Friday, November 20, 2009

  1. Two ropes hold a boat at a dock. The tension in the ropes can be represented by 40 + 10j N and 50 – 25j N. Find the resultant force.
  2. The total power P (in watts) transmitted by an AM radio station is given by

    P = 500 + 250 m^2 ,
    where m is the modulation index. Find the instantaneous rate of change of P with respect to m for m = 0.92
  3. Find the derivative of the following polynomials:
    i. y = 4 x^2 + 7 x + 3
    ii. y = 4 x^ -6 - 5 x^3 + x
  4. Find the derivative of the following trigs:
    a. y = 6 sin Ө
    b. y = 7 sin Ө + 3 cos Ө
  5. Determine the gradient of each of the given functions at the given point.
    a) s = 2t^3 - 5t^2 + 4 (-1,-3)
    b) y = 5 sin Ө where Ө = 38◦
  6. Find the area under the curve y = x^3 that is between the lines x = 1 and x = 2.
  7. Integrate the following
    i) 4x^6 + 3x^ -4 + 1
    ii) 2 cos Ө
  8. The electric field E at a distance r from a point charge is E = k/(r2), where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.
  9. The blade of a saber saw moves vertically up and down, and its displacement y (in cm) is given by y = 1.85 sin t, where t is the time (in s). Find the velocity of the blade for t = 0.025 s.
  10. An open-top container is to be made from a rectangular piece of cardboard 24 cm by 38 cm. Equal squares of side x cm are to be cut from each corner, then the sides are to be bent up and taped together. Find the instantaneous rate of change of the volume V of the container with respect to x for x = 4 cm.
  11. An analysis of a company’s records shows that in a day the rate of change of profit (in dollars) in producing x generators is dp/dx = 60(30 – 4x^5). Find the profit in producing x generators if a loss of $500 is incurred if none are produced.
  12. The vertical displacement y (in cm) of the end of an industrial robot arm for each cycle is y = 2t^1.5 – cos t, where t is the time (in s). Find its vertical velocity for t = 15s.

127 comments:

  1. (3)firstly the derivative of the equation is the differential of the equation. the first thing we do is multiply the power by the base and then minus 1 from the power.

    (1)y= 4x^2 +7x +3
    dy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
    dy/dy= 8x+7
    (2) y=4x^-6 - 5x^3 +x
    dy/dx= (4*-6)x^(-6-1) - (5*3)x^(3-1) + x^(1-1)
    dy/dx= -24x^-7 - 15x^2

    ReplyDelete
  2. 2) P = 500 + 250 m^2
    dP/dm = 2(250)(m^2-1)
    dP/dm = 500m
    dP/dm = 500 x 0.92
    dP/dm = 460

    ReplyDelete
  3. 3) i. y = 4 x^2 + 7x + 3
    dy/dx = 2(4)(x^2-1) + 1(7)(x^1-1)
    dy/dx = 8x + 7

    ii. y = 4 x^-6 - 5x^3 + x
    dy/dx = -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)
    dy/dx = -24x^-7 - 15x^2 + 1

    ReplyDelete
  4. 4)(a) y=6 sin Ө
    dy/dx= 6 cos Ө

    (b) y= 7 sinӨ + 3 cosӨ
    dy/dx= 7 cosӨ - 3 sinӨ

    ReplyDelete
  5. Find the derivative of the following polynomials:
    i. y = 4 x^2 + 7 x + 3

    ANSWER
    y = 4x^2 + 7x + 3
    dy/dx = 8x + 7

    This is done by multiplying the base (4x) by the power (2) and then minusing 1 from the power

    The second term 7x, here x is to the power of one so the differential of 7x is 7

    The third term 3 is a constant so the differatial is 0

    ReplyDelete
  6. Find the derivative of the following polynomials:
    ii. y = 4x^ -6 - 5x^3 + x

    y = 4x^ -6 - 5x^3 + x

    dy/dx = -24x^-7 - 15x^2 + 1

    ReplyDelete
  7. 9)displacement given by
    y = 1.85sin t
    differentiating the above equation gives an expression for velocity
    dy/dx = 1.85cos t
    velocity = 1.85cos t
    t = 0.025
    v = 1.85cos (0.025)
    =1.85
    velocity = 1.85cm/s^2

    ReplyDelete
  8. 12)
    displacement given by
    y = 2t^1.5 - cos t
    differentiating the above equation gives velocity
    dy/dx = (1.5)(2)t^0.5 + sin t
    velocity when t = 15
    velocity = (1.5)(2)(15^0.5) + sin15
    11.9 cm/s^2

