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Ive been stuck on this for over 2 hours trying to figure this out but it isn't working so finally I decided to bother you guys by asking you.

I know that in level curve you set equation to constants K but I am not even sure how you draw level curve once you set it to a constant....

for example below is a solution to an equation lnx+lny=f(x,y) where part b is the solution shows how to graph the level curve but i REALLLY cant figure out where or why e^k came from...

I know in part A they isolated y because they are looking to find equation of level curve passing through (1,1) but in part b where did they come up with e^k?

I am having a really hard time graphing or drawing level curves anything would help.

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up vote 1 down vote accepted

For fix $k$ you need to find all pairs $x,y$ so that $\ln(x)+\ln(y)=k$. Applying the exponential to both sides we get $e^{\ln(x)+\ln(y)}=e^k$. The left-hand side is equal to $e^{\ln(x)}\cdot e^{\ln(y)}=xy$. Therefore the level surface is equal to the hyperbola $e^k=xy$.

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There isn't really a simple general procedure for drawing level curves. You just have to use what you know about curve drawing to figure it out. What this means is that we are only good at drawing level curves of the following types:

  1. The graph of a function $y=g(x)$.

  2. A "sideways" graph of the form $x=g(y)$.

  3. A conic section such as a circle, ellipse, or hyperbola.

For the function $f(x,y) = \ln x + \ln y$, the level curves are curves of the form $$ \ln x + \ln y \,=\, C $$ where $C$ is a constant. This isn't of types (1), (2), or (3) listed above, but we can make it of type (1) by solving the equation for $y$. This gives $$ y \,=\, \frac{e^C}{x}. $$ Then the level curves look the graphs of this equation for different values of $C$. This is a family of hyperbolas with asymptotes along the $x$ and $y$ axes.

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oooooo i got it! what u said fixed with azerel makes sense to by making C constant we remove 1 unknown variable and then we graph it like normal xy plane. – Raynos Apr 15 '12 at 23:35

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