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