Suppose we want to write the vector:
$$\vec{A}=4\hat{x} +3\hat{y}$$
in polar coordinates $(\rho, \phi)$, by making use of the identities:
$$
\hat{x}=\cos\phi\,\hat{\rho} -\sin{\phi}\,\hat{\phi}, \quad \hat{y}=\sin\phi\,\hat{\rho} +\cos{\phi}\,\hat{\phi}
$$
which simply yield:
$$
\vec{A}= \left(4\cos{\phi}+3\sin{\phi}\right)\hat{\rho} + \left(-4\sin{\phi}+3\cos{\phi}\right)\hat{\phi}
$$
Now expressing $\vec{A}$ as above does not make much sense to me because $\vec{A}$'s orientation is fixed so there should appear no free parameter such as the angle $\phi$. If $\phi$ indeed represents the orientation of the given vector then:
$$
\cos\phi=\frac{A_x}{||A||}=\frac{4}{5}, \quad \sin\phi=\frac{A_y}{||A||}=\frac{3}{5}
$$
If I plug these in $\vec{A}$, I end up with:
$$A=5\hat{\rho}$$
By using the same reasoning above, a generic vector's, $\vec{A}=A_x \hat{x} + A_y\hat{y}$, representation in polar coordinates appears to be
$$\vec{A}=||A|| \hat{\rho}, \quad ||A||=\sqrt{A_x^2+A_y^2}$$
But on the other hand, a generic vector in polar coordinates is given in the form:
$$
\vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}
$$
If my steps of converting $\vec{A}$ from cartesian to polar coordinates is correct, then why does the $\hat{\phi}$ component seems to disappear ?
EDIT
Thank you all for your answers. I need to clear up few more things. In one of the answers below it is stated the representation (although I have seen such notation on the web):
$$
\vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}
$$
with constant coefficients is wrong. Is this also true for the representation of
say:
$$
\vec{A}=4\hat{x} +3\hat{y} + 2\hat{z}
$$
in cylindrical and spherical coordinates ? If so, then the proper conversion of $\vec{A}$ must read
$$
\vec{A}=5\hat{\rho} + 2\hat{z}, \quad \rho=\sqrt{x^2 + y^2}
$$
in cylindrical coordinates and
$$
\vec{A}=\sqrt{29}\hat{r}, \quad r=\sqrt{x^2+y^2+z^2}
$$
in spherical coordinates. Therefore the representation of a generic vector with arbitrary but constant coefficients:
$$
\vec{A}= A_{\rho} \hat{\rho} + A_{\phi}\hat{\phi} + A_{z}\hat{z}\\
\vec{A}= A_{r} \hat{r} + A_{\phi}\hat{\phi} + A_{\theta}\hat{\theta}
$$
is wrong. Is this conclusion correct ?
Best Answer
The formula $$\underline{v}=v_r \hat{\underline{r}}+v_\theta \hat{\underline{\theta}}$$ Is wrong. The radius unit vector is defined such that the position vector $\underline{\mathrm{r}}$ can be written as $$\underline{\mathrm{r}}=r~\hat{\underline{r}}$$ That's what makes polar coordinates so useful. Sometimes we only care about things that point in the direction of the position vector, making the theta component ignorable.