Simple differentiation mistake?

1. Dec 16, 2015

influx

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. Dec 16, 2015

Samy_A

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
3. Dec 16, 2015

SteamKing

Staff Emeritus
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

4. Dec 16, 2015

Samy_A

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.