Why is the derivative of Vector equal to Derivative of its rectilinear components?

Question

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 $\mathbf k$?

Answer 1

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

$\displaystyle \frac {d\vec A}{dt} \ne \left(\frac{d|A|} {dt} \right) \vec k$

Answer 2

$\frac{dA}{dt}=\lim_{\delta t\rightarrow 0} \frac{A(t+\delta t)-A(t)}{\delta t}=\lim_{\delta t\rightarrow 0}\frac{(t+\delta t)^4 i + (t+\delta t)^2 j-t^4 i -t^2j}{\delta t}=\lim_{\delta t\rightarrow 0}\frac{[(t+\delta t)^4 i -t^4 i] + [(t+\delta t)^2 j-t^2j]}{\delta t}=\lim_{\delta t\rightarrow 0}\frac{(t+\delta t)^4 i -t^4 i}{\delta t} + \lim_{\delta t\rightarrow 0}\frac{(t+\delta t)^2 j-t^2j}{\delta t}=4t^3i + 2t j$

Answer 3

$$\vec v=\begin{bmatrix} a(t) \\ b(t) \\ \end{bmatrix}$$

$$\hat{\vec{v}}=\frac{1}{\sqrt{a(t)^2+b(t)^2}}\begin{bmatrix} a(t) \\ b(t) \\ \end{bmatrix}=\vec{k}$$

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

$$\frac{d}{dt}\left(|v|\,\vec{k}\right)=\frac{d}{dt}\,(|v|)\,\vec k+|v|\,\vec{\dot{k}}= \dot{v}$$

Answer 4

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 $\mathbf j$).