Suppose that the height of a hill above sea Suppose that the height of a hill above sea
level is given by $z=8x^2y^3+x^3y+6$.  If you are at the point (2,1,46) in what direction is the elevation changing fastest? 
What i have done? 
The gradient vector at this point$(16xy^3+3x^2y,24x^2y^2+x^3)=(44,104)$.
The problem, that the answer is given as vector $(a,b,c)$, what is $c$ in this case and how should I to compute $c$.
Thanks!
 A: The gradient tells you that if you move on the tangent plane one unit step in the $x$ direction you go up by $44$,
and if you move in the $y$ direction by $104$ (definitely, it is not a “hill”).
So the vectors $(1,0,44)$ and $(0,1,104)$ are parallel to the plane.
Then if you move by $cos(\alpha)$ and $sin(\alpha)$ on the horizontal plane, you move up by 
$$
h = 44\cos \alpha  + 104\sin \alpha  = \sqrt {44^2  + 104^2 } \left( {\cos \phi \cos \alpha  + \sin \phi \sin \alpha } \right) = 4\sqrt {797} \cos \left( {\alpha  - \phi } \right)
$$
where $\phi  = \arctan \left( {{{104} \over {44}}} \right)$
So the slope is max = $4\sqrt{797}$ in the direction $\phi$ on the $x,y$ plane.
Or otherwise,
 take the vector normal to the plane
$$
{\bf N} = \left( {1,0,{{\delta z} \over {\delta x}}} \right) \times \left( {0,1,{{\delta z} \over {\delta y}}} \right) = \left( {1,0,44} \right) \times \left( {0,1,104} \right) = \left( { - 44, - 104,1} \right)
$$
its horizontal projection will indicate the steepest descent, so
$$
 - \pi  < \phi  = \arctan _4 \left( {{{ - N_y } \over { - N_x }}} \right) \le \pi 
$$
and its angle vs. vertical will give the absolute measure of the slope.
Thus
$$
\eqalign{
  & {\bf V} = \left( { - N_x , - N_y ,{{\sqrt {N_x ^2  + N_y ^2 } } \over {N_z }}\sqrt {N_x ^2  + N_y ^2 } } \right) =   \cr 
  &  = \left( {{{\delta z} \over {\delta x}},{{\delta z} \over {\delta y}},\left( {{{\delta z} \over {\delta x}}} \right)^2  + \left( {{{\delta z} \over {\delta y}}} \right)^2 } \right) \cr} 
$$
is a vector parallel to the tangent plane and orientated along the max slope.
