Polar Coordinates and Graphs

Convert between polar and rectangular coordinates, graph polar equations, and understand polar curves.

Polar Coordinates and Graphs

Introduction to Polar Coordinates

In the polar coordinate system, each point is determined by:

  • rr = distance from the origin (pole)
  • θ\theta = angle from the positive x-axis (polar axis)

A point is written as (r,θ)(r, \theta) in polar form.

Conventions

  • rr can be positive or negative
    • If r>0r > 0, move rr units in the direction of θ\theta
    • If r<0r < 0, move r|r| units in the opposite direction of θ\theta
  • θ\theta is typically measured in radians
  • Multiple representations: (r,θ)(r, \theta), (r,θ+2πn)(r, \theta + 2\pi n), (r,θ+π)(-r, \theta + \pi) all represent the same point

Conversion Formulas

Polar to Rectangular

Given (r,θ)(r, \theta) in polar: x=rcos(θ)x = r\cos(\theta) y=rsin(θ)y = r\sin(\theta)

Rectangular to Polar

Given (x,y)(x, y) in rectangular: r=x2+y2r = \sqrt{x^2 + y^2} θ=arctan(yx)\theta = \arctan(\frac{y}{x}) (adjust for quadrant)

Or use: tan(θ)=yx\tan(\theta) = \frac{y}{x}

Important: When finding θ\theta:

  • Use arctan(yx)\arctan(\frac{y}{x}) as starting point
  • Adjust based on the quadrant of (x,y)(x, y)
  • Quadrant I: θ=arctan(yx)\theta = \arctan(\frac{y}{x})
  • Quadrant II: θ=π+arctan(yx)\theta = \pi + \arctan(\frac{y}{x})
  • Quadrant III: θ=π+arctan(yx)\theta = \pi + \arctan(\frac{y}{x})
  • Quadrant IV: θ=2π+arctan(yx)\theta = 2\pi + \arctan(\frac{y}{x}) or θ=arctan(yx)\theta = \arctan(\frac{y}{x}) (negative)

Common Polar Graphs

Circles

  • r=ar = a: Circle centered at origin with radius aa
  • r=2acos(θ)r = 2a\cos(\theta): Circle with diameter 2a2a on the polar axis
  • r=2asin(θ)r = 2a\sin(\theta): Circle with diameter 2a2a perpendicular to polar axis

Lines

  • θ=α\theta = \alpha: Line through origin at angle α\alpha
  • rcos(θ)=ar\cos(\theta) = a: Vertical line x=ax = a
  • rsin(θ)=ar\sin(\theta) = a: Horizontal line y=ay = a

Limaçons

General form: r=a±bcos(θ)r = a \pm b\cos(\theta) or r=a±bsin(θ)r = a \pm b\sin(\theta)

  • If ab<1\frac{a}{b} < 1: Inner loop
  • If ab=1\frac{a}{b} = 1: Cardioid (heart-shaped)
  • If 1<ab<21 < \frac{a}{b} < 2: Dimpled limaçon
  • If ab2\frac{a}{b} \geq 2: Convex limaçon

Rose Curves

General form: r=acos(nθ)r = a\cos(n\theta) or r=asin(nθ)r = a\sin(n\theta)

  • If nn is odd: nn petals
  • If nn is even: 2n2n petals
  • Length of each petal: a|a|

Lemniscates

  • r2=a2cos(2θ)r^2 = a^2\cos(2\theta): Figure-eight along polar axis
  • r2=a2sin(2θ)r^2 = a^2\sin(2\theta): Figure-eight at 45°45°

Spirals

  • r=aθr = a\theta: Archimedean spiral
  • r=aebθr = ae^{b\theta}: Exponential spiral

Symmetry in Polar Graphs

Test for symmetry to help sketch graphs:

  1. Symmetry about the polar axis (x-axis):

    • Replace (r,θ)(r, \theta) with (r,θ)(r, -\theta)
    • If equation unchanged, symmetric about polar axis
  2. Symmetry about θ=π2\theta = \frac{\pi}{2} (y-axis):

    • Replace (r,θ)(r, \theta) with (r,πθ)(r, \pi - \theta) or (r,θ)(-r, -\theta)
    • If equation unchanged, symmetric about θ=π2\theta = \frac{\pi}{2}
  3. Symmetry about the pole (origin):

    • Replace (r,θ)(r, \theta) with (r,θ)(-r, \theta) or (r,θ+π)(r, \theta + \pi)
    • If equation unchanged, symmetric about the pole

Graphing Strategy

  1. Identify the type of polar curve
  2. Check for symmetry
  3. Create a table of values for θ\theta from 00 to 2π2\pi
  4. Plot key points and note special values
  5. Sketch the curve connecting points smoothly
  6. Consider domain restrictions if r2r^2 appears (must have r20r^2 \geq 0)

📚 Practice Problems

1Problem 1easy

Question:

Convert the point (3,2π3)(3, \frac{2\pi}{3}) from polar to rectangular coordinates.

