7.2 Calculus of Parametric Curves

[STARTS AT 9:43]

Instructor: I'm going to go ahead and write an equation that is quite important for finding the second derivative. So d squared y over dx squared is equal to ddx dy dx. Now that is just the standard definition. But the way we're going to write this is a little different. This is ddt of dy dx divided by dx dt. 

So we'll take the derivative of-- and if you noticed we basically added in a dt there, the differential. We can sort of think of it that way. Again, that's not exactly what's happening, but that's pretty close. We effectively added in a dt down here. Instead of a ddx, we have a ddx and a dt. We're taking the derivative with respect to t, because dy dx is going to be in terms of t. We have some crazy things happen without doing that. 

All right. So from the last question we know that dy dx is 1 over t. So d squared y dx squared, is going to be ddt of 1 over t. Derivative of that, with respect to t, divided by dx dt. Which we know, dx dt is just 2t. And the derivative of 1 over t is negative 1 over t squared. And so our second derivative is negative one half. But we are going to have a t cubed there. There is a formula for our second derivative. 

[ENDS AT 11:37]