Help finding an equation for the level curve

  • #1
Hey guys, I tried it out, but I just don't get it. I have to find the equation for the level curve f(x, y)=(x^2 + y^2)e^(xy); that contains the point P(1,0). By the way, e^(xy) is read e to the x times y, just in case.


What I did, which looks wrong the whole way was:

(x^2 + y^2)e^(xy) ---> (x^2)(e^(xy)) + (y^2)(e^(xy)) = 0

---> ln(x^2) + ln(e^(xy)) = (-1) ln(y^2) + ln(e^(xy))
---> 2ln(x) + xy = (-1) (2ln(y) + xy)
---> 2ln(x) = (-1)(2ln(y))


...and i'm stuck there. Could anyone help correct this, or if possible, help continue? Thanks a bunch.
 

Answers and Replies

  • #2
CarlB
Science Advisor
Homework Helper
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I thought that the equation you need to solve would be written as:

[tex]f(x,y) = f(1,0)[/tex]

Or

[tex](x^2+y^2)e^{xy} = e[/tex]

From here, my instinct would be to convert to polar coordinates.

Carl
 
  • #3
ehild
Homework Helper
15,543
1,914
SigmaCrisis said:
I have to find the equation for the level curve f(x, y)=(x^2 + y^2)e^(xy); that contains the point P(1,0).

What I did, which looks wrong the whole way was:

(x^2 + y^2)e^(xy) ---> (x^2)(e^(xy)) + (y^2)(e^(xy)) = 0
...

Why did you write that

[tex]f(x,y)=(x^2+y^2)e^{xy}=0[/tex]?

You want a level curve. That means that f(x,y) is the same for all points of the curve.The level curve should contain the point with x=1, y=0. Plug in these values and see what you get for f(x,y).

ehild
 

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