    ReplyDelete
  9. (1)

    find the two equations:
    40 + 10j N .............1
    50 – 25j N..............2

    using the elimination method; eq1 + eq2;
    =>(40 + 50) + (10jN - 25jN)
    =>90 + (-15jN)
    =>90 - 15jN

    ReplyDelete
  10. (2)
    P = 500 + 250 m^2
    dP/dm = 2(250)m^2-1
    = 500m

    when m = 0.92;
    dP/dm = 500 (0.92)
    = 460

    ReplyDelete
  11. (3)

    i) y = 4 x^2 + 7 x + 3
    dy/dx = (2)4x^(2-1) + 7
    = 8x + 7

    ReplyDelete
  12. (3)

    ii)
    y = 4 x^ -6 - 5 x^3 + x
    dy/dx= (-6)4x^(-6-1) -(3)5x^(3-1) + 1
    = -24x^-7 - 15x^2 + 1

    ReplyDelete
  13. (4)
    a. y = 6 sin Ө
    dy/dӨ = 6 cos Ө

    ReplyDelete
  14. (4)
    b. y = 7 sin Ө + 3 cos Ө

    dy/dӨ of sin Ө = cos Ө
    dy/dӨ of cos Ө = -sin Ө

    dy/dӨ = 7 cos Ө - 3 sin Ө

    ReplyDelete
  15. (1)
    For bornagain16
    You cannot use the elimination method of solving simultaneous equations to find the resultant force because the two statements are vectors and not equations (there is no equality sign).

    To find the resultant force you need to add the two forces (tension) which are represented by vectors.

    40 +10j
    50 -25j
    We simply add all the common terms together

    Resultant force in N = 90 - 15j

    so even though your final answer was accurate your working or your arrival to your answer made little or no sense. sorry.

    ReplyDelete
  16. nu(2) p= 500+250m^2
    m= .92
    500(.92)= 460

    ReplyDelete
  17. ques 6
    y = x^3
    integral y = { x^4/4} between x = 1 and x =2
    [(1^4/4) - (2^4/4)]
    0.25 - 4
    -3.75
    Area of curve bounded between x=1 & x=2
    = -3.75 squared units
    The negative value just indicates that the area is under the x-axis.

    ReplyDelete
  18. no.3(i)
    y=4x^2+7x+3
    dy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
    dy/dy= 8x+7

    ReplyDelete
  19. no.3(ii)
    y=4x^-6-5x^3+x
    dy/dx=(4*-6)x^(-6-1)-(5*3)x^(3-1)+x^(1-1)
    dy/dx= -24x^-7-15x^2

    ReplyDelete
  20. 2) P=500+250m^2
    dp/dm=500m
    =500(0.92)
    =460

    ReplyDelete
  21. 3)Find the derivatives of polynomials
    i)y=4x^2+7x+3
    dy/dx=8x+7

    ii)y=4x^-6-5x^3+x
    dy/dx=-24x^-7-15x^2+1
    or
    dy/dx=-24/x^7-15/x^-2+1

    ReplyDelete
  22. 4)Derivative of foll trigs
    a) y=6sinQ
    dy/dx=6cosQ

    b)y=7sinQ+3cosQ
    dy/dx=7cosQ-3sinQ

    ReplyDelete
  23. 5)Determine the gradient
    a)s=2t^3-5t^2+4 (where t is -1,-3 respectively)
    ds/dt=6t^2-10t (-1)
    =6(-1)^2-10(-1)
    =6+10
    =16
    or
    =6(-3)^2-10(-3)
    =54+30
    =84

    b)y=5sinQ whereQ=38degrees
    dy/dx=5cosQ
    =5cos38
    =3.94

    ReplyDelete
  24. question 2

    P=500 + 250m^2

    dP/dm=500m

    when m=0.92

    dP/dm=500(0.92)
    dP/dm=460

    ReplyDelete
  25. For question 8

    E=k/(r2) or k(2r^-1)
    dE/dr=k(-2r^-2) or k/-2r^2

    ReplyDelete
  26. Question 9

    y=1.85 sin t
    dy/dx=1.85cos t where t is 0.025s
    =1.85 cos (0.025s)
    v=1.85cm/s

    ReplyDelete
  27. Coming to a conclusion, we have
    l=(38-2x), w=(24-2x) and h=(2x)
    v(Is given by multiplying l*w*h)

    v=(912-48x-76x+4x^2)(2x)
    =1824x-96x^2-152x^2+8x^3
    dv/dx=24x^2-496x+1824x

    where x=4cm
    dv/dx=24(4)^2-496(4)+1824(4)
    =384-1984+7296
    =5696

    ReplyDelete
  28. Question 12

    y=2t^1.5-cos t
    dy/dt=3t^0.5+sin t where t=15s
    =3(15)^0.5+sin15
    =11.6 + 0.3
    V=11.9cm/s

    ReplyDelete
  29. This comment has been removed by the author.

