Polar functions derivatives | Advanced derivatives | AP Calculus BC | Khan Academy

What we have here is the graph of r is equal to sine of two theta in polar coordinates. 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 precalculus section. But I'll give you a little bit of a primer here. Let's just familiarize ourselves with why this graph looks the way it does.

When we look at this graph, we can specify points in terms of x and y coordinates or in terms of an angle and a radius. 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.

To familiarize ourselves with this curve, let's explore it intuitively. When theta is 0, r is going to be 0; sine of 2 times 0 is just 0. So our r is at the origin. As theta gets larger, our r increases, allowing us to trace out this petal of this flower or clover-looking shape. It starts looking like that, and we could keep going all the way.

What happens when theta is equal to pi over 4? Well, when theta is equal to pi over 4, sine of 2 times pi over 4 is sine of pi over 2, which means r is equal to 1. We reach a kind of maximum r there. Then, as theta increases, our r once again starts to get smaller.

Now, in a calculus context, you might wonder how 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? There’s really nothing new here; you have one variable as a function of another. You just use the chain rule and take the derivative with respect to theta.

The derivative of sine of two theta with respect to two theta is going to be cosine of two theta. Then you multiply that times the derivative of 2 theta with respect to theta, which is 2. So we could just keep the times 2 here or write a 2 out front.

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. A review from precalculus is that when you want to go between the polar world and, I guess you could say, rectangular world, you have to remember the transformations: y is equal to r sine of theta, and x is equal to r cosine of theta.

Now, just as a really quick primer, why does that make sense? Let's take one of these angle-r combinations right over here. So let's say this is theta and that is our r. The height of that side will be our y, and the length of this side will be our x. From trigonometry, we know sine of theta is opposite over hypotenuse, meaning sine of theta is equal to y over r, and cosine of theta is equal to the adjacent or x over r. You just have to multiply to rearrange these equations to get to what we have right over there.

Once again, if this is going too fast, this is just a review of polar coordinates from precalculus. Now we can express these purely in terms of theta. We know that r is equal to sine of two theta. So, we just have to replace these r's with sine of two theta.

Then y would be equal to sine of two theta times sine of theta, and x is going to be equal to sine of two theta times cosine of theta. Just like that! But we can also use these expressions to find the rate of change of y with respect to theta. Let’s find a general expression for it. Pause the video and see if you can do that.

Alright, let’s work through it together. Again, we’re just going to use our derivative techniques. I could write y prime of theta, the derivative of y with respect to theta. We’re going to use the product rule right over here. The derivative of this first expression is 2 cosine of 2 theta. We’ve already seen that coming out of the chain rule, and this times the second expression sine of theta plus the first expression sine of two theta times the derivative of the second expression, which is cosine of theta. Fair enough!

Now, we do the same thing for x: x prime of theta is the derivative of the first expression, which is going to be 2 times cosine of 2 theta times the second expression cosine of theta. Then you’ll have the first expression sine of 2 theta times the derivative of the second expression, which is negative sine of theta.

We could evaluate these at, for example, when theta is equal to pi over 4. When theta is pi over 4, we are going to be at this point right over there. Let’s evaluate it. If y prime of pi over 4 is equal to 2 cosine of pi over 2, that’s going to be zero because cosine of pi over 2 is 0.

So if that’s 0, all of this stuff is going to be 0. Here, sine of pi over 2 is 1, and cosine of pi over 4 is square root of 2 over 2. This is going to be equal to square root of 2 over 2.

We could do the same exercise with x. x prime of pi over 4 will still have this first part be equal to 2 times cosine of pi over 2, which is 0 as well. Thus, we conclude - sine of pi over 2 times sine of pi over 4, which is equal to square root of 2 over 2.

Let’s see why that makes sense. If you increase theta a little bit from pi over 4, your y coordinate continues to increase, which makes sense; you have a positive slope. But what happens to your x coordinate? As theta increases a little bit, your x coordinate starts to decrease. That’s why it makes sense that you have a negative rate of change here.

Now, the next question you might ask is, "How do I 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 1.

How would we actually calculate it? One way to think about it is the derivative of y with respect to x is going to be equal to the derivative of y with respect to theta divided by the derivative of x with respect to theta. So at theta equal to pi over 4, this is going to be positive square root of 2 over 2 over negative square root of 2 over 2, simplifying to negative 1.

This makes sense; this does look like a tangent line that has a slope of negative 1. Hopefully, this puts it all together, and you’re feeling a little bit more comfortable. You got to review a little bit of polar coordinates, but we've augmented that knowledge by starting to take some derivatives.