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Math Help - Finding the instantaneous rate of change

  1. #1
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    Finding the instantaneous rate of change

    A soccer ball is kicked into the air such that its height, in m, after t seconds, can be modelled by the function h(t)=-4.9t^2+12t+0.5. Determine the instantaneous rate of change of the height of the ball after 1 s.

    I tried doing it using derivatives, but I got this unfactorable numerator.
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  2. #2
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    Re: Finding the instantaneous rate of change

    Find h'(t) then, h'(1)
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  3. #3
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    Re: Finding the instantaneous rate of change

    Quote Originally Posted by Dragon08 View Post
    A soccer ball is kicked into the air such that its height, in m, after t seconds, can be modelled by the function h(t)=-4.9t^2+12t+0.5. Determine the instantaneous rate of change of the height of the ball after 1 s.

    I tried doing it using derivatives, but I got this unfactorable numerator.
    \lim_{t \to 1} \frac{(-4.9t^2+12t+0.5) - (-4.9+12+0.5)}{t-1}

    \lim_{t \to 1} \frac{-4.9t^2+12t - 7.1}{t-1}

    \lim_{t \to 1} \frac{(-4.9t+7.1)(t-1)}{t-1}

    \lim_{t \to 1} (-4.9t+7.1) = 2.2 m/s
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  4. #4
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    Re: Finding the instantaneous rate of change

    There are other ways to find a derivative than using the definition!

    However, the definition is, of course,
    \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}
    Obviously, the denominator is going to 0. In order that the limit exist, the numerator must also go to 0. And if a polynomial goes to 0 as h goes to 0, The polynomial must be 0 at h= 0- and if a p(a)= 0, x- a is a factor.
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  5. #5
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    Re: Finding the instantaneous rate of change

    Oh, okay. Thanks!
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