# Multivariate Functions

• Apr 2nd 2008, 12:36 PM
kdogg121
Multivariate Functions
I am having trouble with this problem, directions indicate to sketch the indicated level curve f(x,y) = C for each choice of constant C.

f(x,y)= x+2y ; C=1, C=2, C=-3

I am not too concerned with the actual sketching, but I am having trouble finding the points to graph. The answer in the book shows 3 separate lines with line C=1 having points of (0,1/2), (1,0) line C=2 having points of (0,1), (2,0) and line C=-3 having points of (-3,0), (0,-3/2). My professor showed us an example where we set the C number equal to the function. So I did for C=1 1=x+2y. I then isolated Y by itself as he did in the example and divided by 2, and got Y= -x+1 Is this right? How do I find the other coordinates? Thanks...
• Apr 2nd 2008, 12:44 PM
TheEmptySet
Quote:

Originally Posted by kdogg121
I am having trouble with this problem, directions indicate to sketch the indicated level curve f(x,y) = C for each choice of constant C.

f(x,y)= x+2y ; C=1, C=2, C=-3

I am not too concerned with the actual sketching, but I am having trouble finding the points to graph. The answer in the book shows 3 separate lines with line C=1 having points of (0,1/2), (1,0) line C=2 having points of (0,1), (2,0) and line C=-3 having points of (-3,0), (0,-3/2). My professor showed us an example where we set the C number equal to the function. So I did for C=1 1=x+2y. I then isolated Y by itself as he did in the example and divided by 2, and got Y= -x+1 Is this right? How do I find the other coordinates? Thanks...

Setting $f(x,y)=C$ gives

$x+2y=C \iff y=-\frac{1}{2}x+\frac{C}{2}$

This gives us a family of lines with slope -1/2 and y intercepts C/2

so if we plug in different values of C we get all of the level curves of the function.

I hope this helps.
• Apr 2nd 2008, 12:52 PM
kdogg121
Thank You for your help. So I would apply the same technique to this problem: f(x,y) = x^2-4x-y ; C= -4 , C=5

Y= x^2-4x-C ?? How do I get points out of this?
• Apr 2nd 2008, 12:57 PM
TheEmptySet
Quote:

Originally Posted by kdogg121
Thank You for your help. So I would apply the same technique to this problem: f(x,y) = x^2-4x-y ; C= -4 , C=5

Y= x^2-4x-C ?? How do I get points out of this?

Complete the square and you get

$y=x^2-4x+4 -4 -C \iff y= (x-2)^2-(C+4)$

So our level curves are Parabola's with the vertex at $(2,-(C+4))$