Why shouldn't we move the graph forward to x = 2?

[chetzread]

Homework Statement

Find the slope of tangent line to curve that is intersection to the surface z= (x^2) - (y^2) with plane x =2 , at point (2,1,3)
The ans given by the author is only∂z /∂y = -2

Homework Equations

The Attempt at a Solution

Is my diagram correct ?
I'm wondering , why shouldn't we move the entire graph 'forward ' to x = 2 ? [/B]

 

[fresh_42]

Your diagram is o.k. so far. Except you are right that it should be at $x=2$.
Which thoughts brought you to draw it this way. i.e. which equation did you use?
Next you have to find out, how to calculate a slope at a point. Do you know what this means?

 

[chetzread]

Your diagram is o.k. so far. Except you are right that it should be at $x=2$.
Which thoughts brought you to draw it this way. i.e. which equation did you use?
Next you have to find out, how to calculate a slope at a point. Do you know what this means?

by looking at the the equation z = 4-(y^2) alone , the whole graph is at x=0 axis alone , right ?

 

[fresh_42]

No, it should be at $x=2$ as you've said earlier. But that doesn't change the question, because you have eliminated $x$ in the equation anyway. So you have only a curve in a plane, the same as the usual $(x,y)$ case. The variables simply have different names and all is $(y,z)$ instead. And the slope of a tangent at the curve in a point $(2,1,3)$ is now simply at $y=1$ and $z=z(y)=4-y^2$.
You are correct, except you didn't say how to calculate this slope.

 

[chetzread]

The variables simply have different names and all is (y,z)(y,z) instead and z= 4-(y^2) .

so, the the whole graph is at x = 0 , right ?
since in z= 4-(y^2) , everything is in z and y , how we know that x = 2 by looking at z= 4-(y^2) ?

 

[fresh_42]

so, the the whole graph is at x = 0 , right ?

No. Originally it's in 3D space with $(x,y,z)$. The intersection is at $x=2$. It is like cutting the whole thing along $x=2$.
That does not change. But once you have cut it, there is no $x$ anymore. However, you must not forget $x=2$ if further investigations on the original surface would be made, e.g. by comparing the result with a cut at another $x$.
As long as the calculations take place with $x=2$ fixed, you may substitute all $x$ by $2$ as you did and forget (for the moment) that there is an $x$ at all. But it stays $x=2$. We simply do not consider it.

since in z= 4-(y^2) , everything is in z and y , how we know that x = 2 by looking at z= 4-(y^2) ?

By looking only at the plane, our cut, there is no $x$ anymore. You can think of it as a parabola with an $y-$ and $z-$ axis. On top of this parabola drawing you note "$\text{Intersection along }x=2$" as its label. Or the more complicated way in a 3D picture like yours, with the $x-$coordinate $2$. (As you also already mentioned in your first post.)

 

[chetzread]

No. Originally it's in 3D space with $(x,y,z)$. The intersection is at $x=2$. It is like cutting the whole thing along $x=2$.
That does not change. But once you have cut it, there is no $x$ anymore. However, you must not forget $x=2$ if further investigations on the original surface would be made, e.g. by comparing the result with a cut at another $x$.
As long as the calculations take place with $x=2$ fixed, you may substitute all $x$ by $2$ as you did and forget (for the moment) that there is an $x$ at all. But it stays $x=2$. We simply do not consider it.By looking only at the plane, our cut, there is no $x$ anymore. You can think of it as a parabola with an $y-$ and $z-$ axis. On top of this parabola drawing you note "$\text{Intersection along }x=2$" as its label. Or the more complicated way in a 3D picture like yours, with the $x-$coordinate $2$. (As you also already mentioned in your first post.)

do you mean for z= (x^2) - (y^2) , when x =2 , z= 4 - (y^2) , we draw the curve at x = 0 first , then , extend the line along x-axis and draw another same curve at x = 2?

 

[fresh_42]

do you mean for z= (x^2) - (y^2) , when x =2 , z= 4 - (y^2) , we draw the curve at x = 0 first , then , extend the line along x-axis and draw another same curve at x = 2?

I mean we draw it at $x=2$ in the first place. The question asks about the point $(2,1,3)$ and this point isn't part of anything with $x=0$.
Also at $x=0$ the parabola becomes $z=-y^2$ which is shifted by $4$ compared to $z=4-y^2$.

In your computer graphic you set $x=2$ as a constant dimension of a 3D space which it is not. A point $(0,1,-1)$ is on the original surface, however, not on the surface of your computer graphic. Have a look how it really looks like:

http://www.wolframalpha.com/input/?i=z=x^2-y^2

 

[chetzread]

I mean we draw it at $x=2$ in the first place. The question asks about the point $(2,1,3)$ and this point isn't part of anything with $x=0$.
Also at $x=0$ the parabola becomes $z=-y^2$ which is shifted by $4$ compared to $z=4-y^2$.

In your computer graphic you set $x=2$ as a constant dimension of a 3D space which it is not. A point $(0,1,-1)$ is on the original surface, however, not on the surface of your computer graphic. Have a look how it really looks like:

http://www.wolframalpha.com/input/?i=z=x^2-y^2

ok , for the curve at x = 2 , the value of x doesn't change , so ∂z / ∂x = 0 ?

 

[fresh_42]

ok , for the curve at x = 2 , the value of x doesn't change , so ∂z / ∂x = 0 ?

For the curve with $x=2$ there is no $x$ anymore. We substituted it. $z$ is a function of $y$ alone.