Level curve of (x² + y²)e^(xy) through point (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.Help finding an equation for the level curve...

--------------------------------------------------------------------------------

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.

 

[SigmaCrisis]

sorry bout the double pasting there.

 

[Edwin]

I believe that level curves are values of f(x,y) where f(x,y) = a constant.

To find the level curve of f(x, y)=(x^2 + y^2)*e^(x*y) that contains the point p(1,0):

Step 1: Evaluate f(x,y) at p(1,0) that is, f(1,0) = constant k

Step 2: your level curve that contains the point p(1,0) is then just
f(x,y) = k, the constant you found in step one.

I hope this helps, if not, let me know and I'll work an example similar to the one you posted above.

Best Regards,

Edwin

 

[Tide]

Why don't you use polar coordinates? Also, why are you setting the function to 0? You should be setting it to f(1, 0) -- as edwin suggested!

 

[SigmaCrisis]

Thanks a bunch guys...