0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Well, are your displacements from the point going to be given as differences in the polar or Cartesian coordinates? $\endgroup$
    – amd
    Apr 25 '19 at 18:44
  • $\begingroup$ @amd Well since my function is given in polar coordinates lets say the differences are in polar coordinates. $\endgroup$
    – Mad Max
    Apr 25 '19 at 18:53
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Yes I agree with what you wrote. But the question remains: Is the map $T$ that I gave the total derivative or not? $\endgroup$
    – Mad Max
    Apr 25 '19 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.