Polar Coordinates and Polar Derivatives

A polar curve $r = f(\theta)$ is traced as $\theta$ turns. Write it in $x$ and $y$ and it becomes a parametric curve with parameter $\theta$:

$$x = r\cos\theta = f(\theta)\cos\theta, \qquad y = r\sin\theta = f(\theta)\sin\theta.$$

So every idea from Topics 9.1 and 9.2 carries over. This page adds two things polar questions ask about: the slope in polar form, and the meaning of $\frac{dr}{d\theta}$.

Derivatives of $x$ and $y$ with respect to $\theta$

Both $x$ and $y$ are products, so each derivative has two terms:

$$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta.$$

Do not memorize these as new formulas. Write $x = r\cos\theta$, apply the product rule, and they appear. The signs tell you how the point moves: $\frac{dy}{d\theta} > 0$ means the point is moving up as $\theta$ increases, $\frac{dx}{d\theta} < 0$ means it is moving left.

The slope of a polar curve

As for any parametric curve, where $\frac{dx}{d\theta} \ne 0$:

$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}.$$

Horizontal and vertical tangents. A horizontal tangent needs $\frac{dy}{d\theta} = 0$ with $\frac{dx}{d\theta} \ne 0$; a vertical tangent needs $\frac{dx}{d\theta} = 0$ with $\frac{dy}{d\theta} \ne 0$. On the cardioid at $\theta = \frac{\pi}{3}$: $r = 3$, $\frac{dr}{d\theta} = -\sqrt{3}$, so $\frac{dy}{d\theta}$$= -\sqrt{3}\cdot\frac{\sqrt{3}}{2} + 3 \cdot \frac{1}{2}$$= 0$ while $\frac{dx}{d\theta}$$= -\frac{\sqrt{3}}{2} - \frac{3\sqrt{3}}{2}$$= -2\sqrt{3}$. That is the top of the curve.

At the pole. If $r = 0$ at $\theta = \alpha$ and $\frac{dr}{d\theta} \ne 0$ there, the formula simplifies to $\frac{dy}{dx} = \tan\alpha$: the curve leaves the pole along the ray $\theta = \alpha$.

What $\frac{dr}{d\theta}$ tells you

$\frac{dr}{d\theta}$ is the rate of change of $r$, the signed distance from the pole. It is not the slope. Its sign, read together with the sign of $r$, says whether the point is moving toward or away from the pole:

The short rule: the same sign means away, opposite signs mean toward. On the exam, give both signs as your reason: "$r(2) > 0$ and $r'(2) < 0$, so the point is moving toward the origin."

Rates in time and distances from the axes

If $\theta$ depends on time, the chain rule converts: $\frac{dr}{dt} = \frac{dr}{d\theta} \cdot \frac{d\theta}{dt}$, and likewise for $x$ and $y$. Distance from the $y$-axis is $|x| = |r\cos\theta|$; distance from the $x$-axis is $|y| = |r\sin\theta|$. On calculator questions, write the setup ("$x = r(\theta)\cos\theta$") before the number, use radian mode, and round to three decimals only at the end.