A polar curve $r = f(\theta)$ is really a parametric curve with $\theta$ as the parameter. Once you write it that way, slopes and lengths come straight from the parametric formulas, and they pair well with polar area.
Por qué funciona
Write $x = r\cos\theta$ and $y = r\sin\theta$ with $r = f(\theta)$. The product rule gives
$$\frac{dx}{d\theta} = r'\cos\theta - r\sin\theta,$$ $$\frac{dy}{d\theta} = r'\sin\theta + r\cos\theta.$$
Squaring and adding, the cross terms $\pm 2rr'\sin\theta\cos\theta$ cancel and $\sin^{2} + \cos^{2} = 1$, leaving
$$\left(\frac{dx}{d\theta}\right)^{2} + \left(\frac{dy}{d\theta}\right)^{2} = r^{2} + (r')^{2}.$$
That makes sense: as $\theta$ changes, the point moves out along the ray at rate $r'$ and around the circle at rate $r$, two directions at right angles.
Idea clave
$$\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta},$$ $$L = \int_{\alpha}^{\beta}\sqrt{r^{2} + \left(\frac{dr}{d\theta}\right)^{2}}\,d\theta.$$
Ejemplo resuelto: A slope on a limaçon
Find the slope of $r = 1 + \sin\theta$ at $\theta = 0$.
- At $\theta = 0$: $r = 1$, $r' = \cos 0 = 1$, $\sin\theta = 0$, $\cos\theta = 1$.
- $\frac{dy}{dx} = \frac{(1)(0) + (1)(1)}{(1)(1) - (1)(0)} = 1$.
Respuesta: $1$
Ejemplo resuelto: The length of a cardioid
Find the length of $r = 1 + \cos\theta$.
- $$\begin{aligned} &r^{2} + (r')^{2} \\ &= (1 + \cos\theta)^{2} + \sin^{2}\theta \\ &= 2 + 2\cos\theta \\ &= 4\cos^{2}\frac{\theta}{2}. \end{aligned}$$
- So the integrand is $2\left|\cos\frac{\theta}{2}\right|$. By symmetry, double the length for $0 \le \theta \le \pi$, where $\cos\frac{\theta}{2} \ge 0$.
- $L = 2\int_0^{\pi}2\cos\frac{\theta}{2}\,d\theta = 2\left[4\sin\frac{\theta}{2}\right]_0^{\pi} = 8$.
Respuesta: $8$
Ejemplo resuelto: A spiral
Find the length of $r = \theta^{2}$ for $0 \le \theta \le \sqrt{5}$.
- $r^{2} + (r')^{2} = \theta^{4} + 4\theta^{2} = \theta^{2}(\theta^{2} + 4)$, so the integrand is $\theta\sqrt{\theta^{2} + 4}$.
- $$\begin{aligned} L &= \frac{1}{3}\left[(\theta^{2} + 4)^{3/2}\right]_0^{\sqrt{5}} \\ &= \frac{1}{3}(27 - 8) \\ &= \frac{19}{3}. \end{aligned}$$
Respuesta: $\frac{19}{3}$
Error común
Using the Cartesian formula $\int\sqrt{1 + (r')^{2}}\,d\theta$. Test on the circle $r = 3$ for $0 \le \theta \le 2\pi$, whose length is $6\pi$: the correct integrand is $\sqrt{9 + 0} = 3$, giving $6\pi$; the wrong one gives $\sqrt{1} = 1$ and only $2\pi$. In polar form the "1" is replaced by $r^{2}$, because moving around a circle of radius $r$ covers distance $r\,\Delta\theta$, not $\Delta\theta$.
At the origin ($r = 0$) the slope formula simplifies to $\frac{dy}{dx} = \tan\theta$: the curve leaves the pole along the ray at that angle. For the rose $r = \cos 3\theta$, the petals touch the pole along $\theta = \pm\frac{\pi}{6}$, $\frac{\pi}{2}$.