     # EE107 SP 02B Directional Derivatives Directional Derivatives

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7-1

EE107 Lec2B

Directional Derivatives

7.1 DIRECTIONAL DERIVATIVES

Directional derivatives give the rate of change of a function in a given direction. Consider

the following figure.

i,jand kare unit vector in the direction of positive x, y, zaxis respectively. A vectorrcan

be defined as zyxzyx ,,kjir vector from 0 to zyx ,, .

Therefore, we have

ir hzyhx ,, , jr hzhyx ,, , kr hhzyx ,, .

By using the limit definition, we have

h

fhf

h

zyxfzyhxfr

x

fzyx

x

f

hh

)()(lim

,,,,lim,,

00~

rir

h

fhf

h

zyxfzhyxfr

y

fzyx

y

f

hh

)()(lim

,,,,lim,,

00~

rjr

h

fhf

h

zyxfhzyxfr

z

fzyx

z

f

hh

)()(lim

,,,,lim,,

00~

rkr

x

f

,

y

f

and

z

f

measure the rate of change of f in the direction of i,jand k

respectively.

z

xy

(x,y,z)

(0,0,1)

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

0ij

r

k

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7-2

DEFINITION 7.1. Let kjiu cba be any given unit vector and point zyxp ,, .

Then, the directional derivative of fin the direction of u at the point p, denoted by

pfDu , is given by

h

zyxfhczhbyhaxfpfDh

u,,,,lim

0

pfDpx

fi

, pfDp

y

fj

, pfDp

z

fk

.

It is difficult to find the directional derivatives by Definition 7.1. We need a TOOL.

DEFINITION 7.2. Iff is a function, of which the partial derivatives exist at the point

zyxp ,, , then the gradient off at the pointpis defined by

kji pz

fp

y

fp

x

fpf

.

Recall that a unit vectoruhaving the same direction as r is given by rr

u1

.

THEOREM 7.1. If

zyxf ,, is a differentiable function at any given point

zyxp ,, , then the directional derivative offat point pin the direction of any given

unit vector kjiu cba is

zyxfczyxfbzyxfa

pfpfD

zyx

u

,,,,,,

u

THEOREM 7.2. The maximum value of pfDu at the point yxp , is pf

and it occurs in the direction of pf .

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7-3

EXAMPLE 7.1. (A) Find the directional derivative of xyezzyxf 2,, at the point

3,2,1 in the direction of the vector kjir 53 . (B) What is the maximum rate of

increase at the point 3,2,1 ?

SOLUTION. (A):

22122 18233,2,1,, eefyezzyxf xxy

x

22122 9133,2,1,, eefxezzyxf yxy

y

221 6323,2,12,, eefzezyxf yxy

z

By Definition 7.2, kjikji 236369183,2,12222

eeeef . A unit

vectoruhaving the direction of kjir 53 is

kjirr

u 5335

11

.

By Theorem 7.1,

35

15

35

30

35

9

35

54

5335

1

69183,2,13,2,1

2222

222

eeee

eeeffDu kjikjiu

(B) By Theorem 7.2, the maximum rate of increase offat 3,2,1 is

2444 2136813243,2,1 eeeef .

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7-4

EXAMPLE 7.2. Suppose a rectangular coordinate system is located in space such that the

temperature Tat the point zyx ,, is given by 222100 zyxT .

(a) Find the rate of change of T at the point 2,3,1 in the direction of the

vector kjir .

(b) In what direction from the point 2,3,1 does T increase most rapidly?

What is the maximum rate of change ofTat 2,3,1 ?

SOLUTION.

(a) 222100 zyxT

4910023122002,3,1200

4915023132002,3,1200

495023112002,3,1200

22222222

22222222

22222222

zz

yy

xx

TzyxzT

TzyxyT

TzyxxT

By Definition 7.2, 50 150 100 50

1,3, 2 3 249 49 49 49

T i j k i j k . A unit

vectoruhaving the direction of kjir is

kjirr

u

3

11 .

By Theorem 7.1,

349

10010015050

349

1

3

1

49

100

49

150

49

502,3,12,3,1

kjikjiuTTDu

(b) The maximum rate of change of Tat 2,3,1 occurs in the direction of the

gradient, that is, in the direction of the vector kji 23 . By Theorem 7.2, the

maximum rate of increase ofTat 2,3,1 is

2500 22500 10000

1,3, 2 3.82401 2401 2401

T .

