# Derivatives of vector functions

1. Mar 5, 2008

### tony873004

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

$$\overrightarrow r \left( t \right) = \left\langle {t^2 ,t} \right\rangle$$.

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

My effort so far:

$$\begin{array}{l} \overrightarrow r \left( 1 \right) = \left\langle {1^2 ,\,1} \right\rangle = \left\langle {1,\,1} \right\rangle \\ \overrightarrow {r'} \left( t \right) = \left\langle {2t,\,1} \right\rangle ,\,\,\,\,\overrightarrow {r'} \left( 1 \right) = \left\langle {2 \cdot 1,\,1} \right\rangle \,\, = \,\left\langle {2,\,1} \right\rangle \\ \\ \overrightarrow T \left( 1 \right) = \frac{{\overrightarrow {r'} \left( 1 \right)}}{{\left| {\overrightarrow {r'} \left( 1 \right)} \right|}} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt {2^2 + 1^2 } }} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt {4 + 1} }} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt 5 }} \\ \end{array}$$

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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 5, 2008

### jhicks

In general, if a 2-coordinate vector is given by $<a_1,a_2>$ then a normal vector (my best guess is as good as yours) can be given by $<-a_2,a_1>$ or $<a_2,-a_1>$. There might be an actual convention as to which to use, but the normal vector typically points "out" away from the curvature I think. I didn't normalize the result for brevity.

Last edited: Mar 5, 2008
3. Mar 5, 2008

### bob1182006

Looking over my notes:

$$\overrightarrow N = \frac{\overrightarrow T'}{|\overrightarrow T' |}$$

N is the Unit Normal Vector, a special one since T and T' are orthogonal

Don't know if that's what your problem asks for, but the T's are defined the exactly the same, so guessing this definition is about right.