I am currently reviewing basic vector analysis and trying to understand every single detail, however, I got stuck in some derivation.
What I want to show is the following:
Given the del operator (i.e., vector differential operator) in
Cartesian coordinates $(x,y,z)$$$\nabla=\frac{\partial }{\partial x}\mathbf{a}_x+\frac{\partial
}{\partial y}\mathbf{a}_y+\frac{\partial }{\partial z}\mathbf{a}_z$$show that the corrseponding operator in Cylindrical coordinates
$(\rho, \phi ,z)$ is given by$$\nabla=\frac{\partial }{\partial\rho}\mathbf{a}_\rho+\frac{1}{\rho}\frac{\partial }{\partial
\phi}\mathbf{a}_\phi+\frac{\partial }{\partial z}\mathbf{a}_z$$
I tried one approach. However, for curiosity I tried a different method but I couldn't get it right.
Approach #1:
From the point-to-point transformation
$$\rho=\sqrt{x^2+y^2}, \; \phi=\text{tan}\frac{y}{x}$$
partial differentiation with respect to $x$ and $y$ yields
\begin{align} \frac{\partial \rho}{\partial x} &=\frac{x}{\sqrt{x^2+y^2}}=\frac{\rho \, \text{cos}\phi}{\rho}=\text{cos}\phi \\
\frac{\partial \rho}{\partial y}&=\frac{y}{\sqrt{x^2+y^2}}=\frac{\rho \, \text{sin}\phi}{\rho}=\text{sin}\phi
\end{align}
and
\begin{align} \frac{\partial \phi}{\partial x}&=\frac{-y}{x^2}\frac{1}{1+(\frac{y}{x})^2}=\frac{-y}{x^2+y^2}=\frac{-\rho \, \text{sin}\phi}{\rho^2}=\frac{-\text{sin}\phi}{\rho} \\
\frac{\partial \phi}{\partial y}&=\frac{1}{x}\frac{1}{1+(\frac{y}{x})^2}=\frac{x}{x^2+y^2}=\frac{\rho \, \text{cos}\phi}{\rho^2}=\frac{\text{cos}\phi}{\rho}
\end{align}
Now, plugging these in the chain rule differentiation formulas
\begin{align} \frac{\partial }{\partial x}&=\frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial x}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial x} \\
\frac{\partial }{\partial y}&=\frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial y}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial y}
\end{align}
and making use of the unit vector transformation from Cartesian to Cylindrical
\begin{align} \mathbf{a}_x&=\text{cos}\phi\;\mathbf{a}_\rho-\text{sin}\phi\;\mathbf{a}_\phi\\
\mathbf{a}_y&=\text{sin}\phi\;\mathbf{a}_\rho+\text{cos}\phi\;\mathbf{a}_\phi
\end{align}
We get
\begin{align} \nabla&=\frac{\partial }{\partial x}\mathbf{a}_x+\frac{\partial
}{\partial y}\mathbf{a}_y+\frac{\partial }{\partial z}\mathbf{a}_z \\
&=\left (\frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial x}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial x} \right )\left ( \text{cos}\phi\;\mathbf{a}_\rho-\text{sin}\phi\;\mathbf{a}_\phi \right )\\
&+\left ( \frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial y}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial y} \right )\left ( \text{sin}\phi\;\mathbf{a}_\rho+\text{cos}\phi\;\mathbf{a}_\phi \right )+\frac{\partial }{\partial z}\mathbf{a}_z \\
&=\left (\frac{\partial }{\partial \rho}\;\text{cos}\phi+\frac{\partial }{\partial \phi}\;\frac{-\text{sin}\phi}{\rho} \right )\left ( \text{cos}\phi\;\mathbf{a}_\rho-\text{sin}\phi\;\mathbf{a}_\phi \right )\\
&+\left ( \frac{\partial }{\partial \rho}\;\text{sin}\phi+\frac{\partial }{\partial \phi}\;\frac{\text{cos}\phi}{\rho} \right )\left ( \text{sin}\phi\;\mathbf{a}_\rho+\text{cos}\phi\;\mathbf{a}_\phi \right )+\frac{\partial }{\partial z}\mathbf{a}_z\\
&=\left ( \text{sin}^2\phi+\text{cos}^2\phi \right )\frac{\partial }{\partial \rho}\mathbf{a}_\rho+\frac{1}{\rho}\left ( \text{sin}^2\phi+\text{cos}^2\phi \right )\frac{\partial }{\partial \phi}\mathbf{a}_\phi+\frac{\partial }{\partial z}\mathbf{a}_z\\
&=\frac{\partial }{\partial\rho}\mathbf{a}_\rho+\frac{1}{\rho}\frac{\partial }{\partial
\phi}\mathbf{a}_\phi+\frac{\partial }{\partial z}\mathbf{a}_z
\end{align}
which is the desired result.
