Topic 3.13A

Trigonometry and Polar Coordinates

Rectangular coordinates $(x, y)$ describe a point by how far it is right and up. But for anything that turns — a pedal, a wheel, a radar sweep — it is more natural to say how far the point is from the center and in which direction. That is the idea of polar coordinates. Part A introduces them; Part B uses the same idea for complex numbers.

6 MIN READ4 IDEAS42 PROBLEMS10 flashcards

Practice 42 problems Next: Chapter 3.13B Trigonometry and Polar Coordinates

Plotting Polar Points

CONCEPT

Polar coordinates $(r, \theta )$

The pole is the origin. The polar axis is the positive x-axis.

$\theta$ is the angle in standard position (3.2A), and $r$ is the directed distance from the pole along the terminal ray of $\theta$ .

To plot $(r, \theta )$ : face the direction $\theta$ , then walk $r$ units. If $r$ is negative, walk $|r|$ units BACKWARD, to the opposite side of the pole.

Two different polar names for the same point: $(3, 5\pi /6)$ and $(-3, -\pi /6)$ .

On a polar grid, the circles mark distances $r = 1$ , 2, 3, … and the spokes mark angles. A point has infinitely many polar names, because adding $2\pi$ to $\theta$ lands on the same ray, and adding $\pi$ to $\theta$ while changing the sign of $r$ also lands on the same point.

Worked example

Example 1. Give four different polar names for the point $(3, 5\pi /6)$ , two with $r > 0$ and two with $r < 0$ .

01
Positive $r$ : add or subtract full turns. $(3, 5\pi /6 + 2\pi ) = (3, 17\pi /6)$ and $(3, 5\pi /6 - 2\pi ) = (3, -7\pi /6)$ .
02
Negative $r$ : the opposite ray is $5\pi /6 + \pi = 11\pi /6$ or $5\pi /6 - \pi = -\pi /6$ . Walking backward 3 units along either one reaches the original point: $(-3, 11\pi /6)$ and $(-3, -\pi /6)$ .
03
Check one by converting (Section 2): $(-3, -\pi /6)$ gives $x = -3 \cos (-\pi /6) = -3\sqrt{3}/2$ and $y = -3 \sin (-\pi /6) = 3/2$ , the same as $(3, 5\pi /6)$ ✓.

COMMON MISTAKE

" $(-3, 5\pi /6)$ is the same point as $(3, 5\pi /6)$ , just written differently."

Changing only the sign of $r$ sends you to the opposite side of the pole. $(-3, 5\pi /6)$ is the point ( $3\sqrt{3}/2$ , −3/2), in Quadrant IV. To keep the same point, a sign change in $r$ must come with a change of $\pi$ in $\theta$ .

Quick check

Give another polar name for $\left(3, \frac{\pi}{4}\right)$ with $r < 0$ .

Polar to Rectangular

Remember from 3.3: the point at distance $r$ along the terminal ray of $\theta$ is $(r \cos \theta , r \sin \theta )$ . That is exactly the conversion formula.

x=r\cos\theta\qquad y=r\sin\theta

Worked example

Example 2. Convert $(6, 5\pi /3)$ and $(8, 3\pi /4)$ to rectangular coordinates.

01
$(6, 5\pi /3)$ : $\cos (5\pi /3) = 1/2$ and $\sin (5\pi /3) = -\sqrt{3}/2$ (Quadrant IV). $x = 6 \cdot 1/2 = 3$ and $y = 6 \cdot (-\sqrt{3}/2) = -3\sqrt{3}$ . The point is (3, $-3\sqrt{3}$ ).
02
$(8, 3\pi /4)$ : $\cos (3\pi /4) = -\sqrt{2}/2$ and $\sin (3\pi /4) = \sqrt{2}/2$ . $x = -4\sqrt{2}$ and $y = 4\sqrt{2}$ . The point is ( $-4\sqrt{2}$ , $4\sqrt{2}$ ).

REAL-LIFE EXAMPLE

The pedal in polar coordinates

Remember the bicycle crank from 3.1: 17 cm long, center 28 cm above the ground. Put the pole at the center of the crank. Then the pedal is always at polar point $(17, \theta )$ , where $\theta$ is the crank's angle — only $\theta$ changes as the rider pedals.

At $\theta = 7\pi /6$ , the pedal is at $x = 17 \cos (7\pi /6) = -17\sqrt{3}/2 \approx -14.72$ and $y = 17 \sin (7\pi /6) = -17/2 = -8.5$ , measured from the crank center.

