Wednesday, March 4, 2009

Trapezoidal Rule

  1. A plate cam for opening and closing a valve has an irregular shape. The widths of the face of the cam are 2cm intervals. Find the area of the face of the cam if the length respectfully are 3, 4, 3, 3, 2 and 0. Draw a diagram in a paper then explain your answer.
  2. From a satellite photograph of a lake, the widths were 26 kmm intervals. The lengths were 0, 45, 50, 60, 61, 66, 74, 87, 76, 66, 86, 77, 0 respectfully. Draw a diagram in a paper then explain your answer.
  3. The widths of a kidney-shaped swimming pool were measured as 20m intervals. Calculate the surface area of the pool given the respective lengths were 0, 6, 7, 8, 6, 5, 4, 5 and 0.
  4. The widths of a plane wing swimming pool were measured as 0.30m intervals. Calculate the surface area of the pool given the respective lengths were 0, .16, .23, .32, .35, .3 and .2.
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38 comments:

  1. WeezyMarch 5, 2009 at 5:02 PM

    solution 1

    As miss explained the trapeziodal rule to us:
    brief-

    Appox. Area = (width of strip)[(1/2)(sum of first and last oridnate) + (sum of other ordinates)].
    so
    from question 1:
    width= 2
    first ordinate= 3
    last ordinate= 0
    sum of other ordinates = 12

    approx. area = 2[(3+0) + (4+3+3+2)]
    approx. area = 2(1.5 + 12)
    approx. area = 2 * 13.5
    approx. area = 27

    answer = 27 sq. cm

    ReplyDelete
  2. WeezyMarch 5, 2009 at 6:05 PM

    solution 2

    using the same trapeziodal rule from above:

    Appox. Area = (width of strips)[(1/2)(sum of first and last oridnate) + (sum of other ordinates)].
    so
    approx. area = (26)*[(1/2)(0+0) + (748)
    approx. area = 26 * 748
    approx area = 19 448
    answer = 19 448 sq. kmm

    ReplyDelete
  3. WeezyMarch 5, 2009 at 6:15 PM

    solution 3

    again trapezoidal rule for this question:

    approx. area = (width of strips)[(1/2)(sum of frist and last oridnate) + (sum of other ordinates)

    approx. area = (20)[(1/2)(0+0) + (41)
    approx. area = (20)[ (41)]
    approx. area = 20 * 41
    approx. area = 820

    answer = 820 sq. m

    ReplyDelete
  4. WeezyMarch 5, 2009 at 7:27 PM

    solution 4

    i will use the trapezoidal rule to determine the solution to this problem also which states briefly:

    approx. area = (width of strips)[(1/2)(sum of first last ordinate) + (sum of other ordinate)]

    approx. area = (.30)[(1/2)(0+.2) + (1.36)]
    approx. area = (.30) * [(0.1 + 1.36)]
    approx. area = .30 * 1.46
    approx. area = .438 M

    ANSWER = 0.438 m

    ReplyDelete
  5. WeezyMarch 5, 2009 at 7:29 PM

    if anyone has problems or questions talk to me...word!

    ReplyDelete
  6. sky captainMarch 6, 2009 at 7:39 AM

    solution for Q.1.

    As miss explained in class, the trapeziodal rule:


    Area = (width of strip)[(1/2)(sum of first and last oridnate) + (sum of other ordinates)].

    from question 1:
    width= 2
    first ordinate= 3
    last ordinate= 0
    sum of other ordinates = 12

    approx. area = 2[(3+0) + (4+3+3+2)]
    approx. area = 2(1.5 + 12)
    approx. area = 2 * 13.5
    approx. area = 27

    answer = 27 sq. cm

    ReplyDelete
  7. sky captainMarch 6, 2009 at 7:53 AM

    solution for Q.2.

    using the same trapezoidal rule from question 1 above:

    Area = (width of strips)[(1/2)(sum of first and last ordinate) + (sum of other ordinates)].

    area = (26)*[(1/2)(0+0) + (748)
    area = 26 * 748
    area = 19 448
    answer = 19 448 sq. km

    ReplyDelete
  8. sky captainMarch 6, 2009 at 7:57 AM

    solution For Q.3.

    again using back the same trapezoidal rule from above:

    area = (width of strips)[(1/2)(sum of first and last ordinate) + (sum of other ordinates)

    area = (20)[(1/2)(0+0) + (41)
    area = (20)[ (41)]
    area = 20 * 41
    area = 820

    answer = 820 sq. m

    ReplyDelete
  9. sky captainMarch 6, 2009 at 8:01 AM

    solution for Q.4.
    using the trapezoidal rule to determine the solution.

    area = (width of strips)[(1/2)(sum of first last ordinate) + (sum of other ordinate)]

    area = (.30)[(1/2)(0+.2) + (1.36)]
    area = (.30) * [(0.1 + 1.36)]
    area = .30 * 1.46
    area = .438 M

    ANSWER = 0.438 m

    ReplyDelete
  10. WeezyMarch 6, 2009 at 4:07 PM

    hmmm!!

