# Curve and tangent of a surface intersected by a plane

1. Oct 7, 2009

### Ral

1. The problem statement, all variables and given/known data
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 $$\stackrel{\rightarrow}{n}$$?

2. Relevant equations

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 7, 2009

### Staff: Mentor

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?

3. Oct 7, 2009

### Ral

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

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

4. Oct 7, 2009

### 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.

5. Oct 8, 2009

### Ral

$$z = \sqrt{x^2 + 1}$$

Then would that be the curve?

6. Oct 8, 2009

### Staff: Mentor

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