Mathematical Immaturity

14.13 Exercises

• (a) When a circle rolls (without slipping) along a straight line, a point on the circumference traces out a curve called a cycloid. If the fixed line is the x-axis and if the tracing point $(x,y)$ is originally at the origin, show that when the circle rolls through an angle $\theta$ we have $$x = a(\theta - \sin \theta), \quad y = a(1 - \cos \theta)$$ where $a$ is the radius of the circle. These serve as parametric equations for the cycloid. (b) Referring to part (a), show that $dy/dx = \cot \frac{1}{2}\theta$ and deduce that the tangent line of the cycloid at $(x,y)$ makes an angle $\frac{1}{2}(\pi - \theta)$ with the x-axis. Make a sketch and show that the tangent line passes through the highest point on the circle.
• (a) To show that the position of the point is defined by the components $$x = a(\theta - \sin \theta), \quad y = a(1 - \cos \theta)$$ let us first consider the position of the point as the circle rotates in place. The circle has a radius of length $a$ centered at $(0, a).$ As the circle rotates clockwise, the angle $\theta$ swept out by the radius can be described as the angle between $A = (0, -a)$ and $B = (-x, -a + y),$ whose cosine is defined as: $$ \cos\theta = \frac{A \cdot B}{\|A\|\|B\|} $$ But $A$ and $B$ are both points on the circle, so their norms are both $a,$ giving us: $$ \cos\theta = \frac{a^2 - ay}{a^2} $$ In other words, we can parameterize $y$ as a function of $\theta:$ $$ y = a(1 - \cos\theta) $$ To express $x$ as a function of $\theta,$ we note once again that $B$ is a point on the circle with radius $a.$ Substituting the above equation for $y,$ we get $$ \begin{align*} \\ \|B\|^2 &= a^2 \\ &= x^2 + \left[-a + (a - a\cos\theta)\right]^2 \\ &= x^2 + a^2\cos^2\theta \end{align*} $$ Which means that $x^2 = a^2(1 - \cos^2\theta) = \sin^2\theta$ or that $|x| = a\sin\theta.$ But we know from before that $B = (-x, y - a),$ which implies that $a\sin\theta = -x,$ hence $$x = -a\sin\theta$$ Now, suppose the circle rolls along the positive $x$-axis. This simply means that the circle's horizontal position is displaced by $a\theta$ units of length as it rotates $\theta$ radians. Thus, our $x$-component is displaced by $a\theta,$ which gives us the parametric equations for position: $$x = a(\theta - \sin \theta), \quad y = a(1 - \cos \theta)$$ as requested. $\blacksquare$ (b) To show that $dy/dx = \cot(\theta/2),$ we first take the derivative of our positional components with respect to $\theta:$ $$ \frac{dx}{d\theta} = a(1 - \cos\theta), \quad \frac{dy}{d\theta} = a\sin\theta $$ Then, we use the double-angle identities for sine and cosine: $$ \begin{align*} \\ 1 - \cos\theta = 2\sin^2\frac{\theta}{2} \quad \text{and} \quad \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{align*} $$ giving us: $$ \begin{align*} \\ \frac{dx}{d\theta} = 2a\sin^2\frac{\theta}{2}; \quad \frac{dy}{d\theta} = 2a\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{align*} $$ Then, we see that $$ \begin{align*} \frac{dy}{dx} &= \frac{2a\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2a\sin^2\frac{\theta}{2}} \\ &= \cot\frac{1}{2}\theta \end{align*} $$ (valid for $0 < \theta < \pi$) Now, we note that $\cot(\frac{1}{2}\theta) = \tan(\frac{\pi}{2} - \frac{\theta}{2}),$ which means the tangent line is parallel to the unit vector $\mathbf{u}$ defined by: $$ \mathbf{u} = \cos\left(\frac{\pi}{2} - \frac{\theta}{2}\right)\mathbf{i} + \sin\left(\frac{\pi}{2} - \frac{\theta}{2}\right)\mathbf{j} $$ The angle $\phi$ between $\mathbf{u}$ and the $x$-axis satisfies: $$ \begin{align*} \cos\phi = \frac{\mathbf{u} \cdot \mathbf{i}}{\|\mathbf{u}\|\|\mathbf{i}\|} =\cos\left(\frac{\pi}{2} - \frac{\theta}{2}\right) \end{align*} $$ where $0 \leq \phi \leq \pi$ But this implies that the line tangent to the cycloid at $(x, y)$ makes an angle of $(\frac{\pi}{2} - \frac{\theta}{2})$ with the $x$-axis. $\blacksquare$