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How to plot parametric equations by using "for loop"

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Wajahat 2017-5-18

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编辑： KSSV 2017-5-19

在 MATLAB Online 中打开

If we have,

t= linspace (-30,-1.6);
X= ((3*t)./(1+t.^3);
Y= ((3*t.^2)./(1+t.^3));
plot(X,Y);

How we plot X and Y by using for loop?

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Torsten 2017-5-18

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What do you want to do within the for-loop ?

Best wishes

Torsten.

KSSV 2017-5-18

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编辑：KSSV 2017-5-18

Why you want a loop, when the given code plots the data?

Wajahat 2017-5-18

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Actually, i want to plot a matrix equations. To plot matrix equations we use for loop.

KSSV 2017-5-18

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How?

Wajahat 2017-5-18

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编辑：Guillaume 2017-5-18

在 MATLAB Online 中打开

t= linspace (-30,-1.6); 
q=linspace (-30,-1.6);  
X= ((3*i*t)./(1+t.^3);  
Y= ((3*i*t.^2)./(1+t.^3));  
a=[X Y; -Y X];  
b=[X Y; -conj(Y) conj(X)];  
c=[X X; -Y Y];  
q1=q-(det(a)./det(b));  
u1=-(det(c)./det(b));  
plot(q1,u1)

now we have matrix a, b and c. How we plot it using for loop?

KSSV 2017-5-18

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This gives error because, det exists for square matrices...in this a is not square matrix.

Wajahat 2017-5-18

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No, a is a square matrix a=[X Y; -Y X]

Guillaume 2017-5-18

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a is a 2x200 matrix. It's definitively not square.

Rik 2017-5-18

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a is not square with the above code, it is 2x200.

And when you want to plot something, it is at least as important to think about how you want to plot as what you want to plot. What do you want to be on what axis? Do you want to show more information using marker sizes or colors?

With a, b and c, you could make a 3D scatter plot (or surface plot) with the values in a, b and c as the x, y and z values. Is that what you want?

Wajahat 2017-5-18

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without using for loop, it gives error (i.e. one of the matrix would not be a square), my question is, how to plot parametric equations with for loop?

KSSV 2017-5-18

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And I would like to point put that determinant of a=[X Y; -Y X] would be always zero, even though you make it a square. We cannot help you unless your question is clear.

Guillaume 2017-5-18

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Wajahat, We really have no idea what it is you're asking. You started with:

if we have [parametric equation and code that plots it given], how we plot X and Y by using for loop?

To which you were told: It's already working, why do you want to use a loop?

You then changed your question to say that you had a matrix instead. Again, said you needed to use a loop without explaining why. When asked why you wanted to use a loop, you gave some code that did not even work.

You've now changed your question again back to the original question. Our answer is still the same: It's already working, why do you want to use a loop?

We have no idea why you insist on using a loop. You haven't showed us why a loop is needed and we have no idea what it is you want looping over.

Wajahat 2017-5-19

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编辑：Wajahat 2017-5-19

在 MATLAB Online 中打开

Guillaume, i have written a code (below) in which a matrices a and b are involved and get the plot of a solution q1 by using loop.

clear all;

close all;

x=-10:.1:10;

t=-10:.1:10;

q=.2;

mu1=4-.3*i;

mu2=.3+i;

for k=1:length(x)

for l=1:length(t)

x11=exp(i.*x(k)./(2*mu1)).*exp(-i.*mu1.*t(l)./2);

x21=exp(i.*x(k)./(2*mu2)).*exp(-i.*mu2.*t(l)./2);

y11=exp(-i.*x(k)./(2*mu1)).*exp(i.*mu1.*t(l)./2);

y21=exp(-i.*x(k)./(2*mu2)).*exp(i.*mu2.*t(l)./2);

a=[y11 y21; mu1.*x11 mu2.*x21];

b=[x11 x21; y11 y21];

q1(l,k)=q-(det(a)./det(b));

end

surf(x,t,abs(q1));

This is a plot of q1, where a and b are square matrices.

Now i write another solution i.e.

u1=-(det(c)./det(b));

where c=[x11 x21; mu2.*y11 mu1.*y21];

and plot a parametric curve between q1 and u1 (i.e. plot(q1,u1)). How can we do it by using loop? If we don't use loop then the dimension of the matrices does'nt match, i.e. one of the matrix would not be a square.

KSSV 2017-5-19

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在 MATLAB Online 中打开

Still, there is something missing....First change your above code following: Replace i with 1i, and preallocate q1.

q1 = zeros(length(x),length(t)) ;

Now coming to a,b,c which are

a=[X Y; -Y X];  
b=[X Y; -conj(Y) conj(X)];  
c=[X X; -Y Y];

You can make them a square matrix, using meshgrid to X,Y. But the determinants of them would be zero.

Wajahat 2017-5-19

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KSSV, can you send me a complete code with changing as you mentioned?

KSSV 2017-5-19

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在 MATLAB Online 中打开

clear all;
close all;
x=-10:.1:10;
t=-10:.1:10;
q=.2;
mu1=4-.3*1i;
mu2=.3+1i;
q1 = zeros(length(x),length(t)) ;
for k=1:length(x)
    for l=1:length(t)
        x11=exp(1i.*x(k)./(2*mu1)).*exp(-1i.*mu1.*t(l)./2);
        x21=exp(1i.*x(k)./(2*mu2)).*exp(-1i.*mu2.*t(l)./2);
        y11=exp(-1i.*x(k)./(2*mu1)).*exp(1i.*mu1.*t(l)./2);
        y21=exp(-1i.*x(k)./(2*mu2)).*exp(1i.*mu2.*t(l)./2);
        a=[y11 y21; mu1.*x11 mu2.*x21];
        b=[x11 x21; y11 y21];
        q1(l,k)=q-(det(a)./det(b));
    end
end
surf(x,t,abs(q1));