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Third derivative and polar coordinates

  1. Mar 8, 2010 #1
    I'm studying for a maths test.
    I know that the second derivative of the position R(t) of a particle moving in the plane, in polar coordinates, is (r''-r([tex]\vartheta[/tex]')2)er + (r[tex]\vartheta[/tex]''+2r'[tex]\vartheta[/tex]')eo. o = [tex]\vartheta[/tex]

    How to differentiate this to find R'''(t), in polar coordinates and in turn find R'''(1) in polar coordinates if R(t) has polar coordinates r(t) = t2, [tex]\vartheta[/tex](t) = t2
     
  2. jcsd
  3. Mar 9, 2010 #2

    tiny-tim

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    Hi squenshl! :smile:

    (have a theta: θ :wink:)

    Hint: what is (er)' ? what is (eθ)' ? :smile:
     
  4. Mar 10, 2010 #3
    er' = [tex]\vartheta[/tex]'e([tex]\vartheta[/tex])
    e([tex]\vartheta[/tex])' = -[tex]\vartheta[/tex]'(er)
     
  5. Mar 10, 2010 #4

    tiny-tim

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    (what happened to that θ i gave you? :confused:)

    ok, so (rer)' = r'er + rθ'eθ,

    and similarly, (r'er + rθ'eθ)' = … ?

    (r'er + rθ'eθ)'' = … ? :smile:
     
  6. Mar 13, 2010 #5
    I keep getting different answers everytime.
    I must be doing something wrong.
    Can I just do the fact that er' = [tex]\vartheta'e_\vartheta[/tex] & e[tex]\vartheta[/tex] = -[tex]\vartheta'e_r[/tex]
    and then just use the product rule on R''(t) to get R'''(t).
     
    Last edited: Mar 13, 2010
  7. Mar 14, 2010 #6

    tiny-tim

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    Yes. :smile:

    (But if you keep getting something wrong, it's a good idea to find out why, so show us what you got anyway :wink:)
     
  8. Mar 14, 2010 #7
    I got R'''(t) = (r'' - r([tex]\vartheta'[/tex])2)'er + er'(r''-r([tex]\vartheta'[/tex])2) + (r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta'[/tex])'etheta + etheta'(r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta[/tex]') = (r'' - r([tex]\vartheta'[/tex])2)'er + [tex]\vartheta'[/tex]etheta(r''-r([tex]\vartheta'[/tex])2) + (r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta'[/tex])'etheta - [tex]\vartheta'[/tex]er(r[tex]\vartheta[/tex]'' + 2r'[tex]\vartheta[/tex]')
     
  9. Mar 15, 2010 #8

    tiny-tim

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    Looks ok so far.

    But it'll save a lot of repetition if you use the formula (ab)''' = a'''b + 3a''b' + 3a'b'' + ab''', with a = r and b = er :wink:
     
  10. Mar 15, 2010 #9
    That's handy, never seen it before.
    R'''(t) = r'''er + 3r''er' 3r'er'' + rer'''.
    Does this help to R'''(1) in polar coordinates if R(t) has polar coordinates r(t) = t2 & [tex]\vartheta(t)[/tex] = t2.
    How do I go about finding R'''(1) then.
     
  11. Mar 16, 2010 #10

    tiny-tim

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    You still use er' = θeθ' and eθ' = -θer' , with the particlar formulas for (t2)' and (t2)''
     
  12. Mar 17, 2010 #11
    Since r = t2 [tex]\vartheta[/tex] = t2
    r' = 2t [tex]\vartheta[/tex]' = 2t
    r'' = 2 [tex]\vartheta[/tex]'' = 2
    r''' = 0 [tex]\vartheta[/tex]''' = 0

    R'''(t) = 0er +3(2)er + 3(2t)er' + t2er
    = 6er + 6ter' + t2er

    Then R'''(1) = 6er + 6(1)er + (1)2er
    = 6er + 6er' + er
     
    Last edited: Mar 17, 2010
  13. Mar 18, 2010 #12

    tiny-tim

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    No, 6er' + 6ter'' + t2er''' :smile:
     
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