Source: Paper Homework Assignment 8, Texas A&M University
Tags: polar coordinates, Cartesian conversion, polar to rectangular, rectangular to polar, r theta, unit circle, Math 2414, calculus II
Polar coordinates describe a point by its distance from the origin (r) and the angle from the positive x-axis (θ). You can convert freely between polar and Cartesian using x = r cos θ, y = r sin θ going one way, and r = √(x² + y²), θ = arctan(y/x) going the other. Negative r values and equivalent angles (adding or subtracting 2π) mean multiple polar representations can describe the same point.
Polar coordinates (r, θ)
A system for locating points in the plane using a radial distance r from the origin (the pole) and an angle θ measured counter-clockwise from the positive x-axis (the polar axis).
Pole
The origin of the polar coordinate system, equivalent to (0, 0) in Cartesian coordinates.
Polar axis
The positive x-axis, serving as the reference direction from which θ is measured.
Negative r
When r is negative, the point is plotted in the opposite direction from the angle θ. Effectively, (−r, θ) is the same point as (r, θ + π).
Coterminal angles
Angles that differ by a full rotation (2π radians or 360°) and therefore point in the same direction. Adding or subtracting 2π to θ does not change the point's location.
Each point (r, θ) is plotted by:
Moving along the direction of angle θ
Marking the point at distance |r| from the origin
If r is negative, go in the opposite direction of θ
Common angle–radian equivalences to have cold:
0° = 0, 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2
120° = 2π/3, 135° = 3π/4, 150° = 5π/6, 180° = π
210° = 7π/6, 225° = 5π/4, 240° = 4π/3, 270° = 3π/2
300° = 5π/3, 315° = 7π/4, 330° = 11π/6, 360° = 2π
Example points from the assignment:
A(2, π/2): go to 90°, distance 2 from origin. Lands on the positive y-axis at (0, 2) in Cartesian.
B(3, π/3): go to 60°, distance 3.
F(5, −11π/6): the negative angle −11π/6 is equivalent to −11π/6 + 2π = π/6 (30°). So this is 5 units from the origin at 30°.
G(−1, π/4): negative r means plot 1 unit in the opposite direction of 45°, which is 45° + 180° = 225°.
H(−3, 2π/3): plot 3 units opposite to 120°, so at 300°.
When dealing with awkward negative angles or angles larger than 2π, always reduce by adding or subtracting 2π until you land in [0, 2π). For example, −10π/3 + 4π = −10π/3 + 12π/3 = 2π/3.
The conversion formulas:
x = r cos θ
y = r sin θ
Worked example (a): (3, π/4)
x = 3 cos(π/4) = 3 · (√2/2) = 3√2/2
y = 3 sin(π/4) = 3 · (√2/2) = 3√2/2
Cartesian point: (3√2/2, 3√2/2)
Worked example (b): (2, −π/2)
Note that −π/2 + 2π = 3π/2, so this angle points straight down.
x = 2 cos(−π/2) = 2 cos(3π/2) = 2 · 0 = 0
y = 2 sin(−π/2) = 2 sin(3π/2) = 2 · (−1) = −2
Cartesian point: (0, −2)
The conversion formulas:
r = √(x² + y²)
θ' = arctan(|y/x|), then adjust for quadrant
The quadrant adjustment matters. arctan gives you the reference angle θ'. The true angle θ depends on which quadrant (x, y) sits in:
Quadrant I (x > 0, y > 0): θ = θ'
Quadrant II (x < 0, y > 0): θ = π − θ'
Quadrant III (x < 0, y < 0): θ = π + θ'
Quadrant IV (x > 0, y < 0): θ = 2π − θ'
Special case: if x = 0, the point lies on the y-axis, so θ = π/2 (if y > 0) or θ = 3π/2 (if y < 0).
Worked example (a): (−2, 2)
r = √((−2)² + 2²) = √(4 + 4) = √8 = 2√2
Reference angle: θ' = arctan(|2/(−2)|) = arctan(1) = π/4
The point is in Quadrant II (x negative, y positive), so θ = π − π/4 = 3π/4
Polar point: (2√2, 3π/4)
Worked example (b): (0, 3)
r = √(0² + 3²) = √9 = 3
x = 0 and y > 0, so the point is on the positive y-axis: θ = π/2
Polar point: (3, π/2)
Polar → Cartesian: x = r cos θ, y = r sin θ
Cartesian → Polar: r = √(x² + y²), θ = arctan(y/x) (adjusted for quadrant)
Negative r equivalence: (−r, θ) = (r, θ + π)
Coterminal equivalence: (r, θ) = (r, θ ± 2nπ) for any integer n
⚠️ Negative r is a classic source of errors. Always remember: flip the direction by adding π to the angle.
⚠️ When converting Cartesian → polar, do not blindly use θ = arctan(y/x). You must check the quadrant and adjust. arctan alone only gives values in (−π/2, π/2).
⚠️ The problem may ask for the "smallest positive θ," so always reduce your angle into [0, 2π).
⚠️ Multiple polar representations exist for every point. Exam questions sometimes test whether you recognise that (2, π/3) and (2, π/3 + 2π) and (−2, π/3 + π) are the same point.
Q: Convert the polar point (4, 5π/6) to Cartesian coordinates.
A: x = 4 cos(5π/6) = 4 · (−√3/2) = −2√3. y = 4 sin(5π/6) = 4 · (1/2) = 2. The Cartesian point is (−2√3, 2).
Q: Convert the Cartesian point (−3, −3) to polar coordinates with positive r and smallest positive θ.
A: r = √(9 + 9) = 3√2. Reference angle θ' = arctan(3/3) = π/4. The point is in Quadrant III, so θ = π + π/4 = 5π/4. Polar point: (3√2, 5π/4).
Q: What Cartesian point does (−5, π/3) represent?
A: Negative r, so equivalent to (5, π/3 + π) = (5, 4π/3). x = 5 cos(4π/3) = 5 · (−1/2) = −5/2. y = 5 sin(4π/3) = 5 · (−√3/2) = −5√3/2. The point is (−5/2, −5√3/2).
Q: Give two other polar representations of the point (3, π/4).
A: (3, π/4 + 2π) = (3, 9π/4) and (−3, π/4 + π) = (−3, 5π/4).
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