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Vector analysis

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Let the curve C be paramatized into polar coordinates given by:

    [tex]\[r\left( t \right)=\left( r\left( t \right)\cos \theta \left( t \right),\,\,\,\,\,r\left( t \right)\sin \theta \left( t \right) \right),\,\,\,\,\,a\le t\le b\][/tex]
    where r and theta is continuous derivatives and r(t) > 0.

    Show that:

    [tex]\[F\left( r\left( t \right) \right)\cdot r'\left( t \right)=\theta '\left( t \right)\,\,\,\,\,and\,\,\,\,\,\int_{C}{F\cdot dr=\theta \left( b \right)}-\theta \left( a \right)\][/tex]


    2. Relevant equations

    F is given by:

    [tex]\[F\left( x,y \right)=\frac{-y\,i+x\,j}{{{x}^{2}}+{{y}^{2}}}\][/tex]


    3. The attempt at a solution

    No idea... I get stupid when it changes into polar coords.
    So anyone with a little hint maybe ?


    Regards
     
  2. jcsd
  3. Nov 19, 2009 #2
    Start by calculating what F is in polar coordinates: x = r cos theta, y = r sin theta.
     
  4. Nov 19, 2009 #3
    Then I get:

    [tex]F\left( r,\theta \right)=\frac{\cos \left( \theta \right)j-\sin \left( \theta \right)i}{r}[/tex]
     
    Last edited: Nov 19, 2009
  5. Nov 19, 2009 #4

    Dick

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    So far, so good. Now what's R'(t) where R(t)=r(t)cos(theta(t))i+r(t)sin(theta(t))j? Try not to confuse the scalar r(t) with the vector R(t). It's R'(t) you want to dot with F.
     
  6. Nov 19, 2009 #5
    Well, that's what confuses me :)
    How do I differentiate a function with functions in it ? And with respect to what ? r or theta ? If I do it with respect to both, then I get to different R'(t), dont I ?
     
  7. Nov 19, 2009 #6

    Dick

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    Differentiate with respect to t. You'll need to use the product rule.
     
  8. Nov 19, 2009 #7
    Hmmm...
    But there is no t in R(t) ? :S
     
  9. Nov 19, 2009 #8

    Dick

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    There are two functions r(t) and theta(t)!!
     
  10. Nov 19, 2009 #9
    Ehhh, I feel pretty stupid right now :S
    Do I know the functions r(t) and theta(t) ? :S
     
  11. Nov 19, 2009 #10

    Dick

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    No, but you know dr(t)/dt=r'(t) and dtheta(t)/dt=theta'(t). Just write out R(t)' in terms of them.
     
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