\(r = \sqrt{x^2 + y^2}\)
\(\theta = \tan^{-1}(\frac{y}{x})\) with quadrant adjustments:
Quadrant I: Use \(\theta\) as calculated
Quadrant II: Add \(\pi\) to \(\theta\)
Quadrant III: Add \(\pi\) to \(\theta\)
Quadrant IV: Add \(2\pi\) to \(\theta\)
\(x = r\cos(\theta)\)
\(y = r\sin(\theta)\)
Remember: Multiple polar coordinates can represent the same point
If \((r,\theta)\) represents a point, then \((-r,\theta + \pi)\) represents the same point
A circle can be represented in several ways:
\(r = a\) (circle centered at origin with radius a)
\(r = 2a\cos(\theta)\) (circle through origin, center at (a,0))
\(r = 2a\sin(\theta)\) (circle through origin, center at (0,a))
A cardioid is a heart-shaped curve given by \(r = a(1 \pm \cos(\theta))\) or \(r = a(1 \pm \sin(\theta))\)
Rose curves are given by \(r = a\cos(n\theta)\) or \(r = a\sin(n\theta)\)
If n is odd, the curve has n petals
If n is even, the curve has 2n petals
Limaçons have the general form \(r = a \pm b\cos(\theta)\) or \(r = a \pm b\sin(\theta)\)
If a > b: convex curve
If a = b: cardioid (special case)
If a < b: curve with inner loop
Substitute given θ values directly into the equation
Solve for r
Check for extraneous solutions
Replace r with \(\sqrt{x^2 + y^2}\)
Replace \(\cos(\theta)\) with \(\frac{x}{\sqrt{x^2 + y^2}}\)
Replace \(\sin(\theta)\) with \(\frac{y}{\sqrt{x^2 + y^2}}\)
Solve the resulting Cartesian equation
Convert solutions back to polar form if needed
Group terms with r
Factor when possible
Use trigonometric identities to simplify
Consider domain restrictions:
r can be negative
θ typically restricted to [0, 2π) or [-π, π)
Check for periodicity in θ
1. Solve r = 2cos(θ)
This is a circle through the origin
Center at (1,0), radius = 1
θ ranges from -π/2 to π/2
2. Solve r = 2 + 2cos(θ)
This is a limaçon
r is always positive (2 + 2cos(θ) ≥ 0)
Maximum r = 4 at θ = 0
Minimum r = 0 at θ = π
Test your understanding of polar functions and coordinate conversions.