calculusrotationsvectors
Let $\vec(i)$ be a unit vector fixed in a rotating frame of reference. It has a constant magnitude but changes its direction at an angular velocity of $\Omega$.
How can I then derive the following?
I know that the magnitude of $|d\vec(i)|=d\theta=\Omega dt$ via the arc length formula. From here,
$$\frac{d\vec(i)}{dt}=\Omega$$
What am I failing to understand?
Thanks.
If $\Omega$ is not constant, the equation still holds. But except in special cases, you will have to solve it by numerical integration.
If you have a certain rotation motion, and you found that $f(t)$ fits the equation $dr/dt=f(t)\times r$, it is not necessarily your angular velocity, but it will be true that $\Omega(t)=f(t)+\lambda(t) r(t)$, with $\lambda$ real. If you prescribe a rotation by giving $f(t)$, it surely defines a rotation motion, and can be recovered by numerical integration.
Maybe one way that will help the understanding is to consider the integral $$\int_0^{\theta_0} {\hat \theta} d{\theta}$$
I am assuming $\hat \theta$ defined in this way: $${\hat \theta} = \begin{bmatrix}-\sin(\theta)\\\cos(\theta)\end{bmatrix}$$
It will be unit vector because of the trigonometric identity: $\sin(\theta)^2+\cos(\theta)^2 = 1$. We can also convince ourselves that it will be orthogonal to $\begin{bmatrix}r\cos(\theta)\\r\sin(\theta)\end{bmatrix}$ which is often defined to be the radial component.
If we approximate the integral using a computer, we see a relation between the angle integrated, the unit circle and the line integral result:
The end points of the blue line are angles $0$ and $\theta_0$ on the unit circle. And the red vector is the vector from point $(1,0)$ pointing onto the unit circle at angle $\theta_0$. (I could not find any arrowhead to indicate the direction).
Best Answer
Be careful with what you write because
$$\frac{d\vec{i}}{dt}\neq\Omega \; .$$ You can show however that $$\bigg|\frac{d\vec{i}}{dt}\bigg|=\Omega$$ as you suggest using the arclength formula. Now all we need to do is find the direction of $\frac{d\vec i}{dt} \; .$ You could argue geometrically that in the limit of the triangle of sides $\vec i, d\vec i, \text{ and } \vec i + d\vec i$ that $d\vec i$ tends to $\vec j .$ We would then get the correct result you have listed that $\frac{d\vec i}{dt}=\Omega \vec j \; .$
More rigorously we can say that the limit definition of the derivative is that
$$\lim_{dt\rightarrow 0} \frac{\vec{i}(t + \Omega dt) - \vec{i}(t)}{dt}$$
$$\text{thus}$$
$$\lim_{dt\rightarrow 0} \frac{\begin{bmatrix} 1 \\ \Omega dt \end{bmatrix} - \begin{bmatrix} 1 \\ 0 \end{bmatrix}}{dt} = \lim_{dt\rightarrow 0} \begin{bmatrix} 0 \\ \Omega \frac{dt}{dt} \end{bmatrix} = \lim_{dt\rightarrow 0} \begin{bmatrix} 0 \\ \Omega \end{bmatrix} =\Omega \vec j(t) \; .$$
We can now say that $$\frac{d\vec i}{dt}=\Omega \vec j \; .$$
You can adjust these arguments then to say that $$\frac{d\vec j}{dt}=-\Omega \vec i \; .$$