Addmath Differentiation

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  • 8/12/2019 Addmath Differentiation

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    Topic : DIFFERENTIATION

    Unit E : Determine the first derivative of composite function using chain rule.

    Example Exercise

    1. 2)2( += xy

    )2(2

    1)2(2 1

    +=

    +=

    x

    xdx

    dy

    a. 4)3( += xy

    [4(x+3)3]

    b. 5)2( += xy

    [5(x+2)4]

    c. 3)8( += xy

    [3(x+8)2]

    2. 2)23( += xy

    )23(6

    3)23(2 1

    +=

    +=

    x

    xdx

    dy

    a. 4)32( += xy

    [8(2x+3)3]

    b. 5)24( += xy

    [20(4x+2)4]

    c. 3)85( += xy

    [15(5x+8)2]3. 2)2(3 += xy

    )2(6

    1)2(23 1

    +=

    +=

    x

    xdx

    dy

    a. 4)2(5 += xy

    [20(x+2)3]

    b. 5)24(3 += xy

    [60(4x+2)4]

    c. 3)82(2 += xy

    [12(2x+8)3]

    4.

    2)2(

    2

    +

    =

    xy

    3

    3

    3

    2

    )2(

    4

    )2(4

    1)2(22

    )2(2

    +

    =

    +=

    +=

    +=

    x

    x

    x

    dx

    dy

    xy

    a.

    4)2(

    5

    +

    =

    xy

    5

    20

    (x+2)[- ]

    b.

    5)2(

    3

    +

    =

    xy

    6

    15

    (x+2)[- ]

    c.

    3)8(

    2

    +

    =

    xy

    4

    6

    (x+8)[- ]

    5.

    2)2(5

    2

    +

    =

    xy

    2)2(5

    2

    += xy

    1)2(252 3 += xy

    3)2(5

    4

    +

    =

    xy

    a.

    3)5(4

    3

    +

    =

    xy

    4

    9

    4(x+5)[- ]

    b.

    6)32(5

    4

    =

    xy

    7

    24

    5(2x-3)[- ]

    c.

    4)43(2

    5

    =

    xy

    5

    10

    (3x-4)[ ]

    Differentiation 9

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    Topic : DIFFERENTIATION

    Unit F : Determine the Gradient of a Tangent and a Normal at a point on a Curve.

    Example 1 : Find the gradient of the tangent to

    the curve 5732 23 += xxxy

    at the point (-2,5)

    Solution: 5732 23 += xxxy

    766 2 += xxdx

    dy

    At point (-2,5), x=-2

    Hence, the gradient of tangent at the point (-2,5)

    5

    7)2(6)2(6

    2when766

    2

    2

    2

    =

    +=

    =+=

    ==

    xxx

    xwhendx

    dymT

    Example: Given )32()( = xxxf and the gradient

    of tangent at point P on the curve

    y = f(x) is 29, find the coordinates of thepoint P.

    Solution: )(xfy =

    y = 2x2 3x since f(x) = x(2x-3)

    34 = xdx

    dy

    At point P, 29=dx

    dy

    4x 3 = 29

    x = 8y = 104

    The coordinates of P is (8 , 104)

    (1) Given that the equation of a parabola is2241 xxy += , find the gradient of the tangent

    to the curve at the point (-1,-3)

    8=Tm

    (2) Find the gradient of the tangent to the curve

    ( )( )32 += xxy at the point (3,6).

    7=Tm

    (3) Given that the gradient of the tangent at point P on

    the curve ( )252 = xy is 4, find the coordinatesthe point P.

    P(2 , 1)

    (4) Given2

    4)(

    xxxf = and the gradient of tangent

    is 28. Find the value of x.

    32

    =x

    Differentiation 10

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    Topic : DIFFERENTIATION

    Unit G : Determine the Equation of a Tangent and a normal at a Point on a Curve.

