13.2 Polar Coordinates, Polar-Rectangular Conversion & Polar Graphs

13.2 Polar Coordinates, Polar-Rectangular Conversion & Polar Graphs

The Polar Coordinate Framework and Coordinate Non-Uniqueness

While the Cartesian coordinate system identifies a planar location via orthogonal directed horizontal and vertical distances $(x, y)$, the polar coordinate system identifies a point $P$ using a directed radial distance $r$ from a fixed central origin called the pole (coincident with the Cartesian origin $O$) and a directed angular displacement $\theta$ from a horizontal ray called the polar axis (coincident with the positive Cartesian x-axis). The location is written as an ordered pair: $P = (r, \theta)$

In standard convention, $\theta$ is positive when measured counterclockwise from the polar axis and negative when measured clockwise. The radial coordinate $r$ represents a directed distance along the terminal ray of $\theta$:

• If $r > 0$, the point lies $r$ units from the pole directly along the terminal ray of $\theta$.
• If $r < 0$, the point lies $|r|$ units from the pole in the direction directly opposite the terminal ray (reflected through the pole along the ray $\theta + \pi$).
• If $r = 0$, the point resides at the pole regardless of the value of $\theta$: $(0, \theta) = O$.

A vital distinction between Cartesian and polar systems is that Cartesian coordinates are unique, whereas polar coordinates are infinitely non-unique. Any given geometric point $P(r, \theta)$ in the plane can be represented by infinitely many distinct polar coordinate pairs: $(r, \theta) = (r, \theta + 2k\pi) \quad \text{for any integer } k \in \mathbb{Z}$ $(r, \theta) = (-r, \theta + (2k+1)\pi) \quad \text{for any integer } k \in \mathbb{Z}$

For example, the point with Cartesian coordinates $(0, -3)$ can be expressed in polar form as $\left(3, \frac{3\pi}{2}\right)$, $\left(3, -\frac{\pi}{2}\right)$, $\left(-3, \frac{\pi}{2}\right)$, or $\left(3, \frac{7\pi}{2}\right)$.

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Coordinate and Equation Conversion Mechanics

Superimposing the polar plane onto the standard Cartesian grid allows right-triangle trigonometry and the Pythagorean theorem to establish the four fundamental bridge formulas linking $(x, y)$ and $(r, \theta)$: $x = r\cos\theta, \quad y = r\sin\theta$ $r^2 = x^2 + y^2 \implies r = \pm\sqrt{x^2 + y^2}$ $\tan\theta = \frac{y}{x} \quad (x \neq 0)$

Converting Points: Rectangular to Polar

When converting a Cartesian point $(x, y)$ to polar coordinates $(r, \theta)$:

1. Calculate $r = \sqrt{x^2 + y^2}$ (choosing $r > 0$).
2. Determine $\theta$ using $\tan\theta = \frac{y}{x}$, adjusting for the appropriate quadrant:
   • Quadrant I ($x > 0, y > 0$): $\theta = \arctan\left(\frac{y}{x}\right)$
   • Quadrant II ($x < 0, y > 0$): $\theta = \arctan\left(\frac{y}{x}\right) + \pi$
   • Quadrant III ($x < 0, y < 0$): $\theta = \arctan\left(\frac{y}{x}\right) + \pi$
   • Quadrant IV ($x > 0, y < 0$): $\theta = \arctan\left(\frac{y}{x}\right) + 2\pi$ (or $-\arctan\left|\frac{y}{x}\right|$)
   • Axis Points: If $x = 0$ and $y > 0$, $\theta = \frac{\pi}{2}$; if $x = 0$ and $y < 0$, $\theta = \frac{3\pi}{2}$.

Converting Equations Between Forms

• Rectangular to Polar: Substitute $x = r\cos\theta$ and $y = r\sin\theta$ into the Cartesian relation, then solve for $r$ in terms of $\theta$ where possible.
• Polar to Rectangular: Strategically manipulate the polar expression to introduce the bridge terms $r^2$, $r\cos\theta$, and $r\sin\theta$. Common techniques include:
  • Multiplying both sides of the equation by $r$.
  • Taking trigonometric identities (such as squaring both sides or expanding sum/difference formulas).
  • Replacing $\cos\theta$ with $\frac{x}{r}$ and $\sin\theta$ with $\frac{y}{r}$, followed by multiplying by $r = \sqrt{x^2 + y^2}$.

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Classification Matrix of Classic Polar Curves

Polar equations produce distinctive geometric loci that are often cumbersome to write in Cartesian coordinates.

1. Circles and Lines

• Concentric Circles: $r = a$ ($a \neq 0$) describes a circle centered at the pole with radius $|a|$.
• Off-Center Circles Passing Through the Pole:
  • $r = 2a\cos\theta$: Circle of radius $|a|$ centered at $(a, 0)$ on the polar axis, tangent to the vertical line $x = 0$.
  • $r = 2a\sin\theta$: Circle of radius $|a|$ centered at $\left(0, a\right)$ on the line $\theta = \frac{\pi}{2}$, tangent to the horizontal polar axis.
• Lines:
  • $\theta = \alpha$: Straight line passing through the pole with Cartesian slope $m = \tan\alpha$.
  • $r\cos\theta = d \iff x = d$: Vertical line located $d$ units from the pole.
  • $r\sin\theta = d \iff y = d$: Horizontal line located $d$ units from the pole.

