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QuestionMathematics

The arc length $L$ of a curve given parametrically by $(x(t), y(t))$ for $a \leq t \leq b$ is given by the formula: L=\int_{a}^{b} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t A path of a point on the edge of a rolling circle of radius $R$ is a cycloid, given by: \begin{aligned} & x(t)=R(t-\sin t) \\ & y(t)=R(1 -\cos t) \end{aligned} where $t$ is the angle (in radians) the circle has rotated. Find the length $L$ of one "arch" of this cycloid. That is, find the distance traveled by a small stone stuck in the tread of a tire of radius $R$ during one revolution of the rolling tire.
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Answer

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Step 1:
I'll solve this cycloid arc length problem step by step:

Step 2:
: Find the derivatives of x(t) and y(t)

x'(t) = R(1 - \cos t)
y'(t) = R \sin t

Step 3:
: Square the derivatives

[x'(t)]^{2} = R^{2}(1 - \cos t)^{2}
[y'(t)]^{2} = R^{2} \sin^{2} t

Step 4:
: Add the squared derivatives

[x'(t)]^{2} + [y'(t)]^{2} = R^{2}(1 - \cos t)^{2} + R^{2} \sin^{2} t

Step 5:
: Use trigonometric identity \sin^{2} t + \cos^{2} t = 1

Rearranging, $$\sin^{2} t = 1 - \cos^{2} t

Step 6:
: Simplify the integrand

\sqrt{[x'(t)]^{2} + [y'(t)]^{2}} = R\sqrt{2(1 - \cos t)}

Step 7:
: Determine integration limits

One complete revolution occurs from $$t = 0$$ to $$t = 2\pi

Step 8:
: Set up the arc length integral

L = \int_{0}^{2\pi} R\sqrt{2(1 - \cos t)} dt

Step 9:
: Solve the integral

This integral evaluates to $$8R

Final Answer

The arc length of one complete arch of a cycloid is 8 times the radius of the generating circle.