Curve and tangent of a surface intersected by a plane

[Ral]

Homework Statement

a.)\sqrt{x^2+y^2}
Find the equation of the tangent plane at the point given by: x = 1, y = 1
Draw the 3d-graph of the surface and the tangent plane.
\stackrel{\rightarrow}{n} = the normal vector to the tangent plane.

b.) If the surface is intersected with the plane y = 1, what curve do you get?
Find the intersection of the tangent line to the resulting curve when x = 1.
Draw a 2d graph of the curve and the tangent line.

c.) Is the tangent line obtained in part (b) orthogonal to \stackrel{\rightarrow}{n}?

Homework Equations

The Attempt at a Solution

I have part a done.

For the tangent plane, I have:
z= \frac{2}{\sqrt{2}}(x-1)+\frac{2}{\sqrt{2}}(y-1)+\sqrt{2}

and \stackrel{\rightarrow}{n}=(\frac{2}{\sqrt{2}},\frac{2}{\sqrt{2}},-1)

I'm not quite sure what to do for part b. I was also trying to use Maple to do the 3d graph, but that was also confusing me, if possible, can I also see what it's suppose to look like.

 

[Mark44]

Homework Statement

a.)\sqrt{x^2+y^2}
Find the equation of the tangent plane at the point given by: x = 1, y = 1
Draw the 3d-graph of the surface and the tangent plane.
\stackrel{\rightarrow}{n} = the normal vector to the tangent plane.

b.) If the surface is intersected with the plane y = 1, what curve do you get?
Find the intersection of the tangent line to the resulting curve when x = 1.
Draw a 2d graph of the curve and the tangent line.

c.) Is the tangent line obtained in part (b) orthogonal to \stackrel{\rightarrow}{n}?

Homework Equations

The Attempt at a Solution

I have part a done.

For the tangent plane, I have:
z= \frac{2}{\sqrt{2}}(x-1)+\frac{2}{\sqrt{2}}(y-1)+\sqrt{2}

and \stackrel{\rightarrow}{n}=(\frac{2}{\sqrt{2}},\frac{2}{\sqrt{2}},-1)

I'm not quite sure what to do for part b. I was also trying to use Maple to do the 3d graph, but that was also confusing me, if possible, can I also see what it's suppose to look like.

Is your surface z = \sqrt{x^2 + y^2}?
You didn't state that in your post. Part b asks you to find the curve where the plane y = 1 intersects your surface. Just substitute y = 1 in the equation of your surface. What do you get?

 

[Ral]

Yeah, z = \sqrt{x^2 + y^2} is the surface.

So the curve would then be \sqrt{x^2 + 1}?

 

[LCKurtz]

That isn't an equation; it's just an expression. What goes on the other side of the = sign?

Once you answer that, put it in a form you can recognize and identify it.

 

[Ral]

<br /> z = \sqrt{x^2 + 1}<br />

Then would that be the curve?

 

[Mark44]

That would be the equation of the curve, which is what the problem is really asking for.