Polar Coordinates and Derivatives

Rectangular coordinates locate a point by going across and then up. Polar coordinates locate it by turning to face a direction and then walking a distance. Curves that are messy in $x$ and $y$, like flowers and heart shapes, often have very simple polar equations. Because $x$ and $y$ can be written in terms of $\theta$, finding slopes is just parametric differentiation with $\theta$ as the parameter.

Key ideas

Polar coordinates

The point $(r, \theta)$ is found by turning through angle $\theta$ from the positive $x$-axis (counterclockwise is positive) and moving a distance $r$ from the origin, which is called the pole. Angles are in radians: $\dfrac{\pi}{2}$ is a quarter turn and $\pi$ is a half turn.

If $r$ is negative, you face direction $\theta$ and then walk backwards. So $(-2, 0)$ is the same point as $(2, \pi)$. Every point has many polar names: adding $2\pi$ to $\theta$ gives the same point.

Converting

Polar to rectangular	Rectangular to polar
$x = r\cos\theta$	$r^2 = x^2 + y^2$
$y = r\sin\theta$	$\tan\theta = \dfrac{y}{x}$ (check the quadrant)

To convert an equation, the substitutions $r\cos\theta = x$, $r\sin\theta = y$ and $r^2 = x^2 + y^2$ are the main tools. Multiplying both sides by $r$ often helps.

A gallery of polar curves

Equation	Shape
$r = a$	circle of radius $\lvert a \rvert$ centred at the pole
$\theta = \alpha$	line through the pole
$r = 2a\cos\theta$, $r = 2a\sin\theta$	circle of radius $\lvert a \rvert$ through the pole (traced once for $0 \le \theta \le \pi$)
$r = a + b\cos\theta$ or $a + b\sin\theta$	limaçon: a cardioid (heart) if $\lvert a \rvert = \lvert b \rvert$, an inner loop if $\lvert a \rvert \lt \lvert b \rvert$
$r = a\cos(n\theta)$ or $a\sin(n\theta)$	rose: $n$ petals if $n$ is odd, $2n$ petals if $n$ is even; each petal has length $\lvert a \rvert$

When in doubt, make a table of $\theta$ and $r$ values and plot the points, or graph it on your calculator in polar mode.

Four standard polar curves: a circle through the pole, a cardioid, a limaçon with an inner loop, and a three-petal rose.

Slopes of polar curves

A polar curve $r = f(\theta)$ is a parametric curve with parameter $\theta$:

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

Differentiate each with the product rule, then divide:

$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}$$

You don’t need to memorize the big fraction: just write $x$ and $y$ in terms of $\theta$ and differentiate. Horizontal and vertical tangents work as before: horizontal where $\dfrac{dy}{d\theta} = 0$ and $\dfrac{dx}{d\theta} \ne 0$, vertical where $\dfrac{dx}{d\theta} = 0$ and $\dfrac{dy}{d\theta} \ne 0$.

Watch out: $\dfrac{dr}{d\theta}$ is not the slope of the curve.

What dr/dθ tells you

The distance from the pole is $\lvert r \rvert$. So $\dfrac{dr}{d\theta}$ tells you whether the point is moving toward or away from the pole as $\theta$ increases:

• $r \gt 0$ and $\dfrac{dr}{d\theta} \gt 0$: moving away from the pole.
• $r \gt 0$ and $\dfrac{dr}{d\theta} \lt 0$: moving toward the pole.
• If $r \lt 0$, it’s reversed. In short: same signs means away, opposite signs means toward.

Similarly, the sign of $\dfrac{dy}{d\theta}$ says whether $y$ is increasing (moving up), and the sign of $\dfrac{dx}{d\theta}$ says whether $x$ is increasing (moving right). AP questions often ask you to interpret these in a sentence.

Worked examples

Example 1: Converting points and equations

• (a) Convert the polar point $\left( 4, \dfrac{2\pi}{3} \right)$ to rectangular coordinates.
• (b) Convert the rectangular point $(-3, 3)$ to polar coordinates with $r \gt 0$ and $0 \le \theta \lt 2\pi$.
• (c) Write $r = 6\sin\theta$ in rectangular form and describe the curve.

