# Arc Length for Parametric Functions

We just learned how to chop up a simple herb like a chive. They are just simple lines that we know how to describe in terms of the variables *x* and *y*. Sometimes, we need to cut up more intricate herbs like rosemary. These lines may be described in different ways, including in parametric equations.

These parametric equations are just a different language for describing lines. They are nice because it's easier to describe the curve using this other language, and it can be easier to integrate this way too.

The process for finding the length of a curve described by parametric equations is the same. We are just using a different language. It's not even an African click language.

Suppose we want to find the length of the curve described by parametric equations *x*(*t*) and *y*(*t*), on the interval *a* ≤ *t* ≤ *b*.

We break up the curve into little pieces, as usual:

The Pythagorean Theorem says that the length of a little piece is (approximately)

We need to get Δ *x* and Δ *y* in more useful terms, preferably in terms of *t*, so we can end up with an integral with respect to *t*.

The amount *x* changes over a little time interval is approximately multiplied by the length of the time interval.

Similarly, for *y*,

This means the length of a little piece is approximately

This simplifies to

When we integrate over the values that make sense for *t*, we get the length of the curve:

This equation is just like the one we derived in the previous section, but now the equation is parameterized with variable *t*. We're using the notation instead of the notation *y*'(*t*) to make it clear that *t* is the independent variable. Since there are so many variables floating around, using prime notation here isn't the best idea. It's best not to get confused with dots and quotation or prime marks floating around a page of mathematical soup.

Here is a recap video of Pythagorean Theorem.

### Sample Problem

Find the length of the parametric curve described by

*x*(*t*) = *t*^{2}, *y*(*t*) = *t*^{3}

for 0 ≤ *t* ≤ 2.

You may use a calculator to evaluate the integral.

Explain why your answer is reasonable.

Answer.

First, we find the derivatives

The length of the curve is

If we graph the curve, we get

This curve is close enough to the straight line between (0,0) and (4,8) that we would expect the length of the curve to be close to the length of that line.

The length of the line is

Since that's close to our answer, our answer makes sense.

When we're asked to explain why an answer is reasonable, we should compare the curve to a line or circle whose length you know how to find. Even when we're not explicitly asked to do this, it's a good idea as a way to check yourself. If a curve is as long as a green bean, we don't want to trust an answer that says it's as long as Route 66.

With parametric functions we have to take a little extra care with the limits of integration. The limits of integration are values of *t*. If we aren't given the values of *t*, we have to find them ourselves.

Fun fact: "normal" functions are just a special case of parametric functions. The function *f*(*x*) can be parametrized as

*x*(*t*) = *t*

*y*(*t*) = *f*(*t*).

Since *x* and *t* are the same thing, the curve *f*(*x*) for *a* ≤ *x* ≤ *b* is the same as the parametrized curve for *a* ≤ *t* ≤ *b*. We can find the length of the curve by working with the parametrized version.

The derivative of

*x*(*t*) = *t* is

and the derivative of *y*(*t*) = *f*(*t*) is

Then the length of the curve is

Since *x* and *t* are the same, we could just as well write this

which is the same thing as the formula we had earlier.

This means you don't need to remember multiple formulas for arc length. Having to memorize less is better. It'll keep you from making mistakes between formulas, or from mistaking a panda from a polar bear.

If you remember the version for parametric equations, you can find the arc length of a normal function by using the parametrization *x* = *t*.

We prefer the parametric version. It looks just a bit more complicated, but it's more fun. The integral

Looks like it's using the Pythagorean Theorem, which makes it easier to remember.