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Intense Logarithmic Diferentiation Question

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey guys, this is my first post and of course first question to ask of calculus



    Alright well I had this test and we had this very difficult question that I could not solve, it was in the hardest section.



    The question is as follows:

    A projectile thrown over level ground, at an angle x to the ground, has a range R given by R = (v^2 / g)(sin2x), where v is the initial speed, in meters per second, and g = 9.8m/s^2. Determine the angle of proection x for which the range is maximum.


    2. Relevant equations

    So I began to isolate for v^2 and then use log differentiation.



    R(g) = v^2(sin2x)

    v^2 = Rg / sin2x

    3. The attempt at a solution


    R = (v^2 / g)(sin2x)

    R = ((v^2)(sin2x) / g)

    ln R = ln v^2 + ln Sin2x - ln g

    dR / dx = [(1 / v^2) + (1 / sin2x) - (1 / g)] (v^2(sinx) / g)



    So im pretty sure that derrivative is right but as of this i am clueless on what to do. Any help is really appreciated, even a guideline so I could figure out the rest myself. Thanks.
     
  2. jcsd
  3. Apr 5, 2009 #2

    Dick

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    Welcome to the forums, vexon. You are making this problem much harder than it is. You don't need to take a log. And your differentiation is not correct. You have to use the chain rule. You just want to find dR/dx and set it equal to 0. What is the derivative of sin(2x)?
     
  4. Apr 5, 2009 #3
    the derivative of sin2x is cos2x(2) = 2cos2x
    but we were informed that log differentiation ln was to be used ?
    But i'm still stuck. Could you help me further
     
  5. Apr 5, 2009 #4
    oh i think i see it,
    so since v and g are constants you dont differentiate them
    therefore,

    dR/dx = (v^2/g)(2cos2x)
    and then
    0 = v^2/g(2cos2x)
    so in order to make the equation equal 0 the angle has to be 30 degrees is that correct ?
    would this be the right answer?
     
    Last edited: Apr 5, 2009
  6. Apr 5, 2009 #5

    Dick

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    The only thing that could be zero is cos(2x). And I don't think that happens at 30 degrees.
     
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