2016 H2 Mathematics Paper 2 Question 3

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(i)

(−1,0),{(-1,0), } (2π−1,0){(2\pi-1,0)}
(π+1,2){(\pi + 1,2)}

(ii)

a−sin⁡a−cos⁡a+14cos⁡2a+34{a - \sin a - \cos a}\allowbreak {+ \frac{1}{4} \cos 2a + \frac{3}{4}}

(iii)

14(π+1)2 units2{\frac{1}{4} (\pi+1)^2 \textrm{ units}^2}

Full solutions

(i)

Where D{D} meets the x-{x\textrm{-}}axis,
y=01−cos⁡t=0cos⁡t=1\begin{gather*} y = 0 \\ 1 - \cos t = 0 \\ \cos t = 1 \\ \end{gather*}
t=0ort=2πx=0−cos⁡0x=2π−cos⁡2π=−1=2π−1\begin{align*} t &= 0 &\textrm{or}&& t &= 2 \pi \\ x &= 0-\cos 0 &&& x &= 2 \pi - \cos 2\pi \\ &= -1 &&& &=2\pi - 1 \end{align*}
Hence the coordinates of the points where D{D} meets the x-{x\textrm{-}}axis are:
(−1,0)  ■  (2π−1,0)  ■\begin{align*} & (-1,0) \; \blacksquare \; \\ & (2\pi-1, 0) \; \blacksquare \end{align*}
At maximum point, dydx=0{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=0}
dydt÷dxdt=0sin⁡t1+sin⁡t=0sin⁡t=0\begin{gather*} \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} = 0 \\ \frac{\sin t}{1+\sin t} = 0 \\ \sin t = 0 \end{gather*}
Since t≠0{t\neq 0} and t≠2π{t \neq 2\pi} at the maximum point,
t=πx=π−cos⁡π=π+1y=1−cos⁡π=2\begin{align*} t &= \pi \\ x &= \pi - \cos \pi \\ &= \pi + 1 \\ y &= 1 - \cos \pi \\ &= 2 \end{align*}
Coordinates of maximum point on the curve:
(π+1,2)  ■(\pi + 1, 2) \; \blacksquare

(ii)

Area under D=∫x1x2y  dx=∫0a(1−cos⁡t)(1+sin⁡t)  dt=∫0a(1−cos⁡t+sin⁡t−cos⁡tsin⁡t)  dt=∫0a(1−cos⁡t+sin⁡t−12sin⁡2t)  dt=[t−sin⁡t−cos⁡t+14cos⁡2t]0a=a−sin⁡a−cos⁡a+14cos⁡2a−(0−0−1+14)=(a−sin⁡a−cos⁡a+14cos⁡2a+34) units2  ■\begin{align*} & \textrm{Area under } D \\ & = \int_{x_1}^{x_2} y \; \mathrm{d}x \\ & = \int_0^a (1- \cos t) (1+\sin t) \; \mathrm{d}t \\ & = \int_0^a (1- \cos t + \sin t - \cos t \sin t) \; \mathrm{d}t \\ & = \int_0^a \left(1- \cos t + \sin t - \frac{1}{2} \sin 2t \right) \; \mathrm{d}t \\ & = \left[ t- \sin t - \cos t + \frac{1}{4} \cos 2t \right]_0^a \\ & = a - \sin a - \cos a + \frac{1}{4} \cos 2a - \left( 0 - 0 - 1 + \frac{1}{4} \right) \\ & = \left(a - \sin a - \cos a + \frac{1}{4} \cos 2a + \frac{3}{4} \right) \textrm{ units}^2 \; \blacksquare \end{align*}

(iii)

When t=12π,{t = \frac{1}{2}\pi, }
x=12πy=1dydx=sin⁡12π1+12π=12\begin{align*} x &= \frac{1}{2}\pi \\ y &= 1 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\sin \frac{1}{2}\pi}{1 + \frac{1}{2}\pi} \\ &= \frac{1}{2} \\ \end{align*}
Hence the gradient of normal is −2{-2}
Equation of normal:
y−1=−2(x−12π)\begin{gather*} y - 1 = -2 \left( x - \frac{1}{2}\pi \right) \end{gather*}
At E,y=0{E, y=0}
0−1=−2x+πx=π+12\begin{gather*} 0 - 1 = -2x + \pi \\ x = \frac{\pi + 1}{2} \end{gather*}
At F,x=0{F, x=0}
y−1=πy=π+1\begin{gather*} y - 1 = \pi \\ y = \pi + 1 \end{gather*}
Area of triangle OEF=12(π+1)(π+12)=14(π+1)2 units2  ■\begin{align*} & \textrm{Area of triangle } OEF \\ & = \frac{1}{2} (\pi+1) \left( \frac{\pi+1}{2} \right) \\ &= \frac{1}{4} (\pi+1)^2 \textrm{ units}^2 \; \blacksquare \end{align*}