Assume that the radius ,r , of a sphere is expanding at a rate of 10in./min. The volume of a sphere is V=43πr^3. Determine the rate at which the volume is changing with respect to time when r=5in. The volume is changing at a rate of how many in^3/min.?

You mean V = (4/3) pi r^3

dV/dt = (4 pi r^2) dr/dt

(You really do not need calculus. It is the surface area times dr/dt)

12.566

To find the rate at which the volume is changing with respect to time, we can use the chain rule of differentiation.

Given: The rate of change of the radius with respect to time is dr/dt = 10 in./min.

The volume of a sphere is given by V = (4/3)πr^3.

First, let's differentiate the volume equation with respect to time (t):

dV/dt = dV/dr * dr/dt

To find dV/dr, we differentiate the volume equation with respect to the radius (r):

dV/dr = 3 * (4/3)πr^2 = 4πr^2

Now, substituting the given value of r = 5 in.:

dV/dr = 4π(5)^2 = 100π in^2

Next, substitute the given rate of change dr/dt = 10 in./min:

dV/dt = 100π * 10 = 1000π in^3/min

So, the volume is changing at a rate of 1000π in^3/min.