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  30. This comment has been removed by the author.

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  31. 2. P = 500 + 250 m^2 ,
    m = 0.92

    dP/dm = 500 m
    dP/dm for (m=0.92) = 500 (0.92)
    dP/dm = 460

    ReplyDelete
  32. 3. the derivatives can be found by integrating the equations

    ReplyDelete
  33. i. y = 4 x^2 + 7 x + 3
    y = (4x3)/3 + (7x2)/2 + 3x

    ReplyDelete
  34. ii. y = 4 x^ -6 - 5 x^3 + x
    y = (4x^-5)/-5 - (5x^4)/4 + (x^2)/2

    ReplyDelete
  35. 4.a. y = 6 sin Ө
    derivative y = 6 cos Ө

    ReplyDelete
  36. 4b. y = 7 sin Ө + 3 cos Ө
    derivative y = 7 cos Ө - 3 sin Ө

    ReplyDelete
  37. 5 a) s = 2t^3 - 5t^2 + 4 (-1,-3)
    ds/dt = 6t^2 - 10t
    ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
    ds/dt = 84

    ReplyDelete
  38. b) y = 5 sin Ө where Ө = 38◦
    dy/d sinӨ = 5 cos Ө
    dy/ d sinӨ where Ө= 38 = 5 cos(38)
    dy/ d sinӨ = 3.94

    ReplyDelete
  39. This comment has been removed by the author.

    ReplyDelete
  40. for ques 3
    y = 4 x^2 + 7 x + 3
    dy/dx = (2)4x^(2-1) + 7
    =8x + 7

    ReplyDelete
  41. #2)
    P = 500 + 250 m^2
    m= 0.92

    dP/dm = 2(250)(m^2-1)
    dP/dm = 500m

    dP/dm = 500 x 0.92
    dP/dm = 460

    ReplyDelete
  42. for ques 3 ii)
    y=4x^-6-5x^3+x
    dy/dx=-24x^-7-15x^2+1 or
    dy/dx=-24/x^7-15/x^-2+1

    ReplyDelete
  43. 3)i)
    y = 4 x^2 + 7x + 3
    dy/dx = 2(4)(x^2-1) + 1(7)(x^1-1)
    dy/dx = 8x + 7

    ReplyDelete
  44. 3)ii.
    y = 4 x^-6 - 5x^3 + x
    dy/dx = -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)
    dy/dx = -24x^-7 - 15x^2 + 1

    ReplyDelete
  45. 4)(b)
    y= 7 sinӨ + 3 cosӨ
    dy/dx= 7 cosӨ - 3 sinӨ

    ReplyDelete
  46. 5) a)
    s = 2t^3 - 5t^2 + 4 (-1,-3)

    ds/dt = 6t^2 - 10t
    ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
    ds/dt = 84

    ReplyDelete
  47. 5)b)
    y = 5 sin Ө where Ө = 38◦

    dy/d sinӨ = 5 cos Ө
    dy/ d sinӨ where Ө= 38 = 5 cos(38)
    dy/ d sinӨ = 3.94

    ReplyDelete
  48. 6)
    y = x^3
    integral y = { x^4/4} between x = 1 and x =2
    [(1^4/4) - (2^4/4)]
    0.25 - 4
    -3.75

    Area of curve between x=1 & x=2
    = -3.75 ^2 untis

    ReplyDelete
  49. 2) p= 500 +250 m^2
    dp/dm= (2*250)m^(2-1)
    dp/dm= 500m

    sub m=0.92 to find the instanteous rate of change of P

    dp/dm= 500(0.92)
    dp/dm= 460

    ReplyDelete
  50. 9)
    y=1.85 sin t
    dy/dx=1.85cos t where t is 0.025s
    =1.85 cos (0.025s)
    v=1.85cm/s

    ReplyDelete
  51. 2)
    p= 500 +250 m^2
    dp/dm= (2*250)m^(2-1)
    dp/dm= 500m
    sub m=0.92
    dp/dm= 500(0.92)
    dp/dm= 460

    ReplyDelete
  52. 3(i)
    y=4x^2+7x+3
    dy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
    dy/dy= 8x+7

    ReplyDelete
  53. hmmmm..bornagain16 u cannot use simultaneous equations when you are asked to find the resultion force..your method is wrong for number 1.

