Addison Trujillo

Answered

2022-07-14

Finding the volume of the pyramid and also determining the increase and decrease of its volume

The volume of a pyramid with a square base x units on a side and a height of h is $V={\displaystyle \frac{1}{3}}{x}^{2}h$.

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.

The volume of a pyramid with a square base x units on a side and a height of h is $V={\displaystyle \frac{1}{3}}{x}^{2}h$.

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.

Answer & Explanation

furniranizq

Expert

2022-07-15Added 20 answers

Step 1

$V=\frac{1}{3}{e}^{2t}{e}^{-2t}=\frac{1}{3}{e}^{2t-2t}=\frac{1}{3}{e}^{0}=\frac{1}{3}$

Thus ${V}^{\prime}(t)=0$. But if you insist on using the chain rule then, we apply the product rule and then the chain rule

Step 2

${V}^{\prime}=-\frac{2}{3}{e}^{2t}{e}^{-2t}+\frac{2}{3}{e}^{2t}{e}^{-2t}=0$

Clearly the volume of the pyramid clearly is a constant so doesn't increase or decrease.

$V=\frac{1}{3}{e}^{2t}{e}^{-2t}=\frac{1}{3}{e}^{2t-2t}=\frac{1}{3}{e}^{0}=\frac{1}{3}$

Thus ${V}^{\prime}(t)=0$. But if you insist on using the chain rule then, we apply the product rule and then the chain rule

Step 2

${V}^{\prime}=-\frac{2}{3}{e}^{2t}{e}^{-2t}+\frac{2}{3}{e}^{2t}{e}^{-2t}=0$

Clearly the volume of the pyramid clearly is a constant so doesn't increase or decrease.

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