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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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