計算拋物線y²=2px(p>0)從頂點到點(p/2,p)的一段曲線弧長.
對y²=2px取導數(shù)得 2yy′=2p,故y′=p/y=p/√(2px)
于是弧長S=[0,p/2]∫[√(1+y′²)]dx=[0,p/2]∫√[1+(p/2x)]dx
令1+(p/2x)=u²,p/2x=u²-1,2x/p=1/(u²-1),x=p/2(u²-1),當x=0時,u=+∞；當x=p/2時,u=√2.
dx=-2udu/2(u²-1)²=-udu/(u²-1)²
故S=[0,p/2]∫√[1+(p/2x)]dx=[-∞,√2]-∫[u²/(u²-1)²]du=[-∞,√2]-∫[u²/(u+1)²(u-1)²]du
=[-∞,√2]-(1/4)∫{[u/(u-1)²]-[u/(u+1)²]du}=[-∞,√2]-(1/4)∫{[1/(u-1)²+1/(u-1)]-[1/(u+1)-1/(u+1)²]}du
=-(1/4)[-1/(u-1)+ln︱u-1︱-ln︱u+1︱-1/(u+1)]︱[-∞,√2]
=(1/4)[1/(u-1)-ln︱(u+1)/(u-1)︱+1/(u+1)]︱[-∞,√2]
=(1/4){1/(√2-1)-ln[(√2+1)/(√2-1)]+1/(√2+1)}
=(1/4)[(√2+1)+(√2-1)]=(√2)/2y′=p/y=p/√(2px)于是弧長S=[0，p/2]∫[√(1+y′²)]dx=[0，p/2]∫√[1+(p/2x)]dxy′=p/√(2px)應該是s=[0，p/2]∫√[1+(p^2/2x)]dx吧，不過我現(xiàn)在會了，所以仍然很謝謝你，嘻嘻！