Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 13 / b

For the graph surface $Z=z(x,y)$, an upward normal is $(-z_x,-z_y,1)$. At $x=y=1$, let $q=e^{-1/2}$; then $z_x=q$ and $z_y=0$. Normalizing gives the unit normal vectorIts negative is the equally valid downward normal.

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.

The horizontal direction of steepest descent on a graph surface is parallel to $-\nabla z$. On the line $y=1$, the component $z_y=X(1-y^2)H$ vanishes identically, so that line is invariant under the downhill path. Starting at $X=0$, $z_x=(1-X^2)e^{-(X^2+1)/2}>0$ for $-1<X<0$, so $x$ decreases towards zero. The path reaches the minimumAt the minimum no strictly downhill direction remains. A constant-speed parametrization reaches it after the finite arc length divided by $u$; the unnormalized gradient flow approaches the same endpoint only asymptotically in its own time parameter.