Polar Coordinates and Polar Graphs

Q: What are polar coordinates?

A pair (r, θ) that locates a point by its distance r from the origin and the angle θ it makes with the positive x-axis.

Q: How do I convert from rectangular to polar?

r = √(x² + y²), and θ satisfies tan θ = y/x. Choose the θ that points into the same quadrant as the point.

Q: What does a negative r mean?

Go the distance |r| in the direction opposite to θ. So (−2, π/4) is the same point as (2, 5π/4).

Q: How many petals does a rose r = cos nθ have?

n petals when n is odd and 2n petals when n is even. So r = cos 3θ has 3 and r = cos 2θ has 4.

Quick answer

Polar coordinates locate a point by its distance r from the origin and its angle θ from the positive x-axis. The conversions come from a right triangle: x = r cos θ and y = r sin θ, and in reverse r² = x² + y² and tan θ = y/x, with θ in the point's quadrant. A point has many polar names, since adding 2π to θ, or negating r and adding π, lands on the same spot. Polar equations describe curves around the origin: r = 2 is a circle, r = 1 + cos θ a cardioid and r = cos 2θ a four-petal rose.

What you'll learn

Plot points given in polar coordinates, including negative r
Convert points and equations between polar and rectangular form
Graph polar equations from a table of values
Recognize circles, cardioids, limaçons and roses

A distance and a direction

Rectangular coordinates say how far to go across and how far up. Polar coordinates $(r, \theta)$ say which way to face and how far to walk. Turn through the angle $\theta$ from the positive $x$ -axis, counterclockwise, then go a distance $r$ from the origin.

Polar coordinates: a distance and a direction

The circles of constant $r$ and the rays of constant $\theta$ form the polar grid, the way vertical and horizontal lines form the rectangular one.

Converting

The point at distance $r$ and angle $\theta$ is the corner of a right triangle with hypotenuse $r$ , so

x = r\cos\theta \qquad y = r\sin\theta \qquad\qquad r^2 = x^2 + y^2 \qquad \tan\theta = \frac{y}{x}

Going from rectangular to polar, $\tan\theta = \tfrac{y}{x}$ has two solutions between $0$ and $2\pi$ , pointing in opposite directions. Pick the one in the point’s quadrant.

Why a point has many polar names

Directions to a place are not unique. Turning a full extra circle, $2\pi$ , faces the same way, so $\left(2, \tfrac{\pi}{3}\right)$ and $\left(2, \tfrac{7\pi}{3}\right)$ are the same point. A negative $r$ means walking backward: face $\theta$ , then go $\lvert r \rvert$ in the opposite direction. So $\left(-2, \tfrac{4\pi}{3}\right)$ is that point too. Polar coordinates are directions to a point, and many sets of directions lead to the same place. Rectangular coordinates never have this ambiguity.

Polar graphs

A polar equation $r = f(\theta)$ gives a distance for every direction. Its graph is traced as $\theta$ turns: at each angle, mark the point at distance $f(\theta)$ .

r = 2
r = 1 + cos θ
r = cos 2θ

Three polar graphs

Some families to know:

Equation	Graph
$r = a$	circle of radius $a$ around the origin
$r = 2a\cos\theta$ or $r = 2a\sin\theta$	circle of radius $a$ through the origin
$r = a + a\cos\theta$	cardioid
$r = a + b\cos\theta$ , $a \ne b$	limaçon, with an inner loop when $b > a$
$r = a\cos n\theta$	rose with $n$ petals if $n$ is odd, $2n$ if $n$ is even

Worked examples

Example 2 — A cardioid from a table

Graph $r = 1 + \cos\theta$ .

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

Starting $2$ units to the right, the curve swings up through $(0, 1)$ , shrinks to the origin at $\theta = \pi$ , and returns symmetrically below: the heart-shaped curve in the graph.

Common mistakes

Practice problems

Convert $\left(4, \tfrac{\pi}{6}\right)$ to rectangular coordinates.

Answer

$\left(2\sqrt{3}, 2\right)$

Full solution

$x = 4\cos\tfrac{\pi}{6} = 4 \cdot \tfrac{\sqrt{3}}{2}$ and $y = 4\sin\tfrac{\pi}{6} = 4 \cdot \tfrac{1}{2}$ .
Convert $(3, \pi)$ to rectangular coordinates.

Answer

$(-3, 0)$

Full solution

$x = 3\cos\pi = -3$ and $y = 3\sin\pi = 0$ .
Convert $(0, 5)$ to polar coordinates.

Answer

$\left(5, \tfrac{\pi}{2}\right)$

Full solution

The point is $5$ units straight up, so $r = 5$ and $\theta = \tfrac{\pi}{2}$ .
Convert $(-\sqrt{3}, -1)$ to polar coordinates with $0 \le \theta < 2\pi$ .

Answer

$\left(2, \tfrac{7\pi}{6}\right)$

Full solution

$r = \sqrt{3 + 1} = 2$ , and $\tan\theta = \tfrac{1}{\sqrt{3}}$ . The point is in quadrant III, so $\theta = \pi + \tfrac{\pi}{6}$ .
Give two other polar names for $\left(2, \tfrac{\pi}{4}\right)$ , one with a negative $r$ .

Answer

For example, $\left(2, \tfrac{9\pi}{4}\right)$ and $\left(-2, \tfrac{5\pi}{4}\right)$

Full solution

Adding $2\pi$ to the angle faces the same way. Facing the opposite way, $\tfrac{5\pi}{4}$ , and walking backward $2$ units also reaches the point.
Write $r = 6\sin\theta$ in rectangular form and identify the curve.

Answer

$x^2 + (y - 3)^2 = 9$ : a circle of radius $3$ centered at $(0, 3)$

Full solution

Multiply by $r$ : $x^2 + y^2 = 6y$ . Completing the square gives $x^2 + (y - 3)^2 = 9$ .
Write $x^2 + y^2 = 25$ in polar form.

Answer

$r = 5$

Full solution

$x^2 + y^2 = r^2$ , so $r^2 = 25$ , and $r = 5$ traces the whole circle.
Write the line $y = x$ in polar form.

Answer

$\theta = \tfrac{\pi}{4}$

Full solution

Every point on the line makes an angle of $\tfrac{\pi}{4}$ with the $x$ -axis, or the opposite direction, which negative values of $r$ cover.
How many petals do $r = \sin 4\theta$ and $r = \cos 5\theta$ have?

Answer

$8$ and $5$

Full solution

An even multiple, $4$ , gives $2 \cdot 4 = 8$ petals; an odd one, $5$ , gives $5$ .
A student converts $(-2, -2)$ to polar coordinates as $\left(2\sqrt{2}, \tfrac{\pi}{4}\right)$ . What went wrong?

Hint

Which quadrant does the angle $\tfrac{\pi}{4}$ point into?

Answer

The angle points into quadrant I, but the point is in quadrant III. The answer is $\left(2\sqrt{2}, \tfrac{5\pi}{4}\right)$ .

Full solution

$r = \sqrt{4 + 4} = 2\sqrt{2}$ is right. But $\tan\theta = 1$ holds for both $\tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$ , and only $\tfrac{5\pi}{4}$ points down and to the left.

The student’s point, $\left(2\sqrt{2}, \tfrac{\pi}{4}\right)$ , is $(2, 2)$ .

Frequently asked questions

What are polar coordinates?