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

Explanation:

The volume of a sphere is given by

`V=(4)/(3)\pi r^3`

taking the derivative of both sides gives

`(dV)/(dt)=(d((4)/(3)\pi r^3))/(dt)`

`(dV)/(dt)=(4)/(3)\pi(d(r^3))/(dt)`

using the product rule gives

`(d(r^3))/(dt)=r^2(dr)/(dt)+r(dr^2)/(dt)`
`=r^2(dr)/(dt)+2r^2(dr)/(dt)`
`=3r^2(dr)/(dt)`

Thus, the rate of change of volume is

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

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