A-Level Mathematics May/June 2024 Q9(a): A container in the shape of a cuboid has a square base of side x and a height of (10-x)…

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  The volume, $V$, of the cuboid is given by the area of the base multiplied by the height. $V = x^2(10-x) = 10x^2 - x^3$

  The container decreases in volume at a rate inversely proportional to $t$. We can write this as: $\frac{dV}{dt} = -\frac{k}{t}$, where $k$ is a positive constant of proportionality.

  Next, we find the rate of change of volume with respect to $x$ by differentiating $V$: $\frac{dV}{dx} = \frac{d}{dx}(10x^2 - x^3) = 20x - 3x^2$

  Using the chain rule for related rates, we have: $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$

  Substituting the expressions for $\frac{dV}{dt}$ and $\frac{dV}{dx}$: $-\frac{k}{t} = (20x - 3x^2) \frac{dx}{dt}$

  Rearranging to make $\frac{dx}{dt}$ the subject: $\frac{dx}{dt} = -\frac{k}{t(20x - 3x^2)}$

  We are given that when $t = \frac{1}{10}$, $x = \frac{1}{2}$. We are also told that the rate of decrease of $x$ is $\frac{20}{37}$, which means $\frac{dx}{dt} = -\frac{20}{37}$.

  Now, we substitute these values into our equation for $\frac{dx}{dt}$ to find the constant $k$.

  First, evaluate the term $(20x - 3x^2)$ at $x = \frac{1}{2}$: $20(\frac{1}{2}) - 3(\frac{1}{2})^2 = 10 - 3(\frac{1}{4}) = 10 - \frac{3}{4} = \frac{40-3}{4} = \frac{37}{4}$

  Now substitute everything into the differential equation: $-\frac{20}{37} = -\frac{k}{(\frac{1}{10})(\frac{37}{4})}$

  $\frac{20}{37} = \frac{k}{\frac{37}{40}}$

  $k = \frac{20}{37} \times \frac{37}{40} = \frac{20}{40} = \frac{1}{2}$

  Finally, substitute $k = \frac{1}{2}$ back into the general equation for $\frac{dx}{dt}$: $\frac{dx}{dt} = -\frac{\frac{1}{2}}{t(20x - 3x^2)}$

  $\frac{dx}{dt} = -\frac{1}{2t(20x - 3x^2)}$

  This is the required differential equation. (AG)

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  How the marks are awarded

  • B1 — Correctly translating the statement 'decreases in volume at a rate which is inversely proportional to t' into the equation $\frac{dV}{dt} = -\frac{k}{t}$.
  • B1 — Finding the volume $V = 10x^2 - x^3$ and correctly differentiating with respect to $x$ to obtain $\frac{dV}{dx} = 20x - 3x^2$.
  • M1 — Applying the chain rule, $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$, using the expressions for $\frac{dV}{dt}$ and $\frac{dV}{dx}$.
  • A1 — Correctly rearranging the chain rule to obtain an expression for $\frac{dx}{dt}$ in terms of $k$, $t$, and $x$, such as $\frac{dx}{dt} = -\frac{k}{t(20x - 3x^2)}$.
  • A1 — Using the given values $t = 1/10$, $x = 1/2$ and correctly interpreting 'rate of decrease' as $\frac{dx}{dt} = -20/37$ to find $k=1/2$, and then substituting this back to show the final given answer.

  Common mistakes

  • Sign errors: Forgetting the negative sign for a rate of decrease, e.g., writing $\frac{dV}{dt} = \frac{k}{t}$ or using $\frac{dx}{dt} = +\frac{20}{37}$. This is the most common and costly error.
  • Incorrect chain rule application: MUDDLING the terms, for example by multiplying rates that should be divided, such as writing $\frac{dx}{dt} = \frac{dV}{dx} \times \frac{dV}{dt}$.
  • Differentiation errors: Making a mistake when differentiating the volume expression, for instance, forgetting the power rule and getting $\frac{dV}{dx} = 20x - x^3$.
  • Arithmetic errors: Making a slip when substituting fractions and solving for $k$, which is common under time pressure.

  Examiner tip: Carefully translate verbal descriptions of rates, especially 'increasing' or 'decreasing', into correct mathematical statements with appropriate signs before applying the chain rule.

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