
#1
Jun1209, 06:55 AM

P: 48

1. The problem statement, all variables and given/known data
Sketch the level curve of the surface [tex]z = \frac{x^2  2y + 6}{3x^2 + y}[/tex] belonging to height z = 1 indicating the points at which the curves cut the y−axis. 2. Relevant equations 3. The attempt at a solution I put [tex]1 = \frac{x^2  2y + 6}{3x^2 + y}[/tex] but then don't know how to proceed. The answer shows an inverted parabola at y = 2, but I don't know how to get that. 



#2
Jun1209, 07:19 AM

Sci Advisor
HW Helper
P: 4,301

From
[tex] 1 = \frac{x^2  2y + 6}{3x^2 + y} [/tex] it follows that [tex] 3x^2 + y = x^2  2y + 6 [/tex] Can you cast this in the form y = f(x) ? 



#3
Jun1209, 10:03 AM

P: 48

Do you mean like this...
[tex]3x^2 + y = x^2  2y + 6[/tex] [tex]3y = 2x^2 + 6[/tex] [tex]y = \frac{2x^2 + 6}{3}[/tex] [tex]y = \frac{2x^2}{3} + 2[/tex] ??? 



#4
Jun1209, 10:11 AM

Sci Advisor
HW Helper
P: 4,301

Sketching level curves of f(x,y)
Yes.
Now, that's a parabola, isn't it? It's of the general form y = ax^2 + bx + c with a = 2/3, b = 0, c = 2; you can see that it is inverted (like a mountain top) because a < 0. 



#5
Jun1209, 11:04 PM

P: 48

Ok, that explanation helped, thankyou.



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