The radius of a sphere is decreasing at a rate of 2 centimeters per second. At the instant when the radius of the sphere is 3 centimeters, what is the rate of change, in square centimeters per second, of the surface area of the sphere? (The surface area S of a sphere with radius r is S = 4πr2.)

Using implicit differentiation, it is found that the rate of change of the surface area of the sphere is of -151 square centimetres per second.

The surface area of an sphere of radius r is given by:

Applying implicit differentiation, the rate of change is given by:

For this problem:

The rate of change is of:

The rate of change of the surface area of the sphere is of -151 square centimetres per second.

A similar problem is given at link

The rate of change of the surface area of the sphere is;

-48π cm^2/s

Step-by-step explanation:

Here, we want to calculate dA/dt

Mathematically that will be;

dA/dt = dA/dr * dr/dt

From the question;

dr/dt = -2 cm/s

dA/dr = 8πr

So dA/dt will be;

-2cm/s * 8πr = -16 πr cm/s where r is 3 cm

So;

dA/dt = -48π cm^2/s

b. hoped i'm right lol

step-by-step explanation: