Wednesday, March 4, 2009

Simpson's Rule questions

  1. The widths of a plane wing 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. Draw the diagram on paper then write and explain your answer.
  2. The ariel view of a lake with 1.5km intervals had lengths respectfully of 0, 4.8, 5.7, 10.5, 15.2, 18.5, 18.8, 17.9, 11.3, 8.8 and 3.1 Draw the diagram on paper then write and 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. Draw the diagram on paper then write and explain your answer.
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17 comments:

  1. AnonymousMarch 7, 2009 at 10:24 AM

    Q3 given w=20m L's= 0,6,7,8,6,5,4,5,0A=20/3{(0+0)+(4*(6+8+5+5)+(2*(7+6+4))}=20/3(0+4*24+2*17)=866.7m

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  2. AnonymousMarch 7, 2009 at 10:43 AM

    Q2 given interval= 1.5km
    L's= 0, 4.8, 5.7, 10.5, 15.2, 18.5, 18.8, 17.9, 11.3, 8.8, 3.1
    substituting these values into the simpson's rule equation
    A= (1.50/3)*[(0+3.1)+(4*(4.8+10.5+18.5+17.9+8.8)+(2*(5.7+15.2+18.8+11.3)]
    ={(0.5)*[(3.1)+(242)+(102)]}=173.55 km

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  3. AnonymousMarch 7, 2009 at 10:52 AM

    Q1 given w= 0.3m
    L's= 0, 0.16, 0.23, 0.32, 0.35, 0.3, 0.2
    substituting these values into simpson's rule equation
    A= (0.3/3)*[(0+0.2)+(4*(0.16+0.32+0.3)+(2*(0.23+0.35)]
    ={(0.1)*[(0.2)+(3.12)+(1.16)]=0.448m

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  4. starMarch 9, 2009 at 8:31 AM

    since w=20ms
    L's=0,6,7,8,6,5,4,5,0
    A=20/3
    {0+0)+(4*(6+8+5+5)+(2(*(7+6+4)}
    Answer= 20/3(0+4*24+2*17)
    Therefore i get 866.65

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  5. starMarch 9, 2009 at 9:07 AM

    sool 1
    L's=0,0.16,0.23,0.32,0.3,0.2
    Using simpson's rule
    A=(0.3/3)*(0+0.2)=(4*(0.16+0.32+.3)+(2*(.23+.35)
    =(0.1)*(0.2)+(3.12)+(1.16)
    answer gives 0.45m

    ReplyDelete
  6. ravi bMarch 9, 2009 at 6:07 PM

    Q.3
    length =0, 6, 7, 8, 6, 5, 4, 5,0
    using simpson's rule =20/3[(0+0)+2*(7+5+5)+4*6+8+4)
    =112.67m

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  7. shottaMarch 9, 2009 at 6:48 PM

    solution 1
    widths(m) = 0,.16,.23,.32,.35,.3 and.2
    itdth intreval = 0.3m
    using simpsons rule
    A =0.3/3[(0+0.2)+(2*(.23+.35)+(4*(.16+.32+3)]
    =0.448m^2

    ReplyDelete
  8. R@ptorMarch 11, 2009 at 3:52 PM

    Qes 1
    w= 0.3m
    Ls= 0,0.16, 0.23, 0.32, 0.35, 0.3, 0.2
    now sust values into simpson's rule
    gives:

    A= (0.3/3)*[(0+0.2)+(4*(0.16+0.32+0.3)+(2*(0.23+0.35)]
    =(0.1)* (0.2 + 3.12 + 1.16) =0.448m

    ReplyDelete
  9. computationMarch 12, 2009 at 5:08 PM

    question 1........
    w=.30m
    given the lengths= 0,0.16,.23,.32,.35,.3,.2
    now using simpsons rule to slove the equation..
    we get
    A= 0.3/3 (0+0.2)+(2*(.23+.35)+(4*(.16+.32+3)
    when calculated we should get
    = 0.448m^2

    ReplyDelete
  10. computationMarch 12, 2009 at 5:23 PM

    given that a lake with 1.5km intervals has lengths of 0, 4.8, 5.7, 10.5, 15.2, 18.5, 18.8, 17.9, 11.3, 8.8 and 3.1

    using simpson's rule
    A= 1.5/3(0+3.1)+(4*(4.8+10.5+18.5+17.9+8.8)+(2*(5.7+1.52+18.8+11.3)
    when cal..
    = (0.5)*(3.1+242+102)
    =173.55km

    ReplyDelete
  11. computationMarch 12, 2009 at 5:24 PM

    or sorry i forgat to put the number of the question........
    anyways its #2

    ReplyDelete
  12. computationMarch 12, 2009 at 5:33 PM

    question 3.........
    a swimming pool measured at 20m intervals has lengths of 0,6,7,8,6,5,4,5,0

    using simpson's rule
    A= 20/3 (0+0)+ 4(6+8+5+5)+2(7+6+4)
    = 6.667(0+96+34)
    =866.71m

    ReplyDelete
  13. Hellscream!!!March 15, 2009 at 11:15 PM

    This comment has been removed by the author.

    ReplyDelete
  14. Hellscream!!!March 15, 2009 at 11:31 PM

    pretty much all the questions were done, i reviewed some, i really don't like the way u guys brought the like terms together, it got me confused at first because it didn't look much like the simpson's rule equation.

    ReplyDelete
  15. DarkyMarch 17, 2009 at 5:36 AM

    The simpson's rule states that the area A is found by:
    A = 1/3 [first and last ordinates] + 4 [sum of even ordinate] + 2 [sum of the odd ordinates]

    so from the question above the interval is 0.30m
    and the lengths were 0, .16, .23, .32, .35, .3 and .2

    1/0.30 ( 0 + .2) + 2 (.23 + .35 + .3) + 54 ( .16 + .32)

    ReplyDelete
  16. DarkyMarch 17, 2009 at 5:43 AM

    that is not 54 above it was a typing error it is 4 * (.16 + .32)

    Someone correct me if the above steps for the solution is wrong I am not 100 percent certain.

    ReplyDelete
  17. starApril 21, 2009 at 4:53 PM

    question 2)
    width of lake=1.5km

    length"s= 0, 4.8, 5.7,10.5,15.2,18.5,18.8,17.9, 11.3,8.8,3.1

    putting values in simpson's rule.

    A=(1.5/3)*(0+3.1)+(4*(4.8+10.5+18.5+17.5+8.8)+(2*(5.7+15.2+18.8+11.3)

    = 173.6 km

    ReplyDelete

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