as for G of two that is given to us we know that g of 2 equals 5 so the only thing left to check is G of 7 which we can find just like we found G of -3 we will find it by beginning G of 2 and then integrating the rates of change so G Prime of x from xal 2 to the desired point at xal 7 so let's head back up to our graph and find this integral using some geometry we need the area under the curve from 2 to 7 so that area there as well as this area now we are moving forward so the areas should be positive and negative in the ways that we would expect this first area underneath the x-axis will be counted as negative what is its area well it's 1 12 time the base the base is 1 2 3 4 multiplied by the height which is -2 and then we need to add the area of this little triangle here it's 12 multipli by the base of 1 multiplied by the height of one now we can go ahead and just simplify this real quick before we finish up the integral 12 * 4 * -2 is -4 and then 1 12 * 1 * 1 is just + 12 so this is -3.5 so back down here with our integral G of 2 we know is 5 the integral we just found is -3.5 so in total G of 7 is postive 1.5 now remember the question for Part B was to find the absolute maximum after having checked all the candidates we see the absolute maximum is 7.5 which occurs at x = -3 so I'll just write that a Max of 7.5 which occurs at x = -3 moving on then to part C part C says find the average rate of change of G of X on the interval from -3 to pos7 the average rate of change of G of X can be found by looking at the slope of the secant line through the end points thankfully we just found all the information we need in order to figure this out so let's put part C down here and we will refer back to our work in Part B the average rate of change of G from -3 to POS 7 is G of 7 minus G of -3 that's the change in G over this interval divided by the amount of distance that is passed so 7 minus -3 that's just the width of the interval G of 7 we figured out is 1.5 so we'll have 1.5 minus G of -3 we figured out is 7.5 so 7.5 and then IDE 7 - -3 is 10 that's the interval width 1.5 - 7.5 is -6 so this is -6 / 10 which equal -35 or 0.6 and that is the average rate of change of G over the interval in question finally Part D find the average rate of change of G Prime of X so we just found an average rate of change of G but part D is asking for the average rate of change of G Prime on this interval from -3 to 7 and then we need to answer this does the mean value theorem applied on the interval from -3 to 7 guarantee a value of C between -3 and 7 such that the derivative of the derivative GP Prime of C is equal to this average rate of change why or why not let's just start with the average rate of change of G Prime the average rate of change of G Prime just like with G is just going to be the values of G Prime at the end minus the beginning divided by the interval width so we need to calculate G Prime of 7 minus G Prime of -3 that's going to be the change in G Prime and then divide by the interval width 7 - -3 of course we already know that's 10 now these values G Prime of 7 and G Prime of -3 we can just grab from the graph G Prime of 7 is 1 and G Prime of -3 is -4 so we have 1us -4 over 10 1 --4 over 10 this is 5 over 10 which is 12 now does the mean value theorem guarantee that the second derivative takes on this average rate of change of the first derivative this value of2 at some point the answer is no because the exam next are problems involving tables and reman sums for these problems make sure you read those tables very carefully so that you understand exactly what information is being presented to you you might also find it helpful to sketch a rough picture of a function that agrees with the information in the table in this way you encode the table information in a picture which can be easier to read than looking at a table good luck this is free response question three from the 1998 AP Cal AB exam you can see we are given a graph which is showing us the velocity of a car with respect to time we also have a table telling us exactly what some of the Velocity values are at given times let's read the question the graph of the Velocity V oft in feet per second of a car traveling on a straight road for T between 0 and 50 is shown above a table of values for V of T at 5-second intervals of time T is shown to the right of the graph part A says during what intervals of time is the acceleration of the car positive the acceleration being positive means that the velocity should be increasing we need to give a reason of course for our answer so this is the graph of the Velocity we need to ask when is this increasing that's going to be when the acceleration is positive because acceleration is the derivative of velocity so looking at the velocity we see that it is actually increasing quite a bit it's increasing all the way up to here where t equal 35 that whole time the velocity's increasing and then it's also increasing it appears from 45 to 50 it is only here between 35 and 45 where the acceleration is negative because the velocity is decreasing all right so let's write our answer for part A so here is our answer for part A acceleration is positive on the interval from 0 to 35 and from 45 to 50 because this is where vft T the velocity is increasing whether or not you chose to include the end points in these intervals would not affect how your answer is scored for this question now Part B find the average acceleration of the car in feet perss squared over the interval from 0 to 50 the average acceleration is the average rate of change of the Velocity so we need to find the average rate of change of the Velocity which is like the slope of a secant line in particular it's the slope of the secant line from tal 0 to tal 50 so that line would look something like this and we're looking for the slope of this line that will tell us the average rate of change of the velocity over this interval which is the average acceleration because acceleration is change in velocity so for Part B like we said average acceleration that we're looking for is the average rate of change of velocity so how do we find it Well we need to take the velocity at tal 50 that's the terminal velocity or I should say the velocity at the end of the interval in question and then subtract the initial velocity V of0 then just divide by the change in time over this interval so 50 - 0 now V of 50 we can see from the table is 72 V of 0 is 0 so this is just 72 - 0 over 50 - 0 50 - 0 of course is 50 and 72 - 0 is 72 so this is 72 / 50 and you could use a calculator to find that this is about 1.44 feet per second squared now on to part C part C says find one approximation for the acceleration of the car in feet perss squared at tal 40 and show the computations we use to arrive at our answer given the information we have there are several different approximations we could use for the acceleration at tal 40 remember acceleration is the change in velocity so to approximate acceleration we need to look at an average rate of change in velocity on an interval that contains T = 40 so the options we could use are from 35 to 40 or we could use from 40 to 45 or we could actually have 40 be the middle of the interval and go from 35 to 45 any of those would be valid answers we could use tal 40 as an end point or we could use it as a midpoint for our solution let's use an approximation going from 35 to 40 so the acceleration at tal 40 well it should be about the average rate of change of the Velocity from T = 35 to tal 40 if we had the information to get a closer approximation that was a little bit more narrowed in on tal 40 we would and we could use the graph perhaps to get a closer approximation but we're going to use the table that is what the College Board answer key expects you to do so let's go ahead and do that for T = 35 the velocity is 81 and for T = 40 the velocity is 75 so we have 75 minus 81 that's the change in velocity divided by the change in time so 40 - 35 that's a 5-second interval this turns out to be -6 over 5 which is -1.2 ft/s squared again this is an approx oximation of acceleration at tal 40 so we use the information that we have getting us as close to tal 40 as possible and looking at what the rate of change in the velocity was because that is acceleration let's move on to Part D approximate the integral from 0 to 50 of the Velocity function with a reman sum using the midpoints of five sub intervals of equal length so we're going from 0 to 50 and we're using five sub intervals so they'll each have to have a length of 10 then using correct units explain the meaning of the integral so we are looking for the integral from 0 to 50 of the Velocity function V of tdt and we are going to approximate this using a remen sum and we have to use a midpoint rean sum specifically and we're using five sub intervals of equal length of course so let's use our table to break this down we're going from 0 to 50 and we need to use five sub intervals so each sub interval should have a length of 10 since they all need to be equal so we can go from 0 to 10 and then we can go from let me write that differently 0 to 10 and then we can go from 10 to 20 and then we can go from 20 to 30 and then from 30 to 40 and then from 40 to 50 and you can see what our midpoints will be the points that we use to determine the heights of the rectangles we have five as the midpoint 15 as the midpoint 25 35 and 45 so the heights of our rectangles are 12 30 70 81 and 60 so the areas of all these rectangles will be base time height the base of each one is 10 each rectangle will have a base of 10 because these intervals all have lengths of 10 and then the heights are the highlighted numbers so let's just jot this information down and instead of multiplying each one individually by 10 let's imagine we Factor the 10 out so we just have 10 multipli by the sum of the heights 12 + 30 + 70 + 81 + 60 here that information is back with our integral and if you do this addition and multiplication this equals 2,530 and the units are feet because we are ACC accumulating a change in feet because the velocity is in feet per second if you accumulate feet per second over a span of seconds you get accumulated feet more specifically in this case these feet are describing the change in the car's position from tals 0 to tal 50 the change in the car's position is about 2530 ft this is displacement when you accumulate velocity you get displacement and in this case 2530 ft is actually how far the car traveled as well it's not just the change in position it's actually the total distance traveled that's only the case here because the velocity is never negative if the velocity were negative then when we integrate it we wouldn't necessarily get the total distance traveled we would just get the change in position but in this case because the velocity is positive the whole time we get the change in position which is equal to the the distance traveled and that completes our solution to free response question three from the 1998 AB exam this is problem 3 from the 2005 AP Cal AB exam it is part of the graphing calculator section let's read it a metal wire of length 8 cm is heated at one end the table above gives selected values of the temperature T ofx in De C of the wire X CM from the heated end so for example 5 cm from the heated end the wire has a temperature of 70° Celsius the function T is decreasing which we see in the table and it is twice differentiable which means its derivative is differentiable and thus its derivative is also continuous let's read part a estimate T Prime of 7 show the work that leads to your answer and indicate unit of measure so T remember is the temperature of the wire X CM from the heated end so T Prime is going to be the rate of change of the temperature and here we are asked to estimate it estimate T Prime of seven we can do that with a secant line or more precisely the slope of a secant line in other words we need to find the slope between these two points because x equal 7 is between x = 6 and x = 8 so the best approximation we're going to be able to get with this data the best approximation of the rate of change of T at xal 7 will be the rate of change in this space near xal 7 the rate of change from x = 6 to x = 8 to calculate that that average rate of change over that interval from 6 to 8 we just need to take the change in the temperature and divide by the change in distance so we have that t Prime of 7 is approximately the average rate of change from x = 6 to x = 8 and that is 555us 62 that's the change in temperature and divide by 8 - 6 the change in distance this is equal to7 hves but remember we also need to indicate units of measure in this case this is how the temperature degrees Celsius is changing as we travel more centim across the wire so this is -7° Celsius per CM when x = 7 the temperature of the wire is changing by about -7° C per CM as we go forward another centimeter we're dropping about -7° moving on to Part B write an integral expression in terms of t ofx for the average temperature of the wire for a continuous function like T ofx to find the average we just need to add up all the values from 0 to 8 which we do with an integral and then divide by the length of that interval which is 8 then we need to estimate the average temperature of the wire using a trapezoidal sum with the four sub intervals indicated by the data in the table so the interval from 0 to 1 from 1 to 5 5 to 6 and 6 to 8 so we'll have four trapezoids for that and again we'll need to indicate units of measure all right let's start with the integral expression for the average temperature of the wire and we'll do that down here this is just a basic definition you should know in order to find the average temperature of the wire we need to integrate the temperature over the applicable interval which is from 0 to 8 from the heated end to the other end we need to add up all those temperatures and then we have to divide that by the interval length so we'll just put a 1 over8 in front that's dividing by 8 and that is an integral expression in terms of t ofx for the average temperature of the wire all right now we're going to estimate this using a trapezoidal sum so I've just copied the integral representing the average temperature of the wire and this is approximately 1/8 time our approximation for this integral we're going to approximate that integral using a trapezoidal sum so let's come up to this data and recall the area of a trapezoid and how it applies in this situation I'll just do a little sketch representing this situation the first interval is from 0 to 1 so we can kind of sketch this base that has a length of one on the left the trapezoid goes up to a height of 100 and on the right it goes up to a height of 93 since this function is just decreasing let me make this highest value of 100 a little bit bigger all right and then this other side goes up to about 93 okay so we have a trapezoid that looks something like that the next trapezoid goes from 1 to 5 I'll do that in a different color say blue so maybe we go from 1 to 5 and this thus would have a length of four going from 1 to five means this base length is four on the left again it's going up to 93 and on the right it's going up to 70 so maybe right there it's not perfectly to scale but it's a decent sketch and we have that trapezoid the next trapezoid goes from 5 to six and the right side goes up to 62 and so on I'll sketch the rest of this so that's just a rough sketch of our situation remember each of these numbers on the bottom is representing the length of these intervals in the context of trapezoids these interval lengths are actually the heights of the trapezoids the bases of the trapezoids are these vertical sides 93 93 and 70 70 and 62 those are the parallel sides of the trapezoids and that by definition is the base of the trapezoid or the two bases to find the area of a trapezoid we just need to take the average of the bases B1 plus B2 take the average so divide that by two and multiply that by the height so we need to find the areas of all these trapezoids using that formula and add them all together for example the area of this red trapezoid is the average of the bases so 100 plus 93 those parallel sides or the bases take the average so divide it by two and then multiply by the height which is the interval length which is one this blue trapezoid we calculate the area of similarly average the bases so 93 + 70 IDE by 2 because we're taking the average and multiply by the height which is four the interval length and we can proceed with the rest of these areas in a similar manner all right there it is after writing all of the areas out and there's another look at the sketch in the table just so you can make sure you understand where everything is coming from here now this here is our approximation for this integral using a trapezoidal sum so when we multiply that by 1/8 we're going to get an approximation for the average temperature of the wire and this turns out to be about 75688 8° C and that completes Part B let's move on to part part C find the integral of T Prime of X DX from 0 to 8 and indicate the units of measure then we need to explain the meaning of this in terms of the temperature of the wire at a glance we can say well this is the integral of the rates of change of the temperature when we accumulate or integrate the rates of change of the temperature we're going to get the total change from 0 to 8 but before we write that interpretation out let's actually find this integral the integral of T Prime of x from 0 to 8 is very straightforward if we integrate T Prime we are just going to get the original function T of X and we're evaluating this from 0 to 8 so the value of the integral is T of 8 minus t of 0 and we can find those values by Consulting the table T of 8 we see is 55 and T of 0 is 100 so the value of this integral is 55us 100 which is -45 we should specify this is deg Celsius so that is our value of the integral and then here is our interpretation of what this means this means that the temperature of the wire drops 45° C from the heated end at xal 0 to the other end at x = 8 that's because this is the accumulation of all the rates of change so we get the total change let's move on to Part D are the data in the table consistent with the assertion that the second derivative of T is positive for every X in the interval from 0 to 8 and of course we should explain our answer so for the second derivative to be positive means that this whole time the first derivative T Prime should be increasing now we know the derivative T Prime is negative the temperature T is decreasing so the derivative is negative so for the derivative to be increasing it should be getting less negative which means that even though the temperature is decreasing it should decrease by less and less as X goes on and that should always be true that's what it would mean for the second derivative to be positive the first derivative is increasing the first derivative happens to be negative so it should be getting less negative and the amount that our function is decreasing should be getting less extreme because the first derivative is getting less negative this is rather tricky because we don't actually know any values of the second derivative or any values of the first derivative so how can we say that the second derivative is always positive or that the first derivative is always increasing actually we can use the mean value theorem to get a definitive answer to this question our function T is continuous and differentiable on this interval from 0 to 8 so the mean value theorem tells us that whatever the average rate of change is over some interval that must be the same as the instantaneous rate of change at some point in that interval if we look at the average rate of change from 1 to 5 the instantaneous rate of change must equal that average rate of change at some point in the interval using this we'll be able to find two points that must exist where the derivative actually has decreased and that's going to show that the second derivative can't always be positive so let's check this out what's the average rate of change from 0 to 1 well that would be 93 - 100 which is -7 div by 1 - 0 so it would be -7 so the average rate of change from 0 to 1 is -7 since that's the average rate of change there must be some point from 0 to 1 where the actual derivative equals -7 now let's look at this interval from 5 to 6 what's the average rate of change there well let's write it down T of 6 minus t of 5 / 6 - 5 the average rate of change over this interval is going to be 62 - 70 / 1 the interval length 62 - 70 is -8 so this is just -8 so we can see the derivative has actually gone down by the mean value theorem there has to be some point in this interval from 5 to 6 where the derivative actually does equal that average rate of change8 so at some point the derivative is -7 and at some other point in this interval the D dtive is8 so at some point it must decrease which means it isn't always increasing which means the second derivative isn't always positive so no the data in the table is not consistent with this assertion that the second derivative is always positive and there is our reasoning written out by the mean value theorem there must be some point C1 from 0 to 1 such that the derivative at C1 is -7 and similarly a point C2 in the interval from 5 to 6 where the derivative is -8 and since our original function T is twice differentiable we know that the first derivative must be continuous thus it couldn't have just jumped from -7 to8 it actually had to decrease somewhere and so T Prime isn't positive for every X in that interval from 0 to 8 and that completes our solution to free response question three from the 2005 AP Cal AB exam this is problem three from the 2007 AP Cal AB exam this is part of the calculator section so we may need our calculator for this problem let's read it the functions f and g are differentiable for all real numbers and G is strictly increasing so it's not just that it doesn't go down but it only ever goes up the table above gives values of the functions and their first Der derivatives you can see some FS some fimes some G's some G primes at selected values of x 1 2 3 and 4 the function H is given by h of xal f of g ofx a composite function minus 6 part A is to explain why there must be a value R between 1 and three between 1 and three such that H this function of R equals -5 this is a fair Fair L straightforward question we just need to use the intermediate value theorem or ivt I'll leave a link in the description to my lesson on the intermediate value theorem if you need to review that the idea though is that H is a continuous function because f and g are continuous functions we know that because they're differentiable and differentiability implies continuity so f and g are continuous the composition of continuous functions is continuous and subtracting a constant from a continuous function well you'll still have a continuous function if you do that okay so H is continuous which means what we need to do is look at h of one because remember we're concerned about what happens between 1 and three we want to show that H must equal -5 at some point between 1 and 3 so we're going to look at h of one and we're going to look at H of3 if we evaluate these correctly what we're going to see is that one of these values will be less than neg5 and one will be greater than neg5 so5 is an intermediate value which must have been taken on at some point in the interval because H is continuous it can't just skip over something it has to hit all of those intermediate values so let's evaluate h of 1 and H of 3 using the definition of h and our table of values h 1 by definition of h is f of G of 1 - 6 looking at the table we can see that g of 1 is 2 then we need to plug that into the function f so we're plugging two into f f of two we see is n so this should be 9 - 6 which is pos3 and then we'll do a similar process to evaluate h of 3 by Definition h of 3 is f of G of 3 looking at the table we see that g of 3 is 4 and then we need to plug that into F and we see that F of 4 is -1 thus this is going to be -1 - 6 sorry I forgot to write the minus 6 a minute ago so this is -1 and then subtract 6 that's going to be7 and so there is our conclusion since H is continuous and H of 3 we found was -7 is less than -5 Which is less than h of 1 we found to be three by the intermediate value theorem there must exist sorry I wrote the exists there exists a number R between 1 and 3 so that h of R is equal to5 the function h must have passed ne5 at some point because between 1 and 3 it is less than -5 and then it becomes greater than5 so it must have passed neg5 at some number R between 1 and 3 moving on to Part B explain why there must be a value C between 1 and 3 such that H Prime of C equals -5 in this case we're trying to show that the derivative takes on a certain value so for this we're going to use the mean value theorem the mean value theorem tells us that since H is a differentiable function since it's a composition of differentiable functions whatever the average rate of change is of H on an interval in particular right now we're interested in the interval between 1 and three whatever the average rate of change of H is on that interval the derivative must equal that average rate of change at some point in the interval so we need need to calculate the average rate of change on the interval to find the average rate of change of a function over an interval we just need to take the change in the function so h of 3 minus h of 1 how much did H change from 1 to 3 and subtract by the length of the interval so 3 - 1 now h of 3 we already found is -7 so we have -7 minus h of 1 we already found that's 3 / 3 - 1 is of course a division by 2 so this is -7 - 3 which is -10 / 2 which is5 look at that and there is our conclusion since H is continuous and differentiable we can use the mean value theorem the mvt which tells us since the average rate of change of H over this interval from 1 to 3 is -5 that means that there must exist a point C between 1 and 3 where the derivative of H at C is exactly -5 the instantaneous rate of change at some point must equal the average rate of change over the interval all right moving on to part C part C says let W be the function given by W ofx equals the integral from 1 to g ofx f of T DT and we want to find the value of w Prime of 3 so this might look kind of intimidating but all we actually have to do here is use the fundamental theorem of calculus and the chain rule those things together tell us how to take the derivative of an integral function like this which has a function in the bounds so let's take the derivative and then we'll plug three into the derivative to find W Prime of 3 W Prime of x equals the derivative essentially undoes the integration but how it works is this the upper bound needs to get plugged in this upper bound which is a function so it's F of G of X but then by the chain rule we need to multiply by the derivative of that function so we have to multiply by G Prime of X then to find W Prime of 3 we can just plug that in in use our table we know that g of three from the table is equal to let's see here G of three right there it's four so this is going to be F of 4 and what is f of four F of 4 is1 okay so this part here is1 but remember it's also getting multiplied by G Prime of three which from the table we see is positive2 all right so that right there is positive2 so w Prime of 3 is -2 their product remember that the lower bound of this integral would just yield the constant which is why taking the derivative makes that part totally go away but this function in the upper bound does have an impact on the rate of change so that's part C finally the final part is Part D if G inverse is the inverse function of G write an equation for the line tangent to the graph of y equal G inverse of X at x = 2 okay to write the equation of a tangent line we need need a couple things right we need a point for starters so let's jot down our point we're going to need a point and a slope of course the point has an x coordinate of two so let's put that there what is the y-coordinate well we want a line that's tangent to G inverse so the y-coordinate should be G inverse of the x coordinate 2 now if we look at our table we see that g has an output of 2 when x = 1 G of 1 = 2 by definition of an inverse function since G of 1 equal 2 that means that g inverse of 2 equals 1 so to evaluate G inverse of 2 all I had to do was look at what input makes G give an output of two then I know G inverse of two gives an output of one 1 and so that's going to be my y coordinate thus I've got my point now I just need my slope so I need to find the slope of G Prime at this point all right now to find the slope we're going to have to take the derivative of G inverse it be helpful if you knew the formula for the derivative of an inverse function but it's not the end of the world if you don't have that memorized we can figure it out from scratch pretty easily by definition of an inverse function we know that g of G inverse of x equals x because inverse functions cancel each other out what we're going to do next is take the derivative on the left with respect to X and the derivative on the right with respect to X to take the derivative on the left we have to use the chain rule the outside function is G so we have to start with G Prime and leave the inside function G inverse unchanged but then we need to multiply this by the derivative of the inside function the inside function is G inverse of X so we have G inverse of X Prime the derivative of the inside function on the right the derivative of x is just one and then remember we're trying to find a formula for G inverse Prime so let's solve for G inverse Prime by dividing both sides by G Prime of G inverse of X again ideally you have this formula we're going to get memorized but this is how you can figure it out if you don't have it memorized all right there's our formula for the derivative of G inverse so so we can use that now to find the slope of G inverse at this point let me write this part in Orange so now we are looking for G inverse of 2 Prime right we want the slope where x equal 2 and we know that this is equal to 1 over G Prime of G inverse of 2 and let's see what is g inverse of 2 well we already figured that out G inverse of 2 is 1 because G of 1 is two okay so coming back down here G inverse of 2 is 1 so this is going to be 1 over G Prime of 1 and what is g Prime of 1 going back to the table we see that g Prime of 1 is five so this should be 1 over five I'm going to have to move everything to the left a little bit so I have room for this this is 1 over five and that inverse derivative formula really comes in handy all right so finishing the tangent line we have a point 2 one we just figured out the slope 1 over 5 so we can go ahead and write the equation of the tangent line using point slope form Yus the y coordinate equals the slope multiplied xus the x coordinate and there you go that is our answer to Part D and that completes our solution to free response question three from the 2007 AP Cal AB exam this is free response question five from the 2014 AP Cal AB exam let's read it the twice differentiable functions f and g are defined for all real numbers X values of f f Prime G and G Prime for various values of X are given in the table above so here we have a big table giving us some information about F fime G and G Prime we can see for example that the derivative of f evaluated at x = -1 is 0er we also have some information about their signs on various intervals we can see for example that g Prime is positive between Nega - 1 and positive 1 which means that g is increasing on that interval these sorts of table problems often involve a little bit more abstraction than usual really got to know your stuff let's read part A find the x coordinate of each relative minimum of f on the interval from -2 to 3 and justify our answer now if we're going to have a relative minimum on this interval we're going to have to look at the critical points where the derivative is zero we know the derivative isn't undefined anywhere in this interval because we're told that f and G are twice differentiable so we can take the derivative we got to see where's the derivative zero and again we're talking about F Well by looking at the table we can see that the derivative is 0 when X is -1 and the derivative is 0 when X is positive one so how do we determine if these critical points are the locations of minimums or maximums or neither well we have to think about the sign of the derivative we see that the derivative goes from negative to the left of x = -1 to still negative to the right of x = -1 so the function is decreasing and decreasing and then it flattens out a little and then it continues decreasing so this would not be the location then of a minimum or a maximum remember that we're looking for minimums this would look something like this the function's decreasing flattens out and then just keeps decreasing on the other hand this other critical point to the left the derivative is negative the function is decreasing and then to the right the derivative is positive the function starts increasing so certainly at this critical point where X is one we will have a relative minimum F has a relative minimum at X = pos1 and that is the only one so there's our answer F has a relative minimum at x = 1 because this is the only critical point where the derivative F Prime changes from negative to positive let's move on then to Part B explain why there must be a value C between Nega 1 and postive 1 such that the second derivative of C fpre of C equals zero now we don't have information in our table about the second derivative so we're not going to be able to use say the intermediate value theorem to say that the second derivative goes from negative to positive so it must cross zero at some point we can't do that but we do have information about the first derivative fime we have some values of fime so what we're actually going to use is the mean value theorem we know that f is twice differentiable which means that its derivative F Prime is also differentiable and so it's also continuous we can use the mean value theorem because we know that F Prime is continuous on this closed interval and also that it's differentiable we can apply the mean value theorem to say that whatever the average rate of change of fpre is on this interval its derivative FP Prime must equal that average rate of change at some point on the interval so what is the average rate of change of f Prime on this interval well what we have to do to find that is take F Prime of the ending point on the interval so frime of one and then subtract fime of the beginning Point negative 1 this tells us how much fpre changed on the interval and then we have to divide by the interval length which is two 1 - -1 that's the interval length that's going to tell us the average rate of change now frime of 1 and frime of neg one we can check our table to see what those values are FR Prime of one we know is let's see where is it right there it is zero and frime of negative 1 we see is also zero okay so coming back