find the first and second derivatives of the parametric functions

Question

$x=e^t$ $y=te^(-t)$

$\frac{dy}{dx}= \frac{e^(-t)(1-t)}{e^(t)}$

$\frac{d^2y}{dx^2}= \frac{\frac{dy}{dx}}{\frac{dx}{dt}}= \frac{e^(-t)(1-t)}{e^t}$

any t's without proper enclosement are meant to be to the power...I don't know why its giving me this trouble. I entered these answers into my homework and it said it could not understand my answers and could not be graded.... Also it asked the interval over which it is is concave upward and if im not mistaken from my work it would be from 0 to infinity.

Answer 1

The second derivative is wrong (see here).

If you have $x(t)$ and $y(t)$, then effectively

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ but (you will find the proof in the Wikipedia page) $$\frac{d^2y}{dx^2}=\frac{\frac{dx}{dt} \frac{d^2y}{dt^2}-\frac{dy}{dt} \frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^3 }$$ So, for your case $$x=e^t\implies \frac{dx}{dt}=e^t\implies \frac{d^2x}{dt^2}=e^t$$ $$y=te^{-t}\implies \frac{dy}{dt}=-e^{-t} (t-1)\implies \frac{d^2y}{dt^2}=e^{-t} (t-2)$$ This leads to $$\frac{dy}{dx}=-e^{-2 t} (t-1)\qquad , \qquad \frac{d^2y}{dx^2}=e^{-3t} (2 t-3)$$ Iam sure that you can take it from here.

Answer 2

Since, $$\frac{dy}{dx}=\frac{e^(-t)\,(1-t)}{e^(t)}=0$$ We have, $$t=1$$ This is the point of inflection because by substituting t=1 in $\frac{d^2y}{dx^2}$ , we get zero.

For $t>1$, we have $\frac{d^2y}{dx^2}<0$,

and for $t<1$ we get $\frac{d^2y}{dx^2}>0$

So, by the Concavity Theorem, the function is concave up for $t<1$ and concave down for $t>1$.

Concavity Theorem and Second Derivative Test