Tangent vector on the intersection of surfaces

1. Mar 1, 2017

Mr Davis 97

1. The problem statement, all variables and given/known data
The surfaces $x^2+y^2 = 2$ and $y=z$ intersect in a curve $C$. Find a unit tangent vector to the curve $C$ at the point $(1,1,1)$.

2. Relevant equations

3. The attempt at a solution
So I'm thinking that we can parametrize the surfaces to get a vector for the curve $C$.

Let $z=t$. Then $y=t$. Then $x = \sqrt{2-t^2}$. So we have a vector for the curve $C$, $\vec{r} (t) = \langle \sqrt{2-t^2}, t,t \rangle$. Then $\vec{r}' (t) = \langle \frac{-t}{\sqrt{2-t^2}}, 1,1 \rangle$, and $\vec{r}' (1) = \langle -1, 1,1 \rangle$. Then re-scaling to get a unit vector, we get $\langle \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \rangle$. Is this the correct answer?

2. Mar 2, 2017

Staff: Mentor

I don't see anything wrong with your work.

It might be helpful to sketch a graph of the curve C. Since it is formed by the intersection of a right circular cylinder whose central axis is along the z-axis, with the plane y = z, the curve of intersection is an ellipse.

3. Mar 2, 2017

Mr Davis 97

Also, one thing. What's the difference between letting $x = -\sqrt{2-t^2}$ and letting $x = \sqrt{2-t^2}$?

4. Mar 2, 2017

Staff: Mentor

You'll get one side or another of the ellipse I mentioned, which is symmetric across the y-z plane. With the positive root, you get points in the first octant (really, on the "front" side of the y-z plane; if z < 0, the curve isn't in the first octant any more). With the negative root, you get points on that back side of the y-z plane.