Question

find the rate of change of the radius of a sphere at the point in time when the radius is 6ft if the volume is increasing at the rate of 8 pie cubic feet per second.

Answer 1

Answer:

Explanation:

The volume of a sphere is given by

$V=\frac{4}{3}\pi r^3$

taking the derivative of both sides gives

$\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}$

$\frac{dV}{dt}=\frac{4}{3}\pi\frac{d(r^3)}{dt}$

using the product rule gives

$\frac{d(r^3)}{dt}=r^2\frac{dr}{dt}+r\frac{dr^2}{dt}$$=r^2\frac{dr}{dt}+2r^2\frac{dr}{dt}$$=3r^2\frac{dr}{dt}$

Thus, the rate of change of volume is

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Now we know that at a certain time, r = 6 ft and dV/dt = 8; therefore,

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