$dr/d\theta$ is not a slope BC only

A polar pair $(r, \theta)$ locates a point by a distance and a direction. Dropping the point onto the axes gives the conversion directly from right-triangle trigonometry:

$$x = r\cos\theta, \qquad y = r\sin\theta, \qquad r^{2} = x^{2} + y^{2}, \qquad \tan\theta = \frac{y}{x}.$$

Two features have no rectangular analogue and both cause trouble later:

1. A point has infinitely many polar names. $(2, \frac{\pi}{3})$, $(2, \frac{\pi}{3} + 2\pi)$ and $(-2, \frac{\pi}{3} + \pi)$ are the same point. Adding $2\pi$ to the angle changes nothing, and a negative $r$ means "go backwards along the ray".
2. The last conversion loses a quadrant. $\tan\theta = \frac{y}{x}$ has two solutions per revolution, so the arctangent has to be adjusted by checking which quadrant the point is in.

A polar curve $r = f(\theta)$ is then just an instruction: at each angle, travel that far out. The curve $r = 2\cos\theta$, for instance, is a circle of radius $1$ centred at $(1, 0)$, which is not at all obvious from the equation and is immediate once a few points are plotted.

$\frac{dr}{d\theta}$ measures how fast the distance from the origin changes as the angle sweeps. That is a genuine quantity and it answers a genuine question. It is simply not the question "how steep is this curve".

A slope is $\frac{dy}{dx}$, a rate of one rectangular coordinate against the other. Neither $y$ nor $x$ is $r$, so no amount of differentiating $r$ produces one.

The cleanest demonstration is a curve where $r$ never changes. On the circle $r = 3$, $\frac{dr}{d\theta} = 0$ everywhere. If that were the slope, the circle would be horizontal at every point, which is not merely wrong but impossible for a closed curve.

Two consequences that follow directly:

1. Setting $\frac{dr}{d\theta} = 0$ does not find horizontal tangents. It finds where the distance from the origin stops changing, which is where the curve is momentarily circular rather than momentarily flat.
2. $\frac{dr}{d\theta}$ can be zero where the curve is steep, and nonzero where the curve is flat. The two quantities are not related by any inequality, let alone an equality.

Treat $\theta$ as a parameter. Substituting $r = f(\theta)$ into the conversion gives a parametrisation of the curve:

$$x(\theta) = f(\theta)\cos\theta, \qquad y(\theta) = f(\theta)\sin\theta,$$

and 9.1's quotient applies unchanged:

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

Both numerator and denominator need the product rule, since each is a product of $f(\theta)$ with a trigonometric function. That is the step CA9 skips, and skipping it is what leaves $\frac{dr}{d\theta}$ sitting where a slope should be.

Work the running example, $r = 2\cos\theta$. Then $x = 2\cos^{2}\theta$ and $y = 2\cos\theta\sin\theta = \sin 2\theta$, so $\frac{dx}{d\theta} = -2\sin 2\theta$ and $\frac{dy}{d\theta} = 2\cos 2\theta$, giving

$$\frac{dy}{dx} = \frac{2\cos 2\theta}{-2\sin 2\theta} = -\cot 2\theta.$$

At $\theta = \frac{\pi}{6}$ the slope is $-\cot\frac{\pi}{3} = -\frac{1}{\sqrt{3}} \approx -0.577$, while $\frac{dr}{d\theta} = -2\sin\frac{\pi}{6} = -1$. Two different numbers, and only one of them is a slope.

Tangents are then read exactly as in 9.1: horizontal where $\frac{dy}{d\theta} = 0$ with $\frac{dx}{d\theta} \neq 0$, vertical where the roles swap.

Area in polar form is not $\int y\,dx$ and cannot be, because the natural slice is not a vertical strip. Sweeping the angle by $d\theta$ sweeps out a thin circular sector of radius $r$, and a sector of angle $d\theta$ has area $\frac{1}{2}r^{2}\,d\theta$. Adding the sectors gives

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

Both constants are load-bearing and both go missing independently:

1. The $\frac{1}{2}$ comes from the sector, exactly as a triangle's $\frac{1}{2}$ does. Integrating $r^{2}$ alone doubles the answer.
2. The square comes from the sector's area depending on the radius twice. Integrating $r$ alone accumulates lengths, not areas, and does not even have the units of an area.

And, as in every integral since 8.9, the squaring happens inside. $\left(\int r\,d\theta\right)^{2}$ is not $\int r^{2}\,d\theta$, and the difference is not a constant factor.

Topic 9.8 takes this up properly, where the hard part turns out not to be the formula but the limits.