MAT495_CHAPTER_7

Part 1 PARTIAL DERIVATIVES MAT 295

89

Chapter 7

Gradient and

Directional

Derivatives At the end of this module, students should be able to:

Define and evaluate gradient of function

Define directional derivatives

Evaluate directional derivatives

Evaluate directional derivatives using gradient

7.1 Introduction

Imagine standing at a point 0xx . The slope of the ground in front of where

we are standing will depend on the direction we are facing. It might slope

steeply in one direction and be relatively flat in another direction.

In earlier modules, we have discussed that xf is the slope of f in the positive

x-direction (i.e. slope parallel to x-axis) and yf is the slope of f in the positive

y-direction (i.e. slope parallel to y-axis). In this module we are going to learn

to find slopes at other directions often called as directional derivatives. There

are several ways to evaluate directional derivatives in which one is of the

methods is by using gradient vectors.

7.2 The Gradient Vectors

The gradient of a function of two variables is a vector-valued function of two

variables.


90

Definition

Let ),( yxfz be a function of x and y such that xf and yf exists. Then the

gradient of f at any point ),( yx , denoted by ),( yxf or ),( yxfgrad , is the

vector such that

yx

yx

ff

jy

fi

x

f

jyxfiyxfyxf

,

),(),(),(

The gradient ),( yxf is a vector in the plane (not a vector in space) as in

Figure 7.1.

Figure 7.1

Example 1

Let yyxyxf 32),( 2 . Find the gradient of f at any point ),( yx and the

gradient of f at the point (1, 2).

Identify ),( yxf

yyxyxf 32),( 2

Find xf and yf

Solution

Steps : Gradient

Identify ),( yxf

Find xf and yf

Write gradient as yx ffyxf ,),(

x

z

y

(x,y)


91

xyyxfx 4),( ,

32),( 2 xyxfy

Find the gradient of f at any point

32,4

)32()4(

),(),(),(

2

2

xxy

jxixy

jyxfiyxfyxf yx

Evaluate the gradient of f at the point (1,2)

5,8

58

)3)1(2()2)(1)(4(

)2,1()2,1()2,1(

2

ji

ji

jfiff yx

Example 2

Find the gradient of 2ln),( xyxyyxf at the point (1,3).

Identify ),( yxf

2ln),( xyxyyxf

Find xf and yf

2),( yx

yyxfx , xyxyxfy 2ln),(


jyxfiyxfyxf yx

),(),(),(

xyxyx

y

jxyxiyx

yyxf

2ln,

)2(ln)(),(

2

2


jfiff yx

)3,1()3,1()2,1(

6,12612

))3)(1)(2(1(ln)31

3()2,1( 2

orji

jif

Solution


92

Let xyxyyxf 33),( 33 .

(i) Compute xf and yf .

(ii) Find )1,2( f .

7.3 Directional Derivatives

There are infinitely many directional derivatives of a surface at a given point.

In order to find the directional derivative at that point, we first need to specify

the direction in which to compute the slope. We write the directional derivative

of f in the direction of the unit vector u

at the point ),( yx as ),( yxfDu i.e.

1u

. The directional derivative ),( yxfDu is simply the slope of ),( yxf when

standing at the point ),( yx and facing the direction given by u

. For instance if

x and y were given in meters, then ),( yxfDu would be the change in height

per meter as you moved in the direction given by u

when you are at the point

),( yx .

Definition

The directional derivative of ),( yxf at the point ),( ba and in the direction of

21 u,uu

is given by

h

hubhuafbafD

hu

),(lim),( 21

0

Note that ),( yxfDu is a constant value (representing a slope). In fact, the

directional derivative is the same as a partial derivative if u

points in the

positive x or positive y direction.

Figure 7.2

Warm up exercise

x

z

y

u


93

Theorem

Directional derivative

If f is a differentiable function of x and y, then the directional derivative of f in

the direction of the unit vector jiu

sincos (or sin,cosu

) is

sin),(cos),(),( yxfyxfyxfD yxu

Alternatively, directional derivative may be evaluated by using gradient as in

the following theorem.

Theorem

Directional derivative

If f is a differentiable function of x and y, then f has a directional derivative in

the direction of any unit vector jbiau

(or, bau ,

) and

byxfayxf

bayxfyxf

uyxfyxfD

yx

yx

u

).,().,(

,),(),,(

ˆ),(),(

In this case, the directional derivative is calculated by taking the dot product of

the gradient vector of f and the unit vector in the direction or parallel to the

slope needed.

Steps : Directional derivative

Identify ),( yxf

Compute the gradient ),( yxf

Identify direction v

Compute magnitude v

: v

Compute unit vector v

vu

Compute directional derivative : fu


94

Example 3

Let 4

4),(2

2 yxyxf . Find the directional derivative of f at (1,2) in the

direction of the unit vector u

3sin,

3cos

.

Find xf and yf

xyxfx 2),( , 2

),(y

yxfy

Find the directional derivative of f at point (1,2) in the direction of the

unit vector u

3sin)

2(

3cos)2(

sin),(cos),(),(

yx

yxfyxfyxfD yxu

Thus,

866.1

2

3)1(

2

1)2(

3sin)

2

2(

3cos)1)(2(

sin)2,1(cos)2,1()2,1(

yxu fffD

Example 4

Find the directional derivative of yxyxf 3),( at (-1, 5) in the direction of

the vector jiv

3 .

