Question

The volume of a sphere is decreasing at a constant rate of 9988 cubic centimeters per minute. At the instant when the volume of the sphere is
3131 cubic centimeters, what is the rate of change of the surface area of the sphere? The volume of a sphere can be found with the equation
V=4/3πr^3 and the surface area can be found with
S=4πr^2 Round your answer to three decimal places (if necessary).

Answer 1

Using implicit differentiation, it is found that the rate of change of the surface area of the sphere -2201.293 square centimeters per second.

What is the volume of a sphere?

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

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

What is the surface area of a sphere?

• The surface area of a sphere, of radius r, is given by:

$S = 4\pi r^2$

What is the rate of change?

Applying implicit differentiation, the rate of change of both the volume and the surface area can be found, as follows:

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

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

Volume of 3131 cubic centimeters, hence:

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

$\frac{4\pi r^3}{3} = 3131$

$4\pi r^3 = 3(3131)$

$r = \sqrt[3]{\frac{3(3131)}{4\pi}}$

$r = 9.0754$

The volume of a sphere is decreasing at a constant rate of 9988 cubic centimeters per minute, hence:

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

$-9988 = 4\pi (9.0754)^2\frac{dr}{dt}$

$\frac{dr}{dt} = -\frac{9988}{4\pi (9.0754)^2}$

$\frac{dr}{dt} = -9.651$

Then, the rate of change of the surface area of the sphere is given by:

$\frac{dS}{dt} = 8\pi (9.0754)(-9.651) = -2201.293$

The rate of change of the surface area of the sphere -2201.293 square centimeters per second.

To learn more about implicit differentiation, you can take a look at brainly.com/question/25608353