$(4$ pts$)2)$ Briefly explain what is wrong with the following attempt to find the concavity of the parametric

curve $x=e^t$

$y=t^2$

$x^{\text{'}}(t)=e^t$ and $x^{\text{'}\text{'}}(t)=e^t$

$y^{\text{'}}(t)=t^2$ and $y^{\text{'}\text{'}}(t)=2t$

So the second derivative is $\frac{y^{\text{'}\text{'}}(t)}{x^{\text{'}\text{'}}(t)}=\frac{2t}{e^t}$

$e^t$ is always positive, but $2t$ is only positive if $t>0$ so if $t>0$ then the second derivative is positive and the curve is concave up$.$

But if $t<mn>0</mn>$ then $2t$ would be negative and then the second derivative would also be negative and the curve would be concave down.