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I have a path V[t_]:={3t^2,2,6t}, and I define F[t_]:=V[t]/Norm[V[t]]. I simply want the derivative of F, which is a 3-vector function of the variable t. Somehow I did not manage to find any (simple?) way to calculate it.. Ideas?

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  • $\begingroup$ You might want to use V[t]/Sqrt[V[t].V[t]] instead. $\endgroup$ Commented Apr 19, 2013 at 8:35
  • $\begingroup$ Alternatively, you have to use Assuming or the Assumptions when you will simplify the derivative of the function F $\endgroup$ Commented Apr 19, 2013 at 8:43
  • $\begingroup$ Thanks... I also didn't notice in my attempts.., that Mathematica was still complaining about an ancient error, hidden behind my window... $\endgroup$ Commented Apr 19, 2013 at 8:48
  • $\begingroup$ @Dror And you are sure you want this instead of the normalized derivative of v, which is the direction vector along the path? $\endgroup$ Commented Apr 19, 2013 at 9:28
  • $\begingroup$ @halirutan , thanks for the nice tip. Though, In my case (which, regretfully, I did not specify), V is actually the derivative of my original path, and I wanted to find It's normal. $\endgroup$ Commented Apr 19, 2013 at 10:25

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You don't specified what your actual problem is and what exactly you have tried. Isn't this simple enough?

v[t_] := {3 t^2, 2, 6 t}; f[t_] := v[t]/Sqrt[v[t].v[t]] D[f[t], t]
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  • $\begingroup$ Yes, that works. Somehow It didn't work when using Norm[]. Thanks! (In my next questions, all will be specified! :) ) $\endgroup$ Commented Apr 19, 2013 at 10:26

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