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Differentation of ac sources

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Chain/product or quotient

    2. Relevant equations

    v=4sin^3 (4t)


    3. The attempt at a solution

    NB:@ Means teta(spelling might be wrong)

    4sin^3 (4t) = 4sin (4t).sin (4t).sin (4t)

    using sin^2@=1-cos 2@ /2


    v= 4sin (4t). (1-cos 2@ /2)


    = 4sin (4t)/2 . 4sin (4t). cos (8t)/2

    = 2 sin (4t) - 2 sin (4t) cos (8t)




    then I done some calculation containing product and chain rule with a final answer of


    dv/dt = 8 cos (4t) + 16 sin (4t) sin (8t) - 8 cos (4t) cos (8t)



    Thanx for your help
     
  2. jcsd
  3. Mar 12, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi maali5!
    no, v= 4sin (4t). (1-cos 2@)/2

    but anyway you should have used the chain rule, with g(t) = sin(4t) :wink:
     
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