Parametric curve arc length | Applications of definite integrals | AP Calculus BC | Khan Academy

Let's say we're going to trace out a curve where our x-coordinate and our y-coordinate that they're each defined by, or they're functions of a third parameter T. So we could say that X is a function of T and we could also say that Y is a function of T. If this notion is completely unfamiliar to you, I encourage you to review the videos on parametric equations on Con Academy.

But we're going to think about, and I'm going to talk about it in generalities in this video. In future videos, we're going to be dealing with more concrete examples. But we're going to think about what is the path that is traced out from when T is equal to a. So this is where we are when T is equal to a. So in this case, this point would be X of a, comma, Y of a. That's this point.

And then as we increase from T equals a to T is equal to B, our curve might do something like this. So this is when T is equal to b. T is equal to B, so this point right over here is X of B, comma, Y of B.

Let's think about how do we figure out the length of this actual curve, this actual arc length from T equal to a to T equal to B. Well, to think about that, we're going to zoom in and think about what happens when we have a very small change in t.

So a very small change in t, let's say we're starting at this point right over here and we have a very small change in t. So we go from this point to, let's say, this point over that very small change in t. It actually would be much smaller than this, but if I drew it any smaller, you would have trouble seeing it. So let's say that that is our very small change in our path, in our arc that we are traveling.

And so we want to find this length. Well, we could break it down into how far we've moved in the x-direction and how far we've moved in the y-direction. So in the x-direction, right over here, we would have moved a very small change in x.

And what would that be equal to? Well, that would be the rate of change with which we are changing with respect to t, with which x is changing with respect to t, times our very small change in t. And this is a little hand-wavy; I'm using differential notation, and I'm conceptually using the idea of a differential as an infinitesimally small change in that variable.

But this, and so this isn't a formal proof, but it's to give us the intuition for how we derive arc length when we're dealing with parametric equations. So this will hopefully make conceptual sense that this is our dx. In fact, we could even write it this way: dx/dt, that's the same thing as x prime of t times dt.

And then our change in y is going to be the same idea. Our change in y, our infinitesimally small change in y when we have an infinitesimally small change in t, well you could view that as your rate of change of y with respect to t times your change in t, your very small change in t, which is going to be equal to—we could write that as y prime of t dt.

Now based on this, what would be the length of our infinitesimally small arc length right over here? Well that we could just use the Pythagorean theorem. That is going to be the square root of—that's the hypotenuse of this right triangle right over here. So it's going to be the square root of—I'm going to give myself a little bit more space here because I think I'm going to use a lot of it.

So the stuff in blue squared, dx²; we could rewrite that as x prime of t dt² plus this squared, which is y prime of t dt². And now let's just try to simplify this a little bit. Remember, this is this infinitesimally small arc length right over here, so we can actually factor out a dt². It's a term in both of these, and so we can rewrite this as—let me so I can rewrite this, my big radical sign.

So I'm going to factor out a dt² here. So we could write this as dt² times (x prime of t)² plus (y prime of t)². And then to be clear, this is being multiplied by all of this stuff right over there. Well, now we can actually, if we have this dt² under the radical, we can take it out, and so we will have a dt.

And so this is all going to be equal to the square root of—so the stuff that's still under the radical is going to be (x prime of t)² plus (y prime of t)². And now we took out a dt, and now we took out a dt. I could have written it right over here, but I'm just writing it on the other side. We're just multiplying the two.

So this is once again just rewriting the expression for this infinitesimally small change in our arc length. Well, what's lucky for us is in calculus we have the tools for adding up all of these infinitesimally small changes. That's what the definite integral does for us.

So what we can do if we want to add up that plus that plus that plus that, and remember these are infinitesimally small changes—I'm just showing them as not infinitesimally small just so you can kind of think about them. But if you were to add them all up, then we're essentially taking the integral, and we're integrating with respect to T.

And so we're starting at T is equal to a all the way to T is equal to B. And just like that, we have been able to at least get a conceptual—get feel good conceptually for the formula of arc length when we're dealing with parametric equations. In the next few videos, we'll actually apply it to figure out arc lengths.