Calculus (Texas A&M University) | Source: MATH 308 Lecture Notes, Ch. 10.3
Tags: polar coordinates, polar curves, rectangular to polar, polar to rectangular, cardioid, rose curve, limacon, polar equations, MATH 308
Polar coordinates describe points in the plane using a distance r from the origin and an angle θ from the positive x-axis, rather than the usual (x, y) system. The conversion formulas x = r cos θ and y = r sin θ (and their inverses) let you move between the two systems. Polar equations like r = f(θ) produce a rich family of curves, including circles, cardioids, roses, and limacons, that would be awkward to describe in Cartesian form.
Polar coordinates (r, θ)
A system for locating points in the plane where r is the distance from the origin (the pole) and θ is the angle measured anticlockwise from the positive x-axis. Unlike Cartesian coordinates, polar representation is not unique: the same point can be described by infinitely many (r, θ) pairs.
Pole
The origin in polar coordinates, the point r = 0.
Polar curve
The set of all points satisfying a given equation in r and θ. Often written as r = f(θ).
Cardioid
A heart-shaped polar curve of the form r = a ± a cos θ or r = a ± a sin θ. The curve passes through the origin and has a single cusp.
Limacon (limaçon)
A generalisation of the cardioid, with the form r = a + b cos θ or r = a + b sin θ, where a and b need not be equal. When |a| < |b| the curve has an inner loop; when |a| > |b| it is a dimpled or convex shape without a loop; when |a| = |b| it is a cardioid.
Rose curve
A polar curve of the form r = a cos(kθ) or r = a sin(kθ). If k is odd, the curve has k petals. If k is even, it has 2k petals.
Every point in the plane has a unique Cartesian representation (x, y). In polar coordinates, the same point is described by (r, θ), but this representation is not unique.
Adding 2π to θ gives the same point
Negating r and adding π to θ also gives the same point
The convention is usually r ≥ 0 and 0 ≤ θ < 2π, but negative r values are allowed
The conversion formulas:
x = r cos θ
y = r sin θ
Examples:
(r, θ) = (2, π/4): x = 2 cos(π/4) = √2, y = 2 sin(π/4) = √2. Rectangular: (√2, √2)
(r, θ) = (3, π): x = 3 cos π = –3, y = 3 sin π = 0. Rectangular: (–3, 0)
(r, θ) = (–4, π/2): x = –4 cos(π/2) = 0, y = –4 sin(π/2) = –4. Rectangular: (0, –4). A negative r flips the point through the origin
(r, θ) = (5, 8π/3): since 8π/3 = 2π + 2π/3, this is equivalent to θ = 2π/3. So x = 5 cos(2π/3) = –5/2, y = 5 sin(2π/3) = 5√3/2
The conversion formulas:
r = √(x² + y²), with r ≥ 0
tan θ = y/x, then solve for θ in the correct quadrant
A key subtlety: arctan(y/x) always returns a value in (–π/2, π/2), but you typically want 0 ≤ θ < 2π. If the point lies in the second or third quadrant, you need to add π to the arctan result.
Examples:
(√3, 1): r = √(3 + 1) = 2. tan θ = 1/√3, so θ = π/6. Polar: (2, π/6)
(0, 1): r = 1. The point is on the positive y-axis, so θ = π/2. Polar: (1, π/2)
(–1, 1): r = √2. tan θ = –1, which gives arctan(–1) = –π/4. Since the point is in the second quadrant, θ = π – π/4 = 3π/4. Polar: (√2, 3π/4)
Use x = r cos θ, y = r sin θ, and x² + y² = r².
Example: y = a (a horizontal line). r sin θ = a, so r = a / sin θ = a csc θ.
Example: x² + y² = 6y (a circle). Replace: r² = 6r sin θ, so r = 6 sin θ. This is a circle of radius 3 centred at (0, 3).
Example: x² – y² = 1 (a hyperbola). r²cos²θ – r²sin²θ = 1, so r²cos(2θ) = 1, giving r² = sec(2θ).
Example: r = a. This means √(x² + y²) = a, so x² + y² = a². A circle of radius a centred at the origin.
Example: θ = a (a constant angle). This is the ray from the origin at angle a. In rectangular form: y = x tan a (a line through the origin, though the ray is only the half in the correct direction).
