
#1
Oct305, 08:48 PM

P: 15

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. 



#2
Oct305, 08:49 PM

P: 15

sorry bout the double pasting there.




#3
Oct305, 11:21 PM

P: 167

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 



#4
Oct305, 11:26 PM

Sci Advisor
HW Helper
P: 3,149

Help finding an equation for the level curve...
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!




#5
Oct405, 12:21 PM

P: 15

Thanks a bunch guys...



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