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Math Help - Polynomial and derivative problem

  1. #1
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    Polynomial and derivative problem

    Hello everyone,
    this is the exact question



    what I have so far is that:
    P'(x)= 3ax^3 + 2bx^2 + c

    Then from what I understand P'(0)=0
    so then:
    P'(0)=0=3a0^3 + 2b0^2 + c
    so c=0

    however from here on I'm stuck. Can you please help me with the rest.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by momentum View Post
    Hello everyone,
    this is the exact question



    what I have so far is that:
    P'(x)= 3ax^3 + 2bx^2 + c

    Then from what I understand P'(0)=0
    so then:
    P'(0)=0=3a0^3 + 2b0^2 + c
    so c=0

    however from here on I'm stuck. Can you please help me with the rest.

    Thanks in advance
    P'(x) = 0 implies that the rate of change of speed in the y direction is constant, not that the rate of change of speed in the x direction is constant, so this isn't correct.

    Condition 1 requires that P(0) = 0 and P(L) = h. The diagram appears to also require that P'(0) = 0 and P'(L) = 0. So we have:
    P(0) = d = 0
    P'(0) = c = 0

    So P(x) = ax^3 + bx^2

    P(L) = aL^3 + bL^2 = h
    P'(L) = 3aL^2 + 2bL = 0

    From the P'(L) condition we get that b = -(3L/2)a

    So:
    P(L) = aL^3 - (3L/2)a*L^2 = h

    or
    a = -2h/L^3

    b = -(3L/2)*(-2h/L^3) = 3h/L^2

    Thus
    P(x) = -(2h/L^3)x^3 + (3h/L^2)x^2
    has the required properties.

    -Dan
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  3. #3
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    thank you very much Dan

    I have another question
    how would you use conditions 2 and 3 to prove that
    6hv^2/L^2 < (or = ) k

    so then v= h/L
    and
    P(x)= -2vx^3/L^2 + 3vx^2/L
    since v and L are constants
    P'(x)= -6vx^3/L^2 + 6vx^2/L

    but I don't understand how to find the vertical acceleration....the whole condition #3 is really confusing to me
    Last edited by momentum; March 21st 2007 at 11:10 PM.
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