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Above is the trajectory of a baby fly, whose position vector at time t is given by

[tex]\overrightarrow r \left( t \right) = \left\langle {t^2 ,t} \right\rangle[/tex].

1. On the graph, carefully sketch , [tex]\overrightarrow r \left( 1 \right),\,\,\overrightarrow {r'} \left( 1 \right),\,\,\overrightarrow T \left( 1 \right),\,\overrightarrow N \left( 1 \right),{\rm{ and }}\overrightarrow {r''} \left( 1 \right)[/tex] in their appropriate positions. Use a straight edge and make sure to pay attention to the lengths of the vectors.

My effort so far:

[tex]\begin{array}{l}

\overrightarrow r \left( 1 \right) = \left\langle {1^2 ,\,1} \right\rangle = \left\langle {1,\,1} \right\rangle \\

\overrightarrow {r'} \left( t \right) = \left\langle {2t,\,1} \right\rangle ,\,\,\,\,\overrightarrow {r'} \left( 1 \right) = \left\langle {2 \cdot 1,\,1} \right\rangle \,\, = \,\left\langle {2,\,1} \right\rangle \\

\\

\overrightarrow T \left( 1 \right) = \frac{{\overrightarrow {r'} \left( 1 \right)}}{{\left| {\overrightarrow {r'} \left( 1 \right)} \right|}} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt {2^2 + 1^2 } }} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt {4 + 1} }} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt 5 }} \\

\end{array}[/tex]

Our book doesn't cover N. I imagine that's the normal vector, which is orthogonal to the tangent vector. But how do I find that? I know that if N and T are orthogonal, then their dot product should equal 0. So what times <2,1>/sqrt(5)=0? <0,0> would work, but that's the zero vector. I don't imagine that's what I'm trying to find.

Any clues?

nitpick: There's no such thing as a baby fly. Fly is the adult stage of an insect that was once a maggot.