Math 251 Section 12.6
Definition: The gradient of
Directional Derivatives and the Gradient
f  f  f 
f ( x , y , z ) is the vector function  f 
i
j
k.
x
y
z


The directional derivative of
f ( x , y , z ) in the direction of the vector v is  f 
v

.
v



Explanation: Let
( a , b , c ) be a given point , v a vector, and u a unit vector in the direction of v .
The line through
( a , b , c ) with direction u is x  a  u1t ,

f ( x, y, z )  f (a  u1t , b  u2t , c  u3t )  g (t )
chain rule, g ' (t )

f
f
f
u1 
u2 
u3
x
y
z

More simply,
y  b  u 2t ,
z  c  u3t .
is the function along this line.Then by the

g ' (0)   f ( a , b, c )  u


f ( x , y , z ) along the line is f ( r (t ))  g (t ) and g ' (t )   f ( r (t ))  r ' (t ) from
the chain rule.
The maximum rate of change of
f ( x , y , z ) from the point ( a , b , c ) is  f ( a , b, c ) and occurs in
the direction of the gradient.

This is because
 f  u   f cos  where θ is the angle between the two vectors.
Level Surfaces and their Tangent Planes.
Let
F ( x , y , z ) be a differentiable function and K a constant. The surface F ( x , y , z )  K is called a
level surface of
Let
F ( x, y , z ) .
F ( a , b, c )  K
 F is orthogonal to every curve on S through the point ( a , b , c ).

Why: Let

r (t ) be any curve on this surface and r ( 0 )  ( a , b , c ).

Since

r (t )
is always on the surface,
F ( r (t ))  K

By the chain rule this says

In particular
. So
d  

 F ( r (t ))   0 .
dt 


 F ( r (t ))  r ' (t )  0
which means these two vectors are orthogonal.

 F ( r ( 0 ))  r ' ( 0 )  0
. Since we used any curve on the surface, the gradient is
orthogonal to the surface.
This shows
 F ( a , b , c ) is a normal vector to the tangent plane to the surface at the given point.
The tangent plane equation is
Example:
a) Find
 F ( a , b, c )  x  a , y  b, z  c  0
F ( x , y , z )  x 2 y 3  xyz 2
F .
b) Find the rate of change of F in the direction of the vector
c) Find the tangent plane to the surface
2, 1,1
F ( x , y , z )  x 2 y 3  xyz 2  10
at the point
(1, 2,  1) .