 Solution

A tangent displacement has the form $(dx,dy,z_{x}dx+z_{y}dy)$. At the specified point, $z_y=0$ and $z_x=q>0$, so the fastest decrease of height per unit surface distance has $dy=0$ and $dx<0$. The unit downhill tangent is $(-1,0,-q)/\sqrt{1+q^2}$. Therefore, for speed $u$ measured along the surface,Its Euclidean norm is $u$, and it is orthogonal to the surface normal. The general construction is steepest descent on a graph surface; using a horizontal speed equal to $u$ would overestimate the full three-dimensional speed.