Water is pumped at 2 m³/min into a tank shaped like a cone with its vertex down. The tank is 12 m deep and its top radius is 4 m. How fast is the water level rising when the water is 6 m deep?
How to lay out the answer. Follow the frame, then write it in your own words.
1 · Write what is given and what is to be found. Given: dV/dt = 2 m³/min, tank depth 12 m, top radius 4 m. To find: dh/dt when h = 6 m.
2 · Draw the figure. Draw the cone with its vertex down. Mark the water surface at depth h with radius r. Label 12 m and 4 m.
3 · Write the formula, rule or theorem. V = (1/3)πr²h. By similar triangles, r/h = 4/12, so r = h/3.
4 · Write V in terms of h only. V = (1/3)π(h/3)²h = πh³/27
5 · Differentiate with respect to t. dV/dt = (πh²/9) × dh/dt
6 · Substitute and solve.
2 = (π × 36/9) × dh/dt = 4π × dh/dt
dh/dt = 2/(4π) = 1/(2π) m/min
7 · Check and box the answer. The rate is positive, so the level is rising: 1/(2π) m/min, about 0.16 m/min.
Scoring tip: Remove r with similar triangles before you differentiate. Then V has only one changing quantity, h.