Its height above the ground is $28 - 8.5 = 19.5$ cm — the same answer as the 3.3 model $28 + 17 \sin \theta$ . That model was a polar-to-rectangular conversion all along.

Rectangular to Polar

Going the other way, $r$ comes from the Pythagorean theorem and $\theta$ comes from the slope of the ray (3.8).

r^2=x^2+y^2\qquad \tan\theta=\frac{y}{x}\ \ (\mathrm{check\ the\ quadrant})

Worked example

Example 3. Convert $(-5, 12)$ to polar coordinates with $r > 0$ and $0 \le \theta < 2\pi$ .

01
$r = \sqrt{(-5)^{2} + 12^{2}} = \sqrt{169} = 13$ .
02
$\tan \theta = 12/(-5)$ . A calculator's $\arctan (-12/5) \approx -1.176$ , but that angle is in Quadrant IV, and the point $(-5, 12)$ is in Quadrant II.
03
Use the reference angle $\alpha = \arctan (12/5) \approx 1.176$ . In Quadrant II, $\theta = \pi - \alpha \approx 1.966$ .
04
Answer: $(13, 1.966)$ approximately. Check: $13 \cos (1.966) \approx -5$ and $13 \sin (1.966) \approx 12$ ✓.

Draw the point first: the picture shows which quadrant $\theta$ must be in.

COMMON MISTAKE

" $\theta = \arctan (y/x)$ , always."

$\arctan$ only returns angles between $-\pi /2$ and $\pi /2$ (3.9), which cover Quadrants I and IV. For points with $x < 0$ (Quadrants II and III), add $\pi$ to the $\arctan$ result. Here $\arctan (-12/5) + \pi \approx -1.176 + 3.142 = 1.966$ ✓.

Worked example

Example 4. Convert $(4, -4)$ and $(-3, -7)$ to polar coordinates with $r > 0$ and $0 \le \theta < 2\pi$ .

01
$(4, -4)$ : $r = \sqrt{32} = 4\sqrt{2}$ . The point is in Quadrant IV with reference angle $\pi /4$ , so $\theta = 2\pi - \pi /4 = 7\pi /4$ . Answer: ( $4\sqrt{2}$ , $7\pi /4$ ).
02
$(-3, -7)$ : $r = \sqrt{58} \approx 7.616$ . The point is in Quadrant III with reference angle $\arctan (7/3) \approx 1.166$ , so $\theta = \pi + 1.166 \approx 4.307$ . Answer: about $(7.616, 4.307)$ .

Quick check

Convert $(-1, \sqrt{3})$ to polar form with $0 \le \theta < 2\pi$ .

Common slips

" $(-3, 5\pi /6)$ is the same point as $(3, 5\pi /6)$ , just written differently."

Changing only the sign of $r$ sends you to the opposite side of the pole. $(-3, 5\pi /6)$ is the point ( $3\sqrt{3}/2$ , −3/2), in Quadrant iv. To keep the same point, a sign change in $r$ must come with a change of $\pi$ in $\theta$ .
" $\theta = \arctan (y/x)$ , always."

$\arctan$ only returns angles between $-\pi /2$ and $\pi /2$ (3.9), which cover Quadrants I and iv. For points with $x < 0$ (Quadrants ii and iii), add $\pi$ to the $\arctan$ result. Here $\arctan (-12/5) + \pi \approx -1.176 + 3.142 = 1.966$ ✓.

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10 cards · Polar, Polar points and complex numbers

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Recap card

6 lines to re-read the night before.

01
A polar point $(r, \theta )$ means: face angle $\theta$ , walk $r$ units (backward if $r < 0$ ).
02
Every point has infinitely many polar names: $(r, \theta + 2\pi k)$ and $(-r, \theta + \pi + 2\pi k)$ .
03
Polar to rectangular: $x = r \cos \theta$ , $y = r \sin \theta$ .
04
Rectangular to polar: $r = \sqrt{x^{2} + y^{2}}$ , and $\tan \theta = y/x$ — but pick $\theta$ in the point's quadrant, adding $\pi$ to $\arctan$ when $x < 0$ .
05
Rotating objects are natural in polar form: the pedal is always $(17, \theta )$ from the crank center.
06
Coming up in 3.13B: the same conversions for complex numbers.

Practice 42 problems Next: Chapter 3.13B Trigonometry and Polar Coordinates

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Trigonometry and Polar Coordinates

Plotting Polar Points

Polar to Rectangular

Rectangular to Polar

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