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Math Help - Differentiation - Could some one check my work

  1. #1
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    Differentiation - Could some one check my work

    hi, im quite new to calculus and find it taxing but enjoyable. ive been given 4 questions to answer and have had a good go at them. how ever other class mates have got completely different answers to me , could some one pleae just check over them to make sure im going in the correct direct. many thanks
    questions.doc
    Last edited by mr fantastic; March 12th 2011 at 12:24 PM. Reason: Re-titled.
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  2. #2
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    Hi

    If find the same answers but you haven't finished the last exercice
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  3. #3
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    Hi, thanks for the reply, as for the last question. What else do I have to do to it? Do I have to differentiate the previous differentiated formula and put my x values into that formula and use that answer as the x vale for the original formula? Many thanks.
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  4. #4
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    For future reference, many helpers on this forum will not open Word documents, because of the possibility of virus-containing macros. You're better off converting to pdf or jpg.
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  5. #5
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    Quote Originally Posted by djstar View Post
    Hi, thanks for the reply, as for the last question. What else do I have to do to it? Do I have to differentiate the previous differentiated formula and put my x values into that formula and use that answer as the x vale for the original formula? Many thanks.
    You have found the volume V(x) = (2-2x).(2-2x).(x)
    You must determine the value of the piece to be cut out to give the maximum volume
    This can be done by diffrentiating V(x)
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  6. #6
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    Thanks Ackbeet for the advice, ill remember that one.

    ok so ive worked out the volume for the trough is

    Volume = 4x3 8x2 +4x



    [IMG]file:///C:/DOCUME%7E1/LIAMST%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]now i have found dy/dx = 12x2 8x +4


    This has left me with a quadratic so i have put this into the quadratic equation



    and have came up with two x answer's these are x=1 and x =0.33


    do i put the 0.33 into the orignal equation 4(0.333 ) 8(0.332 )+4(0.33) to find the volume which i worked out as 0.592m2


    Or do i have to differentiate the equation again





    d2y/dx2 = 24x - 8


    and substitue 0.33 into the euqation to give the answer -0.08


    and then use this in the original formula 4(-0.083 ) 8(-0.082 )+4(-0.08) = -1.28 m2
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  7. #7
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    V(x) = (2-2x).(2-2x).(x) = 4x(1-x)
    When differentiating V(x), you should not expand it
    V'(x) = 4(1-x)-8x(1-x) = 4(1-x)(1-3x)
    It is easier to find the roots
    V is defined for x between 0 and 1 therefore 1-x > 0
    The sign of V'(x) is the same as 1-3x
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  8. #8
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    thanks for the reply running-gag , im sorry if im a pain for all the replies, but im really new to calculus and im still at secondary schoollearning it so all of these notations and formulas i can find confusing.

    could i just check with you that the correct answer for the volume of the trough is 0.592 metres cubed?
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  9. #9
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    You are right
    The maximum volume is obtained when x=1/3 and V(1/3) = 16/27 = 0.593 m^3
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