down to our average rate of change calculation we have two in the denominator and in the numerator we just have 0- zero of course then this average rate of change is zero just like we wanted because remember Part B is to show that there's some value C where F Prime of C is zero there must be by the mean value theorem some point C where this instantaneous rate of change equals the average rate of change and there's our conclusion by the mean value theorem there has to be at least one value C on this interval where fpre of C equals 0 because we see the average rate of change of fime on this interval is zero so there must be a point where the instantaneous rate of change of f Prime which is FP Prime is also zero moving on then to part C the function H is defined by H ofx equals the natural log of f ofx we want to find H Prime of three and show the computations that lead to our answer so this is just going to be a little bit of chain rule let's put part C over here if H ofx is Ln of f ofx then let's just quickly find H Prime we have to use the chain rule which begins with the derivative of the outside function the outside function in this case is Ln the derivative of that is one over the input we don't want to change the inside function f ofx right leave that inside function alone and then multiply by the derivative of the inside function so in this case that's frime of X that is H Prime which makes it fairly straightforward then to evaluate H Prime of 3 H Prime of 3 is going to be 1 over F of 3 multiplied by F Prime of 3 now hopefully all of these numbers are in our table if we go up to the table what is f of three F of three we see is 7 that's over here and the other value we needed was fime of 3 and frime of 3 is/ 12 so this calculation is simply 1/ F of 3 so 1 over 7 time frime of 3 so times a half so 17th * 12 which is 114th that's H Prime of 3 moving on to our final problem Part D evaluate the integral of frime of g ofx g Prime of X DX from -2 to 3 now if you don't really understand U substitution this might look kind of intimidating if you do understand U substitution well this should look pretty trivial because when you're trying to do a u substitution integral what you're looking for is an inside function like G of X it's inside fime in this case you're looking for an inside function whose derivative is represented outside like elsewhere in the integral these are the sorts of situations where you want to use U substitution in this one we see l X and the derivative of Ln X is this other thing 1/x so this is a prime candidate for U substitution this integral we have S of x^2 the derivative of x^2 is 2x which is just a multiple of 3x which we see out here also a perfect candidate for U substitution when you have an inside function and its derivative elsewhere in the integral that's when you should be thinking U sub and that's all this is so here's the integral for Part D Rewritten let's proceed with the U substitution we're going to let U as usual be the inside function in this case G of X and then we have that du equals g Prime of X DX and we see how nicely that substitution is going to work what is this integral equal to then well we have the integral of let's worry about the bounds in a minute this is the integral of fime of G ofx X but G of X is U and G Prime of X DX is Du so it's frime of U du now what about the bounds well the bounds should be let's see U equals g of X so if we're integrating with respect to you the lower bound should be G of -2 and the upper bound should be G of pos3 before it was X going from -2 to pos3 but when X goes from -2 to posi3 U goes from G of -2 to G of positive3 so these are our bounds since we're rewriting this in terms of U and we can evaluate these using the table let's go look at the table G of -2 is let's see right here G of -2 is1 and G of posi3 is right here it's positive 1 so the lower bound is negative 1 the upper bound is positive 1 that works out pretty Nic ly here's our integral from - 1 to postive 1 frime of U duu and now the integral of frime from -1 to 1 is just F of U evaluated from -1 to 1 the integral UND does the derivative so this is going to equal F of 1 minus F of1 and of course we can consult the table to find these values going back to the table F of one we see is is 2 right there and F of -1 is 8 right there so this is going to be 2 - 8 so with a little bit of U substitution a little bit of fundamental theorem of calculus we get the answer is -6 pretty cool little U sub problem and that completes our solution to free response question five from the 2014 AP Cal AB exam next up are problems on rates and accumulation for these problems recall that ating a rate of change produces a net change if we integrate velocity we get the net change in position also keep your eye out for units a lot of these problems or I should say most of them request that you include units with many of your answers good luck this is free response question one from the 2013 AP Cal AB exam this is part of the calculator section we will have to consult our graphing calculator a couple times throughout this problem let's read it on a certain workday the rate in tons per hour at which unprocessed gravel arrives at a gravel processing plant is modeled by G of t equal 90 + 45 cosine of t^2 18 where T is measured in hours between 0 and 8 at the beginning of the work day which is T equals 0 so that's not midnight necessarily it's just the beginning of the workday the plant has 500 tons of UN unprocessed gravel that's like our initial condition for the amount of unprocessed gravel during the hours of operation so from tal 0 to tal 8 the plant processes gravel at a constant rate of 100 tons per hour and remember that function G of T is telling us the rate at which unprocessed gravel arrives now part A is to find G Prime of 5 the derivative of G evaluated at tal 5 using correct units we should interpret our answer in the context of the problem so it's really important to note here that g is already a rate G is already the rate in tons per hour at which unprocessed gravel arrives at the gravel processing plan so when we take the derivative of this to find G Prime here for part A that's going to be the rate of change of the rate of change it would be straightforward enough to find the derivative of G on our own but since we can use the calculator and we're going to have to evaluate the derivative anyway let's just use the calculator to evaluate the derivative at five and then we'll worry about the interpretation so over on my calculator I'm going to press the math button and then I'm going to press number eight this is the derivative function I'm taking the derivative with respect to x what function am I differentiating well I'm differentiating G which is 90 + 45 * cosine of t^ 2 but I'm just going to use x as our variable divided by 18 and that is our function then at the end we put where we want this derivative evaluated we want it evaluated at five and we get our answer it's about -4.5 188 we could say now to interpret this what does it mean that g Prime of 5 is -4.5 A8 well remember G is the rate at which unprocessed gravel is arriving so G Prime is the rate at which the rate the gravel is arriving is changing so this means the rate at which gravel is arriving is decreasing because this is a rate of change of the rate of change so how fast the gravel is arriving that's decreasing 5 hours into the workday and it is decreasing by about 24.5 A8 tons per hour and there is the interpretation written out we have our units tons per hour we have the negative CED by the word decreasing which is why you don't see a negative here and we also have our interpretation of this being at tal 5 that means we are 5 hours into the work day when this sentence applies when the rate at which gravel is arriving is decreasing by 24588 tons per hour all right let's move on to Part B find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday the work day goes from tal 0 to tal 8 this function G of T tells us the rate at which unprocessed gravel is arriving so to find the total amount of unprocessed gravel that arrives during the workday we're just going to have to integrate G of T from 0 to 8 that's like accumulating all of the rates of change of unprocessed gravel so for Part B we just add up all of the rates of gravel arriving over the entire workday from t = 0 to T = 8 so that will give us our answer and we'll go ahead and use the calculator to evaluate this integral so again I'll press the math button and then this time press function 9 the integration function now we are integrating from tals 0 to tal 8 or in this case we'll use the variable X so from xal 0 to xal 8 and now I just have to put in my function which of course is G of T so we'll type that into the calculator that is 90 + 45 * cosine of x^2 / 18 and of course we're integrating with respect to X and we find this is about 8255 51 tons of unprocessed gravel that arrive during the workday now part C is the amount of unprocessed gravel at the plant increasing or decreasing at time tal 5 hours and we need to show the work that leads to our answer now of course we know that unprocessed gravel is arriving at the plant during the entire workday and we have the function that describes that but we also know that gravel is being processed at a rate of 100 tons per hour so to answer this question of whether or not the amount of unprocessed gravel is increasing or decreasing at time tal 5 let's write a function for the amount of gravel at the plant and consider it derivative so here for part C I said we would write a function for the amount of gravel I should have just said name a function say we call it a the amount of unprocessed gravel at the processing plant is a so then what is a prime the change in the amount of unprocessed gravel well certainly it would have to be G of T the amount of unprocessed gravel that is arriving at the plant but then minus 100 which is the amount of gravel being processed each hour then to answer part we have to figure out what a prime of 5 is if this is positive that means the amount of unprocessed gravel at the plant is increasing if it's negative then the amount of unprocessed gravel is decreasing a prime of 5 of course is going to be G Prime of 5 minus 0 so all that we really have to determine is if the derivative of G is positive or negative at tal 5 I'll just go ahead and erase the minus 0 we don't really need that let's open up the calculator so to answer part C we will plug T = 5 into a prime this of course is going to be G of 5 - 100 is this positive or is it negative well let's open up the calculator and get a value for G of 5 so here again I'll just type in our function but plug in five so 90 + 45 * cosine of 5^ sared so 25 / 18 and this is about 98141 so a prime of 5 is going to be about 98141 minus 100 this number of course is negative it is less than zero and this means that the amount of unprocessed gravel at the plant is decreasing 5 hours into the workday and again that's because we know that the derivative of the amount of unprocessed gravel at the plant is the rate at which unprocessed gravel arrives minus the rate at which it is processed and when we plug in five we get a negative so the amount of unprocessed gravel is decreasing at this time moving on to Part D the final part what is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday and we need to justify our answer now since we're talking about this workday t is between 0 and 8 so we basically will have to conduct a candidates test we'll have to plug in zero plug in 8 and then use our derivative to find critical points and if there are any critical points we'll have to plug those in as well and see where the function is maximized at what time is the amount of unprocessed gravel the highest now for Part D we will actually need to figure out what our function a of T the amount of unprocessed gravel at time T is and that's pretty straightforward the amount of unprocessed gravel at the plant at time T is 500 which is the starting amount of unprocessed gravel that was given to us in the question that what do we have right here at the beginning of the workday T equals 0 the plant has 500 tons of unprocessed gravel so it's 500 that initial Value Plus the integral of G of s minus 100 so that's the rate at which the amount of unprocessed gravel changes we have to integrate that amount for from 0 to T that way we get the starting amount 500 plus whatever the accumulated change is from 0 to T that will tell us what the actual amount of unprocessed gravel is at time T take that starting amount and accumulate the change up to time T although again this only applies on this closed interval from 0 to 8 now we already know the derivative a prime of T is G of T minus 100 and we need to set this equal to zero to find critical points there are no critical points where the derivative doesn't exist because G of T exists for all values of t on this closed interval there's no forbidden value that we can't plug into cosine so to find the critical points where this derivative equals zero we will plug it into our calculator so I just pressed y equals to get here where I can put in a function and then we can graph it and see where it hits zero in the interval that we're interested in so we've got 90 Plus 45 * cine of x^2 again we're just using X instead of t for convenience divided 18 that is g of T but remember we also have to subtract 100 that's the rate at which gravel is being processed all right now let's press the graph button and take a look at the graph I'll have to zoom out to make sure we can get a good view here and let's scroll over to the positive X values and here we can see is where it passes zero on the interval we're actually interested in it passes zero again over here but that's after T equals 8 that's after the end of the workday that is not important for us it's only this zero that's important for us where T is about 4923 all right now we have all of our candidates the end points of the interval 0 and 8 and the one critical point we just found we need to evaluate the a function the amount of unprocessed gravel at each of these times and see which one is maximum a of zero we already know that was given to us as 500 as for the other two a of the critical point and a of 8 we are going to need our definition of a using the integral here in order to figure those out and so we'll have to consult the calculator once more all right I did this calculation once but forgot to subtract the 100 now you can see in my calculator I got the minus 100 in the integral and our answer for the amount of unprocessed gravel at this critical point time is about 635 376 tons and I can just press second entry in order to get this integral again and just go edit the upper bound the upper bound now we need to change to eight to find the amount of unprocessed gravel at t equal 8 and we get about 52555121 so again all of this was found using the function we wrote for the amount of unprocessed gravel at the plant at time T and the logic behind this is that we're taking that initial amount of unprocessed gravel 500 at T equals 0 and then adding the accumulation of the rate of change of unprocessed gravel from 0 to T the time that we are interested in all right now we have evaluated our function a at all of the candidates and thus we can make our conclusion we see the maximum amount was 635374 so our conclusion is that the maximum amount of unprocessed gravel at the plant during this workday was 635374 point9 2 348 hours into the workday and that completes our solution for free response question one from the 2018 or excuse me 2013 AP Calculus AB exam this is free response question one from the 2016 AP Cal AB exam this is part of the graphing calculator section we will use a calculator a couple times throughout this video water is pumped into a tank at a rate modeled by W of T = 2 e t^2 divided by 20 L per hour for T between 0 and 8 and T is measured in hours water is removed from the tank at a rate modeled by R of T L per hour where R is differentiable and decreasing from tal 0 to tal 8 selected values of the function R of T are shown in this table and at time T equals z there are 50,000 L of water in the tank all right part A is to estimate R Prime of 2 and to show the work that leads to our answer and to indicate units of measure remember that R is the rate in lers hour at which water is being removed from the tank so in estimating R Prime of 2 we're estimating how fast that removal rate is changing at tals 2 now all we have are a few values of R of T so in order to estimate RP Prime of 2 we're going to have to estimate the rate of change near tal 2 the closest we can really get to that is looking at how R of T changes from tal 1 to tal 3 if we find the average rate of change from T = 1 to T = 3 that will give us the best approximation we could get for R Prime of two let's go ahead and put part A down here we are trying to estimate R Prime of 2 and we know that R Prime of 2 is about we need to do some division here it's about the change in R from T = 1 to T = 3 so R of 3 minus r of 1 that's going to tell us how much R changed on that interval and then divide by the width or length of the interval so 3 - 1 this tells us the average rate of change on the interval the interval contains two so it's a rough approximation of the rate of change at tal 2 R of3 we know is 950 and R of 1 is 1190 so our numerator is 950 minus 1190 the denominator of course 3 - 1 the interval length is just two and this is -120 don't forget we also need units of measure R the original function R was in lers per hour so -120 L per hour but this is an estimation of the rate of change of R so this is actually lers per hour per hour you could also write this as lers per hour squared this is how fast the rate of change that rate of removal how fast that rate of removal is changing all right moving on to Part B use a left reman sum with the four sub intervals indicated by the table so 0: 1 1: 3 3: 6 6 to 8 those four sub intervals use this to estimate the total amount of water removed from the tank during the 8 hours is this an overestimate or an underestimate of the total amount of water removed and give a reason for your answer okay so if we look at the table up here again this is giving us values of R of t a few values for how fast water is being removed at different times if we were to integrate this function R of T from 0 to 8 that would accumulate all of those removal rates and thus tell us how much water is removed in total now we don't know what R oft is as a function exactly so we can't actually do that but we can approximate that integral using a left reman sum and again what we're approximating is the integral from tal 0 to T = 8 of our function R of T so this is the integral of R of T DT from 0 to 8 and we're going to approximate it using a left rean sum a left rean sum means that we're adding up rectangles whose Heights are determined by the left end points of our intervals so if this was an X axis for example our first rectangle starting at tal 0 would have a height determined by this left end point so it would be a height of 1340 so it' go up here to 1340 and it' have a width of one because we're going from 0 to one so let's maybe draw a little bit less wide it has a width of one and it has a height determined by the left end point so 1340 the next rectangle in our reman sum would be going from T = 1 to T = 3 and its height is determined by the left end point 1190 the height of this rectangle is 1190 but the width is 2 because this interval width from 1 to 3 is 2 so that rectangle might look something like this it's a little bit lower cuz its height is 1190 but it's twice as wide so I'll put a two down there and then 1190 there the third rectangle I'll write in Orange is over this interval from tal 3 to tal 6 its height is determined by the left end point because we're doing a left R and sum so it goes down some starting there maybe and the width of this rectangle is three because it goes from from 3 to 6 so let me sketch this it starts at about 950 its width is about 3 so maybe it looks something like that again the height is determined by the left end point and there's the last rectangle with a width of two from tal 6 to tal 8 it's height determined by the left end point of 740 so to find the approximation of the integral we just have to add up the areas of all of these rectangles the area of a rectangle is base time height so that would be 1 * 1340 + 2 * 1190 + 3 * 950 and so on those are the areas of the rectangles and these add up to 8,50 L so we would approximate that about 8,50 L of water were removed from the tank from tal 0 2 T = 8 but remember we are not done because part B also asks us to determine if this approximation is an overestimate or an underestimate of the actual total amount of water remain moved and we're able to determine the answer to this question based on knowing that R of T is decreasing so R of T if we were to sketch it in say this reddish color on the same graph where we sketch the rectangles because R of T is decreasing it might look something like this and so you can see that all of the rectangles are overestimates because they lie above the curve because it's decreasing we know that this left rean sum is an over approximation all right moving on to part C use your answer from Part B to find an estimate of the total amount of water in the tank to the nearest liter at the end of 8 hours so we need we need to consider how much water was in the tank to begin with 50,000 L add in the amount that is pumped remember that W of T this function is telling us how much water is pumped into the tank over this interval of time and then we need to subtract the amount which is removed and we know a rough approximation for that that came from Part B about 8,50 lers are removed so our answer for part C is going to be 50,000 the amount that was there to begin with plus the accumulation of all the water that's pumped in from tal 0 to tal 8 minus the approximate amount that is removed again this is just an approximation of the total amount of water at tal 8 now to integrate W of T this function here from tal 0 to Tal 8 we will crack open the calculator so I'll press the math button to access my functions press nine for the integration function we are integrating from 0 to 8 the function that we are integrating is W of T which is 2000 * e to the negative we'll use x as our variable for convenience so x^2 / 20 and this integral is taken with respect to X and this is about 78361 195 that's how much water is pumped into the tank during this time so we have 50,000 plus this integral was about 78361 195 minus that water that's removed 850 this turns out to be about 49,7 186 L of water in the tank at time T equals 8 all right moving on to Part D for T between 0 and 8 is there a time T when the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank explain why or why not so to answer this question we have to consider the amount of water being pumped into the tank W of T minus the amount of water that is being removed which is R of T that difference should be Zer at some point between tal 0 and tal 8 in order for the functions to be equal in order for the rate at which water is being pumped in and removed to be the same their difference needs to be zero now the difference of those functions is continuous and differentiable because we know that both W is differentiable we can look at it we know that's a differentiable function and R of T is differentiable because we're told as much so we can apply the intermediate value theorem and here is all of that logic written out W of 0 minus r of 0 the initial difference in the pumping and removing function is 2us 1340 which is positive we know this because if we just plug Z into W of T we're going to get 2,000 * 1 so that's 2,000 and we're also given that R of 0 is 1340 so that's where those numbers come from we see the initial value of this difference is positive the ending value though W of a minus r of8 is negative we know that because we can just plug 8 into W of T just do that on the calculator you get about 81.5 and then subtract 700 that was given to us R of 8 is given as 700 clearly this is negative so this difference between W and R the pumping in rate and the removal rate was initially negative but at the end it was no sorry it was initially positive but at the end it was negative because this difference is continuous on the interval we know that the intermediate value theorem guaranteed at least one time T between 0 and 8 for which the difference of the functions had to equal zero it was positive then it was negative so it must have crossed zero at some point and at that point in time W of T the pumping in rate was the same as R of T the removal rate so at this time T the rate at which water is pumped into the tank equals the rate at which water is removed from the tank that's how you solve this part using the intermediate value theorem and that completes our solution to free response question one from the 2016 AP Cal AB exam this is free response question one from the 20122 AP Cal AB exam it is part of the graphing calculator section let's read it from 5:00 a.m. to 10:00 a.m. the rate at which vehicles arrive at a certain toll plaza is given by this function AF oft where T is the number of hours after 5: a.m. so tal 0 is 5: a.m. AF of T is measured in vehicles per hour that's the rate at which the vehicles are arriving traffic is Flowing smoothly at 5:00 a.m. with no vehicles waiting in line part A says write but do not evaluate an integral expression that gives the total number of vehicles that arrive at the toll plaza from 6:00 a.m. to 10: a.m. and it even gives us the appropriate T values there from tal 1 1 hour after 5: a.m. at 600 a.m. to tals 5 5 hours after 5:00 a.m. which which is 10: a.m. now an expression for the total number of vehicles that arrive in this time is going to require us to accumulate the rates at which the vehicles arrive we just have to integrate that rate of arrival from tal 1 to tal 5 so for part A the integral expression we're looking for is just the integral of the rate at which vehicles are arriving from tal 1 to T equal 5 if we want to we could rewrite this to be a little more specific by just replacing a of t with what the function actually is since it's given to us as 450 times the square root of let's just double check square root of s of 62t so s of 0.62 T integrating of course with respect to T So this integral if we were to calculate it would give us the number of vehicles that arrived at the toll plaza between tal 1 and tal 5 which is from 6: a.m. to 10: a.m. and we don't have to evaluate this that is it for part A Part B says find the average value of the rate in vehicles per hour at which vehicles arrive at the toll plaza from 6:00 a.m. to 10: a.m. so we're concerned with that same window from tal 1 to t equal 5 this is just asking for the average of the rate of arrival so we just have to add up the rates of arrival like we did in part A and then just divide it by the time that passes the amount of time that has passed from tal 1 to tal 5 is 4 hours so we just need this integral expression again but divide it by four to find the average rate again it's just adding up all the rates from 1 to five and then dividing by the interval width dividing by 5 - 1 we'll consult the calculator for this but let's write out the expression first here's the appropriate integral and then we just have to divide it by 4 which is the same as multiplying by 1/ 4 so this would be the average value of the rate of arrival and we'll bring up the calculator to find this so on the calculator I'll type 1/4 or 0.25 multiplied by math button option nine the integration function we're integrating from 1 to 5 the function we're integrating is 450 multiplied the square otk of s of 62t but it's more convenient to use X on the calculator there we go this is about 37554 I'll use three decimal places so 37553 7 and remember what this actually is this is the average value of the rate of arrival during this time from tal 1 to tal 5 so the units here would be vehicles per hour this is the average of that rate the rate at which vehicles are arriving obviously changes as time passes but on average between 6:00 a.m. and 10:00 a.m. vehicles are arriving at about this rate on to part C is the rate at which vehicles arrive which again is AF of is that rate at 6 a.m. increasing or decreasing that's a really easy question we just have to find the derivative of a of T that will tell us how the rate is changing if the derivative at T equal 1 is positive then the rate is increasing and otherwise the rate would be decreasing so we will write part C over here we'll again use the calculator to evaluate this derivative so what we're looking for is the derivative with respect to time of a of T and we are evaluating this at tal 1 all right so the derivative of a of T at tal 1 is about let's consult the calculator I'll press the math button option 8 for the derivative function we will use x as our variable and we are taking the derivative of a of T so here's a of T which we will type into the calculator 450 multiplied by the square OT of s of 62 * X and we are evaluating this derivative at x = 1 the question tells us that right X or in this case T equals 1 this is about 148 .95 so we'll write that here 14895 since this is positive the derivative of the rate is positive that means that the rate at which the cars are arriving is actually increasing and there our explanation is written out because the derivative of a at tal 1 is positive the rate at which vehicles arrive at the toll plaza is increasing at this time finally Part D a line forms whenever AF of T is at least 400 the number of vehicles in line at time t for T between a and 4 is given by this integral expression where a is the time when a line first begins to form to the nearest toll number we must find the greatest number of vehicles in line at the toll plaza in the time interval where T is between a and 4 and to justify our answer there's a decent amount of notation going on here but don't let that intimidate you we will find the extreme values just the same as always we start by taking the derivative of n oft the number of people in the line or the number of vehicles in the line because remember we're looking for the greatest number of vehicles in line so what is n Prime of T well because n of T is actually the integral of this expression n Prime of T is just the expression being integrated which would be a of T minus 400 now in order to determine when this number of vehicles is maximum on this interval we'll have to check the end points of the interval so that's tal a and tal 4 and we will also have to check the critical points for the critical points we need to see where is this derivative a of tus 400 undefined or equal to zero it's not undefined anywhere but clearly it's going to be equal to zero when a of t equal 400 100 that's not a super easy equation to solve for T but thankfully we have our calculator all right now on the graphing calculator you would press y equals I've already entered our function so we can press the graph button and here our function is we can zoom in on the interesting part here we can see that a of T minus 400 is going to hit the x-axis which means it's equal to zero when X or in our case T is about 1 4694 so 1.4 694 and it's also going to come back to zero when X is about let's look we'll just scroll over here it hits zero again when X is about 35977 now the question asked us to determine when there is a maximum number of people in the line so we're going going to have to check the values of n oft at the possible candidates where there could be a maximum the candidates are the critical points which are these Zer of the derivative we just found where T is 1.4 694 T could be 35977 or we also have to check the end points of the interval which were a and four now four obviously we can add to the table but what the heck is a well a is when the line actually starts to form there isn't a line unless the function a of T is at least 400 from the graph we just looked at we saw that a of T is actually less than 400 which is why the graph was negative up until this first critical point that's when AF of T finally matched 400 and then proceeded to surpass it so at tal 1.46 N4 that actually is the lower end point of this interval when a line first starts to form now n of T is this integral from a to T so if we let t equal this lower bound which actually is a then n of T will clearly be zero because this is right when the line starts to form but technically hasn't actually formed yet if you're thinking about this integral as area under a curve this would be no area at all because no time has passed for the other T values we will consult our calculator so we'll press math and then the integration function we are integrating from a when the line starts to form which as we just discussed is actually 1. 