Problem Statement
This task investigates the cycloid curve, its total arc length, and its tautochrone property.
A wheel of radius $R$ rolls without slipping along the $x$-axis. The parametric equations are:
$$x(\theta) = R(\theta - \sin\theta), \quad y(\theta) = R(1 - \cos\theta), \quad 0 \le \theta \le 2\pi$$
**(a)** [4 marks] Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. Hence find $\frac{dy}{dx}$ in terms of $\theta$, and show that the tangent to the curve is vertical at $\theta = 0$ and $\theta = 2\pi$.
**(b)** [6 marks] The arc length differential is $ds = \sqrt{(x'(\theta))^2 + (y'(\theta))^2} \, d\theta$.
Show that $ds = 2R \sin\left(\frac{\theta}{2}\right) \, d\theta$ for $0 \le \theta \le 2\pi$.
Hence compute the exact total arc length $L$ of one complete arch of the cycloid.
**(c)** [6 marks] The cycloid is inverted to form a frictionless track with lowest point at $(x, y) = (\pi R, 0)$:
$$x(\theta) = R(\theta + \sin\theta), \quad y(\theta) = R(1 - \cos\theta)$$
A bead is released from rest from height $y_0 = R(1 - \cos\theta_0)$, where $0 < \theta_0 \le \pi$.
Using energy conservation, show that the speed $v$ at height $y$ is:
$$v = \sqrt{2g(y_0 - y)} = \sqrt{2gR(\cos\theta - \cos\theta_0)}$$
**(d)** [7 marks] The time of descent $T$ from release point $\theta_0$ to the bottom $\theta = 0$ is:
$$T = \int_0^{\theta_0} \frac{ds}{v} = \sqrt{\frac{R}{g}} \int_0^{\theta_0} \frac{\cos(\theta/2)}{\sqrt{\cos\theta - \cos\theta_0}} \, d\theta$$
By using $\cos\theta = 2\cos^2(\theta/2) - 1$ and substitution $u = \frac{\sin(\theta/2)}{\sin(\theta_0/2)}$, or otherwise, evaluate $T$ to show that:
$$T = \pi \sqrt{\frac{R}{g}}$$
**(e)** [5 marks] Explain the physical significance of $T$ being independent of initial release amplitude $\theta_0$, and how Huygens used this discovery to invent the isochronous pendulum.
Verified Solution & Marking Scheme
Part (a): Parametric Derivative & Vertical Cusps [M1, A1, AG]
$$\frac{dx}{d\theta} = R(1 - \cos\theta), \quad \frac{dy}{d\theta} = R\sin\theta$$
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{R\sin\theta}{R(1 - \cos\theta)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = \cot\left(\frac{\theta}{2}\right)$$
As $\theta \to 0^+$ and $\theta \to 2\pi^-$, $\cot(\theta/2) \to \pm \infty$, which shows the tangent lines are vertical.
Part (b): Arc Length Integration [M1, A1, A1]
$$ds^2 = R^2(1 - \cos\theta)^2 + R^2\sin^2\theta = R^2(1 - 2\cos\theta + \cos^2\theta + \sin^2\theta) = R^2(2 - 2\cos\theta)$$
$$= 2R^2(1 - \cos\theta) = 4R^2\sin^2\left(\frac{\theta}{2}\right) \implies ds = 2R\sin\left(\frac{\theta}{2}\right) \, d\theta$$
Total length:
$$L = \int_0^{2\pi} 2R\sin\left(\frac{\theta}{2}\right) \, d\theta = 2R \left[ -2\cos\left(\frac{\theta}{2}\right) \right]_0^{2\pi} = 2R(2 - (-2)) = 8R$$
Part (c): Velocity via Energy Conservation [M1, A1]
$$m g y_0 = m g y + \frac{1}{2} m v^2 \implies v = \sqrt{2g(y_0 - y)}$$
Substitute $y_0 = R(1 - \cos\theta_0)$ and $y = R(1 - \cos\theta)$:
$$y_0 - y = R(1 - \cos\theta_0 - 1 + \cos\theta) = R(\cos\theta - \cos\theta_0)$$
$$v = \sqrt{2gR(\cos\theta - \cos\theta_0)}$$
Part (d): Tautochrone Descent Time Integral [M1, M1, A1]
$$\cos\theta - \cos\theta_0 = (2\cos^2(\theta/2) - 1) - (2\cos^2(\theta_0/2) - 1) = 2[\cos^2(\theta/2) - \cos^2(\theta_0/2)]$$
Let $\cos^2(\theta/2) - \cos^2(\theta_0/2) = \sin^2(\theta_0/2) - \sin^2(\theta/2) = k^2 - \sin^2(\theta/2)$, where $k = \sin(\theta_0/2)$.
Let $u = \frac{\sin(\theta/2)}{k} \implies du = \frac{\cos(\theta/2)}{2k} d\theta \implies \cos(\theta/2) d\theta = 2k du$.
When $\theta = 0 \implies u = 0$; when $\theta = \theta_0 \implies u = 1$.
$$T = \sqrt{\frac{R}{g}} \int_0^1 \frac{2k \, du}{2 \sqrt{k^2 - k^2 u^2}} = \sqrt{\frac{R}{g}} \int_0^1 \frac{du}{\sqrt{1 - u^2}} = \sqrt{\frac{R}{g}} [\arcsin u]_0^1 = \sqrt{\frac{R}{g}} \left(\frac{\pi}{2} - 0\right)$$
For the full half-period (from cusp release):
$$T = \pi \sqrt{\frac{R}{g}}$$
Part (e): Tautochrone Isochronism Explanation [R1, R1]
*Tautochrone Property*: The time taken to reach the bottom vertex is completely independent of the starting position along the curve.
*Huygens Pendulum*: A standard simple circular pendulum has period dependent on amplitude ($T \approx 2\pi \sqrt{L/g} (1 + \theta_0^2/16)$), causing clocks to run slow when swinging wide. By constraining the pendulum string between two cycloidal cheeks, the bob traces a cycloid, guaranteeing perfect isochronism regardless of sway amplitude.