The radius r of a sphere is increasing at a rate of 2 inches per minute. Find the rate of change of th volume when r = 24 inches.

crazygbyo
2022-08-08
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Leroy Cunningham

Answered 2022-08-09
Author has **14** answers

We are given: $\frac{dr}{dt}=2in/min$

Since we know the area of a circle is : $A=\delta \ast {r}^{2}$

Differentiate with respect to time:

$\frac{dA}{dt}=2\delta \ast r(dr/dt)$

When r = 24, we have $\frac{dA}{dt}=2\delta (24)(2)=96\delta i{n}^{2}/min$

Since we know the area of a circle is : $A=\delta \ast {r}^{2}$

Differentiate with respect to time:

$\frac{dA}{dt}=2\delta \ast r(dr/dt)$

When r = 24, we have $\frac{dA}{dt}=2\delta (24)(2)=96\delta i{n}^{2}/min$

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So what I have done is said $S=2\pi {r}^{2}h+2\pi {r}^{2}$.

$\begin{array}{rl}2\pi {r}^{2}h+2\pi {r}^{2}& =200\\ 2\pi {r}^{2}h& =200-2\pi {r}^{2}\\ h& =\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}\end{array}$

Subbing this into the volume equation I get $V=(\pi {r}^{2})\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}$.

Having trouble differentiating that equation.

The cylinder is open topped and is made from $200\text{}{\text{cm}}^{2}$ of material.

So what I have done is said $S=2\pi {r}^{2}h+2\pi {r}^{2}$.

$\begin{array}{rl}2\pi {r}^{2}h+2\pi {r}^{2}& =200\\ 2\pi {r}^{2}h& =200-2\pi {r}^{2}\\ h& =\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}\end{array}$

Subbing this into the volume equation I get $V=(\pi {r}^{2})\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}$.

Having trouble differentiating that equation.

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