14694 and we're integrating two first let's do this one t equal 35977 so we're integrating to 35977 what we are integrating by definition of n of T is a of xus 400 so we'll have to type in the a function here and then subtract 400 so this is 450 multiplied the Square t of s of 62 again on the calculator we're going to use x instead of T and then we are subtracting 400 and we are integrating here with respect to X and there we go so n of 35977 is about 71.2 at this time there's about 71.2 vehicles in line now we'll do the same exact thing but with tal 4 plugging in the integral again but with t equal 4 you'll get about 62.3 4 so we see the greatest number of vehicles in line between when the line starts to form and T equals 4 the greatest number of vehicles in line is about 71 you can't have 71.2 vehicles so our final answer would be 71 and that completes our solution to free response question one from the 2022 AP Cal AB exam next up are problems on areas and volumes for these problems make sure that you use a picture a lot of these problems provide a bit of a picture that you can start with but not all of them regardless a picture is an invaluable tool for making sure that you represent the information that you have correctly and are more easily able to interpret that information to write correct integral Expressions this is problem one from the 1998 AP Cal exam and as we can see this is from the graphing calculator section so we may find ourselves in need as we go through this problem let's read it let R be the region bounded by the x axis the graph of y equal rootx and the line x = 4 part A is to find the area of the region R so that should be pretty straightforward we'll just start out with a sketch of the function yalun X and the line x = 4 and we just need to find the area of the region enclosed you should know by now that y = < tkx looks something like that and the line x = 4 I'll draw in just a second we should probably just go ahead and label this Y = rootx and then in Orange we will write the line x = 4 and maybe that's right here so there is x = 4 it's a vertical line and so we are looking for the area of this enclosed region because remember um this region is also bounded by the xais right so it's this region here that we're looking for this is the region R we need to find its area so that's just the area under the curve y = < tkx from x = 0 2 x = 4 so very simple for part A we're just going to integrate our function sare < TK of x from 0 to 4 to find the area underneath it now Square < TK of X is the same as X2 so to integrate this we're just using the power rule we need to increase the power by 1 so x^ the2 goes up to x to the 3es and then divide by that new power which is the same as multiplying by its reciprocal that's the integral and we're evaluating this from 0 to 4 now when we plug in zero the lower bound we're just going to get zero so all we have to worry about is plugging in the four and we're going to have 2/3 * 4 the^ of 3/ 2 now remember a fractional exponent is kind of like two separate operations it tell tells us we need to take the square root of four because of that denominator of two and then we need to cube the result the order of course doesn't actually matter that's just the way I like to think of it so first we take the square root of 4 which is two and then we have to cube that so this ends up being 2/3 * 2 cubed 2 cubed is 8 so it's 2 * 8 over 3 and so the answer is 16 thir that is the area that's the answer for part A I'm going to shrink our work for part A and then we're going to move on to Part B Part B says that we are to find the value of H such that the vertical line x equals H divides the region R so the same region as before into two regions of equal area so this sketch that we had before is still useful but what we're looking for now is what line xal H would we have to draw so that the area to the left is the same as the area to the right so we'll have to set up an integral for the area on the left and an integral for the area on the right and then we can solve that for the unknown which is where does this vertical line have to be to make these two regions equal so I've just copied the picture and made it a little bigger there is our line xal H we're not sure exactly where it has to be we're just putting it somewhere on the picture to help us set up our integrals so the first integral is going to be the area of the region of R that's to the left of xal H that's this part over here so for that we're just going to integrate the square < TK of x from 0 to H and this has to equal the right side of the region which is the integral from H to 4 we're still stopping at xal 4 and so these things need to equal each other we can just go through and evaluate this and then solve for H this integral on the left is a lot like the integral we did in part A except instead of going from 0 to 4 we're going from 0 to H in the part A integral you may remember that we ended up with 2/3 * 4 to the 3 by the same process on the left side here we would end up with 2/3 * H the upper limit of this integral to the power of three Hales now on the right side similar stuff will happen we'll integrate the square < TK of X which again becomes 2/3 * x 3 applying the reverse power rule so to speak and we're evaluating this from H to 4 now on the left we have 2/3 H to the 3es and then evaluating the right side first we plug in the upper limit and we already saw before that when we plug four into this integral what we get is 16 over3 so I'm just going to write that as 16 over 3 and then we subtract plugging in the lower limit which in this case is H so minus 2/3 * H to the 3es then we can go ahead and add 2/3 H to the 3es to both sides to collect all of the H's together so we now have 4/3 H to 3 equal = 16/3 let's go ahead and multiply both sides by 34s to get rid of this coefficient in front of the H that's going to give us H to the 3es equals multiply both sides by 34s so we're multiplying by 3/4s over here threes cancel out a factor of four will cancel out and so on the right side we're just going to be left with four now we can square both sides to get rid of that power of 12 thus we have H cubed = 16 and then take the cube root of both sides and that is the answer to our question remember the question here was to find the value of H such that the vertical line x equals H divides the region R into two regions of equal area so if we were to draw the line xal cubot of 16 that would split the region into two regions of equal area let's move on then to part C find the volume of the solid generated when R is revolved about the xais so this is a pretty standard volume of a solid of Revolution problem so let's set up the appropriate integral I'm just going to shrink our work for Part B and then we can get on with it so moving on to part C we are revolving this curve about the x axis and what that will do I'll draw in a different color I'm sure you know but it's helpful to sketch these solids of Revolution especially in more complicated cases it's going to look like this right because we're creating this solid by revolving the curve the cross-sections are circles which makes finding the volume of this solid rather easy and notice that those uh circular cross-sections have radi which are the values of the function root X at those x coordinates so this this is a very classic example nothing weird going on here what we're doing is adding up infinitely many circles which is why in this type of situation we have to integrate PIR sar from in this case 0 to 4 right from xal 0 out to 4 now the pi we can take out of the integral the r remember is the radius of a crosssectional circle and we are integrating this by the way with respect to X so how do we express the radius as a function of X well like we just said from the sketch too we can see this the radius of each circle is just the value of the function rootx it's like a height so we can rewrite this integral writing R SAR in terms of X this is let me take the pi out so this is pi * the integral from 0 to 4 of the the radius squared the radius is the function that's what gives us the height that's root X and we have to square that and then this is getting integrated with respect to X this of course works out pretty nicely because the square < TK of x^2 is just X so this is pi times the integral of x from 0 to 4 the integral of x applying the reverse power rule is just 12 x^2 this is from 0 to 4 plugging in the upper bound is going to give us 12 of 4^ 2ar so that's 1 12 of 16 which is 8 plugging in zero will just give us zero and so our final answer for the volume is 8 pi and we could say units cubed finally we can move on to Part D this is very similar to Part B the vertical line xal K divides the region R into two regions such that when these two regions are revolved about the x axis they generate solids with equal volumes we need to find the value of K that causes this so it's just the three-dimensional version of problem B we need to find a vertical line so that the solids of revolution on either side of that line will be equal I just shrank our work for problem C here's a copy of the picture and remember we're now looking for a vertical line line um so I'll just draw it like that this is the vertical line xal K and we need to make sure that this volume on the left is the same as this volume on the right still going from 0 to four because that's how the region R is created so just like with problem B we'll set up two integrals representing the volume of the side on the left and the volume of the side on the right set them equal and then we can solve for K now the volume of the solid to the left of xal K which perhaps I should have drawn as a curve to more accurately U be consistent with the 3D sketch but regardless we're integrating from 0 to K and we can take the pi out the radius squared we know is just X DX so that's going to be the volume of this part of the solid that's to the left of the vertical line x = k and this is going to equal pi multiplied the integral from K to 4 of the radius squared the radius again is just root X so when we square that we get X now we can evaluate these integrals and then just solve for K so on the left we're going to have pi multiplied by 12 x^2 evaluated from 0 to K this equals on the right piunti 12 x^2 evaluated from K to 4 on the right we now have pi multiplied by plug in the upper limit so 1 half k^ SAR and then plugging in zero won't do anything on the right we have pi multiplied by plugging four in will give us eight and then plugging in k for our lower bound that's going to be - 12 k^ 2 and now we'll just have to solve this for K on the right you can see that we have2 piun k 2 so let's add that to both sides on the right we have 1 half pi k^ 2 and we're going to add another half pi k 2 from the right so that's going to give us a full PK k^ 2 on the left which equals 8 piun divide both both sides by pi and we have that K is equal to the square otk of 8 right divide Pi take the square root of both sides k equal the < TK of 8 which we can simplify because 8 is 4 * 2 so you can take this uh < TK 4 out giving us 2 < tk2 if we were to draw the vertical line x = 2 < tk2 the volume of the solid to the left would be the same as the volume of the solid to the right this is problem one from the 2002 AP Cal AB exam this is part of the graphing calculator section so we will be using a graphing calculator throughout this problem let's read it let f and g be the functions given by F ofx = E to X and G ofx = lnx I'm going to refer to lnx as log X if that's okay with you part A is to find the area of the region enclosed by the graphs of f and g e to X and log X between x = a half and x = 1 so we need to integrate the difference of the functions from 1/2 to 1 that's how you find the area between curves you may want to sketch these functions e to the 0 is 1 and from there it grows exponentially and as X gets very negative e to the x approaches zero on the other hand log X for values of X that are very close to zero log X is very big and very negative and then as X gets larger log X grows indefinitely although it does it very slowly and is certainly less than e to X and remember we want to find the area between x = 12 and X = 1 so maybe that's right around here this is x = 12 and then x = 1 is going to be right here and we know that x = 1 is where log X intersects the X AIS because log one is zero all right so we can see that log is the lower function and e to the x is the upper function so for our integral we're going to integrate from 1 12 to 1 that's given and we're going to integrate the upper function e to x minus the lower function log X this is going to give us the area of that region and we can use our graphing calculator to figure this out so I'll open up the graphing calculator press the math button to get access to all these functions and I'm going to use the integration function where integrating from 1/2 or 0.5 up to 1 and the thing we're integrating is e xus log X and then of course we're integrating this with respect to X and we can enter this to get our answer about 1.22 so that is the area of this region all right let's move on to Part B find the volume of the solid generated when the region enclosed by the graphs of f and g between X = 12 and x = 1 that's this region here which we just found the area of we want to find the volume of the solid generated when that region is revolved about the line Y = 4 so that's a line that's up here which is above this region and we're going to revolve the region about that horizontal line and we need to find the resulting solid now hopefully you can picture if we revolve this region about that line Y = 4 you can see that it is the green curve which is log X which will be tracing the exterior of this surface the interior is going to kind of be carved out by e to the X and so this is what we call a washer problem to find the volume of this solid of Revolution we need to integrate or ADD up the areas of all of these washers so what are we going to do for Part B well again we need an integral from 1/ 12 to 1 and what we're integrating is the areas of these washers we can find those areas by taking the areas of the Outer Circle which is traced by log X and subtract the area of those inner circles which is traced out by e to the A X now the area of the bigger circles will of course be p piun r 2 and R the radius of this bigger Circle which is traced out by revolving log X about the line Y = 4 the radius of that circle is this distance here and that distance is just four minus log X I just redrew it a little more nicely once more the radius of those bigger circles is the distance from y = 4 to that lower function and that is 4 - log X as for the smaller inner circles the radius of those is the distance from yal 4 to that upper function and that is e to the x that distance is 4 minus E X those are our smaller radiuses of the inner circles which are actually being carved out of this solid okay so coming over to set up our integral both both of these areas have a factor of Pi in them so we'll just take the pi out of the integral the radius squared for the larger circles is going to be 4 - lnx sared and then we're subtracting the areas of those smaller inner circles which is 4 - e to x^2 we're integrating this with respect to X and we can just calculate this on our graphing calculator so on the calculator again I press the math button to access these functions press the integration function we're integrating from 1/2 up to one what we are integrating is the difference of the squares of the radi we'll multiply by pi after the fact so let me enter in these radi squared so there's the difference of the squares of the radi we're integrating this with respect to X and there is our answer about 7.52 but don't forget we need to multiply this by Pi so finally the volume of this solid is about 23.6 finally moving on to part C let H be the function given by h ofx = f ofx minus G ofx so just the difference of e and log find the absolute minimum value of H ofx on the closed interval from 1/ 12 to 1 and find the absolute maximum value of H ofx on the closed interval from2 to one and show the analysis that leads to our answers okay to find maximums and minimums of course we're going to need to take the derivative of H that way we can find any critical points we also know that we're going to have to check the end points because remember this is defined on a closed interval between 1/2 and one so this is sometimes called the candidates test what are the candidates for the extreme values well the those would be the end points 1/2 and one so we will need to check those and then also the critical points but to find the critical points we have to take the derivative so let's just jot this down what do we need to check well we'll have to check 1/2 and one and by check them I mean we need to plug them into the function H ofx and see what those values are in order to establish minimums and maximums and then we'll also have to find any critical points and check those so let's find H Prime of X and then we'll be able to find the critical points now remember h of X is just f ofx - G ofx which is e to xus log X so H Prime of X is just FR Prime minus G Prime F Prime is e to X because the derivative of e to X is e to X and then we subtract G Prime G is log X so G Prime is 1/x all right so what are the critical points well the critical points are where this derivative doesn't exist it doesn't exist at xal 0 but 0 is not in this interval so that doesn't matter and also where the derivative is equal to Zer so let's set this derivative equal to zero and then we can use our graphing calculator in order to solve that so I'm going to enter this function e x - 1 /x into my calculator and then then we can look at the graph pressing the graph button we see a function that looks like this and we need to see where does it pass zero in our interval and that would be right about here let's see if we can get the calculator to give us this number it appears to be about. 567 that's when Y is zero when X is about 567 so I'll write that over here that is our critical point so we need to check xal 12 x = 1 and our newly discovered critical point at X = 567 we need to plug all three of these values into the function h of X and C which is the maximum and which is the minimum plugging them in these are the values we get and we can see the lowest value is at that critical point where X was about. 567 here the value of H is about 2 2.33 so that would be our minimum and the maximum we see is at x = 1 where H takes on a value of 2.72 and remember these were discovered on a closed interval so on that closed interval this is actually an absolute maximum and an absolute minimum and that completes our solution to problem one of the 2002 AP Cal AB exam this is free response question two from the 2004 AP Cal AB exam this is part of the graphing calculator section so we will be using a calculator throughout this problem let's read it let f and g be the functions given by F ofx = 2x * 1 - x and G of x = 3 * x -1 * < tkx for X between 0 and 1 and we're given a graph of these curves for free you can see they go from 0 to 1 we have f ofx as the upper function that's right there and G of X as the lower function part A is to find the area of the Shaded region enclosed by the graphs of f and g which is what we have a picture of you're not going to get a much easier part A than this we can already see a picture of this situation we're trying to find the area of this region and we can use a calculator so all we have to do is take the integral of the upper function f ofx so I'll write the integral of f ofx minus the lower function which is G of X this is going to give us the area between the curves and we want this area from 0 to 1 because this shaded region goes from 0 to 1 now F ofx of course is 2x * 1 - x and G of X is 3 * X - 1 * otk X so let me just write those things in place of f and g all right so that there is the integral that we're calculating and we will just use our graphing calculator to do that so I'm going to press the math button and that's going to give me access to all these functions I'm going to go with function n the integration function and we are integrating from 0 to 1 what we're integrating is 2 x * 1 - x - 3 * X - 1 multiplied by the Square < TK of X and we see from the calculator this area is about 1.33 so let's write that this is about 1.33 and that completes part a moving on to Part B find the volume of the solid generated when the Shaded region enclosed by the graphs of f and g so the Shaded region here is revolved about the horizontal line Y = 2 now we know from the picture the horizontal line Y = 2 is somewhere up here right because y = 1 is right there so we're revolving the Shaded region about that horizontal line way up there I've just done a rough redrawing of the figure that gives us a little bit more room to work with this is our shaded region that we found the area of in part A and for Part B we're creating a solid of Revolution by revolving this shaded region about this horizontal line yal 2 so that's going to look something like this you can see how the lower curve which is G of X the lower curve is going to trace out the exterior of this solid of Revolution the interior is going to kind of be cut out by the upper function which was f ofx as we do this revolution it would look some something like that we have these middle circles cut out by that upper function so this is what we call a washer problem to find the volume of this solid we're going to have to add up the areas of all of these washers I said add up and of course that means we are integrating still from 0 to one because we are still talking about this shaded region from 0 to one and what happens when we revolve it about a horizontal line what we need to add up are the areas of all of these washers and the way we find those areas is by taking the area of the bigger Circle traced out by the lower function and subtract the area of the smaller Circle which is traced out by this upper function each circle has an area of Pi R 2 let's just take the pi out so we have a pi in front of the integral and then we're going to need the radius squared of the bigger circles and subtract the radius squar of the inner circles now what is the radius of these big circles well that would just be the distance from the line yal 2 which is the center down to the curve that's tracing out the circle that curve is G of X the lower curve so this radius here is just two the horizontal line minus G ofx because it's the distance between the line and the curve G of X so we'll put that radius in here this is 2 - G of X that's the radius of the big circle the radius of the small circle is similar it's the distance from the horizontal line yal 2 to the curve that's tracing out the smaller inner circles that curve of course is f ofx so in here we'll put the radius of the smaller circles which is being subtracted and that's 2 - F ofx again remember from the picture we see that g of X is that lower curve F of X is that upper curve which is actually tracing out the smaller circles and here's what that integral looks like once we plug in the pieces we still have the pi out front we're integrating from 0 to 1 2us G of X and this is G of X this is getting squared minus 2 - F ofx and this is f ofx and this is getting squared let's jump over to the calculator and find the value of this integral again I'll press the math button and then go to function 9 which is the integration function and then I will go through the long process of typing this whole thing out you want to make sure you're careful when you're typing these long expressions with parentheses and squaring into your calculator this looks right to me so I'm going to go ahead and press enter and we see this is about 5 and 32ths but we don't want to forget that we took a pi out of the integral so we also have to go ahead and multiply this by pi once we do that we get about 16.1 18 I'll round to three decimal places and call it 16179 that completes our solution to Part B so let's go ahead and move on to part C let H be the function given by H ofx = KX * 1 - x for X between 0 and 1 notice that h of X is just like our function f ofx except instead of being scaled by two it's being scaled by K where where K is positive so it's either being shrunken down a little bit and maybe looks like that or it's being stretched and look something like that finishing the question for each K greater than zero the region which is not shown because remember this is f ofx it's not H ofx the region enclosed by the graphs of H and G is the base of a solid with square cross-sections that are perpendicular to the xais there's a Val value of K for which the volume of this solid is equal to 15 so we could stretch this function h of X by some K that's going to make the volume of the described solid equal to 15 we need to write but not solve an equation involving an integral expression that could be used to find the value of K this problem is actually pretty easy but there is a lot going on here we need to read it very carefully let's come back to our picture and draw what H ofx could look like like we said before it looks just like f ofx but scaled up by some factor or potentially scaled down but let's draw it as if it has been scaled up so this is yal H ofx now the question is asking us about a solid whose base is enclosed by the graphs of H and G that is the base so this region between H and G F has nothing to do it this at this point remember this region between H and G is the base of the solid in question I'm going to erase this shading just so our picture isn't too cluttered but just remember that F ofx isn't important right now it's this region between h of X and G of X that is important the problem is saying that we have a solid and the base of this solid is this region and the way that this solid is constructed is by basically taking squares that lie in this region and that are perpendicular to the x-axis that means it looks kind of like this if we were to take a cross-section that's perpendicular to the x-axis it lies in this region and the crosssections as it says have to be squares they're Square cross-sections so just to give you some of this 3D perspective they might look something like this if we we were to do kind of a rough perspective sketch of the cross-sections that's coming up out of the screen it is a square cross-section and its base the base of this Square we can clearly see is just the distance between the upper function H ofx and the lower function G of X so this base the side length of this square is just H ofx minus G ofx now what about the height of the square well it's a square so of course this other length is just another side length of the square it also has a length of H of xus G ofx because that's the base length and this is a square so it's equilateral thus the area of the square is H ofx minus G ofx squared to find the volume of this solid which we know is 15 for some value of K remember K is part of the definition of h ofx to find the volume we just have to add up all of the areas of the square cross-section so let's set up this integral we have to integrate from 0 to 1 the areas of the square cross-sections the areas of the square cross-sections are the squares of the side lengths each side length is H ofx minus G ofx we have to square those to get the area we're adding this up this integral gives us the volume which is equal to 15 for some value of K now where does the K come in here well let's just rewrite this integral one more time replacing h of X and G ofx with what those functions actually are H ofx is KX * x -1 and G of X the lower function here is 3 * x -1 * < tkx and remember this whole thing is the side length of the square cross-sections and so we need to square it to get those areas this integral is equal to 50 for some value of K and we could use this equation to solve for K that's the answer to part C and that completes our solution to free response question 2 from the 2004 AP Cal AB exam this is free response question 5 from the 2019 AP Cal AB exam let R be the region enclosed by the graphs of G of x = -2 + 3 cosine ofun / 2X and H ofx = 6 - 2 2 * x - 1^ 2 the Y AIS and the vertical line x = 2 as shown in the figure above so this was just a description of the region R enclosed by these curves and these vertical lines part A is to Simply find the area of R and note this is problem five this would have been part of the no calculator section so we will not be using a calculator for this all right so for part A we're just going to have to do an integral the integral should go from X = 0 to x = 2 we know that because the problem told us this region R is bounded by the Y AIS which is xal 0 and the vertical line xal 2 so we need to integrate from here to here then what we're integrating is the upper function minus the lower function you could totally disregard which is upper and which is lower and just take the absolute value of your answer for this integral uh but at a you may be able to notice this is clearly the cosine function that's G ofx that's the lower function this one up here is the quadratic that's the upper function so we're going to be integrating that upper function 6 - 2 * x - 1^ 2 minus the lower function the lower function is G that's -2 + 3 cosine Etc it'd be great to have a calculator for this but let's not fret we can do this by hand without too much difficulty let's begin the integral integrating 6 just gives us 6 x as for -2 x -1 2 we'll have to keep the -2 and then the power of two in the x - 1^ 2 needs to be increased by one but then you also have to divide by that new power so divide by three so we end up having -2/3 as the coefficient since we have a function inside the parenthesis we also have to divide by its derivative but its derivative is just one so we're actually done with that already moving on to this next part -2 is pos2 so that integrates to + 2x as for -3 cosine of P piun / 2x we know that we're going to need a -3 sin of PK / 2x the derivative of sin of < / 2x is going to give us that cosine of < / 2X X but by the chain rule it would also give us a factor of Pi / 2 we need to divide that out since we're doing integration dividing by piun / 2 is the same as multiplying by 2 over piun so -3 * 2 over piun is -6 over Pi so let's just put that Min - 6 over pi and there we go we're integrating this from 0 to two you might have to watch that again if you haven't practiced integration much and if this part with the cosine was difficult you might want to just try using Uub hopefully with practice you won't use U substitution in situations where the inside function is linear because in those situations you just have to divide by the derivative it's pretty straightforward but it does take some practice so use the U sub if you need to all right let's start now by plugging in this upper bound and then we'll subtract plugging in the lower bound plugging in the upper bound of two gives us 12 at first so we have 12 and then here we'll have 2 - 1 cubed which is just 1 cubed so we'll just be left with the coefficient of -2/3 then plus 2 * 2 so that's going to be + 4 and then with the S function < / 2 * 2 is pi and sine of Pi is zero so this whole last term is gonzo all right now we will subtract plugging in the lower bound plugging in zero plugging in zero is going to get rid of that and that and the sign function because s of 0 is 0 so all that's left is this term here 0 - 1 is -1 and then when we Cube that we get min-1 so this should just be -23 * -1 which is positive 2/3 all right now we just have to evaluate this let's give everything a denominator of three so that we can combine all of these fractions 12 is the same as 36 over three then we're subtracting 2 over three and then 4 is the same as 12 over 3 and then we're subtracting another 2 over 3 all right now what does this add up to 36/3 + 12/3 is 483 - 2/3 - 2/3 is - 4/3 so this is a total of 443 and that is our answer for part a all right let's move on then to Part B region R so the same region we've been talking about is the base of a solid so it's like we have a solid coming up from the paper for this solid at each X the crosssection perpendicular to the x axis has Area 1 /x + 3 we need to find the volume of the solid this is actually a really straightforward question if the area of the cross-sections perpendicular to the X AIS are 1 / x + 3 we just need to integrate those areas from X = 0 up to x = 2 so setting up this integral all we're doing is integrating from 0 to 2 the areas of the cross-sections that's how we find the volume of this solid the areas of the cross-sections are given to us as 1/ x + 3 sometimes in these types of problems we're just told that the cross-sections are squares and then we we could use the base length to figure out what the areas of the squares are but this example is honestly even easier because we're just given the areas of the cross-sections outright now let's go ahead and integrate this the integral of 1 /x + 3 is the natural log of x + 3 and we don't have to worry about absolute values around the inputs because we're evaluating this from 0 to 2 where the input of the log function will always be positive here so let's just plug in the Upp bound that's going to be Ln of 5 and then subtract the lower bound which is going to be Ln of 3 and that is the volume of the solid described in Part B again notice this is a situation where the inside function x + 3 is linear so no Uub is really required we would just have to divide by the derivative of this inner function in this case that derivative is one which is why it didn't really come up let's move on then to part C write but do not evate oh good we don't actually have to do the integral this time right but do not evaluate an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = 6 coming up to the picture provided y equal 6 is this horizontal line here so what's happening is our region R is getting revolved about this horizontal line and if that happens then hopefully you can see that this lower function or what was the lower function the cosine function is going to be the outer function in the revolution and this what was the upper function the quadratic is going to be the inner function in the revolution about this horizontal line the cross-sections of this solid of Revolution will be these washer shapes where the outside is traced by the cosine function revolving and the inside hole is carved out by the quadratic function revolving so what we need to do is take the areas of the big circles traced out by the cosine function and then subtract the areas of the inner circles which are removed by the quadratic function this is a washer problem so let's set up the big picture here we are integrating from 0 to 2 the area of some circles minus the area of some other circles both of those will have a pi in them right it's like Pi Big R 2 minus PI Little R 2 let's just pull a pi out of the whole thing and put the pi in front of the integral okay and then we're going to need that big radius squared minus the little radius squared and this integral will be taken with respect to X we can go back up to our picture to see what these radi are the radius of the big circles is the distance from the Line yal 6 down to thee cosine function G ofx that distance I'll write it over here that distance is just 6 minus G ofx that's the distance between them which is the radius of the outer circles or the big circles the Inner Circle has a smaller radius it's let's do it in I guess purple it's the distance from yal 6 down to the quadratic function that little distance right there now that distance is found in a similar way it's just six the horizontal line minus the quadratic function H ofx that's the distance between them which is the radius of one of these smaller circles with that we can finish our integral setup the Big Radius Big R S we can replace that with the radius of the outer circles which we found to be 6 minus G of x^2 and then we're subtracting the squared radius of these smaller circles which are cut out that's why we're subtracting them they are cut out from the crosssection those radi we found to be 6 minus that inner function H ofx again squared and don't forget the DX and there we go that is our integral expression which we don't need to evaluate but just to make it very clear that all of this is in the integral let's put it in one big set of parentheses and that will complete our solution to free response question five from the 2019 AP Cal AB exam next up are problems on differential equations call that we can think of a slope field as a graph of a bunch of tiny tangents to the solutions of the differential equation also recall that for separation of variables you're trying to get the DX term with the x's and the Dy term with the Y's you don't want the DX in the denominator you don't want the Dy in the denominator once you've separated variables you can integrate both sides of the equation this is problem five from the 2006 AP Cal AB exam consider the differential equation dydx = 1 + Y overx where X is non zero part A says on the axes provided sketch a slope field for the given differential equation at the eight points indicated so when you sketch a slope field of course you've got to decide how many slopes should you draw and where should you draw them this is telling us how many to draw and where to draw them at these eight points so we'll just have to use the differential equation to find the slopes at those points and sketch