Question:
Gas is being pumped into a spherical balloon at the rate of $ 30 \, \text{ft}^3/\text{min} $. Then the rate at which the radius increases when it reaches the value $ 15 \, \text{ft} $ is:
Gas is being pumped into a spherical balloon at the rate of $ 30 \, \text{ft}^3/\text{min} $. Then the rate at which the radius increases when it reaches the value $ 15 \, \text{ft} $ is:
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In related rates, always identify your "knowns" and your "unknowns" before differentiating. Remember that $ \frac{dV}{dt} = (\text{Surface Area}) \times \frac{dr}{dt} $ for a sphere.
Updated On: May 12, 2026
- $ \frac{1}{30\pi} \, \text{ft}/\text{min} $
- $ \frac{1}{15\pi} \, \text{ft}/\text{min} $
- $ \frac{1}{20\pi} \, \text{ft}/\text{min} $
- $ \frac{1}{25\pi} \, \text{ft}/\text{min} $
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The Correct Option is A
Solution and Explanation
Concept:
This is a related rates problem. We are given the rate of change of volume ($ dV/dt $) and need to find the rate of change of the radius ($ dr/dt $). The volume $ V $ of a sphere is given by:
$$ V = \frac{4}{3}\pi r^3 $$
Step 1: Differentiating the volume formula. We differentiate both sides of the volume equation with respect to time $ t $ using the chain rule: $$ \frac{dV}{dt} = \frac{4}{3}\pi \cdot \frac{d}{dt}(r^3) $$ $$ \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} $$ $$ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $$
Step 2: Plugging in given values.
We are given:
• $ \frac{dV}{dt} = 30 \, \text{ft}^3/\text{min} $
• $ r = 15 \, \text{ft} $ Substitute these into our differentiated equation: $$ 30 = 4\pi (15)^2 \frac{dr}{dt} $$ $$ 30 = 4\pi (225) \frac{dr}{dt} $$ $$ 30 = 900\pi \frac{dr}{dt} $$
Step 3: Solving for $ dr/dt $.
$$ \frac{dr}{dt} = \frac{30}{900\pi} $$ Simplify the fraction by dividing both numerator and denominator by 30: $$ \frac{dr}{dt} = \frac{1}{30\pi} \, \text{ft}/\text{min} $$
Step 1: Differentiating the volume formula. We differentiate both sides of the volume equation with respect to time $ t $ using the chain rule: $$ \frac{dV}{dt} = \frac{4}{3}\pi \cdot \frac{d}{dt}(r^3) $$ $$ \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} $$ $$ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $$
Step 2: Plugging in given values.
We are given:
• $ \frac{dV}{dt} = 30 \, \text{ft}^3/\text{min} $
• $ r = 15 \, \text{ft} $ Substitute these into our differentiated equation: $$ 30 = 4\pi (15)^2 \frac{dr}{dt} $$ $$ 30 = 4\pi (225) \frac{dr}{dt} $$ $$ 30 = 900\pi \frac{dr}{dt} $$
Step 3: Solving for $ dr/dt $.
$$ \frac{dr}{dt} = \frac{30}{900\pi} $$ Simplify the fraction by dividing both numerator and denominator by 30: $$ \frac{dr}{dt} = \frac{1}{30\pi} \, \text{ft}/\text{min} $$
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