Example: r = 2 cos θ. Multiply both sides by r: r² = 2r cos θ. Substitute: x² + y² = 2x. Rearrange: (x – 1)² + y² = 1. This is a circle of radius 1 centred at (1, 0).
Circles:
r = a is a circle of radius a centred at the origin
r = a cos θ is a circle of diameter a, centred at (a/2, 0), passing through the origin
r = a sin θ is a circle of diameter a, centred at (0, a/2), passing through the origin
Examples: r = 3 cos θ and r = –4 sin θ
Cardioids (r = a ± a cos θ or r = a ± a sin θ):
r = 3 + 3 sin θ: a cardioid with the cusp at the origin, extending upward to r = 6
r = 2 – 2 cos θ: a cardioid with the cusp at the origin, opening to the left
Roses (r = a cos(kθ) or r = a sin(kθ)):
r = cos 2θ: 4 petals (k = 2, even, so 2k = 4 petals)
r = cos 3θ: 3 petals (k = 3, odd, so k = 3 petals)
r = cos 4θ: 8 petals
r = sin 2θ: 4 petals
r = sin 3θ: 3 petals
The petal count rule: odd k gives k petals, even k gives 2k petals.
Limacons (r = a + b cos θ or r = a + b sin θ, where a ≠ b):
r = 1 + 2 cos θ: since |a| < |b| (1 < 2), this limacon has an inner loop
r = 2 + cos θ: since |a| > |b| (2 > 1), this is a convex limacon without a loop
To sketch r = f(θ), it helps to:
Identify the curve family (circle, cardioid, rose, limacon) from the equation's form
Find key values: where r = 0, where r is maximum, and any symmetry
Plot a few points at standard angles (0, π/6, π/4, π/3, π/2, etc.)
Trace the curve as θ increases, watching for portions where r becomes negative
Polar to rectangular: x = r cos θ, y = r sin θ
Rectangular to polar: r = √(x² + y²), tan θ = y/x (adjust quadrant)
Useful identity: x² + y² = r²
Multiply-by-r trick: to convert r = f(θ) to rectangular, multiply both sides by r and use r² = x² + y², r cos θ = x, r sin θ = y
Rose petal count: r = a cos(kθ) or r = a sin(kθ) has k petals if k is odd, 2k petals if k is even
⚠️ Polar representations are not unique. The same point can be written with different (r, θ) pairs. Watch for this when finding intersections of polar curves.
⚠️ When converting rectangular to polar, always check the quadrant. The arctan function does not distinguish between quadrants II/IV or I/III on its own.
⚠️ Negative r is allowed. A point (–r, θ) is the same as (r, θ + π). This matters when tracing rose curves and limacons.
⚠️ Know the curve families by sight: circles (r = a, r = a cos θ, r = a sin θ), cardioids (r = a ± a cos θ), roses (r = a cos kθ), limacons (r = a + b cos θ). Exams often ask you to identify or sketch these quickly.
⚠️ The multiply-by-r trick (multiply both sides of a polar equation by r) is the fastest route to a rectangular conversion. It works because r² = x² + y², r cos θ = x, and r sin θ = y are all clean substitutions.
Q: Convert the polar coordinates (2, π/4) to rectangular.
A: x = 2 cos(π/4) = √2, y = 2 sin(π/4) = √2. The point is (√2, √2).
Q: Convert the rectangular point (–1, 1) to polar coordinates with r ≥ 0 and 0 ≤ θ < 2π.
A: r = √(1 + 1) = √2. arctan(1/(–1)) = –π/4, but the point is in the second quadrant, so θ = π – π/4 = 3π/4. Answer: (√2, 3π/4).
Q: Convert the polar equation r = 2 cos θ to rectangular form and identify the curve.
A: Multiply by r: r² = 2r cos θ. Substitute: x² + y² = 2x. Complete the square: (x – 1)² + y² = 1. It is a circle of radius 1 centred at (1, 0).
Q: How many petals does the rose r = sin 5θ have?
A: Since k = 5 is odd, it has 5 petals.
Q: What form of polar equation produces a cardioid?
A: r = a ± a cos θ or r = a ± a sin θ (the coefficients of the constant term and the trig term must be equal).
Q: A limacon r = a + b cos θ has an inner loop when what condition holds?
A: When |b| > |a|, i.e. the coefficient of the cosine term exceeds the constant term in absolute value.
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