Welcome to AP Pre-calculus Unit 3. This is the last video of required course content for the AP pre-calculus AP exam. So, I have a lot of things to say. First thing is that there is a special message at the end of this video about my plans for next year at this time stamp. So, please watch it if you get the chance. Okay, next. There will likely be live reviews for this course before the AP exam on this channel. Since I know a lot of you would want to interact with me live as I go over AP questions for this course for each unit. Once the AP exams are closer, I will release the dates and times of these live streams on my Instagram. So, go follow that to know when I post it. And now, a quick message from our sponsor. Hey there. This video is sponsored by me, Max Allen, because a certain button has appeared next to the subscribe button on this channel that allows you to become a channel member. If you or your loved ones decide to become a channel member, beyond getting my undying love, you also get access to an embarrassing video I made for my 11th grade chemistry class, access to member specific posts I might do, and of course, you get a special badge next to your name in the comments, making it so I will probably always respond to your comments, whether you need to ask a question or just want to say hi. It would mean the world to me if you'd become a channel member. And I'll even sweeten the pot. A lot of you might be wondering if I will be reviewing unit 4 for this course since it's technically not required on the AP exam. So, what I'll do is if I get two new channel members off of this video, I will make videos for the entire unit 4. All it takes is two people. All right, I've stalled you long enough. Let's get in to unit 3. So, unit 3 is here, and with it brings a fun spectacle of severely long topics of great difficulty. I mean, that isn't going to stop me from making each topic in under 3 minutes, but still, this topic is a nice and easy start to this unit, talking all about periods and math. So for a graph to be periodic it has to have a continuous cycle of the same pattern happen over and over over equal length intervals. This unit staple is s and cosine graphs which are by all accounts periodic graphs because they continue going up and down all the way to positive and negative infinity. This means the graphs are continuously repeating a cycle of the same pattern. A period is how long it takes a graph to complete one cycle. The way to calculate this on a graph is to find the horizontal distance between two maximum points next to one another or calculate the horizontal distance between two minimum points next to each other. So for example on the standard sign graph the period would be 2 pi. This is because in the length of 2 pi no matter where you look on the graph it is completing one full cycle of its pattern. But don't be fooled. Graphs don't look all perfect and nice like this all the time. Some period graphs can look like this. It's still a periodic graph though because though the pattern might be strange, it still repeats the same pattern over equal length intervals. And if you were given a period length of a periodic graph and asked to continue drawing the graph, you can do this because you already know the pattern that needs to be repeated. Just remember that a period is a full cycle of a graph's pattern and you'll be fine. So trigonometry, here we go. Trig kind of has to start setup here with a circle. Putting a graph into the circle, we can start at the angle 0°, then 90, 180, 270, and 360. In this course, we will be working with a new angle measure known as radians. The entire circle is equal to 2 pi radians, meaning 180° is pi, 90° is p<unk> /2, and 270° is 3 p<unk> over2. This would mean the conversion of degrees to radians would be this. Now that we have the start set up, the trick idea goes something like this. If you make an angle in the circle, you are creating a terminal side equal to the radius of the circle or r. You have the initial side, which is just the x-axis and an angle called theta. The terminal ray will keep going forever, but where it intersects the circle, it makes a point. We call that point point P. The coordinates of point P are X comma Y. A unit circle is not complex. It simply means the same exact circle when the radius is equal to 1. Now if we introduce S, cosine and tangent, we see they also function differently on the circle. S deals with the vertical displacement which is y / the radius r. In the unit circle, the radius is 1, making s the y-coordinate. Cosine deals with the horizontal displacement, which is x / the radius r. Again, in the unit circle, since the radius is one, it would make cosine the x coordinate. Tangent deals with the slope meaning rise over run or y /x. In the unit circle, it would mean tangent is equal to