Tangents with Polar Coordinates

A tangent to a polar curve is simply the straight line that matches the curve's direction at one point, just like any other tangent line; only the way the curve is described has changed. The trick is that a polar curve is a parametric curve where the angle $\theta$ plays the role of time, so we find its slope exactly the way we did for parametric curves. It's good for finding the highest point of a petal, where a pattern turns straight up, or which way a curve leaves the origin. The big fraction looks awful, but it's just two pieces: the top is $dy/d\theta$ (how fast the point rises as the angle turns) and the bottom is $dx/d\theta$ (how fast it moves sideways). Inside each, $f(\theta)$ is the distance from the origin and $f'(\theta)$ is how fast that distance is changing. It's the same rise over run, written out with the product rule; bear with it.

Polar is parametric in disguise. Here's the question: what's the slope of a curve like $r = 1 + \cos\theta$? It's tempting to think it's $\frac{dr}{d\theta}$, but that only tells you how fast the curve moves away from the origin, not which way it's heading on the page. To get an honest slope we need $x$ and $y$.

And we already have them. Combine $r = f(\theta)$ with the conversion formulas and you get

That's a parametric curve, with $\theta$ playing the role $t$ played before. So everything from the parametric tangents lesson carries over. Differentiate both with the product rule (since $r$ and $\cos\theta$ both change with $\theta$), then divide $dy/d\theta$ by $dx/d\theta$. That's where the formula above comes from: it isn't new, just the old formula written out.

The special tangents are the same as before, too:

• horizontal tangents where $\dfrac{dy}{d\theta} = 0$ (and $\dfrac{dx}{d\theta} \ne 0$),
• vertical tangents where $\dfrac{dx}{d\theta} = 0$ (and $\dfrac{dy}{d\theta} \ne 0$).

At the origin. Something nice happens when the curve passes through the pole. If $f(\theta_0) = 0$ and $f'(\theta_0) \ne 0$, every term with $f$ in it drops out, and the formula reduces to $\dfrac{f'\sin\theta_0}{f'\cos\theta_0} = \tan\theta_0$. That makes sense: when $r$ is passing through $0$, the curve is moving straight out along the ray at angle $\theta_0$, so it leaves the origin along the line $\theta = \theta_0$.

Satellite dishes and car headlights use curves whose tangents bounce rays toward a single point. Designing them means controlling exactly these slopes.