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MathGroup Archive 2002

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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
>
>




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