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

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

    Differentiating both sides with respect to time gives:

    -\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:
    -\frac{300}{11}\left(\frac{dR}{dt}\right) = -\frac{1}{60^2}(-3) - \frac{1}{50^2}(5)

    Then we solve for \frac{dR}{dt} to obtain the answer.
    Last edited by james121515; June 20th 2010 at 04:56 PM.
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  2. #2
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    finish. solve for \frac{dR}{dt} ...

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