Exercise 5.6 · Q5: $\frac{dy}{dx}$ when $x=\cos\theta-\cos2\theta$⁠, $y=\sin\theta-\sin2\theta$

Basics you need

Solution

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

   Difference rule, then Basic 1 for $\cos 2\theta$⁠:

   $$\begin{aligned} \frac{dx}{d\theta} &= -\sin\theta - (-2\sin 2\theta) \\ &= 2\sin 2\theta - \sin\theta. \end{aligned}$$
   The derivative of a difference is the difference of the derivatives. The term $-\cos 2\theta$ gives $-(-2\sin 2\theta) = +2\sin 2\theta$⁠: two minus signs make a plus, and the 2 comes from the chain rule. We write the positive term first, as the book's answer does.
   $\frac{d}{d\theta}\cos\theta = -\sin\theta$ (Table 5.3); chain rule, Theorem 4, Section 5.3.1. Lesson: chain rule

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

   $$\frac{dy}{d\theta} = \cos\theta - 2\cos 2\theta.$$
   Again term by term: $\sin\theta$ gives $\cos\theta$⁠, and $\sin 2\theta$ gives $2\cos 2\theta$ by the chain rule, subtracted.
   $\frac{d}{d\theta}\sin\theta = \cos\theta$ (Table 5.3) and Basic 1.

3. Find where $\frac{dx}{d\theta} = 0$

   Write $\sin 2\theta = 2\sin\theta\cos\theta$ and take out the common factor $\sin\theta$⁠:

   $$\begin{aligned} \frac{dx}{d\theta} &= 4\sin\theta\cos\theta - \sin\theta \\ &= \sin\theta\,(4\cos\theta - 1). \end{aligned}$$

   A product is zero only when a factor is zero, so $\frac{dx}{d\theta} = 0$ exactly when

   • $\sin\theta = 0$⁠, that is $\theta = n\pi$⁠; or
   • $\cos\theta = \frac14$⁠, that is $\theta = 2n\pi \pm \alpha$ with $\alpha = \cos^{-1}\frac14 \approx 1.3181$ radians (Basic 4).

   The formula may be used at every other $\theta$⁠.

   We must know exactly where we are not allowed to divide. In the form $2\sin 2\theta - \sin\theta$ the zeros are hidden because the two terms have different angles; rewriting $\sin 2\theta$ in the single angle $\theta$ exposes the factor $\sin\theta$ and turns the question into two simple equations.
   $\sin 2\theta = 2\sin\theta\cos\theta$ (Basic 3); $\sin\theta = 0 \iff \theta = n\pi$⁠.

4. Divide

   For $\sin\theta \neq 0$ and $\cos\theta \neq \frac14$⁠,

   $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\cos\theta - 2\cos 2\theta}{2\sin 2\theta - \sin\theta}.$$

   This is the final answer, and it is exactly the book's form. Top and bottom have no common factor to cancel (the top is not zero at the zeros of the bottom, as Step 5 shows), so nothing simpler is hiding here.

   The $y$-rate goes on top, the $x$-rate underneath. You may also write it as $\frac{2 + \cos\theta - 4\cos^2\theta}{\sin\theta\,(4\cos\theta - 1)}$⁠, using $\cos 2\theta = 2\cos^2\theta - 1$ on top and Step 3 underneath, but that is not simpler.
   Parametric formula, Section 5.6. Lesson: parametric differentiation

5. What happens at the excluded values: vertical tangents

   At each excluded value, compute $\frac{dy}{d\theta} = \cos\theta - 2\cos 2\theta$⁠:

   • $\theta = 2n\pi$⁠: $\cos\theta = 1$⁠, $\cos 2\theta = 1$⁠, so $\frac{dy}{d\theta} = 1 - 2 = -1$⁠. The point is $(1 - 1,\ 0 - 0) = (0, 0)$⁠.
   • $\theta = (2n+1)\pi$⁠: $\cos\theta = -1$⁠, $\cos 2\theta = 1$⁠, so $\frac{dy}{d\theta} = -1 - 2 = -3$⁠. The point is $(-1 - 1,\ 0) = (-2, 0)$⁠.
   • $\cos\theta = \frac14$⁠: $\cos 2\theta = 2\cdot\frac{1}{16} - 1 = -\frac78$⁠, so $\frac{dy}{d\theta} = \frac14 + \frac74 = 2$⁠. Here $\sin\theta = \pm\frac{\sqrt{15}}{4}$ and $\sin 2\theta = \pm\frac{\sqrt{15}}{8}$⁠, so the point is $\big(\frac98,\ \pm\frac{\sqrt{15}}{8}\big)$⁠.

   At all of them $x$ momentarily stops changing while $y$ keeps changing, so the point moves straight up or down: the tangent is vertical, and a vertical line has no slope. So $\frac{dy}{dx}$ does not exist there; the exclusions are genuine, not just a gap in the method.

   In fact the two rates are never zero together: squaring and adding,

   $$\begin{aligned} &\Big(\tfrac{dx}{d\theta}\Big)^2 + \Big(\tfrac{dy}{d\theta}\Big)^2 \\ &= 5 - 4(\cos 2\theta\cos\theta + \sin 2\theta\sin\theta) \\ &= 5 - 4\cos\theta \geq 1, \end{aligned}$$

   using $\sin^2 + \cos^2 = 1$ twice and $\cos(A - B) = \cos A\cos B + \sin A\sin B$ with $A = 2\theta$⁠, $B = \theta$⁠. So the moving point never stops, and the curve has no sharp tips.

   The curve crosses itself at $(1, 0)$⁠, which it passes at $\theta = \frac{\pi}{3}$ and at $\theta = \frac{5\pi}{3}$⁠. The formula gives the slope $\sqrt3$ at $\theta = \frac{\pi}{3}$ and $-\sqrt3$ at $\theta = \frac{5\pi}{3}$⁠: the two passes go through that point in different directions. This is why a parametric $\frac{dy}{dx}$ is written in terms of the parameter rather than the point.

   Working out the excluded points on the picture shows that each one is a place where the curve runs straight up or down. The crossing point is a nice reminder that a parametric curve can visit the same point twice with two different tangents.
   A vertical line has no slope. Basics: slope at a point

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