Applications of Derivatives
Rate of change
Grade 12
Question:
<p>A spherical balloon is being inflated at a constant rate. The volume of the balloon after 49 minutes of leakage is \(4500\pi - 49(72\pi) = 972\pi\). The rate (in meters per minute) at which the radius of the balloon decreases 49 min after the leakage began is:</p>
<p>\(\dfrac{2}{9}\)</p>
<p>\(\dfrac{1}{9}\)</p>
<p>\(\dfrac{4}{9}\)</p>
<p>\(\dfrac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the relationship V = (4/3)πr³ to connect volume and radius, then differentiate with respect to time to relate dV/dt to dr/dt. The rate of change of radius depends on the current radius at that instant.
<p><strong>Step 1:</strong> Start with the volume formula for a sphere: V = (4/3)πr³</p><p><strong>Step 2:</strong> Differentiate both sides with respect to time t:</p><p>dV/dt = (4/3)π · 3r² · dr/dt = 4πr² · dr/dt</p><p><strong>Step 3:</strong> Find the radius at t = 49 minutes using the given volume:</p><p>V(49) = 972π = (4/3)πr³</p><p>972 = (4/3)r³</p><p>r³ = 729</p><p>r = 9 meters</p><p><strong>Step 4:</strong> Determine dV/dt. Since leakage occurs at a constant rate and the volume decreases by 72π per minute:</p><p>dV/dt = -72π m³/min (negative because volume is decreasing)</p><p><strong>Step 5:</strong> Substitute into the differentiated equation:</p><p>-72π = 4π(9)² · dr/dt</p><p>-72π = 4π(81) · dr/dt</p><p>-72π = 324π · dr/dt</p><p>dr/dt = -72π/(324π) = -72/324 = -2/9 m/min</p><p><strong>Step 6:</strong> The rate at which the radius decreases is the absolute value:</p><p>|dr/dt| = 2/9 m/min</p><p>∴ Answer: A</p>
Correct Answer: A