# Total derivative in polar coordinates

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?

• Well, are your displacements from the point going to be given as differences in the polar or Cartesian coordinates?
– amd
Apr 25 '19 at 18:44
• @amd Well since my function is given in polar coordinates lets say the differences are in polar coordinates. Apr 25 '19 at 18:53

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.
• Yes I agree with what you wrote. But the question remains: Is the map $T$ that I gave the total derivative or not? Apr 25 '19 at 18:55