[Music] hi everyone welcome to the methods one through September lecture um my name is m and I'll be taking the lecture today but just before we begin um i' like to talk to everyone about HR notes so HR notes have been offering heaps of free resources to students across Australia since 2007 um and the lecture that you're watching today um we've been offering lectures like these since 2015 um because the lectures that we offer are in line with our mission to help students as much as possible so that they have the best opportunity to succeed in their studies um we also offer um a bunch of other resources such as study notes lectures online discussions um engaging online revision videos newsletters HR calculators articles and heaps heaps more so if you're interested in checking out these resources please have a look at the info do Under the resources section of this lecture page and if you have any other questions regarding these resources please feel free to put it into the Q&A so that I can answer them for you okay let's get started with today's lecture so we have a lot to go through today um but just before we begin um I like to quickly introduce myself so my name is mot bull um and I graduated VC with an atar of 99.80 and I've been studying Bachelor of biomedical science at monach um and I'm a maths and science shter at truth smart and the subjects that I personally did throughout VC were chemistry maths methods specialist maths English biology and further maths um so yep let me know if you have any other questions otherwise we'll get started with today's content so today we'll be going over functions and transformation s calculus and also a little bit of probability so there will be no designated Break um in today's lecture so pause whenever you need um and we will be using the Q&A software so if you have any questions please pop them through so that I can answer them um as I will be monitoring the chat during the course of the presentation so feel free to ask me any questions um so that I can answer them as they come up okay so let's get started with the function side of things so for functions we're going to be looking at a review of linear and coordinate geometry um quadratic cirular functions and inverse function so the overall um aim of this lecture is to help you get started with your revision for the exam and also um act as a little bit of a head start for year 12 okay let's get started with linear equations which is our first topic so as you might already know linear equations are uh a polom of degree 1 so the X variable here will have a power of one which will be the highest degree in the linear equation and this is the general sort of equation that you will see y = mx plus C where m is the gradient of the graph the gradient or the slope of the graph which describes how steep a graph is and C is the Y intercept so the Y intercept is essentially what the y value is when the x value is zero okay so there's a couple of formulas that you might need to remember a lot of them are actually um found in your formula sheets so you will have to have them written down somewhere or memorized for your exam one so the first Formula is finding the distance between two points so on a linear plane if we are given two discrete points we can view use this formula which is a distance formula to find the distance the physical distance between these two points so the distance formula is square < TK X2 - X1 2 + Y2 - y1 all 2 and the way that it sort of works is it uses the principles of the Pythagorean theorem so remember Pythagoras is y oh sorry a 2 + b^ 2 is equal to c^ 2 it's sort of using that principle to sort of come up with this formula here where a in this case is the difference in the X values and B is the difference in the Y values and instead of finding c^ 2 we're finding d 2 and then to find D we just square root whatever a 2 + B2 is so that's essentially how this formula is dered so just have to memorize this and apply this whenever you get a question asking about the distance between two points this um this sort of concept does come on the exam fairly often so yeah just definitely something that you just have to remember okay sorry another formula that um everyone needs to know is the midpoint formula the midpoint formula is essentially where you're finding the midpoint between true points so when you given two points on the linear plane um and you essentially have to find the X midpoint and the Y midpoint and the way that this formula sort of works is it sort of finds the average of the Y values and the average of the X values and that's essentially where the midpoint of the two points lies so as you can see here the x coordinate is just going to be the midpoint or the average of the true X Valu so X1 plus X X2 / 2 that gives you the average or the midpoint of the x v of the X values and if you look at the um vertical column we can see we have y + + Y 2 / two which is the average of the V variable so again just another formula that comes up um um all the time in the exam um the next Formula is the gradient formula so if you're given true discre points on the linear plane you need to be able to determine the gradient of the line and ultimately the equation of the line so the formula for the gradient um first of all what does the gradient represent well the gradient is just the rise over run or essentially looking at how do the Y values change as my X values increase or decrease that's essentially what the gradient is referring to or the slope of the graph and basically we use this sort of principle of rise over run alternatively change in y over change in X which can be found by using this formula so when you have two discrete points so point a and point B to find the gradient between these two points you will have a coordinate system of X1 x uh X1 y1 and X2 Y2 and basically when you have this you can then find out the um gradient between these two points by using this formula over here and Y 2- y1 just gives you how much is the Y values changing um by doing Y2 - y1 and in the denominator X2 - X1 will tell you how much your X values are changing but in order to use this formula you need to be uh you need to know two points on the graph alternatively if they don't give you any points they just give you a graph then you need to identify two sets of points and then go from there okay so so here is just um another sort of way of thinking about the gradient which is tan Theta so we will come back to Circle functions a bit later on but as you might all remember tan Theta is equal to opposite over adjacent so if we have an angle over here the opposite is going to be the rise whereas the adjacent is going to be the run and essentially um tan Theta is going to be opposite over adjacent which is going to be rise over run and we know that rise over run is equal to the gradient or the M value so that's why tan Theta is often equal to the gradient so if you know an angle you know instead of knowing two points you know the angle that the sort of graph makes with the you know a parallel axis sorry a perp um of horizontal or vertical axis or um you might or you might know this angle instead you can then use that and tan Theta to determine what the rise of a run or the gradient is equal to so another sort of concept that gets tests quite often okay so now we'll move on to pols in general so um linear equations was a polom um of degree 1 you'll also be required to look at um and also sort of do calculations with pols of higher degrees usually it doesn't get much um greater than aquatic I've never really seen a question on the exam where they use um you know anything above aquatic or um a degree above four so you'll usually be seeing these three um pols really often so quadratics where the highest degree is true cubics where the degree is three and quartic where the degree is four so all of these equations um will look Alo all of these um pols will look a little bit different when you graph them but they sort of have a general pattern which will come back to in a second but there are three different forms in which you can write the equation for all the pols so you either have the expanded form as shown here where you have you know a b c so you don't have any brackets essentially so you have all the coefficients listed down in um in um order of degree so you have the highest degree and then um you know sequentially going to the lowest degree which is X to the power of0 so if something is missing and that's still a polinomial it's still going to be a cubic if the highest degree is three for example let's just say we didn't have bx^ S it will still be a cubic but this time we just won't have anything with B X2 so we only look at the highest degree to determine whether it's a cubic quadratic or a um or a quatic and the only information that the expanded for will tell you is the Y intercept so any constants for example C D and E are the Y intercepts of that you know particular graph so this is the um information that we can get from the expanded form next we have the factorized form so the factorized form is when we have you know we factorize the polinomial either using a manual method or using a calculator and this will look something like this so we'll usually have a dilation factor and then we'll have X plus something x minus something etc etc and this is called factorized form because um essentially you're finding the factors of the graph so X plus b x plus C are the factors of whatever this quadratic is similarly X plus b x plus c x plus C are the factors of this particular cubic and what we can determine from from this is the X intercepts by using the null factor law so essentially let's say if we're just looking at the quadratic we have a X plus b x plus C and when we let that equal to zero to find the X intercepts oops sorry that will essentially just be X um x + b equal to Zer thereby x equal to B so that's one of the X intercepts and we also have x + C = to Z and that will give you X is equal to C so the factorized form is probably the hardest one to get into if you're just given um an expression in for example expanded form so most of the time you will be using a calculator but you need to also be able to do it by uh manually and that's usually through the um polinomial division methods which which we'll have a look at a bit later on um and then the other form so here is um graphing so here's a little bit of information about what the graph will look like when you're graphing it so if you have a linear Factor so when you put it in factorized form and you have a linear factor x - A what that means is that at that point so let's say um we solved it and we get an x and. xals a so at the point where the graph crosses the x- axis at the point xal a we're just going to have a linear line going through it so that's what the a linear Factor tells you a linear factor is basically where you have the power of one if you have a quadratic Factor what that tells you is that the graph actually forms sort of like a quadratic at that point so it can either be a quadratic like this or it can be a quadratic like this but essentially you approach that value and then you return essentially so whereas the lar you're just going straight through it like a linear line um if it's a power of two for the factor then you are um doing a quadratic you're forming a quadratic sort of thing and then if if you have a cubic the graph essentially forms a point of inflection or a stationary point of inflection to be more specific at that particular point so what this will look like is if you have a really big graph something like you know y = a let's say x - b x - c cubed let's just say x - D to the power of 6 so another thing that you need to know is if it's an even power for example squared or to the power of six we are going to produce something like a quadratic at that particular point when it crosses the x- axis if we have an e odd power for example cubed um to the power of 5 to the power of s so on and so forth we're going to have a stationary point of inflection except for positive one for example here so positive one is the only odd power that won't give you a um point of inflection instead it will just give you a linear line or or a value that or a line that sort of just crosses at that point so this is how we um sort of graph pols when we're given something in factorized form another form that we usually use is the stationary Point form so a stationary Point form is essentially