Ploting a implicit equation using plot

Basically, I want to plot the equation:

x^2+y^2=1+4.5sin^2(xy)

Normally I would use ezplot, but I was explicitly (not implicitly) told to use the plot command. How would I go about doing this?

Answer 1

[John D'Errico]

If you cannot use ezplot (must use plot), then you just pick a range of values for x, then solve for y at each x. Plot the results. WTP?

help fzero

Of course, if you were feeling creative, and chose NOT to follow your directions, then you could just use contour. Generate points for the function

f(x,y) = x^2+y^2 - 1 - 4.5sin^2(xy)

then plot the zero contour.

help meshgrid
help contour

[Benjamin Midtvedt]

Problem is that one can't easily solve for y. Unless you mean that I should rewrite the equation as

 x.^2+y.^2-1-4.5.*sin(x.*y)=0

Then create some x-values, and somehow use fzero to generate corresponding y values. Wouldn't that require me to create som kind of for-clause. Such as

for x=linspace('some arbitrary range')    % code
y=fzero(x^2+y^2-1-4.5*sin(x*y), y0)
end

But I don't think that's how fzero works, could be mistaken though. And I have no idea where to search for y0. There must be a better way. Am I just overcomplicating things?

[John D'Errico]

No. I meant what I said!

Pick a value for x, for example 2.

Create a function that fzero can use. I'll write it in a general form

fxy = @(x,y) x.^2+y.^2 - 1 - 4.5*sin(x.*y).^2;

fy = @(y) fxy(2,y);
fzero(fy,1)
ans =
       0.966612095479693

Lather, rinse, and repeat for any given value of x. I'd suggest a loop over the possible values for x where you have some interest.

You can get more adventuresome, and recognize that there will be multiple solutions to this problem. That would require finding all solutions in y for a given x. Classic ways of doing this are to simply check a sequence of values for y, looking for sign changes. Wherever you do find one, then use fzero.

Yes, you are just over-complicating things. If you want to be more creative, as I said, try it using contour, but that will not use plot, as you were explicitly told to do.

Finally, IF you want to get some general idea of what a complete plot would look like, try using ezplot.

[David Schörling]

Wouldn't substituting x for r.*cos(Ө) and y for r.*sin(Ө) make it easier?

[Benjamin Midtvedt]

Wasn't that basically what I said? This was my attempt.

f=@(x,y) x.^2+y.^2-1-4.5.*sin(x.*y).^2;
xv=linspace(-3,3);
yv=zeros(size(xv));
for i=1:length(xv);
  yv(i,1)=fzero(@(y) f(xv(i),y), 2);
  yv(i,2)=fzero(@(y) f(xv(i),y), -2);
end
plot(xv, yv)

However, as you mentioned, it doesn't solve for ever y, as there is up to 4 solutions for some values of x. Also, I seem to get artifact zeros in my "yv".

[Benjamin Midtvedt]

I am currently experimenting with that substitution, but I still can't find the complete plot.

[John D'Errico]

For this specific case, possibly. However, the sin(x*y) would still be a bit messy. Then we would have

r^2 - 1 - 4.5*sin(r^2*sin(theta)*cos(theta)) = 0

So, one might choose to play with this more, but it is not even obvious that for any given value of theta, there is a unique r. In fact, ezplot suggests that for some values of theta, this might be multi-valued, with multiple values for r. (That would require some more play.)

Regardless, for the completely general case such a substitution will be of no value. And so often, when someone gives a function they claim to want to plot, the true function they actually want to plot is so much more messy. This appear to be homework, so probably not the case.

In the end, ezplot is the simple solution, in one line. meshgrid and contour, only slightly more work, doable in two lines.

[Benjamin Midtvedt]

So, how would you suggest I'd move forward?

Since the equation obviously makes no distinguishment between x and y, the resulting plot has to be completely symmetrical. I could use this as a argument to simply rotate the graph to find the missing pieces. But this is very unsatisfying, probably equally difficult, and it only works in the special case.