Question

A spherical balloon is filled with (4500\pi) cubic meters of helium gas. If a leak causes gas to escape at (72\pi) m(^3)/min, then the rate at which the radius decreases 49 minutes after leakage began is

• A. (\frac{9}{7})
• B. (-\frac{2}{9})
• C. (\frac{9}{2})
• D. [suspicious link removed]

Hint:

Always calculate the specific value of the independent variable (like radius) at the requested time first.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the rate of decrease of the radius of a spherical balloon given the constant rate of volume decrease.
Step 2: Key Formula or Approach:
Volume of a sphere $V = \frac{4}{3} \pi r^3$.
Rate of change of volume $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Step 3: Detailed Explanation:
1. Initial volume $V_0 = 4500\pi$.
2. Rate of leak $\frac{dV}{dt} = -72\pi$ m$^3$/min.
3. Volume after $t = 49$ minutes:
$V = V_0 + (\frac{dV}{dt})t = 4500\pi - 72\pi(49) = 4500\pi - 3528\pi = 972\pi$ m$^3$.
4. Find radius $r$ at this volume:
$\frac{4}{3} \pi r^3 = 972\pi \implies r^3 = \frac{972 \times 3}{4} = 243 \times 3 = 729 \implies r = 9$ m.
5. Now calculate $\frac{dr}{dt}$:
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \implies -72\pi = 4 \pi (9)^2 \frac{dr}{dt}$
$-72 = 4 \times 81 \times \frac{dr}{dt}$
$-72 = 324 \frac{dr}{dt} \implies \frac{dr}{dt} = -\frac{72}{324} = -\frac{2}{9}$ m/min.
6. The rate of decrease is the absolute value, which is $\frac{2}{9}$.
Step 4: Final Answer:
The rate at which the radius decreases is $\frac{2}{9}$ meters per minute.