# f(x,y,z) surfaces/parametrization

• Mar 1st 2007, 03:00 AM
flash101
f(x,y,z) surfaces/parametrization
We know that if we have 3 variables in a function, its graph will be (usually) a four-dimensional surface.

Suppose we are given a function that has 3 variables f(x,y,z) = 16x^2 - y + 4z^2. What are all the level surfaces of the func.? For EVERY kind of level surface, choose examples and determine a parametrization for each surface that you come up with. Plot these results.

Uh, where to start?
• Mar 1st 2007, 05:55 AM
CaptainBlack
Quote:

Originally Posted by flash101
We know that if we have 3 variables in a function, its graph will be (usually) a four-dimensional surface.

Suppose we are given a function that has 3 variables f(x,y,z) = 16x^2 - y + 4z^2. What are all the level surfaces of the func.? For EVERY kind of level surface, choose examples and determine a parametrization for each surface that you come up with. Plot these results.

Uh, where to start?

The level surfaces are the surfaces (if any) defined by the equations:

f(x,y,z) = C,

for some C, This will ge the level surface corresponding to C.

RonL
• Mar 1st 2007, 11:02 AM
ThePerfectHacker
Quote:

Originally Posted by flash101
We know that if we have 3 variables in a function, its graph will be (usually) a four-dimensional surface.

Suppose we are given a function that has 3 variables f(x,y,z) = 16x^2 - y + 4z^2. What are all the level surfaces of the func.? For EVERY kind of level surface, choose examples and determine a parametrization for each surface that you come up with. Plot these results.

The level surfaces are,
C=16x^2-y+4z^2
Thus,
16x^2+4z^2-C=y
Note, that,
y=16x^2+4z^2
Is always a paraboloid (elliptic) opening along the positive y-axis.
However,
y=16x^2+4z^2-C
Is still a elliptic paraboloid opening along the positive y-axis. However, it is shifted along y either positive or negative that many units depending on the sign of C.
• Mar 1st 2007, 12:24 PM
flash101
Aha! Sorry, my mistake. No wonder that seemed so easy. I made a mistake in the question.

It should be:

f(x,y,z) = 16x^2 - y^2 + 4z^2 ...

Now adding the square makes it 10 times harder.

Also, according to my prof. it's choosing examples and doing the parametrization of each example that is hard, if you're familiar with that at all.

Many thanks again!
• Mar 1st 2007, 01:11 PM
ThePerfectHacker
Quote:

Originally Posted by flash101
Aha! Sorry, my mistake. No wonder that seemed so easy. I made a mistake in the question.

It should be:

f(x,y,z) = 16x^2 - y^2 + 4z^2 ...

Again consider,
16x^2-y^2+4z^2=C

There are three cases.

Case C>0
Then divide through by C,
(16/C)x^2-(1/C)y^2+(4/C)z^2=1
This is a hyperbolid of 1 sheet.
Because, 16/C,1/C,4/C>0

Case C=0
We have,
16x^2+4z^2=y^2
This is a double (elliptic) cone.

Case C<0
We have upon divison by C and multiplication of (-1),
(-16/C)x^2+(1/C)y^2+(-4/C)^2=-1
Now this is a hyperboliod of 2 sheets.
Because (-16/C)>0, (1/C)<0, (-4/C)>0.
• Mar 2nd 2007, 08:29 PM
flash101
Thanks, TPH. How do I go about parametrizing that? And how do I show that it really is a parametrization for that surface?