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Tangent vector on the intersection of surfaces

  1. Mar 1, 2017 #1
    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. jcsd
  3. Mar 2, 2017 #2

    Mark44

    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.
     
  4. Mar 2, 2017 #3
    Also, one thing. What's the difference between letting ##x = -\sqrt{2-t^2}## and letting ##x = \sqrt{2-t^2}##?
     
  5. Mar 2, 2017 #4

    Mark44

    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.
     
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