Question

the radius of the right circular cylinder shown below is growing at a rate of 2ft/min while it's height is shrinking at 3ft/min. At what rate is the volume of the cylinder changing, with respect to time, when the radius is 4ft and the volume is 32 ft cubed.

Answer 1

Answer:

The volume is decreasing at a rate of about 118.8 cubic feet per minute.

Step-by-step explanation:

Recall that the volume of a cylinder is given by:

$\displaystyle V=\pi r^2h$

Take the derivative of the equation with respect to t. V, r, and h are all functions of t:

$\displaystyle \frac{dV}{dt}=\pi\frac{d}{dt}\left[r^2h\right]$

Use the product rule and implicitly differentiate. Hence:

$\displaystyle \frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right)$

We want to find the rate at which the volume of the cylinder is changing when the radius if 4 feet and the volume is 32 cubic feet given that the radius is growing at a rate of 2ft/min and the height is shrinking at a rate of 3ft/min.

In other words, we want to find dV/dt when r = 4, V = 32, dr/dt = 2, and dh/dt = -3.

Since V = 32 and r = 4, solve for the height:

$\displaystyle \begin{aligned} V&=\pi r^2h \\32&=\pi(4)^2h\\32&=16\pi h \\h&=\frac{2}{\pi}\end{aligned}$

Substitute:

$\displaystyle\begin{aligned} \frac{dV}{dt}&=\pi\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right)\\ \\ &=\pi\left(2(4)\left(\frac{2}{\pi}\right)\left(2\right)+(4)^2\left(-3\right)\right)\\\\&=\pi\left(\frac{32}{\pi}-48\right)\\&=32-48\pi\approx -118.80\frac{\text{ ft}^3}{\text{min}}\end{aligned}$

Therefore, the volume is decreasing at a rate of about 118.8 cubic feet per minute.