Question

The surface area of a sphere is increasing at a rate of 14\pi14π14, pi square meters per hour. At a certain instant, the surface area is 36\pi36π36, pi square meters. What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?

Answer 1

Answer:

$21\pi$ cubic meters per hour

Step-by-step explanation:

We know that the surface area of a sphere is,

$A=4\pi r^2$     ...........(1),

Where,

r = radius of the sphere,

We have, A = $36\pi$ square meters,

$36\pi = 4\pi r^2$

$9 = r^2$

$\implies r = 3\text{ meters}$

Now, differentiating equation (1) with respect to t ( time ),

$\frac{dA}{dt} = 8\pi r\frac{dr}{dt}$

We have,

$\frac{dA}{dt}=14\pi\text{ square meters per hour}, r = 3\text{ meters}$

$14\pi = 8\pi (3) \frac{dr}{dt}$

$\frac{14}{24}=\frac{dr}{dt}$

$\frac{7}{12} = \frac{dr}{dt}$

Also, the volume of a sphere is,

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

Differentiating with respect to t(time)

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

By substituting the values,

$\frac{dV}{dt}=4\pi (3)^2 (\frac{7}{12})=36\pi \frac{7}{12}=21\pi\text{ cube meters per hour}$

Hence, the rate of change of the volume of the sphere at that instant would be $21\pi$ cubic meters per hour