Math Core

A polar curve $r = f(\theta)$ describes a point by its distance from the origin and its direction. To find the slope of such a curve, you rewrite it as a parametric curve with $\theta$ as the parameter, and then everything from the start of this unit applies.

Polar curves are parametric curves

Recall that a point with polar coordinates $(r, \theta)$ has rectangular coordinates $x = r\cos\theta$ and $y = r\sin\theta$ . If $r = f(\theta)$ , substitute:

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

These are parametric equations with parameter $\theta$ . So the slope is $\dfrac{dy/d\theta}{dx/d\theta}$ , and each of those derivatives needs the product rule.

Slope of a polar curve

For $r = f(\theta)$ , write $r' = \dfrac{dr}{d\theta}$ . Then

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

and

\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}.

You don't need to memorize the last formula if you remember where it comes from: write $x$ and $y$ in terms of $\theta$ , differentiate each with the product rule, and divide.

Worked example: Slope of a cardioid

Find the slope of $r = 1 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .

At $\theta = \dfrac{\pi}{2}$ : $r = 1 + 0 = 1$ and $r' = -\sin\theta = -1$ .

\begin{aligned} \frac{dy}{d\theta} &= (-1)(1) + (1)(0) = -1 \\ \frac{dx}{d\theta} &= (-1)(0) - (1)(1) = -1 \end{aligned}

So $\dfrac{dy}{dx} = \dfrac{-1}{-1} = 1$ . The point is $(x, y) = (0, 1)$ , and the cardioid crosses the positive $y$ -axis there with slope $1$ .

The cardioid r = 1 + cos θ. The dashed line is the tangent at θ = π/2, through (0, 1) with slope 1. The other marked point, at θ = π/3, has a horizontal tangent.Open in grapher →

Horizontal and vertical tangents

Exactly as for parametric curves:

Horizontal tangent where $\dfrac{dy}{d\theta} = 0$ and $\dfrac{dx}{d\theta} \ne 0$ .
Vertical tangent where $\dfrac{dx}{d\theta} = 0$ and $\dfrac{dy}{d\theta} \ne 0$ .

Worked example: A horizontal tangent on the cardioid

Show that $r = 1 + \cos\theta$ has a horizontal tangent at $\theta = \dfrac{\pi}{3}$ , and find the point.

At $\theta = \dfrac{\pi}{3}$ : $r = 1 + \dfrac{1}{2} = \dfrac{3}{2}$ and $r' = -\sin\dfrac{\pi}{3} = -\dfrac{\sqrt{3}}{2}$ .

\begin{aligned} \frac{dy}{d\theta} &= \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{3}{2}\right)\left(\frac{1}{2}\right) = -\frac{3}{4} + \frac{3}{4} = 0 \\ \frac{dx}{d\theta} &= \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{3}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \ne 0 \end{aligned}

So the tangent is horizontal. The point is $x = \dfrac{3}{2}\cos\dfrac{\pi}{3} = \dfrac{3}{4}$ , $y = \dfrac{3}{2}\sin\dfrac{\pi}{3} = \dfrac{3\sqrt{3}}{4} \approx 1.299$ . That's the highest point of the cardioid.

Common mistake

$\dfrac{dr}{d\theta}$ is not the slope of the curve. It measures how fast the distance from the origin changes as $\theta$ increases. In the first example, $\dfrac{dr}{d\theta} = -1$ but the slope $\dfrac{dy}{dx}$ was $+1$ . Always convert to $x$ and $y$ before talking about slope.

What dr/dθ does tell you

Although $\dfrac{dr}{d\theta}$ isn't the slope, it has its own meaning that AP questions ask about directly.

If $\dfrac{dr}{d\theta} > 0$ (and $r > 0$ ), the curve is moving away from the origin as $\theta$ increases.
If $\dfrac{dr}{d\theta} < 0$ (and $r > 0$ ), the curve is moving toward the origin.

When $r$ is negative, the point is on the opposite side of the pole, and the distance from the origin is $|r|$ . Then $r$ increasing toward $0$ means the point is getting closer. So in general, the distance from the origin is increasing when $r$ and $\dfrac{dr}{d\theta}$ have the same sign.

Worked example: Moving toward or away from the origin

For $r = 3 - 2\cos\theta$ , is the curve getting closer to or farther from the origin at $\theta = \dfrac{\pi}{2}$ ?

$r' = 2\sin\theta$ , so $r'\left(\dfrac{\pi}{2}\right) = 2$ . Also $r\left(\dfrac{\pi}{2}\right) = 3 > 0$ . Both are positive, so the distance from the origin is increasing: the curve is moving away from the origin at a rate of 2 units per radian.

Tangent lines to polar curves

To write a tangent line, you need a point in $(x, y)$ form and the slope. Convert the point with $x = r\cos\theta$ , $y = r\sin\theta$ , find $\dfrac{dy}{dx}$ from the key formula, and use point-slope form.

Tip

When the slope formula gets messy, compute $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$ as two separate numbers first, then divide. Plugging the angle in early is much less error-prone than simplifying the general formula.

Practice

Practice 1

Find the slope of the circle $r = 4\sin\theta$ at $\theta = \dfrac{\pi}{6}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find $\dfrac{dy}{dx}$ for $r = 2 + \sin\theta$ at $\theta = 0$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

The curve $r = 1 + \cos\theta$ passes through which point $(x, y)$ when $\theta = \dfrac{\pi}{2}$ ?

Enter a point like (2, -3)

Practice 4

For the curve $r = 2 + \cos(2\theta)$ , which statement is true at $\theta = \dfrac{\pi}{4}$ ?

Practice 5

Find the slope of the spiral $r = 2\theta$ at $\theta = \dfrac{\pi}{2}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

The circle $r = 2\cos\theta$ has a horizontal tangent at exactly one point with $0 \lt \theta \lt \dfrac{\pi}{2}$ . Find that point in rectangular coordinates $(x, y)$ .

Enter a point like (2, -3)

Practice 7

Find an equation of the line tangent to $r = 3 + 2\sin\theta$ at $\theta = 0$ . Give $y$ in terms of $x$ .

Enter an expression, e.g. 3x^2 - 2x + 1