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 relationship between area of a circle (A = πr²) and apply the chain rule to find dA/dt in terms of dr/dt. The rate of change of area depends on both the current radius and the speed at which the radius is expanding.
<p><strong>Step 1:</strong> Identify the given information.</p><p>Speed of wave = dr/dt = 5 cm/sec (rate of radius increase)</p><p>Radius at the instant = r = 8 cm</p><p><strong>Step 2:</strong> Write the area formula for a circle.</p><p>A = πr²</p><p><strong>Step 3:</strong> Differentiate both sides with respect to time using the chain rule.</p><p>dA/dt = d(πr²)/dt = 2πr(dr/dt)</p><p><strong>Step 4:</strong> Substitute the known values at r = 8 cm and dr/dt = 5 cm/sec.</p><p>dA/dt = 2π(8)(5) = 80π cm²/sec</p><p><strong>Step 5:</strong> Calculate the numerical value.</p><p>dA/dt = 80π ≈ 251.33 cm²/sec (or exact answer: 80π cm²/sec)</p><p>∴ Answer: D</p>
Correct Answer: D