Question

If the volume of a sphere is increasing at the rate of 12 \( \text{cm}^3/\text{sec} \), then the rate (in \( \text{cm}^2/\text{sec} \)) at which its surface area is increasing when the diameter of the sphere is 12 cm is:

• A. \( 2 \)
• B. \( 3 \)
• C. \( 4 \)
• D. \( 6 \)

Hint:

For related rates problems, differentiate both sides and substitute known rates carefully to find the required value.

Answer 1

The Correct Option is C

The volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] Differentiating, \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] Given \( \frac{dV}{dt} = 12 \) and \( r = \frac{12}{2} = 6 \), \[ 12 = 4\pi (6)^2 \frac{dr}{dt} \] \[ \frac{dr}{dt} = \frac{12}{144\pi} = \frac{1}{12\pi} \] The surface area of a sphere: \[ S = 4\pi r^2 \] Differentiating, \[ \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \] Substituting \( r = 6 \) and \( \frac{dr}{dt} = \frac{1}{12\pi} \), \[ \frac{dS}{dt} = 8\pi (6) \times \frac{1}{12\pi} \] \[ = \frac{48\pi}{12\pi} = 4 \] Thus, the correct answer is \( 4 \).