derivative

**alexandros** · January 30th 2010, 09:48 AM

We know that the position vector in physics is defined as:

r(t) = x(t)i +y(t)j +z(t)k ,where i =(1,0,0) ,j=(0,1,0) and k=(0,0,1).

Prove that :

$\frac{dr(t)}{dt} = \frac{dx(t)}{dt}$ i + $\frac{dy(t)}{dt}$ j + $\frac{dz(t)}{dt}$ k

**Defunkt** · January 30th 2010, 10:02 AM

What have you tried? Where are you getting stuck?

**alexandros** · January 30th 2010, 12:10 PM

Originally Posted by Defunkt

What have you tried? Where are you getting stuck?

From the very beggining i don't know what the limit :

$lim_{\Delta t\to 0}\frac{r(t+\Delta t)-r(t)}{\Delta t}$ is equal to

**CaptainBlack** · May 31st 2010, 07:55 PM

Originally Posted by alexandros

From the very beggining i don't know what the limit :

$lim_{\Delta t\to 0}\frac{r(t+\Delta t)-r(t)}{\Delta t}$ is equal to

Just replace $r(t+\Delta t)$ and $r(t)$ by their expressions in terms of the basis vectors and collect components:

$\lim_{\Delta t\to 0}\frac{r(t+\Delta t)-r(t)}{\Delta t}$ $= \lim_{\Delta t\to 0}\left(\frac{x(t+\Delta t)-x(t)}{\Delta t}\textbf{i} \right.$ $+ \frac{y(t+\Delta t)-y(t)}{\Delta t}\textbf{j}$ $+ \left. \frac{z(t+\Delta t)-z(t)}{\Delta t}\textbf{k} \right)$

CB

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