A-Level Mathematics October/November 2024 Q10(b): It is given that when t = 0, h = 50. Find the time taken for the depth of water in the…

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  The differential equation is given by $\frac{dh}{dt} = \frac{250-3h}{200}$.

  Step 1: Separate the variables.

  $\int \frac{1}{250-3h} \,dh = \int \frac{1}{200} \,dt$

  Step 2: Integrate both sides.

  $-\frac{1}{3} \ln|250-3h| = \frac{1}{200}t + C$

  Step 3: Use the initial condition (t=0, h=50) to find the constant of integration, C.

  When $t=0$, $h=50$: $-\frac{1}{3} \ln|250 - 3(50)| = \frac{1}{200}(0) + C$ $-\frac{1}{3} \ln|250 - 150| = C$ $C = -\frac{1}{3} \ln(100)$

  Step 4: Write the particular solution and solve for t when h=80.

  The equation is now: $-\frac{1}{3} \ln|250-3h| = \frac{1}{200}t - \frac{1}{3} \ln(100)$

  Substitute $h=80$: $-\frac{1}{3} \ln|250 - 3(80)| = \frac{1}{200}t - \frac{1}{3} \ln(100)$ $-\frac{1}{3} \ln|250 - 240| = \frac{1}{200}t - \frac{1}{3} \ln(100)$ $-\frac{1}{3} \ln(10) = \frac{1}{200}t - \frac{1}{3} \ln(100)$

  Rearrange to solve for $t$: $\frac{1}{3} \ln(100) - \frac{1}{3} \ln(10) = \frac{1}{200}t$ $\frac{1}{3} (\ln(100) - \ln(10)) = \frac{t}{200}$ $\frac{1}{3} \ln(\frac{100}{10}) = \frac{t}{200}$ $\frac{1}{3} \ln(10) = \frac{t}{200}$ $t = \frac{200}{3} \ln(10)$ $t = 153.505...$

  Step 5: Give the answer to the required accuracy.

  $t = 150$ (to 2 significant figures)

  The time taken for the depth to reach 80 cm is 150 seconds.

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  How the marks are awarded

  • M1 — Correctly separating the variables to get ∫ 1/(250-3h) dh = ∫ 1/200 dt and integrating one side correctly, for instance, the right-hand side to t/200.
  • A1 — Obtaining the correct integrated expression -1/3 ln|250-3h| = t/200 (+C). This requires correct integration of both sides, including the -1/3 factor from the chain rule.
  • M1 — Substituting the initial conditions t=0 and h=50 into an expression containing ln|250-3h| and t to find the value of the constant of integration, C.
  • A1 — Correctly calculating the constant of integration as C = -1/3 ln(100) or an equivalent form.
  • A1 — Substituting h=80 into the particular solution, correctly solving for t to get a value of 153.5..., and providing the final answer correctly rounded to 2 significant figures as 150.

  Common mistakes

  • Forgetting the chain rule when integrating 1/(250-3h), leading to ln|250-3h| instead of the correct -1/3 ln|250-3h|. This is a very common integration error.
  • Incorrectly handling the constant of integration, either by adding it inside the logarithm or by making an algebraic slip when finding its value.
  • Making errors with logarithm laws when solving for t, such as calculating ln(100) - ln(10) as ln(90) instead of the correct ln(100/10) = ln(10).
  • Failing to round the final answer to the required 2 significant figures. An answer of 154 or 153.5 would not receive the final mark.

  Examiner tip: When integrating expressions of the form 1/(ax+b), always remember to divide by the coefficient 'a' as part of the reverse chain rule.

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