# Section11.5Larsonnot.. ```1
Section 11.5: Directional Derivatives and Gradients
Practice HW from LarsonTextbook (not to hand in)
p. 701 # 1-7 odd, 11, 13, 21-25 odd,
The Directional Derivative
Recall that
f x ( a, b) 
Slope of the tangent line to the surface at the
z

point (a, b, f (a, b)) in the x direction
x ( a,b)
f y ( a, b) 
z
y

Slope of the tangent line to the surface at the
point (a, b, f (a, b)) in the y direction
( a ,b )
Instead of restricting ourselves to the x and y axis, suppose we want to find a method for
finding the slope of the surface in any desired direction.
z
( x0 , y 0 , z 0 )
z  f ( x, y )
f y ( x 0 , y 0 )  Slope in y dir
f x ( x0 , y 0 )  Slope in x dir
Du ( x0 , y 0 )  Slope in u dir
y0
y

x0
x
( x0 , y 0 ,0)
u  a, b 
2
Let u = &lt; a, b &gt; be the unit vector (a vector of length one) on the x-y plane which
indicates the direction we are moving. Then we define the following:
Definition of the Directional Derivative
The directional derivative of a function z = f (x, y) in the direction of the unit vector
u = &lt; a, b &gt;, denoted by Du f ( x, y ) , is defined the be the following:
Du f ( x, y )  f x ( x, y )a  f y ( x, y )b
Notes
1. Geometrically, the directional derivative is used to calculate the slope of the surface
z = f (x, y). That is, to calculate the slope of the surface at the point ( x0 , y0 , z 0 ) ,
where z 0  f ( x0 , y0 ) , we compute the following:
Slope of Surface at po int ( x0 , y0 , z 0 )
 Du f ( x0 , y0 )  f x ( x0 , y0 )a  f y ( x0 , y0 )b
in direction of unit vector u  a.b 
2. The vector u = &lt; a, b &gt; must be a unit vector. If we want to compute the directional
derivative of a function in the direction of the vector v and v is not a unit vector, we
compute
v
1
u

v.
|| v || || v ||
3. The direction of the unit vector u can be expressed in terms of the angle  between
the vector u and the x-axis. In this case, u  cos  , sin   (note, u is a unit vector
since || u || cos 2   sin 2   1  1 ) and the directional derivative can be
expressed
as
Du f ( x, y )  f x ( x, y ) cos   f y ( x, y ) sin  .
4. Computationally, the directional derivative represents the rate of change of the
function f in the direction of the unit vector u.
3
Example 1: Find the directional derivative of the function f ( x, y )  3 y  4 xy  6 x at the

point (1, 2) in the direction of the unit vector that makes an angle of   radians with
3
the x-axis.
Solution:
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4
Example 2: Find the directional derivative of the function f ( x, y )  3 y  4 xy  6 x at the
point (-3, -4) in the direction of the vector v  2i  3j .
Solution:
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5
Example 3: Find the directional derivative of the function f ( x, y )  e x
P = (1, 1) in the direction of the point Q = (0, 0).
2
 y2
at the point
Solution: We first need to find a unit vector u that travels in the same direction is the
vector with an initial point P and terminal point Q. First, we see that

v  PQ   0  1,0  1    1,  1   i  j
Then, since || v || (1) 2  (1) 2  2 , we can find the unit vector u to be
u
Hence, a  
v
 1,  1 
1
1


i
jaibj
|| v ||
2
2
2
2
2
1
1
and b  
. For f ( x, y )  e x  y , we have that
2
2
f x ( x, y )  2 xex
2
 y2
and f x ( x, y )  2 ye x
2
 y2
.
Hence,
Directiona l
Derivative
 Du f ( x, y )  f x ( x, y )a  f y ( x, y )b  2 xe x