where the graph is giving us information about the stationary point of inflection so for example if we have a x + b^ 2 and like I said if it's squared then the graph needs to form a sort of like a you know like a um a quadratic at that particular point so it's at b c so therefore the turning point will actually be at B comma C and we'll explain why that's the case when we go to transformation so therefore we will have a turning point at this point depending on what a is so that's what that will look like so um so this point essentially just describes the turning point of the formula so if a p if a polinomial can be expressed in this form so not all pols can be expressed in this Turning Point form only some can but if they can be expressed in this form the polinomial will only have one stationary point which will at negative BC so no other you know turning points or anything like that if it can be written in this form similar here we have a point of inflection at negative BC just draw that again so if we look at the cubic now we have a turning point at negative BC so that will look something like this so it's not really a turning point it's more more of a stationary Point whereas for the CCH it will look something like this but it will usually be very sort of narrowish but again when you're graphing it doesn't really matter they will look the same so this will also look like this um because of what I just mentioned on the previous slide if you have an even power it looks like a quadratic if you have an odd power it looks like a cubic okay so now we'll look at the point um at the process of completing the square so completing the square is essentially just a process that can be used to get something that is in the expanded form into the Turning Point form and why we might need to use this um is because let's say we have an expandion form and we want more information we don't know how to graph that function exactly so we need to convert it into the complete the square form and we can only do that for quadratics we can technically do it for cubics and other polinomial but it's not required for the VC level so here we have a turning point formula um so it's just going to be x + b/ 2^ 2 - b^2 4 + C so what is B and what is C well if you have a polinomial function in the expanded form Y = 2 a x 2 + b x + C well this the number or the coefficient in front of the X and the constant are going to be your B and C so essentially you can just put it into this formula and you should get your complete the square um sort of value and a really good thing about when you complete the square is not only will you find the turning point but it will also make um finding the X intercepts really straightforward and simple by just using simple algebra because you can't solve a quadratic equation by using simple algebra when it's in this form where however when it's in this form you can use Simple algebra okay so Express the following quadratic in the follow in the form y = a x + B2 + C so the first first thing that you need to do is you actually need to divide out the three here sorry so we need to take out the three before we do anything else need to factorize the three out so it's going to be 3 x² - 2x um - 1 over 3 so now that we factorized oops now that we factorized that out what we need to do now we can actually use the turning point formula so when we actually when we actually do the turning point we can actually just straight away use a turning point formula or we can look at so what we need to do is we need to sort of make this a perfect squar so perfect squar is x^2 - 2 BX - b^ 2 or sorry yeah plus b squ yeah um yeah sorry plus b square so therefore this in that formula so x^2 - uh BX or B here is actually true oh sorry negative true so therefore it will be X - 1 which is b/ 2^ 2ar minus b^ 2 um over 4 so b^ 2 is going to be um 4 over 4 which is just 1 minus C so that is essentially how you can put this into the Turning Point form and then if you simplify it further you will get Y is equal to um 3 x - 1^ 2ar -1 - 1 over 3 is actually going to be 4 over3 so therefore it will be 3 X - 1 2ar - 4 so now that we have put it into the Turning Point form let's go ahead and actually determine what the turning point will be well it will just be this so this is B it will just be negative B so netive B is1 which is postive 1 and this c will just be C so4 so this is essentially how you can determine the Turning Point by um using the expanded form of a quadratic for cubics and other types of um pols you will have to use your calculator for that okay so next we'll look at the quadratic equation so quadratic equation is just a equation that is derived from the quadratic formula so I've put the derivation on the right hand side which you can have a look at but it's not required so derived from the quadratic formula you can straight away find the X intercepts of a um equation of a quadratic equation if you know what the a b and c values is and you can only determine that when it's written in the um expanded form so the quadratic formula is just x = b + - b^ 2 uh s < TK b^ 2 - 4 a c over 2 a pretty straightforward just knowing that um so this formula is usually not given in any type of formula sheet so you have to memorize this or have it written down somewhere in your um summary book um okay so here is an example of how we can use the quadratic formula so basically if we're given a function for example in this case 2x^2 - 4x - 3 we can easily determine the um X intercepts by just subbing in the relative values a b and c into the quadratic formula so a so the main thing I think about this formula is first of all the algebra itself so going from here to here that's the main important thing that a lot of students mess up if they're doing it by hand another thing that the students mess up is actually finding what a b and c is so a here is going to be true B is going to be -4 and C is going to be -3 so now that we can determine it it's so much more easier to just substitute those values into the formula but obviously don't do it blindly um you know if um know that sometimes you can make algebraic mistakes mes and if you get a really weird looking value then you might have made a mistake but most of the time the values are weird so sometimes it can be hard to judge but just look at your algebraic um algebra to see if you've made any superficial mistakes okay so another thing that can be driven from the quadratic formula is the fact that if we have a square root the number inside the square root will really determine if we have Solutions or not because a property that you need to know about square roots is that any number inside a square root has to be greater than zero if we have a sorry greater than or equal to zero because if we have a number that's less than zero for example -2 we can't take a square root of a negative number so in that case it's really important um so what that sort of um derives is something called the discriminant so the discriminant can straight away allow you to determine how many solutions a value might have and what I mean by Solutions is if you look at these graphs over here Solutions refers to how many times does the graph Cross or intersect or yeah so Cross or intersect with the x-axis because that's essentially what an X intercept is when is it Crossing when and where is it crossing the x axis so if we have no solution the graph might be completely above the x-axis um for something like let's say X2 + 2 that is completely above the xaxis or completely below the AIS it will never even cross the x intercept so therefore there will be Zero Solutions or no Solutions and we have determined that in this case the determinant or the discriminant so discriminant is essentially just the number inside the square bracket of the quadratic formula or B ^ 2 - 4 a c and we have determined that if there is no solution this must be less than zero because what that will do is if we put this make less than zero into the quadratic formula we will get an undefined answer which is what we want we don't have any solutions to begin with if we are getting an answer then that means that we've done something wrong uh something wrong either in graphing or in terms of the algebra itself however if the graph has a turning point exactly at the x-axis then we will have exactly one solution and for in that case the discriminant will be equal to zero because if it is equal to zero our values are just going to be b + - 0 over 2 a which is essentially just B over 2 a because plusus 0 is the exact same thing so therefore only one solution if discriminant is equal to zero alternatively if we have two solution what that means is the graph is crossing the x-axis so it's going to Cross or intersect at two separate points and if they and if it is intersecting at two separate points the discriminant will actually be greater than zero which makes sense as well if it's greater than zero we can have negative v+ minus square root of that number or so it can be plus that number or minus that number over choice so two different solutions so a lot of the time um questions will give you random equation that will ask you to determine whether there will be no Solutions one solution or two solutions or they might give you a linear um expression with a random variable for example and they might ask you determine what that variable is if there is one solution no solution or two solutions so here's an example illustrating what I just said so here we have a linear sorry a quadratic equation Y = x^2 + mx + 4 where m is a real constant and we are trying to determine the values of M such that the graph has exactly one x intercept so what that means is the discriminant so first you always relate these types of questions with the discriminant so therefore if it is exactly one intercept the discriminant should be equal to zero so X intercepts will occur when Y is equal to Z so let's just first of all find what the X um x- axis is uh sorry the X inter are so if you want only one x intercept then we want the above equation to only have one solution this means we want the discriminant to be equal to zero so that's what we need to first of all identify you know if it says one solution two solutions three uh you know no Solutions we need to determine what the discriminant is going to be in that case and then we just solve the equation as you normally would so we have x^2 + mx + plus 4 so you have your a value which is 1 your B value which is M and your C value which is four sub it into the discriminant formula and let that equal to zero and you should get m is equal to pos4 as your answer so I hope that makes sense again let me know if you have any questions for the functions part of today's lecture we're now going to be moving on to circular functions which is again another topic that a lot of students struggle with throughout year 11 and your 12 methods so um a circular function is basically looking at a circle with a unit of um one or basically a radius of one sorry not a unit of one a radius of One units so there are 360° in a full circle so there are two ways of measuring angles so we're mostly used to degrees up to year 11 but now we will be introduced to another measure measure of angle which is called radians so radians instead of measuring the angle between two points it instead measures the radius um that sort of constitutes that angle so if we have an angle of let's say 40° so that's essentially just describing this sort of angle over here whereas a radian actually describes this circumference distance over here so what does this length actually measure so that's the differences between a um radian and a degree although um this is sort of like very preliminar um and you don't really need to know that um for any type of sort of assessment the main thing that you should know is what's going to be on the next slide and that's how to convert from degrees to radians but just a little bit of an introduction to what an radiant is a radiant is actually donated by the small symbol um small C the subscript but often it's going to be ignored so you're rarely going to see radians written with the c on the top and they usually have Pi in them as well just because they're measuring the circumference so for a whole circle so 360° the radian equivalent for that would be 2 pi because the circumference of a circle is essentially 2 pi r and r in this case is going to be one so unless specifically T you answer in