Applications of Derivatives
Rate of Change of Quantities
Grade 12
Question:
<p>If the volume of a spherical ball is increasing at the rate of \(4\pi\) cc/s, then the rate of increase of its radius (in cm/sec), when the volume is \(288\pi\) cc, is</p>
<p>\(\dfrac{1}{6}\)</p>
<p>\(\dfrac{1}{9}\)</p>
<p>\(\dfrac{1}{36}\)</p>
<p>\(\dfrac{1}{24}\)</p>
Step-by-Step Solution
Key Concept: Differentiate the sphere volume formula V = (4/3)πr³ with respect to time using chain rule to relate dV/dt to dr/dt, then substitute the given conditions to find dr/dt.
<p><strong>Step 1:</strong> Write the volume formula for a sphere: V = (4/3)πr³</p><p><strong>Step 2:</strong> Differentiate both sides with respect to time t using chain rule:<br/>dV/dt = (4/3)π · 3r² · dr/dt = 4πr² · dr/dt</p><p><strong>Step 3:</strong> Find the radius when V = 288π cc:<br/>(4/3)πr³ = 288π<br/>r³ = 216<br/>r = 6 cm</p><p><strong>Step 4:</strong> Substitute dV/dt = 4π cc/s and r = 6 cm into the differentiated equation:<br/>4π = 4π(6)² · dr/dt<br/>4π = 4π(36) · dr/dt<br/>4π = 144π · dr/dt<br/>dr/dt = 4π/(144π) = 1/36 cm/s</p><p>∴ Answer: C (1/36 cm/s)</p>
Correct Answer: C