Question

Find the arc length of the curve on the interval \([0,2 \pi]\). Cycloid arch: \(x=a(\theta-\sin \theta), y=a(1-\cos \theta)\)

Short answer

The arc length of the given curve on the interval \([0,2 \pi]\) is \(16a\).

Step-by-step solution

Compute the derivatives

Firstly, find the derivatives of the parametric equations. Differentiate \(x\) with respect to \(\theta\), \(dx/d\theta = a(1-\cos \theta)\). Differentiate \(y\) with respect to \(\theta\), \(dy/d\theta = a \sin \theta\).

Apply the Arc Length Formula

Then, apply these derivatives to the arc length formula for parametric equations. This gives: \(L = \int_{0}^{2 \pi} \sqrt{( a(1-\cos \theta) )^2 + (a \sin \theta )^2} d\theta\). After simplification this is: \(L = \int_{0}^{2 \pi} \sqrt{a^2 - 2a^2 \cos \theta + a^2} d\theta = \int_{0}^{2 \pi} a \sqrt{2(1- \cos \theta)} d\theta\). The term under the radix, \(2(1-\cos \theta)\) is known in trigonometry to be equivalent to \(2 \sin^2 (\theta/2)\), \(L = \int_{0}^{2 \pi} 2a |\sin(\theta/2)| d\theta\). Note: the absolute value is used because the \(\sin \) function yields negative values in the interval \([ \pi, 2\pi]\), but length can't be negative.

Evaluate the Integral

Finally, solve this integral. As this integral is over the interval \([0, 2 \pi]\) where \(\sin x\) is first positive then negative, it makes sense to split this into two parts, \(L = \int_{0}^{\pi} 2a\sin(\theta/2) d\theta + \int_{\pi}^{2 \pi} 2a(-\sin(\theta/2)) d\theta \). Evaluate these integrals separately then, the two results are added together, resulting in \(8a + 8a = 16a\).