degrees your answer in any questions in math methods always always always have to be in radians so it's just an assumption that your answer is in radians and um in stand as a um standard value or an exact value you never write your answer in decimals unless told to do so so here are two really important formulas can also be done in your calculator but essentially if we're going from radians through degrees we multiply the the angle that we have so whatever angle we have in radians we multiply that by 180 over Pi so the thing about this is if you don't remember the formulas which ones which well you know that radians usually have 2 pi r in them oh sorry Pi um in there um in them so you if you want a degree you want to remove that Pi so you need multiply by something which has pi in the denominator to effectively eliminate that so that's essentially how I memorize which formula to use and when because you will never end up with something like Pi squar so you should never end up with something like this that means you've done something wrong and going from degrees to radians we multiply the angle in degrees by pi over 180 okay so I'm just going to spend not a lot of time but just a decent amount of time on the unit circle so the unit circle is basically a circle which essentially has a radius of one and so on the number plane so the XY axis we have a radian of one oh sorry not a radian um a radius of one so therefore this point over here is going to be 1 comma 0 cuz the y- value is zero and the x value is one and essentially what the goal of this unit circle is true let's say determine any point along the circle so the coordinates of this point along the circle given that the only thing we know is the angle that it makes with the origin so we only know what Alpha is and we have to determine what X and Y are what the coordinates that the angle would make with this point over here on the unit circle so the way that we need to think about this is um first of all finding what you know this vertical length so this vertical length is y for this coordinate point and the X length is you know obviously X another way to think about it is in terms of the angle so this side over here is the opposite side for this angle whereas this side is the adjacent side for this Angle now if we think about it in terms of s and cos we know that s Alpha is equal to opposite over hypotenuse and what do you know hypotenuse is one so therefore it's just going to be opposite and what do we know about the opposite side well it's just y so therefore s of theta is equal to the y coordinate at this particular point and for the cause we can think about in terms of adjacent over one so um again sorry we can think about it in terms of adjacent over hypotenuse and hypotenuse is just one so just going to be equal to the adjacent and we know that the adjacent is equal to the x value so therefore s of the angle is equal to the vertical length because of the angle is equal to the horizontal length in a unit circle and 10 is just going to be sign or the Y value over the x value at that particular point so again I hope that really made sense but sorry but essentially the unit circle can be sort of simplified by this diagram over here so basically if we know the angle we can determine what these coordinates are on the number plane by using COS of Alpha and S of alpha so just quickly talking about the symmetry properties of the unit circle so let's just say if we're trying to measure this um an angle at this point over here so at this point the x value because we're on the negative side of the y- axis oh sorry the negative side of the x axis my x value is actually going to be negative so it's going to be negative X but my y value is going to be positive still it's still going up or above the Y the x- axis so essentially if I were to find what s is equal to and remember s is equal to y or the vertical length so it's just going to be equal to positive y so therefore at this point s is still going to be positive whereas cos which was the X length is going to be negative because equal tox and tan which is equal to sin / cos or y/x is going to be y netive X which is going to give you a negative value so therefore we've come up with an acronym to determine which um of the three um trig functions is going to be positive or negative in that quadrant so a means everything is positive in the first quadrant f means that only s is positive in this quadrant t means that only tan is positive in this quadrant and c means that only cos is positive in that quadrant so all of the other so s and tan will be negative in this quadrant over here because if we think about it the X values are positive but the Y values are negative and Y and and tangent is just y/x so therefore it will be negative whereas in this case both signs of the X values and also the Y values will be negative however since tan is y/x and both of them are negative the negatives will cancel out to just give you y x which is 10 okay and another thing that you know about the UN Circle actually repeats itself so it will go 2 pi and then 2 pi and then 2 pi and then 2 pi and so on and so it can also go NE negative so -2 Pi -4 Pi - 6 Pi so on and so forth so essentially if we have found this value over here for the angle in the first rotation so let's just say it is 150° for example so if we add 360° to the 150° we end up at the same point so therefore these quadrants or sorry these coordinates will be the exact same as well so therefore cos and S will not change regardless of how many rotations you do even if you're going backwards as well so you'll get Negative angles but your s and cosine value will still be positive so that's something really important to just identify um when you're doing questions like this so now we're going to be having a look at exact values quickly so there are actually three ways in which we can memorize the exact values so first is the hand method sorry the first is the table method which I personally prefer we also have the um triangle method which is kind of neutral I I don't really use that method and we also have the hand method so the hand method I personally don't know how to use um in the most accurate manner a lot of people do use it but I haven't really seen many people use it so table method is probably the best way of memorizing the unit circle and I've included it here so essentially what you need to do is for S and cosine start off with or um actually when you're doing the table just start off with sign you don't need to do anything with cos and tan so what you need to do is need to write some um just leave the numerator blank just write numerator which is blank over two for each of them so 0 Pi on 6 Pi 4 Pi on 3 Pi on 2 so just leave the numerator blank for now and write over two over two over two over two now what we need to do is going from left to right you need to do root 0 root 1 root2 root3 root4 um and stop at pi/ 2 so this way you you will get the exact value that it is equal to so I hope that made sense let me know if you need me to explain that again for C it's essentially just the reverse so you start off with the highest value so < TK 4 over 2 then you go < tk3 over2 < tk2 over 2un 1 over two and then zunk 0 over two and then tan is just going to be sign Top Value over the bottom value so for example for Pi / 6 it's going to be half over < tk3 and 2 which is going to be half * 2 < tk3 which is equal to 1k3 so just s/ 10 so the main one that I would say is probably memorize just one of these Co should pretty much be the opposite and then tan is just this over this so that's essentially how you can use the table method to memorize the um exact values but I would say if not um uh by the end of year 11 then definitely before the beginning of year 12 have this sort of memorized so that you're not struggling next year and you can focus more on the more harder aspects of circular functions so here is the triangle method so again you need to memorize two of these triangles you need to me memorize this triangle and then this me uh this triangle and also the values and essentially the way that this will work is for example for pi over 4 you can easily determine what cosine or tan pi/ 4 is because you know what the opposite why does it keep on going back what the opposite what the hypotenuse and what the adjacent is for example if I was to determine what s Theta is going to be pretty straightforward is just going to be opposite over hypotenuse or one over < tk2 but again you will need to memorize this triangle as well as this one over here as well so here is just an example of how we might do a question that asks us to evaluate um a circle function so let's say we're trying to evaluate sin 7 piun on 6 so the first thing that we will need to determine is what quadrant that lies in and 7 pi and 6 can actually be written as pi plus pi on 6 so I already know that Midway of the circle is pi so it's more or a little bit more than Pi so therefore this is actually going to be in the third quadrant and since we're finding the sign value and we look at the cost rule s is actually negative in the third quadrant so then what we need to do is from 7 Pi on 6 we need to find the base uh the base equ of the principal value and that's pretty straightforward so if you have 7 Pi on 6 is just going to be pi over whatever the denominator is so if you had like 7 pi and 3 it would just be pi and 3 if you had 7 pi and 4 it would just be pi and 4 for all of the values that's the principal value so essentially what you're finding now is just s pi on 6 but instead of it being positive s pi and 6 it's actually going to be negative because this value actually lies in the third quadrant so again I hope that makes sense so you're going to get half and for that you need to use your exact value table or your triangle method similarly if we have cos 2 pi on 3 first step is to determine what quadrant that is in so cos 2 pi and 3 is actually in the second quadrant and the second quadrant C is actually negative so therefore if I'm trying to determine what cos 2 pi and 3 will be equal to I need to first of all determine what the principal value is and like I said it's just going to be um Pi on the top over whatever the principal or whatever the denominator value is it's just going to be Pi on 3 and again since cos is negative in that quadrant it's going to be negative cos Pi on 3 to give megative half so again I hope that makes sense um otherwise do pop any questions into the Q&A okay so now we're going to have a look at a quick look at um sketching cyc functions so when you're sketching Circle functions there's a few key processes that you need to follow um so the first step is always to determine the amplitude period and GAP this will help you identify the shape of your graph you know how wide or how narrow it is which is given by the the period or how tall or um and short it is which is determined by the amplitude and also the Gap like I'll show you in the next slide will tell you where to place your important points and basically help you determine like where how your shape is going to look so you need to label the axes so once you've determined the Gap you need to label the axes with The Gap points if the Gap points are like pi and four you need to label the axes as Pi and four 2 pi pi and 4 3 pi and 4 4 pi and 4 5 pi and 4 so on and so forth depending on what the domain is and you and you'll always be given a domain because otherwise your graph um repeats itself for negative Infinity to positive Infinity next determine the starting point if not at 0 0 so if it's not at 0 there's been some form of translation you need to determine the starting point either by subbing it directly into the the function for example and then going from there and then you need to calculate the X intercepts so the X intercepts are really important to label as well as the Y intercepts um yeah so you need to be able to not only calculate the X intercepts by solving which we'll have a look at in a few slides but also you should be able to um uh uh sort of plot it onto the graph and lastly you need to smooth a uh sketch a smooth curve that passes through all the main points so once you have your Gap point something like you know maybe something like this you need to connect it in like a smooth graph oops my bad oops where am I sorry um I am over here so you have no one two three four so just bear with me while I draw this yep just make it really uh make sure it's