A-Level Mathematics October/November 2024 Q10(a): A large cylindrical tank is used to store water. The base of the tank is a circle of ra…

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  Let $V$ be the volume of water and $h$ be the depth of water in the tank at time $t$. The radius of the cylindrical tank is $r=4$ metres.

  The volume of water is given by the formula for a cylinder: $V = \pi r^2 h$ $V = \pi (4)^2 h = 16\pi h$

  Water flows out at a rate proportional to the square root of the volume. Since the volume is decreasing, the rate of change is negative. This can be written as: $\frac{dV}{dt} = -k\sqrt{V}$, where $k$ is a positive constant of proportionality.

  We need to find an expression for $\frac{dh}{dt}$. We can use the chain rule to relate the rates of change: $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$

  First, we differentiate the volume equation $V = 16\pi h$ with respect to $h$: $\frac{dV}{dh} = 16\pi$

  Now, we rearrange the chain rule to make $\frac{dh}{dt}$ the subject: $\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}$

  Substitute the expressions for $\frac{dV}{dt}$ and $\frac{dV}{dh}$: $\frac{dh}{dt} = \frac{-k\sqrt{V}}{16\pi}$

  To get the expression in terms of $h$, substitute $V = 16\pi h$ into the equation: $\frac{dh}{dt} = \frac{-k\sqrt{16\pi h}}{16\pi}$

  Now, simplify the expression: $\frac{dh}{dt} = \frac{-k \times \sqrt{16} \times \sqrt{\pi} \times \sqrt{h}}{16\pi}$ $\frac{dh}{dt} = \frac{-4k\sqrt{\pi}}{16\pi} \sqrt{h}$ $\frac{dh}{dt} = -\left(\frac{k}{4\sqrt{\pi}}\right) \sqrt{h}$

  Let $\lambda = \frac{k}{4\sqrt{\pi}}$. Since $k$ and $\pi$ are positive constants, $\lambda$ is also a positive constant.

  Therefore, we have shown that: $\frac{dh}{dt} = -\lambda\sqrt{h}$

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  How the marks are awarded

  • B1 — Correctly translating the problem statement into the differential equation $\frac{dV}{dt} = -k\sqrt{V}$.
  • M1 — Correctly finding $\frac{dV}{dh} = 16\pi$ from $V=16\pi h$ and applying the chain rule to connect $\frac{dh}{dt}$ and $\frac{dV}{dt}$.
  • A1 — Correctly substituting all parts into the chain rule to obtain the intermediate expression $\frac{dh}{dt} = \frac{-k\sqrt{16\pi h}}{16\pi}$.
  • A1 — Correctly simplifying the algebraic expression, defining a new constant $\lambda = \frac{k}{4\sqrt{\pi}}$, and reaching the final required form $\frac{dh}{dt} = -\lambda\sqrt{h}$ with a concluding statement.

  Common mistakes

  • Sign error: Stating that $\frac{dV}{dt} = k\sqrt{V}$, which neglects the fact that the water volume is decreasing.
  • Incorrect volume formula or differentiation: Using an incorrect formula for the volume of a cylinder or making an error when differentiating $V=16\pi h$, such as getting $\frac{dV}{dh} = 16$.
  • Chain rule confusion: Incorrectly assembling the chain rule, for example by multiplying rates that should be divided, such as writing $\frac{dh}{dt} = \frac{dV}{dh} \times \frac{dV}{dt}$.
  • Algebraic simplification error: Making a mistake when simplifying $\frac{\sqrt{16\pi h}}{16\pi}$, for example by incorrectly cancelling terms inside and outside the square root.

  Examiner tip: For related rates problems, always establish the static equation linking the variables (e.g., volume and height) before using the chain rule to connect their rates of change.

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