B To calculate the polar unit vectors using rotated coordinates

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    Polar Unit Vectors
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The discussion centers on the challenge of deriving polar unit vectors from rotated Cartesian coordinates. The equations for rotated coordinates are provided, but the connection to polar coordinates is not straightforward. Attempts to manipulate the equations lead to contradictions, indicating that the derivation process is more complex than initially thought. A request for hints suggests focusing on specific angles to simplify the transition from Cartesian to polar coordinates. Ultimately, the discussion highlights the intricacies involved in this mathematical transformation.
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To derive the polar unit vectors $$\boxed{\begin{align*}\hat{\rho}&=\cos\theta \;\hat{i}+\sin\theta \;{\hat j}\\

\hat{\phi}&=-\sin\theta \; \hat i+\cos\theta \; \hat j

\end{align*}}\quad\quad\large{\mathbf{(b)}}$$

using coordinates of a point under rotation : $$\begin{align*}x'&=\cos\theta \;x+\sin\theta \;y\\

y'&=-\sin\theta \; x+\cos\theta \; y

\end{align*}\quad\quad\large{\mathbf{(a)}}$$
1726582830959.png
We know that if cartesian coordinates ##(x,y)## (see figure alongside) are rotated to ##(x',y')## about the origin by an angle ##\theta## counter-clockwise as shown, the rotated coordinates are given by $$\begin{align*}x'&=\cos\theta \;x+\sin\theta \;y\\
y'&=-\sin\theta \; x+\cos\theta \; y
\end{align*}\quad\quad\large{\mathbf{(a)}}$$

##\small{\texttt{Can these (the above) be used to derive the (familiar) unit vectors using polar coordinates}}## :


1726582890040.png
$$\boxed{\begin{align*}\hat{\rho}&=\cos\theta \;\hat{i}+\sin\theta \;{\hat j}\\
\hat{\phi}&=-\sin\theta \; \hat i+\cos\theta \; \hat j
\end{align*}}\quad\quad\large{\mathbf{(b)}}$$ I ask because the equations look similar. However, as I show in my workings below, it is far from straightforward. Their (actual) derivation is quite different and doesn't use the rotated co-ordinates shown in ##\mathbf{(a)}## above.



Discussion : The way to go from ##\text{Fig. (2)}\rightarrow\text{Fig. (1)}## is to let ##\text{A}'\rightarrow\text{P}##. That would make ##y'=0\Rightarrow x\sin\theta=y\cos\theta## from ##\mathbf{(a)}## above. But this doesn't lead me anywhere towards deriving ##\mathbf{(b)}##.?

Can I use the equations of ##\mathbf{(a)}## to write ##x'\hat{i}'=\cos\theta x\hat i+\sin\theta y\hat j?## If so, then dividing throughout by ##x'##, we get : ##\hat{i}'=\frac{x}{x'}\cos\theta\hat i+\frac{y}{x'}\sin\theta \hat j##. This would imply ##\frac{x}{x'}=1=\frac{y}{x'}## which is clearly not true.

Request : A hint as to how to derive ##\mathbf{(b)}## from ##\mathbf{(a)}##.
 
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In Fig.1 choose ##\theta## so that P is on x' axis then x',y' are ##\hat{\rho}##, ##\hat{\phi}##, with renaming ##\theta## with ##\phi##.
 
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