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Mathematics 18 Online
OpenStudy (baldymcgee6):

Describe what the graph of the level curve looks like at f(x,y) = 2 for the function: f(x,y) = (sqrt(x-y))/(sqrt(x+y))

OpenStudy (baldymcgee6):

I think that the level curve is just \[2 = \frac{ \sqrt{x-y} }{ \sqrt{x+y}}\] Is this correct?

OpenStudy (baldymcgee6):

@CliffSedge?

OpenStudy (anonymous):

I'll have to dig up my old books for this one . . .

OpenStudy (baldymcgee6):

i know the domain is x>=y and x>-y

OpenStudy (tkhunny):

First, consider Domain issues. AS WRITTEN, we have: \[x \ge y\] from the numerator and \[x + y > 0\] from the denominator. We need to see THAT before thinking anything else. We haven't even thought about the '2'.

OpenStudy (tkhunny):

Second. we can start to deconstruct the confused nature of the whole expression. Once we know everything is positive (because of the Domain Considerations), square both sides. That might simplify our lives.

OpenStudy (baldymcgee6):

okay, \[4 = \frac{{x-y} }{ {x+y}}\]

OpenStudy (tkhunny):

Next. We know x + y is not zero (again because of our Domain Considerations). Multiply by that. Then, Solve for y.

OpenStudy (baldymcgee6):

okay so y = -3x/5

OpenStudy (tkhunny):

And what is that?

OpenStudy (baldymcgee6):

a line

OpenStudy (baldymcgee6):

is this what a level curve is?

OpenStudy (tkhunny):

There you have it, excepting one thing. It's not a line after we consider the Domain Issues. It's only a Ray.

OpenStudy (baldymcgee6):

right, okay, thanks for your help.

OpenStudy (tkhunny):

It is a little surprising that the level curves of that expression would be rays. That was not my first guess. This is where confidence comes in. Check your work. Make sure you thought of everything. Move on with confidence to the next problem, having learned that our first guess it not always right. That's why we go through the exercise. Mathematics rather insists that we go through the exercise, even if we are sure we are thinking correctly. Good work.

OpenStudy (baldymcgee6):

I'm not exactly sure what a "level curve" is... is it just equating the function to some constant? What does it do for us?

OpenStudy (tkhunny):

Ever see a topographical map? It has curves that indicate equal elevation or equal depth. They may be called isoclines in other applciations. They are a projection of a higher dimensional object onto a lower dimensional surface. In this case, we have a 3D fellow we want to understand so we flatten it out on to the x-y plane and try to get a handle on it. Essentially, if we take a slice out of it, parsllel to the x-y plane, what will it look like?

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