really curved and everything otherwise um if it looks like you know something like this that's um you won't get full marks okay so here's an example of the function true Y = 2 sin 2x - 1 so first step determine amplitude period and the and the Gap value as well as the Y intercept so once you've determined those then you need to determine what your maximum and minimum values are and to determine your maximum minimum values all you do is you take the constant and then you plus minus the amplitude so -1 + 2 is postive 1 -1 - 2 is is -3 so then that you've done that you determined your initial starting point you determine your lower values um your Max and your Min then you need to label the data points using your gaps your Gap in this case was Pi um pi over 4 then you should label the axes using those points so you know pi and 4 2 pi and 4 3 pi and 4 4 pi and 4 so on and so forth until you fulfill the entire domain and then the way that your circle function is actually going to work is actually going to start off from the middle of the Max and minum data points then it's going to go up at One D um Gap point then back to middle in the next stop point then bottom in the next stop point and then up again in the next stop point so that's essentially why we find we find the Gap so the Gap is so useful in determining how my graph actually looks um and how it like structured and oriented okay so um now we'll have a look a quick look at asmt um at the tangent function so the tangent function again looks very similar in form to um coose and S functions however in this case the period is actually going to be pi/ n because if you remember back to the unit circle 10 hand was positive here and also here so if this angle is the same as this angle over here or relatively similar there's no point um in you know doing plus 2 pi because you'll miss some of the values so then we just do add pi to get to that value and then add pi to get to the next value and so on and so forth now another thing that you need to know about the tangent function since it's defined at some of the values namely at Pi on 2 we will have ASM tootes and ASM tootes are essentially a value that your graph is approaching but will never reach and the way that you can determine where the first asmor is if there have been some form of Transformations is you let whatever is inside the brackets equal to that um new first ASM toote oh sorry sorry not new your original ASM tootes your original asmp tootes are going to be at Pi on Two And also any odd numbers so you need to let it equal to the original so whatever's inside the brackets let it equal to whatever is inside the original and that will give you the new can uh the um the new first ASM toote so here is what the tangent function will look like so for sketching tan functions there a it's important that we have vertical distance between the um period and the first ASM toote that we also sketch ASM tootes add or subtract the periods of time to find more ASM tootes and that we also draw for the entire given domain we also need to make sure that the excepts will be halfway between ASM tootes so your turning points wherever that may lie have to be halfway between you um between the Asm tootes and lastly you need to sketch the graph so in terms of sketching the graph what's important here is the curvature so having the curvature and also the fact that it's approaching the ASM tootes and doesn't intersect to touch the ASM tootes so here is a quick example that I like everyone to have a go at I'll just give every in a minute so if you're watching this online have a go um have a go at this question and then unpause to go through the solutions okay so hopefully you've had a chance of this question so if we have a function y equal 10 x + P um x / 2 and we need to sketch that function in the given domain pi to 4 Pi first at first is to determine the amplitude so the amplitude for the 10 function is actually a little bit IR irrelevant because the graphing extends from negative Infinity to positive Infinity every single time so I'm actually going to ignore the amplitude but what I will find is the period the period is going to be so since it's tan is just going to be Pi / n so n was x/ 2 so if you do this you should get um let me see oh sorry not pi over what am I doing it's pi over whatever number is in front of the tangent so it will be pi over half which is equal to 2 pi so again I hope this makes sense once you found the period the next step is to then determine um the Gap so the Gap will just be the period to pi over four and this will help me determine where my data values are going to be on the graph and the next step is to determine the ASM tootes so the ASM toote remember we just let it equal to the original ASM toote whatever is inside the bracket so X over 2 is equal [Music] to Pi on 2 so that was um that was the original asmt if there were no Transformations so therefore X is going to be uh 2 Pi / 2 which is essentially just Pi so again I hope that makes sense so this is the first asmor in this graph so if I were to sketch it somewhere okay so now that I've drawn Pi I'm just going to LEL my axes quickly I need to draw it for the given domain so netive pi to 4 Pi so my ASM tootes actually one Pi unit or one period apart so for the next one will be at 2 pi and then the next one will be at 3 pi and 4 pi and so on and so forth similarly if we go backwards it will be at Pi then it will be at Pi minus Pi which will be sorry I'm doing the Gap correctly 2 pi 4 or two Pi on two sorry it was Pi on two all time so it's Pi + Pi on 2 which is 3 Pi on 2 3 Pi 2 can then further be converted into 5 Pi on 2 but we'll go back so Pi - 3 piun on 2 so this will give us um X = um Pi - pi um Pi on [Music] 2 that will be 3 Pi on 2 and then we can also go further back if we need to but I believe that we don't need to go back and in terms of the X intercepts they're usually um also the periodart so we have a rat here oh sorry what am I doing a point here to begin with so this will be one of those X intercepts graph will look something like this and then I add Pi units to get to the next one so 0 plus pi is going to be um sorry the period sorry not the not the P not the pi the period again really confused sorry so we add 2 pi so we'll get 0 + 2 pi will give you 2 pi I think I've done this wrong let's check there um okay sorry why don't know why I just did Pi so it's actually period uh sorry whatever the value is plus the period which is 2 pi not just Pi so again a really Source um a big source of confusion so have that cleared as much as possible and try to work through evidence to see if it makes sense so therefore it will be X = to 3 Pi remember we have a two Pi Catalyst and then we have this value right in the middle which will give me you know this graph over here so again I hope it makes sense and we actually have't finished so we need to have something over here oops my badge remember it has to be approaching that value so something along the lines of this okie dokie so that's essentially it for um graphing of the tangent functions um sorry did I draw this wrong no sorry that should not be correct it should be like this instead the other one that I draw was if you um if there was someone sitting in um yeah so that would have been if the gradient was you know 3 Pi on 2 to 4 Pi but if it isn't just do PI um and so on and so forth so here is what the graph might look like again like I said amplitude is not important what is important is the first asmor the period um potentially the Y intercept as well okay so now we'll move on to functions so first we'll have a look at one: one functions so there are many different types of functions but a onetoone function is essentially a value which um or sorry an equation which does not repeat any y values and the way that we determine that um Al not algebraically but graphically is by looking at the horizontal line test so the horizontal line test is essentially where we are determining um whether the function is one: one so if we draw a horizontal line at any point during the graph it should give us a um only one intersection such as is shown in this graph over here so essentially what I'm saying is it should have each x value corresponding to one y value not multiple alternatively we can have a not one: one graph where we do the horizontal line test but it does not give us a um one a single intersection there's going to be multiple intersections does that make sense um I hope it does otherwise please pop it into the Q&A so another definition of the inverse function besides swapping of the Y Y and X is the fact that it is actually a reflection in the graph y = x so if we look at this line over here these two graphs are actually going to be symmetrical so actually going to form a plane of symmetry this is the plane of symmetry so what this is going to do is if I flip it on one side is actually going to give me the inverse graph because essentially what's been happened is the X and the Y values have been swapped another way to think about the inverse function is that a function takes X values and turns them into y values the inverse of that function could then be um taken um from those y values and turn them back into um original X values but the way that I like to think of it is sort of like a machine so so we have the X values chugged into a machine to give us y values and those yv values get chugged into the inverse machine to give us X values so that's essentially how the um how the inverse functions essentially work and also what they mean but in terms of VC you need to be really aware of their sort of notation and Convention so for a function f the inverse function is notated as F inverse a lot of people or a lot of students get this confused with 1 / f that is completely wrong 1 f is 1 f f ^1 or F inverse is f inverse so when you have been given a short answer question which asks you to you know prepare dinner or H sorry what am I doing um yeah so when you have been given a question that ask you to determine the inverse function of F what you will need to do is first of all let y = to F ofx so once you let yal F ofx you need to actually say let yal F ofx if it's not already given in the form yal F ofx but once you have um written this sort of notation the next thing that you need to write is to invert swap X and Y so again something you need to write just in terms of notation and then you need to rewrite the equation y = f ofx where the Y X and Y y are swapped so for example let's say the original function was y = x² + um six my new value will now look something on the lines of X is = to y^ 2 + 6 so that is what my function will now look like um when I swap it but I will obviously need to rearrange to make why subject so that's the last step to make sorry that's the fourth step rearrange the equation to make why subject um so should be pretty straightforward obviously most of the questions will allow you to use your C but if they don't it can be easily done on your um by hand as well and lastly you might want to Define what F ofn inverse is by say something along something along the lines of Y is equal F inverse of X um and so on and so forth so another thing that's important in this case is sometimes a question we ask you to Define define the function so if you're defining remember this specific notation so we have F inverse then the two dots then you have the domain to the co- domain and then you have what the f ofx f of inverse of X actually equal to so just make sure you really read the question if it hasn't asked I would still probably give my answer into this form as well because sometimes they are assessing for the domain okay so here is an example consider the function um g 1 to Infinity um and here we have an equation um G of X is equal to 4x - 3 X -1 so what we have to do is determine the inverse function of G so we're finding G inverse of X so first step for inverse or we write these two um steps first so we write this notation and once we've done that then we can start doing the algebra so here I've actually written down what I what I what I've done to solve this question and get to the final answer so first you have x = 2 4 my- - 3 over y 1 so I've swapped the X and Y variables alternatively what you could have done is simplify this expression first before you know substituting X or swapping X and Y so in this case what has been done next is we've split the fraction so we get X is = to 4 y - 4 + 4 - 3 