Defining polar coordinates and differentiating in polar form — AP Calculus BC
1. Introduction to Polar Coordinates ★★☆☆☆ ⏱ 15 min
Unlike Cartesian coordinates, which assign a unique $(x,y)$ pair to every point, polar coordinates can have multiple valid representations for the same point. For example, $(r, \theta) = (-r, \theta + \pi)$ describes the exact same point.
To convert between polar and Cartesian coordinates, we use the following core trigonometric relationships:
x = r \cos \theta \\ y = r \sin \theta \\ r^2 = x^2 + y^2 \\ \tan \theta = \frac{y}{x}, \quad x \neq 0
2. Deriving the Polar Derivative Formula ★★★☆☆ ⏱ 20 min
A polar curve $r = f(\theta)$ can be treated as a parametric curve with parameter $\theta$. We use the parametric derivative rule $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ to find the slope of the tangent line.
\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}{\frac{dr}{d\theta} \cos \theta - r \sin \theta}
3. Horizontal and Vertical Tangents ★★★☆☆ ⏱ 15 min
We use the components of the polar derivative to find where polar curves have horizontal or vertical tangents, following these rules:
- Horizontal tangents occur when $\frac{dy}{d\theta} = 0$ **and** $\frac{dx}{d\theta} \neq 0$
- Vertical tangents occur when $\frac{dx}{d\theta} = 0$ **and** $\frac{dy}{d\theta} \neq 0$
- If both derivatives are zero, the slope is indeterminate and requires further analysis
Common Pitfalls
Why: $r$ is a function of $\theta$, so product rule is required for both derivatives
Why: When both derivatives are zero, $\frac{dy}{dx}$ is indeterminate, not zero, so the slope is not guaranteed to be zero
Why: Both derivatives follow similar product rules, making it easy to mix up their positions
Why: Many students think negative $r$ requires changing $\theta$ first, but this leads to extra work and errors