On Mon, Jun 25, 2012 at 7:39 AM, Slumberland <[email protected]> wrote:
> Okay, that's maybe not the answer you're looking for.
>
> What he means is that you can't plot it explicitly without solving for y.
> "Implicit" is another way of saying
> "not in the form y = "
> (or z= , x =   .... etc)
>
> But it can be instructive to figure out how to plot the functions which
> define a circle.  Is that what you want?
>
> Sage will solve the equation for you.  You can type
>
> sage: x,y = var('x,y')
> sage: b = x^2 + y^2 == 4
> sage: c = solve(b,y)
> sage: c
>
> [y == -sqrt(-x^2 + 4), y == sqrt(-x^2 + 4)]
>
>
> The command "solve(b,y)" solves your equation for the variable y.
>
> Plotting this way is awkward but you can do it.  In this case, we need each
> of the functions
>
>     -sqrt(-x^2+4),
>
>     sqrt(-x^2+4)
>
>
> Both of these are stored in c.
>
> To get them, you can tell python what parts you want:
>
> c[0].rhs() gives you the right-hand-side{ ;-) of the first equality stored
> in c
>
> c[1].rhs() grabs the second.
>
> So, you could do this:
>
>
> sage: bottom = plot( c[0].rhs(), xmin=-3, xmax=3, aspect_ratio=1)
>
> sage: top = plot(c[0].rhs(), xmin=-3, xmax=3)
>
> sage: show(bottom+top)
>
>
> and the graph will show, but Python complains that it can't evaluate the
> function at some points.
>
>
>  implicit_plot() is easier to use, that's all.

Thanks for explaining the things . :-) It executed the code and it
gives half circle. Anyways if it gives warnings then we should not do
this way.
Thanks once again.



-- 
Priyanka Kapoor
priyankacool10.wordpress.com
Linux User Group, Ludhiana

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