which gives me Y is = or X is = to 1/ 4 y or 1 / y - 1 + 4 next what we can do is we can move the four to the other side and then we can um move the x -1 in the denominator and swap it with the y - one and rearrange to make y subject to give you g inverse of X so really um a really crucial process that comes again and again in exam one so you should be really familiar with how to go from the um normal to the inverse okay so let me again let me know if you have any questions for that part of this um today's content but otherwise we'll move on to transformation so just to Briefly summarize a transformation essentially where we are changing the x or y variables of a graph um in order to um sorry yeah so essentially we just change the x or y variables over graph either by reflection translation or dilation where dilation is just stretching or Contracting the graph and this could have a variety of um uses in terms of using mats for modeling for example so there's quite a few methods for doing the um Transformations I personally use the coordinate method so the coordinate method um will take a little bit longer but you can be 100% confident that if you've done all the steps correct you will get to your corre correct answer so the way that you do the coordinate method is once you've been given a transformation you write it in this sort of like coordinate method so XY gives you or um transforms into you know something something and then you start applying the transformations in the order in which they are given so first it's a dilation from the y- AIS so if it's a dilation by a factor of a from the Y AIS we know that if it's a delation from the y- axis actually impacts the X values because if you think about it um I don't really um let's just do a cubic oops that's not a good cubic actually let's do a quadratic so for this quadratic if I have a dilation from the waxes which means my graph is being stretched this way or contracted this way my y values remain the same it's my X values that are changing at those particular points so therefore we put it in front of the whatever the dilation factors we put it in front of x simly if it's from the x axis then we put it in front of the Y it's a bit of an opposite effect next if we have reflection same thing here if it's in the X then we put in the Y if it's in the Y we put it in the X and then we finally Dre the translation so so C units up so we put um so up is changing the Y values so therefore we put it as plus C at the end and D to the right so therefore we do plus d next step is to write what x- and Y Das is equal to so whatever is in this coordinate is essentially just x dash and Y dash so this is essentially what x d and Y dash are so what we can do is we can let x- is um equal to this coordinate y D equal to this coordinate and rearrange for X and Y so once I've rearranged for X and Y such as shown in the last step over here then I can immediately sub that in into the equation and once I substitute X and Y into the equation I should get my new transformed equation okay so now we're essentially done with the Transformations the transation was pretty short it was just looking at the coordinate method and that's pretty much all that's required for year 11 but now we'll move on to calculus and specifically looking at average and instantaneous rates of change as well as turning points and um stationary points so first let's look at look um first let's have a look at what a rate of change actually is so rate of change is basically a way of describing the gradient of a function um so sort of of similar to linear equations look um that we looked at before but for a linear equation the gradient was actually uniform so what I mean by that is if we have a linear equation the gradient at this point is the same as a gradient at this point or the slope however graphs like these the gr is constantly changing you know throughout the graph cuz it's not linear so a rate of change is going to be a way of describing you know how my yv values change as my X values change which is essentially what the gradient or the M represents so there are two types of gradients that that we need to be familiar with so we have the average rate of change which investigates the rate rate uh the average gradient between two points or we can find the instantaneous rate of change which will'll come back to in the next few slides so the average rate of change pretty much works in the same way as the um uh the gradient formula for linear expression so Y2 - y1 X2 - X1 the only difference here um so that's only works when you're given two discrete points sometimes what you might be given is the X values and the equation of the line so if you have the equation of line you can EAS substitute it in to find what the Y value and um yeah what the y- values are going to be and then from there on you can use the average rate of change formula which is just the gradient formula so pretty straightforward for average rate of change next we'll look at tangents and also instantaneous rates of change so if you think about a tangent a tangent is just a straight line that touches a curve once but a property of the tangent is that it's going to have the same gradient of the point that it touches um at that particular point so at this particular point the gradient of the tangent and the gradient of the circle is actually the same if it wasn't the same so let's say the gradient of this line was you know something else it will it might look like this um no sorry that's not a good representation it might look oops so if it wasn't the same it might look something like this but are we determining the average rate of change here or the instantaneous rate of change because remember the tangent here will have the same gradient and also will touch the curve once but if we deviate this gradient you know even by a little bit it's going to now potentially touch the gradient at that point which is essentially what we don't want so if a gradient at Point X1 y1 is M then the gradient at that point is going to be so pretty straightforward just another form formula derived from linear Expressions to determine what the Y value is going to be if we know the x value uh if we know the X values and the gradient or we know a pair of coordinates so just one coordinate and the gradient so the uh and this formula I always use if I know what the gradient is what one of the X values and what one of the Y values is I can just use y of t or Y of the tangent or the gradient of the tangent sorry not the gradient the equation of the tangent is equal to the gradient bracket x - X1 + y1 so once we've actually determine what the gradient is going to be at that tangent point we can easily go ahead and determine the equation of the tangent so the equation of the actual line that is going to be you know intersecting the graph at one point but what you need to know about this is is not only going to show the coordinates at this point it's also going to show the gradient so that's going to be the same okay so next we'll look at normals so normals are Essen just the opposite of tangents so basically a normal line is perpendicular to the tangent and passes through the same point as the tangent so what that means is at that point instead of having a line that has the same gradient at that point we now have a line that goes perpendicularly through that point with a gradient that is a negative reciprocal of the tangent gradient okay so what that might look like is let's say we have um M1 being the gradient of the normal but we've already found the gradient of the tangent by finding the instantaneous of change or calculus which we'll have a look at in a second once I've determined that what I need to do is use this formula over here so M2 or the gradient of the um normal is going to be the gradient of the parallel line or the tangent um it's just going to be the negative reciprocal of that so-1 over the gradient of that line of the tangent line sorry and the way that you determine the equation is pretty much the same same sort of process um y n - y1 is = to1 m x - X1 so the instantaneous s of change tells us the exact gradient at an exact point on a curve like I've mentioned before so before when we were looking at the average rate of change it was between two discrete points on the graph now from the average rate of change we've actually defined the instantaneous range so it's the rate of change at you know let's say this point by itself now that's really hard to determine because if we have two points we have two sets of coordinates to work with but if we just know one point and we know where it lies on the graph there's literally nothing else that we can do to work out what the grain is going to be at a particular Point unless using something called calculus and differentiation so that was just a brief difference between the instantaneous and average rates of change and also their properties okay so now we're going to be looking at the concept of a limit so a limit is essentially a value that a function approaches as the input approaches some other value so essentially so a knowledge of limits is um is not required in too much extensive detail in the method's course however essentially what this expression or this limit expression is essentially telling you is that if my X values approach P for the graph y f ofx um my y values are approaching l so that is what that is going to tell you um and essentially that's what we want so if we are trying to DET the instantaneous rate of change from the average rate of change we want to find you know two really close points and we want to determine the grade in between them because that's going to be much more accurate as opposed to having really far away points so essentially my difference between these two points is approaching zero and that is essentially the derivative however there is the first principles method that um you do need to know in a little bit of detail but not too much and it really won't come that often in the exam but just know that it is actually derived from the quadratic form uh sorry not the quadratic it is derived from the average rate of change formula so really just be aware of that but what we're doing here is let's just say we have uh we want to determine the instantaneous rate of change so the rate of change at that particular point for y for X is equal to a so what we might do is we might work out the gradient just how we normally would so we're just going to let this distance between a and you know another point A Plus H so a plus an arbitrary number um and just find the gradient so just going to be F of a plus h so whatever the Y value is at that point minus the y- value A F of a over this point a + H minus a but remember that a plus h minus a is just going to be H so that's how we get the limit formula and what we want is the H value to essentially approach zero so we want it to be as close to zero as possible so that we can find what the instantaneous rate of change is so that's why we let the limit hit approaches zero of the average rate of change function so again I hope that made sense um in terms of how the instantaneous rate of change has been derived by first principles from the average rate of change okay so so now we're going to be looking at the different types of notations of derivatives so first we have just the normal Langer lras um notation which is just yal F ofx and the derivative being y equals f-x so pretty straightforward you're going to see this um you know 99% of the time other times you can also see Le uh leb's notation which is y = f ofx and then um Dy on DX gives you the derivative so as opposed to f-x is going to be Dy and DX and then we lastly have Newton's notation which you will not see in methods whatsoever but I've just put it there just in case so the Newton notation is essentially where you have a function and the gradient is listen um is written like y dot essentially um is just another type of uh so all three are different types of notations but depending on the question is given you so if the question is given you you know f ofx is defined as blah blah blah then you will use l lr's notation so F the derivative will be f-x however if your answer is given in the form of y equal you know something then it'll be in the form of leb's notation which is dy on DX so I have included an example for the um differentiation by my first principles but I won't be going through this example in too much detail