Answer is: option2

0.12 in/sec

Solution:

We need to determine the rate at which the radius of the balloon is changing at \( t = 6 \) seconds, given the volume function:

\[ V = \frac{4}{3} \pi r^3 \]

Differentiating both sides with respect to \( t \):

\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]

Rearrange to solve for \( \frac{dr}{dt} \):

\[ \frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^2} \]

We are given:

• \( V(6) = 4\pi \)
• \( V'(6) = \pi \)

First, solve for \( r(6) \) using:

\[ 4\pi = \frac{4}{3} \pi r^3 \]

Cancel \( 4\pi \) from both sides:

\[ 1 = \frac{1}{3} r^3 \]

\[ r^3 = 3 \]

\[ r = \sqrt[3]{3} \approx 1.44 \]

Substituting values into the equation:

\[ \frac{dr}{dt} = \frac{\pi}{4\pi (1.44)^2} \]

Cancel \( \pi \):

\[ \frac{dr}{dt} = \frac{1}{4 (2.07)} \]

\[ \frac{dr}{dt} = \frac{1}{8.28} \]

\[ \frac{dr}{dt} \approx 0.12 \]

Final Answer is 0.12 in/sec.