2012 H2 Mathematics Paper 1 Question 11

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(i)

Gradient of C{C} at the point where θ=π{\theta=\pi} is 0{0}
As θ→0{\theta \to 0} and θ→2π,{\theta \to 2\pi, } the tangents to C{C} are parallel to the y-{y\textrm{-}}axis

(ii)

(iii)

3π units2{3 \pi \textrm{ units}^2}

Full solutions

(i)

dxdt=1−cos⁡θdydt=sin⁡θ\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 1- \cos \theta \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= \sin \theta \\ \end{align*}
dydx=dydt÷dxdt=sin⁡θ1−cos⁡θ=2sin⁡12θcos⁡12θ1−(1−2sin⁡212θ)=2sin⁡12θcos⁡12θ2sin⁡212θ=cos⁡12θsin⁡12θ=cot⁡12θ  ■\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{\sin \theta}{1- \cos \theta} \\ &= \frac{2\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta}{1-\left( 1 - 2\sin^2\frac{1}{2}\theta \right)} \\ &= \frac{2\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta}{2\sin^2\frac{1}{2}\theta} \\ &= \frac{\cos\frac{1}{2}\theta}{\sin\frac{1}{2}\theta} \\ &=\cot \frac{1}{2} \theta \; \blacksquare \end{align*}
When θ=π,{\theta = \pi, } cos⁡π2=0{\cos \frac{\pi}{2} = 0} and sin⁡π2=1{\sin \frac{\pi}{2} = 1} so the gradient of C{C} at the point where θ=π{\theta=\pi} is 0  ■{0 \; \blacksquare}
As θ→0{\theta \to 0} and θ→2π,{\theta \to 2\pi, } sin⁡12θ→0{\sin \frac{1}{2} \theta \to 0} so dydx→∞{\frac{\mathrm{d}y}{\mathrm{d}x} \to \infty}
Hence the tangents to C{C} as θ→0{\theta \to 0} and θ→2π{\theta \to 2\pi} are parallel to the y-{y\textrm{-}}axis ■{\blacksquare}

(ii)

(iii)

Area=∫02πy  dx=∫02π(1−cos⁡θ)(1−cos⁡θ)  dθ=∫02π1−2cos⁡θ+cos⁡2θ  dθ=∫02π1−2cos⁡θ+cos⁡2θ+12  dθ=∫02π32−2cos⁡θ+cos⁡2θ2  dθ=[32θ−2sin⁡θ+sin⁡2θ4]02π=32(2π)−0+0−0+0−0=3π units2  ■\begin{align*} & \textrm{Area} \\ & = \int_0^{2\pi} y \; \mathrm{d}x \\ & = \int_0^{2\pi} (1 - \cos \theta) (1 - \cos \theta) \; \mathrm{d}\theta \\ & = \int_0^{2\pi} 1 - 2 \cos \theta + \cos^2 \theta \; \mathrm{d}\theta \\ & = \int_0^{2\pi} 1 - 2 \cos \theta + \frac{\cos 2 \theta + 1}{2} \; \mathrm{d}\theta \\ & = \int_0^{2\pi} \frac{3}{2} - 2 \cos \theta + \frac{\cos 2 \theta}{2} \; \mathrm{d}\theta \\ & = \left[ \frac{3}{2}\theta - 2 \sin \theta + \frac{\sin 2 \theta}{4} \right]_0^{2\pi} \\ & = \frac{3}{2} (2 \pi) - 0 + 0 -0 + 0 - 0 \\ & = 3 \pi \textrm{ units}^2 \; \blacksquare \end{align*}

(iv)

At point P{P} with parameter p{p}
dydx=cot⁡12p\frac{\mathrm{d}y}{\mathrm{d}x} = \cot \frac{1}{2}p
Hence the gradient of normal is −tan⁡12p{-\tan \frac{1}{2}p}
Normal at P:{P:}
y−(1−cos⁡p)=−tan⁡12p(x−(p−sin⁡p))y−(2sin⁡212p)=−tan⁡12p(x−p+sin⁡p)\begin{gather*} y - \left( 1 - \cos p \right) = -\tan {\textstyle \frac{1}{2}p} \Big( x - \left( p - \sin p \right) \Big) \\ y - \left( 2\sin^2 {\textstyle \frac{1}{2}p} \right) = -\tan {\textstyle \frac{1}{2}p} ( x - p + \sin p ) \\ \end{gather*}
When the normal to C{C} at P{P} crosses the x-{x\textrm{-}}axis
0−(2sin⁡212p)=−tan⁡12p(x−p+sin⁡p)x−p+sin⁡p=−2sin⁡212p−tan⁡12px−p+2sin⁡12pcos⁡12p=2sin⁡12pcos⁡12p\begin{gather*} 0 - \left( 2\sin^2 {\textstyle \frac{1}{2}p} \right) = -\tan {\textstyle \frac{1}{2}p} ( x - p + \sin p ) \\ x - p + \sin p = \frac{-2\sin^2 {\textstyle \frac{1}{2}p}}{-\tan {\textstyle \frac{1}{2}p}} \\ x - p + 2 \sin {\textstyle \frac{1}{2}p} \cos {\textstyle \frac{1}{2}p} = 2 \sin {\textstyle \frac{1}{2}p} \cos {\textstyle \frac{1}{2}p} \\ \end{gather*}
x=px = p
Hence the normal to C{C} at P{P} crosses the x-{x\textrm{-}}axis at the point with coordinates (p,0)  ■{(p,0) \; \blacksquare}