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Thread: Basic Related rates question

  1. #1
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    Basic Related rates question

    Hi guys,

    I have a basic Cal I related rates question and am unsure how to correctly solve. Can you check my solution? The part I'm a little confused about is where I differentiated both sides. Can you do as I did?
    THanks for any help,
    james

    If two resistors of $\displaystyle R_1$ and $\displaystyle R_2$ are connected in parallel in an electric circuit to make an $\displaystyle R$-ohm resister, the value of $\displaystyle R $ can be found from the equation:
    $\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$
    If $\displaystyle R_1$ is decreasing at the rate of $\displaystyle 3$ ohms/sec and $\displaystyle R_2$ is increasing at a rate of $\displaystyle 5$ ohms/sec at what rate is $\displaystyle R$ changing when $\displaystyle R_1 = 60$ ohms and $\displaystyle R_2 = 50$ ohms

    Substituing $\displaystyle R_1 = 60$ and $\displaystyle R_2 = 50$ gives that $\displaystyle R = \frac{11}{300}$

    Differentiating both sides with respect to time gives:

    $\displaystyle -\frac{1}{R^2}\left(\frac{dR}{dt}\right) = -\frac{1}{R_1^2}\left(\frac{dR_1}{dt}\right) -\frac{1}{R_2^2}\left(\frac{dR_2}{dt}\right)$

    Substituting all known values into the above equation give:
    $\displaystyle -\frac{300}{11}\left(\frac{dR}{dt}\right) = -\frac{1}{60^2}(-3) - \frac{1}{50^2}(5)$

    Then we solve for $\displaystyle \frac{dR}{dt}$ to obtain the answer.
    Last edited by james121515; Jun 20th 2010 at 03:56 PM.
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  2. #2
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    finish. solve for $\displaystyle \frac{dR}{dt}$ ...

    $\displaystyle \frac{dR}{dt} = \frac{11}{300}\left(-\frac{3}{60^2} + \frac{5}{50^2}\right)$
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