Bipolar cylindrical coordinates (u, v, w) are defined by the transformation equations
x =a sinh v/(cosh v − cos u), y =a sin u/(cosh v − cos u), z = w
where a is a positive constant.
(a) Find the basis unit vectors eu, ev and ew for this coordinate system
Answer is:eu =(− sin u sinh v, cos u cosh v − 1, 0)/cosh v − cos u ,ev =(1 − cos u cosh v, − sin u sinh v, 0)/cosh v − cos u ,ew = (0, 0, 1) But not sure how to get these.
If $\underline{r} = \underline{r}(u_1, u_2, u_3)$ then If we change $u_1$ by a small amount, $du_1$, then $\underline{r}$ moves to position $(\underline{r} + d\underline{r})$, where
\begin{align} d\underline{r} &= \frac{\partial \underline{r}}{\partial u_1}\,d u_1 \\ &= h_1 \,e_1 \,du_1 \end{align} Where scale factor $h_1$ is given by $$h_1 = \bigg|\frac{\partial \underline{r}}{\partial u_1}\bigg|$$
and the base unit vector $e_1$ is given by $$e_1 = \frac{1}{h_1} \frac{\partial \underline{r}}{\partial u_1}$$
As an example (from Spherical Polar Coordinates), let $u_1 = r, u_2 = \theta, u_3 = \phi$ with
$$\underline{r} = r \sin \theta \cos \phi \textbf{i} + r\sin \theta \sin \phi \textbf{j}+r \cos \theta \textbf{k}$$
It's easy to show that \begin{align} \frac{\partial \underline{r}}{\partial r}&= \sin \theta \cos \phi \textbf{i} + \sin \theta \sin \phi \textbf{j}+\cos \theta \textbf{k} \implies h_r=1 \\ \frac{\partial \underline{r}}{\partial \theta}&= r \cos \theta \cos \phi \textbf{i} + r \cos \theta \sin \phi \textbf{j}-r \sin \theta \textbf{k} \implies h_\theta=r \\ \frac{\partial \underline{r}}{\partial \phi}&= -r \sin \theta \sin \phi \textbf{i} + r \sin \theta \cos \phi \textbf{j}\implies h_\phi= r \sin \theta \end{align}
Thus \begin{align} e_r &= \sin \theta \cos \phi \textbf{i} + \sin \theta \sin \phi \textbf{j}+\cos \theta \textbf{k} \\ e_\theta &= \cos \theta \cos \phi \textbf{i} + \cos \theta \sin \phi \textbf{j}-\sin \theta \textbf{k} \\ e_\phi &= -\sin \phi \textbf{i} + \cos \phi \textbf{j} \end{align}
This method can then be applied to your example above.