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## Differential Calculus

### Course: Differential Calculus>Unit 6

Lesson 5: Polar functions

# Polar functions derivatives

Finding derivatives of 𝑟, 𝘹, and 𝘺 of a function given in polar coordinates.

## Want to join the conversation?

• If anyone is struggling to find the intro to polar coordinates like I did, here's a link https://www.khanacademy.org/deprecated/dep-math/dep-trig/v/polar-coordinates-1
• This is what I learned on this video and just want to verify if they're correct.

1) Calculating y' in terms of theta will give you the rate of change of the y-value as theta changes,
2) Calculating x' in terms of theta will give you the rate of change of the x-value as theta changes, and
3) The rate of change of y with respect to x will give you the slope of the graph defined by the polar function at any point.
• Why don't you get the same result for x' and y' if you differentiate r, and then convert the result to Cartesian coordinates?
• Because at that point you aren't calculating x' and y' but rather the x and y values of r' which isn't the change that x and y go through as theta changes, but rather a cartesian representation of the change r goes through. The simplest way to get x' and y' is to follow this method Sal used as it shows clearly that first you are obtaining functions x(theta) and y(theta) and from there you are taking their change for the change in theta (i.e.: taking their derivatives).
• So, to differentiate a polar function, we first express x and y as functions of the third parameter theta, which is in fact "parametrization", and then differentiate the x- and y-components separately. Do I got this right?
• Yes, then you divide the derivative of y with respect to theta by the derivative of x with respect to x to find out the tangent, which is the derivative of y with respect to x(dy/dx).
• I am confused why evaluating the derivative of the polar expression--r'(theta) = 2 cos(2 theta)) -- at pi/4 equals zero, while the dy/dt / dx/dt evaluation of r(theta)=sin(2theta) equals negative 1. Its seems like the dy/dt / dx/dt derivative of the initial expression should equal xy expression of the polar derivative!
• The dy/dt/dx/dt evaluation is describing the change in y of the function with respect to x. The evaluation of r'(theta) is describing the change in the radius of the function, the distance from the point on the function the the origin, with respect to theta. These two evaluations describe change in the function at the same point, just using different variables.
• What does the derivative of r'(theta)= 2cos(2theta) actually mean in itself (without respect to x or y)? I thought it would refer to the slope of the r-vector, but clearly that isn't the case (or at pi/4 the slope would be zero).
• r'(theta) = dr/[d(theta)]

This is the rate of change of r with respect to theta.
• i dont know how to calculate horizontal distances
• In polar? If you have two polar coordinates you can simply convert them to rectangular (Cartesian) form and then subtract the 𝑥-coordinates to find the horizontal distance between the two coordinates. If you're not sure how to do this, I suggest watching the videos on the polar coordinate system on KA.
• If you were to find the magnitude of the x'(theta) and y'(theta) components, would it be equal to r'(theta)?
• I tried using sin(2x)=2*sinx*cosx to turn r=sin(2θ) into x,y form. I got (x^2 + y^2)^3/2 but then I graphed to check if I did correct, and it did not have the same graph as r=sin(2θ). Where did I go wrong? Any ideas why what I did has error?
• I am not sure what steps you took here, but whatever you did, you started with an equation (r=sin(2θ)) and ended up with an expression (there's no = sign in (x²+y²)^(3/2)).

If you apply the identity, we get r=2sin(θ)cos(θ). Multiplying by r² gives
r³=2(rsinθ)(rcosθ)
r³=2(y)(x)

Now from the identity r²=x²+y², we can raise both sides to the 3/2 to get r³=(x²+y²)^(3/2).

If we back-substitute this expression for r³ into our equation, we get (x²+y²)^(3/2)=2xy.

If you graph r=sin(2θ), you get a four-petaled rose. If you graph the equation above, you get exactly half of this graph, the petals in the first and third quadrants. This is because we took a square root when we raised the equation to the 3/2 power, so we are missing a ± sign.

Indeed, if we replace 2xy by -2xy, we get the other two petals of the graph instead, the second- and fourth-quadrant petals. So if we took the union of these graphs (algebraically, if we include the ± sign), we get exactly the same curve in both cases.
• 1. at , what makes the line from the origin to a point on the graph of r=sin(2θ) be r? Is it because it's defined that way?

2. sort of related to the above question. at , for r=sin(2θ), how do we know it's y=rsinθ? I know y=rsinθ works for the unit circle, but how do we know that it also works for the graph of r=sin(2θ)? thanks.
• 1. Yes, it is the way it’s defined. The r stands for “radius”. Polar functions work by taking in an angle and outputting a distance/radius at that angle.