2. Rose Curves

Rose curves are sinusoidal loops radiating symmetrically from the pole, defined by: $r = a\cos(n\theta) \quad \text{or} \quad r = a\sin(n\theta) \quad (a \neq 0, n \in \mathbb{Z}^+)$

• Petal Count Rule:
  • If $n$ is odd, the curve possesses exactly $n$ petals, completely traced as $\theta$ spans $[0, \pi)$.
  • If $n$ is even, the curve possesses exactly $2n$ petals, completely traced as $\theta$ spans $[0, 2\pi)$.
• Petal Length: The maximum length of each petal is $|a|$.
• First Petal Orientation:
  • For $r = a\cos(n\theta)$, a petal crest occurs at $\theta = 0$.
  • For $r = a\sin(n\theta)$, the first petal crest occurs where $n\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{2n}$.

3. The Limaçon Family

Limaçons are heart-shaped or looped curves governed by: $r = a \pm b\cos\theta \quad \text{or} \quad r = a \pm b\sin\theta \quad (a > 0, b > 0)$ The geometric morphology depends entirely on the dimensionless ratio $\frac{a}{b}$:

• Inner Loop Limaçon ($\frac{a}{b} < 1$): Because $a < b$, $r$ changes sign when $a \pm b\cos\theta = 0$. The curve passes through the pole twice and loops inside itself.
• Cardioid ($\frac{a}{b} = 1$): A heart-shaped curve with a sharp pointed cusp at the pole ($r = 0$).
• Dimpled Limaçon ($1 < \frac{a}{b} < 2$): The curve never passes through the pole ($r > 0$ for all $\theta$), but features an indented "dimple" or flattened contour toward the pole.
• Convex Limaçon ($\frac{a}{b} \ge 2$): A flattened oval curve with no indentation; the curvature remains everywhere convex.

4. Lemniscates

Lemniscates are propeller- or figure-eight-shaped loops centered at the pole, governed by: $r^2 = a^2\cos(2\theta) \quad \text{or} \quad r^2 = a^2\sin(2\theta)$ Because $r^2 \ge 0$, the function exists only where the trigonometric factor is non-negative. Each loop has length $a$, and the curves exhibit symmetry through the pole.

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Polar Curve Classification Matrix

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Worked Exemplar: Converting and Analyzing an Off-Center Polar Curve

Problem: A polar curve is defined by the equation $r = 6\cos\theta - 8\sin\theta$.

1. Convert this equation into rectangular Cartesian coordinates $(x, y)$.
2. Identify the specific geometric locus, including the center coordinates and radius.
3. Determine the Cartesian coordinates of all points where the curve intersects the coordinate axes.

Step 1: Algebraic Conversion to Cartesian Coordinates

Begin with the polar equation: $r = 6\cos\theta - 8\sin\theta$ Multiply both sides by $r$ to create terms matching the bridge formulas $r^2$, $r\cos\theta$, and $r\sin\theta$: $r^2 = 6r\cos\theta - 8r\sin\theta$ Substitute $r^2 = x^2 + y^2$, $r\cos\theta = x$, and $r\sin\theta = y$: $x^2 + y^2 = 6x - 8y$

Step 2: Complete the Square to Determine Center and Radius

Rearrange all variable terms to the left-hand side of the equation: $(x^2 - 6x) + (y^2 + 8y) = 0$ Complete the square for both the $x$ and $y$ quadratic binomials:

• For $x$: $\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$
• For $y$: $\left(\frac{8}{2}\right)^2 = (4)^2 = 16$

Add $9 + 16 = 25$ to both sides of the equation: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 25$ $(x - 3)^2 + (y + 4)^2 = 25$

This is the standard Cartesian equation of a circle: $(x - h)^2 + (y - k)^2 = R^2$.

• Geometric Locus: Circle
• Center: $(h, k) = (3, -4)$
• Radius: $R = \sqrt{25} = 5$

Because the distance from the origin $(0, 0)$ to the center $(3, -4)$ is $\sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$, which exactly equals the radius $R$, the circle passes directly through the pole $(0, 0)$.

Step 3: Determine Coordinate Axis Intercepts

• Find x-intercepts (set $y = 0$): $(x - 3)^2 + (0 + 4)^2 = 25$ $(x - 3)^2 + 16 = 25 \implies (x - 3)^2 = 9$ $x - 3 = \pm 3 \implies x = 0 \quad \text{or} \quad x = 6$ The x-intercepts are $(0, 0)$ and $(6, 0)$.

• Find y-intercepts (set $x = 0$): $(0 - 3)^2 + (y + 4)^2 = 25$ $9 + (y + 4)^2 = 25 \implies (y + 4)^2 = 16$ $y + 4 = \pm 4 \implies y = 0 \quad \text{or} \quad y = -8$ The y-intercepts are $(0, 0)$ and $(0, -8)$.