Solution. (a) $x = 4\cos\dfrac{2\pi}{3} = 4\left( -\dfrac{1}{2} \right) = -2$ and $y = 4\sin\dfrac{2\pi}{3} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$. The point is $\left( -2, 2\sqrt{3} \right)$.

(b) $r = \sqrt{9 + 9} = 3\sqrt{2}$. $\tan\theta = \dfrac{3}{-3} = -1$, and the point is in quadrant II, so $\theta = \dfrac{3\pi}{4}$ (not $-\dfrac{\pi}{4}$, which points into quadrant IV). The point is $\left( 3\sqrt{2}, \dfrac{3\pi}{4} \right)$.

(c) Multiply both sides by $r$, then substitute:

$$r^2 = 6r\sin\theta \quad\Rightarrow\quad x^2 + y^2 = 6y \quad\Rightarrow\quad x^2 + (y - 3)^2 = 9$$

It’s a circle with centre $(0, 3)$ and radius $3$, passing through the pole.

Example 2: Sketching a cardioid

Sketch $r = 1 + \cos\theta$ for $0 \le \theta \le 2\pi$.

Solution. Make a table:

$\theta$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$
$r$	$2$	$1$	$0$	$1$	$2$

Start at $(2, 0)$ on the positive $x$-axis. As $\theta$ goes to $\dfrac{\pi}{2}$, $r$ shrinks to $1$, reaching the point $(0, 1)$. At $\theta = \pi$, $r = 0$: the curve reaches the pole. Then it repeats the same shape below the $x$-axis and returns to $(2, 0)$. The result is the heart-shaped cardioid in the figure, with its point at the pole.

Example 3: A tangent line to a polar curve

Find the slope of $r = 1 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$, and the equation of the tangent line there.

Solution. Write $x$ and $y$ in terms of $\theta$:

$$x = (1 + \cos\theta)\cos\theta = \cos\theta + \cos^2\theta, \qquad y = (1 + \cos\theta)\sin\theta = \sin\theta + \sin\theta\cos\theta$$

Differentiate:

$$\frac{dx}{d\theta} = -\sin\theta - 2\sin\theta\cos\theta, \qquad \frac{dy}{d\theta} = \cos\theta + \cos^2\theta - \sin^2\theta$$

At $\theta = \dfrac{\pi}{2}$ (where $\sin\theta = 1$, $\cos\theta = 0$):

$$\frac{dx}{d\theta} = -1 - 0 = -1, \qquad \frac{dy}{d\theta} = 0 + 0 - 1 = -1, \qquad \frac{dy}{dx} = \frac{-1}{-1} = 1$$

The point is $r = 1$ at $\theta = \dfrac{\pi}{2}$, which is $(x, y) = (0, 1)$. The tangent line is $y = x + 1$.

Example 4: Toward or away from the pole?

For $r = 2 + 3\cos\theta$, is the point on the curve moving toward or away from the pole at (a) $\theta = \dfrac{\pi}{2}$, (b) $\theta = \dfrac{3\pi}{4}$? Give a reason.

Solution. $\dfrac{dr}{d\theta} = -3\sin\theta$.

(a) At $\theta = \dfrac{\pi}{2}$: $r = 2 \gt 0$ and $\dfrac{dr}{d\theta} = -3 \lt 0$. Since $r \gt 0$ and $r$ is decreasing, the distance from the pole is decreasing: the point is moving toward the pole.

(b) At $\theta = \dfrac{3\pi}{4}$: $r = 2 - \dfrac{3\sqrt{2}}{2} \approx -0.121 \lt 0$ and $\dfrac{dr}{d\theta} = -\dfrac{3\sqrt{2}}{2} \lt 0$. Now $r$ is negative and getting more negative, so $\lvert r \rvert$ is increasing: the point is moving away from the pole. (Same signs, so away.) This part of the curve is the limaçon’s inner loop.

Common mistakes

Thinking dr/dθ is the slope. $\dfrac{dr}{d\theta}$ is how fast the distance from the pole changes, not $\dfrac{dy}{dx}$. For the slope you must go through $x = r\cos\theta$ and $y = r\sin\theta$.

