The discussion revolves around a potential mistake in the differentiation process related to the expression for velocity, v, in a physics context. Participants are examining the implications of the derivative of the arctangent function and its effect on the equations involving variables such as m, r, theta, and Uinfinity. The scope includes mathematical reasoning and conceptual clarification regarding differentiation and trigonometric identities.
Participants express disagreement regarding the correct formulation of the velocity expression and the implications of the differentiation process. Multiple competing views remain, and the discussion does not reach a consensus on the correct interpretation of the mathematical steps involved.
There are limitations related to the assumptions made about the differentiation rules and the dependence on the definitions of the variables involved. The discussion highlights unresolved mathematical steps and the need for clarity in the application of trigonometric identities.
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.influx said:
##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).influx said:
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:influx said:
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?
Oops, the stated reason is wrong.Samy_A said: