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Homework Help: Finding the Maximum

  1. Sep 5, 2006 #1
    Hi all,

    I am having a problem understanding how to find a certain maximum of this equation and am not sure if I am going about this the proper way.

    [tex] v^x_s = \frac{v \sin{\theta}}{1-v \cos{\theta}} [/tex]

    Find an expression for the angle [tex] \theta_{max} [/tex] at which [tex] v^x_s [/tex] has its maximum value for a given speed v. Show that this angle satisfies the equation [tex] \cos{\theta_{max}}=v [/tex].

    Answer: Using my knowledge of calculus, I believe I should take the derivative of the above equation with respect to [tex] \theta [/tex]

    Using the quotient rule, this gives me [tex] \frac{v \cos{\theta}-v^2 (\sin{\theta})^2}{1-v \cos{\theta}} [/tex]

    I should now find values of theta that make the equation 0 or undefined. However, I do not know what v is, which is throwing me off on what value theta should be. Any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 5, 2006 #2
    You're on your way set the numerator to zero, and solve for [tex] \theta [/tex] You'll see the answer makes sense.
     
  4. Sep 6, 2006 #3

    HallsofIvy

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    However, I do not get ][tex] \frac{v \cos{\theta}-v^2 (\sin{\theta})^2}{1-v \cos{\theta}} [/tex] as the derivative.
    For one thing, the denominator must be [itex](1- vcos \theta)^2[/itex]. For another, the numerator will involve both [itex]sin^2 \theta[/itex] and [itex]cos^2 \theta[/itex] which will then combine.
     
  5. Sep 6, 2006 #4
    Opps! I found my error and problem solved. Thanks
     
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