Topic 9.7 Defining Polar Coordinates and Differentiating in Polar Form: convert between polar and Cartesian coordinates and find dy/dx for a polar curve r = f(theta) (BC).

What this topic is asking

The College Board (Topic 9.7, BC only) introduces polar coordinates, in which a point is located by a distance $r$ from the origin and an angle $\theta$ from the positive $x$-axis, rather than by $(x, y)$. A curve is then given as $r = f(\theta)$. The calculus task is to find the slope $\frac{dy}{dx}$ of such a curve, which you do by viewing the polar curve as a parametric curve in $\theta$.

Coordinates, conversion, and the curve as parametric

Why the slope is not dr/dtheta

A common misconception is that the slope of a polar curve is $\frac{dr}{d\theta}$. It is not: $\frac{dr}{d\theta}$ measures how the radius changes with angle, which is a different quantity from the slope $\frac{dy}{dx}$ of the traced curve in the plane. Because both $x$ and $y$ depend on $\theta$ through $r = f(\theta)$ and through the trigonometric factors, you must apply the product rule:

The slope is the ratio of these. So $\frac{dr}{d\theta}$ appears inside the formula but is not the whole story; the angle's own rotation contributes the $-r\sin\theta$ and $+r\cos\theta$ terms.

A worked polar slope

Horizontal and vertical tangents in polar form

As with parametric curves, special tangents come from the two derivatives separately. A polar curve has a horizontal tangent where $\frac{dy}{d\theta} = 0$ (and $\frac{dx}{d\theta}\neq 0$) and a vertical tangent where $\frac{dx}{d\theta} = 0$ (and $\frac{dy}{d\theta}\neq 0$). To find them, compute $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ using the product-rule expressions above and solve each for zero. This is a standard free-response task, and it is where forgetting the product rule, or confusing which derivative governs which tangent, costs marks. Sketching the curve from a few sampled angles helps confirm that the tangents you compute make geometric sense.

Recognizing common polar curves

Knowing the shapes behind the equations speeds up reasoning. The line through the origin is $\theta = c$; a circle of radius $a$ centered at the origin is $r = a$; circles through the origin are $r = a\cos\theta$ (centered on the $x$-axis) and $r = a\sin\theta$ (centered on the $y$-axis). The curves $r = a \pm b\cos\theta$ and $r = a \pm b\sin\theta$ are limaçons (a cardioid when $a = b$), and $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$ are rose curves. You do not need to memorize every shape, but recognizing that $r = 2\sin\theta$ is a circle (as the worked example's conversion suggests) lets you predict and check tangent lines and prepares you for the area topics 9.8 and 9.9, where identifying the region matters as much as the integral.