Question

Find the traces for the multivariable equation y=z^2

Answer 1

I am mainly confused about how you know which variable to make the trace equal to

Answer 2

The trace of a surface in a plane is the intersection of the surface with that plane. While we can discuss traces in any plane, for surfaces in the form x = f(y,z) we are particularly interested in traces in planes parallel to the yz plane. Note that the yz plane has the equation x = 0, and planes parallel to the yz plane have equations in the form x = a. To obtain the trace of x=f(y,z) in the plane x= a, we set f(y,z) =a
hence x = f(y, z) = y -z^2
the trace in the plane x =1 is y - z^2 =1
the trace in the plane x =2 is y -z^2 =2
so far and so on.

Answer 3

Whoa. f(y,z)? I'm not actually in multivariable. We are just doing an intro to that stuff in case any of us decides to to that, and we have not done any of that notation

Answer 4

And we are only dealing with xy,xz, and yz btw

Answer 5

If you have z = f(x,y) , why can't you have x = f(y,z)??

Answer 6

Obviously, in 3 D , the equation y = z^2 implies x =0, right?

Answer 7

Yes

Answer 8

So the xy trace is y=0?

Answer 9

hihihi... that's what I thought. No need to follow if you don't study it yet.

Answer 10

ok

Answer 11

xy trace is z =0

Answer 12

Why can't it be y=0

Answer 13

if the function is x,z, then y =0

Answer 14

The function is y=z^2

Answer 15

And it looks like this

Answer 16

that is y, z , since y = z^2 \(\implies\) y -z^2 =0