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 [itex]\pi\sqrt{a/g}[/tex] is the time it takes to reach the point [itex](\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:" />[/itex][/itex]

 

[maverick280857]

Okay I got it.