little tangent lines representing those slopes so let's get into it this should be pretty quick because there's only two parts to this problem it might help to organize our information in a table for part A so we can put the coordinates on the left side and we can put their corresponding derivatives on the right side I've split this up into two tables just so that the table's not super long and I have to scroll way down the page to fill it out so it's just space reasons all right we'll start with the first four points in this table those are -2 0 so I'll put that here -2 0 we've got -1 0 we have -1 1 and we have this point here -1 -1 we'll have to plug these points into the derivative given the differential equation in order to find the slopes at these points plugging in -2 0 gives us a one in the numerator because we'd have 1 plus that y coordinate of 0o divided by that x coordinate of -2 so the slope here should be -2 plugging in the point -1 0 should give us 1 + 0 in the numerator so 1 divided by that x coordinate of 1 so in total the slope would be1 plugging in the third Point -1 1 we should have 1 + 1 or 2 in the numerator divide by -1 in the denominator for a total of -2 being the slope last point-1 1 we should have 1 + -1 in the numerator which is 0o divided by -1 in the denominator but it ends up just being zero we will sketch little tangent lines to represent those slopes but first let's finish calculating the slopes for the other four points which we'll do in this table all right so we've got this point which is 1 0 we've got this point which is 2 0 we've got this point which is 1 one and we have this point here which is 1 1 and we'll plug these in and find the slopes just as we did for the previous four points if we plug in one zero the numerator will be one and the denominator will be one also so the slope will be one if we plug in two Z we'll have one in the numerator and two in the denominator so that will be a slope of 1/2 if we plug in 1 one the numerator will be two and the denominator will be 1 so the slope will be two if we plug in one - 1 the numerator will be zero and the denominator will be one so the slope will be zero all right let's go ahead and plot tangent lines representing these slopes at the appropriate points note if you were actually doing this on the exam for real you would have been asked to use the axes provided in the pink exam booklet but we're going to do it right here and remember these things don't have to be exact right negative half I don't have to draw a slope of exactly Nega half there I'm doing this by hand that's not really possible but I do need to get the big details right so I need it to be going down because it's negative and I would need it to be less steep than the slope of -1 which should be less steep than the slope of -2 got to get those big relative details right but let's start with a real simple one the slope of zero at the point11 at this point here11 we should have a horizontal tangent so I'll just draw a little segment you can see it there representing the slope at that point then going back to the top of the table at the point -2 0 we should have a slope of2 here is -2 0 a slope of2 should be a slight downward slope so maybe something like that then at -10 we should have a slope of1 so it should be downwards and it should be steeper than that slope of2 and again this is at -10 so maybe that looks something like that then at the point -11 we should have a slope of -2 so right here we should again have a downward slope and it should be the steepest yet so maybe it looks something like that now if we had arranged this data more intelligently let's say we swap rows 1 and two we could easily see that the remaining points and their corresponding slopes are just the Opposites of everything we just did these are the same slopes that we have left to sketch that we just did except opposite instead of negative it's positive instead of negative it's positive instead of negative it's positive you see that after noticing that I can very quickly finish this slope field by just drawing a mirror image of everything we just did so I'll try to do that quickly H maybe it looks something like this and there is our slope field remember you just got to get those big details right like this slope of positive one is going upwards and it's less steep than this slope of positive2 which is going upwards let's move on then to Part B we are asked to find the particular solution y = f ofx to the differential equation with the initial condition f of1 = 1 and to State its domain the differential equation in question here of course is this one that we were originally given so we can find a particular solution using this differential equation and this initial condition and we can do it using the separation of variables technique so I've written the differential equation and the given initial condition let's multiply both sides of the differential equation by DX to get the X's together and divide both sides by 1 + y to get the Y's together if we do that on the left we're going to have 1 over 1 + y Dy and on the right we're going to have 1 /x DX after after separating the variables the next step is to integrate we integrate both sides on the left integrating 1 over 1 + y that's just the natural log of 1 + y remember the absolute value is necessary here in the event that y might be negative if we knew for sure y would never be negative we would not have to bother with those absolute values but in this case we do over here on the right side similar situation the integral of 1 /x is the natural log of the absolute value of x in theory there is an arbitrary constant on the left and the right but let's just say we gathered them both together on the right side and called that c now remember we're trying to get y by itself because we want y equals F ofx we want some actual function that is a particular solution to the differential equation given this initial condition so to get this y by itself or to take one more step step towards that goal we will exponentiate both sides of this equation that's going to give us e to the Ln absolute value of 1 + y on the left side equals e to the Ln absolute value of x + C on the right but by our exponent laws that's the same as e to the Ln absolute value of x time e to the C right because we could bring these together by just just adding the exponents but e to a constant is just a constant so let's just agree to call that constant C and move it to the front order of multiplication doesn't matter we just prefer to write it this way finally the E and the LNS cancel out that was the whole point and so now we have the absolute value of 1 + y on the left equals this constant C multiplied by the absolute value of x now we can clean this up sum because remember we're looking for a function yal f ofx that is a solution to our differential equation with this initial condition so we know it can't be the case that 1 + y can take on multiple values sure 0.5 and negative .5 have the same absolute value but if that was allowable then we would not have a function because that would mean you could plug in an X and I could give you a couple valid options for y to make this equation true because of the absolute value so does 1+ y take on negatives or positives because it can't possibly take on both well from the particular solution we know that it must take on positive values because we have a positive y we know that for a fact so this absolute value of 1 + Y Must only take on positive values and thus the absolute value can be dropped because it will not change the positive input so we have that 1 + y = c * the absolute value of x then we can plug in our initial condition in order to solve for the constant C the initial condition gave an x coordinate of -1 so we have C * the absolute value of - 1 and the corresponding y-coordinate was positive 1 so we have 1 + 1 on the left the absolute value of -1 just goes away since it's one and we find that c equal 2 then we can rewrite our equation with this information we have the 1 + Y = 2 multiplied by the absolute value of x all right now what about the domain well let's make the one more step of getting y by itself so Y = 2 * the absolute value of x minus one just moving that one over and now we can figure out the domain by Consulting our initial condition the initial condition said that we have ax value in the domain and looking at the slope field and the differential equation we know that X can't equal zero so our function either lies on this side of the slope field or this side and since we know we have a negative x value it must be on the left side and so so the domain is X is less than zero and since X is negative we know EX exactly what the absolute value is going to do it's going to apply an additional negative to make the X positive so we could also write the function like this y = -- 2x -1 where again the domain is that X is less than 0 as suggested by the initial condition and that completes our solution to free response question five from the 2006 apal AB exam this is free response question four from the 2015 AP Cal AB exam let's read it consider the differential equation dydx = 2x - Y part A on the axis provided sketch a slope field for the given differential equation this one here at the six points indicated all right let's just create a quick table with the XY values that we are interested in see what the derivatives are at those points and then sketch some little tangents so that we end up with a slope field here's my table of XY Pairs and then in this third column we'll put the derivative dydx which is just 2x - Y at let's see this point here where x is 0 and Y is -1 2x - Y is 0 - -1 which is POS 1 at this point here where X is0 and Y is POS 1 2x - Y is 0 - 1 which is -1 at this point 02 2x - Y is 0 - 2 so that's -2 at this point here 1 -1 2x - Y is 2 - -1 so 2 + 1 so 3 at this point here 1 1 2x - Y is 2 - 1 which is 1 finally at this point 1 2 2x - Y is 2 - 2 which is 0 all right now we can sketch the slope field by sketching some little tangents that capture the right idea of what's going on here like for starters the point one two right here has a derivative of zero at that point so we should have a horizontal tangent like that at let's see let's go back to the top so 01 that's right here we should have a slight upward slope because that's a positive slope of one so something like that let me try again just to make that a little more straight then at this point 01 we should have the same sort of slope as before but negative so at 01 right here we'll have a slight negative slope something like that at the point 02 so way up here we should have a steeper negative slope of -2 so something like that perhaps then at one negative 1 we should have our steepest slope of all a slope of three and it's positive that's right here one negative 1 so maybe something like that or maybe that just a little bit steeper at the point 1 one that's our last one to sketch we have a slight positive slope of one so it should be the same sort of slope as this guy here so maybe that looks something like this and there you go remember you're not going to be able to sketch these slopes exactly but you want to keep the general idea correct this guy here with the SL of posi3 should be steeper than any of the other lines and obviously it should be going upwards this one should be horizontal this one here should be steeper than these others that have slopes of plus or minus one just get the general details correct moving on then to Part B find the second derivative of y with respect to X in terms of X and Y and then determine the concavity of all solution curves for the given differential equation in quadrant 2 so the differential equation here we're going to be able to make a statement about the concavity of its solution curves by finding this second derivative also remember in quadrant 2 x is negative but Y is positive so we will consider that once we find this second derivative to find the second derivative we'll just differentiate both sides of the first derivative this is the differential equation that we were given let's differentiate both sides if we do that then on the left we are going to have the second derivative of y with respect to x equals on the right the derivative of 2x is just 2 and the derivative of y is dy DX but of course we know that dydx is this 2x - y so let's just replace that dydx with 2x - Y and then we can distribute the negative here and find that this equals let's put the positive y first it's going to be y - 2x and then + 2 and if we want to factor out the two we could rewrite this as y let's say + 2 * 1 - x then addressing the concavity part of the question in quadrant 2 like we said X is negative and Y is positive so if we look at our second derivative we know that 1 - x is going to be 1 minus a negative in quadrant 2 which is 1 plus a positive so 1 - x is obviously going to be positive in quadrant 2 and that makes it clear that this whole second derivative is going to be positive because the only other thing in question would be why but if we're in quadrant 2 then we know that Y is positive so we can say for sure this second derivative y + 2 * 1 - x is positive which means that in quadrant 2 every solution curve has to be concave up all right moving on to part C let yal F ofx be the particular solution to the differential equation with the initial condition that F of 2 equals 3 does f have a relative minimum a relative maximum or neither at x equals 2 and justify our answer to determine if there's a relative minimum or a relative maximum let's first plug x = 2 into the derivative and see if this is a critical point if it's not a critical point then no further investigation is needed if in fact this is an extreme value the derivative should be zero where X is equal to 2 so let's evaluate this so for part C we're evaluating Dy DX at xal 2 and remember in this problem we're told that when x equals 2 the function is equal to 3 so we could say this is at x = 2 y = 3 now the derivative dydx is 2x - y so this is 2 * X which is 2 - Y which is 3 this is 4 - 3 which is POS 1 which is not equal to zero thus no further investigation is necessary F does not have a relative minimum nor a relative maximum at xal 2 because its derivative there is not zero so there is our conclusion there is no extreme value at xals 2 moving on to Part D find the values of the constants m and b for which the line Y = MX plus b is a solution to the differential equation all right well if y = mx plus b what is dydx the derivative naturally that would just be M the slope so we would need M to equal 2x - y all right let's write this down and start doing some work all right again if y = mx plus b it's derivative is just the slope which is M which if it's a solution to our differential equation we know that this should equal 2x - y because that is the differential equation that it must satisfy now what can we do with this well we might have that M = 2x - y but we also know that y = mx + b so this is the same as M = 2x - MX - B we can then do some factoring to see that M must equal 2 - M * X that's factoring the X out of those two terms and then of course minus B now we can use a method called equating coefficients on the left side we see there are no X's the coefficient of x is zero so 0 must equal the coefficient of x on the right side which is 2 - M 0 must equal 2 - M which means that M must equal 2 on the other hand there's one constant on the last which is M which has to equal the constants on the right B so that gives us this other equation m equals B which if we want to learn about B here we could say that means that b equal m and since we found the m equals 2 this means that b must equal -2 so there is one possibility for m and b in order to have a curve or in this case a line that satisfies the differential equation in question that one pair values is m = 2 and B = -2 which gives us this solution curve y = 2x minus 2 and that completes our solution to free response question 4 from the 2015 AP Cal AB exam this is free response question three from the 2023 AP Cal AB exam a bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed The increased ing function m models the temperature of the milk at time T where M of T is measured in De C and T is the number of minutes since the bottle was placed in the pan M satisfies the differential equation dmdt = a 4 * 40 - M at time t equal 0 the temperature of the milk is 5° C so we have a nice initial condition there it can be shown that M of T is less than 40 for all Valu of T all right part A a slope field for the differential equation dmdt equals Etc is shown right here we are asked to sketch the solution curve through this given initial condition 05 so coming down to the slope field notice that this point 05 is already labeled on the picture for us we just have to sketch the curve that is in agreement with the slope field so that curve should generally follow V path of the tangent lines something like this and then it's got sort of a horizontal ASM toote around m equal 40 and so it would continue to proceed in this manner the rate at which it is increasing is getting very very slow as time goes on when you first started graphing things like lines back in the day your teacher might have often told you to make sure to put arrow heads at the ends so it's perhaps interesting to note that according to the college board's official scoring guidelines you do not need arrows on this graph so this would be a perfectly fine answer all right anyways let's move on to Part B use the line tangent to the graph of M at T equals 0 so using that initial condition that we have to approximate M of 2 the temperature of the milk at time T equals 2 remember what a tangent line is is a linear approximation of a function at a point point we're looking for the tangent line at tals 0 so if I were to sketch that in blue that might look something like this and for values of T near T equals 0 this line is a decent approximation of the function so we'll need to find the equation for this tangent line which requires a point and a slope and then we'll plug tal 2 into that equation to get an approximation of the temperature of the mil let's do Part B down here like we said we need a point and a slope for an equation of a tangent line the slope we'll figure out we'll put it in that box the point is very straightforward we're given the point as the initial condition at time T equals 0 the temperature of the milk is five so the point is y - 5 Yus the y coordinate equals the slope multiplied by x minus the x coordinate or in this case the T coordinate and that is zero and in fact instead of using Y and X let's be consistent with the context here the Y is M the temperature of the milk at time T and the x is T the time since the milk was placed in the pan all that's left for this equation is to find the slope which is dmdt at the time t equal 0 and this is easy to calculate with our differential equation the differential equation given was 1/4 multiplied by 40 - M however we know that when t equal 0 m is 5 that was given to us so we can replace M here with five thus the slope that we're looking for is 1/4 of 35 or 35 over 4 since we're trying to approximate M the temperature of the milk let's solve this tangent line equation for M that's going to give us M = 35 over 4 t + 5 then to approximate the temperature of the milk at time tal 2 we just have to plug two into this tangent line so that's going to be 35 over 4 ultip 2 and + 5 this is 17.5 + 5 or 22.5 and this of course is in De C and there is our written out conclusion the temperature of the milk at time T = 2 is about 22.5 de C we just found the equation of the tangent line using that initial condition that we had and then plug in tal 2 moving on to part C write an expression for the second derivative of M with respect to T in terms of M and then use this expression to determine whether the approximation from Part B that we just got is an underestimate or an overestimate for the actual value of M of2 right here that was given to us so this second derivative is -1/4 * dmdt but dmdt we know is 1/4 * 40 - M and finally we can write this as 1 over 16 combining the4 and the positive 4th multiplied by 40 - M this is an expression for the second derivative purely in terms of M now based on this second derivative is our approximation from Part B an overestimate or an underestimate well if our function is concave up then the tangent lines are underestimates on the other hand if the function is concave down the tangent lines are over estimates so we need to assess the sign of this second derivative as far as the sign goes there is a negative here so the second derivative is going to be negative unless 40 - M is negative because then the two negatives would cancel out so is 4D minus M negative no it's not we know for sure that 40 minus m is positive because we know for sure that m is less than 40 that was given to us in the question it said right here that it can be shown that M of T is less than 40 for all values of T So for sure this part 40 minus m is positive so for sure the second derivative is negative which means the function is concave down and so our tangent line approximation would have been an over estimate and there that reasoning is written out let's move on then to Part D use separation of variables to find an expression for M of T the particular solution to the differential equation dmdt = 4 * 40 - M with the initial condition we've been working with so far M of 0 equals 5 all right so we will use separation of variable and then we will use this initial condition to get the particular solution and let's see if we can fit Part D over here the differential equation we have is DM DT = a 4 * 40 - M now we need to separate the M's and the t's we'll want the M's over on the left with the DM and then we'll move the DT over to the right to get the M's over to the left side we'll divide both sides by 40 - M thus we'll have 1 over 40 - M D M and then multiply both sides by DT and so we have 1/4 DT on the right side now we can go ahead and integrate the left and right sides integrating 1 over 4dus M gives negative natural log of absolute value 40 minus M and on the right we have 1/4 T let's suppose that all of the arbitrary constants we gathered to the right side and so I'll just put a plus c over there on the right notice this negative came from the fact that 4D minus M inside the log has a derivative of -1 so this negative is there to undo that and we need the absolute value bars because you can't have negatives in the natural log although oh wait we already know that 40 minus M isn't negative it's always positive so in fact we don't need those absolute value bars let's just use parenthesis now let's go ahead and use our initial condition to solve for the constant C we know that when T equals 0 the temperature of the milk m is 5 so on the right side of the equation we have 1/4 * 0 + C and on the left side we have negative natural log of 40 - 5 so on the right we're just going to have C and thus this tells us that C equals natural log of 35 that is the value of C so we can move on with this and find our particular Solution by solving for M this is our equation now that we've replaced C with its known value let's go ahead and multiply both sides by Nega 1 to get rid of this negative in front of the log on the left so we have the natural log of 40 - M multiplying the right side by -1 turns it into natural log of 3 term so - 35 e to the /4 T and that completes our solution to free response question three from the 2023 AP Cal AB exam next up are problems on series for these it's useful to know the general form of the mclen series and the Taylor series it's also useful to keep in mind that if you have a tailor series say for a function like s of X you can find a series for similar functions like s of x^2 by replacing the input X in your tailor series with x^2 good luck this is problem six from the 2001 AP Cal BC Exam so this would have been the final question no graphing calculator allowed let's take a look a function f is defined by f ofx equals this power Series where we can see a general and term given there this equation of course is only true for all X in the interval of convergence of the given power Series so that's the domain and part A is to find this domain this interval of convergence for the power series and of course we must show the work that leads to our answer so let's go ahead and get into part A we're going to use the ratio test this is what the series in question looks like the form of course was given to us and we know that n has to start at zero so that we agree with that first term because there shouldn't be any x's in that first term so we need x to the power of0 for that first term now hopefully you remember the ratio test the ratio test involves a limit statement and we take the limit of this expression the absolute value of the n+ oneth term of the series divided by the nth term of the series and this case that's going to work out pretty well because 3 to the n + 2 I say n + 2 because we're plugging in n + 1 so 3 to the n + 2 and 3 to the n + 1 are going to cancel out fairly nicely and n + 2 and n + 1 will also cancel out as n goes to Infinity we have to take this limit as n goes to infinity and when we do that n plus 2 and n + one are barely different right they would approach one so let's go through the details of evaluating this limit again this is the ratio test and we don't need our series to have any special properties to apply the ratio test we can just jump right into the limit so for the ratio test like I said we're going to take the limit as n goes to Infinity of the absolute value of the n + oneth term divided by the nth term we're going to write that division though as multiplication by a reciprocal let's begin with the n+ one term so where you see an N we are plugging in n + 1 so the numerator n + 1 is actually going to be n + 1 + 1 which is n + 2 again that's because we're replacing n with n + 1 we also have x to the n + 1 up here and then in the denominator we have 3 to the n + 1 but we're plugging in n + 1 for n so that's 3 n + 2 then we divide by a n to do that we'll multiply by the reciprocal so the stuff in the numerator n + 1 and x to the N will be in the denominator we have n + 1 and X the n and the denominator 3 the n + 1 because we're multiplying by the reciprocal that's going to go up to the numerator now we just have to see what needs to be true about X to make this limit less than one the ratio test tells us that the series is going to converge so long as this limit equals something less than one so we need to see what condition does this force on X and in order to figure that out we're just going to have to make some simplifying steps and there is a lot of simplification we can do here I just duplicated the limit so we can start Crossing stuff out 3 the n + 1 and 3 the n + 2 almost completely cancel out it just leaves one factor of three in the denominator n + 2 and n + one those cancel out as well in theory we could split this up into a product of limits and when we have just n + 2 over n +1 on its own well as n goes to Infinity these two things are virtually identical and so they would approach one and thus they're not actually having an effect on the limit the the X's will also almost cancel out X the n + 1 and X the N that's just going to leave a single factor of X in the numerator and so we have simplified this considerably at this point we don't even need the limit statement because we've gotten rid of all of the ends so we need the absolute value of x over3 to be less than one this means that X over3 is less than 1 and greater than - 1 multiply everything by three and we get that -3 is less than x is less than pos3 now the ratio test tells us as long as this limit is less than one and so all of this has to be true then our series will converge but it doesn't tell us anything about the end points so we'll have to check these end points separately we know that our series will converge when X is between -3 and 3 but it's possible it also converges at the end points we'll have to check and here's what happens if we evaluate those end points where X is -3 and where X is positive3 in the -3 case this x to the N of course becomes -3 to the N the effect that has is almost canceling out with the denominator just leaving a single factor of three in the denominator and also giving us a -1 to the N up in the numerator this series clearly diverges because it just gets bigger and bigger while flipping from negative -2 positive in the case where X is positive 3 this x to the N is just 3 to the n and so the same thing happens except without that Nega one and again this series will diverge because the terms just get bigger and bigger the terms don't approach zero in either case so both of these series will diverge by what is called the Divergence test if the terms of the series don't converge to zero the series must diverge and so our final answer for the interval of convergence is from -3 to posi3 neither of the end points are valid points where the series converges and so that is our final answer for part A Part B asks us to find the limit of f ofx that's our power series remember minus A3 all overx as X approaches zero so let's write out this limit statement um X approaching Zer is within our interval of convergence right so this is totally valid and also notice as we go to write this that we are subtracting 13 so this 1/3 at the start of our power series we don't have to worry about that that's just going to get canceled out with this minus 1/3 and we'll just be left with f ofx overx but without the 1/3 so this is what the limit statement looks like in the denominator of course we have X in the numerator we just have our series exclude that 1/3 at the beginning that canceled out with the minus 3r so we just have 2 3^ 2 x + 3 3 Cub x^2 and so on in each term the power of X and three both increase by one and this numerator increases by one so what's going to happen is one factor of X in every term will cancel out with the X in the denominator so let's write that step so now we're looking at the limit as X approaches 0 of 2 over 3^ 2ar that factor of X got canceled out plus 3 over 3 cubed x it did have an X squar but one of the X's got canceled out and so on every term of our series just has one less power of X so since X is approaching zero in this limit statement all of these terms that have an x in them are going to go away which just leaves this first term without the X which which is 2 over 3^ 2 and so this limit is equal to 2 over 9 all right I just shrank our work and moved it out of the way so we have a little bit more room for part C which asks us to write the first three nonzero terms and the general term so that's just like they did here right they wrote the first three nonzero terms of the series and then a term representing a general form that's what we're supposed to do WR the first three nonzero terms and the general term for an infinite series that represents the integral of this function f ofx from 0 to 1 so let's go ahead and set this up I'm going to write the integral from 0 to 1 and then I'll put our power Series in here so here's what it looks like the integral of our power series now we can go ahead and integrate each individual term and we'll be evaluating that from 0 to 1 the first term 1/3 integrates to A3 X the second term well we have to increase the power of X by 1 so we would increase it from 1 to two and then divide by that new power of two which is going to cancel out with that numerator thus it's going to be 1/ 3^ 2 x ^ 2 in the third term the power of 2 will again go up to three and when we divide by that new power we're going to cancel out the numerator and so this is going to be 1/ 3 Cub * x ^ 3 then we can go ahead and integrate the general term in the general case you can see increasing the exponent to n + 1 and then dividing by that new exponent will cancel out the numerator so what does an integral of the general term look like well it has one over 3 n * X the n + 1 and this continues we are evaluating this of course from 0 to 1 when we plug in zero all of these terms have an x in them so they will all go away when we plug in zero thus we just have to plug in xal 1 I was just writing out what we get when we plug in xal 1 I'm sorry though I just noticed an error here where I've written 3 to the N we should have had 3 to the n + 1 so I'll just go ahead and correct that now that doesn't really change anything too major it's just that the nth term of the series has a 3 the n + 1 so here where I'm representing the nth term of our series we should have 3 the n + 1 in the denominator all right so that is the series representing this integral of our series from 0 to one and we've written it by writing three nonzero terms and then we've included a general term as well which is exactly what the question asked for and so we are done with part C and where unlock Part D is very easy we just have to find the sum of the series that we just found in part C and you probably recognize that the sum we found this thing uh is a geometric Series so to find the sum of the series will be a very easy task the first term is 1/3 let's just call this well we don't have to call it s we could just represent it as the integral from 0 to one because that's what it is the integral from 0 to one of our function we found is equal to this series which we can evaluate using the geometric series formula that's going to be the first term which was a thir divided 1 minus the common ratio the common ratio of course is a thir each term is 1/3 of the preceding term 1/3 1/3 squar 1/3 cubed and so on so 1 Min - A3 and what is this well 1 is the same as 3/3 so 3/3 - 1/3 in the denominator that's 2/3 so this is 1/3 / 2/3 and that of course is 1/2 those threes cancel out and that is our answer and that completes our solution to free response question 6 from the 2001 apal BC exam this is free response question 6 from the 2002 apal BC Form B exam the mclen series for the natural log of 1 1 - x is this sum with an interval of convergence from NE - 1 to positive 1 part A asks us to find the mclen series for the natural log of 1 / 1 + 3x and determine the interval of convergence now this expression the natural log of this is super similar to this expression for which we already have a mclen series so all we have to do is take this series and replace the X with whatever we would need to replace the X with in this expression to produce the desired expression so for part A what we want is 1 + 3x in the denominator what we were given was 1 - x in the denominator so we just need to replace x with Min - 3x if we plug - 3x in for X we're going to get plus 3x because of the double negative right that's going to give us what we want so in the known muren series that we were given we will just replace x with -3x so the natural log of 1 over 1 + 3x is equal to this mclen series so that's the sum from n = 1 to Infinity of -3x to the N because we're replacing that x with -3x -3x to the n ided by n again all we did was replace X in the given series with what we need to replace x with in the expression in order to produce the expression that we want now where