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MathGroup Archive 2002

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Re: Newbie mathematica question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg37157] Re: [mg37123] Newbie mathematica question
  • From: purcell <chris.purcell at drdc-rddc.gc.ca>
  • Date: Sun, 13 Oct 2002 05:57:03 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

If you use SetDelayed rather than Set, and then Simplify, Mathematica 
returns the answer you expect:

f[t_] := {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} ;
Simplify[Sqrt[f'[t] .f'[t]]]


Out[2]= 2 Sqrt[2]


>Hi, I have defined a function from R -> R^3 as follows: f[t_] = {t + 
>Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} So It's basically a vector 
>whose coordinates are determined based on the values you pass in. Then I 
>took the derivative by just typing f', which outputs {1 + Sqrt[3] Cos[#1], 
>-2 Sin[#1], Sqrt[3] - Cos[#1]}& What I'd like to do is have Mathematica 
>calculate the norm of this as it would any vector, so that I can play with 
>the norm function. As it turns out, the norm in this case is identical to 
>Sqrt[8], so it would be nice if Mathematica could figure that out. Is it 
>possible to do this? Thanks

Christopher J. Purcell
Defence R&D Canada ? Atlantic
9 Grove St., PO Box 1012
Dartmouth NS Canada B2Y 3Z7
Tel 902-426-3100 x389, Fax 902-426-9654
E-mail:    chris.purcell at drdc-rddc.gc.ca



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