however it is in the slide so if you would like to check it out please do so um um from the resources section of this lecture page otherwise we will look at just basic differentiation so there's a few rules for differentiation that you need to follow but um the basic rule here is that the derivative of any polom is essentially um a dropping of the power and then subtracting by the power so therefore the derivative of f ofx is equal to X the^ of n is equal to f- X and here we're using the lr's notation is equal to n so I've dropped the N down X multi to the^ of nus1 so here are some other rules that you might come across when doing um calculus and that is the coefficient rule so if you have a function and then you have a coefficient at the front or you factorized something you can move it to the front without affecting your um your derivative and then when you find what um DDX f ofx is equal to you just multiply the answer by a pretty straightforward just a minor trick um in order to get you know improve your efficiency in the exam next if we have um the addition or subtraction of two functions so essentially what we just do is we derive f of x separately and we derive G of X separately and your answer just going to be f- of X Plus g- of X next we have the derivative of a constant so the derivative of any constant is always going to be zero and this is be we um because we imagine that um we just have a constant line so therefore there's no real gradient whatsoever okay so here's just a really quick practice question to uh find the derivative of the following function so we have 3 to^ 4 sorry 3 over 4 4X ^ 4 - 1 + 5 2 uh 5 * 2 x ^ 2 - 1 + 0 um and this apparently is equal to 3x Cub + 10 x we'll have a look at how we might solve this so we're just going to simplify this so this and this is going to cancel out so we're going to get 3 X x 4 - 1 is going to be 3 and then lastly we have 5 * 2 which is 10 x 2 - 1 is just going to be 1 so therefore the answer here is like the solution I've suggested is just going to be 3x cubed + 10 x okay now we'll look at stationary point so stationary point is just a point on the graph where the gradient is zero essentially so that's why it's called stationary because it's not moving um but 10 but f- of X is going to be equal to zero or Dy DX is going to be equal to zero at the point where we have a stationary point for the derivative function so um tangents at the parallel um at the stationary points are parallel to the x-axis and that makes sense as well if you have a grain of zero um at a stationary Point your tangent is obviously going to just be a straight line so it's going to parallel to the x-axis we can have three main types of stationary points we can either have local minimums local maximums and also points of inflection so most of the time questions will be concerned with local maximums and local minimums so tangents at stationary points are parallel to X xaxis like I've established however if we have a vertical um point it's going to be undefined so again um just really um being familiar with you know um the local Minima and local Maxima and how they look is really important for this um these types of questions so there are three methods mainly for determining the nature of a stationary point so either using the first derivative method and drawing up a nurture table oh sorry nature table or we can do the second derivative method so second derivative method is more so spe but it can be still used in methods to get full marks however I won't be using this method today in order to avoid any confusions but we can also have knowledge of function shape and sketching so just knowing how the function looks or behaves we can also determine the um the shape um of the graph so here we have a derivative um and we're trying to determine the stationary points yeah as long um as well as their nature so what we do straight away is derive the function so we have the function x^2 + 2x - 3 we can derve that function and we will get x2 uh x + 3 x - 1 and then we can solve it to give us negative uh sorry positive three oh sorry -3 and negative um and positive 1 so -3 and positive1 next thing would be to determine what the Great gradient is at that point so if the gradient is point we we use a slope table to give us a positive line at that point however if we have a negative gradient such as that fdh is equal to zero then we will draw a negative gradient like this however if it's equal to zero then we can draw something like this which is a straight um Point oops sorry what did I do here was supposed to be here yeah but overall this is how the nature table works and overall you can sort of determine you know there's a local maximum X3 a local Minima at xal you know pos1 so again I would say um uh dooring up a nature table would probably be one of the best methods to find um the um the local uh Maxima and local Minima as well as any other stationary point so we also have the second der method which I will not be going through as it's not accessible but I've listed down um an example question of how to use the um second derivative method for everyone to review and I've also included some rules for the second derivative okay so now we'll have a look at um the recognition method so recognition method revolves a really good or involves a really good knowledge of the function itself so if you have um function such as this one which I can straight away tell that it's going to look something along those lines I can straight away tell that the coefficient of x cubed is going to be positive and therefore we're going to have a positive cubic function so in a positive cubic the first St Point must be a very relaxed and the SE stry Point must be a local minimum for this particular particular graph so again just using the N uh the knowledge of your graphs is probably a little bit difficult so therefore I usually always use my um nature tables which are the best method to use um on an in an exam scenario to avoid making any mistakes especially if you don't have a calculator for exam one so next we have this concept of absolute Maxima and Minima so absolute Maxima and Minima is just the absolute highest and absolute lowest value of a function so they different to local Maxima and local Minima is because we can have values higher than the local Maxima or values that are lower than Lo uh the local Minima just due to the nature of the graph however absolute Maxima and absolute Minima are essentially just you know the absolute greatest and absolute lowest value values in a particular graph and they can either be local manim local minimums or end points or there could be local maximum or end points as well okay so here is um an illustration of the concept of absolute Maxima and Minima so if you look at the first graph over here we can clearly see that the point that is the highest is this value over here although we have a Max the local maximum at being at this point over here Here Local Bima I can clearly see that it's not going to be this value over here because I have a value that's lower going down here so therefore in this case this is my absolute Maxima and this is my absolute Minima so again I like I said it depends on the question and what you are given um in terms of what the absolute Maxim my is but if you are determinate by hand you need to have determined all the end points and all the uh the local Maxima and Minima in order to judge which one's going to be the greatest or lowest so in the second graph here we have this being the highest point so therefore we have the local maximum being the absolute maximum as well and I can clearly see that there's minimum values going you know further down so therefore that's going to be the absolute Minima so here is a bit of a practice question or an application question more so about calculus an absolute man Maxima and Minima so um if you're online please have a quick go um or if you're not watching this as a live stream have a quick go at this question pause and then we will go through the answers otherwise if you're watching it online we will go to the Sol uh as a live stream you we'll go through the solutions now so we have trig a farmer who is building a fence against a wall he has 20 millit of U 20 M of fencing and is thinking of making a rectangular area so our qu the question asked what are the dimensions that will maximize the area covered entally we want to have the maximum distance covered so we want the greatest X and the greatest y values but overall they should make up the largest area so what we can do is we can formulate an equation so the first equation that I can think of is that X so 2x + y is equal to 20 because the total um fence use is going to be over here we have X on this side so therefore X must be on this side as well and we have Y which should be you know uh just y so therefore 2x + y is equal to 20 so this one is correct next we can formulate a second equation for the area and we know that the area is going to be x * y we already know what x is he's just uh it's just X whereas y we have determined this in this equation over here so Y is going to be 2x - 20 um so therefore area becomes X brackets y has been replaced with 2x - 20 and now we have a much more easier a much easier equation to derive and solve so straight away we will find the second uh the first derivative so that will just be you know 4x - 20 equal and then we let it equal to zero since we're finding the uh the maximum and the minimum values so if we let it equal to Z we get X is equal to 5 and therefore the Y value must therefore be equal to um you know if we put X back into the formula we should get Y is equal to 10 so this is how we can solve an application question like this um but yeah so that was essentially the end of the calculus section of today's lecture we'll now move on to probability and have a look at some of the conventions and definitions so what is prob probability well probability is just the chance of a certain event occurring when you know what the different outcomes are so essentially um consider a scenario first of all and the scenario is that you're throwing a fair six-sided die the diee can um you know uh when you roll it it can give you different numbers but it will be the six different outcomes for each side of the die the sample space therefore describes the possible outcomes that we can get which would be obviously 1 2 3 4 5 6 which are all the numbers in the um in the six fair fair uh in the die and we we donate any values that we're listing in curly brackets if we were listing the sample space as you know 1 2 5 but that will include every single decimal value between 1 and five we want to use this squiggly brackets here so Squigly brackets we only use when we're sort of denoting something or usually listing something as opposed to giving a range so that's actually what a sample space is all the possible outcomes that can happen whereas an event is a group of specific outcomes so an event for um drawing uh or rolling a six D diet is landing on an even number or landing on an odd number for example or Landing in on a number that's um you know the top three or something along those lines so an event is just a group of possible outcomes from your overall sample space so again just basic definitions um here are some notation elements so we have this upside down U symbol which is called intersection and it's actually read as a and b or a intersection B I usually just say A and B because this is essentially what it means and we'll have a look at that in a coming slide a and so essentially what it does is it lists all the outcomes that are common to both A and B so common means it must have a elements of a and elements of B for example rolling a six-sided dice that's a um that's the experiment getting an even number and getting a number that is a multiple of three are two of A and B would be anything that's common to both so that would be you know the number six for example that is common to both of those events so that would be a and b however A or B is just all the elements either in a or in B so if we're using the exact same example it would be all numbers you know 2 4 6 um 3 six all those numbers will be part of this set now as opposed to just being six next we have compliment so compliment is just the list of outcomes that are not in that particular event so for example let's just say I'm finding what probability of a not is it will just be any outcomes that are not