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Homework Help: L in terms of a and b as a maximum

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] L = \frac{a}{\sin{\theta}} + \frac{b}{\cos{\theta}} [/tex]


    find L in terms of a and b where L is a maximum .....


    2. Relevant equations

    place long list of trigonometric identities here??


    3. The attempt at a solution

    so my attempt looks something like this:

    as a cause of the chain rule:

    [tex] \frac{dL}{d\theta} = - \frac{a \cdot \cos{\theta} }{\sin^2{\theta}} + \frac{b \cdot \sin{\theta}}{\cos^2{\theta}} [/tex]


    so with [tex] \frac{dL}{d\theta} = 0 [/tex] ..

    [tex] \sin{\theta} = \cos{\theta}(\frac{a}{b})^{\frac{1}{3}} [/tex]


    [tex] \cos{\theta} = \frac{\sin{\theta}}{ (\frac{a}{b})^{\frac{1}{3}}} [/tex]


    also..

    [tex] L = a\csc{\theta} + b\sec{\theta} [/tex]


    so that by identities..

    [tex] \frac{dL}{d \theta} = a(-\cot{\theta} \cdot \csc{\theta}) + b(-\csc^2{\theta}) [/tex]


    which when [tex] \frac{dL}{d \theta} [/tex] is set to zero and the resultant reduced ....

    [tex] \cos{\theta} = - \frac{b}{a} [/tex]


    substituting the above into the original equation and reducing I have come to

    [tex] = - \frac{a^{\frac{5}{3}}}{b^{\frac{2}{3}}} - \frac{a \cdot b^{\frac{2}{3}}}{b^{\frac{2}{3}}} [/tex]


    which looks like it is approaching the final answer of

    [tex] L = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}} [/tex]


    but is most likely somehow wrong and there is probably a much simpler answer I am overlooking as this seems a bit out of context :/

    if I skipped too many steps forgive me I will add more .... latex is very time consuming for me!
     
    Last edited: Aug 31, 2010
  2. jcsd
  3. Sep 1, 2010 #2
  4. Sep 1, 2010 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You've got tan(theta)=(a/b)^(1/3). Stop there. So theta=arctan((a/b)^(1/3)). Now you need to find sin and cos of the arctan. BTW this isn't necessarily a maximum - so far it's just a critical point.
     
  5. Sep 1, 2010 #4
    Thank you both!

    o~o
     
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