A-Level Mathematics October/November 2024 Q10(b): At time t = 0 the tap is opened. It is given that h = 4 when t = 0 and that h = 2.25 wh…

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  The differential equation modelling the height $h$ of the water at time $t$ is of the form:

  $\frac{dh}{dt} = -\lambda \sqrt{h}$

  To solve this, we first separate the variables:

  $\int \frac{1}{\sqrt{h}} \, dh = \int -\lambda \, dt$

  $\int h^{-1/2} \, dh = \int -\lambda \, dt$

  Now, we integrate both sides:

  $2h^{1/2} = -\lambda t + C$

  $2\sqrt{h} = -\lambda t + C$

  We are given the boundary condition that when $t=0$, $h=4$. Substituting these values into the equation:

  $2\sqrt{4} = -\lambda(0) + C$

  $2(2) = C \implies C = 4$

  So the equation becomes:

  $2\sqrt{h} = -\lambda t + 4$

  We are given a second boundary condition that when $t=20$, $h=2.25$. Substituting these values:

  $2\sqrt{2.25} = -\lambda(20) + 4$

  $2(1.5) = -20\lambda + 4$

  $3 = -20\lambda + 4$

  $20\lambda = 4 - 3$

  $20\lambda = 1 \implies \lambda = \frac{1}{20}$

  Now we substitute the values of $\lambda$ and $C$ back into the integrated equation to find the relationship between $t$ and $h$:

  $2\sqrt{h} = -\frac{1}{20}t + 4$

  We need to make $t$ the subject of the formula:

  $\frac{1}{20}t = 4 - 2\sqrt{h}$

  $t = 20(4 - 2\sqrt{h})$

  $t = 80 - 40\sqrt{h}$

  To find the time taken to empty the tank, we set $h=0$:

  $t = 80 - 40\sqrt{0}$

  $t = 80 - 0$

  $t = 80$

  The time taken to empty the tank is 80 minutes.

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  How the marks are awarded

  • M1* — Correctly separating the variables to get an expression of the form ∫(1/√h) dh = ∫-λ dt and attempting to integrate.
  • A1 — Obtaining the correct integrated equation, 2√h = -λt + C, or an equivalent form. The constant of integration C must be present.
  • DM1 — Using the first boundary condition (t=0, h=4) in the integrated equation to find the value of the constant of integration, C=4.
  • DM1 — Using the second boundary condition (t=20, h=2.25) and the value of C to form an equation and solve for λ, finding λ = 1/20.
  • A1 — Substituting the calculated constants back into the general solution and rearranging to obtain the correct expression for t, t = 80 - 40√h.
  • A1 — Correctly calculating the time to empty the tank by substituting h=0 into the expression for t, giving the final answer of 80.

  Common mistakes

  • Forgetting the constant of integration '+C' after integrating. This makes it impossible to satisfy both boundary conditions and scores zero further marks.
  • Making a sign error during separation or integration, for example integrating to 2√h = λt + C, which leads to an incorrect final answer.
  • Incorrectly integrating h^(-1/2), often getting something like ln(√h) or (1/2)h^(1/2).
  • Making algebraic errors when solving for the constants λ and C, or when rearranging the final equation to make t the subject.

  Examiner tip: For differential equations with boundary conditions, always follow the sequence: separate variables, integrate (remembering the constant '+C'), then use the given conditions one by one to find the unknown constants.

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