Question

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

Short answer

The arc length of the curve on the interval \([0,2 \pi]\) can be calculated by a numerical method, as the integral is not solvable analytically.

Step-by-step solution

Find derivatives

Find the derivatives of \(x(\theta)\) and \(y(\theta)\) with respect to \(\theta\). They are \({dx/d\theta} = -\sin \theta + \sin \theta + \theta \cos \theta = \theta \cos \theta\) and \({dy/d\theta} = \cos \theta + \cos \theta + \theta \sin \theta = \cos \theta + \theta \sin \theta\).

Calculate square of derivatives and their sum

Find the square of each derivative and their sum: \(({dx/d\theta})^2 = (\theta \cos \theta )^2 = \theta^2 \cos^2 \theta\) and \(({dy/d\theta})^2 = (\cos \theta + \theta \sin \theta)^2 = \cos^2 \theta + 2\theta \sin \theta \cos \theta + \theta^2 \sin^2 \theta\). The sum is \(\theta^2 \cos^2 \theta + \cos^2 \theta + 2\theta \sin \theta \cos \theta + \theta^2 \sin^2 \theta = \theta^2 + 1\)

Compute Integral

Plug the calculated values into the formula for the arc length and compute the integral: \(L = \int_{0}^{2\pi} \sqrt{\theta^2 + 1} d\theta\). Since an analytical solution to this integral is not readily available, it can be approximated numerically. Using a method such as Romberg's method or Simpson's rule for numerical integration should give an approximate value.