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$r$ is defined as usual in polar. Can't $r$ just be thought of as a function of $\theta$ and $x$ so that,

$$r (x, \theta)=x \sec \theta$$

$$r_x=\sec \theta$$

What is wrong with this? (The answer is $r_x=\cos \theta$) I know that if I look at $r=f(x,y)=\sqrt{x^2+y^2}$ I can get the answer but I want to know what is wrong with this work.

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The problem is that you're mixing your variables: $(r,\theta)$ are one set of coordinates, $(x,y)$ another. But $\theta$ depends on $x$ and $y$ just as $r$ does, so $\partial_x\theta \neq 0$. What you've calculated is $$\left.\frac{\partial r}{\partial x}\right|_{\theta \text{ const.}},$$ which is not the same as $$\left.\frac{\partial r}{\partial x}\right|_{y \text{ const.}},$$ the (implied) meaning of $\partial r/\partial x$ (partial derivative of new with respect to one old, holding the other olds constant). This is an important distinction to understand, especially in subjects where all the variables get mixed up like thermodynamics, but in changing coordinate systems in simple multivariable calculus, the second one is the usual convention.

What is $\partial_x\theta$ with $y$ constant? At least locally, we have $$ \cot{\theta} = x/y, $$ so differentiating implicitly, $$ -\csc^2{\theta}\left. \frac{\partial \theta}{\partial x}\right|_{y} = 1/y, $$ or $-(\sin^2{\theta})/y = -(\sin{\theta})/r$. Then, using the chain rule, \begin{align} \left.\frac{\partial r}{\partial x}\right|_{y} &= \left.\frac{\partial}{\partial x}\right|_{y} x \sec{\theta} \\ &= \sec{\theta} + x\sec{\theta}\tan{\theta} \left.\frac{\partial\theta}{\partial x}\right|_{y} \\ &= \sec{\theta} - r\tan{\theta} \frac{\sin{\theta}}{r} \\ &= \frac{1}{\cos{\theta}}-\frac{\sin^2{\theta}}{\cos{\theta}} \\ &= \cos{\theta}, \end{align} as it should be. Of course, this is a lot slower than just using $$ \left.\frac{\partial r}{\partial x}\right|_{y} = \left.\frac{\partial}{\partial x}\right|_{y} \sqrt{x^2+y^2} = \frac{x}{\sqrt{x^2+y^2}} = x/r = \cos{\theta}. $$

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  • $\begingroup$ Thanks. What I'm looking for! I'm thinking that because $\theta$ is a function of $x$ and $y$ that when we take the partial of $\theta$ with respect to $x$, then $\theta$ can be thought of as a function of $x$ because $y$ is constant. So we then can apply product rule and chain rule. Is that the right thinking, may you please let me know. $\endgroup$ – Ahmed S. Attaalla Apr 17 '17 at 2:08
  • $\begingroup$ Yes, I think that's pretty much my reasoning. What is constant when taking a partial derivative is just as important as what is not constant! $\endgroup$ – Chappers Apr 17 '17 at 3:11

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