Exercise 5.6 · Q6: $\frac{dy}{dx}$ when $x=a(\theta-\sin\theta)$⁠, $y=a(1+\cos\theta)$

Basics you need

Solution

1. Differentiate $x$ with respect to $\theta$

   $$\frac{dx}{d\theta} = a\left(1 - \cos\theta\right).$$
   $x = a(\theta - \sin\theta)$ is a constant times a difference. The derivative of $\theta$ is $1$ and that of $\sin\theta$ is $\cos\theta$⁠; the constant $a$ stays in front.
   Difference rule $(u - v)' = u' - v'$ (Section 5.3). Basics: derivative rules

2. Differentiate $y$ with respect to $\theta$

   $$\frac{dy}{d\theta} = a\left(0 + (-\sin\theta)\right) = -a\sin\theta.$$
   The constant $1$ has derivative $0$⁠, and $\cos\theta$ has derivative $-\sin\theta$⁠. Here the sign stays negative (in Example 33 the term was $-\cos\theta$ and the sign flipped).
   $\frac{d}{d\theta}\cos\theta = -\sin\theta$ (Table 5.3).

3. Find where the formula may be used

   We need $\frac{dx}{d\theta} \neq 0$⁠. Since $a \neq 0$⁠,

   $$\begin{aligned} a(1 - \cos\theta) \neq 0 &\iff \cos\theta \neq 1 \\ &\iff \theta \neq 2n\pi. \end{aligned}$$

   So from now on $\theta$ is not an even multiple of $\pi$⁠.

   We are about to divide by $\frac{dx}{d\theta}$⁠, so its zeros must be found first. Step 6 shows what the curve does at these values.
   Basic 4: $\cos\theta = 1$ only at $\theta = 2n\pi$⁠.

4. Divide

   For $\theta \neq 2n\pi$⁠,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} = \frac{-a\sin\theta}{a(1 - \cos\theta)} \\ &= \frac{-\sin\theta}{1 - \cos\theta}. \end{aligned}$$
   The $y$-rate goes on top, and $a \neq 0$ cancels. This is already correct; the next step simplifies it.
   Parametric formula, Section 5.6. Lesson: parametric differentiation

5. Simplify with the half-angle formulas

   Rewrite top and bottom using Basic 3:

   $$\begin{aligned} \frac{-\sin\theta}{1 - \cos\theta} &= \frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} \\ &= -\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = -\cot\frac{\theta}{2}. \end{aligned}$$

   Cancelling $2\sin\frac{\theta}{2}$ is allowed: by Step 3, $\theta \neq 2n\pi$⁠, so $\frac{\theta}{2} \neq n\pi$ and $\sin\frac{\theta}{2} \neq 0$⁠. Hence

   $$\frac{dy}{dx} = -\cot\frac{\theta}{2} \qquad (\theta \neq 2n\pi).$$
   In half-angles, both $\sin\theta$ and $1 - \cos\theta$ contain the factor $\sin\frac{\theta}{2}$⁠; cancelling it is the whole simplification. The condition from Step 3 is exactly the condition for $\cot\frac{\theta}{2}$ to be defined, so nothing is lost or gained.
   $\cot u = \frac{\cos u}{\sin u}$⁠, defined when $\sin u \neq 0$⁠. Basics: trigonometric identities

6. Picture the curve and the excluded points

   At $\theta = 2n\pi$ both rates are zero: $a(1 - \cos\theta) = 0$ and $-a\sin\theta = 0$⁠. The point there is $(2n\pi a,\ 2a)$⁠. As $\theta$ approaches $0$⁠, $\cot\frac{\theta}{2}$ grows without bound in size, so the curve comes into the point $(0, 2a)$ steeper and steeper and leaves it the same way: a sharp tip (a cusp), where $\frac{dy}{dx}$ does not exist.

   Halfway between, at $\theta = \pi$⁠, the point is $(\pi a,\ 0)$ and the slope is $-\cot\frac{\pi}{2} = 0$⁠: the lowest point, with a horizontal tangent.

   The excluded values are not a technicality: they are the sharp tips of the curve. The curve is a cycloid (the path of a point on the rim of a rolling wheel) turned upside down: since $y = 2a - a(1 - \cos\theta)$⁠, it is the reflection of the usual cycloid $x = a(\theta - \sin\theta)$⁠, $y = a(1 - \cos\theta)$ in the line $y = a$⁠.
   For $0 \lt \theta \lt 2\pi$⁠, the tangent's slope $-\cot\frac{\theta}{2}$ runs from very negative, through $0$ at $\theta = \pi$⁠, to very positive. Basics: slope at a point

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