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Math Help - Distance formula

  1. #1
    Member
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    Distance formula

    Distance: s=-16t^2+800t

    find t, for s increasing, for s'=0, and for s decreasing

    Am I figuring maximums and minimums here?

    y'=-32t+800 = y'=-t+25

    so I have a critical point at x=25

    y''=-1

    So our singular critical point is a maximum. So t>25 we start to decrease?

    I have a multiple choice solution set

    a.) (0,25]
    increasing
    {0,50}
    s=0

    b.)[25,50]
    decreasing
    {0,50}
    s=0

    c.) {0,50}
    s=0

    d.) None of these

    Keep in mind. He likes to trick us and I've noticed a lot of "none of these" resulting from things like positive and negative signs in the equation. Part of the thing confusing me is how he wrote up the solution set
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  2. #2
    Eater of Worlds
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    As you probably know, it is a concave down parabola with vertex at
    V(25,10000).

    It follows that f'(x) < 0 \;\ if \;\ 25 < x < {\infty}

    and f'(x) > 0 \;\ if \;\ {-\infty}<x<25

    Since f(x) is continuous at x=25:

    Therefore it is decreasing over the interval [25,\infty) and increasing over the interval (-\infty, 25]

    f''(x) < 0---->f''(x) = -32, so it is concave down. We can tell that from the negative coefficient of x^2.
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