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partial differentiation

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Poirot

Banned
Feb 15, 2012
250
I have x=x(t) and y=y(t) and I'm working in polar co-ordinates so $$x=rcos{\theta}$$ and $$y=rsin{\theta}$$.

I want to find ${\theta}'(t)$ so by the chain rule I want $${\theta}'(x)*x'(t)+{\theta}'(y)*y'(t)$$. I know $${\theta}=arctan(y/x)$$ but how do I partially differentiate theta w.r.t x and y?
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have x=x(t) and y=y(t) and I'm working in polar co-ordinates so $$x=rcos{\theta}$$ and $$y=rsin{\theta}$$.

I want to find ${\theta}'(t)$ so by the chain rule I want $${\theta}'(x)*x'(t)+{\theta}'(y)*y'(t)$$. I know $${\theta}=arctan(y/x)$$ but how do I partially differentiate theta w.r.t x and y?
If $\displaystyle \theta(x,y)= \tan^{-1} \frac {y}{x}$ then is...

$\displaystyle \theta^{\ '}_{x}= \frac{- \frac{y}{x^{2}}}{1+ (\frac{y}{x})^{2}}= - \frac{y}{x^{2}+y^{2}}$

$\displaystyle \theta^{\ '}_{y}= \frac{\frac{1}{x}}{1+ (\frac{y}{x})^{2}}= \frac{x}{x^{2}+y^{2}}$

Kind regards

$\chi$ $\sigma$