# How to plot this implicit function

5 views (last 30 days)
Atom on 8 Sep 2019
Commented: madhan ravi on 8 Sep 2019
I need a beta vs eps plot of the curve t^2-d. The expression of t and d depends on x, y where x, y depend on beta. Please help. Ranges of beta and eps are both equal to 0:.01:1,
alpha=1.5; A=0.0207; gamma=0.25;
x=@(beta) (beta-alpha*(beta-gamma)+sqrt((beta-alpha*(beta-gamma)).^2+4*alpha*beta*gamma*A))./(2*beta);
y=@(beta) x.*(1-x).*beta/gamma;
t=@(beta,eps) -x+x.*y.*(alpha-eps*beta)./((A+x+y).^2);
d=@(beta,eps) eps*beta.*x.*y.*(x.^2+(A+x).*x+A*alpha)./((A+x+y).^3);
F=@(beta,eps) (t.^2-d);
figure
ezplot(F,[0,1,0,1]);
Atom on 8 Sep 2019
Inserted dot. still warning exists.
Warning: Function failed to evaluate on array inputs; vectorizing the function may speed up its evaluation and avoid the
need to loop over array elements.
> In ezplotfeval (line 56)
In ezplot>ezimplicit (line 257)
In ezplot (line 153)
In untitled1 (line 10)
darova on 8 Sep 2019
Works for me

darova on 8 Sep 2019
Try symbolic expression
clc,clear
alpha=1.5; A=0.0207; gamma=0.25;
syms beta eps
x = (beta-alpha*(beta-gamma)+sqrt((beta-alpha*(beta-gamma)).^2+4*alpha*beta*gamma*A))./(2*beta);
y = x.*(1-x).*beta/gamma;
t = -x+x.*y.*(alpha-eps*beta)./((A+x+y).^2);
d = eps*beta.*x.*y.*(x.^2+(A+x).*x+A*alpha)./((A+x+y).^3);
% convert symbolic expression to function handle
F = matlabFunction(t.^2 - d,'vars',[beta eps]);
ezplot(F,[0,1,0,1]);
darova on 8 Sep 2019
Can be used without bit-wise ./ and .* (just / and *)
madhan ravi on 8 Sep 2019
There are in-built functions under the name alpha() , beta() , gamma() , eps().

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