A-Level Mathematics May/June 2024 Q4: A curve is defined by the parametric equations x = 4 cos²t, y = √3 sin 2t, for values o…

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  1. Find the derivatives with respect to t.

  The parametric equations are given by: $x = 4 \cos^2t$ $y = \sqrt{3} \sin 2t$

  First, differentiate x with respect to t using the chain rule: $\frac{dx}{dt} = 4 \times 2(\cos t)^1 \times (-\sin t) = -8 \cos t \sin t$ Using the double angle identity $\sin 2t = 2 \sin t \cos t$, this can be simplified: $\frac{dx}{dt} = -4(2 \sin t \cos t) = -4 \sin 2t$

  Next, differentiate y with respect to t: $\frac{dy}{dt} = \sqrt{3} \times (\cos 2t) \times 2 = 2\sqrt{3} \cos 2t$

  2. Calculate the gradient of the tangent.

  The gradient of the curve, $\frac{dy}{dx}$, is found using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\sqrt{3} \cos 2t}{-4 \sin 2t} = -\frac{\sqrt{3}}{2} \cot 2t$

  Now, substitute $t = \frac{\pi}{6}$ to find the gradient at the specific point: $\frac{dy}{dx} = -\frac{\sqrt{3}}{2} \cot(2 \times \frac{\pi}{6}) = -\frac{\sqrt{3}}{2} \cot(\frac{\pi}{3})$ Since $\cot(\frac{\pi}{3}) = \frac{1}{\tan(\pi/3)} = \frac{1}{\sqrt{3}}$, the gradient of the tangent is: $\frac{dy}{dx} = -\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} = -\frac{1}{2}$

  3. Find the gradient of the normal.

  The gradient of the normal is the negative reciprocal of the gradient of the tangent. Gradient of normal, $m_{normal} = -\frac{1}{-1/2} = 2$

  4. Find the coordinates of the point.

  Substitute $t = \frac{\pi}{6}$ into the original parametric equations: $x = 4 \cos^2(\frac{\pi}{6}) = 4 \left( \frac{\sqrt{3}}{2} \right)^2 = 4 \left( \frac{3}{4} \right) = 3$ $y = \sqrt{3} \sin(2 \times \frac{\pi}{6}) = \sqrt{3} \sin(\frac{\pi}{3}) = \sqrt{3} \left( \frac{\sqrt{3}}{2} \right) = \frac{3}{2}$ So the point on the curve is $(3, \frac{3}{2})$.

  5. Determine the equation of the normal.

  Using the point-gradient form of a line, $y - y_1 = m(x - x_1)$: $y - \frac{3}{2} = 2(x - 3)$

  To express this in the form $ax+by+c=0$ with integer coefficients, multiply the entire equation by 2: $2(y - \frac{3}{2}) = 2(2(x - 3))$ $2y - 3 = 4(x - 3)$ $2y - 3 = 4x - 12$

  Rearranging gives the final equation: $4x - 2y - 9 = 0$

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  How the marks are awarded

  • *M1 — Correctly applying the chain rule to differentiate either x or y with respect to t, obtaining a form like k₁cos(t)sin(t) or k₂cos(2t). This is shown by finding dx/dt = -8cos(t)sin(t) or dy/dt = 2√3cos(2t).
  • A1 — Obtaining both correct derivatives: dx/dt = -8cos(t)sin(t) (or -4sin(2t)) AND dy/dt = 2√3cos(2t).
  • *DM1 — Attempting to find the value of dy/dx by substituting t = π/6 into their expression for (dy/dt)/(dx/dt). This mark is dependent on the first M1.
  • A1 — Obtaining the correct gradient of the tangent, dy/dx = -1/2.
  • **M1FT — Correctly finding the gradient of the normal by taking the negative reciprocal of their calculated dy/dx. For the correct tangent gradient of -1/2, this means stating or using a normal gradient of 2. (FT indicates follow-through from their dy/dx).
  • **DM1 — Attempting to find the equation of the normal. This requires finding the (x, y) coordinates at t = π/6 and using them with their normal gradient in the form y - y₁ = m(x - x₁).
  • A1 — Obtaining the final correct equation $4x - 2y - 9 = 0$, or an equivalent form with integer coefficients (e.g., -$4x + 2y + 9 = 0$).

  Common mistakes

  • Differentiation errors, particularly with the chain rule for cos²(t), often forgetting the negative sign from differentiating cos(t), or forgetting the factor of 2 from differentiating sin(2t).
  • Confusing the gradient of the tangent with the gradient of the normal, and using dy/dx directly in the final line equation instead of its negative reciprocal.
  • Arithmetic errors when evaluating trigonometric functions of π/6 and π/3, or when rearranging the final equation to the required form ax+by+c=0.
  • Failing to convert the final equation to one with integer coefficients, for example leaving it as $2x - y - 4.5 = 0$.

  Examiner tip: For parametric curves, consistently apply the two-step process: first find the tangent's gradient using dy/dx = (dy/dt)/(dx/dt), then find the normal's gradient using the negative reciprocal.

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