finding variable of a given equation
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Irfan
el 19 de Oct. de 2024
Comentada: Sam Chak
el 20 de Oct. de 2024
I want to get the value of b given the equation, where AonAstar = 16 and gamma = 1.22
AonAstar = ((b)^(1/gamma)*(1-(b)^((gamma-1)/gamma))^(1/2))/(((gamma-1)/2)^(1/2)*(2/(gamma+1))^((gamma+1)/(2*(gamma-1))));
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Walter Roberson
el 19 de Oct. de 2024
Editada: Walter Roberson
el 19 de Oct. de 2024
Q = @(v) sym(v);
AonAstar = Q(16);
gamma = Q(1.22);
syms b
eqn = AonAstar == ((b)^(1/gamma)*(1-(b)^((gamma-1)/gamma))^(1/2))/(((gamma-1)/2)^(1/2)*(2/(gamma+1))^((gamma+1)/(2*(gamma-1))));
fplot(lhs(eqn) - rhs(eqn), [-2 2])
b_solution = vpasolve(eqn, b)
disp(char(b_solution))
We can see from the graph that there are no solutions over the reals.
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John D'Errico
el 19 de Oct. de 2024
gamma is a really bad choice of variable names, since gamma is itself a very useful function in MATLAB. Try to avoid doing these things, because one day soon you will need to use gamma yourself as a function. I've used g instead, to replace gamma.
AonAstar = 16;
g = 1.22;
syms b
I'll subtract the right hand side from AonAstar, to create a problem where we will look for a zero.
eqn = AonAstar - (b^(1/g)*sqrt(1-b^((g-1)/g)))/((sqrt(g-1)/2)*(2/(g+1))^((g+1)/(2*(g-1))))
fplot(eqn,[0,1])
For a solution to exist, this curve must cross zero, and must do so in the interval [0,1], since it is only real valued in that interval.
As you can see, it never even comes close to zero, being always positive.
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John D'Errico
el 20 de Oct. de 2024
It is a habit we need to train our minds to avoid if we are to use MATLAB successfully. And yes, it is far too easy to name variables alpha, beta, gamma, and yes, I fail at it at times myself. My tendency is to use contractions, like alph, bet, gam, or I might deliberately mispell the greek letter, perhaps as gammuh. Or I'll capitalize the name, using Gamma. At least all of these variations in names remind me of the relation of my code to my thinking.
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