Exercise 5.6 · Q9: $\frac{dy}{dx}$ when $x=a\sec\theta$⁠, $y=b\tan\theta$

Basics you need

Solution

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

   $$\frac{dx}{d\theta} = a\sec\theta\tan\theta.$$
   The constant $a$ stays in front, and $\sec\theta$ differentiates to $\sec\theta\tan\theta$⁠. (Quick derivation: $\sec\theta = (\cos\theta)^{-1}$⁠, so by the chain rule its derivative is $-(\cos\theta)^{-2}(-\sin\theta) = \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta}$⁠.)
   Basic 1; chain rule (Theorem 4, Section 5.3.1). Lesson: chain rule

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

   $$\frac{dy}{d\theta} = b\sec^2\theta.$$
   The other rate needed by the formula, straight from Table 5.3.
   $\frac{d}{d\theta}\tan\theta = \sec^2\theta$ (Table 5.3).

3. Find where the formula may be used

   We need $a\sec\theta\tan\theta \neq 0$⁠. Here $a \neq 0$⁠, and $\sec\theta = \frac{1}{\cos\theta}$ is never zero, so the condition is

   $$\tan\theta \neq 0 \iff \sin\theta \neq 0 \iff \theta \neq n\pi.$$

   Together with $\cos\theta \neq 0$ from the given equations, $\theta$ must not be any multiple of $\frac{\pi}{2}$⁠.

   We are about to divide by $\frac{dx}{d\theta}$⁠, so its zeros come first. A fraction $\frac{1}{\cos\theta}$ can be large but never zero, so the only possible zeros come from $\tan\theta$⁠.
   $\tan\theta = \frac{\sin\theta}{\cos\theta}$ is zero exactly when $\sin\theta = 0$⁠.

4. Divide

   For $\theta$ not a multiple of $\frac{\pi}{2}$⁠,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} = \frac{b\sec^2\theta}{a\sec\theta\tan\theta} \\ &= \frac{b}{a}\cdot\frac{\sec\theta}{\tan\theta}. \end{aligned}$$
   The $y$-rate goes on top. One factor $\sec\theta$ cancels (it is never zero).
   Parametric formula, Section 5.6. Lesson: parametric differentiation

5. Simplify to $\operatorname{cosec}\theta$

   $$\frac{\sec\theta}{\tan\theta} = \frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta} = \frac{1}{\sin\theta} = \operatorname{cosec}\theta.$$

   Hence

   $$\frac{dy}{dx} = \frac{b}{a}\operatorname{cosec}\theta.$$
   Dividing by $\tan\theta = \frac{\sin\theta}{\cos\theta}$ means multiplying by $\frac{\cos\theta}{\sin\theta}$⁠; the $\cos\theta$ (nonzero) cancels. This is the form of the book's answer.
   Basic 3: write everything in sines and cosines. Basics: trigonometric identities

6. Picture the curve and the excluded points

   At $\theta = n\pi$⁠, $\tan\theta = 0$ and $\sec\theta = \pm1$⁠, so the point is $(\pm a,\ 0)$⁠. There $\frac{dx}{d\theta} = 0$ but $\frac{dy}{d\theta} = b\sec^2\theta = b$⁠, which is not zero if $b \neq 0$⁠: the point moves vertically (up if $b \gt 0$⁠, down if $b \lt 0$⁠), the tangent is vertical, and $\frac{dy}{dx}$ does not exist. Indeed $|\operatorname{cosec}\theta|$ grows without bound as $\theta$ approaches $n\pi$⁠.

   The two branches and the vertical tangents at $(\pm a, 0)$ are the familiar features of a hyperbola, as the check below confirms. Far from the centre, $|\operatorname{cosec}\theta|$ approaches $1$ (as $\theta$ approaches $\pm\frac{\pi}{2}$⁠), so the slope approaches $\pm\frac{b}{a}$⁠, the slopes of the dashed asymptotes.
   A vertical line has no slope. Basics: slope at a point

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