🎯 Calculus with Polar Coordinates

Slope of Polar Curves

For a polar curve $r = f(\theta)$, we can express it parametrically:

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

Then use the parametric derivative formula!

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

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Derivatives for Polar Curves

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(r\cos\theta) = \frac{dr}{d\theta}\cos\theta - r\sin\theta$

$\frac{dy}{d\theta} = \frac{d}{d\theta}(r\sin\theta) = \frac{dr}{d\theta}\sin\theta + r\cos\theta$

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Then:

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

Memorize this formula or derive it each time using product rule!

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Example 1: Slope of Cardioid

Find $\frac{dy}{dx}$ for $r = 1 + \cos\theta$ at $\theta = \frac{\pi}{2}$.

Step 1: Find $\frac{dr}{d\theta}$

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

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Step 2: Compute numerator and denominator

At $\theta = \frac{\pi}{2}$:

• $r = 1 + \cos\frac{\pi}{2} = 1 + 0 = 1$
• $\frac{dr}{d\theta} = -\sin\frac{\pi}{2} = -1$

Numerator: $\frac{dr}{d\theta}\sin\theta + r\cos\theta = (-1)(1) + (1)(0) = -1$

Denominator: $\frac{dr}{d\theta}\cos\theta - r\sin\theta = (-1)(0) - (1)(1) = -1$

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Step 3: Calculate slope

$\frac{dy}{dx} = \frac{-1}{-1} = 1$

Answer: The slope at $\theta = \frac{\pi}{2}$ is $1$.

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Horizontal and Vertical Tangents

Horizontal Tangent

When $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \neq 0$:

$\frac{dr}{d\theta}\sin\theta + r\cos\theta = 0$

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Vertical Tangent

When $\frac{dx}{d\theta} = 0$ and $\frac{dy}{d\theta} \neq 0$:

$\frac{dr}{d\theta}\cos\theta - r\sin\theta = 0$

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Example 2: Tangent at the Pole

For most polar curves, when $r = 0$ (at the origin/pole):

The tangent line has slope based on the angle $\theta$ where $r = 0$.

At the pole, the tangent is the line $\theta = \theta_0$ where $r(\theta_0) = 0$.

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Area in Polar Coordinates

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

$A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$

💡 Key Idea: Sum up infinitely many thin "pizza slices" with area $\frac{1}{2}r^2\Delta\theta$!

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Why This Formula?

Each infinitesimal sector has:

• Radius: $r$
• Angle: $d\theta$
• Area: $\frac{1}{2}r^2\,d\theta$ (like area of triangle with arc)

Integrate to get total area!

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Example 3: Area of Circle

Find the area enclosed by $r = 3$ for $0 \leq \theta \leq 2\pi$.

Step 1: Set up integral

$A = \frac{1}{2}\int_0^{2\pi} 3^2\,d\theta = \frac{1}{2}\int_0^{2\pi} 9\,d\theta$

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Step 2: Integrate

$= \frac{9}{2}[\theta]_0^{2\pi} = \frac{9}{2}(2\pi) = 9\pi$

Answer: $9\pi$ (area of circle with radius 3) ✓

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Example 4: Area of Cardioid

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

Step 1: Determine limits

Full cardioid: $0 \leq \theta \leq 2\pi$

By symmetry, can do $0$ to $\pi$ and double.

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Step 2: Set up integral

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

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Step 3: Expand

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

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

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

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

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Step 4: Integrate

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

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

$= \frac{1}{2}\left[3\pi + 0 + 0 - 0 - 0 - 0\right]$

$= \frac{3\pi}{2}$

Answer: $\frac{3\pi}{2}$

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Area Between Two Polar Curves

For $r_1 = f(\theta)$ (outer) and $r_2 = g(\theta)$ (inner) from $\theta = \alpha$ to $\beta$:

$A = \frac{1}{2}\int_{\alpha}^{\beta} \left[r_1^2 - r_2^2\right]d\theta$

Think: Big circle sector minus small circle sector!

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Example 5: Area Between Curves

Find the area inside $r = 2$ but outside $r = 2\cos\theta$.

Step 1: Find intersection points

$2 = 2\cos\theta$ $\cos\theta = 1$ $\theta = 0$

Wait, that's not right. Let me reconsider: when does $2\cos\theta \leq 2$?

When $\cos\theta \leq 1$, which is always true.

Actually, find where $r = 2\cos\theta$ intersects $r = 2$:

Actually the circle $r = 2\cos\theta$ has maximum radius 2 (when $\theta = 0$).