s over cosine. Quadrantal angles means one of the big four angles or any angle that has a multiple of 90. It is pretty simple to solve the sign cosine and tangent of these quadrantal angles. First, you make coordinates for each one. Since the radius is one, we simply make coordinates for each one pretending like it's a graph that only stretches out by one. So 1 comma 0 comma 1 comma 0 and 0 comma 1. Remember these coordinates are x comma y or cosine comma s. Meaning if you were asked to find s of let's say 180 degrees or pi radians it would simply be zero. Or let's say cosine of 270 degrees or 3 pi over2 radians it would be zero. If you were asked to find tangent of a quadrantal angle so let's say 90° or p<unk> over2 radians. We can do this because we know tangent is just s over cosine or yx which would be at p<unk> /2 radians 1 / 0 which is undefined because you can't divide anything by 0. Of course if the radius was something else like three you would just simply change the quadrantal coordinate points to be three instead of one. Okay. So the trig circle, it's cool and all, but let's make it disappear. Think of it this way. Of course, remember the angle measures, but just think of a graph. We already know how to find S, cosine, and tangent of quadrantal angles. But what if we have something like S of 30°? Let's draw a line. Draw the 30° angle, but this time we draw a line to make a triangle. Just so you know, no matter where you draw a line, it needs to go to the x-axis. Then since we know one angle is 30 and the other is 90, this would make a 30 609 triangle. So filling in the sides knowing the 30 60 90 triangle we know s is opposite over hypotenuse meaning 1 /2 meaning s of 30° is 1 /2 and if we had something like cosine of 45° then we can construct the 45 4590 triangle and we know cosine is adjacent over hypotenuse which in this case is 1 / <unk>2 thing is you can't have a square root in the denominator so you must rationalize it by multiplying both the numerator and denominator by <unk>2 and solving this gets us <unk>2 over2 is our answer to cosine of 45° or p<unk> 4 radians and if you wanted to find say tangent 60° or p<unk> over 3 radians, you construct your 30 60 90 triangle. We know tangent is opposite over adjacent, which in this case is roo<unk>3 over 1 or roo<unk>3. So to spare this video from being over 3 minutes, if we were to do this anywhere on the unit circle where any special right triangles are possible, we can get the parentheses answers here that show the answers in the format cosine, sign to each of the angles. And of course, if you wanted to find tangent, it's just s Okay, a handful of other things we need to talk about. If you were given an inverse s cosine or tangent equation, it's not that difficult. So for example, in this one where it's inverse sign of roo<unk>3 /2, all it's asking is what angle will have roo<unk>3 over2 as an answer for the sign of it. So looking at our unit circle, we see it in two places. Thing is dealing with inverse functions, certain inverse functions only exist in certain quadrants of the graph because of domain restrictions. Since inverse sign can only exist in quadrants 1 and 4, even if there is an answer of roo<unk>3 over2 at 120°, it can't be an answer because inverse sign can't exist there. Meaning our only answer would be 60° or p<unk> over 3. Another way to notate inverse function is using arc notation. Just know it means the same exact thing. Angles can also be negative by going clockwise instead of counterclockwise. So 30° is equal to -3°. You would still solve the sign, cosine, and tangent of them the same way. These equations, know them. Here's why. If you were given a question like this, you were asked to find the x and y. You can just plug in r and theta to get these answers. Or in this question where you're asked to find r and theta, you find r by using this formula and theta through algebra. Okay, we figured out how to solve all these s and cosine values on the unit circle from the last few videos. Let's start by taking all the sign values. If we were to plot all of these on a graph, it would look like this. You'll see the sign graph is a curve that keeps going up and down all the way to positive and negative infinity. And since it does go to positive and negative infinity, it makes all sign functions have a domain of all real numbers. Now, if there are no transformations on a sign graph, it will typically go on the y-axis from one to negative 1. This means the range of any sign graph would be whatever the minimum to maximum is put into brackets. Now, let's graph cosine and all of its values. You see that the graph looks practically the same to the sign graph. That is because it basically is. The cosine graph is a sign graph with a horizontal translation of minus p<unk> /2. So, that means that all the