2016 H2 Mathematics Paper 1 Question 2

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(i)

dydx∣x=0=0{\frac{\mathrm{d}y}{\mathrm{d}x} \Big\vert_{x=0} = 0}
dydx∣x=π2=−0.693{\frac{\mathrm{d}y}{\mathrm{d}x} \Big\vert_{x=\frac{\pi}{2}} = -0.693}

(ii)

y=2{y=2}
y=−0.693x+2.09{y = -0.693x + 2.09 }
(0.128,2){\left( 0.128, 2 \right)}

Full solutions

(i)

When x=0,{x=0, } gradient of y=2cos⁡x{y=2^{\cos x}} is
dydx∣x=0=0  ■\frac{\mathrm{d}y}{\mathrm{d}x} \Big\vert_{x=0} = 0 \; \blacksquare
When x=12π,{x=\frac{1}{2}\pi, } gradient of y=2cos⁡x{y=2^{\cos x}} is
dydx∣x=π2=−0.69315=−0.693 (3 sf)  ■\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} \Big\vert_{x=\frac{\pi}{2}} &= -0.69315\\ &= -0.693 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(ii)

Tangent at x=0:{x=0:}
y=2cos⁡0y=2  ■\begin{align*} y &= 2^{\cos 0} \\ y &= 2 \; \blacksquare \end{align*}
Tangent at x=12π:{x=\frac{1}{2}\pi:}
y−2cos⁡12π=−0.69315(x−π2)y−1=−0.69315(x−π2)y=−0.69315x+2.0888\begin{align*} y - 2^{\cos\frac{1}{2}\pi} &= -0.69315 \left( x - \frac{\pi}{2} \right) \\ y - 1 &= -0.69315 \left( x - \frac{\pi}{2} \right) \\ y &= -0.69315x + 2.0888 \end{align*}
y=−0.693x+2.09  ■y = -0.693x + 2.09 \; \blacksquare
Solving the two equations simultaneously,
2=−0.69315x+2.0888x=2−2.0888−0.69315x=0.12810\begin{gather*} 2 = -0.69315x + 2.0888 \\ x = \frac{2-2.0888}{-0.69315} \\ x = 0.12810 \end{gather*}
Coordinates of the point where these tangents meet:
(0.128,2)  ■\left( 0.128, 2 \right) \; \blacksquare