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7-5

7.2 EXTREMA OF FUNCTIONS

DEFINITION 7.3. Iff is A function of two variables is said to have a local maximumat

ba,if there is a region R containing

ba,such that

bafyxf ,, for all other pairs

yx, inR. It follows that if baf , is a local maximum, then 0, bafx and 0, bafy .

DEFINITION 7.4. Iff is A function of two variables is said to have a local minimumat

dc, if there is a region R containing dc, such that dcfyxf ,, for all other pairs

yx, inR. It follows that if dcf , is a local maximum, then 0, dcfx and 0, dcfy .

NOTE.

The local maxima and minima are called the extremaoff.

NOTE.

The converse of Theorem 7.3 is false, that is, if 0,, bafbaf yx , it does notnecessarily follow that ba, is the local extrema off. If 0,, bafbaf yx , then ba, is

called the critical point off. Critical points that are not the local extrema are called the saddle

points.

THEOREM 7.3. Iff has first partial derivatives and ba, is the local extrema off,

then 0,, bafbaf yx .

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7-6

TEST FOR EXTREMA

Let f be a function of two variables which has continuous second partial derivatives on a

regionR and let

2

,,,, yxfyxfyxfyxG xyyyxx

for all yx, in R. If ba, is in R and 0, bafx , 0, bafy , then the following

statements hold.

(i) If 0, baG and 0, bafxx , then baf , is a local maximum off.

(ii) If 0, baG and 0, bafxx , then baf , is a local minimum off.

(iii) If 0, baG , then baf , is not an extremum, that is, ba, is a saddle point.

(iv) If 0, baG , then no conclusion can be made from this test.

ABSOLUTE EXTREMA OF FUNCTIONS

If a function fof two variables is continuous on a closed region R, then f has an absolute

maximum baf , and an absolute minimum dcf , for some ba, and dc, inR. This

means that

bafyxfdcf ,,,

for all yx, inR.

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7-7

EXAMPLE 7.3. Find the extrema offif 242, 32 yxyxyxf .

SOLUTION. 242, 32 yxyxyxf .

234,

44,

yxyxf

yxyxf

y

x

By Theorem 7.3, if ba, is the local extrema off, then 0,, bafbaf yx .

4

3034,

044,

22 y

xyxyxf

yxyxyxf

y

x

Solving these simultaneously gives us the pairs of critical points 0,0 and 34,34 .

Use the Test for Extrema to find the extrema off. The second partial derivatives off

are 4, yxfxx , yyxfyy 6, and 4, yxfxy .

At 0,0 : 40,0 xxf , 00,0 yyf and 40,0 xyf .

016404

,,,,

2

2

yxfyxfyxfyxG xyyyxx

0,0f is not an extremum, that is, 0,0 is a saddle point

At 34,34 : 434,34 xxf , 834,34 yyf and 434,34 xyf .

016484

,,,,

2

2

yxfyxfyxfyxG xyyyxx

034,34 G and 034,34 xxf ,

34,34f is a local minimum off.

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EXAMPLE 7.4. A rectangular box with no top and having a volume of 12 m3 is to be

constructed. The cost per m2 of the material to be used is \$4 for the bottom, \$3 for two of the

opposite sides and \$2 for the remaining pair of opposite sides. Find the dimensions of the box

that will minimize the cost.

SOLUTION. {DIY}

Volume 12

xyz m

3

xyz

12

Cost : 4 3 2 2 2 4 6 4C xy xz yz xy xz yz

12 12 72 484 6 4 4C xy x y xy

xy xy y x

2

484

xC y

x ;

2

724

yC x

y

To determine the possible extrema, we must find the simultaneous solutions of

2

484 0

xC y

x and

2

724 0

yC x

y .

These equation imply that 2 12x y and 2 18xy . Substituting 212y x in 2 18xy

gives us 318 144x . Solving for x we obtain 2x . Since 212y x , the

corresponding value ofy is 3. The Test for Extrema can be used to show that these

values ofx andy determine a minimum value of C. Finally,12

2zxy

.

The minimum cost occurs if the dimensions of the box are 3 m 2 m 2 m as

shown below.

x

y

z

2 m

3 m

2 m

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7.3 LAGRANGE MULTIPLIERS {SKIP P 9-12; READ P13-19: R3 SURFACES }

NOTE.

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