Polar Calculus

Polar Calculus

Part 1 of 7 — Polar Coordinates & Conversions

In polar coordinates, each point is described by $(r, \theta)$: a distance $r$ from the origin and an angle $\theta$ from the positive $x$-axis.

Conversion Formulas

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

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

Key Fact: A point $(r, \theta)$ can also be written as $(-r, \theta + \pi)$ or $(r, \theta + 2\pi)$. Polar representations are NOT unique!

Common Polar Equations

Converting equations:

• $r = 2\cos\theta$: multiply both sides by $r$: $r^2 = 2r\cos\theta \implies x^2+y^2 = 2x$

Practice Problems

Conversion Practice

Compute

Key Takeaways

• Polar coordinates $(r, \theta)$ locate points by distance and angle
• Convert with $x = r\cos\theta$, $y = r\sin\theta$ and $r^2 = x^2+y^2$
• Representations are not unique: $(r, \theta) = (-r, \theta+\pi) = (r, \theta+2\pi)$
• Know the standard curves: circles, cardioids, roses, lima\c{c}ons

Next: Part 2 covers graphing polar curves and symmetry analysis.

Polar Calculus

Part 2 of 7 — Graphing Polar Curves & Symmetry

Graphing Strategy

To sketch $r = f(\theta)$:

1. Make a table of $\theta$ vs. $r$ values
2. Plot points in polar coordinates
3. Check for symmetry to reduce work
4. Note where $r = 0$ (curve passes through origin)
5. Note where $r < 0$ (point is reflected through origin)

Symmetry Tests

Important: These are sufficient but not necessary conditions. A curve may have symmetry that the test does not detect.

Graphing Examples

Cardioid $r = 1 + \cos\theta$:

Symmetric about the $x$-axis (since $\cos(-\theta) = \cos\theta$).

Rose $r = \sin(3\theta)$:

• $r = 0$ when $3\theta = n\pi$, i.e., $\theta = 0, \pi/3, 2\pi/3, \pi$
• $3$ petals (odd coefficient), each spanning $\pi/3$ radians
• First petal: $0 \le \theta \le \pi/3$, max $r = 1$ at $\theta = \pi/6$

Lemniscate $r^2 = 4\cos(2\theta)$:

• Only exists when $\cos(2\theta) \ge 0$: $-\pi/4 \le \theta \le \pi/4$ and $3\pi/4 \le \theta \le 5\pi/4$
• Figure-eight shape through the origin

Practice Problems

Symmetry Analysis

Compute

Key Takeaways

• Graph polar curves by making a $\theta$-$r$ table and plotting
• Use symmetry tests to reduce graphing work
• Know the shapes: circles, cardioids, lima\c{c}ons, roses, lemniscates, spirals
• Negative $r$ values reflect the point through the origin

Next: Part 3 covers polar area with the integral formula.

Polar Calculus

Part 3 of 7 — Area in Polar Coordinates

The Polar Area Formula

The area enclosed by $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$:

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

Why $\frac{1}{2}r^2$? Each thin sector has area $\frac{1}{2}r^2\,d\theta$ (like a triangle with base $r\,d\theta$ and height $r$).

Critical: Choose $\alpha$ and $\beta$ carefully. The curve must trace the region exactly once. Symmetry can reduce the integral.

Example 1: Area of a Cardioid

Find the area enclosed by $r = 1 + \cos\theta$.

Full curve: $0 \le \theta \le 2\pi$. By symmetry about the $x$-axis: $A = 2 \cdot \frac{1}{2}\int_0^{\pi}(1+\cos\theta)^2\,d\theta$.

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

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

Example 2: Area of One Petal

One petal of $r = \sin(2\theta)$: the first petal spans $0 \le \theta \le \pi/2$.

$A = \frac{1}{2}\int_0^{\pi/2}\sin^2(2\theta)\,d\theta = \frac{1}{2}\int_0^{\pi/2}\frac{1-\cos 4\theta}{2}\,d\theta = \frac{\pi}{8}$

Practice Problems

Setting Up Integrals

Compute

Key Takeaways

• Polar area: $A = \frac{1}{2}\int_{\alpha}^{\beta}r^2\,d\theta$
• Between curves: $\frac{1}{2}\int(r_{\text{outer}}^2 - r_{\text{inner}}^2)\,d\theta$
• Use half-angle identities: $\cos^2\theta = \frac{1+\cos 2\theta}{2}$, $\sin^2\theta = \frac{1-\cos 2\theta}{2}$
• Exploit symmetry to simplify calculations

Next: Part 4 covers slopes of polar curves ($dy/dx$ in polar).

Polar Calculus

Part 4 of 7 — Slopes of Polar Curves

To find $dy/dx$ for a polar curve $r = f(\theta)$, treat it as parametric with parameter $\theta$:

$x = r\cos\theta = f(\theta)\cos\theta, \quad y = r\sin\theta = f(\theta)\sin\theta$

The Slope Formula

$\boxed{\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}}$

This comes from the product rule:

• $dx/d\theta = f'\cos\theta - f\sin\theta$
• $dy/d\theta = f'\sin\theta + f\cos\theta$

Worked Example

Find the slope of $r = 1 + \cos\theta$ at $\theta = \pi/3$.

$r = 1 + \cos(\pi/3) = 3/2$, $dr/d\theta = -\sin(\pi/3) = -\sqrt{3}/2$

$\frac{dy}{dx} = \frac{(-\sqrt{3}/2)\sin(\pi/3) + (3/2)\cos(\pi/3)}{(-\sqrt{3}/2)\cos(\pi/3) - (3/2)\sin(\pi/3)}$

$= \frac{(-\sqrt{3}/2)(\sqrt{3}/2) + (3/2)(1/2)}{(-\sqrt{3}/2)(1/2) - (3/2)(\sqrt{3}/2)} = \frac{-3/4 + 3/4}{-\sqrt{3}/4 - 3\sqrt{3}/4} = \frac{0}{-\sqrt{3}} = 0$

The tangent is horizontal at $\theta = \pi/3$!

