6.3 Polar Coordinates and Parametric Equations

6.3 Polar Coordinates and Parametric Equations

Scope note: Polar coordinates and parametric equations do not appear in the College Board's published content outline for CLEP Precalculus. The six tested areas are algebraic expressions/equations/inequalities, functions, representations of functions, analytic geometry (lines, circles, parabolas, ellipses, hyperbolas), trigonometry, and functions as models. Treat this section as optional enrichment — valuable for the calculus course this exam prepares you for, and useful reinforcement of the unit circle and conic forms, but study it only after the six tested areas are solid.

While rectangular (Cartesian) coordinates locate points using horizontal and vertical grid lines, many physical phenomena and non-linear curves are described far more naturally using radial distance and angle, or by tracing positions over an independent parameter such as time.

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1. Polar Coordinate System Fundamentals

In the polar coordinate system:

• The reference point is the pole (equivalent to the origin $(0, 0)$).
• The reference ray is the polar axis (a ray pointing horizontally to the right, coincident with the positive $x$-axis).
• A point $P$ is designated by ordered pair $(r, \theta)$, where $r$ is the directed radial distance from the pole, and $\theta$ is the directed angle measured counterclockwise from the polar axis.

                      P(r, θ)
                     /|
                    / |
                   /  |
                 r/   | y = r sin(θ)
                 /    |
                / θ   |
  Pole (0,0) ──┴──────┴──────── Polar Axis
               x = r cos(θ)

Negative Radial Distances and Multiple Representations

Unlike Cartesian coordinates, every point in the polar plane has infinitely many equivalent polar coordinate pairs:

1. Adding Full Rotations: $(r, \theta) = (r, \theta + 2k\pi)$ for any integer $k \in \mathbb{Z}$.
2. Negative Radial Distance: A negative radius $-r$ means moving $r$ units in the direction opposite to the terminal side of $\theta$ (i.e., along $\theta + \pi$): $(-r, \theta) = (r, \theta + \pi + 2k\pi)$

Worked Example 1: Multiple Representations

Plot $P\left(4, \frac{2\pi}{3}\right)$ and find three equivalent representations satisfying $-2\pi \le \theta \le 2\pi$.

Solution:

1. Same positive radius ($r = 4$), subtract $2\pi$: $\left(4, \frac{2\pi}{3} - 2\pi\right) = \left(4, -\frac{4\pi}{3}\right)$
2. Negative radius ($r = -4$), add $\pi$ to angle: $\left(-4, \frac{2\pi}{3} + \pi\right) = \left(-4, \frac{5\pi}{3}\right)$
3. Negative radius ($r = -4$), subtract $\pi$ from angle: $\left(-4, \frac{2\pi}{3} - \pi\right) = \left(-4, -\frac{\pi}{3}\right)$

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2. Polar $\leftrightarrow$ Rectangular Conversions

Conversion Equations

• Polar to Rectangular: $x = r \cos\theta \quad \text{and} \quad y = r \sin\theta$
• Rectangular to Polar: $r^2 = x^2 + y^2 \implies r = \pm \sqrt{x^2 + y^2}$ $\tan\theta = \frac{y}{x} \quad (x \neq 0)$

Note on Quadrant Selection: When finding $\theta = \arctan\left(\frac{y}{x}\right)$, adjust the angle by adding $\pi$ if $(x, y)$ lies in Quadrant II or III.

Worked Example 2: Equation Conversion Both Ways

1. Convert rectangular equation $x^2 + y^2 - 6x = 0$ to polar form.
2. Convert polar equation $r = \frac{4}{2 - \cos\theta}$ to rectangular form.

Solution 1: Substitute $r^2 = x^2 + y^2$ and $x = r \cos\theta$: $r^2 - 6r \cos\theta = 0 \implies r(r - 6\cos\theta) = 0 \implies r = 6\cos\theta$ (This represents a circle centered at $(3, 0)$ with radius $r = 3$).

Solution 2: Clear the denominator of $r = \frac{4}{2 - \cos\theta}$: $r(2 - \cos\theta) = 4 \implies 2r - r\cos\theta = 4$ Substitute $r\cos\theta = x$ and $r = \sqrt{x^2 + y^2}$: $2\sqrt{x^2 + y^2} - x = 4 \implies 2\sqrt{x^2 + y^2} = x + 4$ Square both sides to eliminate the radical: $4(x^2 + y^2) = (x + 4)^2 \implies 4x^2 + 4y^2 = x^2 + 8x + 16$ Subtract $x^2 + 8x$ from both sides to group terms: $3x^2 - 8x + 4y^2 = 16$ Complete the square for $x$: $3\left(x^2 - \frac{8}{3}x + \frac{16}{9}\right) + 4y^2 = 16 + 3\left(\frac{16}{9}\right) \implies 3\left(x - \frac{4}{3}\right)^2 + 4y^2 = \frac{64}{3}$ Dividing by $\frac{64}{3}$ yields the standard equation of an ellipse.

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3. Classical Polar Graphs & Symmetry

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4. Parametric Equations & Eliminating the Parameter

When a planar curve cannot easily be described by $y = f(x)$, we introduce an independent third variable $t$ (the parameter):

$x = f(t) \quad \text{and} \quad y = g(t) \quad \text{for } t \in I$

As $t$ increases across interval $I$, the point $(x(t), y(t))$ traces out the curve in a specific direction called the orientation.

Eliminating the Parameter

To convert parametric equations into rectangular form:

1. Solve one parametric equation for $t$ algebraically, then substitute into the second equation.
2. Use trigonometric identities (e.g., $\sin^2 t + \cos^2 t = 1$ or $\sec^2 t - \tan^2 t = 1$) when $t$ appears inside trigonometric functions.

Worked Example 3: Trigonometric Parameter Elimination

Eliminate the parameter for $x = 3 + 2\cos t$ and $y = -1 + 5\sin t$ for $0 \le t \le 2\pi$.

Solution:

1. Isolate the trigonometric terms: $\cos t = \frac{x - 3}{2} \quad \text{and} \quad \sin t = \frac{y + 1}{5}$
2. Apply the Pythagorean identity $\cos^2 t + \sin^2 t = 1$: $\left(\frac{x - 3}{2}\right)^2 + \left(\frac{y + 1}{5}\right)^2 = 1 \implies \frac{(x - 3)^2}{4} + \frac{(y + 1)^2}{25} = 1$ This is an ellipse centered at $(3, -1)$ with horizontal semi-axis $b = 2$ and vertical semi-major axis $a = 5$.

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5. CLEP Exam Traps & Common Errors

• Trap 1: Quadrant Misplacement in Polar Angle Calculation
  Calculating $\theta = \arctan\left(\frac{y}{x}\right)$ for point $(-3, -3)$ gives $\arctan(1) = \frac{\pi}{4}$. However, $(-3, -3)$ is in Quadrant III. You must add $\pi$ to obtain the correct angle $\theta = \frac{5\pi}{4}$.
• Trap 2: Rose Curve Petal Count Formula
  For $r = 4\cos(3\theta)$, $n = 3$ (odd) gives $3$ petals. But for $r = 4\cos(4\theta)$, $n = 4$ (even) gives $2(4) = 8$ petals, NOT $4$. Always check whether $n$ is even or odd.
• Trap 3: Domain Restrictions when Eliminating Parameters
  For $x = t^2$ and $y = t$, eliminating $t$ yields $x = y^2$. However, because $x = t^2 \ge 0$, the rectangular graph is only the right half of the parabola ($x \ge 0$). Ignoring parameter domain limits results in false rectangular domains.