Science:Math Exam Resources/Courses/MATH200/December 2011/Question 01 (a)/Solution 1

The level curves of $f(x,y)$ satisfy $f(x,y)=C$ for arbitrary values of $C$ such that a solution to $f(x,y)=C$ exists. Starting with that equation we have

$\displaystyle e^{-x^{2}+4y^{2}}=C.$

We notice that we can take the natural logarithm of both sides giving us

$\displaystyle -x^{2}+4y^{2}=\ln(C),$

where $\ln(C)$ is just an arbitrary constant as well! In other words, the level curves of $e^{-x^{2}+4y^{2}}$ look the same as those of $-x^{2}+4y^{2}=g(x,y)$. The only difference between the level curves in each situation is the values of $C$ corresponding to each specific curve. The level curves of $g(x,y)$ are hyperbolae.

If we choose five different values of $C$: $C=e^{-2},e^{-1},1,e^{1},e^{2}$ and draw the resulting level curves, we get the figure below. (Hint: If you begin drawing the level curves starting with $C=1$, the remaining curves should be easier to draw)

Some level curves of f(x,y)