Graph z=f(x,y): Tracing the x-axis, y-axis, and Lines

[jwxie]

Homework Statement

Sketch the indicated traces, and the graph z = f(x,y)

f(x,y) = x - 2y z = 0; z = 1; x = 0; y = 0

The Attempt at a Solution

z = f(x,y) = 0 = x - 2y -> y = -x/2 the trace is a line

z = f(x,y) = 1 = x - 2y -> y = x-1/2 another line going through x = 1/2

z = f(0,y) = 0 - 2y -> z = -2y a line with slope -1 in zy plane

z = f(x,0) = x - 0 -> z = xI drew the graph, and i wanted to verify. The one on wolframalpha is too hard to verify...

If it is wrong, please tell me what went wrong. Thanks...

When I drew the graph, I noticed that the third and fourth traces were simply x and y axis...

 

[HallsofIvy]

If f(x, y, z)= x- 2yz= 0 then 2yz= x and z= x/2y. That is a "hyperoloid of one sheet".

Your graph doesn't really show much- it looks to me like a part of a plane.

 

[Mark44]

If f(x, y, z)= x- 2yz= 0 then 2yz= x and z= x/2y. That is a "hyperoloid of one sheet".

Your graph doesn't really show much- it looks to me like a part of a plane.

It's actually a plane, but jwxie didn't include punctuation to separate the function from the traces required.

I'm pretty sure this is what was intended:
For this function f(x,y) = x - 2y, sketch these traces: z = 0; z = 1; x = 0; y = 0.

 

[Mark44]

Homework Statement

Sketch the indicated traces, and the graph z = f(x,y)

f(x,y) = x - 2y z = 0; z = 1; x = 0; y = 0

The Attempt at a Solution

z = f(x,y) = 0 = x - 2y -> y = -x/2 the trace is a line

The equation of the line in the x-y plane is y = x/2. Its slope is m = 1/2.

z = f(x,y) = 1 = x - 2y -> y = x-1/2 another line going through x = 1/2

The trace is a line in the plane z = 1. Its slope is 1/2, as before. The line goes through (0, -1/2, 1).

z = f(0,y) = 0 - 2y -> z = -2y a line with slope -1 in zy plane

No, the slope is -2.

z = f(x,0) = x - 0 -> z = x


I drew the graph, and i wanted to verify. The one on wolframalpha is too hard to verify...

If it is wrong, please tell me what went wrong. Thanks...

When I drew the graph, I noticed that the third and fourth traces were simply x and y axis...

No, the 3rd and 4th traces are lines in the y-z plane and x-z plane, respectively.

 

[njama]

First to get sense what figure are you drawing notice:
z=x-2y

x+2y+z=0

That is clearly a plane which passes through (0,0,0) and it is perpendicular of the vector (1,2,1).

You've found

z = f(x,y) = 0 = x - 2y -> y = -x/2 the trace is a line

z = f(x,y) = 1 = x - 2y -> y = x-1/2 another line going through x = 1/2

z = f(0,y) = 0 - 2y -> z = -2y a line with slope -1 in zy plane

z = f(x,0) = x - 0 -> z = x

well. You did not sketch them right.

Regards.

 

[Mark44]

First to get sense what figure are you drawing notice:
z=x-2y

x+2y+z=0

Your algebra is off, here. z = x - 2y <==> x - 2y - z = 0. A normal to this plane is the vector <1, -2, -1>.

That is clearly a plane which passes through (0,0,0) and it is perpendicular of the vector (1,2,1).

You've found



well. You did not sketch them right.

Regards.

 

[jwxie]

Hi, Mark. Sorry for all the mistakes. I just checked what I had on the paper, apparently I should have looked at my paper while posting the message.

I had the exact thing as you mentioned.

I agreed with you that
when z = 0,, we have y = x/2, and thus slope m = 1/2
when z = 1,, we have y = x-1/2, and thus slope m = 1/2 because the equation is 1/2 (x-1)
when z = 0,, we have z = -2y and thus m = -2
when y = 0,, we have z = xi know how the curves each individual trace in R2 (for example, when z = 0, we have xy axis graph, y = x/2) look like.

how do i combine them?
thank you.

 

[Mark44]

You already have the y-z trace. Do another one in the plane x = 2, for example. That should give you a good idea of what your function looks like (the graph is a plane).

 

[jwxie]

You already have the y-z trace. Do another one in the plane x = 2, for example. That should give you a good idea of what your function looks like (the graph is a plane).

Hi. Thank you. I have the xy traces (two lines, x/2 and x-1/2)
i also have the xz trace (when y = k), and this gives me the xz plane

you mean i need to sketch an yz plane with x = k?

 

[Mark44]

Hi. Thank you. I have the xy traces (two lines, x/2 and x-1/2)

There is only one x-y trace: the line in the x-y plane whose equation is y = x/2. The other line is the cross-section (not trace) in the plane z = 1, with equation y = x/2 - 1/2.

i also have the xz trace (when y = k), and this gives me the xz plane

The x-z trace is the plane y = 0, not y = k. This trace is not the x-z plane; it is the line z = x in the plane y = 0.

you mean i need to sketch an yz plane with x = k?

What I mean is that you should sketch the cross section in some plane parallel to the y-z plane, such as in the plance x = 5.

There are only three traces:
x-y trace, the cross section of the graph of your function in the x-y plane (the plane z = 0).
x-z trace, the cross section of the graph in the x-z plane (the plane y = 0).
y-z trace, the cross section of the graph in the y-z plane (the plane x = 0).

Intersections of the graph of the function with other planes are called cross sections. Cross sections in planes parallel to the x-y plane are called level curves.

 

[jwxie]

thank you mark. i got it.