Why is the derivative of Vector equal to Derivative of its rectilinear components? Take a vector $\mathbf A=t^4\mathbf i +t^2\mathbf j$, and call the unit vector along direction of $\mathbf A$ is $\mathbf k$, so the magnitude of this vector $\mathbf A$ along $\mathbf k$ will be $\sqrt{t^8+t^4}$ and thus the vector will be $\sqrt{t^8+t^4}\ \mathbf k$.
So why is its derivative $d\mathbf A/dt=4t^3\mathbf i+2t\mathbf j$? will its magnitude be the same as that of the magnitude of derivative of the vector along $\mathbf k$? How can we distribute its derivative along $\mathbf i$ and $\mathbf j$ and how to prove that their magnitude is same as the magnitude of the derivative of the vector along $\mathbf k$?
 A: If $\vec A = f(t)\vec i + g(t) \vec j$ for some functions $f,g$ then
$\displaystyle \frac {d \vec A} {dt} = \left( \frac {df(t)}{dt} \right)\vec i + f(t) \left(\frac {d \vec i} {dt}\right) + \left(\frac {dg(t)}{dt}\right) \vec j + g(t)\left( \frac {d \vec j} {dt}\right)$
But $\vec i$ and $\vec j$ are constant vectors (they have constant magnitude and constant direction), so $\frac {d \vec i} {dt} = \frac {d \vec j} {dt} = 0$ and we have
$\displaystyle \frac {d \vec A} {dt} = \left(\frac {df(t)}{dt}\right) \vec i + + \left(\frac {dg(t)}{dt}\right) \vec j$
If we express $\vec A$ as $\vec A = |A|\vec k$ where $k$ is a unit vector in the direction of $\vec A$ then we have
$\displaystyle \frac {d\vec A}{dt} = \left(\frac{d|A|} {dt} \right) \vec k + |A|\left( \frac {d\vec k}{dt}\right)$
but in general the direction of $\vec k$ changes with time, so $\frac {d \vec k}{dt} \ne 0$ and
$\displaystyle \frac {d\vec A}{dt} \ne \left(\frac{d|A|} {dt} \right) \vec k$
A: $\frac{dA}{dt}=\lim_{\delta t\rightarrow 0} \frac{A(t+\delta t)-A(t)}{\delta t}=\lim_{\delta t\rightarrow 0}\frac{(t+\delta t)^4 i + (t+\delta t)^2 j-t^4 i -t^2j}{\delta t}=\lim_{\delta t\rightarrow 0}\frac{[(t+\delta t)^4 i -t^4 i] + [(t+\delta t)^2 j-t^2j]}{\delta t}=\lim_{\delta t\rightarrow 0}\frac{(t+\delta t)^4 i -t^4 i}{\delta t} + \lim_{\delta t\rightarrow 0}\frac{(t+\delta t)^2 j-t^2j}{\delta t}=4t^3i + 2t j$
A: $$\vec v=\begin{bmatrix}
    a(t) \\
    b(t) \\
  \end{bmatrix}$$
$$\hat{\vec{v}}=\frac{1}{\sqrt{a(t)^2+b(t)^2}}\begin{bmatrix}
    a(t) \\
    b(t) \\
  \end{bmatrix}=\vec{k}$$
where $~|\hat{\vec{v}}|=1~$
thus
$$\vec{v}=|v|\,\vec{k}$$
where $~|v|=\sqrt{a(t)^2+b(t)^2}$
the time derivative of $~\vec{v}~$ is
$$\vec{\dot{v}}=\begin{bmatrix}
    \dot a(t) \\
    \dot b(t) \\
  \end{bmatrix}$$
the time derivative of $~|v|\,\vec{k}~$ is
$$\frac{d}{dt}\left(|v|\,\vec{k}\right)=\frac{d}{dt}\,(|v|)\,\vec k+|v|\,\vec{\dot{k}}=
\dot{v}$$
A: After @gandalf61's answer above, the following is just tautology, but let's do it just for the sake of the question.
$$\begin{align}
\mathbf A &=t^4\mathbf i +t^2\mathbf j \\
\end{align}$$
Or, in the direction of $\mathbf A$,
$$\begin{align}
\mathbf A &= |A|.\mathbf k \\
\end{align}$$
Where $\mathbf k$ is the unit vector in the direction of $\mathbf A.$
Thus:
$$\begin{align}
\mathbf A &= \sqrt{t^8+t^4}.\frac{t^4\mathbf i +t^2\mathbf j}{\sqrt{t^8+t^4}} \\
\end{align}$$
Now, we can derive $\frac{d\mathbf A}{dt}$:
$$\begin{align}
\frac{d\mathbf A}{dt} &= \frac{d|A|}{dt}.\mathbf k + |A|.\frac{d\mathbf k}{dt} \\
&=\frac{8t^7 + 4t^3}{2\sqrt{t^8+t^4}}.\frac{t^4\mathbf i +t^2\mathbf j}{\sqrt{t^8+t^4}} + \sqrt{t^8+t^4}.\frac{2t^7 \mathbf i - 2t^9\mathbf j}{(t^8+t^4)^{3/2}} \\
&= \frac{(4t^{11} + 2t^7 + 2t^7) \mathbf i + (4t^9 + 2t^5 -2t^9)\mathbf j}{t^8+t^4} \\
&= \frac{t^4[(4t^7 + 4t^3)\mathbf i + (2t^5 + 2t) \mathbf j]}{t^4(t^4+1)} \\
&= \frac{(t^4+1)(4t^3 \mathbf i + 2t\mathbf j)}{t^4+1} \\
&= 4t^3 \mathbf i + 2t\mathbf j
\end{align}$$
So, we see that the derivative of $\mathbf A$ is indeed the same whether we represent $\mathbf A$ in terms of $\mathbf i$ and $\mathbf j$, or in terms of $\mathbf k$ (and subsequently $\mathbf k$ in terms of $\mathbf i$ and $\mathbf j$).
