# Understanding this partial derivative problem

Problem:

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

Solution:

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

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

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

Question:

(1) So first of all, why does differentiating with respect to $$x$$ result in $$\dfrac{\partial r}{\partial x}$$ and $$\dfrac{\partial \theta}{\partial x}$$ (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: $$\dfrac{\partial \theta}{\partial x} = -\dfrac{\sin \theta \frac{\partial r}{\partial x}}{\cos \theta}$$ and $$\dfrac{\partial r}{\partial x} = \dfrac{1+e^{2r}\sin \theta \frac{\partial \theta}{\partial x}}{2e^{2r}\cos \theta}$$

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!

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 $$\frac{dx}{dy}=0$$.

• thank you for helping me with this arturo Jul 1, 2011 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? Jul 1, 2011 at 8:18
• @ghshtalt: $3\sin^2\theta +2\cos^2\theta = \sin^2\theta + 2(\sin^2\theta+\cos^2\theta) = \sin^2\theta + 2(1) = 2+\sin^2\theta$. Jul 1, 2011 at 16:14

(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...

• this definitely does help, thank you Jul 1, 2011 at 7:10