Applications of Derivatives
Rate of change
Grade 12
Question:
<p>A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/sec. At that instant, when the radius of circular wave is 8 cm, the rate of increase of enclosed area is:</p>
<p>(a) \(6\pi\) cm\(^2\)/sec</p>
<p>(b) \(8\pi\) cm\(^2\)/sec</p>
<p>(c) \(\dfrac{8\pi}{3}\) cm\(^2\)/sec</p>
<p>(d) \(80\pi\) cm\(^2\)/sec</p>
Step-by-Step Solution
Key Concept: Use the chain rule to relate the rate of change of area to the rate of change of radius: dA/dt = (dA/dr)(dr/dt), where A = πr² and dr/dt = 5 cm/s.
<p><strong>Step 1:</strong> Identify the given information: dr/dt = 5 cm/s, r = 8 cm at the instant considered.</p><p><strong>Step 2:</strong> Write the area formula: A = πr²</p><p><strong>Step 3:</strong> Differentiate both sides with respect to time using chain rule: dA/dt = 2πr(dr/dt)</p><p><strong>Step 4:</strong> Substitute the values: dA/dt = 2π(8)(5) = 80π cm²/s</p><p><strong>Step 5:</strong> Calculate numerically: dA/dt = 80π ≈ 251.33 cm²/s or 80π cm²/s</p><p>∴ Answer: D (assuming D is 80π cm²/s or approximately 251.33 cm²/s)</p>
Correct Answer: D