💡 Show Solution

Solution:

Given polar coordinates: (r,θ)=(3,2π3)(r, \theta) = (3, \frac{2\pi}{3})

Use conversion formulas: x=rcos(θ)x = r\cos(\theta) y=rsin(θ)y = r\sin(\theta)

Find xx: x=3cos(2π3)=3(12)=32x = 3\cos(\frac{2\pi}{3}) = 3 \cdot (-\frac{1}{2}) = -\frac{3}{2}

Find yy: y=3sin(2π3)=332=332y = 3\sin(\frac{2\pi}{3}) = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}

Answer: (32,332)(-\frac{3}{2}, \frac{3\sqrt{3}}{2})

Verification:

  • Check distance: r=(32)2+(332)2=94+274=9=3r = \sqrt{(-\frac{3}{2})^2 + (\frac{3\sqrt{3}}{2})^2} = \sqrt{\frac{9}{4} + \frac{27}{4}} = \sqrt{9} = 3
  • Check angle: Point is in Quadrant II with correct ratio ✓

2Problem 2medium

Question:

Convert the rectangular equation x2+y2=4yx^2 + y^2 = 4y to polar form.

💡 Show Solution

Solution:

Given: x2+y2=4yx^2 + y^2 = 4y

Use substitutions:

  • x=rcos(θ)x = r\cos(\theta)
  • y=rsin(θ)y = r\sin(\theta)
  • x2+y2=r2x^2 + y^2 = r^2

Substitute: r2=4rsin(θ)r^2 = 4r\sin(\theta)

Simplify: r24rsin(θ)=0r^2 - 4r\sin(\theta) = 0 r(r4sin(θ))=0r(r - 4\sin(\theta)) = 0

This gives r=0r = 0 or r=4sin(θ)r = 4\sin(\theta)

Since r=0r = 0 is included in r=4sin(θ)r = 4\sin(\theta) when θ=0\theta = 0 or π\pi, we can write:

Answer: r=4sin(θ)r = 4\sin(\theta)

Interpretation: This is a circle with diameter 4 on the line θ=π2\theta = \frac{\pi}{2} (the y-axis).

Verification in rectangular:

  • r=4sin(θ)r = 4\sin(\theta) means r2=4rsin(θ)r^2 = 4r\sin(\theta)
  • Substituting back: x2+y2=4yx^2 + y^2 = 4y
  • Completing the square: x2+(y2)2=4x^2 + (y-2)^2 = 4, a circle centered at (0,2)(0,2) with radius 2 ✓

3Problem 3hard

Question:

Identify and sketch the polar curve r=2+2cos(θ)r = 2 + 2\cos(\theta). Describe its key features.

💡 Show Solution

Solution:

Given: r=2+2cos(θ)r = 2 + 2\cos(\theta)

Step 1: Identify the curve type

This is a limaçon of the form r=a+bcos(θ)r = a + b\cos(\theta) where a=2a = 2 and b=2b = 2.

Since ab=22=1\frac{a}{b} = \frac{2}{2} = 1, this is a cardioid (heart-shaped).

Step 2: Check for symmetry

Test symmetry about the polar axis (x-axis): Replace θ\theta with θ-\theta: r=2+2cos(θ)=2+2cos(θ)r = 2 + 2\cos(-\theta) = 2 + 2\cos(\theta)

Equation unchanged, so symmetric about the polar axis

Step 3: Create table of values

| θ\theta | 00 | π3\frac{\pi}{3} | π2\frac{\pi}{2} | 2π3\frac{2\pi}{3} | π\pi | 4π3\frac{4\pi}{3} | 3π2\frac{3\pi}{2} | 5π3\frac{5\pi}{3} | 2π2\pi | |-----------|-----|-------------------|-------------------|--------------------|----|--------------------|--------------------|--------------------|----| | cos(θ)\cos(\theta) | 11 | 12\frac{1}{2} | 00 | 12-\frac{1}{2} | 1-1 | 12-\frac{1}{2} | 00 | 12\frac{1}{2} | 11 | | rr | 44 | 33 | 22 | 11 | 00 | 11 | 22 | 33 | 44 |

Step 4: Key features

  • Maximum rr: r=4r = 4 when θ=0\theta = 0 (rightmost point)
  • Minimum rr: r=0r = 0 when θ=π\theta = \pi (cusp at origin)
  • Axis of symmetry: Polar axis (x-axis)
  • Shape: Heart-shaped curve pointing right
  • Key points in rectangular coordinates:
    • θ=0\theta = 0: (4,0)(4, 0)
    • θ=π2\theta = \frac{\pi}{2}: (0,2)(0, 2)
    • θ=π\theta = \pi: (0,0)(0, 0) (cusp)
    • θ=3π2\theta = \frac{3\pi}{2}: (0,2)(0, -2)

Sketch description: The curve starts at (4,0)(4, 0), curves upward and left, passes through (0,2)(0, 2), continues to the origin at θ=π\theta = \pi forming a cusp, then curves downward through (0,2)(0, -2), and returns to (4,0)(4, 0). The overall shape resembles a heart lying on its side.