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Level Curves

by Grzegorz
Tags: curves
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Grzegorz
#1
Jan19-10, 07:22 PM
P: 1
hi... I am new to this topic and frustrated.

I have a curve f(x,y)= -3y/(x2 +y2 + 1)

I was asked to draw a level curve of this and I'm not getting anywhere with it. If anyone has any pointers, or can help me with solving this question I would be gretfull. The only other thing this question asks is to describe it at the orgin or at (0,3) ( which is steeper).


thanks for any help.
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elibj123
#2
Jan20-10, 01:12 AM
P: 240
Level curves are curves where the function is constant, and equals lets say to a number c.

So you get an equation

[tex]\frac{-3y}{x^{2}+y^{2}+1}=c[/tex]

Rearranging this gives you

[tex]cx^{2}+cy^{2}-3y=-c[/tex]
[tex]cx^{2}+cy^{2}-2c\frac{3}{2c}y=-c[/tex]
[tex]cx^{2}+cy^{2}-2c\frac{3}{2c}y+\frac{9}{4c^{2}}=\frac{9}{4c^{2}}-c[/tex]
[tex]cx^{2}+(cy-\frac{3}{2c})^{2}=\frac{9}{4c^{2}}-c[/tex]

Now you just have to investigate different values of c.
HallsofIvy
#3
Jan20-10, 05:40 AM
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Quote Quote by elibj123 View Post
Level curves are curves where the function is constant, and equals lets say to a number c.

So you get an equation

[tex]\frac{-3y}{x^{2}+y^{2}+1}=c[/tex]

Rearranging this gives you

[tex]cx^{2}+cy^{2}-3y=-c[/tex]
[tex]cx^{2}+cy^{2}-2c\frac{3}{2c}y=-c[/tex]
[tex]cx^{2}+cy^{2}-2c\frac{3}{2c}y+\frac{9}{4c^{2}}=\frac{9}{4c^{2}}-c[/tex]
[tex]cx^{2}+(cy-\frac{3}{2c})^{2}=\frac{9}{4c^{2}}-c[/tex]
That should be [itex]c(y- 3/(2c))^2[/itex]. That is, that leading "c" should be outside the parentheses.

Now you just have to investigate different values of c.


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