Gradient Discussion, 6/23/08 The function 32),( xyyxf = has a gradient

23 6,2 xyyf =∇

The direction of maximum rate of change is (after some simplification)

2222 9

3,

9 xy

x

xy

y

f

fu

++=

∇∇=v

At the point )0,2(− we found that the direction of maximum rate of change will be 1,0 − but that the

actual slope will be 0 (because f∇ evaluated at )0,2(− gives 0,0 which when gives 0=⋅∇ ufv

). This

seemed a bit counter-intuitive – how can the “maximum” rate of change be 0? It turns out not to be so peculiar after all. The graph of this function is:

Note that along the x and y axes, the z-values are zero. In this case, the point )0,2(− sits on the x-axis. If

we make a cross-section along the line 2−=x we get a cubic function 34yz −= . Single-variable calculus

shows that the instantaneous rate of change when 0=y is zero as well: 212yz −=′ , so 0)0( =′z ).

It turns out that the derivative of this surface at the point )0,2(− is zero in all directions! So the direction

of maximum increase is still true: it is 1,0 − , which points into quadrant 3 where the surface does

actually get larger. But at the point )0,2(− , the instantaneous rate of change is still zero. In fact, for all points on the x-axis, this will be the case. Another way to consider this problem: find a vector normal to this plane. We know this will be in general

1,, −= yx ffnv

So for the function 32),( xyyxf = , the normals are

1,6,2 23 −= xyynv

.

Whenever a point is chosen on the x-axis, meaning 0=y , we get a normal vector 1,0,0 −=n

v, which is a

unit vector in the z-direction (the fact that it points down in this case is not relevant). Therefore, any tangent plane to a point on the surface confined to the x-axis will have a tangent plane of 0=z , which has zero slope in all x and y directions.