What does it mean to differentiate a vector?

Question

I didn’t quite understand what does it mean to differentiate a vector, i suspect that the derivative of a vector valued function is just answering the question : What vector should i added to the previous one to get the next one. For example :

Consider the vector $$\vec{X}(t)=\langle 1,t \rangle$$

If you take its derivative, we get $$\vec{X}’(t)=\langle 0,1 \rangle$$

Now if i take a random vector from the vector valued function $\langle 1,t \rangle$, say for example $\langle 1,3 \rangle$ If i add to it the derivative of $\langle 1,t \rangle$ which is $\langle 0,1 \rangle$, i get $\langle 1,4 \rangle$ which is the next vector.

But this is only true for $t\in \mathbb Z$ because in the real numbers the next vector is not defined, it could be $\langle 1,3.0001 \rangle$.

So how can i really understand what is the derivative of a vector.

Answer 1

One way to think of it is by components. A vector in $n$ dimensions has $n$ components, and each of those can be a function of one or more variables. So if you have a vector $\overrightarrow v = <x(t), y(t), z(t)>$, then its derivative would be $\overrightarrow v' = <x'(t), y'(t), z'(t)>$.

Vectors don't have to have integer components, and neither do their derivatives. So you don't have to worry about the "next" vector any more than you have to worry about the "next" number when dealing with $y = f(x)$.

Answer 2

\begin{align*} X'(t) &= \lim\limits_{h\to 0} \frac{X(t+h)-X(t)}{h}\\ \lim\limits_{h\to 0} X(t+h) &= X(t) + \lim\limits_{h\to 0} X'(t)\cdot h\\ \end{align*} or in other words, as $h\to 0$, $$X(t+h)= X(t) + X'(t)\cdot h$$ In your calculation of "next vector", I think you missed multiplying by $h$?