2015 H2 Mathematics Paper 1 Question 11

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(i)

dydx=2cot⁡θ−tan⁡θ{\frac{\mathrm{d}y}{\mathrm{d}x}= 2 \cot \theta - \tan \theta}

(ii)

tan⁡θ=2,{\tan \theta = \sqrt{2},}
(2323,23){\left( \frac{2}{3}\sqrt{\frac{2}{3}}, \frac{2}{\sqrt{3}} \right)}

(iii)

0.884{0.884}

(iv)

a=32{a=\frac{3}{\sqrt{2}}}

Full solutions

(i)

dxdt=3sin⁡2θcos⁡θdydt=6sin⁡θcos⁡2θ−3sin⁡3θ\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 3 \sin^2 \theta \cos \theta \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 6\sin \theta \cos^2 \theta - 3 \sin^3 \theta \\ \end{align*}
dydx=dydt÷dxdt=6sin⁡θcos⁡2θ−3sin⁡3θ3sin⁡2θcos⁡θ=2cot⁡θ−tan⁡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{6\sin \theta \cos^2 \theta - 3 \sin^3 \theta}{3 \sin^2 \theta \cos \theta} \\ &= 2 \cot \theta - \tan \theta \; \blacksquare \end{align*}

(ii)

At turning point, dydx=0{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=0}
2cot⁡θ−tan⁡θ=0tan⁡θ=2tan⁡θtan⁡2θ=2\begin{align*} 2 \cot \theta - \tan \theta &= 0 \\ \tan \theta &= \frac{2}{\tan \theta} \\ \tan^2 \theta &= 2 \end{align*}
Since 0≤θ≤12π,{0 \leq \theta \leq \frac{1}{2}\pi,}
tan⁡θ=2\tan \theta = \sqrt{2}
sin⁡θ=23cos⁡θ=13x=sin⁡3θ=2323y=sin⁡2θcos⁡θ=233\begin{align*} \sin \theta &= \sqrt{\frac{2}{3}} \\ \cos \theta &= \frac{1}{\sqrt{3}} \\ x &= \sin^3 \theta \\ &= \frac{2}{3} \sqrt{\frac{2}{3}} \\ y &= \sin^2 \theta \cos \theta \\ &= \frac{2}{3\sqrt{3}} \end{align*}
Exact coordinates of turning point:
(2323,23)  ■\left( \frac{2}{3}\sqrt{\frac{2}{3}}, \frac{2}{\sqrt{3}} \right) \; \blacksquare
d2ydx2=ddθ(dydx)÷dxdθ=−2cosec⁡2θ−sec⁡2θ3sin⁡2θcos⁡θ<0 when tan⁡θ=2\begin{align*} \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} &= \frac{\mathrm{d}}{\mathrm{d}\theta} \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) \div \frac{\mathrm{d}x}{\mathrm{d}\theta} \\ &= \frac{-2 \cosec^2 \theta - \sec^2 \theta}{3 \sin^2 \theta \cos \theta} \\ &< 0 \quad \textrm{ when } \tan \theta = \sqrt{2} \\ \end{align*}
Since d2ydx2<0,{\displaystyle \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} < 0,} the turning point is a maximum ■{\blacksquare}

(iii)

Area=∫01y  dx=∫012π3sin⁡2θcos⁡θ(3sin⁡2θcos⁡θ)  dθ=∫012π39sin⁡4θcos⁡2θ  dθ  ■\begin{align*} & \textrm{Area} \\ &= \int_0^1 y \; \mathrm{d}x \\ &= \int_0^{\frac{1}{2}\pi} 3 \sin^2 \theta \cos \theta ( 3 \sin^2 \theta \cos \theta ) \; \mathrm{d} \theta \\ &= \int_0^{\frac{1}{2}\pi} 3 9 \sin^4 \theta \cos^2 \theta \; \mathrm{d} \theta \; \blacksquare \end{align*}
Using a GC,
Area =0.884 (3 dp)  ■\textrm{Area } = 0.884 \textrm{ (3 dp)} \; \blacksquare

(iv)

x=sin⁡3θy=3sin⁡2θcos⁡θy=ax\begin{align} && \quad x &= \sin^3 \theta \\ && \quad y &= 3 \sin^2 \theta \cos \theta \\ && \quad y &= ax \\ \end{align}
Substituting equations (1){(1)} and (2){(2)} into (3),{(3),}
3sin⁡2θcos⁡θ=asin⁡3θasin⁡3θsin⁡2θcos⁡θ=3tan⁡θ=3a  ■\begin{align*} 3 \sin^2 \theta \cos \theta &= a \sin^3 \theta \\ \frac{a \sin^3 \theta}{\sin^2 \theta \cos \theta} &= 3 \\ \tan \theta &= \frac{3}{a} \; \blacksquare \end{align*}
At maximum point, tan⁡θ=2.{\tan \theta = \sqrt{2}. } Hence
3a=2a=32  ■\begin{align*} \frac{3}{a} &= \sqrt{2} \\ a &= \frac{3}{\sqrt{2}} \; \blacksquare \end{align*}