■サイクロイドと積分・変分法(その5)
【1】サイクロイド弧長のn等分
2r(−2cos(t/2)+2)=8r/n
−2cos(t/2)+2=4/n
cos(t/2)=1−2/n
y=r(1−cost)=r(2−2cos^2(t/2))=2r(1−(1−2/n)^2)
[1]3等分する場合,
y=2r・8/9,t=2arccos1/3
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[2]4等分する場合,
y=2r・3/4,t=2arccos1/2=2π/3
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【2】内サイクロイド弧長のm等分
4(n−1)/n(−cos(nt/2)+1)=8(n−1)/mn
−cos(nt/2)+1=2/m
cos(nt/2)=1−2/m
r^2=x^2+y^2=(n−1)^2+1+2(n−1)cosnt=(n−1)^2+1+2(n−1){2cos^2(nt/2)−1}=(n−1)^2+1+2(n−1)(1−8/m−8/m^2)
[1]3等分する場合,
r^2=(n−1)^2+1−14(n−1)/9
t=2/n・arccos1/3
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[2]4等分する場合,
r^2=(n−1)^2+1−(n−1)
t=2/n・arccos1/2=2π/3n
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【3】外サイクロイド弧長のm等分
4(n+1)/n(−cos(nt/2)+1)=8(n+1)/mn
−cos(nt/2)+1=2/m
cos(nt/2)=1−2/m
r^2=x^2+y^2=(n+1)^2+1−2(n+1)cosnt=(n+1)^2+1−2(n+1){2cos^2(nt/2)−1}=(n+1)^2+1−2(n+1)(1−8/m−8/m^2)
[1]3等分する場合,
r^2=(n+1)^2+1+14(n+1)/9
t=2/n・arccos1/3
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[2]4等分する場合,
r^2=(n+1)^2+1+(n+1)
t=2/n・arccos1/2=2π/3n
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