Polar Coordinates

📍 Introduction to Polar Coordinates

Part 1 of 7

Instead of locating points by horizontal/vertical distances $(x, y)$, polar coordinates use a distance and angle: $(r, \theta)$.

Polar vs. Rectangular

Key Components

• $r$ = distance from the pole (origin)
• $\theta$ = angle measured counterclockwise from the polar axis (positive $x$-axis)

$\boxed{\text{Point } (r, \theta): \text{ go distance } r \text{ in direction } \theta}$

Important Notes

• $r$ can be negative: $(- r, \theta)$ means go distance $r$ in the opposite direction of $\theta$
• Angles can exceed $360°$ or be negative (clockwise)
• The same point has infinitely many polar representations

🔄 Converting Between Systems

Polar → Rectangular

$\boxed{x = r\cos\theta, \qquad y = r\sin\theta}$

Rectangular → Polar

$\boxed{r = \sqrt{x^2 + y^2}, \qquad \tan\theta = \frac{y}{x}}$

When finding $\theta$, always check the quadrant — $\arctan$ alone may give the wrong angle!

Example 1: Polar → Rectangular

Convert $(4, \frac{\pi}{3})$ to rectangular:

$x = 4\cos\frac{\pi}{3} = 4 \cdot \frac{1}{2} = 2$

$y = 4\sin\frac{\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$

Answer: $(2, 2\sqrt{3})$

Example 2: Rectangular → Polar

Convert $(-3, 3)$ to polar:

$r = \sqrt{9 + 9} = 3\sqrt{2}$

$\tan\theta = \frac{3}{-3} = -1$. Since the point is in QII: $\theta = \frac{3\pi}{4}$

Answer: $(3\sqrt{2}, \frac{3\pi}{4})$

🔁 Multiple Representations

The Same Point Has Many Names

The point $(r, \theta)$ is also represented by:

• $(r, \theta + 2\pi n)$ for any integer $n$
• $(-r, \theta + \pi + 2\pi n)$ for any integer $n$

Example: All Representations of $(3, \frac{\pi}{4})$

• $(3, \frac{\pi}{4})$ — standard
• $(3, \frac{9\pi}{4})$ — add $2\pi$
• $(-3, \frac{5\pi}{4})$ — negate $r$ and add $\pi$
• $(-3, -\frac{3\pi}{4})$ — negate $r$ and subtract $\pi$

Plotting Negative $r$

To plot $(-2, \frac{\pi}{6})$:

1. Face direction $\frac{\pi}{6}$ (30°)
2. Walk backwards 2 units
3. You end up at $(2, \frac{7\pi}{6})$ — same point!

Polar Basics Quiz 🎯

Convert Coordinates 🧮

1) Convert $(5, \frac{\pi}{6})$ to rectangular. What is $x$? Round to 1 decimal. (e.g., $4\cos\frac{\pi}{3} = 4(0.5) = 2.0$)

2) Convert $(5, \frac{\pi}{6})$ to rectangular. What is $y$? (e.g., $4\sin\frac{\pi}{3} = 4(0.866) = 3.5$)

3) Convert rectangular $(0, -4)$ to polar. What is $r$? (e.g., $\sqrt{3^2 + 4^2} = 5$)

Coordinate Matching 🔽

Exit Quiz ✅

🌹 Polar Curves — Basic Shapes

Part 2 of 7

Polar equations define curves using $r$ as a function of $\theta$. The shapes are often strikingly beautiful.

Lines & Circles in Polar

Example: $r = 4\cos\theta$

This is a circle with diameter $4$, centered at $(2, 0)$ in rectangular coordinates.

To verify: $r = 4\cos\theta \implies r^2 = 4r\cos\theta \implies x^2 + y^2 = 4x$

$\implies (x-2)^2 + y^2 = 4$

Circle with center $(2, 0)$ and radius $2$. ✓

🌹 Rose Curves

Standard Forms

$r = a\cos(n\theta) \quad \text{or} \quad r = a\sin(n\theta)$

Petal length = $|a|$

Examples

Why the Odd/Even Rule?

For odd $n$: each petal is traced once as $\theta$ goes from $0$ to $\pi$.

For even $n$: petals in each "half" are traced, and the curve also traces petals when $r$ is negative, doubling the count.

🐌 Limaçons

Standard Forms

$r = a \pm b\cos\theta \quad \text{or} \quad r = a \pm b\sin\theta$

The shape depends on the ratio $\frac{a}{b}$:

Example: $r = 2 + 3\cos\theta$ (Inner Loop)

$\frac{a}{b} = \frac{2}{3} < 1$ → inner loop

• Maximum $r$: when $\cos\theta = 1$, $r = 5$
• Minimum $r$: when $\cos\theta = -1$, $r = -1$ (inner loop!)

