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,
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.
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!
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.