Question

The radius of a sphere is increasing at a rate of 0.5 centimeters per minute. At a certain instant, the radius is 17 centimeters.
What is the rate of change of the volume of the sphere at that instant (in cubic centimeters per minute)?

Answer 1

Using implicit differentiation, it is found that the rate of change of the volume of the sphere at that instant is of 1815.84 cubic centimetres per minute.

The volume of a sphere of radius r is given by:

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

Applying implicit differentiation, the rate of change is of:

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

In this problem, we have that $\frac{dr}{dt} = 0.5, r = 17$, thus:

$\frac{dV}{dt} = 4\pi (17)^2(0.5) = 1815.84$

The rate of change of the volume of the sphere at that instant is of 1815.84 cubic centimetres per minute.

A similar problem is given at brainly.com/question/11496075