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Help finding an equation for the level curve... |
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| Oct3-05, 07:37 PM | #1 |
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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. |
| Oct4-05, 01:14 AM | #2 |
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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 |
| Oct4-05, 01:17 AM | #3 |
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[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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