Specimen 2017 H2 Mathematics Paper 1 Question 9

Maclaurin Series

Answers

(c)

α≈0.000610,{\alpha \approx 0.000610,}
h≈3.87.{h \approx 3.87.}

Full solutions

(a)

By Pythagoras Theorem,
x=(R+h)2−R2=2Rh+h2=2Rh(1+h22Rh)=(2hR)12(1+h2R)12  ■\begin{align*} x &= \sqrt{(R+h)^2 - R^2} \\ &= \sqrt{2Rh + h^2} \\ &= \sqrt{2Rh \left( 1 + \frac{h^2}{2Rh} \right) } \\ &= (2hR)^{\frac{1}{2}} \left( 1 + \frac{h}{2R} \right)^{\frac{1}{2}} \; \blacksquare \end{align*}

(b)

sin⁡θ=xR+h=(2hR)12(1+h2R)12R+h=(2hR)12(1+α2)12(R(1+hR))−1=(2hR)12R(1+α2)12(1+α)−1=(2h)12R12(1+14α+…)(1−α+…)=(2α)12(1+14α−α+…)≈(2α)12(1−34α)  ■\begin{align*} \sin \theta &= \frac{x}{R+h} \\ &= \frac{(2hR)^{\frac{1}{2}} \left( 1 + \frac{h}{2R} \right)^{\frac{1}{2}}}{R+h} \\ &= (2hR)^{\frac{1}{2}} \left( 1 + \frac{\alpha}{2} \right)^{\frac{1}{2}} \left(R\left(1+\frac{h}{R}\right)\right)^{-1} \\ &= \frac{(2hR)^{\frac{1}{2}}}{R} \left( 1 + \frac{\alpha}{2} \right)^{\frac{1}{2}} \left(1 + \alpha \right)^{-1} \\ &= \frac{(2h)^{\frac{1}{2}}}{R^{\frac{1}{2}}} \left( 1 + \frac{1}{4} \alpha +\ldots \right) \left( 1 - \alpha +\ldots \right) \\ &= (2\alpha)^{\frac{1}{2}} \left( 1 + \frac{1}{4} \alpha -\alpha + \ldots \right)\\ &\approx (2\alpha)^{\frac{1}{2}} \left( 1 - \frac{3}{4} \alpha \right)\; \blacksquare \end{align*}

(c)

Solving
(2α)12(1−34α)−sin⁡2∘=0(2\alpha)^{\frac{1}{2}} \left( 1 - \frac{3}{4} \alpha \right) - \sin 2^\circ = 0
using a GC,
α≈0.000610  ■ \alpha \approx 0.000610 \; \blacksquare
h=αR≈0.00060954×6357≈3.87  ■\begin{align*} h &= \alpha R \\ &\approx 0.00060954 \times 6357 \\ &\approx 3.87 \; \blacksquare \end{align*}