Why is $\vec u_x \cdot \vec u_\phi =-\sin\phi $? I am learning the coordinate transform between cylindrical and cartesian,and the picture is the teaching,and i don't understand why is  $\vec u_x \cdot \vec u_\phi =-\sin\phi $ ?
I guess the reason of  $\vec u_x \cdot \vec u_r =\cos\phi $ is $x=r\cos\phi$,but it can't explain why  $\vec u_x \cdot \vec u_\phi =-\sin\phi $,can anyone teach me?

 A: $
\newcommand\PD[2]{\frac{\partial#1}{\partial#2}}
$
$\vec u_x$ is the vector telling you how a point $P(x, y, z)$ moves when you change $x$ a tiny bit; similarly for $\vec u_y$ and $\vec u_z$. Since the point at coordinates $(x, y, z)$ is
$$
  P(x, y, z) = x\vec u_x + y\vec u_y + z\vec u_z
$$
and $\vec u_x, \vec u_y, \vec u_z$ are constant, we see
$$
  \PD Px = \vec u_x,\quad \PD Py = \vec u_y,\quad \PD Pz = \vec u_z.
$$
The partial derivatives of $P$ capture the idea of moving a point a tiny bit and giving the resulting displacement vector. So now write $P$ in cylindrical coordinates:
$$
  P(r, \phi, z) = (r\cos\phi)\vec u_x + (r\sin\phi)\vec u_y + z\vec u_z.
$$
Then
$$
  \vec u_\phi = \PD p\phi = -(r\sin\phi)\vec u_x + (r\cos\phi)\vec u_y,
$$
so we see
$$
  \vec u_x\cdot\vec u_\phi = -(r\sin\phi)\vec u_x\cdot\vec u_x + (r\cos\phi)\vec u_x\cdot\vec u_y = -r\sin\phi.
$$

The coordinate basis $\vec u_r, \vec u_\phi, \vec u_z$ derived in this way does not have to consist of unit vectors, but it is often desirable to normalize and make them into unit vectors. In this particular case
$$
  |\vec u_\phi|
    = \sqrt{\vec u_\phi\cdot\vec u_\phi}
    = \sqrt{r^2\sin^2\phi + r^2\cos^2\phi}
    = \sqrt{r^2}
    = r,
$$
the last equality following since $r$ is usually assumed to be positive. Then
$$
  \hat u_\phi = \frac{\vec u_\phi}{|\vec u_\phi|} = -(\sin\phi)\vec u_x + (\cos\phi)\vec u_y
$$
and so
$$
  \vec u_x\cdot\hat u_\phi = -\sin\phi.
$$
A: Start by looking at the definition of $\hat a_r$ and $\hat a_\phi$. Those are vectors of length $1$, with $\hat a_r$ along the $r$, and $\hat a_\phi$ perpendicular to $\hat a_r$, in the horizontal plane. So now you need to write these in terms of the components along the axes. For $\hat a_r$, you have $$\hat a_r=\frac x{\sqrt{x^2+y^2}}\hat a_x+\frac y{\sqrt{x^2+y^2}}\hat a_y=\cos\phi\hat a_x+\sin\phi\hat a_y$$
You can easily see that the slope is $y/x$. The perpendicular to this will have the slope $-x/y$. Using the fact that it's normalized to $1$, you get $$\hat a_\phi=\pm\sin\phi\hat a_x\mp\cos\phi\hat a_y$$
The convention about the sign is given by the cross product:
$$\hat a_r\times\hat a_\phi=\hat a_z$$
