Clarification of the change in vector A I understand that a change of the vector A can be the result of a rotation and a change in magnitude. (The perpendicular part and the parallel part)
This can be represented by ΔA = ΔA(perpendicular) + ΔA(parallel)
What I am having difficulty understanding is how |ΔA(perpendicular)| = AΔθ 
and how  |ΔA(parallel)| = ΔA (for small angles of θ).
This is the image used in the textbook I am using. Would appreciate if the answer could be related to this image. Thank you.
 A: Since for small angle $\Delta \theta$ we have


*

*$\sin \Delta \theta\approx  \Delta \theta$

*$\cos \Delta \theta \approx 1$
we obtain
$$\|\Delta\vec A_\perp\|=\|\vec A(t+\Delta t)\|\sin \Delta \theta \approx \|\vec A(t+\Delta t)\| \Delta \theta$$
$$\|\Delta\vec A_\parallel\|=\|\vec A(t+\Delta t)\|\cos \Delta \theta -\|\vec A(t)\|\approx \|\vec A(t+\Delta t)\| -\|\vec A(t)\|$$
A: For simplicity reasons, let $\underline{B}=\Delta\underline{A}$ and $B=|\underline{B}|$ (And so on for the other vectors). Then we have that
$$\underline{B}=\underline{B}_{\bot}+\underline{B}_{\parallel}$$
You can see that $\underline{B}$,$\underline{B}_{\bot}$ and $\underline{B}_{\parallel}$ can be the sides of a right triangle (if you move them a bit). So we have that
$$\sin(\theta)=\frac{B_{\bot}}{B}$$
$$\cos(\theta)=\frac{B_{\parallel}}{B}$$
And if $\theta$ is really small (maybe around some degrees), we have that $\sin(\theta) \approx \theta$ and $\cos(\theta) \approx 1$, so
$$\theta \approx \frac{B_{\bot}}{B}$$
$$1 \approx \frac{B_{\parallel}}{B}$$
And finally
$$B\theta \approx B_{\bot}$$
$$B \approx B_{\parallel}$$
Note that $\theta$ is measured in radians, not degrees.