Find xf and yf

yxyxfx23),( , 3),( xyxfy

Since the direction is not given in the form of a unit vector, then we

need to find the unit vector u

in the same direction as v

Solution

Solution

3 6

3

2

1


95

10

3,

10

1

3,1

31

1

22

v

vu

Find the directional derivative of f at point (-1,5) in the direction of

the unit vector u

uyxfyxfDu

),(),(

Thus,

795.310

12

10

3)1(

10

1)15(

10

3,

10

11,15

10

3,

10

1)5,1(),5,1(

)5,1()5,1(

yx

u

ff

uffD

Example 5

Let xyxyxf 32),( 22 . Find the:

a) gradient of f at (1,1).

b) directional derivative at the same point in the direction of the unit

vector u

4sin,

4cos

.

Find xf and yf

32),( xyxfx , yyxfy 4),(


yxorjyix

jyxfiyxfyxf yx

4,32)4()32(

),(),(),(


jfiff yx

)1,1()1,1()1,1(

Solution


96

4,1

4

)1(4)12()1,1(

ji

jif

Find the directional derivative at point (1,1) as a dot product of

gradient of f and the unit vector u

4sin,

4cos4,1

)1,1()1,1(

uffDu

121.22

3

)2

1)(4()

2

1)(1(

2

1,

2

14,1

Example 6

Given xyexyxf sin),( . Find the directional derivative of f at (0,1)

parallel to the vector jiv

43 .

Find xf and yf

xyx yexyxf cos),( , xy

y xeyxf ),(


xyxy

xyxy

yx

xeyex

jxeiyex

jyxfiyxfyxf

,cos

)()(cos

),(),(),(


0,2

02

)1,0()1,0()1,0(

ji

jfiff yx

Since the direction is not given in the form of a unit vector, then we

need to find the unit vector u

in the same direction as v

.

Solution


97

5

4,

5

3

4,3

)4(3

1

22

v

vu

Find the directional derivative of f at point (0,1) in the direction of the

unit vector u

.

5

4,

5

30,2

)1,0()1,0(

uffDu

5

6

)5

4)(0()

5

3)(2(

Example 7

Find the directional derivative of yyxyxf 3),( 2 in the direction ji

34

at the point (1, -1).

Identify ),( yxf

yyxyxf 3),( 2


3,2 2 xxyf


3,4 v

Compute magnitude v

: v

525

916

)3(4 22

v

Compute unit vector v

vu

3,45

1u

Compute directional derivative

Solution


98

3(5)2(45

1

3,25,45

1

),(

2

2

xxy

xxy

fuyxfDu

15855

1 2 xyx

Directional derivative at (1, -1)

5

12

15)1)(1(855

1)1,1(

fDu

Example 8

Determine the directional derivative of the function 22 33),( yxyxf in

the unit vector j

direction.

Identify ),( yxf

22 33),( yxyxf


yxyxf 6,6),(


1,0 jv

Compute magnitude v

1v

Compute unit vector

1,01

jj

v

vu

Compute directional derivative

y

y

yxyxfDu

6

60

6,61,0),(

Solution


99

Example 9

Find the directional derivative of 432),( yyxyxf at (1,2) in the

direction given by the angle 4

.

The gradient of f is given by

3223 43,2),( yyxxyyxf

while the unit vector u

is

.2

1,

2

1

4sin

4cos

jiu

Hence, the directional derivative is determined by

2

143

2

12

2

1,

2

143,2),(

3223

3223

yyxxy

yyxxyyxfDu

Therefore, at (1, 2)

222

4

2

20

2

16

2

1)2(4)2()1(3

2

1)2)(1(2)2,1( 3223

fDu

Note that )2,1(fDu represents the slope or the rate of change of ),( yxf at

point (1, 2) in the direction of 4

sin,4

cos

v

or 2

1,

2

1v

.

Solution

1 4

4

2

1


100

Let 23 43),( yxyxyxf at the point (1, 2) in the direction of

(a) 6

(b) 4,3

(i) Compute the gradient ),( yxf .

(ii) Identify the direction.

(iii) Compute the magnitude of the direction vector.

(iv) Compute the unit vector.

(v) Compute the directional derivative.

Exercise 7

1. Find the gradient for the following functions.

a) xyexyxf sin),( at (0, 1)

b) yyxyxf 4),( 32 at (2, -1)

c) 222),( zyxyxf at (1, 4, 2)

2. For the following functions, points and vectors:

a) find the gradient of f.

b) evaluate the gradient at the given values of (x,y).

c) find the directional derivative of f in the direction of the vector v

.

i) 13

12,

13

5),2,1(,45),( 32 vyxxyyxf

ii) jivxyyxf

5

3

5

4),3,1(,ln),(

iii) 3

),0,2(),sin(),(

xyxyxf

iv) 6

),2,1(,43),( 23 yxyxyxf

v) jivxyxeyxf y

43),0,2(),cos(),(

Warm up exercise


101

3. Find the directional derivatives of yexyyxf ),( at P(1,1) in the direction of

a) the y-axis

b) PQ where Q is (4,5).

4. Compute )1,2(fDu , where u

is a unit vector in the direction of the vector

3,1v

and 324),( yxxyxf .

5. Find the directional derivatives of tyx

tytyxf

),,( at (2,1,-1) in the direction

of kjiv

3 .

6. Find the directional derivative of the function xyyxf ),( at the point P(2, 8) in

the direction of the point Q(5, 4).

7. Find the directional derivative of yx

xyxf

),( in the direction of

jiu sincos when

6

.

8. Find the directional derivative of the function yyxyxf 4),( 32 at the point

P(2, -1) in the direction of the vector jiv

52 .

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