Forgetting the product rule. $y = f(\theta)\sin\theta$ is a product, so $\dfrac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta$. Dropping either term is a frequent error.

Taking θ straight from the calculator’s inverse tangent. $\tan^{-1}\left( \dfrac{y}{x} \right)$ only gives angles between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$. For points in quadrants II and III, add $\pi$.

Ignoring negative r. When $r \lt 0$, the point is on the opposite side of the pole, and the “toward or away” rule reverses. Check the signs of both $r$ and $\dfrac{dr}{d\theta}$.

Degree mode and degree angles. Polar work in AP calculus uses radians throughout. If you learned trig in degrees, convert: $30^\circ = \dfrac{\pi}{6}$, $45^\circ = \dfrac{\pi}{4}$, $60^\circ = \dfrac{\pi}{3}$, $90^\circ = \dfrac{\pi}{2}$.

Practice

1. (Warm-up) Convert $\left( 6, \dfrac{\pi}{6} \right)$ to rectangular coordinates.

Solution

$$x = 6\cos\frac{\pi}{6} = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3}, \qquad y = 6\sin\frac{\pi}{6} = 6 \cdot \frac{1}{2} = 3$$

The point is $\left( 3\sqrt{3}, 3 \right)$.

2. (Warm-up) Convert $\left( -1, \sqrt{3} \right)$ to polar coordinates with $r \gt 0$ and $0 \le \theta \lt 2\pi$.

Solution

$r = \sqrt{1 + 3} = 2$. $\tan\theta = -\sqrt{3}$ with the point in quadrant II, so $\theta = \dfrac{2\pi}{3}$. The point is $\left( 2, \dfrac{2\pi}{3} \right)$.

3. (Warm-up) Write $r = 4\cos\theta$ in rectangular form and describe the curve.

Solution

$$r^2 = 4r\cos\theta \quad\Rightarrow\quad x^2 + y^2 = 4x \quad\Rightarrow\quad (x - 2)^2 + y^2 = 4$$

A circle with centre $(2, 0)$ and radius $2$, passing through the pole.

4. (Core) Write each rectangular equation in polar form.

• (a) $x + y = 4$
• (b) $x^2 + y^2 = 10x$

Solution

(a) $r\cos\theta + r\sin\theta = 4$, so $r = \dfrac{4}{\cos\theta + \sin\theta}$.

(b) $r^2 = 10r\cos\theta$, so $r = 10\cos\theta$ (dividing by $r$ loses nothing, since $r = 10\cos\theta$ already passes through the pole at $\theta = \tfrac{\pi}{2}$).

5. (Core) For the rose $r = 4\sin(3\theta)$:

• (a) How many petals does it have, and how long is each?
• (b) The first petal is traced for $0 \le \theta \le \dfrac{\pi}{3}$. At what angle is its tip?
• (c) How many petals does $r = 4\sin(2\theta)$ have?

Solution

(a) $n = 3$ is odd, so $3$ petals. The largest value of $\lvert r \rvert$ is $4$, so each petal has length $4$.

(b) $r = 0$ at $\theta = 0$ and $\theta = \dfrac{\pi}{3}$, and $r$ is largest when $\sin(3\theta) = 1$, at $3\theta = \dfrac{\pi}{2}$, so $\theta = \dfrac{\pi}{6}$.

(c) $n = 2$ is even, so $2n = 4$ petals.

6. (Core) Find the tangent line to the spiral $r = \theta$ at $\theta = \dfrac{\pi}{2}$.

Solution

$x = \theta\cos\theta$ and $y = \theta\sin\theta$, so

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

At $\theta = \dfrac{\pi}{2}$: $\dfrac{dx}{d\theta} = 0 - \dfrac{\pi}{2} = -\dfrac{\pi}{2}$ and $\dfrac{dy}{d\theta} = 1 + 0 = 1$, so

$$\frac{dy}{dx} = \frac{1}{-\pi/2} = -\frac{2}{\pi}$$

The point is $(x, y) = \left( 0, \dfrac{\pi}{2} \right)$. Tangent line: $y = -\dfrac{2}{\pi}x + \dfrac{\pi}{2}$.