    ReplyDelete
  11. the dawgMarch 7, 2009 at 7:10 PM

    SOLUTION 1
    Width of each strip= 2cm
    1st ordinate= 3cm
    last ordinate= 0cm
    remaining ordinates= 4cm, 3cm, 3cm, 2cm

    TRAPEZOIDAL RULE=
    AREA OF OBJECT= width of strip *[(1/2 the sum of 1st ordinate + last ordinate)+ sum of remaining ordinates]
    = 2*[1/2(3+0)]+ 4+3+3+2
    = 2*1.5+4+3+3+2
    =2*13.5
    =27cm sq

    ReplyDelete
  12. the dawgMarch 7, 2009 at 7:11 PM

    SOLUTION 2

    Width of each strip= 26mm
    1st ordinate= 0mm
    last ordinate= 0mm
    remaining ordinates=45mm,50mm,60mm,61mm,66mm,
    74mm,87mm,76mm,66mm,86mm,77mm

    TRAPEZOIDAL RULE=
    AREA OF OBJECT= width of strip *[(1/2 the sum of 1st ordinate + last ordinate)+ sum of remaining ordinates]

    = 26*[1/2(0+0)
    45+50+60+61+66+74+87+76+66+86+77
    =26*(0+748)
    =26*748
    =19448mm sq

    ReplyDelete
  13. the dawgMarch 7, 2009 at 7:11 PM

    SOLUTION 3

    Width of each strip= 20m
    1st ordinate= 0m
    last ordinate= 0m
    remaining ordinates=6m,7m,8m,6m,5m,4m,5m

    TRAPEZOIDAL RULE=
    AREA OF OBJECT= width of strip *[(1/2 the sum of 1st ordinate + last ordinate)+ sum of remaining ordinates

    =20*[1/2(0+0)+6m+7m+8m+6m+5m+4m+5m
    =20*(0+6m+7m+8m+6m+5m+4m+5m)
    =20*41
    =820m

    ReplyDelete
  14. the dawgMarch 7, 2009 at 7:12 PM

    SOLUTION 4

    Width of each strip= 0.30m
    1st ordinate= 0m
    last ordinate= 0.2m
    remaining ordinates= 0.16m, 0.23m, 0.32m,0.35m, 0.3m

    TRAPEZOIDAL RULE=
    AREA OF OBJECT= width of strip *[(1/2 the sum of 1st ordinate + last ordinate)+ sum of remaining ordinates

    =0.30*(1/2(0+0.2)0.16+0.23+0.32+0.35+0.3
    =0.30*(0.1+0.16+0.23+0.32+0.35+0.3)
    =0.30*1.46
    =0.438m

    All problems were solved using the formula:

    TRAPEZOIDAL RULE=
    AREA OF OBJECT= width of strip *[(1/2 the sum of 1st ordinate + last ordinate)+ sum of remaining ordinates

    If there are any mistakes please show me.

    ReplyDelete
  15. WeezyMarch 8, 2009 at 6:06 PM

    i tink ur gud...

    ReplyDelete
  16. fellaMarch 8, 2009 at 7:25 PM

    SOLUTION ONE

    Trapezoidal Rule = Area=width of strip X (1/2 sum of the first + last ordinate)+(sum of middle ordinates)

    Area=2cm X [1/2(3+0)] + (4+3+3+2)
    =2cm X [1/2(3)] + 12
    =2cm X (1.5 + 12)
    =2cm X (13.5)
    =27cm2

    ReplyDelete
  17. fellaMarch 8, 2009 at 7:27 PM

    hhmmm should i use middle ordinates?

    ReplyDelete
  18. WeezyMarch 8, 2009 at 7:45 PM

    doesn't matter..

    ReplyDelete
  19. Fresh PrinceMarch 9, 2009 at 7:40 AM

    solution 1

    width=2cm

    lengths=3,4,3,3,2&0

    trapezoidal rule=width*(1/2 the sum of the first&last ordinate)+(sum of middle ordinates)

    therefor,
    area of irregular shape= 2*[1/2(3+0)]+(4+3+3+2)

    = 2*[1.5]+(12)
    = 2*(13.5)
    = 27squared cm

    ReplyDelete
  20. starMarch 9, 2009 at 8:38 AM

    Question 1
    Area=(width of strip)(1/2)(sum of first and last ordinate)+sum of rest of ordinates
    First ordinate is 3
    last ordinate is 0
    width is 2
    sum of other ordinates is 12

    approx area=2(3+0)+(4+3+3+2)
    approx area =2(1.5+12)
    approx area =2*13.5
    approx area=27

    ReplyDelete
  21. starMarch 9, 2009 at 8:54 AM

    question 2

    last ordinate=0
    first ordinate=0
    width of each strip=45
    remaining ordinates=45mm,50mm,60mm,61mm,66mm,74mm,87mm,76mm,66,86mm

    trapezodial rule= area of object=width of strip
    (1/2 the sum of first ordinate+ plus sum of last ordinates

