Second derivatives (parametric functions) | Advanced derivatives | AP Calculus BC | Khan Academy

So here we have a set of parametric equations where x and y are both defined in terms of t. If you input all the possible ts that you can into these functions and then plot the corresponding x and y's for each chord for each t, this will plot a curve in the xy plane.

What I want to do in this video is figure out the first derivative of y with respect to x and the second derivative of y with respect to x. In both cases, it's going to be in terms of t. So, let's get to it!

First, let's find the first derivative of y with respect to x. The first derivative of y with respect to x, and we've seen this before in other videos, is going to be the derivative of y with respect to t over the derivative of x with respect to t.

So, this is going to be equal to... well, what is the derivative of y with respect to t? dy/dt is equal to... let's see, the derivative of e to the 3t with respect to 3t is just e to the 3t, and then the derivative of 3t with respect to t is going to be 3. So, I could say times 3 like that or I could put that 3 out front, and then the derivative of negative 1... well, negative a constant doesn't change, no matter what you do with your t, so that's just going to be zero.

So that's dy/dt. It's going to be equal to 3e to the 3t. All of that over... well, what's the derivative of x with respect to t? The derivative of x with respect to t is equal to... well, we're going to have the 3 out front, and so the derivative of e to the 2t with respect to 2t is going to be e to the 2t. Then we're going to take the derivative of 2t with respect to t, which is just 2.

So, this is going to be 6e to the 2t, or 6e to the 2t. And let's see, we could simplify this a little bit. I'll now go to a neutral color; this is equal to... so, e to the 3t minus 2t.

I'm just using exponent properties right over here. But three... if I have three ts and I take away two of those ts, I'm just gonna have a t. So, this is just going to simplify to t right over here.

Now that we've figured out the first derivative of y with respect to x in terms of t, now how do we find the second derivative of y with respect to x? Now, I’ll give you a hint: we're going to use this same idea. If you want to find the rate of change of something with respect to x, you find the rate of change of that something with respect to t and divide it by the rate of change of x with respect to t.

What this is going to be... this is going to be... we want to find the derivative of the first derivative with respect to t. So, let me write this down. We want to find... we want to take the derivative with respect to t in the numerator of the first derivative, which I will put in blue now of dy/dx—all of that over dx/dt.

Now, I want you to, if it doesn’t jump out at you, why this is the same thing that we did before. I encourage you to pause the video and think about it. Think about what we did over here the first time. When we finally wanted to find the derivative of y with respect to x, we found the derivative of y with respect to t and then divided that by the derivative of x with respect to t.

Here, we want to find the derivative... we want to find the second derivative of y with respect to x. Actually, let me just write it down out here a little bit clearer. What we really want to do is find the derivative with respect to x of the first derivative with respect to x. So, everywhere we saw a y here, replace that with the first derivative.

So this is going to be equal to in the numerator the derivative with respect to t of dy/dx. Notice this was the derivative with respect to t of y. In fact, let me write it that way just so you can see it. If I clear this out, we're going to get... this is the derivative with respect to t of y.

So hopefully you see, before we had a y there, now we have dy/dx over dx/dt. Now this might seem really daunting and complicated except for the fact that these are actually fairly straightforward things to evaluate. Taking the derivative with respect to t of the first derivative... well, that's just taking the derivative with respect to t of this, and this is pretty easy.

This is the derivative; it's just going to be one-half, and the derivative with respect to t of e to the t is just e to the t. So that's going to be over the derivative of x with respect to t, which we already saw is 6e to the 2t.

So we can write this... we can write this as, let's see, one-half divided by 6 is 1 over 12 and then e to the t minus 2t, which is equal to... we could write this as 1/12 e to the negative t, or we could write this as 1 over 12 e to the t, and we're done.