Example 31: $\frac{dy}{dx}$ when $x=a\cos\theta$⁠, $y=a\sin\theta$

Basics you need

Solution

1. Name the parameter and the two derivatives we need

   Both coordinates are written in terms of $\theta$⁠:

   $$x = a\cos\theta, \qquad y = a\sin\theta.$$

   So $\theta$ is the parameter, and by Basic 3 we need $\dfrac{dx}{d\theta}$ and $\dfrac{dy}{d\theta}$⁠.

   We could try to get $y$ in terms of $x$ first: from $x^2 + y^2 = a^2$ we would get $y = \sqrt{a^2 - x^2}$ on the upper half of the circle and $y = -\sqrt{a^2 - x^2}$ on the lower half. That means two formulas and square roots. The parametric route avoids all of this: each of $x$ and $y$ is a simple function of $\theta$⁠.
   A parameter is just a third variable that both $x$ and $y$ depend on. Think of $\theta$ as a clock: as it runs, the point $(x, y)$ moves along the curve. Lesson: parametric differentiation

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

   The constant $a$ stays outside (Basic 2), and the derivative of $\cos\theta$ is $-\sin\theta$ (Basic 1):

   $$\frac{dx}{d\theta} = a\,\frac{d}{d\theta}(\cos\theta) = -a\sin\theta.$$
   $\frac{dx}{d\theta}$ measures how fast $x$ changes as $\theta$ increases. For $0 \lt \theta \lt \pi$ (with $a \gt 0$⁠) it is negative: moving anticlockwise over the top of the circle, the point moves to the left.
   $\frac{d}{d\theta}\cos\theta = -\sin\theta$⁠. Forgetting this minus sign is the most common slip in this example. Basics: derivative rules

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

   $$\frac{dy}{d\theta} = a\,\frac{d}{d\theta}(\sin\theta) = a\cos\theta.$$
   This is the same kind of step as Step 2, now for the second coordinate. Both rates are needed because the slope compares the change in $y$ with the change in $x$⁠.
   $\frac{d}{d\theta}\sin\theta = \cos\theta$ (Table 5.3).

4. Find where the formula may be used

   Basic 3 needs $\frac{dx}{d\theta} \neq 0$⁠. Since $a \neq 0$⁠,

   $$\begin{aligned} -a\sin\theta \neq 0 &\iff \sin\theta \neq 0 \\ &\iff \theta \neq n\pi,\ n \in \Z. \end{aligned}$$

   The excluded values $\theta = 0, \pm 2\pi, \dots$ give the point $(a, 0)$⁠, and $\theta = \pm\pi, \pm 3\pi, \dots$ give $(-a, 0)$⁠. These are the two ends of the horizontal diameter of the circle.

   At these two points $x$ momentarily stops changing while $y$ keeps changing, so the curve is heading straight up or down. The tangent is vertical and has no slope, so no formula could give a value of $\frac{dy}{dx}$ there (see the figure).
   $\sin\theta = 0$ exactly when $\theta$ is an integer multiple of $\pi$⁠. Basics: trigonometric identities

5. Divide and simplify

   For $\theta \neq n\pi$⁠, by the parametric formula,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} = \frac{a\cos\theta}{-a\sin\theta} \\ &= -\frac{\cos\theta}{\sin\theta} = -\cot\theta. \end{aligned}$$

   So $\dfrac{dy}{dx} = -\cot\theta$, for every $\theta$ that is not a multiple of $\pi$⁠.

   The $\theta$-rate of $y$ goes on top and the $\theta$-rate of $x$ goes underneath, because $\frac{dy}{dx}$ asks "how much does $y$ change per unit change in $x$". Cancelling $a$ is allowed because $a \neq 0$⁠. The answer is in terms of $\theta$ only, which is normal for parametric problems (the book's Note after Example 33 says the same).
   $\frac{\cos\theta}{\sin\theta} = \cot\theta$ (Basic 4).

6. See what the answer means

   Every point of the curve lies on a circle, because

   $$x^2 + y^2 = a^2\cos^2\theta + a^2\sin^2\theta = a^2.$$

   So the curve is the circle of radius $|a|$ centred at the origin $O$⁠. Where $\cos\theta \neq 0$⁠, the radius $OP$ to the point $P(a\cos\theta, a\sin\theta)$ has slope $\frac{y}{x} = \tan\theta$⁠, and

   $$\tan\theta \times (-\cot\theta) = -1.$$

   Slopes with product $-1$ belong to perpendicular lines: the tangent is perpendicular to the radius, which is exactly the geometry fact about tangents to a circle. (Where $\cos\theta = 0$⁠, the radius is vertical and $-\cot\theta = 0$⁠, so the tangent is horizontal: again perpendicular.)

   A figure turns the formula into something you can see: $-\cot\theta$ is not just an expression, it is the steepness of the circle at the point labelled by $\theta$⁠. It also explains the excluded values of Step 4.
   $\cos^2\theta + \sin^2\theta = 1$ for every $\theta$⁠. Basics: trigonometric identities

7. Check by a second method: implicit differentiation

   Differentiate $x^2 + y^2 = a^2$ with respect to $x$⁠, treating $y$ as a function of $x$ (chain rule on $y^2$⁠):

   $$2x + 2y\,\frac{dy}{dx} = 0, \ \text{ so } \ \frac{dy}{dx} = -\frac{x}{y}$$

   wherever $y \neq 0$⁠.

   Put in $x = a\cos\theta$⁠, $y = a\sin\theta$⁠:

   $$-\frac{x}{y} = -\frac{a\cos\theta}{a\sin\theta} = -\cot\theta.$$

   The same answer, and the condition $y \neq 0$ is the same as $\sin\theta \neq 0$ from Step 4.

   Getting one answer by two independent methods is strong evidence that neither contains a slip. It also shows that the parametric answer $-\cot\theta$ and the answer $-\frac{x}{y}$ in terms of $x$ and $y$ are the same slope, written in different variables.
   Implicit differentiation: $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ by the chain rule. Lesson: implicit differentiation

Every fact used, and where it comes from