The diagram shows a tank for holding water. The tank is in the shape of a cube of side 50 cm. At time t seconds, the depth of water in the tank is hcm. Water is poured into the tank at a rate of 5000 cm³ s⁻¹. Water pours out of the tank through a hole in the bottom at a rate proportional to h². When h = 20, the depth of the water is increasing at a rate of 0.4 cms⁻¹. [Figure 10.1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [1] (a) State dV/dt = 5000-kh². SOI.
2. [1] (a) Use V = 2500h (tank base area = 50x50=2500) and chain rule to obtain dh/dt = (dV/dt) / 2500.
3. [1] (a) Use h = 20, dh/dt = 0.4 to obtain k. E.g., 0.4 = (5000-k(20)²) / 2500 ⇒ k=10.
4. [1] (a) Obtain dh/dt = (500-h²) / 250 from correct working. There needs to be a complete statement for A1.
5. [1] (b) Separate variables correctly. E.g. ∫ 1/(500-h²) dh = ∫ 1/250 dt. Condone missing integral signs or missing dh, dt, but not both.
6. [1] (b) Obtain (1/(2√500)) ln(|(√500+h)/(√500-h)|) or (1/(20√5)) ln(|(10√5+h)/(10√5-h)|).
7. [1] (b) Obtain t/250.
8. [1] (b) Use t=0, h=0 in an expression of the form pt+qln(a+h)+rln(b-h) to obtain the constant of integration or as limits on a definite integral. Need some indication that the conclusion is based on using t = 0, h = 0, e.g. "t=0, h=0 ⇒C=0".
9. [1] (b) Obtain t=16.1 (s). Accept AWRT.