Can the slope in polar coordinate be expressed as $dr/rd\theta$?

Question

Can the slope in polar coordinate be expressed as $\frac{dr}{rd\theta}$?

Answer 1

The slope of a polar function is

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{\frac{\mathrm{d}r}{\mathrm{d}\theta}\cdot\sin\theta + r\cdot\cos\theta}{\frac{\mathrm{d}r}{\mathrm{d}\theta}\cdot\cos\theta - r\cdot\sin\theta}$$

Answer 2

Let $x=r \cos (\theta)$ and $y=r \sin (\theta)$.

We have by the product rule,

$$\frac{dx}{d\theta}=-r\sin( \theta)+\frac{dr}{d\theta}\cos(\theta)$$

We also have,

$$\frac{dy}{d\theta}=r\cos (\theta)+\frac{dr}{d\theta}\sin( \theta)$$

So it follows by the chain rule,

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

Solving for $\frac{dy}{dx}$ gives,

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

Substituting in the values we found we get,

$$\frac{dy}{dx}=\frac{r\cos (\theta)+\frac{dr}{d\theta}\sin( \theta)}{-r\sin( \theta)+\frac{dr}{d\theta}\cos(\theta)}$$