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Related rates of resistors

  1. Feb 24, 2007 #1
    If two resistors with resistances and are connected in parallel, as in the figure, then the total resistance R measured in ohms , is given by 1/R=1/R1+1/R2. If and are increasing at rates of .6 and .7 respectively, how fast is R changing when R1=80 and R2=100?

    so i take the derivative of the function which is (-1/R^2)dR/dt=(-1/R1^2)dR1/dt-(1/R2)dR2/dt

    and then can calculate R to be 1/180 and i know R1, R2, and dR1/dt=.6 and

    i get 5.052E-9 for dR/dt....which is not right

    please help me
  2. jcsd
  3. Feb 24, 2007 #2


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    1/100 + 1/80 is not 1/180.

    Your derivative is correct (except for the typo where you left out the square on the R2 term). Check your algebra, and it should work out.
  4. Feb 24, 2007 #3


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    R is not equal to 1/180.

    1/80 + 1/100 does not equal 1/(80+100) = 1/180

    .. and also the formula says 1/R = 1/R1 + 1/R2 not R = 1/R1 + 1/R2
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