$6\mathrm{.}5($a$)$ In module II$,$ the vector transformational law is: $x_j^{\text{'}}=c_{ij}^{\text{'}}{x}_i,$ while $c_{i{j}^{\text{'}}}=c_{ij}^{\text{'}}=cos(x_i,x_j^{\text{'}})=cos$ of the angle from $x_i$ to $x_j^{\text{'}}$ axis. Use the same law to derive the relationship between the velocity components from the Polar coordinate $u_r$ and $u_{}$ to the velocity components from the Cartesian coordinate $u$ and $v.$ Recall that the transformation is a simple rotation of a counterclockwise with an angle of $.$ Need to show steps to get the full credit. If done correctly, the final result is given as vec$(V)=$uvec$(e_x)+$vvec$(e_y)=u_r$vec$(e_r)+u_{}$vec$(e_{}),$ and in complex component form: $W=\frac{dF}{dz}=u-iv=(u_r-i{u}_{})e^{-i}.$ You can also use matrix transformation form if you want.

$($b$)$ Write the first $($and only$)$ component equation for $j=1,$ the $x-$component of the Navier Stoke's equations:

$\frac{del{u}_j}{\text{d}\text{e}\text{l}\text{t}}+u_i\frac{del{u}_j}{del{x}_i}=-\frac{\text{d}\text{e}\text{l}\text{p}}{del{x}_j}+\frac{\text{d}\text{e}\text{l}}{del{x}_j}(\frac{del{u}_k}{del{x}_k})+\frac{\text{d}\text{e}\text{l}}{del{x}_i}[(\frac{del{u}_i}{del{x}_j}+\frac{del{u}_j}{del{x}_i})]+f_j$

For consistency, use $x,y,z$ as spatial coordinates and $u,v,w$ as velocity components.