Hi! I had some questions regarding this exercise. I was wondering if it is possible for me to do it in two different ways:
First way: I calculate the Jacobians for each of the functions, so then I multiply $D(V(c(t)))\cdot D(c(t))$. But, how do I calculate this: D(V(c(t))), because I know how to get to $D(V(x,y,z))$ but not to $D(V(c(t)))$.
Second way: And the other way I was thinking, but I'm not sure about it, consists on defining: $$ \begin{align} x(t) &= e^{t^2} \\ y(t) &= e^{-t^2} \\ z(t) &= \log(1+t^2) \\ \end{align} $$ And then applying the Chain Rule to $V(x(t),y(t),z(t))$. However, I'm not sure if this is a valid way of doing it.
Thanks a lot !
Both ways work, but (2) gives you immediately that the answer is zero: $x(t) y(t) $ is independent of $t$ so differentiating it gives zero.
For (1) you have \begin{equation} DV(x(t),...) = (2x y^2,2x^2y,0)|_{x=x(t),...}= (2e^{-t^2},2 e^{t^2},0) \end{equation} \begin{equation} DC= (2t e^{t^2}, ,-2t e^{-t^2},...) \end{equation} did not compute the last component as we don't need it: DV is a row vector, DC is a column vector, so multiplying them gives \begin{equation} 4t -4t+0=0. \end{equation}