rules of a sign graph apply to a cosine graph. The domain is all real numbers and the range is minimum to maximum in brackets. The period of a s and cosine graph would be 2 pi. And to write out the basic s and cosine equation, you would say something like f of theta is equal to s of theta or f of theta is equal to cosine of theta. Theta in this case would just be an angle measure. So it would function the same as something like say x or z sinoso functions. Congrats. This is the second to last function you need to know for this course. A sinosoidal function by definition is a periodic function that continuously oscillates between a set minimum and maximum point. The syosoidal functions in this course will just be the basic cosine and s functions. The actual details of the equation will be covered next video. So let's talk about the specificities for the graph. Now first of all as a review for my last video cosine of theta is equal to s of theta + p<unk> / 2. This is [Music] because of course you know how to calculate periods from the 3.1 video and in the parent s and cosine functions it is 2 pi. The frequency is simply whatever the period is with a one on top of it. A middle line of a sinosoidal function is the invisible line that's halfway between the minimum and maximum points. The middle line of the parent sign and cosine graphs is y equals 0. The amplitude of a sinosoidal function is the vertical distance from the middle line to the maximum point on the graph. That means in the parent sign and cosine graphs the amplitude is one. You'll see that on a sign and cosine graph the concavity changes at wherever the line passes the midline. And as a reminder on concavity just memorize this image here. Okay, one more thing to talk about and to understand it let's go back to my unit one topic 5 video where I sounded truly dead inside. A function is even if it satisfies the property f ofx= f ofx or if reflected across the yaxis the graph looks the exact same. A function is odd if it satisfies the property f of negativex= negative f ofx or if rotated across the origin 180 degrees it remains looking the exact same. If we were to apply these properties to the s and cosine functions or try the rotational and reflective symmetry on the graphs you will see that y= s of theta is an odd function and y= cosine of theta is an even function. Okay, manipulation, a tool that can get you really anything you want in life. And of course, I'm talking about this in the context of sinosodal functions. I mean, what else could it mean? This video serves as a continuation of my topic 3.5 video, but instead of asking you to rewatch it, I'll just reuse some clips from the video in this one. And since I said it here, you're not allowed to get mad at me. Take a look at this equation. This is the skeleton sinoso equation. I want you to know this same skeleton equation is used for s and cosine functions. Let's examine it from left to right. A is the amplitude of the function and roll the clip. The amplitude of a sinosoidal function is the vertical distance from the middle line to the maximum point on the graph. That means in the parent sign and cosine graphs the amplitude is 1. B is a little strange. B is equal to 2 pi over the period. And remember whatever the period is will show how long it takes for the graph to complete one cycle of its pattern. C is the horizontal shift or as it's known in the trig world as the phase shift. Just like any horizontal translation, it moves the graph by negative C units left or right. So negative p<unk> /2 would move the graph pi over2 units to the right because it's always the opposite. And finally d is your vertical shift, but you know it better as the middle line. And if you want to know what that is, roll the clip. A middle line of a sinosoidal function is the invisible line that's halfway between the minimum and maximum points. The middle line of the parent sign and cosine graphs is y equals 0. I should also mention d can be on the front of the equation as well and mean the exact same thing. Okay, so let's practice really quick. Take a look at this graph. To construct an equation from it, first we find the midline which is three. Then find the amplitude by finding the distance from the midline to the maximum point and it's two. Then we find the period which is pi and put it under 2 pi which is reduced to two. Then we take either the sign or cosine parent graph and see what the phase shift would be to follow the pattern. We see for cosine it wouldn't have a phase shift but for sign the phase shift would be plus pi over 4. This topic is very important. In other topics of this course we might say oh yeah this will probably be on your AP exam. No, this topic is literally guaranteed to be on your exam. It is written in the exam description that this is a literal