How to translate differential form from polar to cartesian?

212 Views Asked by At

Please tell me how to translate this differntial form from polar to cartesian? I understand that I should suppose $x =r \cos{\varphi}$ and $y = r \sin{\varphi}$. But I do not understand how to fix $\dot{r}$ and $\dot{\varphi}$. Should I solve this equation to do that? I will be grateful if you show me how it works step by step!

$ \dot{r} = (r^2 - 1) (2 r \cos{\varphi} - 1)$

$ \dot{\varphi} = 1$

1

There are 1 best solutions below

1
On BEST ANSWER

$$x=r\cos\varphi$$ $$y=r\sin\varphi$$

$\implies$ Apply the chain and product rules and we have

$$\dot x=-r\dot\varphi\sin\varphi+\dot r\cos\varphi$$ $$\dot y=r\dot\varphi\cos\varphi+\dot r\sin\varphi$$

$\implies$ Substituting in equations for $x,y,\dot r,\dot\varphi$

$$\dot x=-y+(r^2-1)(2r\cos\phi-1)\cos\varphi$$ $$\dot y=x+(r^2-1)(2r\cos\phi-1)\sin\varphi$$

$\implies$ Substituting in equations for $x,r,\cos\varphi,\sin\varphi$

$$\dot x=-y+(x^2+y^2-1)(2x-1)\frac{x}{\sqrt{x^2+y^2}}$$ $$\dot y=x+(x^2+y^2-1)(2x-1)\frac{y}{\sqrt{x^2+y^2}}$$