# Unable to understand vector derivative

1. Nov 4, 2015

### GiuseppeR7

Hi guys...i'm having a bad time understanding the concept of vector derivative with respect to different frames.
Suppose i have the vector displayed in the picture, the frame A and B are in motion with respect of each other, i can understand the concept of the derivative of the vector R in the frame A...but what does it mean the derivative of the vector with respect to B since the frame's origins are not coincident?
Thanks!

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2. Nov 4, 2015

### BvU

Vectors that are time dependent are differentiated according to the definition.
If there is a time-dependent coordinate transformation between two frames of reference, that simply comes in:
So if frame A origin as described in frame B is at $\vec A_0(t)$ and moves with velocity $\vec v_A$,
and the derivative of $\vec R$ in frame A is $\vec v_{\rm frame \ a}$ then $${d\,\vec R_{\rm frame\ b} \over dt} = { d \left ( \vec R(t) + \vec A_0(t) \right ) \over dt } = \vec v_{\rm frame \ a} + \vec v_A$$

(I do hope this isn't in a relativity course context ?)

That the frame origins are not coincident does not appear in the derivative. $\vec A_0(0)$ is a constant and drops out. You can check that by writing out the definition in full.

3. Nov 4, 2015

4. Nov 4, 2015

### BvU

How did it manage to do so ?

5. Nov 4, 2015

### GiuseppeR7

it was using a confusing notation! :p

6. Nov 4, 2015

### lightarrow

You seem to have already solved your problem, however your question is ambiguous: what does "derivative of the vector R in the frame A" means? I know what is a "derivative of a vector with respect to time" or what is a "vector written using different systems of coordinates", but I don't know what it is what you have said.

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lightarrow