Applications of Derivatives
Rate of Change
Grade 12
Question:
<p>The volume of a right circular cone is given by \(V = \frac{1}{3}\pi r^2 h\). When \(h = 20\) cm, \(\frac{dh}{dt} = -4\) cm/s, \(r = 10\) cm and \(\frac{dr}{dt} = 2\) cm/s, find \(\frac{dV}{dt}\) (in cm³/s).</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the volume formula V = (1/3)πr²h with respect to time t, treating both r and h as functions of t. The chain rule gives dV/dt = (1/3)π[2r(dr/dt)·h + r²(dh/dt)].
<p><strong>Step 1:</strong> Start with V = (1/3)πr²h</p><p><strong>Step 2:</strong> Differentiate both sides with respect to t using the product rule on r²h:</p><p>dV/dt = (1/3)π[d/dt(r²h)]</p><p>dV/dt = (1/3)π[r²(dh/dt) + h·d/dt(r²)]</p><p>dV/dt = (1/3)π[r²(dh/dt) + h·2r(dr/dt)]</p><p><strong>Step 3:</strong> Substitute given values: r = 10, h = 20, dr/dt = 2, dh/dt = -4</p><p>dV/dt = (1/3)π[(10)²(-4) + (20)·2(10)·(2)]</p><p>dV/dt = (1/3)π[100(-4) + 20·20·2]</p><p>dV/dt = (1/3)π[-400 + 800]</p><p>dV/dt = (1/3)π[400]</p><p>dV/dt = (400π)/3 ≈ 419.0476 cm³/s</p><p>∴ <strong>Answer: 419.0476</strong></p>
Correct Answer: 419.0476