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# Polar functions derivatives

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.G (LO)
,
FUN‑3.G.1 (EK)
,
FUN‑3.G.2 (EK)

## Video transcript

what we have here is the graph of R is equal to sine of two theta in polar coordinates and if polar coordinates look unfamiliar to you or if you need to brush up on them and encourage you to the search for polar coordinates and Khan Academy or look at our precalculus section but I'll give you a little bit of a primer here let's just for let's just familiarize ourselves why this graph looks the way it does when so what we're doing for any point here we could obviously specify these points in terms of x and y coordinates but we could also specify them in terms of an angle and a radius so for example this would be have some x coordinate and some y coordinate or we could draw a line from the origin to that point right over here and specify it with some angle theta and some R which is the distance from the origin to that point and just to familiarize ourselves with this curve let's just see why it's intuitive so when theta is 0 R is going to be 0 sine of 2 times 0 is just 0 so R we're just going to get to origin and then it's theta gets larger our R gets larger and so we start tracing out this petal of this flower or clover looking thing so it starts looking like that and we could keep going all the way what happens when theta is equal to PI over 4 when theta is equal to PI over 4 right over there well sine of 2 times pi over 4 is sine of PI over 2 R is equal to 1 so we reach a kind of a maximum are there and then it's data increases our R once again starts to get smaller and smaller and smaller now we're going to do this in a calculus context so the first question might be well how do we express the rate of change of R with respect to theta pause this video and see if you can figure it out what is R prime of theta well there's really nothing new here that you just have one variable as a function of another you just use the chain rule a plot take the derivative with respect to theta right over here so the derivative of sine of two theta with respect to 2 theta is going to be cosine of 2 theta and then you take multiply that times the derivative of 2 theta with respect to theta which is 2 so we could just say times 2 here or we could write a 2 OUTFRONT all right that was interesting but let's see if we can express this curve in terms of X's and Y's and then think about those derivatives so one primer a review from precalculus is that when you want to go between the polar world and the I guess you could say rectangular world you have to remember the transformation that Y is equal to R sine of theta and that X is equal to R cosine of theta now this is a really quick primer why does that make sense well let's just take one of these angle R combinations right over here so let's say this is theta and that is our R well this is our this the height of that side is going to be our Y and then the length of this side is going to be our X well we know from trigonometry from our unit circle definition is sohcahtoa definition of our trig functions sine of theta is opposite over hypotenuse sine of theta is equal to Y over our hypotenuse which is R and cosine of theta is equal to the adjacent or x over R and you just have to multiply both sides of these equations by R to get to what we have right over there and once again if this is going too fast this is a review of just polar coordinates from precalculus but now we can use these to express in terms P in terms of theta how do we do that well we know that R is equal to sine of two theta so you can just have to replace these R's with sine of two theta so Y would be equal to sine of two theta sine of two theta times sine of theta times sine of theta and X is going to be equal to sine of two theta sine of two theta times cosine of theta times cosine of theta just like that but now we can use these expressions to find the rate of change of Y with respect to theta find a general expression for it pause the video and see if you can do that work through it together well this just once again we're just going to use our derivative techniques so I could write y prime of theta the derivative of Y with respect to theta just going to use the product rule right over here derivative of this first expression is 2 cosine of 2 theta cosine of 2 theta we've already seen that that's just coming out of the chain rule and then times the second expression sine of theta and then plus plus the first expression sine of 2 theta sine of 2 theta times the derivative of the second expression derivative sine of theta is cosine of theta fair enough now look at the same thing for X X prime of theta derivative of the first expression it is going to be 2 times cosine of 2 theta 2 times cosine of 2 theta times the second expression cosine of theta and then you're going to have the first expression sine of 2 theta times the derivative of the second expression which is negative sine of theta negative sine of theta and we can use this we can actually evaluate these at points for example we could say well what's happening when theta is equal to PI over 4 so when theta is PI over 4 I'll do that in black right over here we are going to be at this point right over there well let's evaluate it so if I were to say Y prime of PI over 4 is equal to let's see this is going to be equal to 2 cosine of PI over 2 2 times pi over 4 times sine of PI over 4 plus sine of 2 times pi over 4 is sine of PI over 2 times cosine of PI over 4 cosine of PI over 4 what is this going to be equal to well cosine of PI over 2 is 0 so if that zero all of this stuff is going to be 0 and here's sine of PI over 2 this is one cosine of PI over four is square root of two over two square root of two over two so this is going to be equal to square root of two over two and actually sort of take up saving some space I'll just write it right over here it's going to be equal to square root of two over two well we could do the same exercise with X we could say X prime of PI over four let's see we're still going to have two times cosine of two times pi over four so that's going to be two times cosine of PI over two this first part right over here is going to look the same so that first term is going to be zero then we're going to have minus so this is all going to be zero so then we going to minus sine of PI over two times pi over four is sine of PI over sine of PI over two and then times sine of PI over 4 sine of PI over four now this is just going to be one and so this is going to be equal to and this is square root of two over two as well so this is negative square root of two over two now let's see why is that that makes sense so let's think about what happens if theta increases here if you increased it a little bit from PI over four if you increase it just a little bit your Y your y coordinate continues to increase so it makes sense you have a positive slope here but what happens to your x coordinate as theta increases a little bit as theta goes from there to there well then your x coordinate starts to decrease when theta increases so that's why it makes sense that you have a negative rate of change right over here now the next question you might say is well I want to find the rate of change of Y with respect to X because I want to figure out the slope of the tangent line right over there and it looks like it has a slope of negative one but how would we actually calculate it well one way to think about it is the derivative of Y the derivative of Y with respect to X is going to be equal to the derivative of Y with respect to theta over the derivative of X with respect to theta and so at that value so at theta is equal to PI over four this is going to be equal to positive square root of 2 over 2 over negative square root of 2 over 2 negative square root of 2 over 2 and this all simplifies to being equal to negative 1 which makes sense this does look indeed like a tangent line that has a slope of negative 1 so hopefully this puts it all together you're feeling a little bit more comfortable you got to review a little bit of the polar coordinates but we've augmented that knowledge by starting to take some derivatives
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