2. On the unit circle, the y-value is found by taking sin(θ). Notice the r isn’t in the formula because on the unit circle r=1. Now, for polar functions, r changes, so to get the y-value you have to multiply r by sin(θ).

Why do you multiply r by the sine? Because sine is opposite over hypotenuse. For polar functions, the “opposite” is the y-value and the hypotenuse is the radius. So, when you multiply the sine by r you are just left with the y-value.

## Video transcript

- [Instructor] What we have here is the graph of R is equal to sin of two theta in polar coordinates and if polar coordinates look unfamiliar to you or if you need to brush up on them I encourage you to do a search for polar coordinates in Khan Academy or look at our pre-calculus section but I'll give you a little bit of a primer here. Let's just familiarize ourself why this graph looks the way it does. 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 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 zero, R is going to be zero, sin of two times zero is just zero, so our R we're just gonna be at the origin and then as theta get larger, our R gets larger and so, we start tracing out this pedal 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 four? When theta is equal to pi over four right over there, well, sin of two times pi over four is sin of pi over two, R is equal to one. So, we reach a kind of a maximum R there and then and as theta 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. You just have one variable as a function of another. You just use the chain rule. Take the derivative with respect to theta right over here. So, the derivative of sin of two theta with respect to two theta is going to be cosine of two theta and then you multiply that, times the derivative of two theta with respect to theta which is two, so we could just say times two here or we could write a two out front. Alright, that was interesting but let's see if we can express this curve in terms of Xs and Ys and then think about those derivatives. So, one primer, a review from pre-calculus is that when you wanna go between the polar world and the, I guess you can say rectangular world, you have to remember the transformation that Y is equal to R sin of theta and that X is equal to R cosine of theta. Now, just as 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, the height of that side is going to by 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, the SOHCAHTOA definition of our trig functions, sin of theta is opposite over hypotenuse, sin 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 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 pre-calculus. But now we can use these to express purely in terms of theta. How do we do that? Well, we know that R is equal to sin of two theta, so you just have to replace these Rs with sin of two theta. So, Y would be equal to sin of two theta, sin of two theta times sin of theta, times sin of theta and X is going to be equal to sin of two theta, sin 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. Right, let's work through it together. Well, this is once again, we're just gonna use our derivative techniques, so I could write Y prime of theta, the derivative of Y with respect to theta, just gonna use the product rule right over here, derivative of this first expression is two cosine of two theta, cosine of two theta, we've already seen that. That's just coming out of the chain rule and then times the second expression, sin of theta and then plus, plus the first expression, sin of two theta, sin of two theta times the derivative of the second expression, derivative of the sin of theta is cosine of theta. Fair enough and we could do the same thing for X. X prime of theta, derivative of the first expression, it is going to be two times cosine of two theta, two times cosine of two theta, times the second expression, cosine of theta and then you're gonna have the first expression, sin of two theta, times the derivative of the second expression which is negative sin of theta, negative sin of theta and we could use this, we could actually evaluate these at points. For example, we could say well, what's happening when theta is equal to pi over four? So, when theta's pi over four, 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 four is equal to, let's see, this is going to be equal to two cosine of pi over two, two times pi over four, times sin of pi over four plus sin of two times pi over four is sine of pi over two times cosine of pi over four. Cosine of pi over four. What is this going to be equal to? Well, cosine of pi over two is zero, so if that's zero all of this stuff's gonna be zero and here's sin of pi over two, 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 just for the sake of 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 gonna 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 gonna look the same, so that first term's gonna be zero. Then we're gonna have minus, so this is all gonna be zero, so then we're gonna have minus sin of pi over two times pi over four is sin of pi over, sin of pi over two, and then times sin of pi over four, sin of pi over four. Now, this is just going to be one and so, this is gonna 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 that that makes sense. So, let's think about what happens as theta increases here. If you increase theta a little bit from pi over four, if you increase it just a little bit, 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 that you might say is say is well, I wanna 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 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 two over two over negative square root of two over two, negative square root of two over two and this all simplifies to being equal to negative one, which makes sense. This does look indeed like a tangent line that has a slope of negative one. So, hopefully this puts it altogether, you're feeling a little bit more comfortable, you got a review a little bit of the polar coordinates but we've augmented that knowledge by starting to take some derivatives.