Clarification of the change in vector A

Question

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.

Answer 1

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.

Answer 2

Since for small angle $\Delta \theta$ we have

• $\sin \Delta \theta\approx \Delta \theta$

• $\cos \Delta \theta \approx 1$

we obtain

$$\|\Delta\vec A_\perp\|=\|\vec A(t+\Delta t)\|\sin \Delta \theta \approx \|\vec A(t+\Delta t)\| \Delta \theta$$

$$\|\Delta\vec A_\parallel\|=\|\vec A(t+\Delta t)\|\cos \Delta \theta -\|\vec A(t)\|\approx \|\vec A(t+\Delta t)\| -\|\vec A(t)\|$$