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I have a Jacobian function:

D[{Sqrt[x^2 + y^2], ArcTan[x, y]}, {{x, y}}]

It gives me a matrix with the formulas I need for my transposition matrix:

{{x/Sqrt[x^2 + y^2], y/Sqrt[x^2 + y^2]}, {-(y/(x^2 + y^2)), x/(x^2 + y^2)}}

How do I turn this into a formula that will give me values (the basis vectors) at a given value of $r$ and $\theta$? (e.g. f[{0.3,0.5}] = {{0.514496, 0.857493}, {-1.47059, 0.882353}})

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  • $\begingroup$ As an aside: CoordinateTransformData["Cartesian" -> "Polar", "MappingJacobian"] returns a pure function you can evaluate on vector arguments. $\endgroup$ – J. M. will be back soon Nov 15 at 0:08
  • $\begingroup$ This is a teach-a-man-to-fish kind of question. But thanks for the fish! $\endgroup$ – Quarkly Nov 15 at 0:19
  • $\begingroup$ Yes, that's why it's a comment and not an answer. ;) $\endgroup$ – J. M. will be back soon Nov 15 at 0:27
  • $\begingroup$ See, that's funny because that was the last thing the dolphins said before they left Earth. $\endgroup$ – Quarkly Nov 15 at 0:57
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ClearAll[f]
f[x0_, y0_] := D[{Sqrt[x^2 + y^2], ArcTan[x, y]}, {{x, y}}] /. {x -> x0, y -> y0}

f[0.3, 0.5] // MatrixForm

enter image description here

I would caution you against using MatrixForm in the definition, as that would leave you with results that cannot be easily used in further computation.

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You can write something like

f[{r_, θ_}] := 
 Module[{M = D[{Sqrt[x^2 + y^2], ArcTan[x, y]}, {{x, y}}]}, 
  Block[{x = r Cos[θ], y = r Sin[θ]}, M]]

I wouldn't be surprised if this has a slightly simpler formulation. The symbolic result agrees with what I expect:

Assuming[r > 0, Simplify[f[{r, θ}]]]
(* {{Cos[θ], Sin[θ]}, {-(Sin[θ]/r), Cos[θ]/r}} *)

Numerically, it doesn't agree with your expectation.

f[{0.3, 0.5}]
(* {{0.877583, 0.479426}, {-1.59809, 2.92528}} *)

I might have misunderstood...

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