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Chapter 8: Problem 54

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

Expert verified
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

01

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 \({dy/d\theta} = \cos \theta + \cos \theta + \theta \sin \theta = \cos \theta + \theta \sin \theta\).
02

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 \(\theta^2 \cos^2 \theta + \cos^2 \theta + 2\theta \sin \theta \cos \theta + \theta^2 \sin^2 \theta = \theta^2 + 1\)
03

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.

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Most popular questions from this chapter

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=1+\frac{1}{t}, \quad y=t-1 $$

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=6 \cos \theta & 0 \leq \theta \leq \frac{\pi}{2} & \text { Polar axis } \end{array} $$

Write a short paragraph describing how the graphs of curves represented by different sets of parametric equations can differ even though eliminating the parameter from each yields the same rectangular equation.

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Determine any differences between the curves of the parametric equations. Are the graphs the same? Are the orientations the same? Are the curves smooth? $$ \text { (a) } \begin{array}{l} x=\cos \theta \\ y=2 \sin ^{2} \theta \\ 0<\theta<\pi \end{array} $$ $$ \text { (b) } \begin{aligned} x &=\cos (-\theta) \\ y &=2 \sin ^{2}(-\theta) \\ 0 &<\theta<\pi \end{aligned} $$
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