Example: $r = 3 + 3\sin\theta$ (Cardioid)

$\frac{a}{b} = 1$ → cardioid

Passes through the origin when $\sin\theta = -1$, i.e., $\theta = \frac{3\pi}{2}$.

Polar Curves Quiz 🎯

Polar Curve Analysis 🧮

1) $r = 6\cos(2\theta)$: how many petals? (e.g., $r = a\cos(3\theta)$ has 3 petals since 3 is odd)

2) $r = 1 + 3\sin\theta$: compute $\frac{a}{b}$ as a decimal. (e.g., for $r = 2 + 4\cos\theta$: $\frac{a}{b} = \frac{2}{4} = 0.5$)

3) $r = 4\sin\theta$ is a circle with what diameter? (e.g., $r = 6\cos\theta$ is a circle with diameter 6)

Curve Identification 🔽

Exit Quiz ✅

🔄 Converting Polar ↔ Rectangular Equations

Part 3 of 7

Converting equations between polar and rectangular form is essential for graphing and analysis.

Key Substitution Relationships

Strategy: Polar → Rectangular

1. Look for $r\cos\theta$ (replace with $x$) or $r\sin\theta$ (replace with $y$)
2. Look for $r^2$ (replace with $x^2 + y^2$)
3. Multiply both sides by $r$ if needed to create these forms

📝 Converting Polar → Rectangular

Example 1: $r = 3$

$r^2 = 9 \implies x^2 + y^2 = 9$

Circle of radius 3.

Example 2: $r = 4\sec\theta$

$r = \frac{4}{\cos\theta} \implies r\cos\theta = 4 \implies x = 4$

Vertical line!

Example 3: $r = 2\sin\theta + 4\cos\theta$

Multiply by $r$: $r^2 = 2r\sin\theta + 4r\cos\theta$

$x^2 + y^2 = 2y + 4x$

$(x-2)^2 + (y-1)^2 = 5$

Circle with center $(2, 1)$ and radius $\sqrt{5}$.

Example 4: $r = \frac{6}{2\cos\theta + 3\sin\theta}$

$r(2\cos\theta + 3\sin\theta) = 6 \implies 2x + 3y = 6$

A straight line! In standard form: $2x + 3y = 6$.

📝 Converting Rectangular → Polar

Example 5: $x^2 + y^2 = 16$

$r^2 = 16 \implies r = 4$

Example 6: $y = x$

$r\sin\theta = r\cos\theta \implies \tan\theta = 1 \implies \theta = \frac{\pi}{4}$

Example 7: $x^2 + y^2 - 6x = 0$

$r^2 - 6r\cos\theta = 0 \implies r(r - 6\cos\theta) = 0$

Since $r = 0$ is just the origin (already on the curve): $r = 6\cos\theta$

Example 8: $y = 3$

$r\sin\theta = 3 \implies r = 3\csc\theta$

Quick Reference

Conversion Quiz 🎯

Convert Equations 🧮

1) Convert $r = 8\cos\theta$ to rectangular. What is the radius of the resulting circle? (e.g., $r = 6\cos\theta \to (x-3)^2 + y^2 = 9$, radius = 3)

2) Convert $x^2 + y^2 = 49$ to polar. What is $r$? (e.g., $x^2 + y^2 = 16$ becomes $r = 4$)

3) Convert $r = 3\sec\theta$ to rectangular. What is the constant $x$ value? (e.g., $r = 5\sec\theta \to x = 5$)

Match the Forms 🔽

Exit Quiz ✅

📊 Graphing Polar Equations by Hand

Part 4 of 7

Graphing polar equations by hand requires building a table of $(r, \theta)$ values and plotting points.

Step-by-Step Process

1. Make a table of $\theta$ values (usually multiples of $\frac{\pi}{6}$ or $\frac{\pi}{4}$)
2. Compute $r$ for each $\theta$
3. Plot each $(r, \theta)$ point on polar grid
4. Connect points with a smooth curve
5. Check symmetry to reduce work

Symmetry Tests

📝 Graphing $r = 2 + 2\cos\theta$ (Cardioid)

Step 1: Table of Values

Step 2: Symmetry Check

Replace $\theta$ with $-\theta$: $r = 2 + 2\cos(-\theta) = 2 + 2\cos\theta$ ✓

Symmetric about the polar axis! Only need to plot $[0, \pi]$ and reflect.

Key Feature

Passes through origin at $\theta = \pi$ (where $r = 0$).

