Question

If the radius of a sphere is increasing at the rate of 2 inches per second, how fast, in cubic inches per second, is the volume increasing when the radius is 10 inches?

(A) 40π
(B) 80π
(C) 800
(D) 800π
(E) 3200π

Answer 1

When the radius of a sphere is increasing at 2 inches per second, the volume is increasing at a rate of 800π cubic inches per second when the radius is 10 inches.

Step-by-step explanation:

To solve how fast the volume of a sphere is increasing, we must relate the rate of change of the radius (dr/dt) to the rate of change of the volume (dV/dt). The volume (V) of a sphere is given by the formula V = (4/3)πr^3. Taking the derivative with respect to time (t) gives us dV/dt = 4πr^2(dr/dt). Given that dr/dt is 2 inches per second and the radius (r) at the instant in question is 10 inches,

we can substitute these values into the equation:

• dV/dt = 4π(10 inches)^2(2 inches/second)
• dV/dt = 4π(100 inches^2)(2 inches/second)
• dV/dt = 800π cubic inches per second

Therefore, the volume of the sphere is increasing at a rate of 800π cubic inches per second when the radius is 10 inches, which corresponds to answer choice (D) 800π.