    ReplyDelete
  54. 1)40 + 10j N
    50 – 25j N
    to find the resulting force you add the like terms

    40+50=90
    10j N+ - 25j N = -15J N

    ANSWER= 90 - 15j N

    ReplyDelete
  55. (2)
    P = 500 + 250 m^2
    dp/dm = 2*250 m^2-1
    = 500 m

    m = 0.92
    dp/dm = 500*0.92
    = 460

    ReplyDelete
  56. 3)
    y= 4 x^2 + 7x + 3

    dy/dx= 2(4)(x^2-1) + 1(7)(x^1-1)

    dy/dx= 8x + 7

    y= 4 x^-6 - 5x^3 + x

    dy/dx= -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)

    dy/dx= -24x^-7-15x^2

    ReplyDelete
  57. 9)
    y = 1.85sin t

    dy/dx = 1.85cost

    velocity = 1.85cos t

    t = 0.025

    velocity = 1.85cos (0.025)
    =1.85

    velocity = 1.85cm/s^2

    ReplyDelete
  58. 2). P=500+250m^2
    dP/dm=500m

    when m= 0.92
    dP/dm=500(0.92)
    = 460

    ReplyDelete
  59. 3) (2)
    y=4x^-6-5x^3+3
    dy/dx= -24x-7-15x^2+1

    ReplyDelete
  60. (4)b) y= 7 sin Ө + 3 cos Ө
    dy/dӨ= 7 cos Ө - 3 sin Ө

    ReplyDelete
  61. (7) (i) 4x^6 + 3x^-4 +1
    ∫ (4x^(6+1))/(6+1) + (3x^(-4+1))/(-4+1) +x +c
    ∫ ( 4x^7)/7 + (3x^-3)/-3 + x +c
    ∫ ( 4x^7)/7 + (-x^-3) +x +c

    ReplyDelete
  62. (8) E= k/ (2r)
    E= k2r^-1
    dE/dr= (2*-1)kr^(-1-1)
    dE/dr= -2kr^-2
    dE/dr= -2k/r^2

    ReplyDelete
  63. question 2

    P=500 + 250m^2
    dP/dm=500m
    m=0.92

    dP/dm=500(0.92)
    =460

    ReplyDelete
  64. This comment has been removed by the author.

    ReplyDelete
  65. This comment has been removed by the author.

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  66. number4
    a.

    y=6 sin Ө
    dy/dx= 6 cos Ө

    ReplyDelete
  67. number4
    b.

    y = 7 sin Ө + 3 cos Ө

    dy/dӨ of sin Ө = cos Ө
    dy/dӨ of cos Ө = -sin Ө

    dy/dӨ = 7 cos Ө - 3 sin Ө

    ReplyDelete
  68. number9

    y=1.85 sin(t)
    dy/dx=1.85cos(t)

    where t is 0.025s
    therefore
    dy/dx=1.85 cos (0.025s)

    ReplyDelete
  69. (1)

    find the two equations:
    40 + 10j N .............1
    50 – 25j N..............2

    ReplyDelete
  70. This comment has been removed by the author.

    ReplyDelete
  71. question 1 cont..
    using the elimination method; eq1 + eq2;
    =>(40 + 50) + (10jN - 25jN)
    =>90 + (-15jN)
    =>90 - 15jN

    ReplyDelete
  72. (2)
    P = 500 + 250 m^2
    dP/dm = 2(250)m^2-1
    = 500m

    when m = 0.92;
    dP/dm = 500 (0.92)
    = 460

    ReplyDelete
  73. 3) i. y = 4 x^2 + 7x + 3
    dy/dx = 2(4)(x^2-1) + 1(7)(x^1-1)
    dy/dx = 8x + 7

    ii. y = 4 x^-6 - 5x^3 + x
    dy/dx = -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)
    dy/dx = -24x^-7 - 15x^2 + 1

    ReplyDelete
  74. 4.a. y = 6 sin Ө
    derivative y = 6 cos Ө
    4b. y = 7 sin Ө + 3 cos Ө
    derivative y = 7 cos Ө - 3 sin Ө

