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## Second derivatives of parametric equations

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# Second derivatives (parametric functions)

AP Calc: CHA‑3 (EU), CHA‑3.G (LO), CHA‑3.G.3 (EK)

## Video transcript

- [Voiceover] So here we have
a set of parametric equations where x and y are both
defined in terms of t. So if you input all the
possible t's that you can into these functions and then plot the corresponding x and y's for each t, this will plot a curve in the x-y plain. What I wanna 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. And in both cases it's
going to be in terms of t. So let's get to it. So first lets find the first derivative of y with respect to x. First derivative of y with respect to x. And we've seen this
before in other videos, where this is going to be the derivative of y with respect to t over the derivative of
x with respect to t. And 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 either 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 three, so I can say times three like that, or I can put that three out front. And then the derivative of negative one, well, a constant doesn't change no matter what you do to your t, so that's just going to be zero. So that's dy, dt. So 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? Derivative of x with
respect to t is equal to, well, we're gonna have
the three out front, and so the derivative of e
to the 2t with respect to 2t is going to be e to the 2t and then we're going to
take the derivative of 2t with respect to t, which is just two, so this is gonna be 6e to the 2t. 6e to the 2t. And lets see, we could
simplify this a little bit. I'll now go to a neutral color. This is equal to, so this is gonna be one half, that's three over six, e to the 3t minus 2t. 3t minus 2t. And I'm just using exponent
properties right over here. But if I have three t's and
I take away two of those t's, I'm just gonna have a t. So this is just going to
simplify to a t right over here. So now that we've now figured
out the first derivative of y with respect to x in terms of t, now how do we find the second derivative? How do we find the second derivative of y with respect to x now? And I'll give you a hint! We're going to use this same idea. If you wanna 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. So what this is going to be, we wanna find the derivative
of the first derivative with respect to t, so
let me write this down. We wanna 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. 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 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 wanna 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 wanna do is we wanna find the derivative with respect,
let me write it this way: When we wanted to find the derivative with respect to x of y, that was equal to derivative
of y with respect to t over the derivative of
x with respect to t. Now we wanna find the
derivative with respect to x of the first derivative with respect to x. And 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 derivative
with respect to t of y. In fact, let me write it that
way just so you can see it. So if I clear this out, if I clear that out we're gonna get, this is the derivative
with respect to t of y. So hopefully you see,
before we had a y there, now we have a dy, dx. Dx, dt. Now this might seem really
daunting and complicated except for the fact
that these are actually fairly straight 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 gonna be one half. And the derivative with
respect to t of e to the t is just e to the t. And so that's going to be over the derivative of x with respect to t, which we already saw is 6e to the 2t. 6e to the 2t. We can write this as, one half divided by six is one over 12, and then e to the t minus 2t, which is equal to, we could write this as 1/12th e to the negative t, or we could write this as one over 12 e to the t. And we're done.

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