- #1

Mr Davis 97

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## Homework Statement

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)##.

## Homework Equations

## 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?