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# Second derivatives (parametric functions)

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

## Video transcript

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 court 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 and in both cases it's going to be in terms of T so let's let's get to it so first let's 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 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 to your T so that's just going to be 0 so that's dy DT so it's going to be equal to 3 e 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 3 out front and so the derivative of e to the 2t with respect to 2 T 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 2 so this is going to be 6 e to the 2t 6 e to the 6 e 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 so this is going to be 1/2 that's three over six e to the 3t minus two t 3t minus 2t and I'm just using exponent properties right over here but three if I have three T's and I take away two of those T's I'm just going to have a T so this is just going to simplify to a T right over here so now that we know though we've now figured out the 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 with respect to X now and 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 so 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 so 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 all of that over the all of that over DX DX DT now I want you to if it didn't if it doesn't jump out at you why this might 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 find 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 we want to find the derivative with respect let me write it this way when we wanted to find the positive 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 want to find the derivative with respect to X of the first derivative with respect to X and so everywhere we saw 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 so if I clear this out so if I clear that 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 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 the of 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 well is this going to be 1/2 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 a 6 e to the 2t 6 e to the 2t and so we can write this we can write this as C 1/2 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