# Application of the Inverse Function Theorem to polar co-ordinate transformation

$U=\{(u,v)\in {R}^2:u>0\}$

Define a function $F:U \rightarrow R^2$ by $F(u,v)= (u\cos(v),u\sin(v))= (x,y)$

a) Show $F$ is an open mapping on $U$. [I've done this.]

b) Calculate $\Large \frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$.

Partial result: I think I can write $u$ in terms of $x$ and $y$ by doing the following: $$x^2+y^2=u^2(\sin^2(v)+\cos^2(v))$$ Thus we have $u=\sqrt{x^2+y^2}$. Then I could differentiate with respect to $x$ of course.

However, I think there is a way to compute these partials directly from the derivative of $F$ using the Inverse Function Theorem, but I don't know how.

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Yes, you could calculate the derivative in the "induced coordinates", but the inverse function theorem would tell you where the change-of-coordinates is valid. Otherwise, I don't know what corollary of IFT you could use, other than $f(f^-1)(x)=x$ , which is not a corollary other than by allowing to talk about a well-defined inverse and its derivative/differential. –  gary Jun 17 '11 at 1:11

## 1 Answer

By the Inverse Function Theorem, the matrices $$\begin{bmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{bmatrix} \qquad\text{and}\qquad \begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{bmatrix}$$ should be inverses. The second matrix is easy to compute: $$\begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{bmatrix} \;\;=\;\; \begin{bmatrix} \cos v & -u \sin v \\ \sin v & u\cos v \end{bmatrix}$$ Inverting yields: $$\begin{bmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{bmatrix} \;\;=\;\; \frac{1}{u} \begin{bmatrix} u\cos v & u \sin v \\ -\sin v & \cos v \end{bmatrix}$$ If we want these partial derivatives in terms of $x$ and $y$, we can substitute $\cos v = x/u$ and $\sin v = y/u$ to get $$\begin{bmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{bmatrix} \;\;=\;\; \begin{bmatrix} x/u & y/u \\ -y/u^2 & x/u^2 \end{bmatrix}$$ where $u=\sqrt{x^2+y^2}$.

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But this is only valid locally; the polar charts are not globally-defined. –  gary Jun 17 '11 at 2:04
@gary: That's not a problem here, since derivatives are inherently local. The resulting formulas work globally on any branch of the inverse map. –  Jim Belk Jun 17 '11 at 2:15