    Example 1 :Find the equation of the tangent at the point (2,7)

    on the curve 53 2 = xy

    Solution:

    126(2)dx

    dy2,when x

    6

    532

    ===

    =

    =

    xdx

    dy

    xy

    Gradient of tangent, mT

    =12

    Equation of tangent is

    ( )11 xxmyy T = ( )

    01712xy24127

    2127

    =+

    =

    =

    xy

    xy

    Example 2 :Find the equation of the normal at the point x = 1

    on the curve2324 xxy +=

    Solution:

    46(1)2dx

    dy,1when x

    62

    3242

    =+==

    +=

    +=

    xdx

    dy

    xxy

    Gradient of normal,4

    1=

    Nm

    when x = 1 , y = 4 2(1) + 3(1)2= 5

    Equation of normal is

    ( )11 xxmyy N =

    ( )

    021-y4x

    )1(1204

    14

    1

    5

    =+

    =

    =

    xy

    xy

    (1) Find the equation of the tangent at the point (1,9)

    on the curve ( )252 = xy

    2112 += xy

    (2) Find the equation of the tangent to the curve

    ( )( )112 += xxy at the point where itsx-coordinate is -1.

    33 = xy

    (3) Find the equation of the normal to the curve

    232 2 += xxy at the point where its

    x-coordinate is 2.

    0225 =+ yx

    (4) Find the gradient of the curve32

    4

    +

    =

    xy at the

    point (-2,-4) and hence determine the equation of

    the normal passing through that point.

    0308;8 == yxmT

    Differentiation 11

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    Topic : DIFFERENTIATION

    Unit I : Determine the Types of Turning Points(Minimum and Maximum Points)

    Example :Find the turning points of the curve

    31812223

    ++= xxxy and determine

    whether each of them is a maximum or aminimum point.

    Solution:

    31812223

    =

    ++=

    dx

    dy

    xxxy

    At turning points, 0=dx

    dy

    ( )( )3or x1

    031034

    018246

    2

    2

    ==

    =

    =+

    =+

    x

    xxxx

    xx

    Substitute the values of x into

    318122 23 ++= xxxy

    When x = 1 , 113)1(18)1(12)1(2 23 =++=y

    When x = 3 , 333)1(18)3(12)3(2 23 =++=y

    Thus the coordinates of the turning points are

    and

    24122

    2

    = xdx

    yd

    When x = 1 , 01224)1(122

    2

    ==

    dx

    yd

    Thus , (3 , 33) is the point

    (1) Find the coordinates of two turning points on the

    curve ( )32 = xxy

    (1 , 2) and (1 , 2)

    (2) Determine the coordinates of the minimum point of

    442 += xxy .

    (2 , 0)

    (3) Given 523

    2 23= xxy is an equation of a

    curve, find the coordinates of the turning points of

    the curve and determine whether each of the turningpoint is a maximum or minimum point.

    min. point = (0 , 5) ; max. point = ),2(323

    Differentiation 13

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    Topic : DIFFERENTIATION

    Unit J : Problems of Rates of Change

    Task 1 : Answer all the questions below.

    (1) Given that xxy 23 2 = andxis increasing at a

    constant rate of 2 unit per second, find the rate ofchange ofywhenx= 4 unit.

    1

    2

    44

    )2)(22(

    22

    2)4(6

    4

    26

    23

    2

    =

    =

    =

    =

    =

    =

    =

    =

    =

    sunit

    dt

    dx

    dx

    dy

    dt

    dy

    dx

    dy

    xWhen

    x

    dx

    dy

    xxy

    dt

    dx

    (2) Given that xxy =24 andxis increasing at a

    constant rate of 4 unit per second, find the rate ofchange ofywhenx= 0.5 unit.

    12 unit s1

    (3) Given thatx

    xv 19 = andxis increasing at a

    constant rate of 3 unit per second, find the rate ofchange of vwhenx= 1 unit.

    30 unit s1

    (4) SPM 2004 (Paper 1 Question 21) [3 marks]Two variables,xandy, are related by the equation

    .2

    3x

    xy += Given thatyincreases at a constant

    rate of 4 unit per second, find the rate of change ofxwhenx=2.

    58 unit s1

    If y =f(x) and x =g(t), then using the chain rule

    dt

    dx

    dx

    dy

    dt

    dy= , where

    dt

    dy is the rate of change of

    y anddt

    dx is the rate of change of x.

    Differentiation 14

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    Task 2 : Answer all the questions below.