part of a and then we have this null set or the mutually exclusive sign so it just indicates that there's nothing in common so for example rolling and odd number and rolling an even number on a six-sided die are mutually exclusive events because we can't end up having element um any outcomes that sort of you know conform to both of those um events and next we have a Lipson The Lipson is used as a symbol to indicate the entire sample space of a random experiment so usually you will see the symbol in V diagrams sometimes Caro Maps usually not but yep definitely in a um um a ven diagram okay so here we have just a quick um example question so we have probability a or sorry event a and the sample um and the outcomes are 135 B is 1 2 3 4 and the sample space is 1 2 3 4 5 six so we have to list the following sets so what is a and b so A and B is just going to be what are the common things between elements A and B so definitely 1 and definitely three so therefore the answer here is going to be 1 over 3 now if we were to determine the probability of a and b instead probability is we'll come back to that in a second but it's just how many you know successful outcomes or favorable outcomes there are over the total number of outcomes so therefore probability a so what are the favorable outcomes that's just going to be one and three so therefore there's two favorable outcomes over the total number of outcomes which is six so the probability of event a becoming true is going to be 1 over 3 now if we look at A or B that just gives you all elements that are in both A or B so not it can be you know either an a just by itself B by itself or both common elements as well so in this case it will be 1 3 5 1 2 3 4 so all of these numbers and and lastly we have a dash and B so what this means is not a that's how you that's how I read it and B so essentially we're just looking at values that are just part of B and are not part of a so we've already determined What A and B is that was 1 and three so we're DET Valu that are just in B so it can be 1 and three so it'll just be two and four as our answer so just elements that are part of B that are not part of a okay so now we'll look at the fundamental properties of probability so what is a probability in the first place so the probability of an outcome occurring or an event occurring is just the number of favorable outcomes over the number of total outcomes so that's essentially what the probability essentially means and probabilities usually have so since this was like a chance sort of um way of determining um you know the chance of something occurring they are usually between or probability of something is usually between zero and one where zero means impossible and one means certain so um the yep so you really need to um know this formula any probability will be between zero and one and another thing they need to know about is that the sum of all the outcomes of the probabilities of all the outcomes is always equal to one so the probability of getting one is um you know probably getting one probably getting two probably getting three probably getting four probably getting five probably getting six in a um when you roll a six side di when you add them all up together it's going to be equal to one so essentially the total probability is always equal to one and individual events are always between zero and one so then diagrams are a really graphical way of representing probabilities so we can visualize the sample space through a vent diagram so what we might do is we will have a box with the umson symbol inside it and we will have two events so one of the circles represents one event one of uh the other Circle represents the other event so here in this case we have a and b so what we do is we put in element so let's just say we're rolling a six-sided die event a is probability of getting an even number event B is let's say getting a mult multiple of three for example so even number so we have two four six what about event B we have one oh sorry not one we have multiples of three so we have three and we also have six anything that's common that's both of them that's in both of them we actually put it in the middle so six we won't write it separately we will actually put it in the middle like so and any events that are not part of any of these events actually go outside so five would be a number that would be outside this so event diagram is really good because depend um depending on what we're trying to find out we can sort of easily determine so let's just say we're trying to determine what event a is going to be so event a will just be this shaded region over here so however many values you have or sometimes also written as probably instead of being written as two or four sometimes it might be just written as 1 over three or something along those lines but most of the time you'll get individual entries like like so but yeah so you can easily determine certain probabilities just by looking at where those events are located so if you're determining just event a it will just be this shaded region over here if you're determining event A and B it will just be the common this Mutual region between a and b B which accounts for all events that are or all outcomes that are part of both A and B at the same time it we're determining um the probability of event a not and B so just event B essentially we're looking at all elements that are part of B which are not part of a so that will just be all of these so essentially we're just removing the intersection between A and B next if we're looking at A or B we're essentially just looking at all of this entire stage region so that will give us event a and also v b we also have an additional probability formula so if we want to determine probability A or B and we know what probably a is probably B is and probably A and B is we can use this formula if the graph or if the vend diagram is not mutually exclusive which we'll look at in the next slide so this formula can allow you to determine probably a or B or probably A and B if you are given the rest of the values it's a really powerful formula that's usually given in your formula sheets but otherwise you do definitely need to know how to use and sort of apply this formula okay so next we have mutually exclusive events so mutually exclusive events um not to confuse with independent events is basically when we have events that cannot occur at the same time so for example it's um raining and um Being Sunny at the same time although that sometimes does happen but in the general sense that raining you know complete thunderstorm and also being you know uh Sunny is you know a mutually you're not going to get that happen at the same time so these events are actually called mutually exclusive events so this how a vent diagram would look if you had two mutually exclusive events so as you can see there are no intersections between these two graphs so there sort of just like two discrete circles so we can also have V diagrams in with three events instead of true so these even become a little bit more complicated but essentially they function in the exact same way so this region over here in the middle represents outcomes that are part of both f g and H this part represents only events that are part of F and H and not g this just part of um f and g and this just part of G and H so let's have a go so we're I'm looking at these probabilities so probability F will just be you know all of these values so how many favorable outcomes do we have so in total we have 20 outcomes so that's why that's in the denominator but the number of favorable outcomes so the probability of six is just going to be six so therefore the probability of of f occurring is going to be 6 over 20 which is 3 over 10 which is 0.3 now if we look at probability of G that also has um so if we just count the number of favorable outcomes that is going to be eight so therefore now it's going to be 8 over 20 which is 2 over 5 which is 0.4 next if we look at probability f and g what we need to do for this one is look at what are the um outcomes that are intersecting you know f and g so f and g so we only have one outcome which is three intersecting both f and g therefore the favorable outcome is one over the total outcomes which is 20 and F or G will essentially look at all of these values so anything that's inside G anything that's inside F which is going to be 13 over 20 so just includ excluding the these values over here but the rest rest will be included so yeah so pretty straightforward these Pro probabilities is you know you're going to get questions on these um and you should be able to um do them fairly easily so another thought of thing that a lot of students get confused with is the concept of conditional probability so conditional probability is essentially where we know that something is occurring already so let's just say we know that that it's raining already and I want to determine what's the probability of me using an umbrella for example so if it's raining already obviously my chances are going to be uh my chances of using an umbrella are going to be much higher as opposed to if it's not raining so in this case what I'm trying to determine is probability of using an umbrella given that it's already raining because if I was just using U looking at the probability of me using an umbrella it's want to be a very low probability because you know it doesn't rain quite often so therefore um finding the probability of me using Umbrella given that it rains is going to be much higher so that's essentially what conditional prob is we're looking at the probability of something occurring if some if we know that another event has already occurred so here is the formula so probability a given B is equal to probability A and B over probability B yep so just a formula that you know you just have to memorize or have it written down somewhere usually it will be given in Formula sheets as well so for example if it has rained the day before there is usually a higher chance that it will rain the next day if the if a card is drawn from the deck is black there is a higher chance that it Spades is before so here we have kind of few scenarios of conditional probability so in the first one the condition is that it has rained the day before in the second one the condition is that the deck is black so another useful way of displaying probability information as opposed to ven diagrams is a caror map really really good and useful um sort of way of representing probability so usually it's only used for two events so we have probability a probability um not a probability B probability not B on the um horizontal and vertical sides and then in the gaps we have these specific probabilities written down so whenever we're finding inter section so the intersection of A and B we're just doing a intersection B so intersection of B KN and a will be a intersection B KN intersection of a KN and B will be probably of a a KN and B and so on and so forth so this is really good um you can either write the number of outcomes that are present so instead of doing probability you can either also write n or the number of events that are in a and b or you can just write the probability so most of um so depending on what the question is um you'll be using various different methods so we'll do our practice questions um just to get sort of familiar with what this sort of represents so the probability that Cynthia walks to school and it is raining is 0.1 so let's just draw up a caral map so we have raining not raining um walks does not walk so this is essentially to set up your table and remember the total probability is always equal to one so the problem that Cynthia walks to school so walks and it is raining so raining as well is 0.1 so therefore 0.1 will go here the probability that it doesn't rain so no rain um is 0.6 so just probability of no rain we're we're just we're not looking specifically for um walking to school or not walking we're just looking at the probability of not rain which is 0.6 so essentially these true gaps should add up to give you one so therefore this is 0.6 so therefore this must be 0.4 and this should add up to give you 0.4 so therefore this must be 0.3 so that's how the Caro map is so useful we can use information that we already been given to easily find other pieces of information that we might need using a cono map what is the probability that on any given day she doesn't walk um and is sorry