20/12/09 19:45:02.50 nSTBriB8.net
>>905
(略証)
L = {1 + (ω/sinω) + (α/sinα)}d,
面積条件
S_o = f(ω)(d/2)^2 = 1,
S_a = f(α)(d/2)^2 = 2,
があるので ω,α もdの陰関数。
⊿ω･d = -2f(ω)/f '(ω)⊿d,
⊿α･d = -2f(α)/f '(α)⊿d,
よって
⊿L = {1 + (ω/sinω) + (α/sinα)}⊿d
+ {(ω/sinω) '・⊿ω + (α/sinα) '・⊿α}d
= {1 + (ω/sinω) - (ω/sinω) '・f(ω)/f'(ω)
+ (α/sinα) - (α/sinα) '・f(α)/f'(α)}⊿d
= (1 + cosω + cosα)⊿d,
∵ f(θ) = (θ-sinθcosθ)/(sinθ)^2 より
(θ/sinθ) - 2(θ/sinθ) '･f(θ)/f '(θ) = cosθ,