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Simple question related to polar curves

  1. Jun 16, 2013 #1
    1. The problem statement, all variables and given/known data
    For a curve in Cartesian form, show that
    [tex]\tan \phi = \frac{xy'-y}{x+yy'}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    According to the book notation, ##\phi## is the angle between the radius vector and tangent at any point of the curve. I know that ##\tan \phi=rd\theta/dr## but how am I supposed to convert this to Cartesian form. I am currently studying polar curves by myself. I thought I can do simple stuff before studies begin so any help is appreciated. Thanks! :)
     
  2. jcsd
  3. Jun 16, 2013 #2

    ehild

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    Stay with Cartesian coordinates and express tan(theta) with them.

    ehild
     
  4. Jun 16, 2013 #3
    And how should I do that? :confused:

    I tried it again and I get the answer now. In Cartesian coordinates, ##x=r\cos \theta##(i) and ##y=r\sin \theta##(ii).
    Differentiating (i) w.r.t ##\theta##
    [tex]\frac{dx}{d\theta}=\cos \theta \frac{dr}{d\theta}-r\sin\theta[/tex]
    [tex]\frac{dx}{d\theta}=\frac{x}{r}\frac{dr}{d\theta}-y[/tex]
    Let ##rd\theta/dr=k##
    [tex]\Rightarrow \frac{dx}{d\theta}=\frac{x}{k}-y[/tex]
    Similarly,
    [tex]\frac{dy}{d\theta}=\frac{y}{k}+x[/tex]
    Dividing the above two equations and solving for ##k##,
    [tex]k=\frac{xy'-y}{x+yy'}[/tex]
    And since ##k=\tan \phi##, hence proved.

    Looks good?
     
  5. Jun 16, 2013 #4

    ehild

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    Why to mix polar coordinates in? Φ is the angle between the radius vector r=xi+yj and the tangent vector 1i+y'j. Express cosΦ with their dot product and sinΦ with the cross product.

    ehild
     
  6. Jun 16, 2013 #5
    Thanks ehild! Your method gives the answer in a few steps. It is really a nice way to solve the problem contrary to mine which took more time. :smile:
     
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