Related Rates Under a Leaking Constraint
A conical tank (vertex down) has height 12 m and top radius 4 m. Water leaks out at a constant rate of 3 m³/min. At time t = 0 the water depth is 9 m. An engineer models the water volume as V = (1/3)π r² h and, because the tank's cross-section is similar at every depth, uses r/h = 4/12 to write V purely in terms of h. She wants to know how fast the depth is falling when h = 6 m, and she also wants to check a claim from a colleague that 'the depth falls at a constant rate because the leak rate is constant.' She is careful to note that the leak rate is the rate of change of the remaining volume, so dV/dt = −3.
1. Express V as a function of h alone, then differentiate to relate dh/dt and dV/dt. Find dh/dt when h = 6 m, with units.
Read the worked answer
Relate r and h: r/h = 4/12 = 1/3, so r = h/3.
Volume in terms of h: V = (1/3)π r² h = (1/3)π (h/3)² h = (1/3)π (h²/9) h = π h³/27.
Differentiate with respect to t: dV/dt = (π/27)·3h²·(dh/dt) = (π h²/3)(dh/dt).
Solve for dh/dt: dh/dt = (3 dV/dt)/(π h²).
With dV/dt = −3 and h = 6: dh/dt = 3(−3)/(π·36) = −9/(36π) = −1/(4π) m/min ≈ −0.0796 m/min.
The negative sign confirms the depth is decreasing at about 0.08 m/min when the water is 6 m deep.
2. Evaluate the colleague's claim that the depth falls at a constant rate. Use the formula for dh/dt to justify your answer quantitatively.
Read the worked answer
The colleague's claim is false. From dh/dt = (3 dV/dt)/(π h²) = −9/(π h²), the rate depends on h.
- At h = 9 m: dh/dt = −9/(81π) = −1/(9π) ≈ −0.0354 m/min.
- At h = 6 m: dh/dt = −9/(36π) = −1/(4π) ≈ −0.0796 m/min.
- At h = 3 m: dh/dt = −9/(9π) = −1/π ≈ −0.318 m/min.
Although dV/dt is constant, the surface area π r² = π h²/9 shrinks as h decreases, so the same volume loss produces a larger depth drop. The depth falls faster as the tank empties - the rate is not constant, and dh/dt is unbounded as h → 0⁺. The correct statement is that the volume decreases at a constant rate, not the depth.