- #1
rebeka
- 43
- 0
Homework Statement
L = \frac{a}{\sin{\theta}} + \frac{b}{\cos{\theta}}find L in terms of a and b where L is a maximum ...
Homework Equations
place long list of trigonometric identities here??
The Attempt at a Solution
so my attempt looks something like this:
as a cause of the chain rule:
\frac{dL}{d\theta} = - \frac{a \cdot \cos{\theta} }{\sin^2{\theta}} + \frac{b \cdot \sin{\theta}}{\cos^2{\theta}}so with \frac{dL}{d\theta} = 0 ..
\sin{\theta} = \cos{\theta}(\frac{a}{b})^{\frac{1}{3}}\cos{\theta} = \frac{\sin{\theta}}{ (\frac{a}{b})^{\frac{1}{3}}}also..
L = a\csc{\theta} + b\sec{\theta}so that by identities..
\frac{dL}{d \theta} = a(-\cot{\theta} \cdot \csc{\theta}) + b(-\csc^2{\theta})which when \frac{dL}{d \theta} is set to zero and the resultant reduced ...
\cos{\theta} = - \frac{b}{a}substituting the above into the original equation and reducing I have come to
= - \frac{a^{\frac{5}{3}}}{b^{\frac{2}{3}}} - \frac{a \cdot b^{\frac{2}{3}}}{b^{\frac{2}{3}}}which looks like it is approaching the final answer of
L = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}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:
