4.6 Directional Derivatives and Gradient Vector
Theorem If is a differentiable function of and , then has a directional derivative in the direction of any unit vector = + and
, = , + ,
EXAMPLE 2 Find the directional derivative , if
, = − 3 + 4
and u is the unit vector given by angle . What is 1,2
SOLUTION
Therefore
The Gradient Vector
Definition If is a function of two variables and , then the gradient of is the vector function defined by
, = , + , = ,
EXAMPLE 3 If , = sin + , then
With this notation for the gradient vector, we can rewrite the directional derivative of a differentiable function as
EXAMPLE 4 Find the directional derivative of the function
, = − 4
at the point (2, −1) in the direction of the vector v = 2 i + 5 j.
SOLUTION We first compute the gradient vector at (2, −1)
Note that is not a unit vector, but since = 29 , the unit vector in the direction of is
Therefore
Functions of Three Variables
EXAMPLE 5 If
, , = sin
(a) find the gradient of .
(b) find the directional derivative of at (1,3,0) in the direction of v = i + 2j − k.
SOLUTION
(a) The gradient of is
(b) At (1,3,0), we have 1,3,0 = 0,0,3 . The unit vector in the direction of v is
u = 1
6 + 2
6 − 1 6 Therefore
1,2,3 = 1,3,0 .
= 3 . 1
6 + 2
6 − 1 6
= 3 . − 1
6 = − 3 2 .
Exercise 9:
(a) Find the gradient of .
(b) Evaluate the gradient at the point .
(c) Find the rate of change of at in the direction of the vector .
, , = − , 2, −1,1 , = 0, 4
5 , − 3 5 . SOLUTION
= 2 − , = − , = − 3
(a) The gradient of is
= 2 − , − , − 3
(b) The gradient at the point (2, −1,1)
2, −1,1 = −3, 2, 2
(c) The rate of change of at in the direction of the vector .
2, −1,1 = 2, −1,1 . = −3, 2, 2 . 0, 4
5 , −3
5 = 8
5 − 6
5 = 2 5
Exercise 12: Find the directional derivative of the function at the given point in the direction of the vector .
, = , 1,2 , = 3,5 . SOLUTION
= 1 + − (2 )
+ = −
+ , = − 2
+
= −
+ , − 2
+
Then
(1,2) = 3
25 , − 4
25
The unit vector in the direction of is
u = 3
34 + 5 34
The directional derivative of at P in the direction of the vector is
1,2 = 1,2 . = 3
25 − 4
25 . 3
34 + 5 34
= 9
25 34 − 20
25 34 = − 11 25 34
Maximizing the Directional Derivative
Theorem Suppose is a differentiable function of two or three variables. The maximum value of the directional derivative
is and it occurs when has the same direction as the gradient vector .
where is the angle between and u. The maximum value of cos is 1 and this occurs when = 0. Therefore the maximum value of is
and it occurs when = 0, that is, when has the same direction as .
EXAMPLE 6
(a) If , = , find the rate of change of at the point (2,0) in the direction from to , 2 .
(b) In what direction does have the maximum rate of change? What is this maximum rate of change?
SOLUTION
(a) We first compute the gradient vector:
= , =
2,0 = , = 1,2
The unit vector in the direction of = − , 2 is
= = 2
5 − 3 2 , 2
so the rate of change of in the direction from to is
(2,0) = (2,0). = 2
5 − 3
2 + 4 = 2 5
5
2 = 1
We see that increases fastest in the direction of the gradient vector
2,0 = , = 1,2 The maximum rate of change is
2,0 = 1,2 = 5.
Exercise 25: Find the maximum rate of change of at the given point and the direction in which it occurs.
, , = + + , 3,6, −2 . SOLUTION
= + + , =
+ + ,
= + +
3,6, −2 = 3 7 , 6
7 , − 2 7
The maximum rate of change is
3,6, −2 = 3 7 , 6
7 , − 2
7 = 1
7 9 + 36 + 4 = 1.
The direction of the gradient vector
3,6, −2 = , , = 3 7 , 6
7 , − 2 7
Tangent Planes to Level Surfaces
The tangent plane to the level surface ( , , ) = at ( , , ) as the plane that passes through P and has normal vector ( , , ). Using the standard equation of a plane, we can write the equation of this tangent plane as
, , − + , , − + , , − = 0
Normal Line
The normal line to S at P is the line passing through P and perpendicular to the tangent plane. The direction of the normal line is therefore given by the gradient vector ( , , ) and so, its symmetric equations are
−
, , = −
, , = −
, ,
EXAMPLE 8 Find the equations of the tangent plane and normal line at the point (−2,1, −3) to the ellipsoid
4 + +
9 = 3.
SOLUTION The ellipsoid is the level surface of the function
, , =
4 + +
9 Therefore we have
, , = , , , = 2 , , , =
−2,1, −3 = −1, (−2,1, −3) = 2, −2,1, −3 = − Then the equation of the tangent plane
−1 + 2 + 2 − 1 − 2
3 + 3 = 0 which simplifies to
3 − 6 + 2 + 18 = 0
The symmetric equations of the normal line are + 2
−1 = − 1
2 = + 3
− 2 3
Exercise 42: Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point.
= − , (4,7,3).
SOLUTION
, , = − +
Therefore we have
, , = −2 , , , = 1, , , = 2
(4,7,3) = −8, (4,7,3) = 1, (4,7,3) = 6 The tangent plane
−8 − 4 + − 7 + 6 − 3 = 0 Thus
8 − − 6 − 7 = 0 The normal line
− 4
−8 = − 7
1 = − 3 6