does this mclen series converge well we know that the expression we used has its mclen series convergent when the input is between negative 1 and positive 1 so our input which is -3x needs to be between -1 and positive 1 and make sure you pay attention to the strictness of the inequality our input -3x needs to be greater than or equal to1 and less than one so the interval of convergence we just need to solve this for X so divide everything by -3 that then produces X is greater than -13 and less than or equal to positive 1/3 and that is our interval of convergence we just took the interval of convergence of the original input changed the input to our new input which is -3x and then solve for x moving on then to Part B Part B asks us to find the value of this infinite series and this is just the given infinite series notice but with -1 plugged in for X is -1 in the interval of convergence of the original series uh yes it is X is greater than or equal to1 so this series in Part B is just the original given series with an input of x = -1 which means it equals the natural log of 1 / 1 - x where X is -1 so for Part B we are evaluating this series from n = 1 to Infinity of -1 to the N / n and we know from the given information that's just this natural log expression but with x = -1 plugged in so the natural log of 1 / 1 -1 and 1 --1 is just two so this is the natural log of 1/ 2 all right moving on to part C give a value of P such that the infinite series as shown here converges but this one where the power of n in the denominator has been doubled diverges and we need to give reasons why our value of p is is correct now note this series is just a p series but instead of P the exponent is 2p now A P Series in general will diverge when this exponent of n is less than or equal to 1 so we would just need P to be less than or equal to 1 12 that way 2 p is less than or equal to 1 and in fact P = 1/2 is a perfectly good answer plugging P = 12 into this other series which needs to converge yes indeed it will converge because it's an alternating series where the terms converge to zero n to the 1/2 is just a square root of n which certainly diverges to Infinity meanwhile the numerator is just alternating between plus one and minus one so yes by the alternating series test this series would converge for P = 1/2 so there's our conclusion for part C for p equal 1/2 this series converges by the alternating series test but this series is the harmonic series and thus diverges right this is the series of 1/ n 2 p if P equals 1 12 then 2 p is just one and so this is actually just the harmonic series which is known to diverge for the alternating series test I should point out it's important that the absolute value of the terms are decreasing which is the case because again the numerator has an absolute value of one the whole time where the denominator is just diverging to Infinity because it's root n in the case that P is 1/2 Part D asks us to give a value of P such that the series 1 over n the P diverges but 1 over n the 2 p converges unlike last time where 1 over n the 2 p had to diverge and of course we need to give reasons why our value of p is correct Now 1 over n the 2p would diverge when the exponent of n is between zero and one inclusive to converge on the other hand we need the exponent of n to be greater than one so that the terms get small enough fast enough for the series to converge this is just a p series we're talking about now if we need 2 p to be greater than one p just needs to be greater than 1/2 we could pick a simple fraction like 3/4s for example choosing P equal 34s will also make this series diverge as necessary because in this case the exponent of n would be between 0 and 1 P 34s but over here the exponent of n would be greater than 1 because 2 * 34s is 64s so here's our answer for p = 3 over 4 the series 1/ N the P diverges because it's a p series with P between 0 and 1 but the series 1/ N the 2 p which is equal to 1/ n 1.5 converges because it's a p series with p greater than one and that completes our solution to free response question six from the 2002 apal BC Form B exam this is free response question six from the 2016 AP Cal BC Exam part of the no calculator section the function f has a tailor series about xal 1 that converges to F ofx for all X in the interval of convergence it is known that F of 1 = 1 frime of 1al -2 and the nth der derivative of F at xal 1 is given by this equation here for n greater than or equal to two so we have all the information we need to write the tailor series part A asks us to write the first four nonzero terms and the general term of the tailor series for f about xals 1 so we'll put part A down here in order to answer this question naturally you have to know what a tailor series looks like in general in general a tailor series will look like this the nth derivative of the function f evaluated at the center a in our case the center is 1 multiplied by x minus the center to the power of n and then divided by n factorial so we want to keep that in mind as we begin to write the first four nonzero terms of the tailor series for our function f ofx the first term of the tailor series is with n equals 0 so the zeroth derivative of the function which is just the function evaluated at the center so that's just the function evaluated at one which is given to us as one so we're going to have a one there and then everything else when nals 0 everything else is just one x - a 0 is 1 and 0 factorial in the denominator is one so that's the first term of the tailor series we can move on to the next term when n equals 1 then we have the first derivative F Prime evaluated at the center which is 1 and we're told that frime of 1 is -2 so we're going to have minus a half and then we have x minus the center of 1 to the power of one divided by 1 factorial we don't have to divide by one factorial because that's just a division by one so we'll just write the x -1 part which is in the numerator x -1 to the^ of 1 I'm not going to bother writing that power though then let's move on to the next term term three where n equals 2 for this term we're going to have to consult the somewhat complicated equation it gave us for evaluating the nth derivative of the function at one let's worry about the simpler Parts first before we do that we're going to have x -1 to the power of 2 because this is Nal 2 the third term and then divide by n factorial again in this case that's 2 factorial all right then all of this is going to have to be multiplied by the second derivative of the function evaluated at one and we are told how to find that we're told up here that to evaluate this we need -1 the^ of 2 multiplied 2 - 1 factorial divided 2 to the 2 now -1 to the 2 is just positive 1 so we don't have to worry about that 2 - 1 factorial is just 1 factorial so we don't have to worry about that either we just have to worry about the two to the two in the denominator so coming back to this expression we actually have more space in the numerator than we need but in the denominator we have to add a 2^ squared that's part of multiplying by the second derivative of the function evaluated at one and there we go those are the first three nonzero terms we need to write one more nonzero term and then we can write the general term so now we're doing this again but with Nal 3 so let's again focus on the easy piece is first x -1 ^ 3 and divide 3 factorial and now let's consult the equation for the derivative again we're looking for the third derivative of the function evaluated at one which is going to be -1 3 which is just - 1 multiplied by 3 - 1 factorial which is 2 factorial ID 2 3 so divide by 2 3 all right so we just have to add all that stuff in here via multiplication the Plus becomes a minus because we have this negative factor and then in the numerator we have 2 factorial and in the denominator we have 2 cubed and that is going to be the fourth nonzero term notice however in this term the factorials cancel out a bit 2 factorial / 3 factorial if you think about what that is everything cancels out except the three in the denominator so we might as well rewrite that in that slightly simplified form 2 factorial is going to be canceled out by the three factorial in the denominator and all that's left behind is that factor of three with that done we can go ahead and move on to writing the general term of this tailor series the general term is going to look just like that so we're going to have the nth derivative of the function evaluated at the center which is 1 multiplied by xus the center to the power of n and then all of this is divided by n factorial now of course we know what the nth derivative of the function is evaluated at one it's given to us here so let's replace this FN of one with the given expression FN of 1 we know is -1 to the n and then multiply by nus1 factorial so we're going to have to make a little bit more room up here in the numerator nus1 factorial and then we also have to divide by 2 the N so we'll also throw a 2 to the N down there in the denominator so that's what a general term of this series looks like but just like in the previous term this one here we have factorial cancellation n minus1 factorial divided n factorial those are going to cancel out completely except for a factor of n in the denominator so we might as well go ahead and simplify this a little bit those factorials are just going to leave n behind in the denom minator and so the general terms of our series will look like that you want to make sure to also include the plus dot dot dot at the end to indicate it's a series it goes on forever let's move on now to Part B Part B says the tailor series for f about xal 1 has a radius of convergence of two we want to find the interval of convergence and show the work that leads to our answer so for Part B we already know this series is centered at x = 1 and we know the radius of convergence is 2 so we could go up to to get to three and we could go back to to get to1 the question is do the end points also result in a convergent series we know everything in here everything between these bounds is going to be convergent if we plug those X values into the series but what about the end points that's what we actually have to figure out for this question does the interval of convergence look like this x is less than or equal to 3 and greater than or equal to1 or maybe X can't equal ne1 it has to be strictly greater than negative 1 or any other combination of inequalities that's what we're asking does the series converge for these endpoints the information about the interval is pretty much given to us right we know the center is is one we know the radius is two that's how we get -1 and three the only lingering question is does the series converge at the end points or not to figure that out let's first consider the lower end point if we plug -1 into our series what's going to happen well let's look at a general term of our series to get a sense of what that's going to look like x -1 if x is -1 then x -1 is -2 and so this is actually -2 to the power of n then the negative with that two would cancel out with this Nega 1 so in fact there would be no negatives what we would have left would be 2 the N / 2 the n * n oh so the 2 the n's cancel out and this is actually just the harmonic series 1/n we just have to be a little bit careful because seeing that this actually is the harmonic series when xal -1 that was using this general term which relied on this derivative formula the derivative formula that applies only when n is at least 2 so what happened for Nal 0 and N = 1 well looking at these terms the first term obviously is one the second term if x is NE - 1 the second term is just -2 * -2 which is also just positive1 so what our series would look like for f of1 is 1 + 1 and then at Nal 2 is when the harmonic series kicks in 1 / 2 + 1 3 and so on again that's because the harmonic series that we found resulted from considering this derivative formula which only applies when n is at least two we were given specific values rather than a formula for the other uh possibilities so for xal -1 this is what our series looks like clearly it's just the harmonic series with one added at the beginning so yeah this is going to diverge it's not actually in the interval of convergence so we can start to write our interval X definitely has to be greater than1 when x equals to -1 we basically just get the harmonic series which diverges all right to finish this up let's consider what happens if we plug in the upper bound what is f of3 again let's come over to the general term of our series in order to figure out what this would look like if we're plugging in x = pos3 then this guy x -1 is going to be 3 - 1 so pos2 so again we're going to have 2 the N / 2 the N the 2 to the n's will cancel out it's just that in this case there was nothing to cancel out the ne1 so we're still going to have that alternating Behavior resulting from the Nega 1 so this is actually just going to be like an alternating harmonic series with the exception again that it starts with those two ones except in this case the second one is negative because 3 - 1 is 2 * half is1 so 1 - 1 plus a half minus A3 and so on it's just an alternating harmonic Series this converges the alternating harmonic series converges and having a plus one at the beginning does not change convergence so the interval of convergence does not include NE -1 but it does include posi3 so that's the interval let's move on then to part C the tailor series for f about xal 1 can be used to represent F of 1.2 as an alternating series use the first three nonzero terms so these three terms to approximate F of 1.2 so you can approximate F of 1.2 by taking as many terms of the series as you like the more terms you take the better approximation you get we're going to use the first three terms because that's what the question asked us to use so what is f of 1.2 well based on our series it's about these three terms which I'll just go ahead and copy and paste but we're going to plug in 1.2 for X so I'll bring these right down here change it so it's color matches and let's just consider plugging in 1.2 for x 1.2 minus 1 that's just going to be 0. 2 so right there we get. 2 1.2 minus 1 that's just going to be. 2 all right so here we go let's just finish with this calculation 12 * 2 is 0.1 so this is 1 minus 0.1 which is .9 as for this fraction let's see what do we have 2^ squared that's going to cancel out with the 2^ squar in the denominator uh it's just going to also leave a factor of 110 squared which is 100 in the denominator there's also a two factorial in the denominator which is just 2 so this is9 + 1/200 just like 1 over 20 is .5 1 over 200 is 0.5 so this sum is 905 that is our approximation which finishes part C so moving on to Part D show that the approximation found in part C this approximation of 095 is within 0.001 or 1,000th of the exact value of f of 1.2 in this case we're dealing with an alternating series and there is a nice way to estimate the error on an alternating series approximation and here is that application of the theorem for the remainder of an alternating series approximation the series for f of 1.2 we looked at the first three terms of that Series in part C but the series the one that goes on forever alternates with terms that decrease in magnitude to zero we know that because this is our series we're thinking about the case where x equal 1.2 so clearly the terms will decrease in magnitude to zero because the numerator is going to be 1.2 minus 1 the N so 0 2 to the N that's just going to get really small and it's getting divided by these massive numbers as n gets larger so yeah it's decreasing in magnitude to zero and the terms are alternating because of the 1 the N thus we can say confidently that the error given any approximation the error is going to be at most the magnitude of the first excluded term in the approximation so in this case when you have an alternating series where the terms are decreasing in magnitude to zero if you approximate something with say n terms the error in that approximation is at most the magnitude of the n+ oneth term of the series for our approximation we use the first three terms so the error is at most the magnitude of the fourth term which looks like this when n equals 3 right because we use the N equals Zer term the N equal 1 term the N equals 2 term the first one we excluded is the N equals um the fingers are weird the first one we excluded is the nals 3 term that's the fourth term of the series that's that's where this is coming from the magnitude of the fourth term looks like this and that is an upper bound on our error now what does this thing actually equal well let's see the negative one just goes away because this is inside of magnitude bars the 2 cubed will just cancel out with the 2 cubed in the denominator but there will also be a 1110th cubed left behind in the denominator so that's 1/ 1,000th and then we also have that factor of three so 1 over 1,000 * 3 we have this in absolute value bars but the absolute value bars don't matter now because everything is positive now remember we were actually asked to show that the approximation from part C is within 1,000th of the exact value of f of 1.2 we've just shown that hey the error has this as an upper bound this is equal to this which is clearly less than one 1000 this is certainly less than 0.001 because this is 1 over a th000 this is 1 over 3,000 so yes by applying the alternating series remainder theorem we found that the approximation from part C is within a thousandth of the exact value of f of 1.2 so pretty darn good approximation using only three terms and that completes our solution to free response question six from the AP Cal BC Exam from 200 16 this is free response question 6 from the 2022 AP Cal BC Exam here is the mark of the beast the mark of the BC Exam an infinite series problem the function f is defined by the power series f ofx equals and you can read this for yourself for all real numbers X for which the series converges part A using the ratio test a very useful test indeed find the interval of convergence of the power series for f and justify our answer all right so let's set up the ratio test what the ratio test has us do is look at the ratio of a term of the series say a n + 1 to the preceding term which would then be a n we look at the absolute value of this ratio the ratio of a term of the series to the preceding term we look at the limit of this as n goes to in infinity and if this ratio is approaching something less than one then the series converges so for part A here's what that looks like setting up the ratio test for our series notice in the numerator we've plugged n+ one into the general form of a term of the series so for example when we plug n +1 in right there that becomes 2 * n + 1 which is 2 2 n + 2 but there's also A+ one over there so that's why we have this 2 n + 3 right there so this whole thing in the numerator is the n + oneth term of the series and then we're dividing by the nth term so here in the denominator what you see looks just like this that's the nth term of the series this of course is all in an absolute value so we will be able to discard those negatives and we're taking the limit of this as n goes to Infinity the idea is that we'll try to simplify and evaluate this limit and whatever ends up equaling we need that thing to be less than one which will typically Place some restriction on X this turns out to be fairly easy to evaluate the ratio test often does work pretty nicely no major difficulties this first step is just discarding those negatives like I said before we have these absolute value bars those negatives then aren't going to do anything then let's simplify the X's X x 2 n + 3 / x 2 n + 1 that's just x^2 most of those factors of X will cancel out but there are two more factors of X in the numerator than there are in the denominator so we're left with an X squ as far as the n's go this division by 2N + 1 in the denominator of a fraction gets sent up to the numerator it's like multiplying by the reciprocal right and then this guy 2 n + 3 that stays in the denominator so we have 2 n + 1 / 2 n + 3 now as n goes to Infinity 2 n + 1 and 2 n + 3 behave pretty much the same their ratio approaches one the addition of a constant at the end doesn't really do anything because as n gets really really big the primary character here is the 2 N so these will have a ratio approaching one and so the limit of this is just the absolute value value of X2 that is the limit of this thing and we need this to be less than one in order for the series to converge then the absolute value of X2 which is this limit being less than one that implies that the absolute value of x is less than one thus the series converges when X is between negative 1 and positive 1 because that's when it has an absolute value less than one unfortunately we do also have to separately consider the case where the absolute value of x equals 1 in that case when this limit equals one the ratio test is not conclusive so although we know it does converge in this interval we do have to check the end points because right now we cannot say anything definitively about the series Behavior at the end points so let's check the end points we'll start with x = -1 and see how the series behaves if we plug in x = -1 the first term of this power series is going to be -1 and then we'll have plus A3 and then minus a fifth and then plus a 7th and so on it looks like this and this is an alternating series which we can quickly see converges by the alternating series test because it's an alternating series right it goes negative positive negative positive and so on and the absolute values of it terms are decreasing to zero the absolute values of the terms are just reciprocals of odd numbers 1 a thir a fifth a seventh and so on so the absolute values are approaching zero and they're decreasing thus by the absolute Sorry by the alternating series test this series converges so the interval of convergence for our series includes -1 now let's check xal pos1 turns out it's going to be practically the same if we go back up to our power series if x = POS 1 then we're going to have 1 - A 3r plus a 5th - a 7th so same idea when x equal pos1 the series looks like this which again is an alternating series with terms whose absolute values decrease to zero thus by the alternating series test the a this series will converge so it actually converges at both of these end points both Nega -1 and and positive 1 so we can state with confidence our final interval of convergence the interval of convergence is X less than or equal to 1 and greater than or equal to positive 1 all right that completes part A which took up quite a bit of space so we will have to shrink this to give us some room for part B Part B is to show that the absolute value of f of 12us a half is less than 1110th let's do Part B over here we know certainly that F of 1/2 is a convergent series because in part A we showed that the interval of convergence is everything from negative 1 to positive 1 so certainly it's convergent to xal 1/2 it's also an alternating series because we could just see from the definition it is alternating thus we can use the alternating series error bound notice that the first partial sum of the series which is just the first term in this case is 1/2 if we plug in 1/2 that first term which is the first partial sum will be 1/2 so the alternating series error bound tells us that the distance between 1/2 the first partial sum and the actual value that the series converges to the distance between them which is what we're trying to prove something about for Part B that distance has to be less than the absolute value of the next term of the series once again what the alternating series error bound tells us in general is that the distance between the true value of an alternating series and its nth partial sum is less than the absolute value of the N plus oneth term if you add all of the terms up to the nth term that has to be within the absolute value of the next term of the True Value that the series converges to lot of words there hopefully this makes sense check out a video introducing the alternating series error bound if you need a review of that but as far as this goes let's see what is the second term of the series we know the distance we're interested in is less than the absolute value of the second term of the series so what is that well if we're plugging in x = 12 the second term of the series is 12 cubed divided by 3 and of course this is in an absolute value so 12 cubed ided by 3 this of course is 1/8 / 3 which is 1 over 24 the absolute value is not going to do anything to it at this point that's 1 over 24 and remember we were trying to show that this distance is less than 1110th we've shown that it's less than 1 24 which is indeed less than 1/10th since this denominator is bigger than this one thus we have completed Part B by the alternating series error bound the distance between F of2 and2 has to be less than10 moving on to part C write the first four nonzero terms and the general term so just like this the first for nonzero terms and a general term for an infinite series that represents the derivative fime of x all right let's do part C down here for convenience I have Rewritten the power Series F ofx we're just going to have to use the power rule to take the derivative of this guy all right so what is f Prime of X the first term would be 1 then we're going to have Min - x^2 because the exponent of three comes down as a factor cancels out the denominator and then the power gets reduced by one so we're just left with an x^ squar the next term will have a derivative of x 4 plus x 4 the next one is - x to the 6 and so on you can see how this works we're getting alternating signs with the even powers of X the general term is going to be -1 to the N that checks out because when n equals 0 this would just be 1 and we do need that first term to be positive so there should be no negative factor when n equals z so we know that works fine and then the power rule applying to the rest of this is just going to give us X 2N of course we've just got an even power there and that is it that is fime of X we've got the first four non-zero terms and a general term this is just using the power rule which we can see behaves very nicely with this function finally we can move on to Part D which says use the result from part C to find the value of f Prime of6 so what is the value of f Prime when x = a 6 well let's plug it in frime of 1 16 based on our previous work is going to be 1 - 6^ 2ar + 16 to the power of 4 plus or excuse me minus 16 to the power of 6 and so on this hopefully you notice is a geometric series it's geometric where each term has a ratio of neg 16 squared with the previous term notice if we multiply 1 by 1 6^ squared let me rewrite that 6 we get here and then if we multiply this by 6^ sared we get here right you've got it positive now because of the two negatives and 16 to the 4 because square and the square and this continues each term has a ratio of negative 16 SAR with the preceding term clearly then this is geometric and again that that ratio is 6^ squared and the absolute value of this is less than 1 which means that it will converge this is a geometric series that converges now what does it converge to well hopefully you remember your geometric series formula frime of 16 it's a geometric series it's going to converge to this the first term which is 1 iD 1 minus the common ratio the common ratio is 16 sared this of course is the same as 1 over 36 over 36 that's because 1 is just 36 over 36 andus 6^ 2ar is - 1 36 but since we're subtracting it it's like we're adding it so it's plus 1 over 36 so this is actually equal to 1 over 37 over 36 and dividing by 37 over 36 is the same as multiplying by the reciprocal so our final answer is 36 over 37 and there you go that's a pretty fun problem isn't it that completes our solution to free response question six from the 2022 AP Cal BC Exam next up are the polar coordinate problems recall that if you're trying to find the area of a polar region you can't just integrate the polar curve this comes from the area of a sector of a circle which you actually have to do is 12 times the integral of your R function squared it's like 1 12 radius squared over whatever your relevant interval is which will be in terms of theta have fun this is problem four from the 2009 AP Cal BC Form B exam so this was part of the no calculator section and this is on Polar curves something which is not in the AB exam the graph of the Polar curve R equals 1 - 2 cosine Theta for Theta between 0 and Pi is shown above we see these beautiful curve here let s be the Shaded region in the third quadrant bounded by the curve and the axis so we can see the region s there part A is to write an integral expression for the area of this shaded region s so we just have to write the expression we don't have to evaluate it this is a pretty straightforward question you just got to remember how to find the area of a polar region it of course is going to be an integral and we have a 1/2 out front and then the radius or r squared in this case r is given to us as 1 - 2 cosine Theta and we need to square this and the integral is taken with respect to Theta remember this 1/2 and the radius squared comes from the fact that when we figured out how to do this originally it comes from the area of a sector of a circle just like traditional integrals are built from rectangles polar integrals are built from sectors of circles and that's why we have 1/ 12 integral r squared that's how we're going to calculate the area of this region the question is what are the bounds the lower bound and the upper bound well we have to think let's look at this picture we see that the region s starts where r equal -1 so we need to figure out what Theta value gives us an r of-1 that will be the lower bound of our integral now the curve or the region I should say is closed the region is closed when the curve comes back to this point the origin that's where R equals 0 so we'll also have to figure out where does Theta make r equal to0 and that's going to be the upper bound of the integral let's begin with -1 so we'll write 1 - 2 cosine Theta that's our R is equal to -1 subtract one from both sides and we find that -2 cosine Theta = -2 divide both sides by -2 and we find that cine Theta = pos1 and thus Theta equal 0 cosine of 0 is 1 so Theta is 0 that's the lower bound of our integral and we'll do a similar thing for Ral 0 we'll say that 1 - 2 cosine Theta equal 0 remember we're trying to find the upper bound of our integral now and uh let's add 2 cosine Theta to both sides so 1 = 2 cosine Theta and then divide both sides by 2 so 12 equal cine Theta and if you remember your unit circle which you should this means Theta has to equal p over 3 cosine takes on a value of 1/2 at the angle pi over 3 so that's the upper bound of our integral and that actually completes the answer to part A that is our integral expression remember this all comes from the standard way that we calculate the areas of polar regions which is integrating from A to B 1 12 out front r^ 2 D Theta this is how you find the area of a polar region and that's all we did for part A all right moving on to Part B write expressions for DX D Theta and Dy D Theta in terms of theta okay so you'll have to remember some formulas for this we know that X is equal to R cine Theta and Y is equal to R sin Theta these are relationships between cartisian coordinates and polar coordinates that you should know now in order to find DX D Theta we're going to have to take the derivative of this equation which is going to involve a Dr D Theta because we're going to have to take the derivative of R so let's just go ahead and find that first let's find Dr D Theta we know that R is 1 - 2 cine Theta and if we take the derivative of that we're just going to get 2 sin Theta so that's what Dr D Theta is all right now we can go ahead and take the derivative of x and the derivative of y let's start with X taking the derivative of this equation both sides we're taking the derivative with respect to Theta on the left we have DX D Theta on the right we have to use the product rule because we have a product of functions R which changes with respect to Theta and cosine Theta which of course is also changing so the product rule is frime G Plus G Prime F in this context that's Dr D Theta cosine theta plus the derivative of cosine Theta which is Nega sin so let's just write minus sin Theta * R Dr D Theta of course we know is 2 sin Theta so this is just 2 sin Theta cosine Theta and then then we are subtracting sin Theta * R but of course R is just 1 - 2 cine Theta so we have - sin Theta * R which is 1 - 2 cosine Theta now notice on the left we have 2 sin Theta cosine Theta if we were to distribute over here on the right we'd have minus sin Theta * - 2 cosine Theta so that would be another 2 sin Theta cosine Theta in total then we have 4 sin Theta cosine Theta the other term we would have is minus sin Theta * 1 which is just minus sin Theta and that is DX D Theta next we'll do a similar procedure in order to find Dy D Theta so we're going to take the derivative on the left and right side of this equation with respect to Theta I'll write that here Dy D Theta what is this going to be well again it's the product rule fime G so Dr D Theta * sin theta plus G Prime F so cine Theta * R Dr D Theta of course we know is 2 sin Theta so Dr D Theta * sin Theta is 2 sin Theta * sin Theta so 2 sin s Theta then we add cosine Theta R but of course we know that R is 1 - 2 cosine Theta so that is our expression for Dy D Theta and thus we have completed Part B which was just to write expressions for DX D Theta and Dy D Theta in terms of theta so we didn't want any RS in there and indeed these are those final answers moving on to part C part C asks us to write an equation in terms of X and Y for the line tangent to the graph of the polar curve this one here of course at the point where Theta equal PK / 2 and we want to show the computations that lead to our answer now to write the equation of a tangent line we're going to need a point and a slope a point is a pair of X and Y coordinates we know that xal R cosine Theta and we know that y equals R sin Theta so since we're looking for the tangent line at Theta = < / 2 we can figure out what the X and Y coordinates have to be when Theta equal piun / 2 x is R * cosine of piun / 2 cosine of piun / 2 though is 0 so X would be R * 0 which is just 0er as for y y is R * sin of < / 2 s of < / 2 is 1 so so y would just equal R but we know R is 1 - 2 cine Theta but Theta is pi 2 so it's 1 - 2 cosine piun / 2 cosine pi/ 2 is 0 so Y is just 1 thus the point on the curve that our tangent line is passing through at pi/ 2 is the point 01 so we can start to write the equation of the tangent line now y - 1 that y coordinate equals the slope which we'll figure out multiplied xus the x coordinate 0 now I shrunk our work in Part B to make a little bit more room to find the slope we need to look at Dy DX because we are trying to write this equation in terms of Y and X so what is dydx well to find it we need to take Dy D