# Level Curves

by Grzegorz
Tags: curves
 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.
 P: 240 Level curves are curves where the function is constant, and equals lets 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.
Math
Emeritus
 Quote by elibj123 Level curves are curves where the function is constant, and equals lets 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.