Exercise 5.6 · Q10: $\frac{dy}{dx}$ when $x=a(\cos\theta+\theta\sin\theta)$⁠, $y=a(\sin\theta-\theta\cos\theta)$

Basics you need

Solution

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

   Sum rule, with the product rule on $\theta\sin\theta$ (Basic 1):

   $$\begin{aligned} \frac{dx}{d\theta} &= a\left(-\sin\theta + \sin\theta + \theta\cos\theta\right) \\ &= a\theta\cos\theta. \end{aligned}$$
   $\cos\theta$ gives $-\sin\theta$⁠. The product $\theta\sin\theta$ gives "derivative of the first times the second plus the first times the derivative of the second", $1\cdot\sin\theta + \theta\cos\theta$⁠. The $-\sin\theta$ and $+\sin\theta$ cancel.
   Product rule (Section 5.3); $\frac{d}{d\theta}\cos\theta = -\sin\theta$ (Table 5.3). Basics: derivative rules

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

   $$\begin{aligned} \frac{dy}{d\theta} &= a\left(\cos\theta - (\cos\theta - \theta\sin\theta)\right) \\ &= a\theta\sin\theta. \end{aligned}$$
   $\sin\theta$ gives $\cos\theta$⁠; the product $\theta\cos\theta$ gives $\cos\theta - \theta\sin\theta$⁠, and it is subtracted, so the bracket is needed. The two $\cos\theta$ terms cancel and $-(-\theta\sin\theta) = +\theta\sin\theta$⁠.
   Basic 1, second line.

3. Find where the formula may be used

   We need $a\theta\cos\theta \neq 0$⁠. Since $a \neq 0$⁠, this means

   $$\theta \neq 0 \quad\text{and}\quad \cos\theta \neq 0,$$

   that is, $\theta \neq 0$ and $\theta \neq (2n+1)\frac{\pi}{2}$⁠.

   A product is zero only when one of its factors is zero, and there are two variable factors here, $\theta$ and $\cos\theta$⁠. It is easy to forget the factor $\theta$⁠.
   Basic 3: the zeros of cosine.

4. Divide

   For $\theta \neq 0$ and $\cos\theta \neq 0$⁠,

   $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta.$$
   The common factor $a\theta$ is not zero (Step 3), so it cancels, and $\frac{\sin\theta}{\cos\theta} = \tan\theta$⁠.
   Parametric formula, Section 5.6. Lesson: parametric differentiation

5. Picture the curve: the involute of a circle

   Two kinds of excluded value behave differently:

   • At $\theta = 0$ both rates are zero. The point is $(a, 0)$⁠. Replacing $\theta$ by $-\theta$ leaves $x$ unchanged and changes $y$ into $-y$⁠, so the part of the curve with $\theta \lt 0$ is the mirror image in the $x$-axis of the part with $\theta \gt 0$⁠, and the two parts meet at $(a, 0)$ in a sharp tip (a cusp). The figure shows only $\theta \geq 0$⁠.
   • At $\cos\theta = 0$⁠, $\theta \neq 0$⁠, we have $\frac{dx}{d\theta} = 0$ but $\frac{dy}{d\theta} = a\theta\sin\theta = \pm a\theta \neq 0$⁠: the point moves straight up or down, the tangent is vertical, and there is no slope.

   The curve is the path traced by the end of a string unwound from a circle of radius $a$⁠. Write $T = (a\cos\theta,\ a\sin\theta)$⁠, a point of the circle. Then

   $$P - T = (a\theta\sin\theta,\ -a\theta\cos\theta),$$

   a segment of length $|a\theta|$⁠, which is the length of the arc of the circle from $(a, 0)$ to $T$⁠: the unwound string. Our answer says the tangent at $P$ has the direction of $(\cos\theta,\ \sin\theta)$⁠, the direction of the radius $OT$⁠, so it is perpendicular to the string. In the figure, at $\theta = \pi$ the string (dashed) is vertical and the tangent horizontal.

   This explains the answer $\tan\theta$⁠: the tangent to the curve is always parallel to the radius to the point where the string leaves the circle. The excluded value $\theta = 0$ is where the string has zero length, so the end of the string starts from rest.
   Slope of the direction $(p, q)$ is $\frac{q}{p}$⁠; here $\frac{\sin\theta}{\cos\theta} = \tan\theta$⁠. Basics: slope at a point

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