    ReplyDelete
  75. 5 a) s = 2t^3 - 5t^2 + 4 (-1,-3)
    ds/dt = 6t^2 - 10t
    ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
    ds/dt = 84

    ReplyDelete
  76. 5 a) s = 2t^3 - 5t^2 + 4 (-1,-3)
    ds/dt = 6t^2 - 10t
    ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
    ds/dt = 84

    ReplyDelete
  77. 9)displacement given by
    y = 1.85sin t
    differentiating the above equation gives an expression for velocity
    dy/dx = 1.85cos t
    velocity = 1.85cos t
    t = 0.025
    v = 1.85cos (0.025)
    =1.85
    velocity = 1.85cm/s^2

    ReplyDelete
  78. 12)
    displacement given by
    y = 2t^1.5 - cos t
    differentiating the above equation gives velocity
    dy/dx = (1.5)(2)t^0.5 + sin t
    velocity when t = 15
    velocity = (1.5)(2)(15^0.5) + sin15
    11.9 cm/s^2

    ReplyDelete
  79. ques 6
    y = x^3
    integral y = { x^4/4} between x = 1 and x =2
    [(1^4/4) - (2^4/4)]
    0.25 - 4
    -3.75
    Area of curve bounded between x=1 & x=2
    = -3.75 squared units
    The negative value just indicates that the area is under the x-axis.

    ReplyDelete
  80. (1)

    find the two equations:
    40 + 10j N .............1
    50 – 25j N..............2

    using the elimination method; eq1 + eq2;
    =>(40 + 50) + (10jN - 25jN)
    =>90 + (-15jN)
    =>90 - 15jN

    ReplyDelete
  81. (2)
    P = 500 + 250 m^2
    dP/dm = 2(250)m^2-1
    = 500m

    when m = 0.92;
    dP/dm = 500 (0.92)
    = 460

    ReplyDelete
  82. (3)

    i) y = 4 x^2 + 7 x + 3
    dy/dx = (2)4x^(2-1) + 7
    = 8x + 7

    ReplyDelete
  83. (3)

    ii)
    y = 4 x^ -6 - 5 x^3 + x
    dy/dx= (-6)4x^(-6-1) -(3)5x^(3-1) + 1
    = -24x^-7 - 15x^2 + 1

    ReplyDelete
  84. (4)
    a. y = 6 sin Ө
    dy/dӨ = 6 cos Ө

    ReplyDelete
  85. (4)
    b. y = 7 sin Ө + 3 cos Ө

    dy/dӨ of sin Ө = cos Ө
    dy/dӨ of cos Ө = -sin Ө

    dy/dӨ = 7 cos Ө - 3 sin Ө

    ReplyDelete
  86. (7) (i) 4x^6 + 3x^-4 +1
    ∫ (4x^(6+1))/(6+1) + (3x^(-4+1))/(-4+1) +x +c
    ∫ ( 4x^7)/7 + (3x^-3)/-3 + x +c
    ∫ ( 4x^7)/7 + (-x^-3) +x +c

    ReplyDelete
  87. (8) E= k/ (2r)
    E= k2r^-1
    dE/dr= (2*-1)kr^(-1-1)
    dE/dr= -2kr^-2
    dE/dr= -2k/r^2

    ReplyDelete
  88. question 1
    40 + 10j N .............1
    50 – 25j N..............2

    elimination method; equation1 + equation2:
    =(40 + 50) + (10jN - 25jN)
    =90 + (-15jN)
    =90 - 15jN

    ReplyDelete
  89. y= 4x^2 +7x +3
    dy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
    dy/dy= 8x+7
    (2) y=4x^-6 - 5x^3 +x
    dy/dx= (4*-6)x^(-6-1) - (5*3)x^(3-1) + x^(1-1)
    dy/dx= -24x^-7 - 15x^2

    ReplyDelete
  90. question 2
    P = 500 + 250 m^2
    dP/dm = 2(250)(m^2-1)
    dP/dm = 500m
    dP/dm = 500 x 0.92
    dP/dm = 460

    ReplyDelete
  91. question 3(i)
    y = 4 x^2 + 7 x + 3
    dy/dx = (2)4x^(2-1) + 7
    = 8x + 7

    ReplyDelete
  92. question 3(ii)
    y = 4 x^ -6 - 5 x^3 + x
    dy/dx= (-6)4x^(-6-1) -(3)5x^(3-1) + 1
    = -24x^-7 - 15x^2 + 1