    (1) The area of a circle of radius rcm increases at aconstant rate of 10 cm

    2per second. Find the rate of

    change of rwhen r= 2 cm. ( Use = 3.142 )

    Answer :

    1

    2

    7957.0

    410

    4

    )2(2

    2

    2

    10

    =

    =

    =

    =

    =

    =

    =

    =

    =

    scmdt

    dr

    dt

    drdt

    dr

    dr

    dA

    dt

    dA

    dr

    dA

    cmrWhen

    rdr

    dA

    rA

    dt

    dA

    (2) The area of a circle of radius rcm increases at aconstant rate of 16 cm

    2per second. Find the rate of

    change of rwhen r= 3 cm. ( Use = 3.142 )

    0.8487 cm s1

    (3) The volume of a sphere of radius rcm increases at aconstant rate of 20 cm3per second. Find the rate ofchange of rwhen r= 1 cm. ( Use = 3.142 )

    1.591 cm s1

    (4) The volume of water , Vcm, in a container is given

    by ,83

    1 3hhV += where hcm is the height of the

    water in the container. Water is poured into the

    container at the rate of .scm10 -13 Find the rate

    of change of the height of water, in ,scm -13 at

    the instant when its height is 2 cm. [3 marks]

    65 cm s1

    Differentiation 15

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    Task 3 : Answer all the questions below.

    Example :

    The above figure shows a cube of volume 729 cm.

    If the water level in the cube, hcm, is increasing at

    the rate of 0.8 cm s 1 , find the rate of increase of

    the volume of water.

    Solution :

    Let each side of the cube bexcm.Volume of the cube = 729 cm

    x = 729x = 9

    Rate of change of the volume of water,

    1364.8

    0.881

    =

    =

    =

    scm

    dt

    dh

    dh

    dV

    dt

    dV

    Hence, the rate of increase of the volume of wateris 64.8 cms 1 .

    (1) A spherical air bubble is formed at the base of apond. When the bubble moves to the surface of thewater, it expands. If the radius of the bubble is

    expanding at the rate of 0.05 cm s1, find the rate

    at which the volume of the bubble is increasingwhen its radius is 2 cm.

    8.0 cm3s1

    (2) If the radius of a circle is decreasing at the rate

    of 0.2 cm s 1 , find the rate of decrease of thearea of the circle when its radius is 3 cm.

    2.1 cm2s1

    h cm

    9cm

    9cm

    9cm

    h cm

    Chain rule

    V = 9 x 9 x h = 81h

    dh

    dV=81

    dt

    dh =rate of increase of

    the water level

    = 0.8 cm s1

    Differentiation 16

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    (3) The radius of a spherical balloon increases at the

    rate of 0.5 cm s1. Find the rate of change in thevolume when the radius is 15 cm.

    450 cm3

    s

    1

    (4) The edge of a cube is decreasing at the rate of3 cm s1. Find the rate of change in the volumewhen the volume is 64 cm3.

    144cm3

    s

    1

    (5) Diagram 1shows a conical container with a

    diameter of 60 cm and height of 40 cm. Water is

    poured into the container at a constant rate of

    1 000-13 scm .

    Calculate the rate of change of the radius of the water

    level at the instant when the radius of the water is 6 cm.

    (Use = 3.142; volume of cone hr2

    3

    1= )

    6.631cm3s1

    (6) Oil is poured into an inverted right circular cone ofbase radius 6 cm and height 18 cm at the rate of

    2-13 scm . Find the rate of increase of the height of

    water level when the water level is 6 cm high.( Use = 3.142 )

    0.1591 cm s1

    60 cm

    Water

    40 cm

    Diagram 1

    Differentiation 17

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    (7)

    The above figure shows an inverted cone with aheight of 20 cm and a base-radius of 4 cm. Water

    is poured into the cone at the rate of 5 cm s1but

    at the same time, water is dripping out from the

    cone due to a leakage a the rate of 1 cm s1.

