Applications of Derivatives
Rate of Change
Grade 12
Question:
<p>A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is ________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: The ice layer forms a spherical shell whose volume V = (4/3)π(R³ - r³) where R is outer radius and r is inner (iron ball) radius. Differentiate with respect to time using chain rule: dV/dt = 4π(R² dR/dt - r² dr/dt), where dr/dt = 0 (iron ball is fixed).
<p><strong>Step 1:</strong> Set up the geometry. Iron ball radius = 10 cm (fixed). When ice thickness = 5 cm, outer radius R = 10 + 5 = 15 cm.</p><p><strong>Step 2:</strong> Volume of ice layer: V = (4/3)π(R³ - r³) = (4/3)π(R³ - 10³)</p><p><strong>Step 3:</strong> Differentiate with respect to time: dV/dt = (4/3)π · 3R² · dR/dt = 4πR² · dR/dt</p><p><strong>Step 4:</strong> Given dV/dt = -50 cm³/min (negative because melting decreases volume) and R = 15 cm at the instant of interest.</p><p><strong>Step 5:</strong> Substitute: -50 = 4π(15)² · dR/dt</p><p>-50 = 4π(225) · dR/dt</p><p>-50 = 900π · dR/dt</p><p>dR/dt = -50/(900π) = -1/(18π) cm/min</p><p><strong>Step 6:</strong> Since thickness h = R - 10, we have dh/dt = dR/dt = -1/(18π)</p><p>dh/dt = -1/(18π) = -0.0177 cm/min</p><p><strong>Step 7:</strong> Rate of decrease of thickness = |dh/dt| = 1/(18π) ≈ <strong>0.0177</strong> cm/min</p><p>∴ Answer: <strong>0.0177</strong> cm/min (or 1/(18π) exactly)</p>
Correct Answer: 0