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Worked solution

Sketch a cycloid and locate a normal intercept

Curve CC has x=θ−sin⁡θx=\theta-\sin\theta, y=1−cos⁡θy=1-\cos\theta, 0≤θ≤2π0\le\theta\le2\pi.

(i) Show that dy/dx=cot⁡(θ/2)dy/dx=\cot(\theta/2) and find the gradient at θ=π\theta=\pi. Describe the tangents as θ→0\theta\to0 and θ→2π\theta\to2\pi.

(ii) Sketch CC, clearly showing its features at θ=0,π,2π\theta=0,\pi,2\pi.

(iii) For PP with parameter pp, 0<p<π/20<p<\pi/2, show that the normal at PP crosses the x-axis at (p,0)(p,0).

[2012/I/11(modified)]

CJC · Discussion · Question 8

Answer

dy/dx=cot⁡(θ/2)dy/dx=\cot(\theta/2); gradient 00 at π\pi, vertical endpoint tangents; normal intercept (p,0)(p,0).

(i) Simplify the gradient with half-angle identities

dx/dθ=1−cos⁡θdx/d\theta=1-\cos\theta and dy/dθ=sin⁡θdy/d\theta=\sin\theta. In the interior, dy/dx=sin⁡θ/(1−cos⁡θ)=2sin⁡(θ/2)cos⁡(θ/2)/(2sin⁡2(θ/2))=cot⁡(θ/2)dy/dx=\sin\theta/(1-\cos\theta)=2\sin(\theta/2)\cos(\theta/2)/(2\sin^2(\theta/2))=\cot(\theta/2). At θ=π\theta=\pi the gradient is 00. As θ→0+\theta\to0^+ it tends to +∞+\infty, and as θ→2π−\theta\to2\pi^- it tends to −∞-\infty. Both endpoint tangents are vertical, namely x=0x=0 and x=2πx=2\pi.

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(ii) Mark the endpoint and maximum features

The curve is a single cycloid arch from (0,0)(0,0) to (2π,0)(2\pi,0), with a maximum at (π,2)(\pi,2). It increases before θ=π\theta=\pi and decreases afterwards, has a horizontal tangent at the maximum and vertical limiting tangents at the endpoints. Since dx/dθ>0dx/d\theta>0 for 0<θ<2π0<\theta<2\pi, it runs left to right.

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(iii) Write the normal and set y = 0

P=(p−sin⁡p,1−cos⁡p)P=(p-\sin p,1-\cos p) and the normal gradient is −tan⁡(p/2)-\tan(p/2). Thus y−(1−cos⁡p)=−tan⁡(p/2)[x−(p−sin⁡p)]y-(1-\cos p)=-\tan(p/2)[x-(p-\sin p)]. At y=0y=0, 1−cos⁡p=tan⁡(p/2)(x−p+sin⁡p)1-\cos p=\tan(p/2)(x-p+\sin p). Using 1−cos⁡p=2sin⁡2(p/2)1-\cos p=2\sin^2(p/2) and tan⁡(p/2)=sin⁡(p/2)/cos⁡(p/2)\tan(p/2)=\sin(p/2)/\cos(p/2), obtain sin⁡p=x−p+sin⁡p\sin p=x-p+\sin p, so x=px=p. Hence the intercept is (p,0)(p,0).

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A cycloid arch from (0,0) to (2 pi,0), with maximum (pi,2), a horizontal tangent at the maximum, and vertical endpoint tangents x=0 and x=2 pi. The arch increases on the left and decreases on the right.
A cycloid arch from (0,0) to (2 pi,0), with maximum (pi,2), a horizontal tangent at the maximum, and vertical endpoint tangents x=0 and x=2 pi. The arch increases on the left and decreases on the right.
A cycloid arch from (0,0) to (2 pi,0), with maximum (pi,2), a horizontal tangent at the maximum, and vertical endpoint tangents x=0 and x=2 pi. The arch increases on the left and decreases on the right.

Common mistakes

At the right endpoint the gradient tends to negative infinity, although the tangent is still vertical.

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