Maximum increase of a function along a surface I'm not quite sure how to approach this question - it's a practice problem, and there is no answer key. Someone told me the answer was to project the gradient onto the surface, but I'm not sure how to do this since using the projection formula with the normal would project it onto the normal. I'm also not sure how this works for the question. I'd really appreciate any insight!

 A: As  the increase of a function in a direction is given its derivative in that direction, one wants to look for the direction of extremal increase of such derivative. However, since we are in a surface we are only interested of the extremal value for directions tangent to the surface (as they come from curves in the surface). Now, 
(i) The directional derivatives of $T$ are (in our point)
$$
dT(x,y,z)(u,v,w)=(y,x,-2z)(u,v,w)^t=(-1,0,-2)(u,v,w)^t=-u-2w.
$$ 
(ii) The tangent space to our surface in the point is given by (with some loosy notation)
$$
d[z^3+zx+y^2-2](u,v,w)=(z,2y,x+3z^2)(u,v,w)^t=(1,-2,3)(u,v,w)^t=u-2v+3w=0.
$$
Now to look for extrema we restrict to the compact set $u^2+v^2+w^2=1$ and we have a Lagrange multipliers problem in the three variables $(u,v,w)$: 
(iii) find the critical points of $h=-u-2w$ with the conditions $f_1=u-2v+3w=0$, $f_2=u^2+v^2+w^2-1=0$.
Quickly, Lagrange says that $\nabla h$ must depend on $\nabla f_1,\nabla f_2$, or in other words, the $3\times 3$ matrix $(\nabla h,\nabla f_1,\nabla f_2)$ must have $\det=0$, that is
$$
0=\det\begin{pmatrix}-1&0&-2\\1&-2&3\\2u&2v&2w
\end{pmatrix}=4w-8u+2v.
$$
Consequently, the direction to give the ant is the solution of the system
$$
-8u+2v+4w=u-2v+3w=u^2+v^2+w^2-1=0.
$$
Since norm 1 is not a condition, only the two first equations matter, and we obtain $(u,v,w)=(1,2,1)$. This gives negative increase, and $(-1,-2,-1)$ positive increase, both maximal in absolute value.
