Question

How do i get the level curves of a function like f(x,y)=(x-y)/(x^2+y^2+1)

Answer 1

I know that for other problems to find the level curves i let the fuction equal constants. But i just dont know how to start it off when its a fraction like this.

Answer 2

Its university maths.

Answer 3

No its calculus i think

Answer 4

@Tutor.Stacey Are you some computerising tutor or something or some helping robot? In every question thread you say the same thing. "May I get to know the level of math?"

"All right I'm working on your question right now".. lol

Perhaps you should start working for Tutor.com. When you go on there, you get redirected to this page where this tutor guy starts to persuade you in to buying a lesson and talks just like you =]

@zzr0ck3r you agree wid me?

Answer 5

Its only just the start of it i would need really and then a brief outline?

Answer 6

lol

Answer 7

yeah she took off on me

Answer 8

now we need to figure out what S.T.A.C.E.Y stands for @genius12

Answer 9

we are playing sorry

Answer 10

joking

Answer 11

lol

Answer 12

anyone have any ideas?

Answer 13

relax lol.

Answer 14

We may have gotten off topic a small bit??!

Answer 15

even if someone knew how to get the gradient of something like this. I know that i get the partials of the equation but when i get them they seem to be wrong?

Answer 16

its the fraction thats putting me off!

Answer 17

can anyone answer mine?!

Answer 18

Hmm, ok I've got an idea.

Answer 19

Let's try setting it equal to a constant, like you normally do.\[\large k=\frac{x-y}{x^2+y^2+1}\]Multiplying through by our denominator,\[\large k(x^2+y^2+1)=x-y\]Multiplying some stuff,\[\large kx^2+ky^2+k=x-y\]Rearranging some stuff,\[\large kx^2-x+ky^2+y=-k\]Dividing both sides by k,\[\large x^2-\frac{1}{k}x+y^2+\frac{1}{k}y=-1\]This 1/k is just a constant, let's write it as something simpler.\[\large x^2-cx+y^2+cy=-1\]Completing the square on x and y gives us something like this,\[\large \left(x-\frac{c}{2}\right)^2-\frac{c^2}{4}+\left(y+\frac{c}{2}\right)^2-\frac{c^2}{4}=-1\]
Which uhhhhhh simplifies to,\[\large \left(x-\frac{c}{2}\right)^2+\left(y+\frac{c}{2}\right)^2=\frac{c^2}{2}-1\]

Lot of messy constants, but it looks like the levels are circles.
Lemme graph it a sec, to make sure I didn't do anything horrendously stupid, cause I probably did.

Answer 20

Hmm yah it doesn't look like circles.. did something stupid.
Hmm... back to the drawing board..

Answer 21

@oldrin.bataku Mr smarty pants can prolly figure this out :) heh

Answer 22

im thinking ...

when x=y
\[z=\frac{0}{2y^2+1}\]

when y=-x
\[z=\frac{2x}{2x^2+1}\]

when x=-y
\[z=\frac{-2y}{2y^2+1}\]

etc ....

Answer 23

letting y = kx, or having some sort of angle setup is another thought in general