# Prove this differential equation of a angle

prove: $d\theta = \frac{1}{r^2} \left( x\;dy - y\;dx \right)\quad\text{where }r^2 = x^2 + y^2$. Full solution would be appreciated. please show your step because i can't follow if you skip the step

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What have you tried? –  mixedmath Oct 15 '11 at 0:14
@mixedmath - i didn't study calculus yet but i try to self study myself, so i really mess up to figure myself. –  Victor Oct 15 '11 at 0:20

let $x=r\cos\theta$ and $y=r\sin\theta$. then differentiate both with respect to $\theta$ then multiply $\frac{\text{d}x}{\text{d}\theta}$ by $y$ and then $\frac{\text{d}y}{\text{d}\theta}$ by $x$. find their difference and send $\text{d}\theta$ to the other side to obtain the result. This should be very easy to follow because $$x\left(\frac{\text{d}y}{\text{d}\theta}\right)-y\left(\frac{\text{d}x}{\text{d}\theta}\right)=x^2+y^2=r^2.$$ Remember that, $\frac{\text{d}x}{\text{d}\theta}=-r\sin\theta=-y$. So, $y\frac{\text{d}x}{\text{d}\theta}=-y^2$. Similarly, $\frac{\text{d}y}{\text{d}\theta}=r\cos\theta=x$. So, $x\frac{\text{d}y}{\text{d}\theta}=x^2$. Then, $$x\frac{\text{d}y}{\text{d}\theta}-y\frac{\text{d}x}{\text{d}\theta}=x^2+y^2=r^2.$$ Thus, $(x\text{d}y-y\text{d}x)/\text{d}\theta=r^2$ and the result follows.