We differentiate polar functions.

and so we can compute the derivative of with respect to using differentials: provided that . With polar functions we have

so

provided that .

- or
- .

Let’s examine the case of first, and restrict to the situation where is between and . If is in the interval , then when . If is in the interval , then when .

The corresponding points in rectangular coordinates are

meaning , and

meaning .

But the numerator was the product of and another term, namely So the numerator is also zero when

At this point we have a choice to make: We can either apply the arcsin function and solve for , or we can work with directly. Perhaps the easier route is to work with directly. To explore , we may choose an algebraic method employing the Pythagorean identity, or a geometric method looking at the unit circle with the Pythagorean theorem. Let’s take the geometric route to compute .

meaning , and

meaning .

In summary, the graph of on the interval has horizontal tangent lines at the points

- ,
- ,
- , and
- .

You can confirm this by looking at the graph below:

We then set this equal to zero and note that the equation is quadratic equation in the variable . The quadratic formula gives that

Now we can find cosine using either geometrically via the unit circle or algebraically with the Pythagorean identity. This time, let’s take the more algebraic route with the Pythagorean identity. When , then

So and when ,

So So now we know that the graph of on has vertical tangent lines at the four points

we see that the graph of on has vertical tangent lines at

You can confirm this visually by looking at the graph below:

When the graph of the polar function intersects the origin (sometimes called the “pole”), then for some angle .

The answer to the question above leads us to an interesting point. It tells us the slope of the tangent line at the pole. When a polar graph touches the pole at , the equation of the tangent line in polar coordinates at the pole is .