Key Concepts and Formulas

• Volume of a sphere: The volume $V$ of a sphere with radius $r$ is given by the formula: $V = \frac{4}{3}\pi r^3$
• Related Rates: If $V$ and $r$ are functions of time $t$, we can differentiate the volume formula with respect to $t$ to relate $\frac{dV}{dt}$ and $\frac{dr}{dt}$.
• Chain Rule: $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$

Step-by-Step Solution

Step 1: Find the volume of the balloon after 49 minutes.

The balloon initially has a volume of $4500\pi$ cubic meters. Gas escapes at a rate of $72\pi$ cubic meters per minute. After 49 minutes, the total volume of gas that has escaped is $(72\pi \text{ m}^3/\text{min}) \times (49 \text{ min}) = 3528\pi$ cubic meters. Therefore, the volume of the balloon after 49 minutes is: $V_{49} = 4500\pi - 3528\pi = 972\pi \text{ m}^3$ We need this volume to find the radius at that time.

Step 2: Find the radius of the balloon after 49 minutes.

We know that $V = \frac{4}{3}\pi r^3$. After 49 minutes, $V = 972\pi$. We can solve for $r$: $972\pi = \frac{4}{3}\pi r^3$ $r^3 = \frac{3}{4} \times 972 = 3 \times 243 = 729$ $r = \sqrt[3]{729} = 9 \text{ m}$ So, the radius of the balloon is 9 meters after 49 minutes.

Step 3: Differentiate the volume formula with respect to time.

Differentiating $V = \frac{4}{3}\pi r^3$ with respect to time $t$, we get: $\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$ This relates the rate of change of the volume to the rate of change of the radius.

Step 4: Plug in the known values and solve for $\frac{dr}{dt}$.

We know that $\frac{dV}{dt} = -72\pi$ (since the volume is decreasing) and $r = 9$ meters after 49 minutes. Plugging these values into the equation from Step 3: $-72\pi = 4\pi (9^2) \frac{dr}{dt}$ $-72\pi = 4\pi (81) \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{-72\pi}{4\pi(81)} = \frac{-72}{4 \times 81} = \frac{-18}{81} = -\frac{2}{9} \text{ m/min}$ The rate at which the radius is decreasing is $\frac{2}{9}$ meters per minute. Since the question asks for the rate at which the radius decreases, we take the absolute value.

Common Mistakes & Tips

• Sign of $\frac{dV}{dt}$: Remember that since the volume is decreasing, $\frac{dV}{dt}$ is negative. Failing to include the negative sign will result in an incorrect answer.
• Units: Always keep track of the units to ensure your answer is in the correct units. In this problem, volume is in cubic meters, time is in minutes, and radius is in meters.
• Careful Calculation: Double-check your arithmetic, especially when dealing with fractions and exponents.

Summary

We used the formula for the volume of a sphere and related rates to find the rate at which the radius of the balloon decreases. First, we calculated the volume after 49 minutes, then found the radius at that time. Next, we differentiated the volume formula with respect to time and plugged in the known values to solve for $\frac{dr}{dt}$. The rate at which the radius decreases is $\frac{2}{9}$ meters per minute.

Final Answer The final answer is \boxed{{2 \over 9}}, which corresponds to option (C).