Question

The radius of a spherical ball is increasing at the rate of 1 m/sec. At the radius equal to 3m, the volume of the ball is increasing at the rate given by:

• A. \(30 \pi m^3/sec\)
• B. \(38 \pi m^3/sec\)
• C. \(36 m^3/sec\)
• D. \(36 \pi m^3/sec\)

Answer 1

The Correct Option is D

To determine the rate at which the volume of the ball is increasing, we can use the formula for the volume of a sphere, which is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. We need to find the derivative of the volume with respect to time, \( \frac{dV}{dt} \). Using the chain rule, we get:

\(\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}\)

First, calculate \(\frac{dV}{dr}\):

\(\frac{dV}{dr} = \frac{d}{dr} \left(\frac{4}{3} \pi r^3\right) = 4 \pi r^2\)

We are given that the radius increases at a rate of \(\frac{dr}{dt} = 1 \, m/sec\). Thus:

\(\frac{dV}{dt} = 4 \pi r^2 \cdot 1 = 4 \pi r^2\)

At \( r = 3 \, m\):

\(\frac{dV}{dt} = 4 \pi \cdot (3)^2 = 36 \pi \, m^3/sec\)

Therefore, the rate at which the volume is increasing when the radius is 3 meters is \( 36 \pi m^3/sec \).