# Find the tangent line between two surfaces

1. Dec 16, 2008

### jheld

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

Let C be the intersection of the two surfaces:
S1: x^2 + 4y^2 + z^2 = 6;
s2: z = x^2 + 2y;
Show that the point (1, -1, -1) is on the curve C and find the tangent line to the curve C at the point (1, -1, -1).

2. Relevant equations
partial derivates, maybe the gradient vector and directional derivatives
though, maybe symmetrical equations like x - x_0/partial derivative with respect to x = y etc...

3. The attempt at a solution
I'm just kind of wondering where to start. I think I should be making these into vectors, but I'm not quite sure how to do so, and of course thinking about partial derivatives.

2. Dec 16, 2008

### NoMoreExams

Well find the intersection, you know that S1 can be written as $$x^{2} = 6 - 4y^{2} - z^{2}$$ and S2 can be written as $$x^{2} = z - 2y$$ so set them equal to each other to find their intersection. Are you sure your 2nd equation is correct?

3. Dec 16, 2008

### Dick

You know that the tangent direction is tangent to both surfaces, and the gradient of each surface is normal to that tangent direction. Use the two gradient directions to deduce the tangent direction.

4. Dec 16, 2008

### jheld

Both equations are written correctly.
I'm trying to find their point of intersection and I thought to complete the square, but it doesn't seem to be working.
I'm unsure of how to find the gradient vector between two surfaces.

5. Dec 16, 2008

### Dick

Find the gradient of each surface separately. That gives you two vectors which are orthogonal to the tangent direction. How can you find a vector that's orthogonal to two given vectors?

6. Dec 16, 2008

### jheld

Okay I found the gradients as:
S1: <2x, 8y, 2z>
s2: <2x, 2 -1>

I went on to find their symmetric equations. But, I'm not sure how to relate them.

7. Dec 16, 2008

### Dick

You are interested in the point x=1, y=(-1) and z=(-1). The vector you want is perpendicular to both those vectors.

8. Dec 16, 2008

### jheld

I'm not quite sure how to show a vector like that. Is that supposed to be the gradient vector? Should I use the dot product between S1 and S2 directional derivatives to get that?

9. Dec 16, 2008

### Dick

You were supposed to say, "Ah ha! I can use the cross product!".

10. Dec 16, 2008

### jheld

I use the cross-product? Oh, well I suppose that could work, haha.
Would I calculate the cross-product before plugging in the values? That leaves me with a bunch of x, y and z's.

11. Dec 16, 2008

### Dick

Yes, the cross product of two vectors is perpendicular to both. It doesn't matter whether you plug in the numbers before or after, does it? Whatever you find easier. When you are done you will have a vector that points in the direction of the tangent line, right?

12. Dec 16, 2008

### jheld

I think I understand it now. I think it would be easier to plug them in before, though, less writing, you know? Yes, it does point in the direction of the tangent line.

Thanks for all your help :)