2021 H2 Mathematics Paper 1 Question 11

Definite Integrals: Areas and Volumes

Answers

(a)

PS=2α(π−β+sin⁡β){PS = 2\alpha( \pi - \beta + \sin \beta )}

(c)

β=1.90{\beta = 1.90}

(d)

Greatest height of arch =22.9 m{=22.9 \textrm{ m}}
Least height of arch =15.1 m{=15.1 \textrm{ m}}

Full solutions

(a)

At point Q,θ=β,{Q, \theta = \beta,}
x=α(β−sin⁡β)x = \alpha(\beta - \sin \beta)
By the symmetry of the curve,
PS=2πα−2α(β−sin⁡β)=2α(π−β+sin⁡β)  ■\begin{align*} PS &= 2\pi\alpha - 2\alpha(\beta - \sin \beta) \\ &= 2\alpha( \pi - \beta + \sin \beta ) \; \blacksquare \end{align*}

(b)

dxdθ=α(1−cos⁡θ)\frac{\mathrm{d}x}{\mathrm{d}\theta} = \alpha(1 - \cos \theta)
Area of shaded region=∫xPxSy  dx=∫β2π−βα(1−cos⁡θ)α(1−cos⁡θ)  dθ=α2∫β2π−β(1−2cos⁡θ+cos⁡2θ)  dθ=α2∫β2π−β(1−2cos⁡θ+cos⁡2θ+12)  dθ=α2∫β2π−β(32−2cos⁡θ+cos⁡2θ2)  dθ=α2[32θ−2sin⁡θ+sin⁡2θ4]β2π−β=α2(32(2π−β−β)−2sin⁡(2π−β)=(32++2sin⁡β+sin⁡(4π−2β)−sin⁡2β4)=α2(3π−3β+4sin⁡β−sin⁡2β2)=12α2(6π−6β+8sin⁡β−sin⁡2β)  ■\begin{align*} & \textrm{Area of shaded region} \\ & = \int_{x_P}^{x_S} y \; \mathrm{d}x \\ & = \int_{\beta}^{2\pi - \beta} \alpha (1- \cos \theta) \alpha (1- \cos \theta) \; \mathrm{d}\theta \\ & = \alpha^2 \int_{\beta}^{2\pi - \beta} (1 - 2\cos \theta + \cos^2 \theta) \; \mathrm{d}\theta \\ & = \alpha^2 \int_{\beta}^{2\pi - \beta} \left(1 - 2\cos \theta + \frac{\cos 2\theta + 1}{2} \right) \; \mathrm{d}\theta \\ & = \alpha^2 \int_{\beta}^{2\pi - \beta} \left( \frac{3}{2} - 2\cos \theta + \frac{\cos 2\theta}{2} \right) \; \mathrm{d}\theta \\ & = \alpha^2 \left[ \frac{3}{2}\theta - 2\sin \theta + \frac{\sin 2\theta}{4} \right]_{\beta}^{2\pi - \beta} \\ & = \alpha^2 \Bigg( \frac{3}{2} (2\pi - \beta - \beta) - 2 \sin (2\pi - \beta) \\ & \phantom{= \Bigg( \frac{3}{2} +} + 2 \sin \beta + \frac{\sin (4\pi-2\beta)-\sin 2\beta}{4} \Bigg) \\ &= \alpha^2 \left( 3 \pi - 3 \beta + 4 \sin \beta - \frac{\sin 2\beta}{2} \right) \\ &= \frac{1}{2}\alpha^2 \left( 6 \pi - 6 \beta + 8 \sin \beta - \sin 2\beta \right) \; \blacksquare \\ \end{align*}

(c)

12α2(6π−6β+8sin⁡β−sin⁡2β)=7.8159α26π−6β+8sin⁡β−sin⁡β−15.6318=0\begin{gather*} \frac{1}{2}\alpha^2 \left( 6 \pi - 6 \beta + 8 \sin \beta - \sin 2\beta \right) = 7.8159\alpha^2 \\ 6 \pi - 6 \beta + 8 \sin \beta - \sin \beta - 15.6318 = 0 \\ \end{gather*}
Using a GC,
β=1.90 (3 sf)  ■\beta = 1.90 \textrm{ (3 sf)} \; \blacksquare

(d)

2α(π−β+sin⁡β)=502\alpha( \pi - \beta + \sin \beta ) = 50
α=25β−sin⁡β=11.427\begin{align*} \alpha &= \frac{25}{\beta - \sin \beta} \\ &= 11.427 \end{align*}
Greatest height of arch (at θ=π)=α(1−cos⁡π)=22.9 (3 sf) m  ■\begin{align*} & \textrm{Greatest height of arch (at } \theta = \pi) \\ & = \alpha (1-\cos \pi) \\ & = 22.9 \textrm{ (3 sf)} \textrm{ m} \; \blacksquare \end{align*}
Least height of arch (at θ=β)=α(1−cos⁡β)=15.1 (3 sf) m  ■\begin{align*} & \textrm{Least height of arch (at } \theta = \beta) \\ & = \alpha (1-\cos \beta) \\ & = 15.1 \textrm{ (3 sf)} \textrm{ m} \; \blacksquare \end{align*}