requirement. So, listen up. For this topic, we're going to take an actual question from the 2024 AP exam. And I urge you to pause the video now and try it on your own. If you don't want to try it, at least pause to read it because I'm not going to read it to you. All right, all good? Let's go. So, the way I like doing this is to make a graph first and then the equation. When you make a graph from one of these scenarios, you will always want to make five points. And I'll show you why. So when the tire touches the ground, it will be our start point at 1 /2. Then it touches the ground again at 5 over two. So let's write down the three points between those two to get up to our fivepoint total. Good. Now let's plot the y-axis. The maximum is 18 making the midline 9. At 1 /2 the tire is on the ground or at zero and it is also at zero at 5 over2. So we just fill in the extra points from there. The way these points are modeled imagines the tire spinning in a circle. Great. So they give us the sign skeleton equation in the problem. So let's make it. The amplitude or a is 9. The period is 2. So we can put that under 2 pi to get pi as our b. The midline or d is 9. And finally c. Here's a trick for c. cosine graphs start at the maximum and go down to the midline. Sign graphs start at the midline and go up to the maximum. So where do we see on the middle line? 8 going up to our maximum. That's right at 2 over two or one. Meaning c would be minus one because it's always the opposite. And this is our fully simplified answer. All right. One more thing to talk about, calculators. If you look through your regressions page, you will find that you also have a sign regression, which is cool. But here's the most important part. Your calculator functions on modes, particularly radians and degrees. If I were to do, say, sine of 90 when I was in degrees mode, it would be doing s of 90°. But if I were in radians mode, I would need to change it to be s of p<unk> /2 radians. Since you only will be dealing with radians in this course, you should always have your calculator be in radians mode. Wow, that was a lot of info. Let's take a quick break. If you're not following my Tik Tok yet, you seriously are missing out on absolute masterpieces such as my AP pre-calculus unit one video has more views than Sabrina Carpenters's video. Oh my goodness, this is very fair. Don't judge me. Just truly marvelous. Now that we're refreshed, let's jump into the second half. So, you might be wondering why did we start this unit solving S, cosine, and tangent, but never talk about the tangent graph. Well, my friends, it's because tangent is a little strange of a graph. Let's look at the unit circle, but only with tangent solutions. We get these solutions off of the original unit circle by knowing tangent is equal to s over cosine. You notice tangent is not defined at p<unk> / 2 and 3 p<unk> /2. That simply means tangent cannot exist at those two points which would make vertical asmmptotes on the graph. If you were to graph all of the tangent points on the graph, it would look like this. Look at all those vertical asmmptotes. You notice that the graph goes to positive and negative infinity and looks real fun. This is the parent tangent graph. f of theta is equal to tangent of theta. The period is not 2 pi but rather now pi. The way you find it is to find the horizontal distance from X intercept to X intercept. The skeleton equation is basically the same from sinosoidal functions. D is your midline. You'll notice the graph changes from concave down to concave up once it passes the midline. C is your phase shift. As for B, since the parent function has a period now that is pi, B is now equal to pi over the period for only tangent functions. And now that gets us to A. You'll notice on the tangent graph that it goes to positive and negative infinity. Since the amplitude is the distance from the middle line to the maximum, what the heck is the amplitude now? Well, here it is. On the horizontal axis, look between two x intercepts. Then divide that quadrant by four and find what the point is. Here it is one. The amplitude is whatever the distance between your midline and the y point here is. So in this example, it is 1. I hope that makes sense. Look from x intercept to x intercept. Divide by 4. Then find the y-coordinate of that point. Find the distance between the middle line and that y point. And that's your amplitude. And though I of course have no knowledge on what might be on your exam this year, when I took the AP precos exam, there was not a single question on the tangent graph. So, you know, it might not be that important to know. Everyone, I honestly forgot this topic existed. The reason I say this is because I thought I needed to teach this earlier, so I taught it in my 3.3 video. I do still have some more things to say on