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Math Help - Connected Rates Of Change (Calculus) - Need an Answer.

  1. #1
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    Talking Approximate Changes (Calculus) - Need right Answer

    I am unable to get the right answer, closest answer : -0.01(r-1)
    Trying again and again it keeps raising a lot more of other questions, that is simply boring for me.
    Would be more than grateful if someone could kindly show me a detailed steps to the solution please.


    (JF Talbert Textbook, Exercise 16D, Q.22)

    For the function A=1/(r-1)^2 , a small change in r when r=2(r-1)^2 produces a 2% reduction in the value of A.
    Find the approximate change in r.
    (Given Answer: 0.01 increase)
    Last edited by zikcau25; September 4th 2012 at 08:48 AM.
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    Re: Connected Rates Of Change (Calculus) - Need an Answer.

    \frac{dA}{dr} = -\frac{2}{(r-1)^3}

    -\frac{dA \cdot (r-1)^3}{2} = dr

    when r = 2(r-1)^2 , r = \frac{1}{2} or r = 2

    note: any info given on the range of values for r ?


    for r = \frac{1}{2} ...

    -\frac{(-0.02) \cdot \left(\frac{1}{2}-1 \right)^3}{2} = dr

    -\frac{1}{800} = dr ... a decrease in r at 1/8 of a percent


    for r = 2 ...

    -\frac{(-0.02) \cdot \left(2-1 \right)^3}{2} = dr

    0.01 = dr

    there is your 1% increase in r
    Thanks from zikcau25
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    Re: Connected Rates Of Change (Calculus) - Need an Answer.

    Thank you so much for that help.

    you put an end to my day of agony and open a day of paradise
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    Exclamation Re: Connected Rates Of Change (Calculus) - Need an Answer.

    I have reviewed the problem, it comprises of 3 stages:

    I. Solving for r from r=2(r-1)^2}, gives \hspace{2mm}r=2\hspace{2mm}or\hspace{2mm} r=\frac{1}{2}


    II. Solving for \delta A from \frac{\delta A}{A}=-2\%, gives:
    \frac{\delta A}{A}=-2\%

    \frac{\delta A}{A}=\frac{-2}{100}

    \delta A=-0.02A


    III. And finally, Solving for \delta r} from \delta A\approx (\frac{\mathrm{d}A }{\mathrm{d} r})_{r=2(r-1)^2}\times \delta r, that is a) \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=2}} or b) \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=\frac{1}{2}}}

    Having given,

    A=(r-1)^{-2}

    \frac{\mathrm{d} A}{\mathrm{d} r}=-2(r-1)^{-3}


    Therefore,
    a) When  r=2, A=(2-1)^{-2}=1

    \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=2}}

    \delta r\approx \frac{-0.02(1)}{-2(2-1)^{-3}}

    r\approx +0.01 (increasing)

    b) when r=\frac{1}{2}, A=(\frac{1}{2}-1)^{-2}=4

    \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=\frac{1}{2}}}

    \delta r\approx \frac{-0.02(4)}{-2(\frac{1}{2}-1)^{-3}}

    r\approx -0.005 (decreasing)


    **********************
    Sir, hope you take into consideration stage II. that lead to a different solution.
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