A-Level Mathematics October/November 2024 Q7(a): The parametric equations of a curve are x = 3 sin 2t, y = tan t + cot t, for 0 < t < 1/…

1 answer

• ▲0
  accepted ✓

  Given the parametric equations: $x = 3 \sin 2t$ $y = \tan t + \cot t$

  First, we differentiate both equations with respect to $t$.

  For $x$: $\frac{dx}{dt} = 3 (\cos 2t) \times 2$ $\frac{dx}{dt} = 6 \cos 2t$

  For $y$: $\frac{dy}{dt} = \sec^2 t - \csc^2 t$

  Next, we use the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ $\frac{dy}{dx} = \frac{\sec^2 t - \csc^2 t}{6 \cos 2t}$

  To simplify the numerator, we express it in terms of $\sin t$ and $\cos t$: $\sec^2 t - \csc^2 t = \frac{1}{\cos^2 t} - \frac{1}{\sin^2 t}$ $= \frac{\sin^2 t - \cos^2 t}{\sin^2 t \cos^2 t}$

  Using the double angle identities $\cos 2t = \cos^2 t - \sin^2 t$ and $\sin 2t = 2 \sin t \cos t$:

  The numerator becomes: $\sin^2 t - \cos^2 t = -(\cos^2 t - \sin^2 t) = -\cos 2t$

  The denominator becomes: $\sin^2 t \cos^2 t = (\sin t \cos t)^2 = \left( \frac{1}{2} \sin 2t \right)^2 = \frac{1}{4} \sin^2 2t$

  So, the expression for $\frac{dy}{dt}$ simplifies to: $\frac{dy}{dt} = \frac{-\cos 2t}{\frac{1}{4} \sin^2 2t} = \frac{-4 \cos 2t}{\sin^2 2t}$

  Now, substitute this back into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{-4 \cos 2t}{\sin^2 2t}}{6 \cos 2t}$ $\frac{dy}{dx} = \frac{-4 \cos 2t}{6 \cos 2t \sin^2 2t}$

  Cancel the $\cos 2t$ terms and simplify the fraction: $\frac{dy}{dx} = \frac{-4}{6 \sin^2 2t} = -\frac{2}{3 \sin^2 2t}$

  Thus, we have shown that $\frac{dy}{dx} = -\frac{2}{3 \sin^2 2t}$.

  ---

  How the marks are awarded

  • B1 — Correctly differentiating x = 3 sin 2t to obtain dx/dt = 6 cos 2t.
  • B1 — Correctly differentiating y = tan t + cot t to obtain dy/dt = sec²t - cosec²t.
  • M1 — Applying the chain rule for parametric equations, by writing dy/dx as the quotient of dy/dt and dx/dt.
  • DM1 — Simplifying the expression for dy/dt by converting to sines and cosines and combining into a single fraction: (sin²t - cos²t) / (sin²t cos²t).
  • A1 — Correctly applying double angle identities to the simplified fraction and completing the proof with full and correct working to show the final given answer.

  Common mistakes

  • A sign error when differentiating cot t, leading to dy/dt = sec²t + cosec²t.
  • Forgetting the chain rule when differentiating x = 3 sin 2t, resulting in dx/dt = 3 cos 2t instead of 6 cos 2t.
  • Incorrectly simplifying the numerator of dy/dt, for example stating sin²t - cos²t = cos 2t instead of -cos 2t.
  • Errors with double angle identities in the denominator, such as writing sin²t cos²t = sin²2t instead of (1/4)sin²2t.

  Examiner tip: Mastering trigonometric identities, especially double angle formulae, is crucial for simplifying complex derivatives in parametric and implicit differentiation.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.