I have an issue with derivatives of vectors

$Assumptions :=  v ∈ Vectors[3, Reals] (*Assuming v is a 3d vector*)

D[Norm[v]^2, v.{0, 0, 1}] (*differentiate with respect to third component*)

Output: 0

But isn't this supposed to be: $\frac{\partial (v_1^2+v_2^2+v_3^2)}{\partial v_3}=2v_3$

  • $\begingroup$ If my answer below answered your question, please accept it so that we can get this question off the unanswered list and so that I can get those meaningless internet points that I so crave! $\endgroup$ – march Nov 6 '15 at 4:09
  • $\begingroup$ @march Ah! Yep here are your well-deserved points :) Thanks a lot for the help! I simply forgot to click ;) $\endgroup$ – Otto Nov 7 '15 at 1:56

Here's how you should probably do this:

vec = Array[v, 3]
D[vec.vec, vec.{0, 0, 1}]
(* {v[1], v[2], v[3]} *)
(* 2 v[3] *)

I'll just make a couple of points about D.

  • It doesn't behave well when given functions like Abs and Norm:

    D[Norm[{a, b, c}]^2, a]
    (* 2 Abs[a] Abs'[a] *)

    Instead, you should typically use more explicit forms of vector norms, which is why I used

    (* v[1]^2 + v[2]^2 + v[3]^2 *)
  • I would guess that Vectors is mainly useful for doing symbolic tensor math, as shown in the documentation. It's not for what you're trying to do with it.

  • Meditate on the result of the following evaluation; it will show you why your result was zero and how D interprets its arguments.

    D[v.{0, 0, 1}, v.{0, 0, 1}]
  • 2
    $\begingroup$ I upvoted this even though your rep counter was just at 4444, and now I messed up that nice number. Sorry for that, I guess. $\endgroup$ – Jens Oct 15 '15 at 5:27
  • $\begingroup$ @Jens I wish I could have stayed there forever. $4444 = 66^2 + 88$; $4444 = 4\times1111$ with $1+1+1+1 = 4$. $\endgroup$ – march Oct 15 '15 at 15:35
  • 2
    $\begingroup$ You are correct that Vectors is meant for symbolic tensor manipulations and simplifications. $\endgroup$ – Stefan R Oct 16 '15 at 17:22

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