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Math Help - simple geometry applied to calculus

  1. #1
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    simple geometry applied to calculus

    hi all, its not the calculus im stuck on but the function A in the following question

    A circular pipe has outer diameter \text{4 cm} and thickness \text{t cm}.

    a) Show that the area of cross-section, \text{A }cm^2, is given by A = \pi(4t-t^2)

    b) Find the rate of increase of A with respect to t when t=\frac{1}{4} and when t=\frac{1}{2}, leaving \pi in the answer.

    As far as i can see

    r=\frac{4-t}{2}

    and

    A=\pi(4-2t+\frac{1}{4}t^2)

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  2. #2
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    sorry i realise

    r=\frac{4-2t}{2}

    so

    A=\pi(4-4t+t^2)

    after thinking about it.
    Last edited by sammy28; August 15th 2010 at 09:43 AM. Reason: forgot to add new A equation
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  3. #3
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    correct ...

    the inner cross-sectional area of the pipe is A = \pi(2-t)^2
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  4. #4
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    hi skeeter, thats what i dont understand the question states A = \pi(4t-t^2)
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  5. #5
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    Quote Originally Posted by sammy28 View Post
    hi skeeter, thats what i dont understand the question states A = \pi(4t-t^2)
    I do not agree with that equation if the problem was written as posted.
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  6. #6
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    thanks skeeter

    the question is in error then.
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  7. #7
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    just thought of this ... the cross-section the problem wants is the cross-section of the physical pipe itself (not the inside).

    A = outer area of pipe - inner area of pipe

    A = \pi \cdot 2^2 - \pi(2-t)^2

    A = 4\pi - \pi(4 - 4t + t^2)

    A = 4\pi - 4\pi + 4\pi t - \pi t^2

    A = \pi(4t - t^2)
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  8. #8
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    Ah so thats what they are talking about. thanks skeeter.
    All makes sense now.
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