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:

[tex]S = 4\pi r^2[/tex]

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

[tex]S = 8\pi r\frac{dr}{dt}[/tex]

For this problem:

The radius of a sphere is decreasing at a rate of 2 centimetres per second, hence [tex]\frac{dr}{dt} = -2[/tex]

It also is of 3 centimetres, hence [tex]r = 3[/tex]

The rate of change is of:

[tex]S = 8\pi(3)(-2) = -48\pi = -151[/tex]

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

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

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.)

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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.)