understanding derivative of a vector function

Question

If I have a vector function $r(t)$ , why is the tangent vector at t
$\lim_{\Delta t \rightarrow 0} (\frac{r(t+\Delta t)-r(t)}{\Delta t})$ and not just simply $\lim_{\Delta t \rightarrow 0} ({r(t+\Delta t)-r(t)})$..I understand that if we divide it by $\Delta t$ and further simplify it , we will get the derivative of the vector function as the derivative of its individual components but i dont understand why we do so..

Answer 1

The tangent at $t$ is the limit case of the secant: $$ \lim_{t' \to t} \frac{r(t') - r(t)}{t' - t} $$ or $t' = t + \Delta t$ thus $$ \lim_{\Delta t \to 0} \frac{r(t+\Delta t) - r(t)}{t+\Delta t- t} $$

Answer 2

You can think of it as calculating the velocity of the moving particle $r$ because the direction of its velocity is the same as the direction of the tangent line of its trajectory.

To calculate its velocity, we divide a certain distance that it passes through by the amount of time it takes to pass through that distance.