Worked example: Parametric arc length | AP Calculus BC | Khan Academy

Let's say that X is a function of the parameter T, and it's equal to cosine of T, and Y is also defined as a function of T, and it's equal to sine of T. We want to find the arc length of the curve traced out, so the length of the curve from T equals 0 to T equals Pi/2.

So pause this video and see if you can work that out based on formulas that we have seen in other videos.

All right, so first I'm going to look at the formula, and then we're going to visualize it and appreciate why what we got from the formula actually makes sense.

The formula tells us that the arc length of a parametric curve is equal to the integral from our starting point of our parameter T equals a to our ending point of our parameter T equals B of the square root of the derivative of x with respect to T squared plus the derivative of y with respect to T squared dT.

This could also be rewritten as this is equal to the integral from A to B of the square root of (dx/dt)² + (dy/dt)² dT, but either way, we can now apply it in this context.

What is dx/dt? So dx/dt is equal to the derivative of cosine of T, which is equal to negative sine of T. And what is dy/dt? The derivative of y with respect to T, the derivative of sine of T, is cosine of T.

So our arc length up here is going to be equal to the integral from T equals 0 to Pi/2. That's what we care about—our parameter is going from 0 to Pi/2—of the square root of the derivative of x with respect to T squared, which is negative sine of T squared.

Well, if you square it, the negative is going to go away. Negative sine times negative sine is positive sine squared. So I could write this as sine²(T) and then (dy/dt) is just cosine of T, plus cosine²(T), and then we have our dT out here.

Now, lucky for us, sine² plus cosine² of some variable is always going to be equal to one. So that just—that's one of our most basic trig identities, coming straight out of the unit circle definition of sine and cosine.

And so we have the square root of one, the principal root of one, which is just going to be one. So all of this thing, everything here, has just simplified to the integral from 0 to Pi/2 dT.

Well, this is going to be equal to—you can view this as a one here—the anti-derivative of one with respect to T is just going to be T. We're going to evaluate that from Pi/2. We're going to evaluate that at Pi/2 and then subtract it evaluated at zero.

So this is going to be equal to Pi/2 - 0. That's going to be equal to Pi/2.

Now let's think about why this actually does make sense. Let's plot this curve. So that is my y-axis; this is my x-axis right over here. When T is equal to 0, you have X of 0, which is cosine of 0. X is equal to 1, and Y of 0 is just zero. So Y is equal to zero, so we're at this point right over here when T is equal to zero.

And then as T increases up to Pi/2, we trace out the top right corner of the unit circle, and we end up right over here when T is equal to Pi/2. You could view T in this case as some type of an angle in radians, and so the arc length is really just the length of a quarter of a unit circle.

Well, we know what the circumference of a circle is: it is 2πr. In the unit circle's case, the radius is one, so the circumference of the entire circle is 2π. One-fourth of that is going to be π/2.

So it's nice when this fancy thing that we feel good about in calculus is consistent with what we first learned in basic geometry.