of the scalar field z which represents the height of the curve.) To picture what the
above equations are describing, let
x=x 0 +scos α and y=y 0 +ssin α
represent the equation of the line of intersection of the plane z= 0 with the plane
normal to z= 0 containing ˆeα.The plane containing ˆeαand the normal to the plane
z= 0 intersects the surface z(x, y )in a curve given by
z=z(x, y ) = z(x 0 +scos α, y 0 +ssin α) = z(s).The directional derivative of the scalar field z(x, y )in the direction αis then
dz
ds =∂z
∂x cos α+∂z
∂y sin α.Observe that the curve of intersection z =z(s) is a two-dimensional curve, and
the methods of calculus may be applied to determine the relative maximum and
minimum values along this curve. However, one must test this curve of intersection
corresponding to all directions α.
One can conclude that at a critical point (x 0 , y 0 )one must have dzds = 0 for all α.
If in addition d
(^2) z
ds^2 >^0 for all directions α, then z^0 =z(x^0 , y^0 )corresponds to a relative
minimum. If the second derivative d
(^2) z
ds^2 <^0 for all directions α, then z^0 =z(x^0 , y^0 )
corresponds to a relative maximum.
Calculate the second directional derivative and show
d^2 z
ds^2 =∂^2 z
∂x^2 cos(^2) α+ 2 ∂^2 z
∂x ∂y sin αcos α+
∂^2 z
∂y^2 sin
(^2) α.
The sign of the second directional derivative determines whether a maximum or
minimum value for zexists, and hence one must be able to analyze this derivative
for all directions α. Let
A=∂^2 z
∂x^2 B=∂^2 z
∂x ∂y C=∂^2 z
∂y^2