A-LevelMathematicsDifferential equations9709/33 · May/June 2020 · Q1012 Marks

A tank containing water is in the form of a hemisphere. The axis is vertical, the lowest point is A and the radius is r, as shown in the diagram. The depth of water at time t is h. At time t = 0 the tank is full and the depth of the water is r. At this instant a tap at A is opened and water begins to flow out at a rate proportional to √h. The tank becomes empty at time t = 14. The volume of water in the tank is V when the depth is h. It is given that V = ⅓π(3rh² – h³).

(a)Show that h and t satisfy a differential equation of the form dh/dt = -B / (2rh^(1/2) – h^(3/2)), where B is a positive constant.[4]
(b)Solve the differential equation and obtain an expression for t in terms of h and r.[8]

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Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[B1] State or imply dV/dt = -k√h
[B1] State or imply dV/dh = 2πrh – πh², or equivalent
[M1] Use dV/dt = dV/dh * dh/dt
[A1] Obtain the given answer correctly
[M1] Separate variables and attempt integration of at least one side
[A3, 2, 1, 0] Obtain terms ⁴/₃ rh^(3/2) - ²/₅ h^(5/2) and -Bt
[M1] Use t=0, h=r to find a constant of integration c
[M1] Use t = 14, h = 0 to find B
[A1] Obtain correct c and B, e.g. c = ¹⁴/₁₅ r^(5/2), B = -¹/₁₅ r^(5/2)
[A1] Obtain final answer t = 14 - ²⁰/₃ (h/r)^(3/2) + ⁶/₅ (h/r)^(5/2), or equivalent

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