inverse functions, but for now, let's get a reminder on what I said back in my 3.3 video. If you are given an inverse s cosine, or tangent equation, it's not that difficult. So for example in this one where it's inverse sign of roo<unk>3 over2 all it's asking is what angle will have roo<unk>3 over2 as an answer for the sign of it. So looking at our unit circle we see it in two places. Thing is dealing with inverse functions certain inverse functions only exist in certain quadrants of the graph because of domain restrictions. Since inverse sign can only exist in quadrants 1 and 4. Even if there is an answer of roo<unk>3 over2 at 120° it can't be an answer because inverse sign can't exist there. Meaning our only answer would be 60° or p<unk> over 3. Another way to notate inverse function is using arc notation. Just know it means the same exact thing. So what the heck do I mean when I say certain inverse functions only exist in certain quadrants of the graph? Essentially because the corresponding trigonometric functions are periodic, they are only invertible if they have restricted domains. The restricted domain of sign is<unk> /2 to p<unk> /2 or the way I like looking at it quadrant 1 and 4. The restricted domain of cosine is 0 to pi or the way I like looking at it quadrant 1 and 2. And the restricted domain of tangent is negative p<unk> over2 to p<unk> /2 or once again quadrants 1 and four. Now I don't want to confuse you. If you see an inverse function, you can assume it will always be solved within their restricted domains. But if you see a trigonometric equation like this, you no longer need to worry about restricted domain. You can solve it and put both of the solutions pi over 3 and 5<unk> over 3. But since these functions are periodic, you need to model all the periodic solutions to the trigonometric equation as well. So that means when you solve trigonometric equations that don't have any domain, you must slap a plus 2 pi k on the end of each solution to showcase all possible co-terminal angles. You will notice that I actually already did this in my topic 3.3 video. So your takeaway is to always use restricted domain if you are solving inverse functions, but find every possible solution with trig equations like this. Okay. Okay. Hey there. Before you watch this video, you need to make sure that you've watched my topic 3.3 and 3.9 video. And as always, my unit 3 playlist is in the description below the link for you to subscribe to my second channel. Okay, take a look at this problem. 2 * cosine of x + 1 is equal to 0. Here's my advice. Get rid of the cosine and add it back later. So now it's just 2x + 1= 0. So 2x =1 and x =1 /2. Great. Now that we've solved for x, we can add back the cosine. So cosine of x is equal to -1 /2. That would be the same thing as saying r cosine of 1 /2, which looking at our unit circle, it's at 2<unk> over 3 and 4<unk> over 3. And since we don't have any domain for the problem, we need to slap the plus 2 p<unk> k on it. And then this is our fully simplified answer. Let's try this problem. It's a calculator problem. Tangent of x - tangent of x - 6 is equal to 0. Let's remove the tangent making it x^2 - x - 6 = 0. Factoring it, we get x - 3 and x + 2. So we get x= 3 and x= -2. So then we add the tangent back and we get these arc equations. These aren't traditional values on the unit circle. So this is where we plug it in on our calculator in radian mode and we get x= 1.249 and x=1.107. And then we slap the plus 2 pi k on it and we are done. Hey wo wo wo. It's not 2 pi k. Tangent has a period of pi. So for this it would actually be plus pi k. It really isn't that bad for these problems. I promise. Let me introduce you to an evil word. A word that should always be hated when in the context of math. Identity. An identity is something that equals something else. Next video is all about proving identities and it is so long so I need to catch a head start in this video. So let's talk about the start of these being the reciprocal identities. This video introduces three functions. Cosecant seeant and cootangent. Cosecant is equal to 1 / s. Seeant is 1 / cosine and cotangent is equal to 1 / tangent. This also is true for the opposite. So 1 over cosecant is equal to s. 1 secant is equal to cosine and 1 / cent is equal to tangent. Oh, and another thing. Since tangent is equal to s over cosine, cotangent is equal to cosine over s. So if you were asked to find the secant, cosecant, or cotangent of an angle, you just find the reciprocals answer to that angle and then put a one on top of it and solve. Now let's talk about the graphs for these functions. Understand that for things like exponential graphs, log graphs, and s and cosine graphs, you need to understand kind of every little thing about it. But the tangent graph and the graphs in this video are