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:

• A. $\frac{1}{30\pi}\text{ ft/min}$
• B. $\frac{1}{15\pi}\text{ ft/min}$
• C. $\frac{1}{20\pi}\text{ ft/min}$
• D. $\frac{1}{25\pi}\text{ ft/min}$

Hint:

In related rates problems involving spheres, remember: \[ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \] The factor $4\pi r^2$ is actually the surface area of the sphere.

Answer 1

The Correct Option is A

Concept: This is a standard related rates problem. The volume of a sphere is related to its radius by: \[ V=\frac{4}{3}\pi r^3 \] Since both $V$ and $r$ change with time, differentiate with respect to time $t$.

Step 1: Differentiating the volume formula with respect to time.
Differentiate both sides: \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) \] \[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \]

Step 2: Substituting the given values.
Given: \[ \frac{dV}{dt}=30\text{ ft}^3/\text{min} \] and \[ r=15\text{ ft} \] Substitute: \[ 30 = 4\pi(15)^2\frac{dr}{dt} \] \[ 30 = 4\pi(225)\frac{dr}{dt} \] \[ 30 = 900\pi\frac{dr}{dt} \]

Step 3: Solving for $\mathbf{\frac{dr}{dt}}$.
\[ \frac{dr}{dt} = \frac{30}{900\pi} \] \[ \frac{dr}{dt} = \frac{1}{30\pi}\text{ ft/min} \] Hence, \[ \boxed{ \frac{dr}{dt} = \frac{1}{30\pi}\text{ ft/min} } \]