弧长的计算公式?
微弧ds=√(dr²+(rdθ)²)=dθ√((dr/dθ)²+r²)
r=1/θ,dr/dθ=-1/θ²代入
ds=dθ√(1/θ^4+1/θ²)
=(1/θ)dθ√(1/θ²+1)
=[√(1/θ²+1)/θ]dθ弧长=∫(3/4,4/3)[√(1/θ²+1)/θ]dθ
设1/θ=tanu,θ=1/tanu=cosu/sinu,dθ=[-sinusinu-cosucosu]/sin²u.du=-1/sin²u.du
u=arctan(1/θ),u1=arctan(4/3),u2=arctan(3/4)
弧长=∫(u1,u2)[secutanu](-1/sin²u.du)
=∫(u2,u1)[sinu/cos²u](1/sin²u.du)
=∫(u2,u1)[1/sinucos²u]du
用万能置换共识:
设t=tan(u/2),u=2arctant,du=2/(1+t²).dt
sinu=2t/(1+t²),cosu=(1-t²)/(1+t²)
弧长=∫(t2,t1)(1+t²)/2t.(1+t²)²/(1-t²)².2/(1+t²).dt
=∫(t2,t1)(1+t²)²/t(1-t²)².dt
=∫(t2,t1)(1+2t²+t^4)/t(t-1)²(t+1)².dt
设(1+2t²+t^4)/t(t-1)²(t+1)²=A/t+B/(t-1)²+C/(t-1)+D/(t+1)²+E/(t+1)
两边恒等,解得A=1,B=1,C=0,D=-1,E=0
弧长=∫(t2,t1)[1/t+1/(t-1)²-1/(t+1)²]dt
=[lnt-1/(t-1)+1/(t+1)](t2,t1)
=[lnt-2/(t²-1)](t2,t1)
=ln(t1/t2)-2[1/(t1²-1)-1/(t2²-1)]
t1=tan(u1/2)=tan[arctan(1/θ1)/2]=tan[arctan(4/3)/2]
t2=tan(u2/2)=tan[arctan(1/θ2)/2]=tan[arctan(3/4)/2]
勾股定理,设α=arctan(3/4),sinα=3/5,cosα=4/5;
sin(α/2)=√[(1-cosα)/2]=√[(1-4/5)/2]=1/√10
cos(α/2)=√(1-1/10)=3/√10
t2=tan(α/2)=1/3;
arctan(4/3)=π/2-α
t1=tan[(π/2-α)/2]=tan(π/4-α/2)=(1-1/3)/(1+1/3)=2/4=1/2代入
弧长=ln(3/2)-2[1/(1/4-1)-1/(1/9-1)]
=ln(3/2)-2[-4/3+9/8]=ln(3/2)+8/3-9/4
=ln(3/2)+5/12
扩展资料:
扇形弧长和计算公式:
半径为R的圆中,n°的圆心角所对弧长
的计算公式为
弧长公式:L=n×π×r/180,L=α×r。
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