just kind of something you need to get a vague idea of what they look like. Okay. Okay. Let's start with seeant. The graph looks weird. It's a bunch of U-shapes. But watch this. If we overlay the reciprocal cosine graph, it suddenly makes sense. Each extremum from the cosine graph matches with the seecant graph. Now for cosecant, it follows the same pattern. If we overlay the sign graph, we see that the cosecant u shapes match with the sign minimum/maximum points. As for co-angent, co-angent is actually really easy. Remember this is what a tangent graph looks like. The co-angent graph looks like this. Mhm. Yeah. It's just the tangent graph, but the lines go the other way and the asmmptotes shift just a little. Just kind of understand the vague idea of what these graphs look like and you'll be fine. My friends, I'd like to preface this video by already apologizing for the content in this video. But I just need to get going. So, we already know the reciprocal identities from last video. Now, we need to add another set of identities known as the Pythagorean identities. All of these identities can be used to prove an identity true. When you see an identity like this and it asks you to rewrite it, in this case, in terms of tangent, let me show you how you do it. Also, this question was taken from the 2024 AP exam. So, the truth is there is no right answer on what to do. You could do things completely different for me and get the same exact answer. I'll just show you how I'd do it. So, first I want to try and simplify each term. In the numerator, we have 1 - sin^ square of x. But here's the thing. We know sin^ square of x + cosine^ 2 of x is equal to 1. So, if we rearrange this, we know 1 - sin^ 2 x is equal to cosine^ of x. From the reciprocal identities, we know is equal to 1 / cosine. Now, we can multiply the two fractions together. We are able to cross out the cosine on the bottom and the square root on the top and that leaves us with cosine of x over sin of x which is equal to cent of x. But we need to simplify it in terms of tangent as the question asked. So we know cent of x is equal to 1 / tangent of x. So this is our final answer. A trig identity like this is guaranteed to show up on your exam. So at the end of this video, there's a handful of them for you to try. But wait, there are more identities to talk about, my friends. We have something called some indifference identities. There are also quite a bit of them. But for this course, you only need to know the sum identity for cosine and s and none of the other ones. You also need to memorize these double angle identities. So that means in total you need to memorize this screen of identities for this exam. And if you're wondering, this is why people say pre-calculus is harder than calcul. But I know you guys are lazy. So I'll just say as someone who took this exam, these ones don't really matter. But definitely know these. Really quick, let me show you how to use the sum identities. If you had an expression like sine of 75°, we can rewrite this as s of 45° plus 30°. And using the identity, we can plug it into it, then simplify, and we get our fully simplified answer. And as for the double angle identities, they really aren't that important. Here are some questions to try if you really care about them so much. And here are the answers. But I wouldn't really care too much for them. All right, my friends. Here's your full identity screen again. Memorize it. Good. Good. Oh my goodness. It's so close. The end is near. Just a few more videos and we can finally be over with this course. Welcome to the last function you need to learn in this course. Polar functions. Here's how I want you to think of polar functions. Think of the unit circle earlier in this unit where we had angles on a graph. That's essentially a polar graph. Now imagine a bunch of invisible circles on the graph. Those are your points. Polar functions don't use the xycoordinates. They instead use r comma theta coordinates. So if we needed to have a point at 2, p<unk> over4, we go to where p<unk> over4 is on the graph. Then we go out to the second circle and plot our point. Isn't that fun? If we had like 3a 4 p<unk> over 3, we go to 4 p<unk> over 3 and then three circles out and plot our point. But what if we had something like -2a 5<unk> over4? Well, here we use something called crossing the pole where we flip the negative point across the origin and we go out by 2. It's really quite a simple coordinate plane to understand. So here's a fun question. If we have a sinosoidal point in the format x comma y like this one roo<unk>3 comma 1, how do we turn it to polar coordinates in the format r theta? Well, you could take the equations from my 3.3 video and use them, but I'll just lay out the basic way here. No matter what, r is equal to the<unk> of x^2 + y^2. And