Sketch contours for f(x,y)=y^2+2x^2-x^4

[mrchartreuse]

The problem is:
. Let f : R^2 → R be given by f(x,y)=y^2 +2x^2 −x^4.
Sketch the contour diagram of f by drawing the contours f = −2, −1, 0, 1 , 1, 2.

I have started with the f = -1 line and managed to get it into a form resembling a hyperbola:

1 = (x^2-1)^2/2 - y^2/2

Is this the right sort of approach? And if so how would i sketch this with the x^2 where the x usually is?

I was thinking of trying to sketch it with all the values along the x-axis square rooted or something so it would look like a sort of squashed hyperbola but since a square root has two possible answers I don't think this is correct... Then there are all the other scary f values...

 

[LCKurtz]

For each value of C you can write

y²+2x²-x⁴=C

y = ± sqrt(C - 2x²+x⁴)

Presumably you have a graphing calculator. For each C plot both the + and - graphs on the same picture. You are right that some of the contours sort of look like hyperbolas, but not all of them do. If you let x go from -2 to 2 you should get good pictures.

 

[mrchartreuse]

Sigh, if only it were that easy. We are supposed to do them all by hand...

 

[LCKurtz]

Sigh, if only it were that easy. We are supposed to do them all by hand...

What can I say? I think graphing calculators are relied on too much in some of the basic math courses, but here we have an instance where a graphing calculator or computer is exactly the proper tool. Does your teacher expect you to calculate the square roots by hand? I wouldn't waste my time plotting those by hand unless I really had nothing better to do with my time.

 

[vela]

You should be able to see the contour plot will be symmetric about the y-axis. It also only has even powers of x, so it'll be symmetric about the x-axis. So really, all you have to do is figure out one quadrant.

Also, the case where C=1 is easy enough to do. :)

 

[vela]

It might help to write the function as [itex]y = \pm \sqrt{(x^2-1)^2+k}[/itex]. If k>0, the function is defined for all x. But if k<0, you have to be a bit more careful because the quantity under the radical will be negative for some values of x.