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Considering $x$ and $y$ as independent variables, find $\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y}$ when $x = e^{2r} \cos \theta, y = e^{3r} \sin \theta$.


First differentiate the given relations with respect to $x$:

$1 = 2e^{2r} \cos \theta \frac{\partial r}{\partial x} - e^{2r} \sin \theta \frac{\partial \theta}{\partial x}$ and $0 = 3e^{3r}\sin \theta \frac{\partial r}{\partial x} + e^{3r} \cos \theta \frac{\partial \theta}{\partial x}$.

Then solve simultaneously to obtain $\frac{\partial r}{\partial x} = \frac{\cos \theta}{e^{2r}(2+\sin^{2} \theta)}$ and $\frac{\partial \theta}{\partial x} = - \frac{3 \sin \theta}{e^{2r}(2+sin^2 \theta)}$


(1) So first of all, why does differentiating with respect to $x$ result in $\frac{\partial r}{\partial x}$ and $\frac{\partial \theta}{\partial x}$ (by the way what are these called?)? Is this because the problem says "$x$ and $y$ as independent variables" ? My initial reaction was to do $r$ and $\theta$ separately while regarding all other variables as constants... This is implicit (partial?) differentiation, right?

How should I understand what is being done here? A lot of times it seems that things like this turn out to really just be a mapping. Can I think of this that way as well? I couldn't even say what the domain and codomain would be...

Whenever I perform this operation, do I just take partial derivatives of both sides treating all independent variables (other than the variable with respect to which I am differentiating) as constants and all non-independent (dependent?) variables as variables that need to be differentiated and will have that partial derivative symbol? Once I accept that I can see how they got the first implicit partial differentiation (if that's what it is called).

(2) What do they mean by "solve simultaneously" ? I tried to solve for each (again I don't know what they're called yet) "partial differential" resulting in: $\frac{\partial \theta}{\partial x} = -\frac{\sin \theta \frac{\partial r}{\partial x}}{\cos \theta}$ and $\frac{\partial r}{\partial x} = \frac{1+e^{2r}\sin \theta \frac{\partial \theta}{\partial x}}{2e^{2r}\cos \theta}$. But I couldn't get the same answer... I tried to substitute this further, but looking at that answer, I was sure I was missing something... Could somebody please show me what to do?

Thank you in advance for any help!

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up vote 4 down vote accepted
  1. That's just the Chain Rule (with partials because you know there are 'really' two variables around). Think of $r$ as a function $r(x,y)$, and $\theta$ as a function $\theta(x,y)$, and you are just doing implicit differentiation/chain rule. The fact that "$x$ and $y$ are independent variables" tells you that $\frac{dy}{dx} = 0$, so that's why you get that differentiating the relation $y=e^{3r}\cos\theta$ with respect to $x$ gives you $0$ on the left hand side.

  2. You have two expressions, both involving the unknown functions $\frac{\partial r}{\partial x}$, $\frac{\partial \theta}{\partial x}$. Think of those two as the "unknowns" in a $2\times 2$ system, and solve the system (like you would a regular system of two equations in two unknowns) to get expressions for each involving only $r$s and $\theta$s, and no partials, $x$s, or $y$s.

    E.g., take the second equation and multiply through by $2\cos\theta$ to get $$0 = 6e^{3r}\sin\theta\cos\theta\frac{\partial r}{\partial x} + 2e^{3r}\cos^2\theta\frac{\partial\theta}{\partial x}.$$ Then take the first equation and multiply through by $-3e^r\sin\theta$ to get $$-3e^r\sin\theta = -6e^{3r}\sin\theta\cos\theta\frac{\partial r}{\partial x} +3e^{3r}\sin^2\theta \frac{\partial\theta}{\partial x}.$$ Adding both equations eliminates $\frac{\partial r}{\partial x}$, so now you can solve for $\frac{\partial\theta}{\partial x}$. Etc.

Then you do the same thing but with partials with respect to $y$, using the fact that $x$ and $y$ are independent to get that $\frac{dx}{dy}=0$.

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thank you for helping me with this arturo – ghshtalt Jul 1 '11 at 7:11
sorry, i had read your answer and thought it was a done deal, but i seem to still be stuck: taking the step you suggest, i get $\frac{\partial \theta}{\partial x} = - \frac{-3 \sin \theta}{e^{2r}} \frac{1}{3\sin^{2}\theta + 2\cos^{2} \theta}$. is there some way that i can change $3\sin^{2}\theta + 2\cos^{2}\theta$ into the expression $2+\sin^{2}\theta$ as in the solution above? – ghshtalt Jul 1 '11 at 8:18
@ghshtalt: $3\sin^2\theta +2\cos^2\theta = \sin^2\theta + 2(\sin^2\theta+\cos^2\theta) = \sin^2\theta + 2(1) = 2+\sin^2\theta$. – Arturo Magidin Jul 1 '11 at 16:14
thank you again! – ghshtalt Jul 1 '11 at 16:43

(1) So first of all, why does differentiating with respect to x result in $\frac{\partial r}{\partial x}, \frac{\partial \theta}{\partial x}$

You're using multidimension chain rule. You want to take a derivative of a function $f(r,\theta)$ with respect to $x$, but both $r$ and $\theta$ depend on $x$. So $\dfrac{d}{dx}x = \dfrac{d}{dx}\left(e^{2r} \cos \theta\right) = $ by chain rule $ \dfrac{\partial}{\partial r}\left(e^{2r}\cos\theta\right)\dfrac{\partial r}{\partial x} + \dfrac{\partial}{\partial\theta}\left(e^{2r}\cos\theta\right)\dfrac{\partial\theta}{\partial x}$.

Similalry $\dfrac{\partial{f(r,\theta)}}{\partial x} = \dfrac{\partial{f(r,\theta)}}{\partial r}\dfrac{\partial r}{\partial x} +\dfrac{\partial{f(r,\theta)}}{\partial \theta}\dfrac{\partial \theta}{\partial x}$

The same is true for $y$. I hope this helps... I'm not sure if this is your question, since you used the chain rule fine. Let me know and I'll revise...

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this definitely does help, thank you – ghshtalt Jul 1 '11 at 7:10

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