# Thread: Derivative of an unit vector

1. ## Derivative of an unit vector

Hello

If I have an unit vector the derivative is an orthogonal vector (orthogonality is enough easy to demonstrate)
How I can demonstrate that the derivative of unit vector is not an unit vectors ?

Thanks !

2. Originally Posted by Gustavis
Hello

If I have an unit vector the derivative is an orthogonal vector (orthogonality is enough easy to demonstrate)
How I can demonstrate that the derivative of unit vector is not an unit vectors ?

Thanks !
It can be a unit vector... Look for instance at $\vec{u}(\theta)={\cos\theta\choose\sin\theta}$. But it does not have to be a unit vector, look at $\vec{u}(\theta)={1\choose 0}$.

3. perfect !

Can I demonstrate in this way?

the versor
$\hat{v}(t) = \cos(\theta(t))\hat{i} + \sin(\theta(t))\hat{j}$

and its derivative
$\hat{v}'(t) = \theta'(t)(-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})$

where :
$(-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})$
is the orthogonal versor

and theta(t)
$\theta'(t)$
in general different to 1 ?

4. Originally Posted by Gustavis
perfect !

Can I demonstrate in this way?

the versor
$\hat{v}(t) = \cos(\theta(t))\hat{i} + \sin(\theta(t))\hat{j}$

and its derivative
$\hat{v}'(t) = \theta'(t)(-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})$

where :
$(-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})$
is a unit orthogonal versor

and
$\theta'(t)$
in general different to 1 ?
That's quite fine.

5. In general $\theta(t)'$ is different from 1 but not always. Your original statement "the derivative of unit vector is not a unit vector" is not true.