$\frac{d\theta}{dt} = \frac{W(x_{1},x_{2})}{|r|^2}$, where $r:= x_{1} i + x_{2} j$ and $\theta$ is the angle between $r$ and $i$,

Question

The title says it all. My biggest problem is: I can't see any part of the Wronskian within the angle; actually I'm having problems to find a angle function. Let $r:= x_{1} i + x_{2} j$ be a curve and $\theta (t)$ the angle between $r$ and the constant vector $i$. The question then asks to show that

$$\frac{d\theta}{dt} = \frac{W(x_{1},x_{2})}{|r|^2},$$ where $r:= x_{1} i + x_{2} j$.

Any help would be appreciated! Thanks in advance!!

Answer 1

The Wronskian is defined as

$ W(x_1, x_2) = \begin{vmatrix} x_1 && x_2 \\ x_1' && x_2' \end{vmatrix} = x_1 x_2' - x_2 x_1' $

If $\theta(t) = \tan^{-1} \left(\dfrac{x_2(t)}{x_1(t)} \right) $

Then

$\dfrac{d \theta}{dt } = \dfrac{ x_2' x_1 - x_1' x_2 }{ x_1^2 \left( 1 + \left(\dfrac{x_2}{x_1}\right)^2 \right) } = \dfrac{ W(x_1, x_2) } { x_1^2 + x_2^2 } = \dfrac{ W(x_1, x_2) }{ | r |^2 } $