Arc Trail

Calculus Level 5

Circular arcs are drawn from a fixed point on the circumference of a unit circle.

What is the area bounded by the locus (shown in dotted blue and red) of the center of masses of the circular arcs?

0 π sin x x d x \displaystyle \int_{0}^{\pi}\dfrac{\sin{x}}{x}\, dx 0 2 π sin x x d x \displaystyle \int_{0}^{2\pi}\dfrac{\sin{x}}{x}\, dx 0 2 π sin 2 x x 2 d x \displaystyle \int_{0}^{2\pi}\dfrac{\sin^2{x}}{x^2}\, dx 0 π / 2 sin 2 x x 2 d x \displaystyle \int_{0}^{\pi/2}\dfrac{\sin^2{x}}{x^2}\, dx

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Joseph Newton
Oct 29, 2018

Let the arc subtend an angle of ϕ \phi and arc length ϕ \phi . Therefore the end points of the arc are ( 0 , 0 ) (0,0) and ( cos ϕ , sin ϕ ) (\cos\phi,\sin\phi) .

For points ( cos t , sin t ) (\cos t,\sin t) along the arc, the average x-coordinate is given by 1 ϕ 0 ϕ cos t d t = sin ϕ ϕ \frac1\phi\int_0^\phi\cos t\,dt=\frac{\sin\phi}\phi while the average y-coordinate is given by 1 ϕ 0 ϕ sin t d t = 1 cos ϕ ϕ \frac1\phi\int_0^\phi\sin t\,dt=\frac{1-\cos\phi}\phi Therefore the centre of mass is the point ( sin ϕ ϕ , 1 cos ϕ ϕ ) \left(\frac{\sin\phi}\phi,\frac{1-\cos\phi}\phi\right) . We will find the locus in polar coordinates: x = sin ϕ ϕ y = 1 cos ϕ ϕ x=\frac{\sin\phi}\phi\qquad y=\frac{1-\cos\phi}\phi θ = tan 1 ( y x ) r = x 2 + y 2 = tan 1 ( 1 cos ϕ sin ϕ ) = sin 2 ϕ + cos 2 ϕ 2 cos ϕ + 1 ϕ 2 = tan 1 ( tan ϕ 2 ) = 2 2 cos ϕ ϕ = ϕ 2 = 2 2 + 4 sin 2 ϕ 2 ϕ = 2 sin ϕ 2 ϕ r = sin θ θ \begin{aligned}\theta&=\tan^{-1}\left(\frac yx\right)&r&=\sqrt{x^2+y^2}\\ &=\tan^{-1}\left(\frac{1-\cos\phi}{\sin\phi}\right)&&=\sqrt{\frac{\sin^2\phi+\cos^2\phi-2\cos\phi+1}{\phi^2}}\\ &=\tan^{-1}\left(\tan\frac{\phi}2\right)&&=\frac{\sqrt{2-2\cos\phi}}\phi\\ &=\frac{\phi}2&&=\frac{\sqrt{2-2+4\sin^2\frac{\phi}2}}\phi\\ &&&=\frac{2\sin\frac{\phi}2}\phi\\ &&\therefore r&=\frac{\sin\theta}\theta\end{aligned} Now to find the area of this curve, we use the formula for area in polar coordinates: A = α β 1 2 r 2 d θ 2 0 π 1 2 sin 2 θ θ 2 d θ 0 π 1 2 ( 1 cos 2 θ ) θ 2 d θ 0 2 π 1 cos x x 2 d x ( by letting x = 2 θ ) [ 1 cos x x ] 0 2 π 0 2 π sin x x d x ( using integration by parts ) = 0 + lim x 0 1 cos x x + 0 2 π sin x x d x = 0 + lim x 0 sin x 1 + 0 2 π sin x x d x ( using L’Hospital’s rule ) = 0 2 π sin x x d x \begin{aligned}A=&\int_\alpha^\beta\frac12r^2\,d\theta\\ &2\int_0^\pi\frac12\frac{\sin^2\theta}{\theta^2}d\theta\\ &\int_0^\pi\frac{\frac12(1-\cos2\theta)}{\theta^2}d\theta\\ &\int_0^{2\pi}\frac{1-\cos x}{x^2}dx&(\text{by letting }x=2\theta)\\ &\left[-\frac{1-\cos x}x\right]_0^{2\pi}-\int_0^{2\pi}-\frac{\sin x}xdx&(\text{using integration by parts})\\ &=0+\lim_{x\to0}\frac{1-\cos x}x+\int_0^{2\pi}\frac{\sin x}xdx\\ &=0+\lim_{x\to0}\frac{\sin x}1+\int_0^{2\pi}\frac{\sin x}xdx&(\text{using L'Hospital's rule})\\ &=\boxed{\int_0^{2\pi}\frac{\sin x}xdx}\end{aligned}

Shouldn't it be d x dx instead of d θ d\theta in the last few integrals?

Digvijay Singh - 2 years, 7 months ago

Log in to reply

Yes, that was a typo. Thanks for letting me know.

Joseph Newton - 2 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...