Theta and divide it by DX D Theta and let's just write this whole time what we're actually doing is evaluating this dydx when Theta = < / 2 so I'll put this vertical line that means we're evaluating the expression at Theta = Pi / 2 to evaluate this let's first focus on the numerator Dy D Theta when Theta is Pi / 2 looking at our previous work this is dy D Theta so what is it when Theta is pi/ 2 well cosine of piun / 2 is just zero so this term would disappear s of piun / 2 is 1 so sin 2 piun / 2 is 1 and so 2 sin^2 piun / 2 is 2 okay so the numerator then is 2 that's Dy D Theta evaluated at Theta = piun / 2 as for the denominator DX D Theta when Theta is Pi / 2 let's look at our previous work right here again cosine of piun / 2 is 0o so this term disappears and all that's left is negative s of piun / 2 s of < / 2 is 1 so that would be -1 and thus the slope is -2 and we can plug that into this box I'll just erase the Box we know what the slope is now it's -2 and that is our tangent line or if you prefer to put it in slope intercept form you could say this is just -2X because a zero doesn't matter and then add one to both sides and you'd find that Y = -2x + 1 and that completes our solution to part C this is free response question two from the 2013 AP Cal BC Exam it's part of the graphing calculator section we will use a calculator throughout this problem the graphs of the Polar curves R = 3 which is a circle and R = 4 - 2 sin Theta are shown in the figure above the curves intersect when Theta equal piun / 6 which we can see right here I'll just go ahead and label that and when theta equals 5i / 6 which is right over here part A asks us to consider this shaded region that's inside both the circle and inside R = 4 - 2 sin Theta so this shaded region here we would like to find the area of s looking at this area we can split it up into two things part of it is just the area under the curve 4 - 2 sin Theta between Theta = < 6 and Theta = 5 piun / 6 we can simply calculate the area that is under the curve using our typical polar integration methods the 1 12 r squared and what that leaves that we still have to calculate is this area that is present in the circle now we know the area of the circle is just p piun r s and since this has a radius of three that would be 9 pi but this is not all of the area of the circle so we need to figure out what fraction of the circle is it well in total a fraction has 2 pi radians how many of the radians have we cut out this part here is what we have cut out well you've got pi/ 6 here here and 5 pi over 6 there so 2 pi 3 Pi 4 Pi 5 Pi / 6 so in total we cut out 1 2 3 4 pi/ 6 4 pi over 6 out of the total 2 pi radians have been cut out so we need to multiply by the proportion of the circle that we actually have the total radian count is 2 pi but if we are removing 4 pi/ 6 of those radians what remains is 8 piun / 6 again this is the area of the circle so we'll go ahead and simplify this to figure out what that area is and then we'll just need an integral to find this area that's underneath the more complicated curve simplifying this we have pies canceling out in the top and bottom so what's left is 9 piun multiplied by 8 over2 now 8 / 12 is the same as 2/3 so this is 9 * 2/3 multiplied by pi then we can cancel a three out of the nine and the denominator thus giving us a final answer of 6 Pi that is the area here in our Circle so for part A we have 6 pi and then the area of our region which we did not already account for we will find by just doing an integral if we imagine a ray opening up along the polar plane you can see that our region is bounded by the circle from here to here which is why we calculated that part just using the circle but then from here to here our region is bounded by 4 - 2 sin Theta which is why we now have to focus on this function for this part of the region to calculate the area of this polar region we have to do 12 multiplied by the integral from < / 6 to 5 piun / 6 of our function squared it's like radius squared so 4 - 2 sin Theta squared D Theta and we'll bust out the calculator for this on the calculator I'll press math and then option nine for the integration function we are integrating from pi/ 6 to 5 pi/ 6 and what we are integrating is 4 minus 2 sin Theta but I'm just going to use sin x because we're on the calculator and this needs to get squared and we're integrating with respect to X notice my careful use of parentheses as we enter this that's the integral don't forget about the 1/2 out front so now on the calculator we'll divide by two and then add the 6 Pi so plus 6 pi so our final answer is about 24787 that's the area moving on then to Part B a particle moves along the polar curve R = 4 - 2 sin Theta so that at time T seconds Theta equal T ^2 find the time T in the interval between 1 and 2 for which the x coordinate of the particles position is -1 so we need to think about cartisian coordinates X we know is equal to R cine Theta now in this context R is 4 - 2 sin Theta so this is 4 - 2 sin Theta multipli cosine Theta but remember our particle is moving along the curve in such a way that theta equals t^2 so we can replace these thetas with t^ SARS so we have 4 - 2 sin of t^ 2 and multiplied by cine of t^2 now the question is what time T in the interval between 1 and two is the x coordinate which is described by this expression is the x coordinate -1 all we have to do is go on a calculator graph y = -1 and graph this function and we will see the intersection so on the calculator like I said we're using Y so instead of writing X as a function of t on the calculator we're writing Y is a function of X but all the numbers are going to work out the same so on the calculator I've gone to graph the X function 4 - 2 sin of x^2 * cosine of x^2 is what it looks like on the calculator and yal-1 and I want to see where these things intersect so I press the graph button to go to My Graph and I can just zoom in and see where these things intersect of course how you find the intersection will depend a bit on your calculator you might have to use a trace function but we can see that this happens when X is about 1. 1428 now in the context of our problem X is actually t on the calculator so T is 1. 1428 we can see that there are other intersections as well but they happen outside of the requested interval which was between 1 and 2 so the answer we're looking for is T = 1. 1428 so I'll write that this the X position equals -1 when T equals about 1 428 finally part C for the particle described in Part B the one that moves such that Theta equal t^2 find the position Vector in terms of T we already have a description of the X position I in terms of T so we'll just have to do the same thing with Y and then find the velocity Vector at time tal 1.5 and so for that we'll just need a derivative all right let's start putting part C down here we know that just like xal R cosine Theta y = r sin Theta but again we're talking about the same particle that's moving along R = 4 - 2 sin Theta so R we will replace with 4 - 2 sin Theta but remember Theta because we're still talking about the same particle Theta is just t^2 so sine of t^2 and then s of theta is s of t^2 remember this s of t^2 comes from R and this s of t^2 comes from sin Theta which is part of the Y equation now part C wanted us to give the position Vector so the position Vector is this x component and this y component so I'll write this with angle brackets to denote that it is a vector and we can just do a little bit of copy and paste the X component is up here in Part B so I'll bring that down here and change its color to green and the Y component we just figured out is right here 4 - 2 sin of T ^ 2 * sin of T ^2 and just put a comma between these and close the vector that is the position Vector now the other part part of the problem asks us for the velocity Vector at time T = 1.5 velocity is the derivative of position so all we have to do is use our calculator to evaluate the derivative of both the X and the Y components at T = 1.5 so I will write V of 1.5 and we'll see what this equals I will do the derivative of the X component on the calculator and show you and then I'll just do the derivative of the Y component by myself so we don't spend all day looking at the calculator so on the calculator I'll press math and then option 8 for the derivative function instead of using t on the calculator we're going to use x so take the derivative with respect to X and now what we're differentiating is 4 - 2 * sin of x^2 and this is getting multiplied by cosine of x s and we are evaluating this derivative at 1.5 let's just double check that that is at yes 1.5 and we get about 8.72 and now I'll do the same thing for the Y component and the calculator gives a derivative of about - 1.67 2 for the Y component at 1.5 and that is our velocity Vector so the particle is both moving to the left and moving moving downwards at tal 1.5 and that completes our solution to free response question two from the 2013 AP Cal BC exam this is free response question five from the 2018 AP Cal BC Exam you know it's from the BC Exam because you can see this beautiful polar graph here let's read it the graphs of the Polar curves are equals 4 that's a circle of radius 4 and R = 3 + 2 cosine Theta which is this cardioid are shown in the figure above the curves intersect at Theta = < / 3 and Theta = 5 piun over 3 part A says let R be the Shaded region that is inside the circle but outside the cardioid that's this shaded region here it's labeled very nicely for us and what we want to do is write an expression involving an integral for the area of this shaded region we don't even have to evaluate the integral we just need to write it let's do part A down here hopefully you recall that the area of a polar region is found by taking 1/2 the integral from A to B whatever those bounds happen to be the integral of R 2 with respect to Theta this comes from the area of a sector of a circle instead of building integrals in polar coordinates up from rectangles like we did with rectangular coordinates in polar coordinates we build integrals from sectors of circles that's where this comes comes from hopefully that's familiar but we're looking for the area specifically between two polar curves so we're going to need to find the area underneath one curve and subtract the area that is underneath the other we can see in the graph that the outside curve is the circle Ral 4 and the inside curve is the cardioid so this is actually going to be very straightforward all we're going to have is 1/2 * the integral from PI over 3 that was given to us to 5 pi over 3 and again that was given to us we were told that the curves intersect at Theta = < over 3 and Theta = 5i over 3 which you can also see there in the figure so we're integrating across that region now the outer curve is the circle Ral 4 so we're going to have 4 squared and then we need to subtract the smaller curve squared or the inner curve squared the inner curve is the cardioid 3 + 2 coine Theta so we subtract 3 + 2 cine Theta squar this is just like finding the area between Curves in rectangular coordinates where you take the upper function minus the lower function but of course this is polar coordinates so there's a little bit more going on and this is taken with respect to Theta and why don't we go ahead and just put this whole thing in parentheses so it's super clear what we are integrating here and that is the correct solution for part A now what you have to know is that the rest of this video was recorded separately because in the first take I accidentally did this integral wrong instead of writing D Theta I wrote Dr but the rest of the first take is correct so this part is actually recorded afterwards but I'm going to send you back to the original recording now you're just going to have to remember that this integral you see on screen is wrong everything else will be correct but part A will be wrong you're going to see Dr instead of D Theta D Theta is what it should be but throughout the rest of the video you're going to see Dr which is wrong all right enjoy the rest of the video moving on then to Part B find the slope of the line tangent to the graph of R = 3 + 2 cosine Theta at Theta = < / 2 in order to do this we can write an equation for x write an equation for y find their derivatives with respect to Theta and then we can use those to find Dy DX that's going to let us find the slope of the tangent line so here for Part B what we are looking for is how much y changes with respect to X at Theta = < / 2 so we're looking for dydx at Theta equal I / 2 that's going to be the slope of the tangent line at that point in order to do this we need an equation for x an equation for y and then we need to take some derivatives now you should know that x equal R cosine Theta and y equal R sin Theta then we can replace R in each of these equations with what R equals in this context we know that R = 3 + 2 cosine Theta so X = 3 + 2 cosine Theta that's R * cine Theta y also equals 3 + 2 cosine Theta that's R multiplied by sin Theta then we'll distribute and differentiate Distributing with the X equation we have 3 cosine Theta + 2 cosine 2 Theta Distributing on the Y equation we have three s thet plus 2 sin Theta cosine Theta now if this is X we can quite easily find DX D Theta DX D Theta is the derivative of this expression with respect to Theta that's going to be -3 sin Theta that's the derivative of 3 cosine and then we're going to need the chain rule over here that's going to be + 4 cosine Theta that's the derivative of the outside function multiplied by sin Theta the derivative of the inside function and if y equals this we can differentiate both sides to find Dy D Theta now on the right we could actually simplify this a little bit to make the derivative easier 2 sin Theta cosine Theta you might recognize a double angle identity in there that is the same as Sin 2 Theta so that will make this derivative a little bit easier let's go ahead and find Dy D Theta Dy D Theta is looking at this expression on the right the derivative of 3 sin Theta is 3 cosine Theta and then the derivative of s of 2 Theta is 2 cine of 2 thet now going back to the original goal to find dydx at Theta = < / 2 the way we're going to calculate this is by evaluating Dy D Theta at Pi / 2 and dividing this by DX D Theta at Pi / 2 so now that we found Dy D Theta and DX D Theta let's evaluate both of them at Pi / 2 and then we can divide them to find the slope of the tangent line that we've been looking for beginning with Dy D Theta evaluated at Theta = < / 2 cosine of < / 2 is 0er so this term goes away 2 * < / 2 is pi and cosine of pi is -1 so this term is 2 * -1 or -2 as for DX D Theta s of piun / 2 is 1 so this is -3 * 1 which is -3 as for the other term cosine of Pi / 2 is 0er so this term goes away thus Dy D Theta is -2 when Theta = > 2 and DX D Theta is -3 when Theta = > / 2 so we can come back up to our fraction and plug in --2 in the numerator for Dy D Theta and -3 in the denominator for DxD Theta so how is y changing with respect to X when Theta equal piun / 2 What would the slope of the tangent line be well it is just 2 over 3 2/3 that completes our solution for Part B finally part c a particle moves along the portion of the curve R = 3 + 2 cosine Theta for Theta between 0 p/ 2 the particle moves in such a way that the distance between the particle and the origin increases at a constant rate of 3 units per second we want to find the rate at which the angle Theta changes with respect to time at the instant when the position of the particle corresponds to Theta = piun / 3 and indicate units of measure for this rate of change of the angle Theta so in this problem we will be looking for D Theta DT T the rate at which the angle is changing we don't know D Theta DT we're trying to find it we do know drdt the problem tells us that the distance between the particle and the origin that's R the radius so to speak is increasing at a constant rate of 3 units per second so we'll use an equation for drdt in order to get D Theta DT involved and then we can solve for D Theta DT so here in part C what is d R DT we happen to know that it's three but if we were just using a formula to try to calculate it drdt would be Dr D Theta because R is a function of theta the radius changes as the angle changes but Theta is a function of time because we're thinking about a particle tracing out our curve the curve in this case being this cardioid Theta is changing as time passes so yes drdt is going to be Dr D Theta but also we have to multiply by D Theta DT because Theta changes with respect to time this is just the chain Rule now we know that R is 3 + 2 cosine Theta that is the curve that the particle is moving along 3 + 2 cosine Theta so Dr D Theta is just the derivative of this with respect to Theta which is -2 sin Theta we were told that drdt is three the distance between the origin and the particle is increasing at 3 units per second so that's 3 and in the denominator we have -2 * sin Theta when Theta is pi over 3 S of pi over 3 hopefully you know your unit circle that's < tk3 / 2 so if we multiply that by -2 we're just going to have < tk3 this simplifies to < tk3 because this is 3 1 this is 3 1/2 just subtract the exponents you get 3 1/2 in the numerator which is < tk3 so it's negative < tk3 and the units because we're talking about how the angle Theta changes with respect to time this is NE < tk3 radians per second and that is our answer to part C and that completes our solution to fre response question five from the 2018 AP Cal BC Exam next next up are parametric and Vector equations recall that when you're asked to find the speed of a particle whose path is described by parametric equations you will have to take the derivative of the X and Y functions in order to find the velocity functions and then you'll have to take the square root of the sum of the squares of those derivatives this is fre response question one from the 2002 AP Cal BC Form B exam it is part of the graphing calculator section we'll use a calculator just a little bit for this problem let's read through it a particle moves in the XY plane so that its position at any time t for T between pi and positive Pi is given by these parametric equations part A asks us to sketch the path of the particle in the XY plane provided and to indicate the direction of motion along the path so for part a here is a recreation of the provided XY plane let's get some basic information about these parametric equations before we start sketching we know that X is a function of T it equals sin of 3T and we know that y of t equal 2T we might as well Begin by eliminating the parameter if y = 2T then T = 1 12 Y and so we can express X as a function of Y rather than having this parameter we don't actually need the parameter so we could say that X as a function of Y is equal to S of 3T but T is half y so three halfs y now if we express X as a function of Y we should ask how small and how big can y get well t T could go from PI to positive pi and Y equals 2T so y can go from -2 pi to positive 2 pi so we will write that here y goes from -2 Pi that's twice the lower bound of T to positive 2 pi which is twice the upper bound of T all right so X is s of 3 Y where y goes from -2 pi to 2 pi this means we're going to have a s function but instead of going along the X AIS the sign function is going to go along the Y AIS also because it's sine of 3 y the frequency is sped up by three Hales so y going from -2 pi to 2 pi that's a total of 4 Pi which would normally be enough space for two full periods of sign but in our case the frequency of the sign function has been sped up by a factor of 1.5 so this space from NE -2 pi to positive 2 pi will actually be enough for three full periods of sign so before even sketching this I know that we should have three full periods of sign in our graph and it should be a sign function going along the Y AIS so All That Remains is to plot some points when y equal 0 0 x is s of 0 which is just 0o so this curve should pass through the origin now this is 2 pi up here which means this is one PI right here so maybe we should label that that's Pi which means this is A3 Pi this is 2/3 pi and so on so right here where Y is 1/3 of Pi if we plug that in we would get PK / 2 and sine of PK / 2 we know is 1 so let's plot that point then right here is 2/3 Pi if we plug y = 2/3 Pi we'll actually get sine of Pi and sine of Pi we know is zero so here it goes to zero and then if we plug in y = Pi we will have sine of 3 Pi sine of 3 Pi is -1 so here at y = piun the x value will be -1 and this pattern will continue in both directions so here at pi and a thir Pi it'll go back to zero and then it'll go back to positive 1 and then at 2 pi it comes back to Zer and this curve goes similarly in the direction of the negative y if we go to - 1/3 Pi the x coordinate is going to be -1 and then when we get to -23 Pi it comes comes back to zero and then if we get to Pi it is positive 1 and then if we get to piun and A3 we come back to zero and then at pi and 2/3 Pi we are back to -1 and then back to zero at the end then we can just do our best to connect these points sketching our S curve All That Remains is to draw a couple arrows indicating the direction now let's think how does y change as T changes well Y is just 2T so as T increases as time passes y also increases which means the direction is moving upwards along the Y AIS so here are just a couple of arrows to indicate that direction and that completes part A Part B says find the range of X of T and the range of Y of T we're actually already halfway done that we already know that y goes from -2 pi to positive 2 pi so we can just copy and paste that interval that we already discussed as for x x is even easier because X is just a sign function which means it's going to be between Nega 1 and positive 1 and as we already saw it does indeed achieve all of those values so it's not just a piece of the sign function it completes as we know three full periods of the sign function 1 2 3 so the range of x X is from -1 to POS 1 x is between POS 1 and 1 now part C says find the smallest positive value of t for which the x coordinate of the particle is a local maximum and then find the speed of the particle at this time now if we're trying to find maximum values of the x coordinate we're going to want to take the derivative of the X function so let's look at X Prime of T well well X of T is s of 3T so the derivative of that is going to be cosine of 3T but then we need to multiply by the derivative of the inside function so throw a factor of three in front there now this we need to set equal to zero to find the critical points the derivative of course exists everywhere so the critical points will just be where the derivative equals zero now we could divide both sides of this equation by three that's just going to get rid of the three and we have cosine of 3T equal 0 where does cosine equal 0 well at infinitely many places but we're looking for the one that produces the smallest T value the first time that cosine is zero is at an input value of pi/ 2 so we would need 3T the argument of the cosine function to equal Pi / 2 that's the first time that cosine is zero this means that tal > / 6 that's the T value we are looking for that of course is when cosine is zero and S is 1 that would be our first Maximum x value which is right here on the graph that is at T = pi over 6 all right the other part of this question part C was to find the speed of the particle at this time to find the speed of the particle we need to look at the magnitude of the Velocity Vector so in general the speed of the particle is going to be the square root of the sum of the square of the derivatives of the position functions so X Prime 2+ y Prime squared all in a square root now we already found what x Prime was so let's start to expand this expression we have the square root of x Prime we know is 3 cosine of 3T if we square that we get 9 cine squar of 3T and then we have to add the square of the derivative of the Y position function y you may recall is 2T and its derivative is just two so if we square that we get four so that is the speed function in general we need to evaluate it at T = < / 6 so let's plug in pi/ 6 and see what this equals plugging in t = < / 6 we have the square < TK of 9 cine s of 3 * < / 6 is < / 2 so we will put < / 2 there and then + 4 now cine of < / 2 is 0 so cosine squ of pi/ 2 is zero this whole term just goes away so it's actually just the square Ro TK of 4 which is two so that is the speed at T = < over 6 finally moving on to Part D is the distance traveled by the particle from PI to Pi greater than 5 pi and we need to justify our answer to find distance traveled we need to accumulate the speed function so we need to take this function that we found for the speed that's the magnitude of the velocity and we need to integrate it accumulating that speed over time is going to give us the distance traveled and for the integral we will use a calculator but let's go ahead and write out the integral first so here for Part D we are integrating it said from NE pi to positive pi and what we're integrating is the speed function which we know is that square root there so it's the square Ro TK of 9 cosine 2ar of 3T + 4 and let's make sure that square root is long enough and this integral is taken with respect to T and we'll just find this on a calculator so on the calculator I'll press the math button and then option nine the integration function we are integrating from PI to positive pi what we are integrating is the square < TK of 9 * cine of 3T but it's more convenient to use X on the calculator so I'm going to use x and that's squared and then plus 4 and we're integrating with respect to X on the calculator we get about 17.9 73 and remember the question also asked is this distance traveled which we just found is about 17973 is this distance traveled greater than 5 Pi the answer is yes it is because Pi which is about 3.14 is obviously less than 3.15 and so 5 Pi then would have to be less than 15.75 so yeah 17.9 73 that's greater than 15.75 which is certainly greater than 5 pi and that completes our solution to free response question one from the 2002 AP Cal BC formb exam this is fre response question two from the 2012 AP Cal BC exam let's read it for T greater than or equal to zero a particle is moving along a curve so that its position at time T is X of t y of T at time T equals 2 the particle is at position 15 it is known that dxdt equals this and dydt equals this and we have a four-part question here and this is part of the graph and calculator section we'll use the calculator quite a bit in this problem to compute some derivatives and whatnot note that the given information tells us the rate of change in the horizontal Direction and the rate of change of the particle in the vertical Direction with that in mind let's move on to part A is the horizontal movement of the particle to the left or to the right at time tal 2 and we need to explain our answer and then find the slope of the path of the particle at time tal 2 beginning with the horizontal movement of the particle which is described by the x of T function is that horizontal movement to the left or to the right at time T equals 2 to answer that question we need to ask how does the horizontal position X of T change at time tals 2 well that's described by the derivative dxdt so in fact all we have to do is plug tal 2 into the derivative if the derivative is positive that means the position is changing to the right the positive Direction and so the movement is to the right if it's negative then the movement is to the left it's in the negative Direction so we will look at DX DT evaluated at tal 2 and this is straightforward because dxdt is given dxdt has a numerator of square < TK of t + 2 T is 2 so this becomes the square < TK of 4 which is just 2 and the denominator is e to the 2 both the numerator and denominator are positive so this is positive and thus the particle is moving to the right at time tals 2 and there that logic is written out at time tal 2 the particle is moving to the right because the rate of change of its horizontal position is positive now the other part of the question asks us to find the slope of the path of the particle at time tal 2 to find that we just have to look at dydt divided by dxdt with T = 2 plugged in this will tell us how the vertical position is changing with respect to the changes in the horizontal position and that's going to give us the slope and we just have to plug these pieces in in the numerator we have dydt but with tals 2 plugged in so that's going to be sin squared of 2 we are dividing by dxdt so the denominator of dxdt will actually move to the numerator with tal 2 plugged in the numerator of dxdt will be in the denominator because we're dividing by it that's going to be the square < TK of 4 which is just two then we'll go ahead and plug this into a calculator so in the calculator we'll Type S of 2^ 2ar multiplied e to the^ 2 and then this is getting divided by two this gives us a slope of about 3.54 so the line tangent to the path that the particle is traveling at time tal 2 that tangent line to the path path has this slope let's move on then to Part B find the x coordinate of the particle's position at time T = 4 we'll write this in blue in order to find the x coordinate of the particle position at time tal 4 we need to take the x coordinate that we are given which is at time tal 2 so I'll write X of 2 and then we need to accumulate the rates of change from tal 2 to the desired ending time of T = 4 so we're going to accumulate DX DT those are the rates of change so X of 2 is given to us as one that's the horizontal position of the particle at time T equals 2 and then we are integrating from T = 2 to T = 4 DX DT which was also given to us as the square < TK of t + 2 / e to the T that was given to us and we are integrating this with respect to T again this is taking the position at the time we know T equals 2 and then accumulating the change in horizontal position from 2 to 4 now we can find this using our trusty graphing calculator so in the calculator I'll press 1+ the math button and then function 9 that's the integration function we are integrating from 2 to 4 and what we are integrating is the square < TK of t + 2 but in the calculator we'll use x because that's more convenient divided by e to the power of T but we're using X integrating with respect to X here on the calculator and we get a position of about 1. 1252 or rounding up this is about 1.25 3 now let's move on to part C find the speed of the particle at time tal 4 and find the acceleration vector of the particle at time tal 4 all right to find the speed of the particle we're going to need to take the magnitude of the Velocity Vector of the particle the velocity Vector is just the vector consisting of the rate of change of the horizontal position dxdt and the rate of change of the vertical position dydt this is basically just the derivative of the position function so then the speed at time T = 4 is just going to be the magnitude of this velocity Vector with tal 4 plugged in it so let me write this big square root so that we have room to fit it all in here and the velocity Vector we know what its components are because dxdt and dydt were given to us so we'll just plug tal 4 into those and you have to square the components right it's just part of the magnitude formula so let's write this out so there is the magnitude of the Velocity Vector at time tal 4 I just plug tal 4 into dxdt and plugged tal 4 into dydt both of them are squared they're getting added together in a square root we of course will calculate this using our calculator got to be careful typing in these big ugly Expressions into calculators so let us be careful we've got the square root of the square < TK of 6 / e to the^ of 4 and this whole thing is getting squared and then we are adding sin 2 of 4^ sared so we have S of 4 squared close the parentheses Square all this is in the square root that looks good it's about 5745 now we also are asked in part C to find the acceleration vector of the particle at time tal 4 so to find the acceleration vector at time tal 4 we're going to need to take the derivative of the Velocity vector and plug in four so V Prime of four that's going to be the acceler a vector all right for this we're definitely going to want to use the calculator no point dealing with more manual derivatives and stuff than we have to so for the horizontal component of the acceleration vector at tal 4 we will take the derivative of dxdt with the evaluation at tal 4 so let's open up the calculator so on the calculator I'll press math and then option 8 that's the derivative function we are differentiating with respect to T but it's more convenient to use X on the calculator now for starters we're going to take the derivative of dxdt which is right there so we have the square Ro T of x + 2 / e to the power of X and we're evaluating this derivative at x = 4 this tells us that the horizontal component of the acceleration vector at tal 4 is about 0411 now we will do a similar thing for the vertical component of the acceleration vector press math option 8 for the derivative function we're differentiating with respect to X and then Dy DT is sin SAR of t or on the calculator we'll just put sin of x^ 2 then we are evaluating this derivative also at x = 4 and we get about 9894 so that's the acceleration vector at time tal 4 just take the derivative of the Velocity Vector with four plugged in and that was the speed at Time tals 4 that completes part C finally Part D asks us to find the distance traveled by the particle from time T = 2 to T = 4 so for Part D distance traveled is found by accumulating speed we need to integrate the speed function from time tal 2 to time tal 4 adding up all those speeds will tell us how far are the particle traveled so we are going to integrate from time T = 2 to T = 4 this same thing that we were using for Speed in part C but without tals 4 plugged in because for the integral of course we want T to be allowed to vary so we're going to have the square root of dxdt 2 plus dydt s so I'll write those things out and there it is this is the integral of the speed function the magnitude of the Velocity Vector from T = 2 to tal 4 We're integrating of course with respect to T let's open up the calculator to figure out this integral so I'll press math option nine the integration function we are integrating from 2 to 4 and now we have to very carefully type in the function we're integrating we have the square root of and then what we have is the sare < TK of x + 2 / e to the power of X and this whole thing is getting squared that's DX dt^ squar and then we need to add dydt ^ 2 so I'll open up parentheses put in s of x s of X is getting squared and then that whole thing is getting squared that's dydt squared we are integrating of course with respect to X and there we go it's about 651 so from T = 2 to T = 4 the particle traveled about 651 units this is not remember displacement displacement would be found by integrating the velocity we integrated the speed so we just get the total distance traveled and that completes our solution to free response question 2 from the 2012 AP Cal BC exam this is free response question two from the 2016 AP Cal BC exam let's read it at time T the position of a particle moving in the XY plane is given by the parametric functions X of t y of T where dxdt = T ^2 + sin of 3T ^2 the graph of y consisting of three line segments is shown in the figure above at time T equals 0 the particle is at position 51 so if you look at the graph at time T equals 0 you can see the y-coordinate which is what this is telling us the y-coordinate is at one this is part of the graphing calculator section we will we will use the calculator throughout this problem part A asks us to find the position of the particle at time tal 3 in order to find the position of the particle at time tal 3 well we know that this is given by X of 3 y of three for the X component we are going to have to use our calculator and the given dxdt function for the Y component we're going to have to use the graph which is given to us using this graph we can find dydt dydt is just the slopes of these three line segments but let's start off with X of 3 to find X of 3 the horizontal position of the particle at time tal 3 we need to take the initial position at time T equals 0 so I'll just write X of0 which we know is five but I'll just write it as X of zero for now and then we need to integrate all of the horizontal rates of change so integrate from 0 to 5 DX DT but dxdt is given to us so let's just write what it is dxdt is T ^2 + sin of 3T ^2 we are integrating of course with respect to T now we'll just use a calculator to figure this out I'll replace X of 0 now with the given initial horizontal position of five so this is is approximately so for part A we are looking for the horizontal position at time 3 x of 3 and the vertical position at time three y of three let's first begin with the horizontal position the horizontal position at time tal 3 is going to be the horizontal position at time T equals 0 plus the accumulation of the horizontal rates of change so we need to integrate from tal 0 to T = 3 the horizontal rates of change dxdt but dxdt is given to us so let's just write it in this integral dxdt is given as T ^2 + sin of 3 T ^2 so this is our integral it is with respect to T and X of 0 over here is also given to us so let's just replace that X of 0 we're told is five so this is five plus this integral and we will go ahead and valuate this using our calculator so here on the calculator I'll press 5 plus the math button option n for the integration function we are integrating from 0 to 3 and what we're integrating is t^2 but we're going to use x in the calculator just because that's more convenient plus sine of 3 * t^ squared and we are integrating on the calculator with respect to X and we get about 14377 so the horizontal position at time tal 3 we know is 14377 in order to find the vertical position y of three we are going to have to use the provided graph we're looking for the y-coordinate right here where tal 3 so let's study this line what is the equation of this line well it goes from the 21 up to the 4 0 so when the x coordinate increased by 2 from 2 to 4 The y-coordinate increased by 1 from - 1 to 0 which means the slope is positive half which means that at tal 3 it must have gone up 12 from 1 1 plus a half is half so y of 3 would be -2 if you wanted to do this really carefully you could just write the equation for this line segment and plug in tal 3 but writing that y component of - one2 that completes part A that's the position of the particle at time tal 3 Part B asks us to find the slope of the line tangent to the path of the particle at tal 3 that's really easy that's just dydt divided by dxdt so what we are looking for is dydt how the Y is changing divided by dxdt how the x is changing and specifically we are evaluating this at tal 3 so this fraction is going to tell us the vertical change with respect to the horizontal change this is equal to well for dydt at tal 3 we already said that the slope of this line which is what's applicable at tal 3 the slope of this line is positive2 we see from here to here it goes forward to and up one so its slope is 1/2 so dydt here is just 1/2 as for the denominator DX DT when T equals 3 we know that DX DT is this so we'll just plug 3 into that using our calculator so here on the calculator we'll just type in dxdt but with tal 3 plugged in so 3^ SAR is what we start off with and then plus s of of 3 t^2 now if T is 3 then t^2 is 9 so 3 T ^2 is just 27 so s of 27 this turns out to be 9. 