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Math Help - Second derivative help!!

  1. #1
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    Second derivative help!!

    Hi there, I've been working this problem forever and I cannot come close to the right answer...

    A function of the position of a car being driven on a straight road, measured in feet from some fixed point is given by:

    s(t)=1/3(t^2+8)^3/2

    BTW: to avoid any confusion, the (t^2+8) is raised to 3/2
    Sorry, I could not get it to look right.
    I am to determine the acceleration of the car when t=1. I know I have to take the second derivative of s(t) to get the right equation to plug t into, but I can't get it for the life of me... Any suggestions?
    Last edited by rust1477; March 2nd 2009 at 05:59 PM. Reason: clarity
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  2. #2
    Member arpitagarwal82's Avatar
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    s'(t) = 1/3 * 2t * (t^2  + 8)^(1/2)

    s''(t) = 2/3 ( (t^2  + 8)^(1/2) + t * 2t * (t^2  + 8)^(-1/2)

    at t = 1, s''(t) = 2/3 ( \sqrt 9 + 2/\sqrt 9)
    =2/3 (3 + 2/3)
    = 2/3 * 11/3
    = 22/9
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  3. #3
    Member arpitagarwal82's Avatar
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    Above solution is assuming (t^2+8)^{3/2} is in numerator of s(t)
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  4. #4
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    Yes, the original s(t) is in the numerator.

    However, any way I interpret your answer is incorrect...

    The choices I'm given are:
    8/3  ft/sec^2
    10/3  ft/sec^2
    3  ft/sec^2
    4  ft/sec^2
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  5. #5
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    Quote Originally Posted by rust1477 View Post
    Hi there, I've been working this problem forever and I cannot come close to the right answer...

    A function of the position of a car being driven on a straight road, measured in feet from some fixed point is given by:

    s(t)=1/3(t^2+8)^3/2

    BTW: to avoid any confusion, the (t^2+8) is raised to 3/2
    Sorry, I could not get it to look right.
    I am to determine the acceleration of the car when t=1. I know I have to take the second derivative of s(t) to get the right equation to plug t into, but I can't get it for the life of me... Any suggestions?
    arpitagarwal has answered
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  6. #6
    Member arpitagarwal82's Avatar
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    Yeah
    I missed some factors
    Here i correct one
    s'(t) = 1/3 * 2t * 3/2 * (t^2  + 8)^(1/2)

    s''(t) =  ( (t^2  + 8)^(1/2) + t * 2t *1/2 * (t^2  + 8)^(-1/2)

    at t = 1, s''(t) =  ( \sqrt 9 + 1/\sqrt 9)
    =(3 + 1/3)
    =10/3
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  7. #7
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    Hey thank you so much for showing me correct answer...

    Can you elaborate a little bit on the computations you did to arrive at the second derivative?

    When I computed the s'(t), I got
    1/3*3/2*(t^2+8)^1/2*2t

    Is this correct for the first derivative, and if so, can you explain how you arrived at the second derivative?
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  8. #8
    Member arpitagarwal82's Avatar
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    Yes yu ave calculated s'(t) correct.
    On simplification s'(t) =  t(t^2 + 8)^{1/2}

    To calculate s''(t) use differentiation by parts.
    s'' = (t^2 + 8)^{1/2} dt/dt + t * d ((t^2 + 8)^{1/2})/dt
    = (t^2 + 8)^{1/2} + t * 2t * 1/2 * (t^2 + 8)^{-1/2}
    =(t^2 + 8)^{1/2} + t^2(t^2 + 8)^{-1/2}

    Is that clear now?
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