    (a) If the height and volume of the water at timets

    arehcm and Vcm respectively, show that

    .75

    1 3hV =

    (b) Find he rate of increase of the water level inthe cone at the moment the water level is 12 cm.( Use = 3.142 )

    (b) 0.2210 cm s1

    (8)

    The above diagram shows a solid which consistsof a cuboid with a square base of side 6xcm,surmounted by a pyramid of height 4xcm. The

    volume of the cuboid is 5832 cm.

    (a) Show that the total surface area of the solid,

    Acm, is given by .3888

    96 2

    xxA +=

    (b) If the value ofx increasing at the rate

    0.08 ,scm-1

    find the rate of increase of the

    total surface area of the solid at the instantx= 4.

    42 cm2s1

    4 cm

    20 cm

    hcm

    B

    H

    F G

    C

    A

    D

    4xcm

    6xcm

    6xcm

    V

    E

    Differentiation 18

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    Topic : DIFFERENTIATION

    Unit K : Problems of Small Changes and Approximations

    Task 1 : Answer all the questions below.

    (1) Given that xxy 42 += , find the smallchange inywhenxincreases from 2 to 2.01.

    08.0

    )01.0)(8(

    8

    24)2(2

    01.0

    201.2

    01.22

    42

    42

    =

    =+=

    =

    ==

    +=

    +=

    y

    y

    xdx

    dyy

    xwhendx

    dy

    x

    x

    xdx

    dy

    xxy

    (2) Given that xxy 32 += , find the smallchange inywhenxincreases from 6 to 6.01.

    0.15

    (3) Given that xxy = 22 , find the smallchange inywhenxdecreases from 8 to 7.98.

    0.62

    (4) Given that xy 4= , finddx

    dy.

    Hence, find the small change inywhenxincreases

    from 4 to 4.02.

    02.0;2 == yxdx

    dy

    xdx

    dyy

    dx

    dy

    x

    y

    xxyy

    inchangesmallinchangesmallwhere

    ==

    Small Changes

    xdx

    dyy

    yyy

    original

    original

    +=

    +=

    new

    Approximate Value

    Differentiation 19

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    Task 2 : Answer all the questions below.

    (1) Given the area of a rectangle , xxA 23 2 += ,wherex is the width, find the small change in the

    area when the width decreases from 3 cm to 2.98cm.Answer :

    12

    2

    scm4.0

    )02.0)(20(

    20

    32)3(6

    02.0

    98.23

    98.23

    26

    23

    =

    =+=

    =

    =

    =

    +=

    +=

    A

    A

    xdx

    dAA

    xwhendx

    dA

    x

    x

    xdx

    dA

    xxA

    (2) A cuboid with square base has a total surface area,

    xxA 43 2 = , wherex is the length of the side ofthe base. Find the small change in the total surfacearea when the length of the side of the basedecreases from 5 cm to 4.99 cm.

    0.26 cm2

    (3) The volume, Vcm3, of a cuboid with rectangular

    base is given by xxxV 32 23 += , wherexcmis the width of the base. Find the small change in

    the volume when the width increases from 4 cm to

    4.05 cm.

    1.75 cm3

    (4) In a pendulum of lengthx meters, the period T

    seconds is given as10

    2 x

    T = . Finddx

    dT.

    Hence, find the small change in Twhenxincreasesfrom 2.5 m to 2.6 m.

    5010; == T

    xdxdT

    second

    Differentiation 20

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    Task 3 : Answer all the questions below.

    Example :

    The height of a cylinder is three times its radius.Calculate the approximate increase in the total

    surface area of the cylinder if its radius increases

    from 7 cm to 7.05 cm.

    Solution :

    Let the total surface area of the cylinder be A cm.

    A = Sum of areas of the top and bottom circularsurface + Area of the curved surface.

    ( )2

    2

    2

    8

    322

    22

    r

    rrr

    rhrA

    =

    +=

    +=

    Approximate change in the total surface area is A

    ( )( )

    2cm6.5

    05.0716

    705.716

    r

    rdr

    dAA

    dr

    dA

    r

    A

    Hence, the approximate increase in the total surface

    area of the cylinder is 5.6 cm .

    (1) A cube has side of 6 cm. If each of the side of

    the cube decreases by 0.1 cm, find theapproximate decrease in the total surface area

    of the cube.

