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Maximum value of E = a.sin(x) + b.cos(x)

  1. Jun 10, 2010 #1
    We've been given the following equation,

    E = 15.sin(50.pi.t) + 40.cos(50.pi.t)

    We've been asked to find a value of E we think at its maximum value (can't quite remember the question at the moment). We know that the second derivative will give us whether the equation gives the maximum or minimum value (think it's giving us the maximum anyway) so we're sure that we are just required to differentiate it once, equal it to zero to find a value for t and then put the value of t back into the original equation.

    dE/dt = 750.pi.cos(50.pi.t) - 2000.pi.sin(50.pi.t)

    The problem is when we transpose the equation because we get really confused. This is as far as we've got:

    750.pi.cos(50.pi.t) - 2000.pi.sin(50.pi.t) = 0
    750.pi.cos(50.pi.t) = 2000.pi.sin(50.pi.t)
    cos(50.pi.t) = [2000.pi.sin(50.pi.t)] / 750.pi
    50.pi.t = cos-1([2000.pi.sin(50.pi.t)] / 750.pi)
    t = [cos-1([2000.pi.sin(50.pi.t)] / 750.pi)] / 50.pi

    Okay, so now what happens? There is still a t on the right side of the equation and we don't know how you can get it over to the other side to get t to the 2 and then square root the right side.

    Any ideas? Are we way off? Any help would be greatly appreciated.

  2. jcsd
  3. Jun 10, 2010 #2


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    You don't want to take cos-1. Divide both sides of your equation by cos(50.pi.t) and remember sin/cos=tan.
  4. Jun 10, 2010 #3
    Okay I think I get what you're saying.


    750.pi.cos(50.pi.t) - 2000.pi.sin(50.pi.t) = 0
    750.pi.cos(50.pi.t) = 2000.pi.sin(50.pi.t)
    750.pi = 2000.pi [sin(50.pi.t)] / cos(50.pi.t)]
    750.pi = 2000.pi.tan(50.pi.t)
    750.pi / 2000.pi = tan(50.pi.t)
    tan-1[750.pi / 2000.pi] = 50.pi.t

    t = (tan-1[750.pi / 2000.pi] / 50.pi
  5. Jun 10, 2010 #4


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    Yes. That's the right way to solve an equation like that for t.
  6. Jun 10, 2010 #5
    That gives us t = 2.284 milliseconds

    So E = 15.sin(50.pi.[2.284x10-3]) + 40.cos(50.pi.[2.284x10-3])

    So E = 42.72
  7. Jun 10, 2010 #6
    Thank you, Dick!!
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