· H2 Mathematics — Ten-Year Series, worked ← All 11 questions

2021 A-Level · Paper 1 · Question 11

2021 H2 Maths Paper 1, Question 11

Parametric equations 13 marks area · greatest and least values
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Part (a)

2 marks

Since PQPQ is parallel to the yy-axis, PP has the same xx-coordinate as QQ, i.e. at θ=β\theta = \beta:

xP=a(β−sin⁡β)x_{P} = a\left( \beta - \sin\beta \right)

Similarly SS has the same xx-coordinate as RR, i.e. at θ=2π−β\theta = 2\pi - \beta:

xS=a[(2π−β)−sin⁡(2π−β)]x_{S} = a\left\lbrack (2\pi - \beta) - \sin(2\pi - \beta) \right\rbrack

Since sin⁡(2π−β)=−sin⁡β\sin(2\pi - \beta) = - \sin\beta,

=a(2π−β+sin⁡β)= a\left( 2\pi - \beta + \sin\beta \right)

Then

PS=xS−xPPS = x_{S} - x_{P}
=a(2π−β+sin⁡β)−a(β−sin⁡β)= a\left( 2\pi - \beta + \sin\beta \right) - a\left( \beta - \sin\beta \right)
=2a(π−β+sin⁡β)= 2a\left( \pi - \beta + \sin\beta \right)

∴\therefore PS=2a(π−β+sin⁡β)PS = 2a\left( \pi - \beta + \sin\beta \right).

Part (b)

6 marks

Approach

The shaded area is ∫y dx\int y\, dx between PP and SS. Converting to the parameter θ\theta via dx=dxdθ dθdx = \frac{dx}{d\theta}\, d\theta turns this into ∫β2π−βy dxdθ dθ\int_{\beta}^{2\pi - \beta}y\,\frac{dx}{d\theta}\, d\theta, which reduces to ∫(1−cos⁡θ)2 dθ\int\left( 1 - \cos\theta \right)^{2}\, d\theta — expand this using cos⁡2θ=12(1+cos⁡2θ)\cos^{2}\theta = \frac{1}{2}\left( 1 + \cos 2\theta \right) before integrating, since (1−cos⁡θ)2\left( 1 - \cos\theta \right)^{2} itself has no elementary antiderivative.

Since dxdθ=a(1−cos⁡θ)\frac{dx}{d\theta} = a\left( 1 - \cos\theta \right) and y=a(1−cos⁡θ)y = a\left( 1 - \cos\theta \right),

Area=∫β2π−βy dxdθ dθ\text{Area} = \int_{\beta}^{2\pi - \beta}y\,\frac{dx}{d\theta}\, d\theta
=∫β2π−βa2(1−cos⁡θ)2 dθ= \int_{\beta}^{2\pi - \beta}a^{2}\left( 1 - \cos\theta \right)^{2}\, d\theta

Expanding, and using cos⁡2θ=12(1+cos⁡2θ)\cos^{2}\theta = \frac{1}{2}\left( 1 + \cos 2\theta \right),

(1−cos⁡θ)2=1−2cos⁡θ+cos⁡2θ\left( 1 - \cos\theta \right)^{2} = 1 - 2\cos\theta + \cos^{2}\theta
=32−2cos⁡θ+12cos⁡2θ= \frac{3}{2} - 2\cos\theta + \frac{1}{2}\cos 2\theta

So

Area=a2∫β2π−β(32−2cos⁡θ+12cos⁡2θ)dθ\text{Area} = a^{2}\int_{\beta}^{2\pi - \beta}\left( \frac{3}{2} - 2\cos\theta + \frac{1}{2}\cos 2\theta \right)d\theta
=a2[32θ−2sin⁡θ+14sin⁡2θ]β2π−β= a^{2}\left\lbrack \frac{3}{2}\theta - 2\sin\theta + \frac{1}{4}\sin 2\theta \right\rbrack_{\beta}^{2\pi - \beta}

Using sin⁡(2π−β)=−sin⁡β\sin(2\pi - \beta) = - \sin\beta and sin⁡(4π−2β)=−sin⁡2β\sin(4\pi - 2\beta) = - \sin 2\beta: at θ=2π−β\theta = 2\pi - \beta the bracket is 3π−32β+2sin⁡β−14sin⁡2β3\pi - \frac{3}{2}\beta + 2\sin\beta - \frac{1}{4}\sin 2\beta, and at θ=β\theta = \beta it is 32β−2sin⁡β+14sin⁡2β\frac{3}{2}\beta - 2\sin\beta + \frac{1}{4}\sin 2\beta. Subtracting,

