Total derivative in polar coordinates

Question

Let's say I have the function $f$, in polar coordinates, $$(r,\theta) \rightarrow (r^2,\theta)$$ and I want to find its total derivative at some point $(r,\theta)$ (i.e. its best linear approximation). So can I say that this will be the map $$T(\rho,\phi)=\begin{pmatrix} 2r&0\\0&1 \end{pmatrix}\begin{pmatrix}\rho\\ \phi\end{pmatrix}$$ or should I first find my function in Cartesian coordinates? If so, then what is the correct formula for the total derivative in polar coordinates?

Answer 1

You were given a function $f:\>{\mathbb R}^2\to{\mathbb R}^2$ by $(r,\theta)\mapsto(r^2,\theta)$, and you have computed the matrix $$\bigl[df(r,\theta)\bigr]=\left[\matrix{2r&0\cr 0&1\cr}\right]\ ,$$ which is the matrix of the total derivative of $f$ at $(r,\theta)$. A posteriori you say that $r$ and $\phi$ are in fact polar coordinates in the $(x,y)$-plane, and that you are actually interested in the function $$\hat f:\>{\mathbb R}^2\to{\mathbb R}^2,\qquad (x,y)\mapsto\ (x',y')$$ engendered by the above $f$ relating to polar coordinates in the $(x,y)$-plane. The given $f$ says that the radius $r=\sqrt{x^2+y^2}$ of points $(x,y)$ is squared, while the argument ${\rm arg}(x,y)$ of the points $(x,y)$ is kept by $\hat f$. This geometric description of $\hat f$ says that $\hat f$ is expressed by $$x'=\sqrt{x^2+y^2}\> x,\qquad y'=\sqrt{x^2+y^2}\>y\ .\tag{1}$$ The Jacobian matrix of $(1)$ with respect to $x$ and $y$ immediately gives you the total derivative of $\hat f$ in terms of $x$ and $y$ – if this is the thing you were actually after.