65d882…: “Inside of a loud fish”

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A curve $C$ in space is described by the vector-valued function: $\vec {p}(t) = \vector {t^2-1,2t,2t^2+2}$ Find a unit vector with positive $x$ -component that is orthogonal to both $\vec {p}(0)$ and $\vec {p}'(0)$ : $\uvec {u} =\vector {\answer {2/\sqrt {5}},\answer {0},\answer {1/\sqrt {5}}}$

Find the point $(x,y,z)$ where $\vec {p}(t)$ is orthogonal to $\vec {p}'(t)$ : $(x,y,z) = \left (\answer {-1},\answer {0},\answer {2}\right )$

Does $\vec {p}$ use arc length as a parameter?

yes no

Just check to see if $|\vec {p}'(t)| = 1$ .

Find all points where the tangent line to the curve at $t=0$ intersects the surface $2x+y-z^2=0$ . $(x,y,z) = \left (\answer {-1},\answer {6},\answer {2}\right )$