Find the equations of the curves that cut the angle between the coordinate lines of the surface in half. Find the equations of the curves that cut the angle between the coordinate lines of the surface in half.
The surface being:
$$x = u \cos v \\ 
  y=u \sin v \\ 
   z=u ;\\ (u>0)$$
So, we have the parameterization immediately: $(u>0)$
$$(u,v)\to(u\cos v, u \sin v, u) $$
The coordinate lines being $\ \ v=v_0; \ u =u_0; \ \ \ \ \ \ v_0,u_0=const.$
 A: The parametrisation sets an orthogonal coordinate system in the surface, so, we look for curves for wich their tangent vector is of the form $T=at_u+bt_v$ and having all along them $a/b=1$, with $t_u$ and $t_v$ unitary vectors tangent to the curves with $v$ constant and $u$ constant respectively.
$T_u=(\dfrac{\partial x}{\partial u},\dfrac{\partial y}{\partial u},\dfrac{\partial z}{\partial u})$; $t_u=\dfrac{T_u}{\vert T_u\vert}=\dfrac{1}{\sqrt{2}}(\cos v, \sin v,1)$
$T_v=(\dfrac{\partial x}{\partial v},\dfrac{\partial y}{\partial v},\dfrac{\partial z}{\partial v})$; $t_v=\dfrac{T_v}{\vert T_v\vert}=\dfrac{1}{u}(-u\sin v, u\cos v,0)$
$\left(t_u·t_v=0\right.$ confirming the orthogonality.$\left.\right)$
The vector tangent to any curve in the surface is:
$T(u(t),v(t))=\dfrac{dr}{dt}=T_u\dfrac{du}{dt}+T_v\dfrac{dv}{dt}=\sqrt{2}u't_u+uv't_v$ ($u'=du/dt$ and $v'=dv/dt$)
And imposing the condition to cut the (right) angle in two:
$\dfrac{uv'}{\sqrt{2}u'}=1$ or $\dfrac{v'}{u'}=\dfrac{\sqrt{2}}{u}$
$\dfrac{dv}{du}=\dfrac{\sqrt{2}}{u}\implies u=Ae^{v/\sqrt{2}}$
The curves are thus,
$r=(Ae^{v/\sqrt{2}}\cos v,Ae^{v/\sqrt{2}}\sin v,Ae^{v/\sqrt{2}})$
