The radius of a sphere is increasing at a rate of 5 meters per hour. When the radius is 12 meters, then how fast is the VOLUME changing?

Question

The radius of a sphere is increasing at a rate of 5 meters per hour. When the radius is 12 meters, then how fast is the VOLUME changing?

Note: Volume of a sphere = (4/3) ⋅ π ⋅ (radius)³

The rate of change of the VOLUME is _____  cubic meters per hour.

(Enter your answer as a decimal number rounded to 2 places.)

Answer 1

Known variables:

1. ^dr/_dt = 5m/hr
2. r = 12m

Solving for ^dV/_dt...

Volume of a sphere equation:

V = (⁴/₃)πr³ Take the derivative of both sides with respect to time

^d/_dt[V] = ^d/_dt[(⁴/₃)πr³] Apply chain rule

1(^dV/_dt) = 4πr²(^dr/_dt) Plugin known variables and solve for ^dV/_dt

^dV/_dt = 4(3.14159)(12)²(5)

^dV/_dt ≈ 9047.78m³/hr