Suppose that I rotate the (traditional) coordinate system $(x,y)$ by an angle $\theta$ to obtain a new coordinate system $(s,n)$. Consider a velocity vector $$v = (v_x,v_y) = v_xe_x + v_y e_y,$$ where $e_x = (1,0)$ and $v_y = (0,1)$, that lies on the $s$-axis of the new coordinate system $(s,n)$.
(1) How does it follow that $v = \|v\|e_s$? Here $e_s = (1,0)$ in the new-coordinate system and $\|v\| = \sqrt{v_x^2+v_y^2}$. It has been a long time, but if remember correctly, every vector in the $(x,y)$ plane with coordinates $(v_x,v_y)$ can be transformed to a vector $(v_s,v_n)$ by $$\left(\begin{array}{ll} v_s \\ v_n \end{array} \right) = \left(\begin{array}{ll} \cos \theta & - \sin \theta \\ \sin \theta& \cos \theta \end{array} \right) \left(\begin{array}{ll} v_x \\ v_y \end{array} \right)$$
(2) My second question is related to PDE's. Given $k_{xx} = (v_x h)/2$ and $k_{yy} = (v_y h)/2$, where $h$ is a constant (a mesh grid). Let $T = T(x,y)$ be function then they state in the course $$k_{xx} \frac{\partial^2 T}{\partial x^2} + k_{yy} \frac{\partial^2 T}{\partial y^2} = k_{ss} \frac{\partial ^2 T}{\partial s^2}+k_{sn} \frac{\partial T}{\partial s \partial n} + k_{nn} \frac{\partial ^2 T}{\partial n^2}, $$ where $$\begin{cases} k_{ss} = \displaystyle \frac{\|v\| h}{2} [(\cos \theta)^3 + (\sin \theta)^3] \\ k_{sn} = \displaystyle \frac{\|v\| h}{2} \sin(2\theta) (\sin \theta - \cos \theta) \\ k_{nn} = \displaystyle \frac{\|v\|h}{2} \cos \theta \sin \theta (\sin \theta + \cos \theta) \end{cases}$$ How is this transformation of the PDE obtained? I assume its due a change of variable rule ...
Thanks in advance!