- #1
SigmaCrisis
- 15
- 0
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...
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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.
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...
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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.