A-Level Mathematics October/November 2024 Q10(c): Solve the differential equation in part (a), obtaining an expression for t in terms of r.

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  The differential equation from part (a) is assumed to be $\frac{dr}{dt} = \frac{50-r}{5r^2}$.

  To solve the differential equation, we first separate the variables:

  $\frac{5r^2}{50-r} dr = 1 dt$

  $\int \frac{5r^2}{50-r} \, dr = \int 1 \, dt$

  From the algebraic division in part (b), we have:

  $\frac{5r^2}{50-r} = -5r - 250 + \frac{12500}{50-r}$

  Substituting this into the integral:

  $\int \left(-5r - 250 + \frac{12500}{50-r}\right) \, dr = \int 1 \, dt$

  Now, we integrate both sides:

  $-\frac{5r^2}{2} - 250r + 12500(-\ln|50-r|) = t + k$

  Assuming $r < 50$, $|50-r| = 50-r$:

  $-\frac{5}{2}r^2 - 250r - 12500\ln(50-r) = t + k$

  We are given the initial condition that $t=0$ when $r=0$. We substitute these values to find the constant of integration, $k$:

  $-\frac{5}{2}(0)^2 - 250(0) - 12500\ln(50-0) = 0 + k$

  $0 - 0 - 12500\ln(50) = k$

  $k = -12500\ln(50)$

  Substitute the value of $k$ back into the integrated equation:

  $-\frac{5}{2}r^2 - 250r - 12500\ln(50-r) = t - 12500\ln(50)$

  Finally, we make $t$ the subject of the expression:

  $t = -\frac{5}{2}r^2 - 250r - 12500\ln(50-r) + 12500\ln(50)$

  This can also be written as $t = -\frac{5}{2}r^2 - 250r + 12500\ln\left(\frac{50}{50-r}\right)$.

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  How the marks are awarded

  • B1FT — Correctly separating the variables to get an expression of the form $\int \frac{5r^2}{50-r} \, dr = \int 1 \, dt$. This is the first line of working.
  • DB1 — Correctly integrating the right-hand side, $\int 1 \, dt$, to obtain the term $t$.
  • M1 — Attempting to integrate the expression from part (b). In the model answer, this is shown by integrating $-5r - 250 + \frac{12500}{50-r}$ to get terms of the form $Ar^2$, $Br$, and $C\ln(50-r)$.
  • A1FT — Obtaining the correct integrated terms on the left-hand side: $-\frac{5}{2}r^2 - 250r - 12500\ln(50-r)$. This mark follows through from the candidate's coefficients found in part (b).
  • M1 — Using the initial conditions $t=0$ and $r=0$ in an integrated expression containing a constant of integration, as shown in the step to find $k$.
  • A1 — Obtaining the final correct expression for $t$ in terms of $r$, with the constant evaluated correctly: $t = -\frac{5}{2}r^2 - 250r - 12500\ln(50-r) + 12500\ln(50)$.

  Common mistakes

  • Sign error when integrating $\frac{1}{50-r}$, leading to $+12500\ln(50-r)$ instead of $-12500\ln(50-r)$ due to forgetting the chain rule derivative of the denominator is -1.
  • Errors with the constant of integration, such as forgetting it completely or making a sign error when calculating its value from the initial conditions.
  • Carrying forward incorrect coefficients from the algebraic division in part (b), which loses accuracy marks (A1) but may still allow for method marks (M1).
  • Incorrectly rearranging the final equation to make $t$ the subject, often by mixing up the signs of the terms being moved.

  Examiner tip: Mastering the integration of functions requiring algebraic division or partial fractions is crucial for solving many first-order differential equations.

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