Polar functions are a special case of parametric functions. These are just equations disguised as nasty trigonometric functions. It's sort of like they are wearing intricate Halloween costumes, like a stick figure man costume made of black cloth and glow sticks. We can't tell who it is, but we can do our best to use our knowledge about a person to figure it out.

The polar function *r* = *f*(*t*) can be parametrized as

*x*(*t*) = *r* cos *t* = *f*(*t*) cos *t*

*y*(*t*) = *r* sin *t* = *f*(*t*) sin *t*

We're using *t* instead of θ so we can talk about instead of , to be consistent with the earlier discussion of parametric functions. To find and , we have to apply the product rule carefully.

### Sample Problem

Find the arc length for one petal of the polar function *r* = sin(3*t*).

Answer.

The function looks like

The first petal corresponds to the interval . We get

*x*(*t*) = *r* cos *t* = sin(3*t*)cos *t*

*y*(*t*) = *r* sin *t* = sin(3*t*)sin *t*.

Using the product rule gives us

We use the arc length formula and get

It looks complicated and drawn out, but we can always just use a calculator to solve the integral. If our teacher asks us to solve this one, they are looking to torture us.

Sometimes, a slightly modified costume simplifies the calculations. We can use the identity

sin^{2} *t* + cos^{2} * t* = 1

to get a nicer formula for the arc length of a polar function.

If we want to make the formula look more like it goes with polar functions, we can put θ in place of *t*:

This formula is nice to keep handy, but we probably don't need to memorize it unless we need to do a lot of polar arc length problems.

## Practice:

Show that the arc length of the polar function *r* = *f*(*t*) for α ≤ *t* ≤ β is

Answer

Since the problem says "show that the arc length... is..." we know what the answer should be at the end. It's always easier to get somewhere if we know where we are going. Let's deal with this like a parametric function. The parametric equations are

*x*(*t*) = *f*(*t*)cos * t*

*y*(*t*) = *f*(*t*)sin * t*

and the corresponding derivatives are

Use FOIL to square the derivatives. It's not as bad as it seems, because the blue terms cancel.

Now group the terms with (*f*'(t))^{2} together and the terms with (*f*(t))^{2} together.

Since sin^{2} *t* + cos^{2} *t* = 1, this simplifies to

Since

the arc length of the curve for α ≤ * t* ≤ β is

This is equivalent to what we were supposed to get.