Can we calculate magnitude of instantaneous velocity by directly differentiating the magnitude of position vector of a particle? Let there be a particle in a plane at any general time instant $t$. Let the coordinates of the particle are $x\mathbf{\hat i} + y\mathbf{\hat j}$, where $x$ and $y$ are functions of time $t$. Then we calculate velocity as $v_x=dx/dt$ and $v_y=dy/dt$. Then we write velocity as
$$\mathbf v= (dx/dt)\mathbf{\hat i} + (dy/dt)\mathbf{\hat j}$$
And its magnitude is given as $\sqrt{(dx/dt)^2 +(dy/dt)^2}$.
But can we do like this to directly find magnitude of velocity:
Magnitude of $r$ vector=$\sqrt{(x^2 + y^2)}$ then magnitude of velocity$=d/dt$ of $\sqrt{(x^2+y^2)}$. I checked it with examples and found the answer to be no. But I am not able to figure out why.
 A: The position vector of any particle in polar coordinates is
$$\mathbf r =r \mathbf {\hat r}$$
where $r=\sqrt{x^2+y^2}$ and $\mathbf{\hat r}$ is the unit vector in the radial direction. Thus differetiating the position vector with respect to time gives us
$$\frac{\mathrm d \mathbf r}{\mathrm d t}=\left(\frac{\mathrm d r}{\mathrm d t}\right)\mathbf {\hat r}+r\left(\frac{\mathrm d\mathbf{\hat r}}{\mathrm dt}\right)$$
The above equation is just the application of product rule of differentiation.
Simplifying the above equation, we get
$$\frac{\mathrm d \mathbf r}{\mathrm d t}=\left(\frac{\mathrm d r}{\mathrm d t}\right)\mathbf {\hat r}+r\omega\, \boldsymbol{\hat {\theta}}$$
where $\omega =\displaystyle \frac{\mathrm d \theta}{\mathrm dt}$ and $\boldsymbol{\hat {\theta}}$ is the unit vector in the tangential direction. If we use the dot notation, then we get
$$\dot{\mathbf r}=\dot r \mathbf{\hat r} + r\dot{\theta}\, \boldsymbol{\hat {\theta}}$$
Thus the magnitude of velocity is
$$|\dot{\mathbf r}|=\sqrt{\left(\dot r\right)^2 + \left(r\dot{\theta}\right)^2}$$
Your expression only includer the term $\dot r \mathbf{\hat r}$ which pertains to the radial velocity. This expression fails to account for the transverse motion which is represented by the term $r\dot{\theta}\, \boldsymbol{\hat {\theta}}$. Thus you are bound to get a wrong answer if you only consider the radial velocity.
Remember that $|\mathbf v|=\displaystyle \left|\frac{\mathrm d\mathbf r}{\mathrm d t}\right|$, but $|\mathbf v|\neq\displaystyle \frac{\mathrm d|\mathbf r|}{\mathrm d t}$.
A: In a planar motion, let us define the position of the particle by a vector $\vec{r}$ or $\mathbf{r}$ as made up of two components i.e. the ordered pair {x,y} in the unit basis vector $\{\hat{\mathbf{i}}, \,\,\hat{\mathbf{j}}\}$ as
$$\mathbf{r}= x \,\,\hat{\mathbf{i}}+ y \,\,\hat{\mathbf{j}}$$
Now the norm or length (i.e. the size or magnitude) of the above vector is
$$ ||r||=\sqrt{x^2+y^2}$$
Suppose, we have the coordinates of the body given by
\begin{eqnarray}
x(t)= 2 t^2 \\
y(t)= 3 t
\end{eqnarray}
then the velocity coordinates would be given by $\frac{d \mathbf{r}(t)}{dt}$ as
\begin{eqnarray}
v_x= 4 t \\
v_y= 3
\end{eqnarray}
Let us take the norm of the $\mathbf{r}(t)$ i.e.
$$  ||r||= \sqrt{4 t^4+ 9 t^2}$$
If we take the derivative with respect to time we get $\frac{8 t^3+9 t}{\sqrt{4 t^4+ 9 t^2}}=\frac{8 t^3+9 t}{||r||}$; and if we compare it to the  $||v||$ i.e. $\sqrt{16 t^2+9}$.
The difference is because they are actually two different vectors having different components, to begin with! Their dependence on the parameter t is not the same. Thus, we find that
$$\left|\left|\frac{d \mathbf{r}}{dt}\right|\right| \neq \frac{d ||\mathbf{r}||}{dt}$$
Since, the L.H.S defines the derivative of a vector quantity, while, the R.H.S is a derivative of a scalar quantity.
The first quantity is the magnitude of $\underline{\textit{tangent vector}}$ ; while the second one is sort of a $\textit{`gradient'}$ (in the sense of a slope) of a scalar.
NB: The vector $\mathbf{r}$ is written as $||r||\, \hat{\mathbf{r}}$. However,the Cartesian coordinates have a special property that the basis is independent of space and time and hence their derivatives vanish when using the product rule.
