Question

A spherical balloon is filled with \( 4500\pi \) cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of \( 72\pi \) cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is:

• A. \( \frac{9}{7} \)
• B. \( \frac{7}{9} \)
• C. \( \frac{2}{9} \)
• D. 9

Hint:

When dealing with rates of change in related quantities, differentiate the formula relating the quantities with respect to time, and substitute the known values to solve for the desired rate.

Answer 1

The Correct Option is C

We are given the following: - The volume of the spherical balloon is \( V = 4500\pi \) cubic meters. 
- The rate at which the gas escapes is \( \frac{dV}{dt} = -72\pi \) cubic meters per minute. - We are asked to find the rate at which the radius decreases 49 minutes after the leakage began. 
Step 1: Volume of the sphere The volume \( V \) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius of the balloon. 
Step 2: Differentiate with respect to time We differentiate the volume formula with respect to time \( t \): \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] Step 3: Substitute the known values We know that \( \frac{dV}{dt} = -72\pi \). Substituting this into the equation: \[ -72\pi = 4\pi r^2 \frac{dr}{dt} \] Canceling \( \pi \) from both sides: \[ -72 = 4r^2 \frac{dr}{dt} \] Simplifying: \[ \frac{dr}{dt} = \frac{-72}{4r^2} \] 
Step 4: Find the radius at \( t = 49 \) minutes At \( t = 0 \), the volume is \( V = 4500\pi \). Using the formula for the volume of the sphere: \[ 4500\pi = \frac{4}{3} \pi r^3 \] Solving for \( r \): \[ 4500 = \frac{4}{3} r^3 \] \[ r^3 = \frac{4500 \times 3}{4} = 3375 \] Taking the cube root: \[ r = 15 \] Thus, the radius of the balloon at \( t = 0 \) is 15 meters. Step 5: Find the rate of change of the radius at \( t = 49 \) We can now substitute \( r = 15 \) into the equation for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{-72}{4 \times 15^2} = \frac{-72}{4 \times 225} = \frac{-72}{900} = \frac{-2}{25} \] Thus, the rate at which the radius decreases 49 minutes after the leakage began is \( \frac{2}{9} \) meters per minute. 
Therefore, the correct answer is Option C.