Yes! Like that. And by gum I understand as well. kappa computes
curvature of parametric curve, and its reciprocal is the radius of
curvature.
mp=: $:~ :(+/ .*)
abs=: [: %: mp
det=: -/ .*
cross=: (1 |. [: , (0 1,:2) det;._3 ,:&(,{.)"1) ([: :)
assert 0 _1 0 -: 1 0 0 cross 0 0 1
dgerund=: 3 :'{.(y`:6 d.1)`*'"0
NB. compute radius of curvature for parametric curve
NB. kappa is curvature
kappa=: 4 :0
Dx=. dgerund x
Ddx=. dgerund Dx
dx=. Dx`:0
ddx=. Ddx`:0
(([: abs (dx cross ddx)) % (3 ^~ [: abs dx)) y
)
NB. curvature of circle test
assert 1 -: (2&o.)`(1&o.)`0:kappa 9
NB. radius of curvature is reciprocal of curvature
rc=: [: % kappa
Date: Tue, 3 Nov 2015 17:05:17 -0500 From: Raul Miller
<[email protected]> To: Programming forum
<[email protected]> Subject: Re: [Jprogramming] derivatives of
a gerund Message-ID:
<CAD2jOU9ia2eJdbJzvhfE1p=9zudbgnaq5elk0zvnx20biwe...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8 Like this? dgerund=:3 :0"0
{.(y`:6 d.1)`'' ) r=:4 :0 dx=. dgerund x ddx=. dgerund dx NB. fill in
the rest... ) Note that if you have verbs which can't be handled by d.
you'll have to use a cover for d.1 which does what you need for those
cases. Thanks,
-- Raul On Tue, Nov 3, 2015 at 4:43 PM, David Lambert
<[email protected]> wrote:
>In algebraic notation I have a parametric curve
>x=u(t), y=v(t), z=v(t)
>and need radius_of_curvature(t)
>
>Would like the verb r taking a gerund argument,
>u`v`w r t
>but requires first and second derivatives.
>
>In LaTeX linear notation, with \times representing cross product, curvature
>is
>\[\frac{\|\vec{r'}\|^3}{\|\vec{r'}\times\vec{r''}\|}\]
>
>This being messy, derivatives of a gerund packed back into a gerund would be
>great, then `:0 would generate the vectors.
>
>Thanks, Dave.
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