What Are the Level Curves for f(x,y)=e^-(2x^2+2y^2)?

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The discussion focuses on calculating level curves for the function f(x,y) = e^-(2x^2 + 2y^2). To derive the level curves, set e^-(2x^2 + 2y^2) equal to a constant c, leading to the equation x^2 + y^2 = -ln(c)/2. This equation is valid only for constants c in the range 0 < c < 1, indicating that the level curves are circular shapes centered at the origin.

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I am being asked to calculate level curves for the following equation:
f(x,y)=e^-(2x^2+2y^2) but I do not know where to start. Any advice on first steps would be greatly appreciated.
 
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Set e^-(2x^2+2y^2) = c, where c is a constant. Solve for x or y.

Once you do that, see if you can come up with a more general solution.
 
One obvious point: if [itex]e^{-(2x^2+ 2y^2)}= c[/itex] then [itex]-(2x^2+ 2y^2)= ln(c)[/itex] so that [itex]x^2+ y^2= -ln(c)/2[/itex] which is only possible if -ln(c)< 0 which means 0< c< 1. What figures will level curves be?
 

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