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Solving Sytem of Equations with Trig

asked 2018-11-24 22:11:01 -0500

jonny_d gravatar image

updated 2018-11-25 17:17:36 -0500

Just trying to solve a system with approximate results, but I am getting an error.

Code:

Fs, Na, Nb, theta = var('Fs Na Nb theta')

eq1 = 0 == Fs + 0.16Na - 10.0sin(theta)

eq2 = 0 == Na - 10.0*cos(theta)

eq3 = 0 == 0.26Nb - 6.0sin(theta) - Fs

eq4 = 0 == Nb - 6.0*cos(theta)

solns = solve([eq1,eq2,eq3,eq4],Fs, Na, Nb, theta, solution_dict=True)

[[s[Fs].n(30), s[Na].n(30), s[Nb].n(30), s[theta].n(30)] for s in solns]

Error:

KeyError Traceback (most recent call last) <ipython-input-27-8201f1ef9ddf> in <module>() ----> 1 [[s[Fs].n(Integer(30)), s[Na].n(Integer(30)), s[Nb].n(Integer(30)), s[theta].n(Integer(30))] for s in solns]

KeyError: Fs

Any advice?? I am new to Sage.

EDIT: Fixed the code I posted so it is more readable. Still looking for help on this. Thanks!

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Comments

1

To display blocks of code or error messages, skip a line above and below, and do one of the following (all give the same result):

  • indent all code lines with 4 spaces
  • select all code lines and click the "code" button (the icon with '101 010')
  • select all code lines and hit ctrl-K

For instance, typing

If we define `f` by

    def f(x, y, z):
        return x * y * z

then `f(2, 3, 5)` returns `30` but `f(2*3*5)` gives:

    TypeError: f() takes exactly 3 arguments (1 given)

produces:

If we define f by

def f(x, y, z):
    return x * y * z

then f(2, 3, 5) returns 30 but f(2*3*5) gives:

TypeError: f() takes exactly 3 arguments (1 given)

Please edit your question to do that.

slelievre gravatar imageslelievre ( 2018-11-26 04:21:19 -0500 )edit

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answered 2018-11-26 08:16:15 -0500

dan_fulea gravatar image

Note that solve is usually used to solve algebraic equations. (The examples of the doc string of solve also contain some trigonometric equations, but mixing them with more variables... I also try it in some specific non-algebraic cases, but do not hope to be successful...)

To get the system solved i am using two variables $s=\sin\theta$, $c=\cos\theta$ instead of the one variable $\theta$. This moves the system into the algebraich world. The code

# i am using s, c for sin(theta), cos(theta)...
Fs, Na, Nb, s, c = var('Fs Na Nb s c')

eq1 = 0 ==  Fs + 0.16*Na           - 10.0*s
eq2 = 0 ==            Na           - 10.0*c
eq3 = 0 == -Fs           + 0.26*Nb -  6.0*s
eq4 = 0 ==                      Nb -  6.0*c
eq5 = 1 == s^2 + c^2

solns = solve( [eq1, eq2, eq3, eq4, eq5]
               , Fs, Na, Nb, s, c
               , solution_dict=True)

for sol in solns:
    for key in sol:
        print "%2s = %s" % ( key, sol[key].n(30) )
    print

then delivers

 c = 0.98104950
 s = 0.19375728
Nb = 5.8862970
Na = 9.8104950
Fs = 0.36789356

 c = -0.98104950
 s = -0.19375728
Nb = -5.8862970
Na = -9.8104950
Fs = -0.36789356

(Getting $\theta$ now should be one more code line.)

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Good solution. Thanks!

jonny_d gravatar imagejonny_d ( 2019-02-08 16:46:03 -0500 )edit

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Asked: 2018-11-24 22:11:01 -0500

Seen: 47 times

Last updated: Nov 26 '18