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Math Help - Is anyone familiar with Rate of change?

  1. #1
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    Is anyone familiar with Rate of change?

    The equation: H(t)=sqr root(3t+5)

    The question: What is the average rate of change of H(t) with respect to t as t changes from 2 to 2.01?

    Can anyone explain this to me?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by jemadd2 View Post
    The equation: H(t)=sqr root(3t+5)

    The question: What is the average rate of change of H(t) with respect to t as t changes from 2 to 2.01?

    Can anyone explain this to me?
    The average rate of change of a function f, between a and b is :

    \frac{f(b)-f(a)}{b-a}

    Try to do it for your problem, please
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  3. #3
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    So this means:

    f(2.01) - f(2) / 2.01-2 ?
    is this correct?
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  4. #4
    Moo
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    Quote Originally Posted by jemadd2 View Post
    So this means:

    f(2.01) - f(2) / 2.01-2 ?
    is this correct?
    Yes it is

    (here, it's not f, it's H but it doesn't matter, you've grasped the concept)
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  5. #5
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    Quote Originally Posted by Moo View Post
    Yes it is

    (here, it's not f, it's H but it doesn't matter, you've grasped the concept)
    okay so now i just fill it in the function.
    so it would be
    1. f(2.01) - f(2) / 2.01-2
    2. sqr root(3(2.01)+5) - sqr root(3(2)+5) / 2.01-2
    3. im confused =/
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  6. #6
    Moo
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    Quote Originally Posted by jemadd2 View Post
    okay so now i just fill it in the function.
    so it would be
    1. f(2.01) - f(2) / 2.01-2
    2. sqr root(3(2.01)+5) - sqr root(3(2)+5) / 2.01-2
    3. im confused =/
    It's correct !

    \frac{\sqrt{3 \cdot 2.01+5}-\sqrt{3 \cdot 2+5}}{2.01-2}=\frac{\sqrt{11.03}-\sqrt{11}}{0.01} \approx 0.45195907277229 (use a calculator)
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  7. #7
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    wow thank you so much....im glad i understand now
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