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文章問題の途中計算
途中計算がわかりません。関係あるかもしれないので念のため全文載せておきます。英文ですがわかる方、宜しくお願い致します。 問題 A beam is to go from the ground over a low wall, height h, to reach a tall wall in order to brace the tall wall. The distance from the low wall to the tall wall is d. Find the length of beam L in terms of the angle Θ , d and h and hence find tan Θ in terms of d and h for the shortest possible beam length that will brae the wall. You may assume that any solution is a minimum. 解答 L=h cosec Θ +d sec Θ これはわかります。 L’= -h cosec Θ cot Θ +d sec Θ tan Θ わかります。 L’ = 0 わかります。 h/(sin Θ tan Θ ) = (d tan Θ) / cos Θ わかります。 tan Θ = 3 √(h/d) ここが分かりません。 （小さい３です。(h/d) ^(1/3) という事です。） h/(sin Θ tan Θ ) = (d tan Θ) / cos Θ → tan Θ = 3 √(h/d) の計算の仕方を教えて頂けますか？
- machikono
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sinθ=s, cosθ=c と略記することにします。 (与式) ⇔ h/(s^2/c)=h*c/s^2 ⇔ h/d=s^3/c^3=(tanθ)^3. よって、tanθ=(h/d)^(1/3) となります。
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- m-take0220
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h/(sinθtanθ) = (dtanθ)/cosθ 両辺にsinθtanθ(≠0)をかけると h = (dsinθtan^2θ)/cosθ sinθ/cosθ = tanθなので、 h = dtan^3θ あとはわかりますね。
お礼
わかりました、こんな風に考えるのですね。 すみません、一番最初に答えて下さった方にベストアンサーを差し上げるべきだと思うのでそうしますが二分出来ないのが残念です。 本当に助かりました、有り難うございました。
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ご回答有り難うございます。 (与式) ⇔ h/(s^2/c)=h*c/s^2 わかります h*c/s^2 ⇔ h/d=s^3/c^3 わかりません s^3/c^3=(tanθ)^3 わかります よって、tanθ=(h/d)^(1/3) わかります。 お時間あれば h*c/s^2 ⇔ h/d=s^3/c^3 の流れを教えて頂けますか？