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