Topic: Engineering Mathematics
A spherical balloon is filled with 3536 * pi cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 70 * pi cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 50 minutes after the leakage began is
- A. 1/9
- B. 1/2
- C. 35/18
- D. -35/18
Answer: D. -35/18
Explanation: The initial volume is V_0 = 3536 * pi. The leakage rate is dV/dt = -70 * pi. After 50 minutes, the volume is V_50 = 3536 * pi - (70 * pi * 50) = 36 * pi. Using the volume of a sphere V = (4/3) * pi * r³, we find the radius at 50 minutes: (4/3) * pi * r³ = 36 * pi, which gives r³ = 27, so r = 3 m. Differentiating the volume formula yields dV/dt = 4 * pi * r² * (dr/dt). Substituting the knowns gives -70 * pi = 4 * pi * (3^2) * (dr/dt). Solving for the rate of change of the radius yields dr/dt = -70 / 36 = -35/18. The negative sign indicates a decrease, so the rate of decrease is 35/18.