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Differentiation: rates of change

  1. Aug 13, 2015 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    1. The combined electrical resistance R of two resistors connected in parallel is ## R = \frac{R1R2}{R1+R2}##, where R, R1 and R2 are measured in ohm. R1 and R2 are increasing at rates of 1 and 1,5 ohm per second respectively. Find the rate of change of R when R1 = 50 ohm and R2 = 75 ohm.

    2. Differentiation: rates of change.

    3. See attached..

    So I'm not sure what the derivative of R1 and R2 would be unless I put in their values of 50 and 75 ohm but then the derivative would be 0? Is it possible to rearrange the given equation?
     

    Attached Files:

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  3. Aug 13, 2015 #2

    Ray Vickson

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    Homework Helper

    You need to calculate the partials
    [tex] \frac{\partial}{\partial R_1} \left( \frac{R_1 R_2}{R_1 + R_2}\right) \; \;\text{and} \;\; \frac{\partial}{\partial R_2}\left( \frac{R_1 R_2}{R_1 + R_2}\right) [/tex]
    as well as ##d R_1 / dt## and ##d R_2 / dt##.
     
  4. Aug 13, 2015 #3
    How do I calculate the partials? Do I put in the value of R1 when doing ##\frac{\partial}{\partial R1}## ##(\frac{R1R2}{R1+R2})## and same for R2?
    ##\frac{dR1}{dt}## = +1ohm and ##\frac{dR2}{dt}## = +1.5ohm
     
  5. Aug 13, 2015 #4

    Mark44

    Staff: Mentor

    No. R1 and R2 are variables (functions of t). It is only at a particular moment that they have the values below. When you take the partial with respect to one of the variables, you treat the other variable as if it were a constant.
     
  6. Aug 14, 2015 #5
    I don't follow. Could you show me the partial for R1?
     
  7. Aug 14, 2015 #6

    Mark44

    Staff: Mentor

    How do I calculate the partials? Do I put in the value of R1 when doing ##\frac{\partial}{\partial R1}## ##(\frac{R1R2}{R1+R2})## and same for R2?
    Finding the partial with respect to R1, you treat R2 as if it were a constant. How would you find this ordinary derivative?

    $$\frac{d}{dx} (\frac{kx}{x+k})$$
     
  8. Aug 14, 2015 #7

    Ray Vickson

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    Are you saying you do not know how to find the derivative
    [tex] \frac{d}{dx} \left( \frac{75 x}{75+x} \right) \; ? [/tex]
     
  9. Aug 16, 2015 #8
    Final answer attached. Thanks to all who helped.
     

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