    26*(1/2 0+0)
    45+50+60+61+66+74+86+66+77
    =26*(0+748
    19448

    ReplyDelete
  22. forzaMarch 9, 2009 at 5:42 PM

    1
    Area = 2/2 (3+8+6+6+4)
    =27 cm ^2

    ReplyDelete
  23. forzaMarch 9, 2009 at 5:47 PM

    2
    Area = 26/2(90+100+120+122+132+148+174+152+132+172+154)
    =19448km^2

    ReplyDelete
  24. forzaMarch 9, 2009 at 5:52 PM

    3
    Area= 10(12+14+16+12+10+8+10)
    =820m^2

    ReplyDelete
  25. forzaMarch 9, 2009 at 5:56 PM

    4
    Area =.15(.2+.32+.46+.64+.7+.6)
    =0.438m^2

    ReplyDelete
  26. AnonymousMarch 9, 2009 at 6:26 PM

    This comment has been removed by a blog administrator.

    ReplyDelete
  27. AnonymousMarch 9, 2009 at 6:30 PM

    This comment has been removed by a blog administrator.

    ReplyDelete
  28. AnonymousMarch 9, 2009 at 6:43 PM

    solution 2

    width=26km
    first lenght=0
    last lenght=0
    area=26/2[(0+0)+(2*45)+(2*50)+(2*60)+(2*61)+(2*66)+(2*74)+(2*87)+(2*76)+(2*66)+(2*86)+(2*77)]
    =19448km^2

    ReplyDelete
  29. AnonymousMarch 9, 2009 at 6:50 PM

    solution 3

    width=20m
    first lenght=0
    last lenght=0
    area=20/2[(0+0)+(2*6)+(2*7)+(2*8)+(2*6)+(2*5)+(2*4)+(2*5)]
    =820m^2

    ReplyDelete
  30. AnonymousMarch 9, 2009 at 6:59 PM

    solution 4

    width=.30m
    first lenght=0
    last lenght=.2
    area=.30/2[(0+.2)+(2*.16)+(2*.23)+(2*.32)+(2*.35)+(2*.3)
    =0.438m^2

    ReplyDelete
  31. AnonymousMarch 9, 2009 at 7:08 PM

    Solution 1

    width=2cm
    first lenght=3
    last lenght=0
    area=2/2[(3+0)+(2*4)+(2*3)+(2*3)+(2*2)]
    =27cm^2

    ReplyDelete
  32. AnonymousMarch 9, 2009 at 7:09 PM

    solution 2

    width=26km
    first lenght=0
    last lenght=0
    area=26/2[(0+0)+(2*45)+(2*50)+(2*60)+(2*61)+(2*66)+(2*74)+(2*87)+(2*76)+(2*66)+(2*86)+(2*77)]
    =19448km^2

    ReplyDelete
  33. AnonymousMarch 9, 2009 at 7:10 PM

    solution 3

    width=20m
    first lenght=0
    last lenght=0
    area=20/2[(0+0)+(2*6)+(2*7)+(2*8)+(2*6)+(2*5)+(2*4)+(2*5)]
    =820m^2

    ReplyDelete
  34. AnonymousMarch 9, 2009 at 7:10 PM

    solution 4

    width=.30m
    first lenght=0
    last lenght=.2
    area=.30/2[(0+.2)+(2*.16)+(2*.23)+(2*.32)+(2*.35)+(2*.3)
    =0.438m^2

    ReplyDelete
  35. R@ptorMarch 10, 2009 at 8:55 AM

    question4

    The trapezoidal rule is used to determine the solution to this problem also which states briefly:

    approx. area = (width of strips)[(1/2)(sum of first last ordinate) + (sum of other ordinate)]

    approx. area = (.30)[(1/2)(0+.2) + (1.36)]
    approx. area = (.30) * [(0.1 + 1.36)]
    approx. area = .30 * 1.46
    approx. area = .438 M

    ANSWER = 0.438 m

    ReplyDelete
  36. computationMarch 12, 2009 at 5:50 PM

    using trapezoidal rule

    w=0.30m
    lenghts= 0, .16, .23, .32, .35, .3, .2

    area = (width of strips)[(1/2)(sum of first last ordinate) + (sum of other ordinate)]

    =.30(1/2(0+0.2)+(.16+.23+.32+.35+.3)

    calcaulating out the brackets first

    =.30*(0.1)+(1.36)
    =.30*1.46
    =.438m

    the surface area of the pool=0.438m

    ReplyDelete
  37. computationMarch 12, 2009 at 8:24 PM

    thats question 4

    ReplyDelete
  38. computationMarch 12, 2009 at 8:36 PM

    using trapezoidal rule
    from the equation
    the width was at 26kmm intervals ...
    The lengths were 0, 45, 50, 60, 61, 66, 74, 87, 76, 66, 86, 77, 0.

    area = (width of strips)[(1/2)(sum of first last ordinate) + (sum of other ordinate)]

    =26(1/2(0+0)+(748)
    =26*748
    =19448
    =19448kmm^2

    ReplyDelete

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