Question

Gas is being pumped into a spherical balloon at the rate of \(30~ft^3/min\). Find the rate at which the radius increases when the radius becomes \(15~ft\).

• A. \(\dfrac{1}{30}\,ft/min\)
• B. \(\dfrac{1}{15}\,ft/min\)
• C. \(\dfrac{1}{30\pi}\,ft/min\)
• D. \(\dfrac{1}{20\pi}\,ft/min\)

Hint:

In related rates problems: \[ \text{Differentiate first, then substitute values.} \]

Answer 1

The Correct Option is C

Concept:
Use related rates and volume formula of sphere. \[ V=\frac43\pi r^3 \]

Step 1: Differentiate with respect to time. \[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \]

Step 2: Substitute given values. \[ \frac{dV}{dt}=30~ft^3/min \] \[ r=15~ft \] Thus, \[ 30 = 4\pi(15)^2\frac{dr}{dt} \] \[ 30 = 900\pi\frac{dr}{dt} \]

Step 3: Find \(\dfrac{dr}{dt}\). \[ \frac{dr}{dt} = \frac{30}{900\pi} \] \[ \frac{dr}{dt} = \frac{1}{30\pi}\,ft/min \]

Step 4: Conclusion. \[ \boxed{ \frac{dr}{dt} = \frac{1}{30\pi}\,ft/min } \] Hence, correct answer is: \[ \boxed{(C)} \]