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Sketching level curves of f(x,y)

by username12345
Tags: curves, sketching
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username12345
#1
Jun12-09, 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.
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CompuChip
#2
Jun12-09, 07:19 AM
Sci Advisor
HW Helper
P: 4,300
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) ?
username12345
#3
Jun12-09, 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]

???

CompuChip
#4
Jun12-09, 10:11 AM
Sci Advisor
HW Helper
P: 4,300
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.
username12345
#5
Jun12-09, 11:04 PM
P: 48
Ok, that explanation helped, thankyou.


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