# I Proof derivative of a vector following precession motion

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1. Apr 10, 2016

### Soren4

I do not get some points of this proof about the time derivative of a unit vector $\hat{u}$ (costant magnitude) which is following a precession motion. The picture is the following.

I want to prove that $$\frac{d\hat{u}}{dt}=\vec{\Omega}\wedge \hat{u}.$$

I'm ok with almost all the proof except the following points.

Consider the red vector $d\hat{u}$.
Since $\hat{u}$ has constant magnitude,$d\hat{u}$ must be orthogonal to $\hat{u}$.That's ok.

But why must $d\hat{u}$ be orthogonal to $\vec{\Omega}$ too (i.e. be tangential to a circle orthogonal to $\vec{\Omega})$?

Secondly how can I geometrically prove that the angle $d\theta$ (the one in purple) that separates $\hat{u}(t)$ and $\hat{u}(t+dt)$ is the same that separates the projection of these two vectors on the circle orthogonal to $\vec{\Omega}$ (the projections are $\hat{u}(t) Sin(\alpha)$and$\hat{u}(t+dt) Sin(\alpha)$)?

2. Apr 10, 2016

### andrewkirk

Define $\vec v$ as the vector from the origin (the lowest point drawn on the diagram) to the centre of the drawn circle, and $\vec w(t),\vec w(t+dt)$ as the two vectors you have drawn from the centre of the circle to its circumference. Then we have $\vec u(t)=\vec v+\vec w(t)$, and $\vec u(t+dt)=\vec v+\vec w(t+dt)$.

A bit of geometry shows that the lengths of $\vec u(t)$ and $\vec u(t+dt)$ are the same. Let that length be $l$. Then $d\hat u=\hat u(t+dt)-\hat u(t)=\tfrac{1}{l}(\vec u(t+dt)-\vec u)=\tfrac{1}{l}(\vec w(t+dt)-\vec w)$. Both $\vec w(t+dt)$ and $\vec w$ are perpendicular to $\vec\Omega$, so their difference must be too, by the linearity of the dot product.