if x is positive, theta is equal to r tangent of yx. And if x is negative, theta is equal to arc tangent of yx +<unk>. In this case, x is positive being<unk>3. So let's plug it into the first equation and solve. Then we plug each value into the r formula and solve. And that my friends is how you get polar coordinates from cartisian coordinates. So if you were given a polar equation like r is equal to 3 + 2 s of theta as an example, we make a table of values for sign and then plot those points on the polar graph. Or you could make a sign cartisian graph and use those points to plot on the polar graph. Honestly, you do you. But graphs are a next video. So polar graphs, they are kind of fun and cool. They make fun drawings, but really they are kind of boring because good luck finding a real life scenario that models this. So, let's just get through this as fast as we can. Polar graphs come in the form r= f of theta. Polar coordinates work in pairs where the input value is the angle and the output value is the radius. So, that means changes in input values correspond to changes in angle measure from the positive x-axis and changes in output values correspond to changes in the distance from the origin. You can restrict the domain of polar functions by limiting the range of theta values which effectively cuts off part of the graph. But let's be real, we only really care about what the main shapes of the graphs are. Polar functions typically come in four forms. Circles, cardioids, limosons, and roses. To understand these, let's model the skeleton equation r= a + b s of theta. Or it can be cosine. It doesn't really matter. A circle happens when you don't have an a. It looks like a circle on a graph. A cardioid happens when a is equal to b. It looks like this on a graph. And I've heard it being compared to a heart before. A limosan happens when b is greater than a. It looks like this on a graph. So kind of like a heart shape but with a loop. And finally, a rose happens when you have something multiplying by theta. It can look like a lot of things, but this is kind of generally what it looks like. All right, this is it. The last video of required course content for the AP exam. Thank you all for sticking with me through this journey. If you want to stay in the loop on what I do next, follow my Instagram. Also, if you have found this content to help you at all this year and would like to support me in a more involved way, consider becoming a channel member by pressing the join button. But for now, for the last time, let's get into it. We've talked about how polar functions work, how to convert between coordinate systems, and how to graph different polar shapes. But what does it actually mean for a polar function to increase or decrease? Think of the polar function r= f of theta like a moving point on a graph. If r is positive and increasing, the point moves away from the origin. If r is negative and decreasing, the point also moves away because it's flipping direction. If r is positive and decreasing, the point moves toward the origin. If r is negative and increasing, the point also moves toward the origin. If r changes from increasing to decreasing or vice versa, that means the function reaches an extremum, which is a maximum or minimum. Average rate of change of r with respect to theta is just how fast the radius is changing per unit of theta. If the rate is positive, the radius is increasing. If the rate is negative, the radius is decreasing. Think of it like driving a car in a circular track. How fast you're moving outward or inward depends on the rate of change of r. You can use the average rate of change to estimate values of r within an interval. It's just like finding the slope of a normal function, except now it tells you how fast the radius is changing per radian. And that's it. The last piece of AP pre-calculus content. It really is it. That is all for your AP exam. And once again, I am eternally grateful for you all. Remember that it only takes two of you guys becoming a channel member for unit 4 to happen. Now for the message I promised. Hey there, I'm Max Allen. This year I covered four AP courses while working a job and maintaining a 4.0 GPA as a full-time college student. Let's be real, that's way too much work. Next year, I'll likely only review one or two courses. So, I'm leaving the decision up to you on what courses I review. Drop a comment below on which AP course you'd like an under three minute series for next year. These are the courses I would be willing to do, and the most liked comments will likely be the ones I choose to review next year. All right, message over. So, I'll finally end this video like this. Remember the Salt Lake City video? Well, did you know that video actually now has an Extra Scenes video to it? Why not watch both of them? [Music]