Good afternoon.

I am trying to find the concavity of the following parametric equations:

$x = e^t$

$y = t^2e^{-t}$

I eventually got the second derivative to be $2e^{-2t}(t^2-3t+1)$. I then solved this equation for y=0 and got two inflection points ($x=0.3819$ and $x=2.6180$). With numbers from this interval I get negative values, which suggests to me that this section must be concave down. However, when I actually plot the original equation, this section seems to be concave up. I don't understand where I went wrong - any help is appreciated.



May be, I did not understand the problem.

If we are considering the concavity of the plot of $y$ as a function of $x$, both of them being defined as a function of a parameter $t$, we need to look at the sign of $$\frac{d^2y}{dx^2}=\frac {\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{dy}{dt}\frac{d^2x}{dt^2} } {\left(\frac{dx}{dt}\right)^3}$$ (see here).

In the considered case $$x=e^t\implies \frac{dx}{dt}=e^t \implies \frac{d^2x}{dt^2}=e^t$$

$$y=t^2e^{-t}\implies \frac{dy}{dt}=-e^{-t} (t-2) t\implies \frac{d^2y}{dt^2}=e^{-t} (t^2-4t+2)$$ All of the above make $$\frac{d^2y}{dx^2}=2 e^{-3 t} (t^2-3 t+1)$$ So,$$\frac{d^2y}{dx^2} > 0 \qquad \text{if}\qquad -\infty < t <\frac{3 - \sqrt5}{2}\implies 0 < x <e^{\frac{3 - \sqrt5}{2} }$$ $$\frac{d^2y}{dx^2} < 0 \qquad \text{if}\qquad \frac{3 + \sqrt5}{2} < t <\infty\implies e^{\frac{3 + \sqrt5}{2} }<x <\infty $$

Just to check, let us eliminate $t$ from $x$, this gives $$t=\log(x)\implies y=\frac{\log ^2(x)}{x}\implies \frac{d^2y}{dx^2}=2\,\frac{ \log^2 (x)-3 \log (x)+1}{x^3}$$ from which the same conclusions are reached.


A workaround is by eliminating $t$,

$$y=\frac{\log^2(x)}x, x>0.$$


$$y''=\frac{2\log^2(x)-6\log(x)+2}{x^3}=\frac2{x^3}(\log(x)-r)(\log(x)-s)$$ where $r$ and $s$ are the two roots of the numerator, which are positive.

The function is concave down between the roots.


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