A-Level Mathematics October/November 2024 Q8(a): The parametric equations of a curve are x = tan²2t, y = cos 2t, for 0 < t < π/4. Show t…

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  Model Answer

  Given the parametric equations: $x = \tan^2{2t}$ $y = \cos{2t}$

  First, we differentiate $x$ with respect to $t$ using the chain rule. Let $u = \tan{2t}$, so $x = u^2$. Then $\frac{dx}{du} = 2u = 2\tan{2t}$. Also, $\frac{du}{dt} = (\sec^2{2t}) \times 2 = 2\sec^2{2t}$. Therefore, $\frac{dx}{dt} = \frac{dx}{du} \times \frac{du}{dt} = (2\tan{2t})(2\sec^2{2t}) = 4\tan{2t}\sec^2{2t}$.

  Next, we differentiate $y$ with respect to $t$. $y = \cos{2t}$ $\frac{dy}{dt} = -\sin{2t} \times 2 = -2\sin{2t}$.

  Now, we use the formula for the derivative of parametric equations: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

  Substituting our derivatives: $\frac{dy}{dx} = \frac{-2\sin{2t}}{4\tan{2t}\sec^2{2t}}$

  To simplify, we express $\tan{2t}$ and $\sec^2{2t}$ in terms of $\sin{2t}$ and $\cos{2t}$. $\frac{dy}{dx} = \frac{-2\sin{2t}}{4 \left( \frac{\sin{2t}}{\cos{2t}} \right) \left( \frac{1}{\cos^2{2t}} \right)}$

  $\frac{dy}{dx} = \frac{-2\sin{2t}}{\frac{4\sin{2t}}{\cos^3{2t}}}$

  $\frac{dy}{dx} = -2\sin{2t} \times \frac{\cos^3{2t}}{4\sin{2t}}$

  Cancelling the $2\sin{2t}$ terms: $\frac{dy}{dx} = -\frac{1}{2}\cos^3{2t}$

  This is the required result.

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  How the marks are awarded

  • M1 — This mark is for correctly applying the chain rule to differentiate $x = \tan^2{2t}$. The working must show a result of the form $k \tan{2t} \sec^2{2t}$.
  • A1 — This mark is for obtaining the fully correct derivative $\frac{dx}{dt} = 4\tan{2t}\sec^2{2t}$.
  • B1 — This mark is for correctly differentiating $y = \cos{2t}$ to get $\frac{dy}{dt} = -2\sin{2t}$.
  • B1 — This mark is for correctly substituting the derivatives into the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ and performing the algebraic and trigonometric simplification to show that the expression equals the given answer, $-\frac{1}{2}\cos^3{2t}$.

  Common mistakes

  • Chain rule error on $x = \tan^2{2t}$: Forgetting to multiply by the derivative of $2t$, leading to $\frac{dx}{dt} = 2\tan{2t}\sec^2{2t}$.
  • Sign error on $\frac{dy}{dt}$: Forgetting the negative sign when differentiating $\cos{2t}$, resulting in $\frac{dy}{dt} = 2\sin{2t}$ and an incorrect final sign.
  • Trigonometric simplification error: Making mistakes when simplifying the fraction, for example, incorrectly cancelling terms or writing $\frac{1}{\tan{2t}\sec^2{2t}}$ as $\cot{2t}\cos^2{2t}$ without further correct steps.
  • Incorrect formula: Using $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$ by mistake, which inverts the final fraction.

  Examiner tip: Master the chain rule for differentiating composite functions, especially those involving trigonometric terms, and be fluent in converting between tan/sec/cosec and sin/cos for simplification.

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