Third derivative and polar coordinates

  • Thread starter squenshl
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
tiny-tim
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Hi squenshl! :smile:

(have a theta: θ :wink:)

Hint: what is (er)' ? what is (eθ)' ? :smile:
 
  • #3
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er' = [tex]\vartheta[/tex]'e([tex]\vartheta[/tex])
e([tex]\vartheta[/tex])' = -[tex]\vartheta[/tex]'(er)
 
  • #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:
 
  • #5
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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:
  • #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:)
 
  • #7
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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]')
 
  • #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:
 
  • #9
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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.
 
  • #10
tiny-tim
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You still use er' = θeθ' and eθ' = -θer' , with the particlar formulas for (t2)' and (t2)''
 
  • #11
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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:
  • #12
tiny-tim
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R'''(t) = 0er +3(2)er + 3(2t)er' + t2er
= 6er + 6ter' + t2er
No, 6er' + 6ter'' + t2er''' :smile:
 

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