Question

The height of a cylinder is increasing at a constant rate of 6 feet per second, and the volume is decreasing at a rate of 171 cubic feet per second. At the instant when the height of the cylinder is 6 feet and the volume is 1254 cubic feet, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V = Tr²h. Round your answer to three decimal places.

Answer 1

Answer:

Step-by-step explanation:

This is related rates in calculus, as far as I can tell from the info you've provided.  It's nice to see students are still trying to conquer its vagueness!

Start with the formula for volume of a cylinder:

$V=\pi r^2h$ and then find its derivative with respect to time:

$\frac{dV}{dt}=\pi (r^2\frac{dh}{dt}+2r\frac{dr}{dt}h)$

From that formula it looks like we need a value for dV/dt, r, dh/dt, and h; dr/dt is our unknown.  We have that

dV/dt = -171

dh/dt = 6

h = 6 and

V = 1254

We need a value for r.  We can find it using the last 2 values listed above in the volume formula:

$1254=\pi r^2(6)$ and

$\frac{1254}{6\pi } =r^2$ and

$66.52676621=r^2$ and

r = 8.156394

NOW we have everything we need to fill in the derivative for the volume:

$-171=66.52676621(6)\pi +6(2)(8.156394)\pi \frac{dr}{dt}$ which simplifies to

$-171=1254+307.4888096\frac{dr}{dt}$ and

$-1425=307.488096\frac{dr}{dt}$ so

$\frac{dr}{dt}=-4.634$ ft/sec