    ReplyDelete
  93. (i) 4x^6 + 3x^-4 +1
    ∫ (4x^(6+1))/(6+1) + (3x^(-4+1))/(-4+1) +x +c
    ∫ ( 4x^7)/7 + (3x^-3)/-3 + x +c
    ∫ ( 4x^7)/7 + (-x^-3) +x +c

    ReplyDelete
  94. question 5(a)
    s = 2t^3 - 5t^2 + 4 (-1,-3)
    ds/dt = 6t^2 - 10t
    ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
    ds/dt = 84

    ReplyDelete
  95. question 5(b)
    y=5sinQ whereQ=38degrees
    dy/dx=5cosQ
    =5cos38
    =3.94

    ReplyDelete
  96. question 12
    y=2t^1.5-cos t
    dy/dt=3t^0.5+sin t where t=15s
    =3(15)^0.5+sin15
    =11.6 + 0.3
    V=11.9cm/s

    ReplyDelete
  97. question 9
    y = 1.85sin t
    dy/dx = 1.85cos t
    v= 1.85cos t
    t = 0.025
    v = 1.85cos (0.025)
    =1.85
    velocity = 1.85cm/s^2

    ReplyDelete
  98. question 8
    E=k/(r2) or k(2r^-1)
    dE/dr=k(-2r^-2)
    or k/-2r^2

    ReplyDelete
  99. 9)
    y = 1.85sin t

    dy/dx = 1.85cost

    velocity = 1.85cos t

    t = 0.025

    velocity = 1.85cos (0.025)
    =1.85

    velocity = 1.85cm/s^2

    ReplyDelete
  100. y = 4 x^2 + 7 x + 3

    ANSWER
    y = 4x^2 + 7x + 3
    dy/dx = 8x + 7

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  101. ii. y = 4x^ -6 - 5x^3 + x

    y = 4x^ -6 - 5x^3 + x

    dy/dx = -24x^-7 - 15x^2 + 1

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  102. This comment has been removed by the author.

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  103. y = 4 x^ -6 - 5 x^3 + x
    = -24x^-7 - 15x^2 + 1
    number 3 part 2

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  104. number 3 part 1
    y = 4 x^2 + 7 x + 3
    = 8x + 7

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  105. 2)

    P = 500 + 250 m^2
    dP/dm = 2(250m)

    m = 0.92
    dP/dm = 2(25(0.92))
    dP/dm = 46

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  106. 8)
    E = k/(r2), where k is a constant
    dE/dr = -k2r^-2

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  107. 12)
    y = 2t^1.5 – cos t
    dy/dt = 3^0.5 + sint

    vertical velocity for t = 15s.
    dy/dt = 3^0.5 + sin(15)
    dy/dt = 1.73 + 0.26
    dy/dt = 1.99

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  108. 2.
    P = 500 + 250 m^2 ,
    m = 0.92

    dP/dm = 500 m
    dP/dm for (m=0.92) = 500 (0.92)
    dP/dm = 460

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  109. 3 part 1
    y = 4 x^2 + 7 x + 3
    = 8x + 7

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  110. y= 5 sin Ө where Ө = 38◦
    dy/d sinӨ = 5 cos Ө
    dy/ d sinӨ where Ө= 38 = 5 cos(38)
    dy/ d sinӨ = 3.94

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  111. Find the derivative of the following polynomials:
    i. y = 4 x^2 + 7 x + 3

    ANSWER
    y = 4x^2 + 7x + 3
    dy/dx = 8x + 7

    This is done by multiplying the base (4x) by the power (2) and then minusing 1 from the power

    The second term 7x, here x is to the power of one so the differential of 7x is 7

    The third term 3 is a constant so the differatial is 0

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  112. 9)displacement given by
    y = 1.85sin t
    differentiating the above equation gives an expression for velocity
    dy/dx = 1.85cos t
    velocity = 1.85cos t
    t = 0.025
    v = 1.85cos (0.025)
    =1.85
    velocity = 1.85cm/s^2

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  113. 2.
    P = 500 + 250 m^2 ,
    m = 0.92

    dP/dm = 500 m
    dP/dm for (m=0.92) = 500 (0.92)
    dP/dm = 460

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  114. y = 4 x^2 + 7 x + 3

    y = 4x^2 + 7x + 3
    dy/dx = 8x + 7

    ReplyDelete