Approach #2:
How can I get the same result starting from the point-to-point transformation
$$x=\rho \, \text{cos}\phi,\; y=\rho \, \text{sin}\phi$$
by using partial differentiation? Maybe implicit differentiation?
Best Answer
The coordinate transformation from polar to rectangular coordinates is given by
$$\begin{align} x&=\rho \cos \phi \tag 1\\\\ y&=\rho \sin \phi \tag 2 \end{align}$$
Now, suppose that the coordinate transformation from Cartesian to polar coordinates as given by
$$\begin{align} \rho&=\rho (x,y)=\sqrt{x^2+y^2} \\\\ \phi&=\phi(x,y) = \begin{cases} \arctan(y/x)&,x>0\\\\ \arctan(y/x)+\pi&,x<0,y\ge 0\\\\ \arctan(y/x)-\pi&,x<0,y<0\\\\ \pi/2&,x=0,y>0\\\\ \pi/2&,x=0,y<0\\\\ \end{cases} \end{align}$$
were unavailable in closed form. We can still proceed to develop a transformation of the gradient operator from Cartesian coordinates to polar.
To do so, we use the differential relationships
$$\begin{align} dx&=\frac{\partial x}{\partial \rho}d\rho+\frac{\partial x}{\partial \phi}d\phi\\\\ dy&=\frac{\partial y}{\partial \rho}d\rho+\frac{\partial y}{\partial \phi}d\phi\\\\ d\rho&=\frac{\partial \rho}{\partial x}dx+\frac{\partial \rho}{\partial y}dy\\\\ d\phi&=\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy \end{align}$$
Defining the Wronskian $W$ as
$$\begin{align} W&=\frac{\partial \rho \cos \phi}{\partial \rho}\frac{\partial \rho \sin \phi}{\partial \phi}-\frac{\partial \rho \cos \phi}{\partial \phi}\frac{\partial \rho \sin \phi}{\partial \rho}\\\\ &=\rho \end{align}$$
we find that
$$\begin{align} \frac{\partial \rho }{\partial x}&=\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \phi}\\\\&= \cos \phi\\\\ \frac{\partial \rho }{\partial y}&=-\frac{1}{W}\frac{\partial \rho \cos \phi}{\partial \phi}\\\\&= \sin \phi\\\\ \frac{\partial \phi }{\partial x}&=-\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \rho}\\\\&=- \frac{\sin \phi}{\rho}\\\\ \frac{\partial \phi }{\partial y}&=\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \phi}\\\\&=\frac{ \cos \phi}{\rho} \end{align}$$
Therefore, we have
$$\begin{align} \hat x \frac{\partial }{\partial x}+\hat y \frac{\partial }{\partial y}&=(\hat \rho \cos \phi -\hat \phi \sin \phi)\left(\frac{\partial \rho}{\partial x}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial x}\frac{\partial }{\partial \phi}\right)+(\hat \rho \sin \phi +\hat \phi \cos \phi)\left(\frac{\partial \rho}{\partial y}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial y}\frac{\partial }{\partial \phi}\right)\\\\ &=\hat \rho \frac{\partial }{\partial \rho}+\hat \phi \frac{1}{\rho}\frac{\partial }{\partial \phi} \end{align}$$
as was to be shown!