Plotting implicit equation with fimplicit

Hello

I have tried to plot this implicit equation. But when I tried it, the plot is showing empty.

Here is the code I used to plot. Could anyone help me with this.

Thanks in advance

clc

syms f(x,y)

n = 8;

a1 = 1.0086*y - 0.9216*(x - y);

b1 = 1.0107*(-x) - 1.0086*(y);

c1 = 0.9216*(x - y) - 1.0107*(-x);

h1 = 0.5877*(161.65);

I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6));

I2 = ((h1.^2)/(3))+((a1.^2 + b1.^2 + c1.^2)/(54));

th = acos(I3/(I2.^(3/2)));

v1 = ((2*th)+pi)/6;

an1 = (abs(2*cos(v1)))^n;

an2 = (abs(2*cos((2*th+3*pi)/6)))^n;

an3 = (abs(2*cos((2*th+5*pi)/6)))^n;

f(x,y) = ((3*I2).^(n/2)) * (an1 + an2 + an3) - (2*(189.32)^8);

fimplicit(f)

Answer 1

[Torsten]

function main
fimplicit (@(x,y)f(x,y))
end

function fun = f(x,y)
n  = 8;

a1 = 1.0086*y - 0.9216*(x - y);
b1 = 1.0107*(-x) - 1.0086*(y);
c1 = 0.9216*(x - y) - 1.0107*(-x);
h1 = 0.5877*(161.65);

I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6));
I2 = ((h1.^2)/(3))+((a1.^2 + b1.^2 + c1.^2)/(54));

th = acos(I3./(I2.^(3/2)));
v1 = ((2*th)+pi)/6;

an1 = (abs(2*cos(v1))).^n;
an2 = (abs(2*cos((2*th+3*pi)/6))).^n;
an3 = (abs(2*cos((2*th+5*pi)/6))).^n;

fun = ((3*I2).^(n/2)).* (an1 + an2 + an3) - (2*(189.32)^8);
end

Resonable limits for plotting are required - no zeros are found in the default range [-5:5] for x and y.

[Jay]

Hello Torsten

I tried this code. But the plot shown is empty.

Best Regards

Jay

[Stephan]

fsurf(@(x,y)f(x,y))

function fun = f(x,y)
n  = 8;
a1 = 1.0086*y - 0.9216*(x - y);
b1 = 1.0107*(-x) - 1.0086*(y);
c1 = 0.9216*(x - y) - 1.0107*(-x);
h1 = 0.5877*(161.65);
I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6));
I2 = ((h1.^2)/(3))+((a1.^2 + b1.^2 + c1.^2)/(54));
th = acos(I3./(I2.^(3/2)));
v1 = ((2*th)+pi)/6;
an1 = (abs(2*cos(v1))).^n;
an2 = (abs(2*cos((2*th+3*pi)/6))).^n;
an3 = (abs(2*cos((2*th+5*pi)/6))).^n;
fun = ((3*I2).^(n/2)).* (an1 + an2 + an3) - (2*(189.32)^8);
end

[Jay]

Hello

I tried it. As expected it is showing 3D surface. But I am interested in 2D curve. Could you please suggest an alternate way.

Best Regards

Jay

[Torsten]

fimplicit (@(x,y)f(x,y),[-200 200 -150 150])

[Jay]

Hello Torsten

It works. But it is not working with other limits. Can you explain why so?

Thank you.

[Torsten]

It "works" as long as the object is contained in the box defined by the specified limits for x and y.

Answer 2

[John D'Errico]

Easy enough. Try this, for example.

vpasolve(f(1,y))
ans =
-80.224189505722446658042301607259

vpasolve(f(-20,y))
ans =
63.634253282860063957543062643774

Hmm. So [1,-80] is roughly a solution. That should be a good hint as to where to have fimplicit look.

fimplicit(f,[-150,150,-150,150])
axis equal
grid on

The problem was fimplicit looks by default in a rather narrow set of limits on x and y. It cannot know where it SHOULD be looking, and computer programs can sometimes be so clueless. Since fimplicit just found no solutions at all in the domain it was looking by default, you saw an empty figure. Sometimes you need to give even a computer a nudge in the right direction.