Polar Coordinates and Graphs — AP® Calculus BC

Part 1 · Hook

Why this matters

A lighthouse beam sweeps around and a boat is spotted 3 miles out along it. That is exactly how polar coordinates name a point: an angle to turn and a distance to go. Some curves, like flower petals and heart shapes, have messy equations in

x

and

y

but simple ones in polar form, and Unit 9 asks for their slopes and areas.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What are $\cos\frac{\pi}{3}$ and $\sin\frac{\pi}{3}$ ?

$\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2}$ and $\frac{1}{2}$
$\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{2}}{2}$
$1$ and $0$

Show the answer

On the unit circle the point at angle $\frac{\pi}{3}$ is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ .

Correct: $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ :
$\frac{\sqrt{3}}{2}$ and $\frac{1}{2}$ :
$\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{2}}{2}$ :
$1$ and $0$ :

2. A curve is given by $x = 2\cos t$ , $y = 2\sin t$ . What shape is it?

A circle of radius 2 centered at the origin
A line through the origin
A parabola
A circle of radius 4 centered at the origin

Show the answer

$x^2 + y^2$ $= 4\cos^2 t + 4\sin^2 t$ $= 4$ : a circle of radius 2.

Correct: A circle of radius 2 centered at the origin:
A line through the origin:
A parabola:
A circle of radius 4 centered at the origin:

Part 4 · See it

See it first

A polar grid with circles of radius 1 to 4 and rays every π/6. Point P has r = 3 and θ = π/3: its legs are x = r cos θ = 1.5 and y = r sin θ = 3√3/2, about 2.598. Point Q has r = −2 and θ = π/4: it lies 2 units from the pole on the ray opposite θ = π/4, at x = −√2, y = −√2. A side panel lists x = r cos θ, y = r sin θ and r² = x² + y². — A point is a turn and a distance: $P$ is 3 units along the ray $\theta = \frac{\pi}{3}$ ; $Q$ , with $r = -2$ , is 2 units along the opposite ray. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Turn to the angle $\theta$ and go a distance $r$ you reach the point $(r, \theta)$
Drop the vertical and horizontal legs of that right triangle $x = r\cos\theta$ and $y = r\sin\theta$
Let $r$ depend on $\theta$ , as $r = f(\theta)$ as $\theta$ turns, the point traces a polar curve
Make a table of $\theta$ and $r$ , then plot each pointyou see where $r = 0$ (the pole) and where $r < 0$ (the opposite ray)

Part 6 · Key ideas

Key ideas

A polar point $(r, \theta)$ : turn to the ray at angle $\theta$ , then go $r$ units. A negative $r$ goes backward, through the pole, onto the opposite ray.
$x = r\cos\theta$ , $y = r\sin\theta$ , $r^2 = x^2 + y^2$ . These connect every polar question to rectangular ones.
A polar curve $r = f(\theta)$ is a parametric curve with parameter $\theta$ : $x = f(\theta)\cos\theta$ , $y = f(\theta)\sin\theta$ .
Shapes to know: $r = a$ (circle), $\theta = c$ (line through the pole), $r = a\cos\theta$ and $r = a\sin\theta$ (circles through the pole), $r = a + b\cos\theta$ (limaçons, cardioid when $a = b$ ), $r = a\sin(n\theta)$ (roses).

Part 7 · Misconception

A common mistake

The wrong idea: A point with $r = -2$ cannot exist: a distance is never negative.

What actually happens: $r$ is a signed distance along the ray. $(-2, \frac{\pi}{4})$ is 2 units from the pole on the ray opposite $\theta = \frac{\pi}{4}$ , the same point as $(2, \frac{5\pi}{4})$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Which rectangular point is the polar point $\left(4, \frac{\pi}{6}\right)$ ?

$\left(2, 2\sqrt{3}\right)$
$\left(2\sqrt{3}, 2\right)$
$\left(4, \frac{\pi}{6}\right)$
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

Show the answer

$x$ $= r\cos\theta$ $= 4 \cdot \frac{\sqrt{3}}{2}$ $= 2\sqrt{3}$ and $y = r\sin\theta = 4 \cdot \frac{1}{2} = 2$ .

$\left(2, 2\sqrt{3}\right)$ : Sine and cosine are swapped: $x$ uses $\cos\theta$ , and $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ .
Correct: $\left(2\sqrt{3}, 2\right)$ : Right: $x = 4\cos\frac{\pi}{6} = 2\sqrt{3}$ and $y = 4\sin\frac{\pi}{6} = 2$ .
$\left(4, \frac{\pi}{6}\right)$ : This repeats the polar pair; convert with $x = r\cos\theta$ and $y = r\sin\theta$ .
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ : This is the point on the unit circle; multiply each coordinate by $r = 4$ .