🌹 Graphing a Rose: $r = 3\cos(2\theta)$

Finding the Petals

Set $r = 0$: $\cos(2\theta) = 0 \implies 2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$

$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

These are the "zeros" between petals.

Petal Locations

Maximum $r = 3$ when $\cos(2\theta) = 1$:

$2\theta = 0, 2\pi, 4\pi \implies \theta = 0, \pi$ (petals along $x$-axis)

Also $r = -3$ when $\cos(2\theta) = -1$:

$2\theta = \pi, 3\pi \implies \theta = \frac{\pi}{2}, \frac{3\pi}{2}$ (petals along $y$-axis, traced with negative $r$)

Result

4 petals along the axes, each of length 3. The curve has both $x$-axis and $y$-axis symmetry, plus origin symmetry.

Graphing Quiz 🎯

Compute $r$ Values 🧮

For $r = 3 - 3\sin\theta$:

1) At $\theta = 0$: $r$ = ? (e.g., for $r = 2 + 2\cos\theta$ at $\theta = 0$: $r = 2+2(1) = 4$)

2) At $\theta = \frac{\pi}{2}$: $r$ = ? (e.g., at $\theta = \frac{\pi}{2}$: $r = 2+2(0) = 2$)

3) At $\theta = \frac{3\pi}{2}$: $r$ = ? (e.g., at $\theta = \pi$: $r = 2+2(-1) = 0$)

Identify Features 🔽

Exit Quiz ✅

📐 Area in Polar Coordinates

Part 5 of 7

The Polar Area Formula

To find the area enclosed by a polar curve $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$:

$A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta = \frac{1}{2}\int_{\alpha}^{\beta} [f(\theta)]^2 \, d\theta$

Why $\frac{1}{2}r^2$?

Think of thin "pie slices" of angle $d\theta$. Each slice is approximately a sector of a circle with area $\frac{1}{2}r^2 d\theta$.

Key Setup Steps

1. Identify the limits $\alpha$ and $\beta$ carefully
2. For a full curve, determine the period (e.g., a rose may complete in $[0, \pi]$ or $[0, 2\pi]$)
3. Use symmetry to simplify: compute part and multiply

📝 Example: Area Inside a Cardioid $r = 1 + \cos\theta$

The full cardioid is traced from $\theta = 0$ to $\theta = 2\pi$.

$A = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta$

Expand: $(1+\cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$

Use $\cos^2\theta = \frac{1+\cos 2\theta}{2}$:

$= 1 + 2\cos\theta + \frac{1}{2} + \frac{\cos 2\theta}{2} = \frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}$

$A = \frac{1}{2}\int_0^{2\pi}\left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta$

$= \frac{1}{2}\left[\frac{3}{2}\theta + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi} = \frac{1}{2}\left(\frac{3}{2}\cdot 2\pi\right) = \frac{3\pi}{2}$

Shortcut: By symmetry about the polar axis, we could compute $2 \cdot \frac{1}{2}\int_0^{\pi}(1+\cos\theta)^2\,d\theta$, getting the same answer.

🔄 Area Between Polar Curves

For the area inside $r_1 = f(\theta)$ and outside $r_2 = g(\theta)$ (where $f(\theta) \geq g(\theta)$):

$A = \frac{1}{2}\int_{\alpha}^{\beta}\left([f(\theta)]^2 - [g(\theta)]^2\right)d\theta$

Example: Area inside $r = 2$ but outside $r = 2(1 - \cos\theta)$.

Find intersections: $2 = 2(1-\cos\theta) \implies \cos\theta = 0 \implies \theta = \pm\frac{\pi}{2}$

By symmetry (both curves are symmetric about polar axis):

$A = 2 \cdot \frac{1}{2}\int_0^{\pi/2}\left(4 - 4(1-\cos\theta)^2\right)d\theta$

Carefully evaluate: $4 - 4(1-2\cos\theta+\cos^2\theta) = 8\cos\theta - 4\cos^2\theta$

This yields $A = 8 - \pi$ after integration.

⚠️ Common Mistake: Always check which curve is "outer" vs "inner" on the integration interval!

