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EE107 SP 02B Directional Derivatives Directional Derivatives

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  • 7/30/2019 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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    7-8

    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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    7-9

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

    NOTE.

    is called the Lagrange multiplier. Theorem 7.4 provides candidates for extreme points.

    SUMMARY.

    To find the extrema offsubject to the constraint 0, yxg , take the following steps.

    STEP 1 : Define a new function yxgyxfyxL ,,, .

    ( is independent ofx andy)

    STEP 2 : Find the critical points for the function yxL , by solving the system

    0, yxLx

    0, yxLy

    0, yxL

    .

    THEOREM 7.4. Iffhas a local extrema at ba, subject to the constraint 0, yxg ,

    then there exists a value c such that at cbayx ,,,, ,

    0, yxLx

    0, yxLy

    0, yxL

    .

    Here, yxgyxfyxL ,,, .

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    7-10

    EXAMPLE 7.5. Find the absolute extrema of yyxyxf 223, on the region

    4|, 222 yxyxE .

    SOLUTION.

    (i) Extrema at 4|,222

    yxyxE {This forms the constraint}

    12,

    6,

    yyxf

    xyxf

    y

    x

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

    2

    1012,

    006,

    yyyxf

    xxyxf

    y

    x

    21,0 is a critical point in 4|, 222 yxyxE .

    Innerregion

    ofE

    BoundaryofE

    y

    x

    y

    x

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    7-11

    (ii) Extrema at 4|, 222 yxyxE

    To find the extrema of f at the boundary, use Theorem 7.4. Let

    04, 22 yxyxg . Hence,

    43, 2222 yxyyxyxL 026, xxyxLx (1)

    0212, yyyxLy (2)

    04, 22 yxyxL

    . (3)

    (1) y : 026 xyxy

    (2) x : 022 xyxxy )

    04 xxy

    014 yx 0 x or 41y .

    Substitute 0x and 41y in (3), we obtain

    2

    04022

    y

    y

    463

    1663

    044122

    x

    x

    The critical points are 2,0 , 2,0 , 41,463 and 41,463 .

    Findfat these critical points to identify the absolute maximum and minimum.

    2 2

    0, 1 2 3 0 1 2 1 2 1 4 1 2 0.25f

    622032,0 22 f

    222032,0 22 f

    125.124141463341,463 22 f

    125.124141463341,463 22 f

    The absolute maximum is 12.125 and the absolute minimum is 0.25 .

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    7-12

    EXAMPLE 7.6. The temperature at any given yx, of the curve 1124 22 yx is

    given by xyxT 2244 22 . Find the points on the curve that give the hottest and coldest

    temperature.

    SOLUTION.

    To find the extrema of f at the boundary, use Theorem 7.4. Let

    xyxyxf 2244, 22 and 01124, 22 yxyxg . Hence,

    11242244, 2222 yxxyxyxL

    0828, xxyxLx (1)

    02448, yyyxLy (2)

    01124, 22 yxyxL

    . (3)

    (1) 3y : 024624 xyyxy

    (2) x : 02448 xyxy

    )

    0624 yxy

    0624 xy 41 x or 0y .

    Substitute 41x and 0y in (3), we obtain

    41

    011241422

    y

    y

    21

    01012422

    y

    x

    The critical points are 41,41 , 41,41 , 0,21 and 0,21 .

    Findfat these critical points to identify the absolute maximum and minimum.

    25.2412412441441,41 22 f

    25.2412412441441,41 22 f

    02120242140,21 22 f

    22120242140,21 22 f

    The hottest temperature occurs at 41,41 and the lowest temperature occurs

    at 0,21 .

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    7-13

    7.4 SURFACES IN R3

    PLANES

    A plane consists of a set of points that satisfies the equation 0 dczbyax where

    dcba ,,, are constant coefficients.

    EXAMPLE 7.7.

    x x

    x

    y y

    y

    z z

    z

    z=1plane

    yzplane

    x = 0

    xyplane

    z= 0

    xzplane

    y = 0

    z=y plane

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    7-14

    QUADRICSURFACES

    The equation 2 2 2 0Ax By Cz Dxy Exz Fyz Gx Hy Iz J in an xyz-coordinate

    system is called a second-degree equation in x, y and z. The graphs of such equations are

    called quadric surfacesor sometimes quadrics. Six common quadric surfaces (cylinder,sphere, cone, ellipsoid, paraboloid and hyperboloid) are discussed in this section.

    CYLINDER

    Cylindrical surface is generated by a straight line which moves along a fixed curve and

    remains parallel to a fixed straight line. A cylinder whose cross section is an ellipse, parabola,

    or hyperbola is called an elliptic cylinder (2 2

    2 21

    x y

    a b ), parabolic cylinder ( 2 2 0x rz ), or

    hyperbolic cylinder (2 2

    2 21

    x y

    a b ) respectively.

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    7-15

    EXAMPLE 7.8.

    (a) 422 yx

    (b) 2yz

    x

    y

    z

    0

    2

    x

    y

    1

    11

    z

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    7-16

    SPHERE

    The surface area of a sphere is given by the formula 2222 rczbyax with

    center cba ,, and radius r.

    EXAMPLE 7.9.

    (a) 2222 221 zyx

    (b) 2222 azyx

    x

    y

    z

    x

    y

    z

    (1, -2, 0)

    a

    (0, 0, 0)

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    7-18

    ELLIPTICPARABOLOID

    The surface area of an elliptic paraboloid is given by the formula czb

    y

    a

    x

    2

    2

    2

    2

    where 0c .

    EXAMPLE 7.12.

    HYPERBOLOID OFONE SHEET

    The surface area of hyperboloid of one sheet is given by the formula 12

    2

    2

    2

    2

    2

    c

    z

    b

    y

    a

    xwhere

    0,, cba .

    EXAMPLE 7.13.

    x

    y

    z

    x

    y

    z

    c

    ab

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    7-19

    HYPERBOLOID OFTWO SHEETS

    The surface area of hyperboloid of one sheet is given by the formula 12

    2

    2

    2

    2

    2

    c

    z

    b

    y

    a

    x

    where 0,, cba .

    EXAMPLE 7.14.

    x

    y

    z

    c

    c


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