Level curve of f(x,y) = (x² + y²)e^(xy) through P(1,0)

[SigmaCrisis]

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.

 

[CarlB]

I thought that the equation you need to solve would be written as:

$$f(x,y) = f(1,0)$$

Or

$$(x^2+y^2)e^{xy} = e$$

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

Carl

 

[ehild]

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

$$f(x,y)=(x^2+y^2)e^{xy}=0$$?

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