Frustrated with Drawing Level Curves: Help Appreciated!

[Grzegorz]

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.

 

[elibj123]

Level curves are curves where the function is constant, and equals let's say to a number c.

So you get an equation

\frac{-3y}{x^{2}+y^{2}+1}=c

Rearranging this gives you

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

Now you just have to investigate different values of c.

 

[HallsofIvy]

Level curves are curves where the function is constant, and equals let's say to a number c.

So you get an equation

\frac{-3y}{x^{2}+y^{2}+1}=c

Rearranging this gives you

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

That should be c(y- 3/(2c))^2. That is, that leading "c" should be outside the parentheses.

Now you just have to investigate different values of c.