In polar coordinates, we can find slopes of tangent lines using derivatives and compute areas enclosed by polar curves. These formulas are essential for analyzing polar curves on the AP Calculus BC exam.
🎯 In this section you will learn
- The formula for $\frac{dy}{dx}$ in polar coordinates
- How to find horizontal and vertical tangents for polar curves
- The area formula in polar coordinates: $A = \frac{1}{2} \int_\alpha^\beta r^2 d\theta$
- How to find the area enclosed by a polar curve or between two polar curves
📌 Derivative in Polar Coordinates
To find $\frac{dy}{dx}$ for a polar curve $r = f(\theta)$, we use parametric equations with $\theta$ as the parameter:
$$ x = r\cos\theta, \quad y = r\sin\theta $$
$$ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta} $$
💡 Derivation
$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta$, $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta$. Then $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
Find $\frac{dy}{dx}$ for $r = 1 + \cos\theta$ at $\theta = \pi/2$.
①
Find $dr/d\theta$
$\frac{dr}{d\theta} = -\sin\theta$
②
Plug into formula
$\frac{dy}{dx} = \frac{(-\sin\theta)\sin\theta + (1+\cos\theta)\cos\theta}{(-\sin\theta)\cos\theta - (1+\cos\theta)\sin\theta}$
③
Evaluate at $\theta = \pi/2$
$\sin(\pi/2)=1$, $\cos(\pi/2)=0$
Numerator: $(-1)(1) + (1+0)(0) = -1$
Denominator: $(-1)(0) - (1+0)(1) = -1$
$\frac{dy}{dx} = \frac{-1}{-1} = 1$
For $r = 1 + \cos\theta$, find where the tangent line is horizontal.
- Horizontal tangent when $dy/d\theta = 0$ (and $dx/d\theta \neq 0$).
- $dy/d\theta = \frac{dr}{d\theta}\sin\theta + r\cos\theta = (-\sin\theta)\sin\theta + (1+\cos\theta)\cos\theta = -\sin^2\theta + \cos\theta + \cos^2\theta$
- Using $\sin^2\theta + \cos^2\theta = 1$, we get $dy/d\theta = -1 + \cos\theta + \cos^2\theta$
- Set $=0$: $\cos^2\theta + \cos\theta - 1 = 0$ → $\cos\theta = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$
- The positive solution gives $\cos\theta \approx 0.618$ → $\theta \approx 0.9046$ (and $2\pi - 0.9046$).
📌 Area in Polar Coordinates
$$ A = \frac{1}{2} \int_\alpha^\beta [f(\theta)]^2 \, d\theta $$
Area enclosed by the polar curve $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$
💡 Derivation
The area of a sector of a circle is $\frac{1}{2}r^2 \Delta\theta$. Summing over small sectors gives the integral.
Find the area of the circle $r = 2$.
- $A = \frac{1}{2} \int_0^{2\pi} (2)^2 d\theta = \frac{1}{2} \int_0^{2\pi} 4 d\theta = 2 \cdot 2\pi = 4\pi$
- This matches the circle area formula $\pi r^2 = 4\pi$.
Find the area enclosed by the cardioid $r = 1 + \cos\theta$.
①
Set up the integral
$A = \frac{1}{2} \int_0^{2\pi} (1 + \cos\theta)^2 d\theta$
②
Expand
$(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta = 1 + 2\cos\theta + \frac{1+\cos2\theta}{2}$
③
Integrate
$A = \frac{1}{2} \int_0^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos2\theta\right) d\theta = \frac{1}{2} \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin2\theta \right]_0^{2\pi}$
④
Evaluate
$= \frac{1}{2} \left( \frac{3}{2} \cdot 2\pi \right) = \frac{1}{2} (3\pi) = \frac{3\pi}{2}$
Find the area inside $r = 2$ and outside $r = 1$.
- $A = \frac{1}{2} \int_0^{2\pi} (2^2 - 1^2) d\theta = \frac{1}{2} \int_0^{2\pi} 3 d\theta = \frac{1}{2} \cdot 3 \cdot 2\pi = 3\pi$
⚠️ Common Mistakes
- Incorrect derivative formula: Remember $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$, not $\frac{dr/d\theta}{1}$.
- Forgetting the $\frac{1}{2}$ factor: Area in polar coordinates is $\frac{1}{2} \int r^2 d\theta$, not $\int r^2 d\theta$.
- Using the wrong limits: For a single loop of a rose, the limits may not be $0$ to $2\pi$.
🔍 Key Takeaways
- $\frac{dy}{dx} = \frac{r' \sin\theta + r\cos\theta}{r' \cos\theta - r\sin\theta}$.
- Area in polar: $A = \frac{1}{2} \int_\alpha^\beta r^2 d\theta$.
- For area between two curves: $A = \frac{1}{2} \int (r_{\text{outer}}^2 - r_{\text{inner}}^2) d\theta$.
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