9564 just about so I'll write that there 9.95 64 and we'll just wrap this up by doing the division on the calculator the numerator of 0.5 divided by dxdt that we just evaluated it's about 0.050 02 so the path that the particle is traveling along has a tangent at tal 3 with this slope that's the slope to the path at tal 3 moving on to part C find the speed of the particle at tal 3 speed is going to use a lot of the same information we've been using because speed is just the magnitude of the Velocity function so we need to take the square root of the horizontal component of the Velocity function at tal 3^ sared the horizontal component of the Velocity function is dxdt and we will need to plug three into that we already did plug three into that for this problem so let's just write that 9.95 64 squared and then we need to add the vertical component of the Velocity function which we already know is just2 and we need to square that so+ 12 squared this is the magnitude again of the velocity function which is the speed now we'll just evaluate this on the calculator so we have the square OT of 9.95 64 SAR + 12 2quared this is about 9969 so that is the speed of the particle at time t equal 3 moving on then to the final part part D part D says find the total distance traveled by the particle from tal 0 to tal 2 so looking on our graph tal 0 to tal 2 is from here to here now to find distance traveled we need to accumulate speed speed depends on dxdt and dydt dxdt doesn't uh change exactly it's the same function right dxdt is just this dydt however does have an Abrupt change here at tal 1 so as we integrate the speed to find the distance traveled we're actually going to have to split it up into two integrals to account for this Sudden Change in the dydt dydt is the slope of the these line segments and it's right there where we see that Sudden Change over here from 0 to one we have that dydt is what what's the slope of the line well when it goes from 0 to 1 it decreases by two so that would be a slope of -2 meanwhile over here from 1 to two the slope is clearly zero it's a horizontal line segment so over there dydt is zero so we are going to have to integrate from 0 to 1 because that's where that dydt change happens from 0 to 1 the speed of the particle which is the square root of DX dt^ 2 now dxdt is given to us as this function so let's write that out and then we need to add the square of dydt now from 0 to 1 like we saw dydt is -2 that's the slope of this line segment so that's dydt squar and we're integrating with respect to T and then separately we have the integral from 1 to two of this same thing but at this point the dydt has changed to zero so I'm just going to duplicate this and then we can change Dy DT to zero so in fact we don't even have to write it the integral though is still with respect to T and now we'll just have to consult the calculator to evaluate this big nasty sum carefully typing this into the calculator with the integration function we get about 4.3 49 that is the distance traveled by the particle from tal 0 to tal 2 again we had to split it up because at tal 1 there's a sudden change in dydt and that completes our solution to free response question 2 from the 2016 AP Cal BC Exam make sure you read these questions carefully it's a lot harder if you don't understand immediately that this graph is just a graph of y of T and X of T is not actually given to us we are only given dxdt explicitly as a function but we don't have any graph relating to X next up are problems on Oilers method for these problems recall that the little steps that you take in X will be given to you in the problem and then the steps that you have to take in y are the changes in X multiplied by the derivatives value at the previous x coordinate good luck this is free response question four from the 1998 AP Cal BC exam this is part of the graphing calculator section consider the differential equation given by dydx = XY / 2 part A says on the axes provided below sketch a slope field for the given differential equation at the nine points indicated so we need to look at each of these nine points and based on their coordinates assess what the derivative dydx should be and sketch a tangent segment that has that slope so let's first go through all of these points and label each one with the appropriate dydx value once we can see what all of their slopes should be we can draw them so that the slopes look accurate right if this point has a slope of two but this point here has a slope of one then this segment should be steeper than this one so we want to go through and find all the slopes before we start sketching let's go from top to bottom so let's start up here the coordinates of this point are -13 so what should the slope be well the numerator of the derivative XY is -1 * 3 so 3 / 2 so -1.5 again we'll just label with the slopes and then we'll go through and sketch the segments now this point here has coordinates 03 so the dydx value would just be zero this point over here has coordinates 13 so the dydx value would be 1.5 because it would just be 3/ 2 then this point here has coordinates -1 2 2 so the dydx value would be -2 over 2 so just -1 this point has an x coordinate of zero so the slope would again be zero this point over here has coordinates 1 2 so it would have a slope of positive 1 this point over here has coordinates -1 1 so its slope would be 0.5 this point has a slope of zero and this point has a slope of positive. five to sketch the slope field then we just need to draw segments that have roughly these slopes and we want to make sure that they look correct relative to each other so this guy over here with a slope of 1.5 should be more steep than this guy with a slope of1 all right let's go through we could do the horizontal lines first because that's really easy slopes of zero at those points so just drawing horizontal lines there and then this point should have a slope of 1 .5 so something like that and then this point over here should have the same slope over here but going up so something like that and then this should be a slope of NE -1 so like 1.5 but a little less steep so maybe like that and then over here same thing but going in the positive direction something like that and then this should be a downward slope at .5 uh but it should be quite shallow so maybe something like this this and then on the other side over here we have the same sort of thing but going up and now that I've sketched the slope field I'm just going to go ahead and erase the slopes that we jotted down because we don't need them anymore and that is our answer for part A there is our sketched slope field moving on to Part B let y equal F ofx be the particular solution to the given differential equation with the initial condition F of 0 = 3 use Oilers method starting at xal 0 with a step size of 0.1 to approximate F of 2 and show the work that leads to our answer for Part B recall that the given differential equation is dydx = XY / 2 now to use Oilers method to approximate F of point2 we need to start by approximating F of 0.1 that's because the initial condition is at xals 0 and we were told to use step sizes of 0.1 so first we step forward 0.1 now what is f of 0.1 well it is approximately the value at F of 0 which is three plus the change in X which is 0.1 our step size multiplied by the derivative the slope at x equal 0 this is our approximation for f of 0.1 so that's three plus what is frime of 0 well if you plug 0 x equals 0 I should say into dydx clearly the slope is 0er so this should just be 3 + 0 and so 3 is our approximation of f of 0.1 now we have to step forward another 0.1 to get our approximation of f of2 what's F of point2 well it should be about F of1 which is three plus the change in X which is 0.1 multiplied by the slope or the derivative evaluated at 0.1 making these incremental steps using tangent line approximations is Oilers method all right now what is fpre of 0.1 well we don't know exactly what frime of 0.1 is because frime of 0.1 depends on Y and we don't know the true value of y when xal .1 the best we can do is use our approximation that when xal .1 Y is about 3 so the value of the derivative which I'll put in parenthesis here the value of the derivative should be about2 that's denominator of two multiplied by X which is 0.1 multiplied by Y which at 0.1 we approximate as about 3 now .1 * 3 is. 3 * 1 12 is 0.15 5 * .1 is 015 so this is 3.15 and that is our answer for Part B using Oilers method starting at xals 0 since that's where the initial condition was and step sizes of 0.1 this is our approximation for f of2 now part C says find the particular solution yal F ofx to the given differential equation with the same initial condition as before and then use our solution to find F of 2 which is what we approximated in Part B so here for part C we are going to need to use separation of variables we know that Dy DX equals X Y / 2 now we will move the DX to the right so all the X's are together and move the Y to the left so all of the Y's are together so we're going to multiply both sides by DX and divide both sides by y then on the left we have 1/ y because of the division by y 1 y Dy equals we multiply both sides by DX so on the right we have x / 2 DX and now we can go ahead and integrate both sides of this equation the integral of 1/ Y Dy is the natural log of the absolute value of y in theory we also have the addition of an arbitrary constant but let's imagine that we just gather all the arbitrary constants on the right side so on the right side we have the integral of x / 2 which is just going to be/ 14 x^2 and then we have all of the arbitrary constants just wrapped up in this 1 + C now we need to solve for y and we also need to use the initial condition to solve for C let's first exponentiate both sides of this equation to get rid of that natural log exponentiating on the left will of course cancel out with the natural log and so we have the absolute value of y and then on the right we would have e to the power of/ 14 x^2 + C which by our exponent laws is the same as e 1/4 x^2 * e c but e to an arbitrary constant is just some other arbitrary constant so we could just say C * e 1/4 x^2 this is being slightly sloppy with notation because e to the C is different from C so using the letter c over and over again is a little bit careless uh but this is typically how we would do it if you want to be a little bit more formal you could call the original constant C1 and then we can call this new constant which is e to the C1 we can just call it C now we actually don't need the absolute value on y because we were told that we're finding the particular solution that passes through this initial condition F of 0al 3 so for this particular solution to the differential equation the Y values are actually positive and so we can drop the absolute value and thus we have that y = c * e to 1/4 x^2 now we can plug in the initial condition which is 03 and in order to solve for C plugging in the initial condition of 03 which I'll just do in a different color produces the equation 3 = C * 1 so c equal 3 there we go thus our particular solution is y = 3 * e 1/4 x^2 so that's our particular solution but don't forget the other part of the problem we're supposed to use this solution to find the value of f of point2 now that we have this particular solution for our function f we can just plug 2 into it so let's write this f of2 equals using this particular solution it equals 3 * e to the 1/4 multiplied by. 2^ sared now let's just go ahead and simplify this is 3 e to the power of 1/4 multiplied 2^ 2 is 0.4 and then 1/4 of 04 is 01 so this is 3E to the .1 plugging this into a calculator this is about 3.03 you can see that our approximation was a bit off but not by too much and that completes our solution to free response question 4 from the 1998 AP Cal BC exam this is free response question six from the 1999 apal BC Exam we will use a calculator throughout this problem let F be the function whose graph goes through the point 36 and whose derivative is given by this equation part A asks us to write an equation of the line that's tangent to the graph of F at x = 3 and to use this tangent line to approximate F of 3.1 a value of the function near the point of tangency all right so let's start off here with part a in order to write an equation of the tangent line we will use point slope form so we need y minus a y-coordinate now at x = 3 we know the y-coordinate of our function is 6 based on the given initial condition so this should be y - 6 equals and then we need the slope to find the slope we just plug xal 3 into fime and fime is given so plugging x = 3 into the given value of the derivative we have 1 + e 3 / 3^ 2 which is 9 that's the slope and this needs to be multiplied by xus the x coordinate the x coordinate in question is three finally we could add six to both sides just to get this in a more standard form so writing this with Y by itself we have that y = 6 + all the stuff from the prior equation and there we go that is our equation of the tangent line at x = 3 and now we need to use it to approximate F of 3.1 in order to do that all we have to do is plug X = 3.1 into the tangent line since 3.1 is pretty close to three the point of tangency this should give this should give us a decent estimate of the function's value so F of 3.1 should be about 6 + 1 + e 3/ 9 multiplied by x - 3 but where X = 3.1 3.1 - 3 is just 0.1 so this is just getting multiplied by 0.1 if you plug this into a calculator you'll find it's about 6.2 3 4 moving on to Part B using Oilers method starting at x = 3 with a step size of 0.5 we must approximate F of 3.1 and then use f Prime to explain why the approximation is less than the true value of f of 3.1 all right so for Oilers method if we're going to approximate F of 3.1 using step sizes of 05 and starting at xal 3 we first need to take our first step away from X = 3 thus we begin with an approximation of f of 3.5 that's taking one step of size 05 away from xal 3 What is the value of f of 3.05 well it should be about the value of F at 3 which we know is six because the function passes through the 36 plus the change in X or the step size 05 multiplied by the derivatives value at three so multiplied by FR Prime of 3 because remember the derivative is giving us a change in y per change in X using the tangent line so if this is the change in X 05 we have to multiply it by the derivative which tells us how much y should change all right so let's just plug this into the calculator we have 6 + 05 multiplied by the derivative evaluated at 3 that's right here right 1 + e 3/ 9 so 1 + e to ^ 3 / 9 this is about 6.17 then we can take the next step going forward another 05 to approximate F of 3.1 it should be about F of 3.05 which is about 6.11 1 7 plus the change in X which is 0.05 the step size multiplied by the value of the derivative at 3.05 to get this approximation we can again use the calculator so we're going to have 6117 + 05 multiplied by the derivative but this time with 3.05 plugged in the derivative of course is 1+ e to the X so 1 + e to the power of 3.05 and then this needs to get divided by 3.05 squared this gives us a approximation of about 6236 now let's not forget the other part of the problem Part B also wants us to use the second derivative of our function we have the first derivative we're supposed to use the second to explain why the approximation we just got is less than the true value of f of 3.1 you may recall that if a function is concave up then its tangent lines lie below the graph now concave up means the second derivative is positive so what we're going to do is take the derivative of the derivative in order to find the second derivative and if we can show that it's positive in the space that we're concerned with which is around 3 and 3.1 if the second derivative is positive there the function must be concave up which means the tangent lines which we're using to construct this approximation that's how Oilers method Works those tangent lines and thus our approximation must be underestimates so let's just write the first derivative which is given to us the first derivative is 1 + e x / x^2 so to take the derivative of this and find the second derivative we are going to need to use the quotient rule so we'll begin with the derivative of the numerator which is e to X multiplied by the denominator x^2 minus the derivative of the denominator so 2x multiplied by the numerator 1 + e to X and then all of this needs to get divided by the denominator squared so x 4 now everything has a fact factor of X so we can cancel out a factor of x from everything so simplifying this a little bit we have x ^ 1 * e x - 2 * 1 + e to x divided Now by x 3 now we have x e to X and - 2 e to X so we can factor out an e to the X from both of those terms which leaves behind X -2 and then we also have -2 * pos1 so let's not forget that and still X cubed is in the denominator so this is our second derivative and the space we are interested in is really between 3 and 3.1 because that's the space within which we were constructing approximations so what is the shape of the graph in this space is it concave up or concave down well if X is at least three this second derivative is clearly positive because if x is at least three then x - 2 is at least 1 and e to the x is a lot bigger than two so the numerator would definitely be positive and it would also be getting divided by a positive number so for X at least three it's clear that e to x * x - 2 is greater than two and thus this second derivative will be positive in the space that we are interested in so yes in fact our approximation of 6.23 6 is an underestimate of the True Value if x is at least three the second derivative is clearly positive so f is concave up on the interval of interest and thus the tangent lines and consequently the oiler approximation that we found is an under estimate again this is because if a function's concave up those tangent lines are underestimates all right let's move on then to part C part C asks us to use the integral of frime of x from 3 to 3.1 to evaluate the value of the function at 3.1 so for part C we want to use this integral of frime of x from 3 to 3.1 to calculate F of 3.1 by the fundamental theorem of calculus we know that the integral of f Prime from 3 to 3.1 is f of 3.1 minus F of 3 and we know that F of three is actually equal to six right that was given as an initial condition in the problem so this on the right is f of 3.1 - 6 and on the left for our integral we'll use a calculator to get an approximation here on the calculator I'll press the math button and then option nine that is the integration function we are integrating from 3 to 3.1 the function we're integrating is f Prime which is this guy here 1 + e x over x^2 so we'll write that careful with our parentheses 1 + e x / x^2 and we are integrating with respect to X this turns out to be about. 2378 we'll say so 23 78 and then finally to solve for f of 3.1 we just add six to this and so F of 3.1 should be about and this will be a very good approximation because it's using a calculator it'll be about 6.23 78 so maybe we don't know what f ofx is but just using its derivative and an initial condition we can figure out a whole lot there's a lot of different ways you can approximate a function's value with this information that completes our solution to free response question 6 from the apal BC Exam from 1999 next our problems on improper integrals recall that the definition of the integral of a function f ofx from a to Infinity is the limit of the integral of the function from A to B as B goes to Infinity this is free response question five from the 2004 AP Cal BC Form B exam to the - 1/2 it will become x to the pos2 and then divide by the new Power dividing by 1/2 is the same as multiplying by two and so we get here 1/3 * 2x^ 1/2 evaluated from 1 to 4 let's start evaluating when we plug in four we get 4 to the2 which is 2 multiplied by 2 so again 4 the 1/2 is 2 multipli 2 then subtract pluging in the lower bound that's just going to be min-2 so finally this is 4 - 2 which is 2 * thir so the average value of the function over the interval in question is 2 over3 Part B says let s be the solid generated when the region bounded by the graph of y = g ofx the vertical lines x = 1 and xal 4 and the xais is revolved about the X AIS we want to find the volume of this generated solid you may find a quick picture helpful in this situation I'm not going to really include the ne X AIS in this picture because it's not super important now the graph of 1/ root X looks very similar to the graph of 1 /x it's going to look something like this and the region that is getting revolved is the region under this curve from x = 1 to x = 4 so if this this was x = 1 and this over here was x = 4 it's this region that is getting revolved about the xaxis so the generated solid maybe I can use purple here the generated solid is going to look something like this and a cross-section of course would just be a circle that looks like that and the radius of the circle would be the value of the function 1 /un x at the horizontal position in question so all we have to do to find the volume of this solid is add up all of the areas of these circular cross-sections so we will integrate from 1 to 4 the area of the circular cross-sections which is pi multiplied by the radius squared the radius is just the value of the function at the point in question so 1 /un x^ SAR and we are integrating with respect to X we can of course take a pi out of this integral so this is pi multiplied the integral from 1 to 4 and then 1un x^2 is just 1/x and this is very easy to integrate this is pi multiplied the natural log of x a valuated from 1 to 4 finally this is pi multiplied the natural log of 4 minus the natural log of one but of course the natural log of one is zero so this is just pi multiplied by the natural log of four that is the volume of the solid moving on then to part C part C says for the solid s given in Part B find the average value of the areas of the cross-sections perpendicular to the x axis that's really easy because the cross-sections perpendicular to the x-axis are precisely the crosssections that we added up in Part B so to find the average area of these cross-sections we need to add all of the areas up which we already did it's pi ln4 and then divide by the length of the interval the length of the interval is Just 4 Min - 1 so all we have to do for part C is take our answer from Part B which was Pi ln4 because this is the sum of all of the areas of the cross-sections and then divide by the length of the interval which is 4 - 1 so our final answer for part C is pi/ 3 multiplied by the natural log of four finally Part D the average value of a function f on the unbounded interval from a to Infinity is defined to be this limit show that the improper integral of G of x from 4 to Infinity is Divergent but the average value as just defined on the interval from 4 to Infinity is finite all right so we must begin with this improper integral and show that it's Divergent the integral from 4 to Infinity of G of X so let's get into that g of X of course is just 1 over the square root of x and we already integrated this function in part A so this is going to be the limit as B goes to Infinity of the integral of 1 /un x which we know from part A is just 2x to the 1/2 the 1/3 was not part of G of X that was just because it was an average value calculation so the integral of G of X is just 2x to the2 so we are going to have 2X to the2 or just 2un X and this is going from 4 to B and B is going to Infinity all right so this is equal to the Limit as B goes to Infinity of 2 multiplied the square < TK of B minus plugging in the lower bound of four that's just going to be 2 * the S < TK of 4 which is 2 * 2 which is four and clearly this limit diverges because the limit of 2 * theun of b as B goes to Infinity is just positive Infinity so indeed this is Divergent now the other part of Part D is to show that the average value of G on the interval from 4 to Infinity is finite so now we've got to look at this limit so using the definition the problem gave us this is the limit representing the average value of the function on the unbounded interval from 4 to Infinity all right so as we proceed with this in the numerator we have the integral of 1 over the square < TK of X and we already know what that is that's 2 * the square < TK of X and in the denominator of course we just have B- 4 so this will be the limit as B goes to Infinity of in the numerator we have the integral which is 2un X from 4 to B so that's 2un B Min - 2 < TK4 which is 4 in the denominator of course we have b - 4 and indeed this is finite in fact this limit is equal to zero because in the numerator we have a root B whereas in the denominator we have plain old b b grows much more quickly than root B its power is one which is greater than B's power of 1 12 in the numerator so that's how we know this converges to zero again if you wanted to rewrite this to put that focus on the powers it's 2B to the 1/2 whereas we have B to the 1 in the denominator so this is going to outpace this the limit is zero and that completes our solution to free response problem five from the 2004 apal BC formb exam this is free response question 5 from the 2017 AP Cal BC Exam let F be the function fun defined by f ofx equals this big fraction part A asks us to find the slope of the line that's tangent to the graph of F at x = 3 to do this we'll just take the derivative of f using the chain Rule and then plug in x 3 you could use the quotient rule but I would say that is a little bit Overkill here because F ofx we can rewrite with a negative power F ofx = 3 multiplied parentheses 2x^2 - 7 x + 5 to the^ of -1 at which point it's clear how we can take the derivative using the chain Rule and the power rule so the derivative frime of x will be bring the exponent down and that will get multiplied by 3 so -3 and then don't change the inside function so I'll just go ahead and duplicate this inside function and we have to reduce the Power by one again we're applying the power rule here so now the power is -2 but we also have to multiply by the derivative of the inside function which is just more power rule 4x - 7 there is our derivative now to finish our answer for part A we just have to plug in x = 3 so this will become -3 * 2 * 9 - 21 + 5 to the^ -2 * 12 - 7 all right now we just have to clean this expression up a little bit the part with a -2 power let's put that in the denominator so we have 2 * 9 which is 18 - 21 is -3 + 5 is pos2 so this is 2 to the power of pos2 the power is positive because we moved it to the denominator in the numerator we have -3 still and then 12 - 7 is 5 so -3 * 5 over 2^ 2 this is5 over 4 because this is the value of the derivative at xal 3 this is the slope of the line tangent to the graph of F at xal 3 moving on to Part B we are asked to find the x coordinate of each critical point of f in the interval from 1 to 2.5 then classify each critical point as the location of a relative minimum a relative maximum or neither and justify our answers now we already found the derivative of f which is certainly useful for critical points we must ask where does the derivative not exist and where does the derivative equal zero here is our derivative that we can use to find critical points first we might ask about points where this doesn't exist this derivative won't exist where this quadratic is equal to zero because due to the negative power this is actually in the denominator so it can't be equal to zero now the problem gives us end points right we're looking in the interval from 1 to 2.5 not including those end points so we might check these end points by plugging them into the derivative and seeing if the derivative exists at those end points it turns out those end points are exactly where this derivative does not exist if you plug in one or if you plug in 2.5 you will get zero for this quadratic and thus the derivative does not exist at those end points but since those end points aren't in the interval we just have to set our derivative equal to zero to find the critical points to set this derivative equal to zero well like we said this part with the negative power will be in the denominator so we just need the numerator which is -3 * 4x - 7 to be 0 -3 * 4x - 7 needs to be 0 of course we could just divide both sides by -3 to just get rid of the -3 then we have 4x - 7 = 0 adding 7 to both sides then dividing by 4 we have that x = 7 over4 as our only critical point so this is a critical point then is it the location of a maximum or minimum or is it neither to answer that question we must ask if the derivative changes sign and if so how at x = 7 over4 turns out the derivative does change sign x = 7 over4 because this part with the negative power is just a square in the denominator so that's always going to be positive that's not going to affect the sign but 4x - 7 does change signs at X = 74s when X is less than 7 over4 4x - 7 will be negative and so the derivative will be positive because it will have this negative and this negative but once X exceeds 7 over4 this term is going to be positive and so the derivative because of the -3 will be negative so at x = 7 over4 the derivative goes from positive so the function is increasing to negative so the function is decreasing and thus at X = 7 over 4 we have a relative maximum and there our logic is written out the only critical point in this interval has an x coordinate of 7 over4 we set the derivative equal to zero to find that the point where the derivative does not exist are the end points of the interval which aren't included since frime changes sign from positive to negative at 7 over4 we know that F has a relative maximum at x = 7 over4 moving on to part C using the identity that 3 over this equals this difference evaluate the integral of f ofx from 5 to infinity or show that the integral diverges so here the expression this nasty expression has been decomposed for us into these two partial fractions these two fractions make up this one fraction we've decomposed the fraction into Parts it is of course not generally easy to integrate rational expressions with quadratics in the denominator so having it split up into a difference of fractions with linear pols in the denominators this makes it a lot easier we'll be able to integrate this because these will just be simple natural log functions and then we'll have to see if those limits exist or not of course this is an improper integral so we'll be looking at limits as a variable goes to Infinity getting started with this integral we want the integral of this rational expression from 5 to Infinity which is equal to the integral of this