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As the title shows this question concerns nothing but chain rule. We now have:

$$\frac{d\theta}{dt}=\frac{x}{r^{2}}\frac{dy}{dt}-\frac{y}{r^{2}}\frac{dx}{dt}$$

I am assuming by chain rule we have $$\frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}+\frac{d\theta}{dy}\frac{dy}{dt}$$

But we have $$\theta=\arccos[\frac{x}{r}]=\arcsin[\frac{y}{r}]$$

Thus taking the derivative we should assume $$\frac{d\theta}{dx}=-\frac{1}{y},\frac{d\theta}{dy}=\frac{1}{x}$$ because $$\frac{d}{dx}\arccos[\frac{x}{r}]=-\frac{1}{r\sqrt{1-\frac{x^{2}}{r^{2}}}}=-\frac{1}{\sqrt{r^{2}-x^{2}}}=-\frac{1}{y}$$

However we know $$-\frac{y}{r^{2}}\not=-\frac{1}{y}$$ I computed this a few times but do not know where I got wrong. The relationship in the title is in Berkeley Problems in Mathematics.

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    Try $\theta=\arctan(y/x)$.2012-07-21
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    @GerryMyerson: Thanks. I still do not know where I had the mistake, but obviously you are right...2012-07-21
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    In the second equality, by the chain rule you have **either** ·$$\frac{{d\theta }}{{dt}} = \frac{{d\theta }}{{dx}}\frac{{dx}}{{dt}}$$ **or** $$\frac{{d\theta }}{{dt}} = \frac{{d\theta }}{{dy}}\frac{{dy}}{{dt}}$$2012-07-21
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    I see. Since I am assuming $\theta=f(r,x)$ with $r$ constant there is no $y$ involved. Thanks.2012-07-22
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    I don't think you computed $\frac{d\theta}{dx}$ correctly.2012-07-22
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    @Peter I assumed the OP meant $\frac{d\theta}{dt}=\frac{\partial\theta}{\partial x}\frac{dx}{dt}+\frac{\partial\theta}{\partial y}\frac{dy}{dt}$2012-07-22
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    @Brian: It should correct. You may compute yourself.2012-07-22
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    @user32240 $r^2 = x^2 + y^2$?2012-07-22
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    @Brian: yeah, standard polar coordinates.2012-07-22
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    I think the question is poorly stated: too thrifty in definitions and details (for example, what is every variable?) The OP should make all this crystal clear, although it is *almost* obvious he means polar coordinates.2012-07-22
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    @DonAntonio: Excuse for my low level, but I assume calculus is common sense on this forum. But thanks for the reminder. I shall write future questions in a cleaner way.2012-07-22
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    @user32240, don't ask for excuses when you don't need them: yes, calculus is common knowledge here, but not necessarily is the notation of a particular thing related to it.2012-07-22
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    @user32240 Maybe you should check your work again because $\frac{d\theta}{dx} = -\frac{y}{x^2+y^2} = -\frac{y}{r^2}$2012-07-22
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    @Brian: No, $\theta=\arccos[x/r$, the computation shows the other way.2012-07-22
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    Can you show your computation?2012-07-22
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    Yeah, updated. You can check.2012-07-22
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    $r$ is not a constant2012-07-22
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    @Brian: yeah, that's why I pointed that it in previous comments.2012-07-22
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    After Gerry told you, you once again told Brian that $\,\theta=\arccos x/r\,$ , and this is incorrect: $$\frac{y}{x}=\frac{\sin\theta}{\cos\theta}=\tan\theta\Longrightarrow \theta=\arctan\frac{y}{x}... $$2012-07-22
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    @DonAntonio: The relationship is not incorrect, but it is incorrect to treat $r$ as a constant, which is why I commented earlier.2012-07-22

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After the clarification (polar coordinates) and summarizing the answers given in the comments above: $$x=r\cos\theta\,\,,\,\,y=r\sin\theta\,\,\,,\,\,r\geq0\,,\,\,\theta\in [0,2\pi]$$ I'm assuming the radius is always non-negative, though not all do this.

From the above, $\,\theta=\arctan\frac{y}{x}\,\,,\,x\neq 0$ (the case $\,x=0\,$ is an easy particular case depending on the sign of $\,y\,$), so if both rectangular coordinates are derivable functions of some parameter $\,t\,$, we'd get: $$\frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}+\frac{d\theta}{dy}\frac{dy}{dt}$$ $$\frac{d\theta}{dx}=-\frac{y}{x^2}\frac{1}{1+\left(\frac{y}{x}\right)^2}=-\frac{y}{x^2+y^2}$$ $$\frac{d\theta}{dy}=\frac{1}{x}\frac{1}{1+\left(\frac{y}{x}\right)^2}=\frac{x}{x^2+y^2}$$

Observe that writing the expressions for $\,x,y\,$ from the beginning you get two differential equations. I'll leave this here as I'm not completely sure whether this already answers your question.

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    thanks a lot, but no need to do that. Gerry's answer is enough.2012-07-22