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Calculus 2nd Derivative

  1. Nov 3, 2007 #1
    A particle moves along the x-axis, its position at time t is given by
    x(t)= 5t/(10+7t^2)

    where t is >= 0 and measured in seconds and x is in meters

    The acceleration of the particle equals 0 at time t = ?

    I took the 1st derivative and got
    i got v(t ) = 5(10-7t^2) / (10+7t^2)^2

    or i got v(t ) = 5(10-7t^2) * (10+7t^2)^-2

    then i took the derivative of that and got:

    a(t ) = 5[(10-7t^2)(-2)(10+7t^2)^(-3) *(14t) + (10+7t^2)^2 *(10-7t^2)]

    so i tried factoring out (10-7t^2), to get one answer which i got as sqrt(10/7)

    and then for the other answer i got:
    (10+7t^2)^5 = 28t

    which i simplified to:
    2401t^8 + 13720t^6 + 29400t^4 + 28000t^2 - 28t + 10000 = 0

    now I'm not sure if I'm on the right track, and if I am, I'm not sure how I can solve for t.

    Any help is much appreciated, I've been working on this (my first problem) for about an hour and I have a midterm in a few days :s

  2. jcsd
  3. Nov 3, 2007 #2


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    Just keep the powers of (10+7t^2)^2 in the denominator like you did when calculating v. The numerator is then easy to factorize.
  4. Nov 3, 2007 #3
    the numerator will still end up being 2401t^8 + 13720t^6 + 29400t^4 + 28000t^2 - 28t + 10000
    if i make a common denomintor with the two terms
  5. Nov 4, 2007 #4


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    Don't write out the brackets, that's gonna be horrible. If you never write out the brackets in your calculations you'll end up with a form that's already quite nicely factorized.

    If you apply to quotient rule to:
    without expanding, you get (10+7t^2)^4 in the denominator right? YOu'r numerator will not be too bad either. Just look for common terms.
  6. Nov 4, 2007 #5
    what do you mean don't write out the brackets?

    applying the quotient rule i got:
    5[(10+7t^2)^2 *(-14t) - (10-7t^2)(2)(10+7t^2)(14t)]/(10+7t^2)^4

    so I can factor out 14t
    and 10+7t^2

    will my two terms be

  7. Nov 4, 2007 #6
    why did I get something different from when I used to product rule?
    or were they equivalent but arranged differently?
  8. Nov 4, 2007 #7


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    You made a small error when using the product rule. The last answer you gave using the qoutient rule is correct and it should be easy finding the zero's now.
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