Applications of Derivatives
Rate of Change
Grade 12
Question:
<p>A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is \(\tan^{-1}(1/2)\). Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10 m, is ________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: Use the semi-vertical angle to establish the relationship between radius and height of water (r = 2h), then apply volume differentiation with respect to time using the chain rule to find dh/dt.
<p><strong>Step 1: Interpret the semi-vertical angle</strong></p><p>Semi-vertical angle α satisfies tan(α) = 1/2. For a cone with height h and radius r at the water surface: tan(α) = r/h = 1/2, so r = h/2.</p><p><strong>Step 2: Express volume in terms of height</strong></p><p>Volume of cone: V = (1/3)πr²h = (1/3)π(h/2)²h = (πh³)/12</p><p><strong>Step 3: Differentiate with respect to time</strong></p><p>dV/dt = (π/12) · 3h² · dh/dt = (πh²/4) · dh/dt</p><p><strong>Step 4: Apply given conditions</strong></p><p>We know dV/dt = 5 m³/min and h = 10 m at the instant in question.</p><p>5 = (π · 10²/4) · dh/dt</p><p>5 = (100π/4) · dh/dt</p><p>5 = 25π · dh/dt</p><p>dh/dt = 5/(25π) = 1/(5π) = 1/15.7080 ≈ 0.0637 m/min</p><p><strong>∴ Answer: 0.0637 m/min (or 1/(5π) m/min)</strong></p>
Correct Answer: 0