numeric solving vs graph - Hewlett Packard

This is a discussion on numeric solving vs graph - Hewlett Packard ; Hello all, I had a quite simple problem, or so I thought. I needed to find the first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t). I put the equation on the stack, LS+Calc ...

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  1. numeric solving vs graph

    Hello all,

    I had a quite simple problem, or so I thought. I needed to find the
    first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).

    I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
    formula filling up the display. Functions in ALG don't really make it
    look any better, so I leave it as is.
    I put 0 on the stack and then =, which gives me dx/dt = 0.

    Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
    seconds is { }.

    Graphing that function, showing f', and getting the roots in Graph
    mode reveal the correct answer of t=0.416 for the first, and from that
    I can deduce t(n) = 0.416+n*PI/3.

    How can I make it work numerically?

    Thanks! Hope this was clear enough....

    Christoph

  2. Re: numeric solving vs graph

    On Sep 20, 11:11*pm, Christoph Koehler
    wrote:
    > Hello all,
    >
    > I had a quite simple problem, or so I thought. I needed to find the
    > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).
    >
    > I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
    > formula filling up the display. Functions in ALG don't really make it
    > look any better, so I leave it as is.
    > I put 0 on the stack and then =, which gives me dx/dt = 0.
    >
    > Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
    > seconds is { }.
    >
    > Graphing that function, showing f', and getting the roots in Graph
    > mode reveal the correct answer of t=0.416 for the first, and from that
    > I can deduce t(n) = 0.416+n*PI/3.
    >
    > How can I make it work numerically?
    >
    > Thanks! Hope this was clear enough....
    >
    > Christoph


    I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.

    The symbolic solver also gave me { }, so I used the numeric solver.
    Use STEQ to store level 1 of the stack into variable EQ, then launch
    the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
    initial guess (I used 1) and I got the root 1.0646... Different
    guesses will result in different roots, though this method is somewhat
    blind.

    If you want the first 3 values where dx/dt = 0, it's probably easier
    to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
    the graphing environment, move the cursor close to the root, press
    FCN, then ROOT. Do this for the first 3 intersections and get t =
    1.0646, t = 2.1118, t = 3.1590.

    S.C.

  3. Re: numeric solving vs graph

    In article
    ,
    sc_usenet@hotmail.com wrote:

    > On Sep 20, 11:11*pm, Christoph Koehler
    > wrote:
    > > Hello all,
    > >
    > > I had a quite simple problem, or so I thought. I needed to find the
    > > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).
    > >
    > > I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
    > > formula filling up the display. Functions in ALG don't really make it
    > > look any better, so I leave it as is.
    > > I put 0 on the stack and then =, which gives me dx/dt = 0.
    > >
    > > Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
    > > seconds is { }.
    > >
    > > Graphing that function, showing f', and getting the roots in Graph
    > > mode reveal the correct answer of t=0.416 for the first, and from that
    > > I can deduce t(n) = 0.416+n*PI/3.
    > >
    > > How can I make it work numerically?
    > >
    > > Thanks! Hope this was clear enough....
    > >
    > > Christoph

    >
    > I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.
    >
    > The symbolic solver also gave me { }, so I used the numeric solver.
    > Use STEQ to store level 1 of the stack into variable EQ, then launch
    > the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
    > initial guess (I used 1) and I got the root 1.0646... Different
    > guesses will result in different roots, though this method is somewhat
    > blind.
    >
    > If you want the first 3 values where dx/dt = 0, it's probably easier
    > to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
    > the graphing environment, move the cursor close to the root, press
    > FCN, then ROOT. Do this for the first 3 intersections and get t =
    > 1.0646, t = 2.1118, t = 3.1590.
    >
    > S.C.


    Curious. For the same dy/dx as you have, I find different zeroes.

    Namely: 0.4163, 1.4635 and 2.5101, to 4 decimal places.

  4. Re: numeric solving vs graph

    On Sun, 21 Sep 2008 09:40:12 -0500:

    > Different guesses will result in different roots,
    > though this method is somewhat blind.


    An aid to systematic searching for numeric ROOTS over an interval:
    http://groups.google.com/group/comp....f2e21bace9dd00

    UserRPL program listing:
    http://groups.google.com/group/comp....f?dmode=source

    [r->] [OFF]

  5. Re: numeric solving vs graph

    On Sep 21, 2:13*pm, Virgil wrote:
    > In article
    > ,
    >
    >
    >
    >
    >
    > *sc_use...@hotmail.com wrote:
    > > On Sep 20, 11:11*pm, Christoph Koehler
    > > wrote:
    > > > Hello all,

    >
    > > > I had a quite simple problem, or so I thought. I needed to find the
    > > > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).

    >
    > > > I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
    > > > formula filling up the display. Functions in ALG don't really make it
    > > > look any better, so I leave it as is.
    > > > I put 0 on the stack and then =, which gives me dx/dt = 0.

    >
    > > > Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
    > > > seconds is { }.

    >
    > > > Graphing that function, showing f', and getting the roots in Graph
    > > > mode reveal the correct answer of t=0.416 for the first, and from that
    > > > I can deduce t(n) = 0.416+n*PI/3.

    >
    > > > How can I make it work numerically?

    >
    > > > Thanks! Hope this was clear enough....

    >
    > > > Christoph

    >
    > > I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.

