<p>Ice forms on a spherical ball of radius 10 cm. If the volume of ice is increasing at the rate of 50 cm³/min, find the rate of decrease of thickness of ice when the thickness of ice is 5 cm.</p>
Step-by-Step Solution
Key Concept: The ice layer forms a spherical shell around the ball. As ice accumulates, the outer radius increases while the inner radius (ball radius) stays constant at 10 cm. Differentiate the volume of the spherical shell with respect to time to relate dV/dt to the rate of change of outer radius, then convert this to the rate of change of ice thickness.
<p><strong>Step 1:</strong> Set up the geometry. Let t = thickness of ice (in cm). The inner radius is fixed at 10 cm, outer radius R = 10 + t.</p><p><strong>Step 2:</strong> Express volume of ice as difference of two spheres:</p><p>V = (4/3)π(R³ - 10³) = (4/3)π[(10+t)³ - 1000]</p><p><strong>Step 3:</strong> Differentiate both sides with respect to time:</p><p>dV/dt = (4/3)π · 3(10+t)² · dt/dt = 4π(10+t)² · dt/dt</p><p><strong>Step 4:</strong> When t = 5 cm, R = 15 cm, and dV/dt = 50 cm³/min:</p><p>50 = 4π(15)² · dt/dt</p><p>50 = 4π(225) · dt/dt</p><p>50 = 900π · dt/dt</p><p><strong>Step 5:</strong> Solve for dt/dt:</p><p>dt/dt = 50/(900π) = 1/(18π) = 1/(18 × 3.14159) ≈ 0.01768 cm/min</p><p>∴ Answer: <strong>0.0177 cm/min</strong> (or 1/(18π) cm/min)</p>
Correct Answer: 0.0177