the probability that she doesn't walk and is so not walk and it is raining so essentially we don't we didn't even need to fill out the rest of the table so she doesn't walk and it is raining it's just going to be 0.3 so entally what we're just Happening Here was the probability of not raining uh sorry raining and not walk which in this case is going to be 0.3 so pretty straightforward if you know how to use a caral map you should be pretty um you know uh pretty familiar with how to use this so if if it is raining today so this is how the questions are going to start if it's a conditional probability question so they're going to give you a condition first it is raining today so that's the condition what is the probability that Cynthia walks and now instead of finding probability what is it raining and she walks we're finding what's the probability that she walks given that it is raining so essentially we just have our formula so probability walking given rain we put it into the formula probability W and R over probability R pretty straightforward just looking straight from the um uh from the cono map we easily get our answer so pretty straightforward again let me know if you have any questions so next we have the independent probability formula so events um so we can have events that are independent dependent or dependent independent events are basically events that don't really have an impact on each other so an example of an independent event would be flipping a coin and the probability that it rains tomorrow so regardless of where you get a head or a tail it shouldn't really impact whether it rains um or doesn't rain tomorrow similarly um however a dependent event might be something along lines of it raining and you taking and you using an umbrella on a particular day because yes if it is raining you are like more likely to use an umbrella so that's a dependent event an independent event should not have effect and there's a formula to confirm if something is an independent event or not so essentially probability A and B is going to be equal to probability a * probability B if it is an independent event so essentially a lot of these questions are going to ask you to prove that something is an independent event and the way that you should go about doing this is you know looking at what probability A and B is um you know what's given to you so what probability A and B is given to you and then finding out your own probabbly A and B and if it matches then it is independent if it doesn't match then it's not independent okay another graphical display of probability is the tree diagram so basically the way that the tree diagram works is it sort of shows the conditional probability so the way that we do is we have true events occur uring in a sequence we can have two events or three events occurring in sequence and it can actually be used to represent both dependent and independent event so we have probability a so we write the probability of a you know above here and therefore the probability of complement of a is just going to be the opposite so 0.4 and then we go along we find the probability of a b probability of B not probably B probably B not and so on and so forth so I'm pretty sure all of you will be very familiar with how to use tree diagrams um otherwise do let me know if you have any questions but we'll have um we'll move on to um a practice question but just before we do that I'm just like to tell um remind everyone that if we are finding the probability that this happens and this happens essentially what we do is we multiply the probabilities so say we're trying to work out what's the probability that um you know be given a curse and be given a occurs essentially we add or sorry multiply oops yep so what's the probability that b occurs and a occurs we just multiply this we Pro we multiply the probability of a multiplied by the probability of B to get probability A and B but if we were determining the probability that something happens or something else happens we add all the probabilities so I'll explain what I mean when I get um when I when we do this example so in blairesville if it rains the previous day the chance of rain today is 0.6 if it doesn't rain the previous day the chance of rain today is 0.3 using a tree diagram if it rains on Monday what is probably that it rains on Wednesday so it has already rained on Monday so for Tuesday the probability that it rains is going to be 0.6 as given in the question step therefore the prob not rain on Tuesday is going to be 0.4 then going along if it has rained on Tuesday the probability of it raining on Wednesday is going to be 0.6 however the probability of it not raining on Wednesday will be 0.4 and then for Wednesday the probability of it raining if it hasn't rained on Tuesday is going to be 0.3 and the probably it not raining is going to be 0.7 so again um uh just like an application of the tree diagram you really should be familiar with drawing a tree diagram based on some information in a question stem now the question is there are two outcomes that result in rain on Wednesday sorry here's the question what is the probability that it rains on Wednesday so there are actually two outcomes for whether it rains on Wednesday so the first outcome is rain on Tuesday then rain on Wednesday or rain on Tuesday and rain on wnes so this is what I mean by all so what is the probability that it rains on Wednesday so we what we're trying to actually determine is the probability that it rains on Wednesday given that it has rained on Tuesday or what's the probability that it rains on Wednesday given that it hasn't rained on Tuesday so essentially what we would do here is we would determine what's probability rain on Tuesday and rain on Wednesday so 0.6 * 0.6 which is equal to 0.36 then probably not rain on Tuesday and rain on Wednesday which would be 0.4 * 0.3 just looking at this stem if you're confused which is equal 0.2 and if we're finally determining the proba that it rains on Wednesday we can just add these two up together because both of them involve raining um on Wednesday so that will just be 0.36 plus 0.12 so this is where we have both and and or in inter playay okay next look at combinations so combinations is basically where we're trying to determine um favorable outcomes when there are so many varable outcomes so let's just say we have a group of 50 students and we're trying to pick you know eight students to you know represent we can have so many combinations or so many different ways in which we can sort of perform a certain perform that particular you know uh application so there's so many ways that we can select eight people from 50 people um you know and therefore we'll have millions of outcomes instead of counting all of them CU that will just not be practicable at that point we use something called a combination it's just a fast way of calculating variable outcomes or the number of total outcomes when we know certain pieces of information so for example um so this is here the formula so entries are will also written as this sometimes so essentially the formula NCR is actually this in um you know in actual like in actual mathematical terms so the N is the number that you're choosing from so the sample size the group size and R is how many you're choosing you or how many successful outcomes there are so for example if you're choosing eight people from a group of 50 N will be 50 and R will be 8 so then you put it into this formula or in your C I'm pretty sure there should be a function where you can just put this NCR straight away um and if you do that you should get your final answer but if you have to do it by hand which usually won't be you can use this formula you don't really need to know why this formula sort of works it's just sort of you know beyond the course but yeah so really important to just you know know the formula and be able to apply it and know when to use the formula as well well so for example if you need to get four questions right out of five on a test 5 C4 will tell you how different ways you can pass you can you might get the first question right second question right third question right fourth question right or you might get the second question right third question right fourth question right fifth question right we might get the first question right third question right you know so many different ways you can do this so this will simplify that entire process and you know solve it for you so if you're confused about the symbol the exclamation mark it just is a factorial so what that means is if it's 5 factorial it's just 5 * 4 * 3 * 2 * 1 so it's basically multiplying that number by all the numbers that come below it sequentially um so here is here was just a quick example so if a coin is flipped seven time uh seven times how many different outcomes are there when three heads are flipped so basically n is going to be seven and three is going to be our R so we just put this into the formula and we should get 35 as our answer so again let me know if you have any questions in terms of this formula but otherwise it should be pretty straightforward okay so now we're actually done with the question uh sorry not the question the uh the content for the um today's lecture i' just like to spend a few minutes just going over the exam because I know many of you will be um you know starting to prepare for year 12 or a little bit overwhelmed you know by what's coming next year and if that's the case um i' just like to you know ease it a little bit um whatever content you've learned today uh and as well as you know throughout the year year 12 will just build on that content so if you have your foundations really clear there's a lot less effort that you will need to put in um you know next year so don't neglect the 11 content it's really important and will actually you know severely impact you um and your overall performance next year but just in terms of your exam at your school although it is up to your um you know um up to the discretion of your school how they would like to structure your exam um you're likely to expect an exam one um worth 1 hour which is no notes and no C and it's usually worth 40 marks and also exam two which is 2 hours and you also allow notes and a c worth 80 marks so although year 11 doesn't technically count towards your VC or your atar you should still be revising hard and you should be preparing yourself you know on the level of you know effort that's required to be put in you know in order to do well in your exam so don't neglect your exams really focus on it try to put in as much effort as you would you know next year um another question um a lot of students will ask around this time is how should they start preparing for unit 3 four methods well you should start preparing for unit 3 for methods by you know really being diligent with you know your your living your year 11 content you know really understanding all the concepts being able to apply them you know if there are any doubts or any uncertainty that you may have in your knowledge you know really get um asking your teacher getting all that cleared off before you go into year 12 so that you're ready to build on your knowledge as opposed to you know having a really weak foundation and working on that Foundation next year study smart um study hard stay organized ask for help all those General things um you know learn your C I would probably say is one of the more more important things know how to do certain functions you C obviously next year going to learn a lot more C sort of um sort of stuff but this year just get familiar with all the basic sort of stuff you know deriving anti- deriving you know defining functions probability function um probability stuff like that and in the exam if you do have an exam this year which most of you will um and if it's um in a very similar structure to the V exam don't waste your times on questions that you do not know how to do and use your reading time very effective so that's just some of the basic tips that I had for you know starting thinking about unit 3 four methods and also how to you know St preparing otherwise if you have any questions please SW them through into the chat I'll be monitoring the chat for a few more minutes um otherwise um thank you everyone for coming to today's lecture and I hope you um um I hope you all the best for the rest of the year um and if you're doing any three four subjects good luck for that um yeah see you