Converting polar equation into cartesian equation to obtain derivatives

Question

If I have a polar equation such as

$r=1-2\cos(\theta)$

How would I convert this into an equation for $x$ and $y$ so that I can get $dx/d\theta$ and $dy/d\theta$ ?

Answer 1

We have $x=r\cos\theta$ and $y=r\sin\theta$. So for example we write $x=(1-\cos 2\theta)(\cos\theta)$ and differentiate.

Answer 2

Very broad hint: $$x = r \cos\theta \Rightarrow \frac{dx}{d \theta} = \frac {\partial x}{\partial r} \frac{dr}{d\theta} + \frac {\partial x}{\partial \theta}\\ =\cos\theta (2 \sin\theta) -r \sin\theta = 2\cos\theta \sin\theta - (1-2 \cos\theta) \sin\theta$$

Do the same for $y$.

Answer 3

Using conversion between coordinate spaces $x = r \cos \theta$, $y = r\sin \theta$ and the chain rule for derivatives.

$\frac{\text{dx}}{\text{d$\theta $}}=\frac{d ( r \cos \theta )}{\text{d$\theta $}}=\frac{\text{dr} \cos \theta}{\text{d$\theta $}}- r \sin \theta =2 \sin (2 \theta )-\sin\theta$

$\frac{\text{dy}}{\text{d$\theta $}}=\frac{d ( r \sin \theta)}{\text{d$\theta $}}=\frac{\text{dr} }{\text{d$\theta $}} \sin \theta+ r \cos \theta=2\sin ^2 \theta-2 \cos ^2 \theta+\text{cos$\theta $}= \cos\theta -2 \cos 2\theta$

If you want to further convert it into x and y, you can use $\tan \theta = \frac{x}{y}$ and $1+\tan^2 \theta = \frac{1}{\cos^2\theta}$