The radius r of a sphere is increasing at the uniform rate of 0.3 inches per second. At the instant when the surface area S becomes 100pi square inches, what is the rate of increase, in cubic inches per second, in the volume V?

V = 4/3 pi r^3

dV/dt = 4pi r^2 dr/dt
= A dr/dt
plug in the values given:

dV/dt = 100pi * .3 = 30pi

what is the volume of a cylinder of mozarella cheese, with radius z, and height a?

(don't use any exponents in your answer)

V = pir^2h

V = pizza
:)

22.5

V = 4/3 pi r^3

dV/dt = 4pi r^2 dr/dt
= A dr/dt ......................... YES
plug in the values given:

dV/dt = 100pi * .3 = 30pi .......... NO A = 4 pi r^2 = 100
so dV/dt = 100 * 0.3 = 30

Sorry, you are right, area = 100 pi

I missed that.

To find the rate of increase in the volume V at the instant when the surface area S becomes 100π square inches, we will use the formulas for the surface area and volume of a sphere.

The surface area S of a sphere is given by the formula:

S = 4πr^2

The volume V of a sphere is given by the formula:

V = (4/3)πr^3

We are given that the radius r of the sphere is increasing at a uniform rate of 0.3 inches per second. This means that the rate of change of the radius with respect to time (dr/dt) is 0.3 inches per second.

Now, let's differentiate both sides of the surface area formula with respect to time (t):

dS/dt = 4π(2r)(dr/dt) (differentiating using the chain rule)

Since we are interested in finding the rate of increase in the volume, we need to find dV/dt. We can do this by differentiating the volume formula with respect to time (t):

dV/dt = (4/3)π(3r^2)(dr/dt) (differentiating using the chain rule)

Now, we have the rate of change of the surface area (dS/dt) and the rate of change of the volume (dV/dt) in terms of the radius (r) and the rate of change of the radius (dr/dt).

At the instant when the surface area S becomes 100π square inches, we can substitute the value of S into the surface area formula:

100π = 4πr^2

Simplifying this equation, we get:

r^2 = 25

Taking the square root of both sides, we find:

r = 5 inches

Substitute this value of r into the rate of change formulas to find the rate of increase in the volume V:

dS/dt = 4π(2r)(dr/dt)

100π = 4π(2*5)(dr/dt)

Simplifying, we find:

100 = 40(dr/dt)

Solving for (dr/dt), we get:

(dr/dt) = 100/40

(dr/dt) = 5/2

So the rate of change of the radius at the instant when the surface area S becomes 100π square inches is 5/2 inches per second.

Finally, substitute the value of (dr/dt) into the rate of change of the volume formula to find the rate of increase in the volume V:

dV/dt = (4/3)π(3r^2)(dr/dt)

dV/dt = (4/3)π(3*(5)^2)(5/2)

Simplifying, we get:

dV/dt = (4/3)π(75)(5/2)

dV/dt = (4/3)(75/2)π

dV/dt = 500π/3

Therefore, the rate of increase in the volume V, at the instant when the surface area S becomes 100π square inches, is 500π/3 cubic inches per second.