AP Tip: Horizontal tangents occur when $dy/d\theta = 0$ and $dx/d\theta \ne 0$. Vertical tangents when $dx/d\theta = 0$ and $dy/d\theta \ne 0$.

Practice Problems

Tangent Line Analysis

Compute

Key Takeaways

• Polar slope: $\frac{dy}{dx} = \frac{(dr/d\theta)\sin\theta + r\cos\theta}{(dr/d\theta)\cos\theta - r\sin\theta}$
• Treat polar as parametric with $\theta$ as the parameter
• Horizontal tangent: numerator $= 0$, denominator $\ne 0$
• At the origin: tangent line has slope $\tan\theta_0$ where $f(\theta_0) = 0$

Next: Part 5 covers arc length in polar coordinates.

Polar Calculus

Part 5 of 7 — Arc Length in Polar Coordinates

The Polar Arc Length Formula

For $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$:

$\boxed{L = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta}$

Derivation: From the parametric formula with $x = r\cos\theta$, $y = r\sin\theta$:

$(dx/d\theta)^2 + (dy/d\theta)^2 = (r')^2\cos^2\theta - 2rr'\cos\theta\sin\theta + r^2\sin^2\theta$ $+ (r')^2\sin^2\theta + 2rr'\sin\theta\cos\theta + r^2\cos^2\theta = (r')^2 + r^2$

Key Fact: The polar arc length formula is simpler than the parametric one because the cross terms cancel!

Example 1: Arc Length of a Circle

$r = a$ (constant). $dr/d\theta = 0$.

$L = \int_0^{2\pi}\sqrt{a^2+0}\,d\theta = a \cdot 2\pi = 2\pi a \checkmark$

Example 2: Cardioid $r = 1 + \cos\theta$

$dr/d\theta = -\sin\theta$

$L = \int_0^{2\pi}\sqrt{(1+\cos\theta)^2 + \sin^2\theta}\,d\theta$

$= \int_0^{2\pi}\sqrt{2 + 2\cos\theta}\,d\theta = \int_0^{2\pi}\sqrt{4\cos^2(\theta/2)}\,d\theta = \int_0^{2\pi}2|\cos(\theta/2)|\,d\theta$

Using symmetry: $= 4\int_0^{\pi}\cos(\theta/2)\,d\theta = 4[2\sin(\theta/2)]_0^{\pi} = 8$

AP Note: The half-angle identity $1 + \cos\theta = 2\cos^2(\theta/2)$ is essential for cardioid problems.

Practice Problems

Setting Up Arc Length

Compute

Key Takeaways

• Polar arc length: $L = \int\sqrt{r^2 + (dr/d\theta)^2}\,d\theta$
• For a circle $r = a$: arc length on $[\alpha, \beta]$ is $a(\beta - \alpha)$
• Half-angle identities simplify cardioid integrals
• Logarithmic spiral $r = e^\theta$: integrand is $e^\theta\sqrt{2}$ (nice!)

Next: Part 6 is a Problem-Solving Workshop with mixed polar problems.

Polar Calculus

Part 6 of 7 — Problem-Solving Workshop

Mixed practice covering polar coordinates, graphing, area, slopes, and arc length.

Workshop Problems

AP FRQ-Style Problem

Consider $r = 2\cos\theta$ for $0 \le \theta \le \pi/2$.

(a) Find the area of the region.

$A = \frac{1}{2}\int_0^{\pi/2}4\cos^2\theta\,d\theta = 2\int_0^{\pi/2}\frac{1+\cos 2\theta}{2}\,d\theta = [\theta + \frac{\sin 2\theta}{2}]_0^{\pi/2} = \frac{\pi}{2}$

(b) Find $dy/dx$ at $\theta = \pi/6$.

$r = \sqrt{3}$, $dr/d\theta = -2\sin(\pi/6) = -1$

$\frac{dy}{dx} = \frac{(-1)(1/2) + \sqrt{3}(\sqrt{3}/2)}{(-1)(\sqrt{3}/2) - \sqrt{3}(1/2)} = \frac{-1/2 + 3/2}{-\sqrt{3}/2 - \sqrt{3}/2} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}$

(c) Find the arc length.

$L = \int_0^{\pi/2}\sqrt{4\cos^2\theta + 4\sin^2\theta}\,d\theta = \int_0^{\pi/2}2\,d\theta = \pi$

Mixed Concepts

FRQ Computation

Workshop Recap

Polar Problem Checklist:

1. Identify the curve type (circle, cardioid, rose, etc.)
2. Determine symmetry and appropriate limits
3. Set up the correct integral (area: $\frac{1}{2}r^2$; arc length: $\sqrt{r^2+(r')^2}$)
4. Use trig identities to evaluate
5. Check your answer against geometric intuition

Coming Up: Part 7 is the Comprehensive Review of polar calculus.

Polar Calculus

Part 7 of 7 — Comprehensive Review

Complete Polar Formula Sheet

Comprehensive Assessment

Curve Classification Summary

Final Concept Check

Final Computation

Polar Calculus Complete!

You have mastered:

• Polar coordinates and conversions
• Graphing polar curves and symmetry analysis
• Polar area formula and area between curves
• Slopes of polar curves via parametric conversion
• Arc length in polar coordinates

AP Exam Note: The polar/parametric FRQ is one of the most predictable on the BC exam. Practice setting up area and arc length integrals with correct limits. Always check for intersections at the origin separately!