Re: Newbie mathematica question

*To*: mathgroup at smc.vnet.net*Subject*: [mg37176] Re: Newbie mathematica question*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Tue, 15 Oct 2002 04:17:51 -0400 (EDT)*References*: <ao3app$n2d$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

[My previous reply seems to have gone astray - at least it has not come back to me. Here is a slightly edited repeat] You have computed f[t_] = {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]}; and then found f' {1 + Sqrt[3]*Cos[#1], -2*Sin[#1], Sqrt[3] - Cos[#1]} & To get the function for the norm of the derivative we can use norm = Evaluate/@(Simplify/@(Sqrt[#.#]&/@(f'))) 2*Sqrt[2] & We map the usual functions for calclulating and simplifying the norm inside Function[.] (which is the full form of (.)& and then map the function Evaluation to make the result evaluate -- this is needed since Function has the attribute HoldAll. Please note that the parentheses are essential. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Anonymous" <not at for.you> wrote in message news:ao3app$n2d$1 at smc.vnet.net... > 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 > >