Area Quiz 🎯

Set Up Area Integrals 🧮

1) Area inside $r = 4\sin\theta$. This is a circle of diameter 4. Its area = ? (Enter as a multiple of $\pi$, like "4pi")

2) One petal of $r = 3\cos(3\theta)$: first petal from $\theta = 0$ to $\theta =$ ? (Enter as a fraction of pi, like "pi/3")

3) Area of one petal of $r = 2\sin(2\theta)$: $A = \frac{1}{2}\int_0^{\pi/2} 4\sin^2(2\theta)\,d\theta =$ ? (Enter as a multiple of $\pi$, like "pi/2")

Area Concepts 🔽

Exit Quiz ✅

🪐 Conic Sections in Polar Form

Part 6 of 7

The Focus-Directrix Form

Any conic section (ellipse, parabola, hyperbola) with one focus at the origin can be written:

$r = \frac{ed}{1 \pm e\cos\theta} \quad \text{or} \quad r = \frac{ed}{1 \pm e\sin\theta}$

where:

• $e$ = eccentricity (determines shape)
• $d$ = distance from focus to directrix

Classification by Eccentricity

Orientation

• $1 + e\cos\theta$: directrix to the right of focus
• $1 - e\cos\theta$: directrix to the left of focus
• $1 + e\sin\theta$: directrix above focus
• $1 - e\sin\theta$: directrix below focus

📝 Example: Identify and Analyze $r = \frac{6}{2 + \cos\theta}$

Step 1: Standard Form

Divide numerator and denominator by 2: $r = \frac{3}{1 + \frac{1}{2}\cos\theta}$

So $e = \frac{1}{2}$ and $ed = 3 \implies d = 6$.

Step 2: Classify

$e = \frac{1}{2} < 1$ → Ellipse

Step 3: Key Points

• At $\theta = 0$: $r = \frac{6}{2+1} = 2$ (closest to directrix)
• At $\theta = \pi$: $r = \frac{6}{2-1} = 6$ (farthest)
• At $\theta = \frac{\pi}{2}$: $r = \frac{6}{2} = 3$

Step 4: Semi-major axis

$a = \frac{r_{\min}+r_{\max}}{2} = \frac{2+6}{2} = 4$

Center is at distance $ae = 4 \cdot \frac{1}{2} = 2$ from the focus (origin).

🎯 Special Case: Parabola ($e = 1$)

$r = \frac{d}{1 + \cos\theta}$

• At $\theta = 0$: $r = \frac{d}{2}$ (vertex)
• At $\theta = \frac{\pi}{2}$: $r = d$ (end of latus rectum)
• At $\theta = \pi$: undefined (approaches infinity — the curve opens left)

Latus rectum: The chord through the focus perpendicular to the axis has length $2d$.

Converting to Rectangular

$r = \frac{d}{1+\cos\theta} \implies r(1+\cos\theta) = d \implies r + x = d$

$\sqrt{x^2+y^2} = d - x \implies x^2+y^2 = d^2 - 2dx + x^2 \implies y^2 = -2dx + d^2$

This is a parabola opening leftward!

Conic Classification 🎯

Analyze Conics 🧮

For $r = \frac{12}{3 + \cos\theta}$:

1) Divide to standard form. The eccentricity $e$ = ? (Enter as a fraction like "1/3")

2) What is $r$ at $\theta = 0$? (Enter a whole number)

3) What is $r$ at $\theta = \pi$? (Enter a whole number)

Conic Properties 🔽

Exit Quiz ✅

🧩 Polar Coordinates — Full Synthesis

Part 7 of 7

Everything Together

This final part combines all polar coordinate skills:

🎓 Problem-Solving Strategies

Identifying a Polar Curve

Flowchart:

1. $r = a$ → Circle centered at origin, radius $a$
2. $r = a\cos\theta$ or $r = a\sin\theta$ → Circle, diameter $|a|$
3. $r = a \pm b\cos\theta$ or $r = a \pm b\sin\theta$ → Limaçon
   • $a = b$: cardioid (passes through origin)
   • $a > b$: dimpled or convex limaçon (no inner loop)
   • $a < b$: limaçon with inner loop
4. $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$ → Rose
   • $n$ odd: $n$ petals
   • $n$ even: $2n$ petals
5. $r^2 = a^2\cos(2\theta)$ or $r^2 = a^2\sin(2\theta)$ → Lemniscate (figure-8)
6. $r = \frac{ed}{1 \pm e\cos\theta}$ → Conic

Common Errors to Avoid

• Forgetting squaring in area formula: it's $r^2$, not $r$
• Wrong limits: always find where $r = 0$ or where curves intersect
• Negative $r$: polar curves can overlap themselves when $r < 0$
• Rectangular conversion: $\tan\theta = \frac{y}{x}$ only in the correct quadrant

Mixed Problems 🎯

Mixed Calculations 🧮

1) Convert $(x, y) = (-3, 3)$ to polar. What is $r$? (Enter exact value like "3sqrt2")

2) For $r = \frac{10}{5-3\cos\theta}$, what is the eccentricity? (Enter as a fraction)

3) How many petals does $r = 4\sin(5\theta)$ have?

Synthesis 🔽

Exit Quiz — Final ✅