    >
    > > The symbolic solver also gave me { }, so I used the numeric solver.
    > > Use STEQ to store level 1 of the stack into variable EQ, then launch
    > > the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
    > > initial guess (I used 1) and I got the root 1.0646... Different
    > > guesses will result in different roots, though this method is somewhat
    > > blind.

    >
    > > If you want the first 3 values where dx/dt = 0, it's probably easier
    > > to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
    > > the graphing environment, move the cursor close to the root, press
    > > FCN, then ROOT. Do this for the first 3 intersections and get t =
    > > 1.0646, t = 2.1118, t = 3.1590.

    >
    > > S.C.

    >
    > Curious. For the same dy/dx as you have, I find different zeroes.
    >
    > Namely: 0.4163, 1.4635 and 2.5107, to 4 decimal places.-


    That's weird. I just did the same thing again and got your roots. Now
    I'm not sure what happened last time.

    S.C.

  6. Re: numeric solving vs graph

    sc_usenet@hotmail.com wrote:

    > I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.
    >
    > The symbolic solver also gave me { }, so I used the numeric solver.
    > ...
    > If you want the first 3 values where dx/dt = 0, it's probably easier
    > to use the grapher.
    > ...
    > Use STEQ, then hold [LS] F1, ERASE DRAW. When in
    > the graphing environment, move the cursor close to the root, press
    > FCN, then ROOT.


    Well, I don't want to disturb this discussion about various more or less
    elegant ways to make the calculator solve the problem, but in this case
    there is a much easier and more straightforward way: simply use your own
    brain, the calculator almost provided the solution anyway:

    > I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.


    So the rest is trivial: a local minimum or maximum of x(t) requires
    dx/dt = 0 => 12 cos(3t) - 4 sin(3t) = 0 => t = 1/3 arctan 3 = 0.4163...

    Since tan x = tan(x + Pi), simply add multiples of Pi/3 for further
    solutions, i.e. t = 0.4163, 1.4635, 2.5107, 3.5579, 4.6051...

    It's that easy.
    Okay, I didn't check the second derivate. <8)

    Maybe this is a good example for combining the benefits of our little
    electronic helpers with good old plain man-made math. The former
    provides the derivate, the latter gives the final result.
    In this case even the correct one. 8-)

    Dieter


  7. Re: numeric solving vs graph

    On Sep 21, 8:07*pm, Dieter wrote:
    > Maybe this is a good example for combining the benefits of our little
    > electronic helpers with good old plain man-made math. The former
    > provides the derivate, the latter gives the final result.
    > In this case even the correct one. 8-)
    >



    Exactly -- the calculator is (or should be) and extension to the
    brain, not a replacement.

    I didn't think about that though, since I was more focused on
    calculator usage with this question.

    S.C.

  8. Re: numeric solving vs graph

    In article
    ,
    sc_usenet@hotmail.com wrote:

    > On Sep 21, 2:13*pm, Virgil wrote:
    > > In article
    > > ,
    > >
    > >
    > >
    > >
    > >
    > > *sc_use...@hotmail.com wrote:
    > > > On Sep 20, 11:11*pm, Christoph Koehler
    > > > wrote:
    > > > > Hello all,

    > >
    > > > > I had a quite simple problem, or so I thought. I needed to find the
    > > > > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).

    > >
    > > > > I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
    > > > > formula filling up the display. Functions in ALG don't really make it
    > > > > look any better, so I leave it as is.
    > > > > I put 0 on the stack and then =, which gives me dx/dt = 0.

    > >
    > > > > Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
    > > > > seconds is { }.

    > >
    > > > > Graphing that function, showing f', and getting the roots in Graph
    > > > > mode reveal the correct answer of t=0.416 for the first, and from that
    > > > > I can deduce t(n) = 0.416+n*PI/3.

    > >
    > > > > How can I make it work numerically?

    > >
    > > > > Thanks! Hope this was clear enough....

    > >
    > > > > Christoph

    > >
    > > > I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.

    > >
    > > > The symbolic solver also gave me { }, so I used the numeric solver.
    > > > Use STEQ to store level 1 of the stack into variable EQ, then launch
    > > > the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
    > > > initial guess (I used 1) and I got the root 1.0646... Different
    > > > guesses will result in different roots, though this method is somewhat
    > > > blind.

    > >
    > > > If you want the first 3 values where dx/dt = 0, it's probably easier
    > > > to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
    > > > the graphing environment, move the cursor close to the root, press
    > > > FCN, then ROOT. Do this for the first 3 intersections and get t =
    > > > 1.0646, t = 2.1118, t = 3.1590.

    > >
    > > > S.C.

    > >
    > > Curious. For the same dy/dx as you have, I find different zeroes.
    > >
    > > Namely: 0.4163, 1.4635 and 2.5107, to 4 decimal places.-

    >
    > That's weird. I just did the same thing again and got your roots. Now
    > I'm not sure what happened last time.
    >
    > S.C.


    We'll probably never know.

  9. Re: numeric solving vs graph

    On Sep 21, 7:07*pm, Dieter wrote:
    > Maybe this is a good example for combining the benefits of our little
    > electronic helpers with good old plain man-made math. The former
    > provides the derivate, the latter gives the final result.
    > In this case even the correct one. 8-)


    I totally agree.
    I can do this by hand. But in a test setting I'd rather concentrate on
    the subject matter and not worry about taking derivatives. This is not
    for a Calculus class but EE.

    Thanks for all the tips

    Christoph

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