The volume of a pyramid with a square base $x$ units on a side and a height of $h$ is $V=\dfrac{1}{3}x^2h$
1) Suppose $x=e^t$ and $h=e^{-2t}$. Use the chain rule to find $V^1(t)$.
2) Does the volume of the pyramid increase or decrease as $t$ increases.
My Try:
From the question $V=\dfrac{1}{3}x^2h$, $x=e^t$ and $h=e^{-2t}$
$V=\dfrac{1}{3}e^{2t}e^{-2t}$
$V^1(t)=\dfrac{1}{3}$
But when I tried it in Symbolab the derivative was $0$ i.e $V^1(t)=0$
How to proceed with the above problem.
Can anyone explain this.
Best Answer
$V=\frac{1}{3}e^{2t}e^{-2t}=\frac{1}{3}e^{2t-2t}=\frac{1}{3}e^{0}=\frac{1}{3}$.
Thus $V'(t)=0$. But if you insist on using the chain rule then, we apply the product rule and then the chain rule
$V'=-\frac{2}{3}e^{2t}e^{-2t}+\frac{2}{3}e^{2t}e^{-2t}=0$.
Clearly the volume of the pyramid clearly is a constant so doesn't increase or decrease.