7. (Core) Find the points on $r = 2\cos\theta$, $0 \le \theta \lt \pi$, where the tangent line is horizontal.

Solution

$x = 2\cos^2\theta$ and $y = 2\sin\theta\cos\theta = \sin(2\theta)$, so

$$\frac{dx}{d\theta} = -4\cos\theta\sin\theta = -2\sin(2\theta), \qquad \frac{dy}{d\theta} = 2\cos(2\theta)$$

$\dfrac{dy}{d\theta} = 0$ when $2\theta = \dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$, so $\theta = \dfrac{\pi}{4}$ or $\dfrac{3\pi}{4}$. There $\dfrac{dx}{d\theta} = -2$ and $2$, both nonzero.

• $\theta = \dfrac{\pi}{4}$: $(x, y) = (1, 1)$.
• $\theta = \dfrac{3\pi}{4}$: $(x, y) = (1, -1)$.

These are the top and bottom of the circle with centre $(1, 0)$ and radius $1$, as expected.

8. (Challenge) For the limaçon $r = 1 + 2\sin\theta$, decide whether the point on the curve is moving toward or away from the pole at (a) $\theta = \dfrac{5\pi}{6}$ and (b) $\theta = \dfrac{4\pi}{3}$. Justify each answer.

Solution

$\dfrac{dr}{d\theta} = 2\cos\theta$.

(a) $r = 1 + 2 \cdot \dfrac{1}{2} = 2 \gt 0$ and $\dfrac{dr}{d\theta} = 2\left( -\dfrac{\sqrt{3}}{2} \right) = -\sqrt{3} \lt 0$. Opposite signs: $\lvert r \rvert$ is decreasing, so the point is moving toward the pole.

(b) $r = 1 + 2\left( -\dfrac{\sqrt{3}}{2} \right) = 1 - \sqrt{3} \approx -0.732 \lt 0$ and $\dfrac{dr}{d\theta} = 2\left( -\dfrac{1}{2} \right) = -1 \lt 0$. Same signs: $\lvert r \rvert$ is increasing, so the point is moving away from the pole.

9. (Challenge) Find all points on the cardioid $r = 1 + \cos\theta$, $0 \le \theta \lt 2\pi$, where the tangent line is vertical.

Solution

From Example 3,

$$\frac{dx}{d\theta} = -\sin\theta - 2\sin\theta\cos\theta = -\sin\theta(1 + 2\cos\theta), \qquad \frac{dy}{d\theta} = \cos\theta + \cos(2\theta)$$

$\dfrac{dx}{d\theta} = 0$ when $\sin\theta = 0$ ($\theta = 0, \pi$) or $\cos\theta = -\dfrac{1}{2}$ ($\theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$). Check $\dfrac{dy}{d\theta}$ at each:

• $\theta = 0$: $1 + 1 = 2 \ne 0$. Vertical tangent at $r = 2$: the point $(2, 0)$.
• $\theta = \dfrac{2\pi}{3}$: $-\dfrac{1}{2} - \dfrac{1}{2} = -1 \ne 0$. Here $r = \dfrac{1}{2}$, so the point is $\left( \dfrac{1}{2}\cos\dfrac{2\pi}{3}, \dfrac{1}{2}\sin\dfrac{2\pi}{3} \right) = \left( -\dfrac{1}{4}, \dfrac{\sqrt{3}}{4} \right)$.
• $\theta = \dfrac{4\pi}{3}$: also $-1 \ne 0$. The point is $\left( -\dfrac{1}{4}, -\dfrac{\sqrt{3}}{4} \right)$.
• $\theta = \pi$: $-1 + 1 = 0$. Both derivatives are $0$ (this is the pole), so this needs more work. Using L’Hôpital’s rule, $\displaystyle\lim_{\theta \to \pi} \frac{dy/d\theta}{dx/d\theta} = 0$, so the tangent at the cusp is horizontal, not vertical.

Vertical tangents: $(2, 0)$, $\left( -\dfrac{1}{4}, \dfrac{\sqrt{3}}{4} \right)$ and $\left( -\dfrac{1}{4}, -\dfrac{\sqrt{3}}{4} \right)$.