Worked example: differentiating polar functions | AP Calculus BC | Khan Academy

Let r be the function given by r if theta is equal to three theta sine theta for theta is between zero and two pi, including zero and two pi. The graph of r in polar coordinates consists of two loops, as shown in the figure above.

So let's think about why it has two loops. As our theta, when theta is zero, r is zero, and then as our theta increases, we start tracing out this first loop all the way until when theta is equal to pi. So we traced out this first loop from theta is equal to zero to theta is equal to pi.

And then the second loop has a larger r, so these are larger r's. This is when we're going from pi to two pi. You might say, well, why doesn't it show up down here? Well, between sine of pi and sine of 2 pi, this part right over here is going to be negative, so it flips it over the r into this side. The magnitude of the r is larger and larger because of this 3 theta. So when we go from pi to 2 pi, we trace out the larger circle. Fair enough, that seems pretty straightforward.

Point P is on the graph of r right over there and the y-axis. Find the rate of change of the x-coordinate with respect to theta at the point P. All right, so let's think about this a little bit. They don't give us x as a function of theta; we have to figure out that from what they've given us.

So, just as a bit of a polar coordinates refresher, if this is our theta right over there, this is our r, and that would be a point on our curve for this theta. Now, how do you convert that to x and y's? Well, you can construct a little bit of a right triangle right over here, and we know from our basic trigonometry that the length of this base right over here is going to be the hypotenuse.

Let me just write that, that's going to be our x-coordinate. Our x-coordinate right over here is going to be equal to our hypotenuse, which is r times the cosine of theta. If we wanted the y-coordinate as a function of r and theta, it'd be y is equal to r sine of theta. But they don't want us to worry about y here, just the x-coordinate.

So we know this, but we want it purely in terms of theta. So how do we get there? Well, what we can do is take this expression for r, r itself as a function of theta and replace it right over there.

So what we can do is we can write, well, x of theta is going to be equal to r, which itself is 3 theta sine of theta, times cosine of theta. And now we want to find the rate of change of the x-coordinate with respect to theta at a point. So let's just find the derivative of x with respect to theta.

So x prime of theta is equal to, well, I have the product of three expressions over here. I have this first expression 3 theta, then I have sine theta, and then I have cosine theta. So we can apply the product rule to find the derivative.

If you're using the product rule with the expression of three things, you essentially just follow the same pattern when you're taking the product of two things. The first term is going to be the derivative of the first of the expressions, 3, times the other two expressions.

So we're going to have 3 times sine of theta, cosine of theta, plus the second term is going to be the derivative of the middle term times the other two expressions. So we're going to have three theta, and then the derivative of sine theta is cosine theta times another cosine theta.

You're going to have cosine squared of theta, or cosine of theta squared, just like that. And then you're going to have the derivative of the last term. The derivative of cosine theta is negative sine theta. So if you multiply negative sine theta times three theta sine theta, you're going to have negative three theta sine squared theta.

And so we want to evaluate this at point P. So what is theta at point B? Well, point P does happen on our first pass around, and so at point P, theta is equal to theta right over here is equal to pi over 2. So pi over 2.

So what we really just need to find is, well, what is x prime of pi over 2? Well, that is going to be equal to 3 times sine of pi over 2, which is 1, times cosine of pi over 2, which is 0. So this whole thing is 0, plus 3 times pi over 2; this is 3 pi over 2, times cosine squared of pi over 2, or cosine of pi over 2 squared. Well, that's just zero.

So, so far, everything is 0, minus 3 times pi over 2, 3 pi over 2, times sine of pi over 2 squared. Well, what's sine of pi over 2? Well, that's one. You square it, you still get one. So all of this simplified to negative 3 pi over 2.

Now it's always good to get a reality check. Does this make sense that the rate of change of x with respect to theta is negative 3 pi over 2? Well, think about what's happening. As theta increases a little bit, x is definitely going to decrease. So it makes sense that we have a negative out here.

So right over here, the rate of change of x with respect to theta is negative 3 pi over 2. As theta increases, our x for sure is decreasing, so at least it does make intuitive sense.