How to plot this implicit function

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]);

[darova]

I don't see here eps

x=@(beta,eps) (beta-alpha*(beta-gamma)+sqrt((beta-alpha*(beta-gamma)).^2+4*alpha*beta*gamma*A))./(2*beta);

What is x? Where is eps?

y=@(beta,eps) x.*(1-x).*beta/gamma;

[Atom]

t and d depend on x and y.

x depends on beta. I have written x=@(beta,eps) although x does not have eps. Should I only write x=@(beta)? Although I am not getting the plot.

Edited...

[darova]

Sorry didn't notice that. You should write input arguments for your function all the time

f1 = @(x,y) x + y;
f2 = @(x,y) f1 + 2*x;
% f2 = @(x,y) f1(x,y) + 2*x;        % should be
f2(1,1)

[Atom]

Not clear... would you please see my problem... y depends on x, & etc

[darova]

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;

has to be

x=@(beta) (beta-alpha*(beta-gamma)+sqrt((beta-alpha*(beta-gamma)).^2+4*alpha*beta*gamma*A))./(2*beta);
y=@(beta) x(beta).*(1-x(beta)).*beta/gamma;

[Atom]

I have modified the code as per your suggestion but getting some warning

alpha=1.5; A=0.0207; gamma=0.25;
syms x y t d fun2
beta=0:.0001:10;
x=@(beta) (beta-alpha*(beta-gamma)+sqrt((beta-alpha*(beta-gamma)).^2+4*alpha*beta*gamma*A))./(2*beta);
y=@(beta) x(beta).*(1-x(beta)).*beta/gamma;
t=@(beta,eps)-x(beta)+x(beta).*y(beta).*(alpha-eps*beta)./((A+x(beta)+y(beta)).^2);
d=@(beta,eps) eps*beta.*x(beta).*y(beta).*(x(beta).^2+(A+x(beta)).*x(beta)+A*alpha)./((A+x(beta)+y(beta)).^3);
fun2=@(beta,eps) (t(beta,eps)*t(beta,eps)-d(beta,eps));
figure
ezplot(fun2,[0,1,0,1])

% Please see this. 

[darova]

Vectorizing

Forgot the dot

Why not use .^2?

fun2 = @(beta,eps) t(beta,eps).^2 - d(beta,eps);

[Atom]

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]

Works for me

Answer 1

[darova]

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]

Can be used without bit-wise ./ and .* (just / and *)

[madhan ravi]

There are in-built functions under the name alpha() , beta() , gamma() , eps().