2018 H2 Mathematics Paper 1 Question 9

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(ii)

x=2α{x=2\alpha}

(iii)

8 units{8 \textrm{ units}}

Full solutions

(i)

dxdt=2−2cos⁡2θdydt=4sin⁡θcos⁡θ\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 2 - 2 \cos 2 \theta \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 4 \sin \theta \cos \theta \\ \end{align*}
dydx=dydt÷dxdt=4sin⁡θcos⁡θ2−2cos⁡2θ=4sin⁡θcos⁡θ2−2(1−2sin⁡2θ)=4sin⁡θcos⁡θ4sin⁡2θ=cot⁡θ  ■\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \\ &= \frac{4 \sin \theta \cos \theta}{2 - 2 \cos 2 \theta} \\ &= \frac{4 \sin \theta \cos \theta}{2 - 2 ( 1 - 2 \sin^2 \theta )} \\ &= \frac{4 \sin \theta \cos \theta}{4 \sin^2 \theta} \\ &=\cot \theta \; \blacksquare \end{align*}

(ii)

At point where θ=α,{\theta = \alpha, }
Gradient of normal=−tan⁡α\begin{align*} & \textrm{Gradient of normal} \\ &= - \tan \alpha \end{align*}
Equation of normal:
y−2sin⁡2α=−tan⁡α(x−(2α−sin⁡2α))\begin{gather*} y - 2 \sin^2 \alpha = - \tan \alpha \left( x - (2\alpha - \sin 2 \alpha) \right) \\ \end{gather*}
At point A,y=0,{A, y=0,}
0−2sin⁡2α=−tan⁡α(x−2α+2sin⁡αcos⁡α)x−2α+2sin⁡αcos⁡α=2sin⁡αcos⁡α\begin{gather*} 0 - 2 \sin^2 \alpha = -\tan \alpha \left( x - 2\alpha + 2 \sin \alpha \cos \alpha \right) \\ x - 2\alpha + 2 \sin \alpha \cos \alpha = 2 \sin \alpha \cos \alpha \\ \end{gather*}
x=2α  ■x = 2\alpha \; \blacksquare

(iii)

Total length of C=∫0π(2−2cos⁡2θ)2+(4sin⁡θcos⁡θ)2  dθ=∫0π(2−2(1−2sin⁡2θ))2+16sin⁡2θcos⁡2θ  dθ=∫0π16sin⁡4θ+16sin⁡2θcos⁡2θ  dθ=∫0π16sin⁡2θ(sin⁡2θ+cos⁡2θ)  dθ=∫0π4sin⁡θ  dθ=[−4cos⁡θ]0π=−4cos⁡π+4cos⁡0=8 units  ■\begin{align*} & \textrm{Total length of } C \\ & = \int_0^\pi \sqrt{ (2-2\cos 2\theta)^2 + (4\sin \theta \cos \theta)^2 } \; \mathrm{d} \theta \\ & = \int_0^\pi \sqrt{ \Big(2 - 2 (1 - 2 \sin^2 \theta) \Big)^2 + 16 \sin^2 \theta \cos^2 \theta } \; \mathrm{d} \theta \\ & = \int_0^\pi \sqrt{ 16 \sin^4 \theta + 16 \sin^2 \theta \cos^2 \theta } \; \mathrm{d} \theta \\ & = \int_0^\pi \sqrt{ 16 \sin^2 \theta (\sin^2 \theta + \cos^2 \theta) } \; \mathrm{d} \theta \\ & = \int_0^\pi 4 \sin \theta \; \mathrm{d} \theta \\ & = \left[ -4 \cos \theta \right]_0^\pi \\ & = -4 \cos \pi + 4 \cos 0 \\ & = 8 \textrm{ units} \; \blacksquare \\ \end{align*}