Rotating Vectors

1. Oct 31, 2006

sherlockjones

Lets say we have a vector $$\vec{A}(t)$$ with a constant magnitude $$A$$. Thus $$\vec{A}(t)$$ can only change in direction (rotation). We know that $$\frac{d\vec{A}}{dt}$$ is always perpendicular to $$\vec{A}$$. This is where I become stuck:

$$\Delta \vec{A} = \vec{A}(t+\Delta t)-\vec{A}(t)$$
$$|\Delta \vec{A}| = 2A\sin\frac{\Delta \theta}{2}$$.

How do we get the trigonometric expression on the right in the second equation? It looks like some type of half/double angle formula. Eventually we are supposed to get $$\vec{A}\frac{d\theta}{dt}$$ or the angular velocity of $$\vec{A}$$

Thanks

2. Oct 31, 2006

HallsofIvy

Staff Emeritus
Interesting! I was sure that couldn't possibly be correct when I started calculating it!

First, what you have is for
$$\Delta \vec{A} = \vec{A}(\theta+\Delta \theta)-\vec{A}(\theta)$$
that is, with $\theta$ not t. And, of course, there is a typo in the second equation: you mean $|\vec{A}|$, not just $A$.

Any plane vector of constant length r can be written $\vec{A}= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}$. Then
$\vec{A(\theta+ \Delta\theta)}- \vec{A(\theta)}= r((cos(\theta+ \Delta\theta)- cos(\theta))\vec{i}+ (sin(\theta+ \Delta\theta)- sin(\theta))\vec{j})$
You can use $cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)$ and $sin(a+b)= sin(a)cos(b)+ cos(a)sin(b)$ to expand those. I won't do all of the calculations here (writing all of that in LaTex would be more tedious than the calculations!) but squaring each those and summing reduces to $2- 2cos(\Delta\theta)$. I was surprised when I saw that the terms involving only $\theta$ rather than $/Delta/theta$ cancel out! Of course, the square root of that does give the half angle formula.

3. Oct 31, 2006

sherlockjones

it was actually the change in $$\vec{A}$$ in the time interval $$t$$ to $$t + \Delta t$$. So wouldnt it be: $$\Delta \vec{A} = \vec{A}(t+\Delta t)-\vec{A}(t)$$?