# Tangent of slope

## 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

## The Attempt at a Solution

Is my diagram correct ?
I'm wondering , why shouldnt we move the entire graph 'forward ' to x = 2 ? [/B]

#### Attachments

• 361.jpg
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fresh_42
Mentor
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?

• 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
Mentor
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.

• 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
Mentor
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.)

• 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?

#### Attachments

fresh_42
Mentor
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

• 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 doesnt change , so ∂z / ∂x = 0 ?

fresh_42
Mentor
ok , for the curve at x = 2 , the value of x doesnt 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.

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