hi everybody it's Jamil sadiki here I teach math at East bridgewire Junior Senior High School in East bridgewire Massachusetts my name is Becky Barn I'm a math teacher from West Springfield High School in Springfield Virginia and today we're gonna be looking at task model number two so as we start we always want to kind of give you guys kind of an overview of what each task model is all about today like I said is going to be free response question number two and that's going to be modeling a non-periodic context so things to keep in mind in this particular task model you're always going to be scored out of six points it's going to require a graphing calculator which always should be set in radiant mode and and let's be clear about this just keep it in radiant mode all the time there's never a reason to take it out of radiant mode when you're in AP precalculus we're going to be focusing on the content from units one and two this is always going to be presented in a real world context it may include the following function types poomi peie wise defined functions exponential functions and logarithmic functions it's always going to include three parts that require constructing a model working with average rates of change and just find conclusions about that model and when you're thinking about those conclusions one of the big things that they're going to focus on is the limitation of the model so you know sometimes we write this model and it only works for a certain region or certain area that's something to keep in the back of your mind as we're going to be looking at these task models so that all being said let's take a look at question number one a biologist began studying the impact of conservation efforts on a certain species of animal in 2015 in 2015 which we're calling T equals z the population was determined to be 230 animals in 2021 when T is equal to 6 the population increased to 580 animals for T greater than or equal to zero the population of the species in a region can be modeled by the function P which is given by P of T equals a + b * the natural log of t + one where P of T is the population T years since 2015 so if we take a look at part A Part One let's give it a read it says use the given data to write two equations that can be used to find the values for constants A and B in expression for p of T so one of the first things I notic is all we're doing is writing two equations we don't actually have to find the values of A and B at this point in time now when Jam wrote the problem I noticed that at a time value of zero there's 230 animals so I can write the expression that P of 0er has to be 230 and at t equal 6 we have 580 animals so I can write the expression that P of six is equal to 580 my goal is to write an equation in the form of P of T = A + B Ln of t + 1 so when I substitute in a value of zero for T I should get an output value of 230 that will let me write my first equation so 230 has to equal a + b Ln of 0 + 1 my second equation I'll get from plugging in six and when I plug in six I should get an output value of 580 so I get 580 = A + B Ln of 6 + 1 and jam I notice I'll mention real fast my six there looks really wonky on your AP exam guys make sure you're writing clear for your AP readers because it's not your normal teacher who's reading this for you so take that extra moment don't like might be like me and make out wonky six that looks like a be make sure you're writing clearly so that they can read what you intend for them to read so now we can go ahead and jump into part two says find the values for a and b as decimal approximations so we already have that equation that 230 is equal to a plus b Ln of now it was 0 plus 1 I can simplify that to just one and then when I my second expression I had 580 equal a + b Ln of we had six plus one I can simplify that to seven now Jam I have a feeling that one of these is going to be a little easier to work with or simplify first which one would you pick yeah I would go with that first one with that Len of one right there I think that's going to be the easiest thing to work with because you know L of one we're looking for that exponent again that's going to make our base one L one is going to be zero that's right so e to the what power gives me one e to the zeroth power I agree so I can get rid of this term that multiplies to zero so I already know that a equals 230 so now I can substitute that a value into my other equation I have 580 is equal to 230 plus b Ln of 7 I want to get that b value by itself so I'm going to start by subtracting that 230 and that's going to give me a 350 now j keep an eye on me sometimes I make arithmetic mistakes so so keep an eye on me here so I've got B time Ln of 7 and to isolate that that B I'm going to divide by Ln of 7 so now I have B is equal to 350 ided Ln of 7 now we have to be careful here I have an exact value for B but this question didn't ask for an exact value it asked for b and a as decimal approximations so this is something that I now need to type into my calculator and when I type this into my calculator I'm going to get a value of 179 864 and I'm going to go ahead and trunk that truncate that or even if I round it I get the same value here here why don't I take a shot at this next part part B1 use the given data to find the average rate of change of the population animals per year from T equals 0 to T equals 6 years express your answer as a decimal approximation remember show the computations that lead to your answer so when I'm looking at this I'm reminded of our earlier sessions when we're talking about the average rate of change keep in mind the average rate of change is always the difference of the outputs over the difference of the inputs mathematically defined as F of B minus F of A over B minus a so when we apply that average rate of change to this problem we're talking about the function P we're looking at P of 6 minus P of 0 over 6 - 0 we know the values of P of 6 and P of 