Applications of Derivatives
Rate of change
Grade 12
Question:
<p>The rate of increase of radius of a sphere when volume is increasing at a constant rate \(4\pi\) is \(\dfrac{dr}{dt}\bigg|_{\text{vol}=288} = \dfrac{dr}{dt}\bigg|_{r=6}\). The rate of increase of radius when \(r = 6\) is:</p>
<p>\(\dfrac{1}{36}\)</p>
<p>\(\dfrac{1}{12}\)</p>
<p>\(\dfrac{1}{r^2}\)</p>
<p>\(\dfrac{1}{6}\)</p>
Step-by-Step Solution
Key Concept: Use the volume formula V = (4/3)πr³ and differentiate with respect to time to relate dV/dt to dr/dt. The rate of change of radius depends inversely on the surface area (4πr²), so it decreases as radius increases.
<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> Given that dV/dt = 4π (constant rate), substitute:</p><p>4π = 4πr² · dr/dt</p><p><strong>Step 4:</strong> Solve for dr/dt:</p><p>dr/dt = 4π/(4πr²) = 1/r²</p><p><strong>Step 5:</strong> When r = 6:</p><p>dr/dt = 1/36</p><p>∴ Answer: C (which equals 1/36)</p>
Correct Answer: C