difference of fractions because we were given that this is equal to this now by definition this improper integral is going to equal the limit as B goes to Infinity of the integral of this same difference of fractions from 5 to B so now let's go ahead and integrate and then evaluate our limit we have the limit as B goes to infinity and then actually performing the integral we are going to get some natural logs so 2 over 2x - 5 the integral of this is going to be the natural log of 2x - 5 but by the chain rule if we were to take the derivative of this function the factor of two would pop out in front just like we want because there's a two in the numerator as for 1X -1 this is just going to be Ln of x -1 so there's our integral this is being evaluated from 5 to B I've copied the limit over here on the left so we have a little bit more room to work with let's rewrite the limit statement the limit as B approaches infinity and then before we do any evaluation let's just simplify this expression the natural log of 2x - 5 minus the natural log of x -1 by our log rules we can write that as the natural log of the quotient of these two things so 2x - 5 / X - 1 just using log rules here and we are evaluating this from 5 to B now when we think about what happens as we plug in b and let B go to Infinity here inside the natural log we have the ratio of two polinomial of the same degree 2x - 5 and X - 1 they both have degree one so the limit of this rational function as B goes to Infinity when we plug in B is going to be the ratio of the leading coefficients it's just going to be 2 / 1 so if we plug in B and then take the limit as B goes to Infinity this is just going to be the natural log of 2 over 1 so the natural log of 2 and then if we plug in 5 we're going to have 10 - 5 in the numerator which is 5 divid four in the denominator so minus the natural log of 5 over 4 doing some final simplification here with our log properties this is equal to the natural log of 2 / 5 over 4 but dividing by a fraction is the same as multiplying by its reciprocal so 2 / 5 over 4 is 2 * 4 5 so our final answer is the natural log of 8 over 5 the integral does converge and it converges to the natural log of 8 fths at this point I'm going to have to go ahead and Shrink our work so we can read Part D and actually have some room to write our answer all right part D determine whether the infinite series as shown here converges or diverges state the conditions of the test used for determining convergence or Divergence since we just did an integral the integral test seems like a pretty good choice here we integrated this exact function right look at that now in order to say that this series from Nal 5 to Infinity converges just because the corresponding integral from 5 to Infinity converges we just need to check a few conditions the function needs to be be continuous positive and decreasing on the interval in question which is from 5 to Infinity so we're looking at this function is it continuous positive and decreasing on the interval in question well it's definitely continuous because there would only be an issue if the denominator were to equal zero but we already checked that its zeros were at these end points from part A 1 and 2.5 the interval we are worried about now is from 5 to Infinity so those zeros are not part of this so certainly the function is continuous now is it positive well the numerator is positive so the only thing that could make it negative would be the denominator but we know that this denominator this quadratic is an upwards facing Parabola and we know that at xal 5 we are well past the second zero of the parabola because the second zero like we said occurs at xal 2.5 so at x equal 5 and all subsequent X's yes this is going to be positive the parabola just keeps going up and since this denominator just keeps increasing on the interval in question the polinomial itself is decreasing because it's 3 divided by an increasing denominator we could also take the derivative of the denominator in which case we would see that we get 4x - 7 which is certainly positive when X is at least five which it is on the interval in question suffice to say we can apply the integral test since our function is continuous positive and decreasing from five to Infinity it must be that the integral we calculated in the previous part has the convergent Behavior which matches up with our series our series must converge because this integral converges and that completes our solution to free response question five from the 2017 AP Cal BC Exam next are problems on the lrange error Bound for these problems it is useful to know the formula for the lrange error bound also in both of these problems I will write the word lrange and accidentally write the second G as a c so please forgive me this is free response question two from the 2004 AP Cal BC Exam it is part of the graph and calculator section but we won't actually need one for this problem let f be a function having derivatives of all orders for all real numbers the third degree tailor polinomial for f about xal 2 is given here part A asks us to find F of 2 and F Prime of 2 because this is a tailor polinomial for f centered at xal 2 it will match the function's value exactly at xal 2 so to find F of two for part A all we have to do is plug two into the tailor polinomial F of 2 will equal T of 2 and if we plug two into this function T of X we see these two terms will Zero out and we're just left with the initial term of 7 so F of 2 equal 7 now for the second derivative of our function evaluated at two all we have to do is look at our tailor polinomial and recall how tailor series are constructed in general the nth term of a tailor series looks like this the nth derivative of the function evaluated at the center divided n factorial multiplied by x minus the center to the power of n so we can find information about the second derivative of our function here where the power of x - 2 is 2 so in fact looking at this term the second derivative of the function evaluated at 2 divided 2 factorial must be9 again the second derivative of the function evaluated at 2 / 2 factorial must equal 9 that's just how these polinomial are constructed okay so then all we have to do is solve for fpre of 2 by multiplying both sides by two factorial 2 factorial is just two and so we have the second derivative of f evaluated at 2 equals -18 moving on then to Part B is there enough information given to determine whether f F has a critical point at xal 2 if not explain why not if so determine whether F of 2 is a relative maximum a relative minimum or neither and justify your answer we certainly can determine if F has a critical point or not because the tailor polinomial this third degree tailor polinomial does give us exact information about the first derivative of the function at xal 2 so we just have to ask is the first derivative of the function equal to 0 at xal 2 based on how tailor polinomial are constructed we know that the first derivative of our function at the center xal 2 will have to equal the derivative of the tailor polinomial at xal 2 they are created so that their derivatives match the derivatives of the function at the center so all we have to do is look at T ofx take its derivative and plug in x = 2 and then we can see if it's zero if so it's a critical point so coming up to the tailor polinomial and taking the derivative to get T Prime the derivative of s is zero and then the derivative of 9 * x - 2^ 2 is -18 * x - 2 and then the derivative of -3 * x - 2 cubed is -9 * x - 2^ 2ar you can see how if we took a second derivative we would get -8 if we were to plug in x = 2 two which of course agrees with our answer from part A but right now we are interested in the first derivative what happens if we plug two into this first derivative we actually get zero right 2 - 2 2 - 2 we would get zero so yeah this is the location of a critical point now is it a maximum or a minimum or neither well it turns out it must be a maximum because we already know the second derivative of the function at this Center of two is negative which which means the function is concave down and so this critical point is the location of a maximum and there is our reasoning written out F does have a critical point at xal 2 and F of 2 is a relative maximum because FR Prime of 2 equals 0 and FP Prime of two is negative so concave down means we've got a maximum on to part C use T ofx to find an approximation for f of0 is there enough information given to determine whether F has a critical point x equals 0 and if not and if so let's just answer the problem all right to find an approximation for f of0 all we have to do is plug zero into the tailor polinomial it won't be a great approximation probably because zero isn't super close to the center of two but should be an all right approximation so for part C we are trying to approximate F of zero and it should be about t of Z at least that's the best approximation we can come up with with plugging in xal 0 to this polinomial will give us what I'll jot it down here and then move it down to where we are writing part C this is going to be 7 - 99 * -2 2 - 3 * -2 cubed and this is equal to 7 - 2^ 2 is + 4 * - 9 is - 36 - 2 cubed is -8 * - 3 is + 24 and this adds up to -5 so that is our approximation for f of 0 as far as whether or not we can determine if there is a critical point at x equals 0 the answer is no because our tailor polinomial is centered to xal 2 and thus only gives us exact values for the function at that Center of xal 2 and here that is written out T of X gives exact information only at the center xal 2 so we can't determine if we've got a critical point at x equal 0 Part D the fourth derivative of f satisfies this inequality that its magnitude is less than or equal to 6 for all X in the closed interval from 0 to 2 notice that closed interval goes from 0 the number we were just discussing to two the center of our tellor polinomial use the lrange error bound on the approximation to F of0 which we found in part C to explain why F of 0 is negative so the approximation we got was -5 we want to use the lrange error bound to show that the True Value has to be negative first let's calculate the lrange error bound this is the formula for the lrange error bound we have X which is the point of our approximation which in this case is zero minus C the center in an absolute value to the power of n + 1 where n is the degree of the tailor polom for us n equals 3 then divide by n+1 factorial and multiply by the maximum value that the absolute value of the n+1 derivative of the function takes on in the interval from the point of approximation to the center of the tailor polinomial so this means that Z is in the interval from 0 to 2 because our approximation is at x equals 0 and the center of our tailor polinomial is at two so the maximum abs absolute value that the n+1 derivative of the function takes on in this interval that's what this is and we're told that the N plus1 derivative in our case the fourth derivative of the function has an absolute value that's bounded above by six on this interval we don't necessarily know if six is actually its maximum value but we do know it's less than or equal to six on this whole interval so plugging in all of our pieces we get that this error bound is less than or equal to the absolute value of 0 - 2^ 4 / 4 factorial * 6 because this maximum value is no more than six and do the math here you will get four as the lrange error bound this means our approximation from part C could be off by no more than four from the true value four more than -5 is -1 which is still negative so for sure the True Value must be negative that's the conclusion written out F of 0 must be less than or equal to the tailor approximation at xal 0 plus four because four is an upper Bound for the error and that's equal to Nega 1 so F of 0 has got to be negative and that will complete our solution to fre response question 2 from the 2004 apal BC exam this is fre response question 6 from the 2011 AP Cal BC Exam let F ofx equal s of x^2 + cine X the graph of y equals the absolute value of the fifth derivative of f is shown above very nice derivative function there part A says write the first four nonzero terms of the tailor series for sin x about x = 0 and write the first four nonzero terms of the tailor series for S of x^2 about xal 0 so if we know the tailor series for for sin x all we'll have to do is replace the X with x^2 for this second part now ideally you have the tailor series for sinx memorized but if not it's not a huge deal as long as you recall the general form of a tailor Series in general the nth term of a tailor series is the nth derivative of the function evaluated at the center which we call a divided by n factorial multiplied by x - a to the power of n so if you don't remember the tailor series for sin x we can just construct it as long as we remember this general form now before we start writing out terms of the tailor series for sin x let's think about what the derivative of sinx will look like when we plug in zero first before we take any derivative so for n equal 0 when we plug in our Center of 0o we would get zer then when we take the first derivative for the Nal 1 term of the tailor series we would have cosine of 0 which is one then when we take the second derivative for the Nal 2 term we would again get 0 then when we take the third derivative for the N = 3 term we would get1 so what's going to happen with the tailor series for sign is that we'll only get odd terms when Nal 1 when Nal 3 when Nal 5 and so on because the even derivative values are just sign functions evaluated at zero which are zero and also the sign as in positive negative sign the sign positive or negative of the odd terms that appear in the series will alternate positive negative the next one will be positive and then negative and and so on and that is enough for us to start writing the tailor series for sin x sinx equals when n equals 0 we're just going to get zero so that first term will not show up when n equal 1 the derivative will evaluate to one because we're plugging in zero and then we'll just have to divide by one factorial one factorial is just one so we don't really have to write the division at all and so we're just left with x minus the center of 0 to the power of 1 but there's no need to write minus 0 so we will just write it as x to the power of 1 then the Nal 2 term will again just be 0 so we don't have to write it for the Nal 3 term the third derivative with zero plugged in will be -1 so we have a minus and then we have to divide by 3 factorial and we're going to have x - 0 which is just X to the^ of 3 so X Cub over 3 factorial and this pattern continues we will then have plus x 5 / 5 factorial and so on we are asked in part A to write the first four nonzero terms so we'll want to write one more minus x 7 ided 7 factorial and plus dot dot dot now the second part of part A asks us to write the first four nonzero terms of the tailor series for S of x^2 about xals 0 so what we'll have to do for sine of x^2 is just use the tailor series that we just found but replace each x with X SAR so the first term just becomes x^2 then x^2 cubed is X to the 6 and we are dividing by 3 factorial this is a bit of a squeeze so let me move this down so we have a little bit more room again it's just like we're plugging X2 into the tailor series we found before so in the next term we had X to 5 but now we're replacing x with X SAR so we have X2 to the 5 which is x 10 ID 5 factorial and then the last term we'll have to write because we need to write four of them is is x^2 to the^ 7 which is x 14 / 7 factorial plus dot dot dot all right so there's our answer for part a moving on to Part B write the first four nonzero terms of the tailor series for cine X about xal 0 and then use this series and the series for sin of x^2 which we just found to write the first four nonzero terms of the tailor series for f because remember f is sin of x^2 + cosine about xal 0 all right if you didn't remember the tailor series for cosine there's a good chance that finding the tailor series for S would jog your memory the tailor series for cosine is just like the tailor series for sign except with the even Powers rather than the odd ones that's because the zeroth derivative of cosine which is just cosine when we plug zero into that we get one so we do have an Nal 0 term then when we take the first derivative so n = 1 this is when we get sign and thus we get zero but then when n equals 2 we take the second derivative we get Negative cosine and this is going to be NE -1 when we plug in zero so still these signs are alternating but this time we are going to have the even end terms rather than the odd end terms so cosine X has the following tailor series X to the 0 which is just 1 and then - x^2 / 2 factorial + x 4 / 4 factorial minus x 6 / 6 factorial and so on next Part B also asked us to write the first four nonzero terms of the tailor series for that function f sin of x^2 + cosine so this tailor series plus this one we'll want to combine like terms and write the resulting first four nonzero terms so there will be a one and then we'll have to combine the x s terms then we'll have the X to the 4 term and then we'll have to combine the X to the 6 terms and we can write that here so F ofx equal 1 the first term and then X minus what is actually just half x^2 x^2 - x^2 is just plus x^2 over 2 and then we'll have + x 4 over 4 factorial and then - x 6 over 3 factorial - x 6 over 6 factorial to make the denominator 3 factorial match the denominator of 6 factorial we would would have to multiply by 6 by 5 by 4 on the top and on the bottom 6 * 5 is 30 * 4 is 120 so this would be minus 120 x 6 over 6 factorial if we were to get common denominators so then in total we will have Min -21 x 6 over 6 factorial so getting common denominators and combining the X to the 6s we have - 121 x 6 over 6 factorial again that's because we'd have to multiply x 6 over 3 factorial by 120 over 120 to make those denominators match up and then this is what we get once we combine and then we can just write plus dot dot dot we do know that the next term will be positive because the next term is going to be x a over 8 factorial from the cine X tailor series because the S of x² tailor series does not have an x to the 8 term at a glance that might see now part C asks us to find the value of the sixth derivative of this function f evaluated at zero that might seem kind of intimidating at first but then recall all these tailor series we've been doing they have information about the derivatives we know that the sixth derivative of this function is going to be divided by n factorial or 6 factorial in the N equals 6 term of the tailor series of this function now if we look down at the tailor series that we wrote in Part B we see that the nals 6 term is right here and the coefficient of x is -11 over 6 factorial just based on how tailor series are constructed this numerator -121 must be the value of the sixth der ative of F at zero plain and simple it's got to be11 finally Part D says let P4 ofx be the fourth degree tailor polinomial for f about xal 0 which is what we just found without the ellipses then using information from the graph of y equals the absolute value of the fifth derivative of f ofx which is shown above we have that picture here should that the absolute value of P4 of 1/4 Minus the True Value F of 1/4 is less than 1 over 3000 so this here you should look at that and recognize as the error of the fourth degree tailor polinomial approximation of the function at 1/4 the tailor polinomial is centered at xal zero so when we plug in other values we're going to get some error this is the error and we want to show that it's less than 1 over 3000 we'll do this using the lrange error bound here is the lrange error bound it's the absolute value of x minus the center C where X is the point we're evaluating at or approximating at in our case it's 1/4 to the power of n + 1 where n is the degree of the tailor polinomial being used in our case it's four divided n +1 factorial multiplied by the maximum of the absolute value of the N +1 derivative of the function over the interval from the point of approximation to the center so I could say that Z here is an element of the interval from zero which is the center of our tailor polinomial to 1/4 which is the point of our approximation so the maximum value that the absolute value of the N Plus one derivative of the function f takes on in this space that's what this is and that's the error bound all right so let's start to plug in some pieces X in our case is 1/4 and C the center of our tailor polinomial is zero so this numerator is going to be the absolute value of 1/4 which is just 1/4 to the power of n + one again n is the degree of the tailor polinomial which in our case is four 1 2 3 4 we're not using any of the other terms so this is going to be raised to the power of 4 + 1 which is 5 divided 4 + 1 factorial so 5 factorial then All That Remains is to put at least an upper bound on the maximum of the absolute value of the fifth derivative of the function over the relevant interval and for that we can use the picture this is a graph of the absolute value of the fifth derivative of the function now the interval we're interested in goes from zero the center of our tailor polinomial to 1/4 and you can see that 1/4 is right here 1/4 24s 3/4s 1 very convenient graph so this is the interval in question and we can see over this entire interval the derivative is or excuse me the absolute value of the fifth derivative is less than 40 that's a very convenient upper bound so coming down to our lrange error bound we can replace this with 40 the maximum of the absolute value of the fifth derivative of our function over this interval is certainly less than 40 so this is a valid upper Bound for the error all right now all that's left is to evaluate this 1/4 to the^ 5 what is that well 4 squared is 16 so 4 cubed is 64 so so 4 the power of 4 is 256 so 4 to the power of 5 is 1,24 so this is 1 over 1,24 now the 40 can be canceled out by some pieces of 5 factorial 5 factorial contains 5 * 4 * 2 and that is 40 so we could cancel those out that would leave in the denominator three and well that's it so let's just put it there ah and look at that this simplifies to 1 over 372 1 over 372 and 1 over 372 is less than 1 over 3000 and thus we've shown that the error in our fourth degree tailor approximation for the value of the function at 1/4 that error is less than 1 over 3,000 which is exactly what we set out to prove and that will complete our solution to free response question 6 from the 2011 AP Cal BC Exam finally we will do a couple of problems on Arc Length recall that the formula for the arc length of a function f ofx from A to B is the integral from A to B of the square < TK of 1 plus the derivative of f squared that's the derivative squar + 1 in the square root good luck this is true response question 4 from the 2008 AP Cal BC formb Exam let F be the function given by F ofx = kx^ 2 - x Cub where K is a positive constant let R be the region in the first quadrant bounded by the graph of F and the x-axis part A asks us to find all values of the constant k for which the area of R equals 2 now we are specifically interested in the first quadrant which allows us to put some restrictions on X and K firstly if we're in the first quadrant of course X has to be at least zero but if we're in the first quadrant we need y or F ofx to also be at least zero now that means that kx^ 2 - x cubed which can be factored as x^2 * K - x that has to be at least zero and thus K has to be at least X otherwise this term would force this to be negative we need we need it to be non- negative so K has to be at least X and we know X has to be at least zero because we want this region to be in the first quadrant with that in mind we can integrate this function from 0 to K to get the total area in the region because it is precisely when x equals K that the function will leave the the first quadrant becoming negative so our region R would look something like this going from0 to xals K so we'll just integrate from 0 to K and then we can set that equal to two and solve for all possible values of the constant K so we are integrating from 0 to K what we're integrating is this function which is easier integrated not in factored form so kx2 minus X cubed and we are integrating this of course with respect to X so we'll just go ahead and use the power rule here that's going to give us K over 3 x cubed minus 1/4 x 4 we are evaluating this from 0 to K when we plug in the lower bound of zero everything's just going to go away so all that's left when we evaluate this is K over 3 * K cubed which is just k 4 over 3 and then minus 1/4 k 4 so in total once we get common denominators this is going to be 4 K 4/ 12us 3 k 4 over 12 which is just going to be k 4 over 12 we need this to equal 2 because we want the area of the region to equal two which then forces K to be the fourth root of 24 if we took the negative of this and then raised that to the 4th power we would also get positive 24 however we were told that K needs to be a positive constant and of course we also know that K is positive because we are in the first quadrant thus k equal 4th root of 24 is the only possibility that was part a moving on then to Part B for K greater than zero right but do not evaluate an integral expression in terms of K for the volume of the solid generated when the region R which we were just discussing is rotated about the xaxis so this will be pretty easy because it's just using this region from the previous part of the problem and we are revolving it about the xais so we would have some shape looking something like this all the cross-sections are circles where the radius of each circle is just the height of the function at the x coordinate in question thus the volume of this solid will just be the integral from 0 to K of the cross-sectional areas pi r 2 like we said the radius of each cross-section each circle is just going to be the value of the function so kx^ 2 - x Cub 2ar because we have to square the radius and we're integrating with respect to X that's it that's the volume finally on on to part C part C says for K greater than zero write but do not evaluate an expression in terms of K involving one or more integrals that gives the perimeter of R so coming back to our silly little sketch here let's erase the blue part because that's not relevant anymore we are looking for the perimeter of this region R so it would be easy to just say let's write the integral for the length of this Arc from here to here but we don't want to forget about this part of the region this part has a length of K because it just goes from Z to K and then this part on the top we'll just have to use an integral for that so the perimeter which we might write is p will be K the bottom of the region R plus hopefully you remember the arc length formula we need to integrate from zero to K and then what we're integrating is the square root of 1 plus the derivative of the function squared so let's just go ahead and find the derivative of the function and that's what we will plug in there we have the function written here so let's just take the derivative of that so D DX of K x^2 - x cubed we'll just use the power rule that's going to be 2 K * x - 3 x^2 and let's see if we can squeeze that into the parentheses and that's it that's an integral expression for the perimeter of the region R again the ark length formula has us integrating across whatever the relevant bounds are in this case from 0 to K and the integrand is the square root of 1 plus the derivative of the function squared that is all in a square root just to make sure that's clear and that's it that will complete our solution to free response question 4 from the 200a AP Cal Form B BC exam this is free response question four from the 2011 AP Cal BC formb exam and it is a pretty big looking problem let's take a look the graph of the differentiable function y equals f ofx with domain from 0 to 10 inclusive is shown in the figure Above This is a graph of f the area of the region enclosed between the graph of F and the X AIS for X between 0 and five we can see that's this region here that area is 10 so I'm just going to jot that down and the area of the region enclosed between the graph of F and the xaxis from 5 to 10 that's all this space here is 27 I'll go ahead and write that as well the arc length for the portion of the graph of f between x = 0 and x = 5 this part here is 11 so I'm going to go ahead and and jot that down as well I'll just put a little green Arc that has a length of 11 and then it also tells us the ark length for the portion of the graph between x = 5 and x = 10 is 18 I will draw that in Red so from x = 5 to xal 10 that arc length is 18 the function f has exactly two critical points that are located at x = 3 right there and x = 8 which we see right there a asks us to find the average value of f on the interval from 0 to 5 so the average value of our function f over this interval we see that average value should certainly be negative let's take an integral and calculate it so for part A the average value of a function is found by just integrating that function so the integral of f ofx over the relevant interval in this case that's from 0 to 5 and then dividing by the interval width the width of this interval is 5 so divide by five or multiply by 1/5 so this will be 1/5 multiplied by the integral of f ofx from 0 to 5 is the area under the curve which was given to us as 10 although we can see that's underneath the x axis so that area should actually be represented as -10 so our final answer is -2 that is the average value of the function over the interval in question Part B says evaluate the integral from 0 to 10 of 3 F ofx + 2 show the computations that lead to your answer this is pretty straightforward we're just going to have to use some basic integral properties the first thing that we can do with this integral of a sum is split it up into a sum of integrals so this is the same as the integral from 0 to 10 of 3F of X although we can take the three out of this so we might as well 3 * the integral of f ofx from 0 to 10 DX and then separately we can add the integral of 2 from 0 to 10 then this will be 3 multiplied by the area underneath F ofx from 0 to 10 and we can split that up into two integrals because we understand there is a bit of negative area from 0 to 5 but then from 5 to 10 we have positive area so let's split that up into these two integrals because we know the values of these individual integrals so we've got that and then the integral of 2 from 0 to 10 that's just a 2x10 rectangle so that's just plus 20 so then this will be 3 multiplied by the area under the curve from 0 to 5 we already know is -10 and we were told the area under the curve from 5 to 10 is 27 and we know that is positive so this would be + 27 and we still have + 20 at the end -10 + 27 is 17 * 3 is 51 and 51 + 20 is 71 so 71 is the final answer all right let's move on then to part C let G of x equal the integral from 5 to X of f of T DT we are asked on what intervals if any is the graph of G both concave up and decreasing and we must explain our reasoning now G being concave up means that the value of its derivative is increasing and it being decreasing means that the value of its derivative is negative now that's useful because the derivative of G ofx by the fundamental theorem of calculus the derivative of G ofx is just our function f ofx which we have a graph of so we're looking for where f ofx is negative and increasing we want it to be negative because that means that g is decreasing and we want F to be increasing because f is the derivative of G so if f is increasing G is concave up so here's this information summarized we need G Prime to be negative and increasing but remember G Prime is just F ofx which we have a graph of so let's see where is G Prime which is f negative well it is negative from 0 to 5 so G Prime is negative on the interval from 0 to 5 then where is G Prime increasing again G Prime is f so let's look at the graph of f we can see that g Prime is increasing from here this critical point at xal 3 to here this critical point at xal 8 so it is increasing on this interval from 3 to 8 so it's going to be the intersection of these intervals which is from 3 to five that's the intersection of these intervals where the necessary requirements are both met so G would be decreasing and concave up and there is our conclusion G is decreasing and concave up when X is between 3 and 5 finally Part D the function H is defined by h ofx = 2 FX / 2 the derivative of H is it's easy to find using the chain rule but it just gives it to us frime of x/2 because you would take the derivative of F and then multiply by the derivative of the inside function the 1/2 would cancel out the two we're asked to find the arc length of the graph of yal H ofx from X = 0 to x = 20 notice that H is based on our function f which is defined from 0 to 20 but the input of f in H's definition is cut in half which is why H can go from 0 to 20 when we plug in xal 20 f is going to receive an input of 10 which is right at the end of its domain I suppose we are going to have to shrink our work for the previous parts of this problem in order to fit the final part so I will start writing Part D over here we are concerned with arc length from 0 to 20 so that's going to be the integral from 0 to 20 and we're finding the Arc Length of H ofx so the way we find arc length of a function is we are integrating the square root of 1 plus the functions derivative squared now we know the derivative of H it's just frime of x / 2 so this is 1 + FR Prime of x / 2^ sared now since we have an inside function X over2 this is a good time to bust out some U substitution If U equals x over 2 then du equal 12 DX and then we can say that DX is equal to 2 duu so then we can replace DX with 2D and write everything else here in terms of U we'll also be changing our bounds if x starts at zero then U also starts at zero if x ends at 20 then U ends at 10 so this is equal to the integral from 0 to 10 now that we have bounds in terms of U and we are integrating 1 + f Prime of x/ 2 which is u^ sared and we're integrating with respect to U but we also need that factor of two let's just put that in front of the integral and now we are basically done because we have written this integral so it's actually just an arc length of f now from 0 to 10 we already know what the Arc Length of f is from 0 to 10 because we have information about these two pieces we know that the arc length from 0 to 5 is 11 and we know that the ark length from 5 to 10 is 18 so coming back down to this integral which is now just the Arc Length of f from 0 to 10 it's going to be 11 + 18 the arc length from 0 to 5 plus the ark length from 5 to 10 11 + 18 is 29 * 2 is 58 all right there we go that completes our solution to free response question 4 from the 2011 AP Cal BC Form B exam well that was fun once again I'd like to thank you for watching and encourage you to donate if you found this video helpful and would like to help me 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