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Simple differentiation mistake?

  1. Dec 16, 2015 #1
    edcacf.png

    The working out states that v = (m/r)sin(theta) but surely this should be (-m/r)sin(theta) ? I mean there is a negative in front of the ∂Ψ/dx ?

    Also, if v=0 then 0 = (m/r)sin(theta), therefore sin(theta) = 0? And from this we know cos(theta) = 1 or -1 which would mean u = Uinfinity + (m/r) or Uinfinity - (m/r) as opposed to u = Uinfinity + (m/x) as they got?
     
  2. jcsd
  3. Dec 16, 2015 #2

    Samy_A

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    When you take the derivative of ##\arctan (y/x)## with respect to ##x##, you also get a ##-1## factor as ##x## is in the denominator. That cancels the first minus sign.

    ##x=r\cos \theta##, so ##u=U_\infty +\frac{m}{r}\cos \theta## becomes ##u=U_\infty+\frac{m}{x}## (the ##\cos \theta## cancels out, so it doesn't matter if it is +1 or -1).
     
    Last edited: Dec 16, 2015
  4. Dec 16, 2015 #3

    SteamKing

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    I think it has to do with the way the arctan is used here. There are two arguments, x and y, instead of one. There is a special rule for the derivative of tan-1(y/x), as shown in this article:

    https://en.wikipedia.org/wiki/Differentiation_rules#Derivatives_of_trigonometric_functions
     
  5. Dec 16, 2015 #4

    Samy_A

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    Oops, the stated reason is wrong.

    Actually, ##u=U_\infty +\frac{m}{r}\cos \theta=U_\infty +\frac{m}{x}\cos² \theta##, and as ##\cos \theta =\pm 1##, we get ##u=U_\infty+\frac{m}{x}##.

    Sorry for the mistake.
     
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