Re: [Maxima-discuss] Symbolic differentiation

[Rupert S. <rsw...@gm...>]

David Brant <bra...@ho...> writes:
> Hello. How might i best implement the following basic operations in
> maxima to get the two solutions ?
>
> m(t)^2= e(t)^2^ - k    (m and e are functions of t, k is a constant)
>
> 1) differentiate wrt t
>
> 2*m(t)*dm/dt = 2*e(t)*de/dt
>
>
> 2) tidy up
>
> dm/dt = e(t)/r(t)*de/dt    (solution 1)
>
>
> 3) differentiate again wrt t
>
> m*d2m/dt2 + (dm/dt)^2 = e*d2e/dt2 + (de/dt)^2
>
>
> 4) tidy up again to finally get
>
> d2m/dt2 = e/m*d2e/dt2 + ((de/dt)^2 - (dm/dt)^2)/m (solution 2)
>
>
> Any help would be very much appreciated.

Something like this?

    (%i34) eq1: m(t)^2 = e(t)^2 - k;
                                    2       2
    (%o34)                         m (t) = e (t) - k
    (%i35) diff(eq1, t);
                                d                    d
    (%o35)              2 m(t) (-- (m(t))) = 2 e(t) (-- (e(t)))
                                dt                   dt
    (%i36) solve(%, 'diff(m(t), t));
                                               d
                                         e(t) (-- (e(t)))
                             d                 dt
    (%o36)                  [-- (m(t)) = ----------------]
                             dt                m(t)
    (%i37) diff (first (%), t);
                                                                2
                                                               d
                                d           d            e(t) (--- (e(t)))
            2             e(t) (-- (e(t))) (-- (m(t)))           2
           d                    dt          dt                 dt
    (%o37) --- (m(t)) = - ---------------------------- + -----------------
             2                        2                        m(t)
           dt                        m (t)
                                                                        d         2
                                                                       (-- (e(t)))
                                                                        dt
                                                                     + ------------
                                                                           m(t)
    (%i38) subst (first (%o36), %);
                               2
                              d
                        e(t) (--- (e(t)))    d         2    2     d         2
            2                   2           (-- (e(t)))    e (t) (-- (e(t)))
           d                  dt             dt                   dt
    (%o38) --- (m(t)) = ----------------- + ------------ - ------------------
             2                m(t)              m(t)              3
           dt                                                    m (t)


In case the last line is a bit cryptic, I'm basically doing something
like

  subst (y=a, expr(y));

to get

  expr(a)

But the equation is coming from the solution that solve found in %o36,
so "y" is m'(t) and "a" is e'(t)*e(t)/m(t).

It occurred to me that all of those "...(t)" bits are a bit
annoying. Another approach is to tell Maxima that e and m depend on t
and then to work from there. For example:

    (%i40) depends([e,m], t);
    (%o40)                           [e(t), m(t)]
    (%i41) eq2: m^2 = e^2 - k;
                                       2    2
    (%o41)                            m  = e  - k
    (%i42) diff(eq2, t);
                                        dm       de
    (%o42)                          2 m -- = 2 e --
                                        dt       dt
    (%i43) solve(%, 'diff(m,t));
                                              de
                                            e --
                                       dm     dt
    (%o43)                            [-- = ----]
                                       dt    m
    (%i44) diff (first (%), t);
                                                 2
                                                d e
                                      de dm   e ---    de 2
                             2      e -- --       2   (--)
                            d m       dt dt     dt     dt
    (%o44)                  --- = - ------- + ----- + -----
                              2        2        m       m
                            dt        m
    (%i45) subst (first (%o43), %);
                                     2
                                    d e
                                  e ---    de 2    2  de 2
                             2        2   (--)    e  (--)
                            d m     dt     dt         dt
    (%o45)                  --- = ----- + ----- - --------
                              2     m       m         3
                            dt                       m

It's exactly the same calculation, but the results look a little less
cumbersome.


Hope this is some use,

Rupert