    7.2 cm2

    (2) The volume of a sphere increases from

    .cm290tocm288 33 Calculate the

    approximate increase in its radius.

    721

    cm

    It is given that

    h=3r

    rdrdA

    rA

    16

    8 2

    =

    =

    New r (7.05)

    Minus old r(7)

    Substitute r with the

    old value of r, i.e. 7

    Differentiation 21

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    Task 4 : Answer all the questions below.

    (1) Given that5

    4

    xy= , calculate the value of

    dx

    dyif

    x= 2.Hence, estimate the values of

    55 98.1

    4(b)

    03.2

    4)a(

    Solution :

    16

    520,2When

    2020

    44

    6

    6

    6

    5

    5

    ===

    ==

    ==

    xdx

    dyx

    xx

    dx

    dy

    xx

    y

    xdx

    dyy

    yyya

    original

    original

    +

    +=

    )( new

    ( )

    .1165250

    320

    3

    32

    4

    203.216

    5

    2

    4

    2

    4

    03.222

    4,

    2

    4

    03.2

    4

    5

    5

    555

    +

    +

    ==+=

    xdx

    dy

    xandywherey

    ( )

    0.13125

    160

    1

    32

    4

    298.1165

    24

    1.984

    (b)

    55

    +

    +

    +

    +=

    xdx

    dyyy

    yyy

    originalnew

    originalnew

    (2) Given3

    27

    xy= , find the value of

    dx

    dywhenx= 3.

    Hence, estimate the value of .03.3

    273

    1=dx

    dy; 0.97

    (3) Given4

    32

    xy= , find

    dx

    dy.

    Hence, estimate the value of .99.1

    324

    5

    128

    xdx

    dy = ; 2.04

    Differentiation 22

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    Task 5 : Answer all the questions below.

    (1) Given thatx

    y

    =2

    20, find the approximate

    change inxwhenyincreases from 40 to 40.5.

    1601

    (2)SPM 2003 (Paper 1 Question 16)

    Given that ,52 xxy += use differentiation tofind the small change inywhenxincreases from 3to 3.01. [3 marks]

    0.11

    (3) Given3

    3

    4rv = , use the differentiation method

    to find the small change in v when rincreases from

    3 to 3.01.

    27.0

    (4) Given that ,53

    xy= find the value of

    dx

    dywhen

    x= 4.

    Hence, estimate the value of

    (a)( )302.45

    (b)( )399.35

    25615=

    dx

    dy; (a) 0.07930 ; (b) 0.07871

    Differentiation 23

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    Topic : DIFFERENTIATION

    Unit L : Problems of Maximum and Minimum Values of a Function.

    Task 1 : Answer all questions below.

    Example 1 :

    Given )540(2 xxL = , find the value ofxforwhichLis maximum.Hence, determine the maximum value ofL.

    Solution:

    27

    10240

    3

    16540

    3

    16

    max3

    16

    080

    803

    163080

    3

    16

    01580,0

    0)1580(

    01580

    0

    3080

    1580

    540

    )540(

    2

    max

    2

    2

    2

    2

    2

    2

    2

    2

    32

    2

    =

    =

    =

    dx

    ydatx=

    x= that satisfies 0=dx

    dywill maximizeyif the value of 0

    2

    2

    =

    +=

    =

    =

    =

    ==

    =

    +=

    =

    +=

    +=

    L

    Limisewillx

    dx

    Ld

    dx

    Ld

    x

    xxKnowing

    xx

    xxxx

    dx

    dLwhen

    xdx

    Ld

    xxdx

    dL

    xxL

    xxL

    (1) Given )128

    (4 2

    xxL += , find the value ofxfor

    whichLis minimum.Hence, determine the minimum value ofL.

    192,4 min == Lx

    (2) Given )283(2

    1 2 += xxy , find the value ofx

    for whichy is minimum.Hence, determine the minimum value ofy.

    35

    min34 , == yx

    (3) SPM 2003(Paper 1, No 15)

    Given thaty=14x(5 x), calculate(a) the value ofxwhenyis a maximum(b) the maximum value ofy [3 marks]

    (a) 2.5 ; (b) 87.5

    Differentiation 25

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