0 that's going to be 580 and 230 so when we set that ratio up we've got 580 minus 230 over 6 - 0 which then turns into a decimal answer of 58333 now guys on your AP exam a lot of times you're start going to start by just finding this average rate of change value and then after that you're going to imply interpret or use this in some way so we're going to add a quick spin-off question for you really here quick here it's going to be to interpret the meaning of your answer from part one in the context of the problem so jam said what we're really doing is a change in the output over a change in the input and the output values are animals and the input value is time in years so what I have here is that on average the population of animals increases by 58.33% on average that my T value increases by one I'm increasing by 58.33% 580 animals we also learned from the previous spin-off question that we did that the average rate of change gave us 58333 animals per year that means every time my T value increases by one we're increasing by 58.33% so I've got P of 6 as my initial or my starting value for this particular problem to get from time 6 to time 7 I would need to add that 8.33 once because as I increase by one year I'm increasing on average by 58.3 33 animals and then to get to time eight I need to add that 58. 333 one more time so if I add that average rate of change two times that will give me an approximation for p of8 if I go ahead and type that into my calculator remember we know that P of six is about uh is 580 I add on that 58333 twice and that gives me to a value of 696 666 dot dot dot if I'm using my exact value from earlier now remember this question is actually asking us to do this to the nearest whole number so I can't leave my answer like this if we were to round to the nearest whole number we would leave this answer as 697 animals but I could also see us leaving this as 696 because we have that partial animal there so in either case you get full credit if you left this as 696 or 696 seven Becky thank you for reading part two of that problem why don't we go ahead and take a look at part three in this part we're asked to consider the average rates of change from P of P from tals 6 to T equals n Years and we're told that n is going to be greater than six for this are these average rates of change less than or greater than the average rate of change from tal 0 to t equal 6 years found in part one explain your reasoning now before we jump to this I'd love to go back and do a rapid review so let's take a look at this next slide where we're going to do the rapid review on exponential functions because the output values of exponential functions in general form are proportional over equal length input value intervals exponential functions are always increasing or always decreasing and their graphs are always concave up or always concave down what that means is if you know you got an exponential function that exhibits increasing behavior in one place it's going to be increasing for its entire domain and the same is true if it's concave up in one place concave up for the entire domain and the converse of that will also be true now let's think about the inverse the logarithmic function now remember how are these related these are reflected over the line yals X compared to what we just saw so because they're the inverse function of exponentials logarithmic functions are also always increasing or always decreasing and their graphs are either always concave up or always concave down with that information at fingertips Becky why don't you go ahead and try to take a look at this question and take us through it so guys as a reminder you have a graphing calculator for this so one of the things you can do is actually look at the graph for this function and you'll notice that our function is logarithmic and it's also increasing and this is partially in part due to the fact that the leading coefficient or the coefficient for my logarithmic term is positive and then the coefficient for the t is positive as well but because p is logarithmic and increasing our graph here is going to be concave down because it's concave down we actually know that as T increases the average rates of change are going to be decreasing over equal length input value intervals and we're going to talk a little bit more about this in our upcoming session this means then that the average rates of change from tal 6 to T equals n are less than the average rate of change from T equals 0 to tals 6 Jam I think we've got one more part do you want to take this one for us sure Becky I'll handle this one in this last part part C the biologist determines that the region can sustain a maximum population of 800 animals explain how this information can be used to determine the domain limitations for the model P well one thing to keep in mind in this problem is that we're given that maximum population right so we know that no matter what we're not going to go above an output value of 800 to me that says let's find out where our model predicts we're going to have 800 animals so we're going to set P of equal to 800 now from our previous sessions we've talked about how we can use our technology to help us one of the great functions we have on that calculator is that intersect function where we can draw two graphs and find out where they cross so using our technology we can see that from this graph those two functions are going to intersect at the input value 22785 so taking that information and thinking about what it means for our problem we are going to say that since 800 is the maximum sustainable population of the animals and because P of T is an increasing function where P of that intersection value 22784 equals 800 that the domain for this particular function is going to be 0 to 22784 and Becky if I remember correctly when I did this on the calculator that's the truncated answer right I think we rounded this you also could have got an answer 22.78%