Let me set them equal: $2\cos\theta = 2$ $\cos\theta = 1$ $\theta = 0$

Hmm, they're tangent. Let me reconsider the problem...

Better approach: $r = 2\cos\theta$ is a circle of diameter 2 centered at $(1, 0)$.

The region "inside $r = 2$ but outside $r = 2\cos\theta$" needs intersection points.

$2 = 2\cos\theta \implies \theta = 0$

$r = 2\cos\theta$ goes negative for $\frac{\pi}{2} < \theta < \frac{3\pi}{2}$.

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For $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, $r = 2\cos\theta \geq 0$ and $\leq 2$.

Setup: $A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} \left[2^2 - (2\cos\theta)^2\right]d\theta$

$= \frac{1}{2}\int_{-\pi/2}^{\pi/2} (4 - 4\cos^2\theta)\,d\theta$

$= 2\int_{-\pi/2}^{\pi/2} (1 - \cos^2\theta)\,d\theta$

$= 2\int_{-\pi/2}^{\pi/2} \sin^2\theta\,d\theta$

Using $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$:

$= 2\int_{-\pi/2}^{\pi/2} \frac{1 - \cos 2\theta}{2}\,d\theta$

$= \left[\theta - \frac{\sin 2\theta}{2}\right]_{-\pi/2}^{\pi/2}$

$= \left(\frac{\pi}{2} - 0\right) - \left(-\frac{\pi}{2} - 0\right) = \pi$

Answer: $\pi$

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Arc Length in Polar

Arc length from $\theta = \alpha$ to $\theta = \beta$:

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

Derivation: From parametric form $x = r\cos\theta$, $y = r\sin\theta$, use:

$\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = r^2 + \left(\frac{dr}{d\theta}\right)^2$

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Example 6: Arc Length

Find the perimeter of $r = 2(1 + \cos\theta)$ (cardioid).

Step 1: Find $\frac{dr}{d\theta}$

$r = 2(1 + \cos\theta)$ $\frac{dr}{d\theta} = -2\sin\theta$

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Step 2: Set up integral

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

$= \int_0^{2\pi} \sqrt{4(1+\cos\theta)^2 + 4\sin^2\theta}\,d\theta$

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

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Step 3: Expand under square root

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

$= 2\int_0^{2\pi} \sqrt{2 + 2\cos\theta}\,d\theta$

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

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Step 4: Use half-angle identity

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

$L = 2\sqrt{2}\int_0^{2\pi} \sqrt{2\cos^2\frac{\theta}{2}}\,d\theta$

$= 2\sqrt{2} \cdot \sqrt{2}\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|d\theta$

$= 4\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|d\theta$

For $0 \leq \theta \leq \pi$: $\cos\frac{\theta}{2} \geq 0$

For $\pi < \theta \leq 2\pi$: $\cos\frac{\theta}{2} < 0$ (needs absolute value)

By symmetry or direct integration:

$= 4 \cdot 2\int_0^{\pi} \cos\frac{\theta}{2}\,d\theta = 8\left[2\sin\frac{\theta}{2}\right]_0^{\pi}$

$= 16[\sin\frac{\pi}{2} - 0] = 16(1) = 16$

Answer: $16$

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⚠️ Common Mistakes

Mistake 1: Forgetting the $\frac{1}{2}$

WRONG: $A = \int r^2\,d\theta$

RIGHT: $A = \frac{1}{2}\int r^2\,d\theta$

Area of sector is $\frac{1}{2}r^2\theta$!

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Mistake 2: Wrong Limits

Always check: what values of $\theta$ trace the desired region?

For full closed curve, often $0$ to $2\pi$, but not always!

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Mistake 3: Squaring the Wrong Thing

For area between curves:

$\frac{1}{2}\int (r_1^2 - r_2^2)\,d\theta$

NOT $\frac{1}{2}\int (r_1 - r_2)^2\,d\theta$!

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Mistake 4: Missing Intersections

When finding area between curves, find ALL intersection points to determine correct limits!

Don't forget the origin (pole) - curves can pass through at different $\theta$ values.

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📝 Practice Strategy

1. For slope: Use $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$, apply product rule carefully
2. For area: Use $A = \frac{1}{2}\int r^2\,d\theta$, don't forget the $\frac{1}{2}$!
3. Expand squares: $(a + b\cos\theta)^2$ and use trig identities
4. Use half-angle formulas for $\cos^2\theta$ or $\sin^2\theta$
5. Check symmetry to simplify integrals
6. Draw the region - visual helps with limits
7. For arc length: Formula is $\sqrt{r^2 + (dr/d\theta)^2}$