1. ## derivative

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 :

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

2. What have you tried? Where are you getting stuck?

3. Originally Posted by Defunkt
What have you tried? Where are you getting stuck?
From the very beggining i don't know what the limit :

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

4. Originally Posted by alexandros
From the very beggining i don't know what the limit :

$\displaystyle lim_{\Delta t\to 0}\frac{r(t+\Delta t)-r(t)}{\Delta t}$ is equal to
Just replace $\displaystyle r(t+\Delta t)$ and $\displaystyle r(t)$ by their expressions in terms of the basis vectors and collect components:

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

CB