1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivatives of vector functions

  1. Mar 5, 2008 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    [​IMG]
    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.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 5, 2008 #2
    In general, if a 2-coordinate vector is given by [itex]<a_1,a_2>[/itex] then a normal vector (my best guess is as good as yours) can be given by [itex]<-a_2,a_1>[/itex] or [itex]<a_2,-a_1>[/itex]. 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
  4. Mar 5, 2008 #3
    Looking over my notes:

    [tex]\overrightarrow N = \frac{\overrightarrow T'}{|\overrightarrow T' |}[/tex]

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Derivatives of vector functions
Loading...