2
 y2
(
2
2
2
xe x  y
2
2
2
1
1
)  2 ye x  y ( 
)
2
2
2
2
2

ye x  y
2
Hence,
Directiona l
2
2
2
2
Derivative  D f (1,1)   2 (1)e (1)  ( 1)  2 (1)e (1)  ( 1)
u
2
2
at (1,1)
2 0
2

e 
( 0) e 0
2
2

2
2
2  2
2 2


(1) 
(0)  
 2


2
2
2
2 2
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6
Given a function of two variables z = f (x, y), the gradient vector, denoted by  f ( x, y ) , is a
vector in the x-y plane denoted by
 f ( x, y )  f x ( x, y ) i  f y ( x, y ) j
1. The directional derivative of the function z = f (x, y) in the direction of the unit vector
u = &lt; a, b &gt; can be expressed in terms of gradient using the dot product. That is,
Du f ( x, y )  f ( x, y )  u
 f x ( x, y ), f y ( x, y )    a, b 
 f x ( x, y )a  f y ( x, y )b
2. The gradient vector  f ( x, y ) gives the direction of maximum increase of the surface
z = f (x, y). The length of the gradient vector is the maximum value of the directional
derivative (the maximum rate of change of f). That is,
Maximum Value of the Directiona l Derivative Du f ( x, y )  || f ( x, y ) ||
3. The negation of the gradient vector  f ( x, y ) gives the direction of maximum decrease.
of the surface z = f (x, y). The negation of the length of the gradient vector is the
minimum value of the directional derivative. That is,
Minimum Value of the Directiona l Derivative Du f ( x, y )   || f ( x, y ) ||
7
Example 4: Given the function f ( x, y )  y cos( x  y ) .
a. Find the gradient of f

b. Evaluate the gradient at the point P ( ,0) .
3
c. Use the gradient to find a formula for the directional derivative of f in the direction of the
3 4
vector u   ,  . Use the result to result to find maximum value of the directional
5 5
derivative at P in the direction of the vector u.
Solution:
8
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9
Directional Derivative and Gradient for Functions of 3 variables
The directional derivative of a function f (x, y, z) of 3 variables in the direction of the
unit vector u = &lt; a, b, c &gt;, denoted by Du f ( x, y, z ) , is defined to be the following:
Du f ( x, y, z )  f x ( x, y, z )a  f y ( x, y, z )b  f z ( x, y, z )c
The gradient vector, denoted by f ( x, y, z ) , is a vector denoted by
 f ( x, y , z )  f x ( x, y , z ) i  f y ( x, y , z ) j  f z ( x, y , z ) k
10
Example 4: Find the gradient and directional derivative of f ( x, y, z)  5x 2  3xy  xyz at
P(1, 2, 4) in the direction of the point Q(-3, 1, 2).
Solution: We first compute the first order partial derivatives with respect to x, y, and z.
They are as follows.
f x ( x, y, z )  10 x  3 y(1)  yz (1)  10 x  3 y  yz
f y ( x, y, z )  0  3 x(1)  xz(1)  3x  xz
f z ( x, y, z)  0  0  xy(1)  xy .
Then the formula for the gradient is computed as follows:
f ( x, y, z )  f x ( x, y, z ) i  f y ( x, y, z ) j  f z ( x, y, z ) k  (10 x  3 y  yz ) i  (3x  xz) j  xy k
Hence, at the point P(1, 2, 4), the gradient is
f (1,2,4)  (10(1)  3(2)  (2)( 4)) i  (3(1)  (1)( 4)) j  (1)( 2) k  12 i  j  2 k  12,1, 2 
To find the directional derivative, we must first find the unit vector u specifying the
direction at the point P(1, 2, 4) in the direction of the point Q(-3, 1, 2). To do this, we


find the vector v  PQ . This is found to be v  PQ  3  1,1  2,2  4  4,1,2  .
This must be a unit vector, so we compute the following:
u
1
1
1
4
1
2
v
 4,1,2 
 4,1,2  
,


|| v ||
21
21
21
21
( 4) 2  ( 1) 2  ( 2) 2
Then, using the dot product formula involving the gradient for the directional derivative
and the results for the gradient at the point P(1,2,4) and u given above, we obtain
The directiona l derivative
 Du f (1,2,4)  f (1,2,4)  u
at the po int P(1,24)
4
1
 12, 1, 2    
,
,
21
21
4
1
 (12)( 
)  (1)( 
)  2(
21
21
48
1
4



21
21
21
53

 11.6
21
2
21
2

21
)
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```