Simple cycloid time calculation problem

[maverick280857]

Homework Statement

Consider a wire bent into the shape of the cycloid

x = a(\theta - \sin\theta)
y = a(\cos\theta -1)

If a bead is released at the origin and slides down the wire without friction, show that \pi\sqrt{a/g}[/tex] is the time it takes to reach the point (\pi a, -2a)[/tex] at the bottom.<h2>Homework Equations</h2><br /> <br /> (See below)<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Energy conservation gives<br /> <br /> \frac{1}{2}mv^{2} = mg(2a)<br /> or<br /> v^{2} = 4ga<br /> <br /> For the point at the bottom, \theta = \pi. So, the arc length is<br /> <br /> s = \int_{0}^{\theta}\sqrt{\left(\frac{dx}{d\theta}\right)^{2} + \left(\frac{dy}{d\theta}\right)^{2}}d\theta<br /> <br /> v = \frac{ds}{dt}<br /> <br /> How do I get rid of the d\theta/dt? I know I'm missing something here...<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f644.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":rolleyes:" title="Roll Eyes :rolleyes:" data-smilie="11"data-shortname=":rolleyes:" />

 

[maverick280857]

Okay I got it.