Area=a2[(3π−32β+2sin⁡β−14sin⁡2β)−(32β−2sin⁡β+14sin⁡2β)]\text{Area} = a^{2}\left\lbrack \left( 3\pi - \frac{3}{2}\beta + 2\sin\beta - \frac{1}{4}\sin 2\beta \right) - \left( \frac{3}{2}\beta - 2\sin\beta + \frac{1}{4}\sin 2\beta \right) \right\rbrack
=a2(3π−3β+4sin⁡β−12sin⁡2β)= a^{2}\left( 3\pi - 3\beta + 4\sin\beta - \frac{1}{2}\sin 2\beta \right)
=12a2(6π−6β+8sin⁡β−sin⁡2β)= \frac{1}{2}a^{2}\left( 6\pi - 6\beta + 8\sin\beta - \sin 2\beta \right)
To check your answer: at β=0\beta = 0 (degenerate case, QQ and RR both at the base points of the cycloid), Area =12a2(6π)=3πa2\text{=}\frac{1}{2}a^{2}(6\pi) = 3\pi a^{2}, matching the known area under one full cycloid arch ✓

∴\therefore Area =12a2(6π−6β+8sin⁡β−sin⁡2β)\text{=}\frac{1}{2}a^{2}\left( 6\pi - 6\beta + 8\sin\beta - \sin 2\beta \right) (shown).

Part (c)

1 mark
β≈1.90\beta \approx 1.90

Explanation: dividing the area equation 12a2(6π−6β+8sin⁡β−sin⁡2β)=7.8159a2\frac{1}{2}a^{2}\left( 6\pi - 6\beta + 8\sin\beta - \sin 2\beta \right) = 7.8159a^{2} through by a2a^{2} leaves a transcendental equation in β\beta alone,

3π−3β+4sin⁡β−12sin⁡2β=7.81593\pi - 3\beta + 4\sin\beta - \frac{1}{2}\sin 2\beta = 7.8159

which has no algebraic solution. By GC, β=1.90\beta = 1.90 (3 s.f.).

∴\therefore β=1.90\beta = 1.90.

Part (d)

4 marks

Approach

With β\beta now known, part (a) gives aa from PS=50PS = 50. The greatest and least heights of the arch above the expressway are then the greatest and least values of y=a(1−cos⁡θ)y = a\left( 1 - \cos\theta \right) for θ\theta between β\beta and 2π−β2\pi - \beta.

From part (a), 50=2a(π−β+sin⁡β)50 = 2a\left( \pi - \beta + \sin\beta \right). Substituting β=1.90\beta = 1.90,

a=25π−1.90+sin⁡1.90≈11.4a = \frac{25}{\pi - 1.90 + \sin 1.90} \approx 11.4

Since dydθ=asin⁡θ\frac{dy}{d\theta} = a\sin\theta, yy is increasing for 0<θ<π0 < \theta < \pi and decreasing for π<θ<2π\pi < \theta < 2\pi. As β≈1.90<π<2π−β\beta \approx 1.90 < \pi < 2\pi - \beta, on the arch QRQR the height yy rises from θ=β\theta = \beta to a maximum at θ=π\theta = \pi, then falls back to the same value at θ=2π−β\theta = 2\pi - \beta (since cos⁡(2π−β)=cos⁡β\cos(2\pi - \beta) = \cos\beta, QQ and RR are at equal height).

Greatest height, at the apex θ=π\theta = \pi:

ymax⁡=a(1−cos⁡π)=2a≈22.9 my_{\max} = a\left( 1 - \cos\pi \right) = 2a \approx 22.9\ m

Least height, at QQ and RR, θ=β\theta = \beta:

ymin⁡=a(1−cos⁡β)=a(1−cos⁡1.90)≈15.1 my_{\min} = a\left( 1 - \cos\beta \right) = a\left( 1 - \cos 1.90 \right) \approx 15.1\ m
To check your answer: ymin⁡≈15.1 m<ymax⁡≈22.9 my_{\min} \approx 15.1\ m < y_{\max} \approx 22.9\ m, as expected for an arch rising from its supports to a central peak ✓

∴\therefore the greatest height of the arch is ≈22.9 m\text{≈}22.9\ m and the least height is ≈15.1 m\text{≈}15.1\ m.

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