2. Where is the polar point $\left(-3, \frac{\pi}{2}\right)$ in rectangular coordinates?

$(0, 3)$
$(-3, 0)$
$\left(-3, \frac{\pi}{2}\right)$
$(0, -3)$

Show the answer

$x = -3\cos\frac{\pi}{2} = 0$ and $y = -3\sin\frac{\pi}{2} = -3$ : the point is on the ray opposite $\theta = \frac{\pi}{2}$ .

$(0, 3)$ : That is $\left(3, \frac{\pi}{2}\right)$ . A negative $r$ puts the point on the opposite ray.
$(-3, 0)$ : That is $(3, \pi)$ ; the ray $\theta = \frac{\pi}{2}$ is vertical, so the point stays on the $y$ -axis.
$\left(-3, \frac{\pi}{2}\right)$ : This repeats the polar pair; convert with $x = r\cos\theta$ and $y = r\sin\theta$ .
Correct: $(0, -3)$ : Right: face straight up, then walk 3 units backward.

Table

Values of $r = 1 + 2\cos\theta$

The polar curve $r = 1 + 2\cos\theta$ at five angles ( $\theta$ in radians)
$\theta$	$r$
$0$	$3$
$\frac{\pi}{3}$	$2$
$\frac{\pi}{2}$	$1$
$\frac{2\pi}{3}$	$0$
$\pi$	$-1$

3. Where is the point of the curve at $\theta = \pi$ , in rectangular coordinates?

$(-1, 0)$
$(1, 0)$
$(0, -1)$
$(-1, \pi)$

Show the answer

$x$ $= r\cos\theta$ $= (-1)\cos\pi$ $= 1$ and $y$ $= r\sin\theta$ $= (-1)\sin\pi$ $= 0$ .

$(-1, 0)$ : That would be $r = 1$ on the ray $\theta = \pi$ . Here $r = -1$ , so the point is on the opposite ray.
Correct: $(1, 0)$ : Right: $r = -1$ on the ray $\theta = \pi$ lands one unit to the right of the pole.
$(0, -1)$ : The ray $\theta = \pi$ is horizontal; with $\sin\pi = 0$ the point stays on the $x$ -axis.
$(-1, \pi)$ : This mixes polar and rectangular. Use $x$ $= r\cos\theta$ $= (-1)(-1)$ $= 1$ and $y = r\sin\theta = 0$ .

4. At which angle in the table does the curve pass through the pole?

$\theta = \frac{2\pi}{3}$
$\theta = 0$
$\theta = \frac{\pi}{2}$
$\theta = \pi$

Show the answer

The pole is the point with $r = 0$ . In the table $r = 0$ at $\theta = \frac{2\pi}{3}$ (check: $1 + 2\cos\frac{2\pi}{3}$ $= 1 - 1$ $= 0$ ).

Correct: $\theta = \frac{2\pi}{3}$ : Right: $r = 0$ there, and $r = 0$ is the pole whatever the angle.
$\theta = 0$ : At $\theta = 0$ the point is 3 units from the pole, the farthest point in the table.
$\theta = \frac{\pi}{2}$ : At $\theta = \frac{\pi}{2}$ , $r = 1$ : the point $(0, 1)$ , one unit from the pole.
$\theta = \pi$ : At $\theta = \pi$ , $r = -1$ : one unit from the pole, on the opposite ray.

5. Find the $x$ -coordinate of the point of the curve at $\theta = \frac{\pi}{3}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$x$ $= r\cos\theta$ $= 2\cos\frac{\pi}{3}$ $= 2 \cdot \frac{1}{2}$ $= 1$ .

Answer: $1$

6. For $\frac{2\pi}{3} < \theta < \pi$ , $r$ is negative. In which quadrant are the points of the curve for these angles?

The second quadrant
The third quadrant
The fourth quadrant
On the negative $x$ -axis

Show the answer

For $\frac{2\pi}{3} < \theta < \pi$ the ray points into the second quadrant, but $r < 0$ , so the point is on the opposite ray, in the fourth quadrant. This is the lower half of the small inner loop.

The second quadrant: The rays point into the second quadrant, but $r < 0$ puts each point on the opposite ray.
The third quadrant: Reflecting through the pole turns the second quadrant into the fourth, not the third.
Correct: The fourth quadrant: Right: these rays point into the second quadrant, and a negative $r$ reflects each point through the pole into the fourth.
On the negative $x$ -axis: Only $\theta = \pi$ gives a point on an axis, and there $r = -1$ gives $(1, 0)$ on the positive $x$ -axis.

Part 9 · Summary

Summary

Polar coordinates name a point by an angle

\theta

and a signed distance

r

. Convert with

x = r\cos\theta

and

y = r\sin\theta

. A polar curve

r = f(\theta)

is plotted from a table of

\theta

and

r

; watch for

r = 0

(the curve meets the pole) and

r < 0

(the point is on the opposite ray).

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of r=1+2cos⁡θr = 1 + 2\cos\theta

Summary

What comes next

Connections

Values of $r = 1 + 2\cos\theta$