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Math Help - Rate of Change of Area of a Circle

  1. #1
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    Rate of Change of Area of a Circle

    For time t between 2 and 3 hours the radius circle C is 5t-t^2-6 feet. At time t=2.3 hours the area of the circle is (to within .001) changing at a rate of ?

    How do I do this? The answers I get are always wrong.
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  2. #2
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    Quote Originally Posted by Velvet Love View Post
    For time t between 2 and 3 hours the radius circle C is 5t-t^2-6 feet. At time t=2.3 hours the area of the circle is (to within .001) changing at a rate of ?

    How do I do this? The answers I get are always wrong.
    f(t) = A = \pi r^2 = \pi (5t-t^2-6)^2

    Use the chain rule to differentiate:

    <br />
f'(t) = 2\pi \,(5t-t^2-6) \times (5-2t)

    f'(2.3) = 2\pi \,(10.6-2.3^2-6)(5-4.6) = -1.734

    (note a negative differential is fine- it means the area is getting smaller)
    Last edited by e^(i*pi); October 14th 2009 at 11:52 AM. Reason: added numeric answer
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  4. #4
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    It says the answer is wrong for some reason.
    I hate this online homework :/
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  5. #5
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    I think you are lookin for dA
    so; dA = A' (dt)
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  6. #6
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    Sorry, but what does that mean? XD. My teacher doesn't teach us terms or notation for some reason
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  7. #7
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    you know that dA/dt is the rate of change with respect to time.

    so aproximately, since dA/dt= A' then dA = A' * dt you can use that to aproximate.

    so you use the f'(x) the gentleman got and plug 2 for t
    then the dt will be .3 since that is the change in time
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  8. #8
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    When I plug 2 in for t, I get 0 for the answer.
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  9. #9
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    do you have the answer?
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  10. #10
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    With the info you gave me, I get the answer to be 0, because when i plug in 2 for t, 5t - t^2 -6 = 0.
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