Find a variable as function of other variable with equations system using Symbolic ToolBox

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Evyatar Sivan 2020-4-4

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评论： Walter Roberson 2020-4-4

I have 3 equations that connects between 4 variables, let's say:

syms a b c d;
r=b+d;
e=pi/2;
eq1 = c - a*sin(d) + b*cos(d) == 0;
eq2 = r - c*tan(e) - a*cos(d) - b*sin(d) == 0;
eq3 = b == e*r + a*tan(e);

Now I would like to find c(d), means, I would like to find what is c as a function of other variable, in this case - d.

Then perform other operations on c(d) such as differntiate etc.

What would be the best/easiest way to do that using Symbolic Toolbox?

Thanks.

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Walter Roberson 2020-4-4

10 radians is an unusual angle, but that is probably not so important for your question.

Evyatar Sivan 2020-4-4

edited

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Answer 1

Walter Roberson 2020-4-4

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syms a b c d;
r = b+d;
e = sym(pi)/10;   %not pi/2, that does not have a solution!
eq1 = c - a*sin(d) + b*cos(d) == 0;
eq2 = r - c*tan(e) - a*cos(d) - b*sin(d) == 0;
eq3 = b == e*r + a*tan(e);
sol = solve([eq1,eq2,eq3], [a, b, c], 'returnconditions', true);
c = symfun(simplify(sol.c), d);

This will be valid except at four sets of d values whose values can be deduced by analyzing sol.conditions (in each case adding an integer multiple of 2*pi gets another forbidden value)

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Evyatar Sivan 2020-4-4

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Thank you so much Walter.

looks like it's working good, but still for some reason I get complex result which I know is not the answer.

I've been over this for hours... wish you might be able to help if there's anything wrong with the code.

my code:

clc;
clear all;
% Unknown Variables
syms Pa N T theta;
% Given Variables
L=5.07/sin(theta);
delta=deg2rad(10);
c=10;
gamma=20;
phi=deg2rad(20);
Q=50.7/tan(theta);
W=((5.07/tan(theta))*0.93+5.07*(5.07/tan(theta))*0.5)*gamma;
% Equations
eq1 = Pa-N*sin(theta)+T*cos(theta) == 0;
eq2 = Q+W-Pa*tan(delta)-N*cos(theta)-T*sin(theta) == 0;
eq3 = T == c*L + N*tan(phi);
% Solution
sol = solve([eq1,eq2,eq3], [N, T, Pa], 'returnconditions', true);
Pa_theta = symfun(simplify(sol.Pa), theta);
eq4 = diff(Pa_theta,theta) == 0;
theta_max = solve(eq4,theta); % gives 4 possible solutions
rad2deg(theta_max(1)) % not quite what I expected... suppose to be 53 degress.

Thank you a lot.

Walter Roberson 2020-4-4

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double(rad2deg(theta_max(3))
ans =
           53.0352986075461 -   6.4856523980211e-71i

The complex portion is due to round-off error.

However, I recommend that you rewrite your code to cut down on the repeated numeric constants, and that you switch into symbolic mode for greater accuracy. Using solve() with equations with floating point constants is always a category mistake: solve() is for exact solutions, and floating point constants are by definition only approximations (except for the ones that are integers.)

Z = @(v) sym(v);
% Unknown Variables
syms Pa N T theta;
% Given Variables
factor507 = Z(5.07);
factor93 = Z(0.93);
L = factor507/sin(theta);
delta = deg2rad(Z(10));
c = Z(10);
gamma = Z(20);
phi = deg2rad(Z(20));
Q = c*factor507/tan(theta);
W = ((factor507/tan(theta))*factor93 + factor507*(factor507/tan(theta))*Z(0.5))*gamma;
% Equations
eq1 = Pa-N*sin(theta)+T*cos(theta) == 0;
eq2 = Q+W-Pa*tan(delta)-N*cos(theta)-T*sin(theta) == 0;
eq3 = T == c*L + N*tan(phi);
% Solution
sol = solve([eq1,eq2,eq3], [N, T, Pa], 'returnconditions', true);
Pa_theta = symfun(simplify(sol.Pa), theta);
eq4 = diff(Pa_theta,theta) == 0;
theta_max = solve(eq4,theta);
rad2deg(theta_max)

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Answer 2

Ameer Hamza 2020-4-4

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try this example

syms a b c d;
r=b+d;
e=pi/2;
eq1 = c - a*sin(d) + b*cos(d) == 0;
eq2 = r - c*tan(e) - a*cos(d) - b*sin(d) == 0;
eq3 = b == e*r + a*tan(e);
sol = solve([eq1 eq2 eq3], [a b c]);
C = sol.c; % c as function of d
% Now find its derivative
dC_dd = diff(C,d); % dc/dd
% you can also integrate it
IC = int(C,d);