Question

计算抛物线y^2=2px(p>0)从顶点到点(p/2,p)的一段曲线弧长。

Answer 1

计算抛物线y²=2px(p>0)从顶点到点(p/2,p)的一段曲线弧长。
解：对y²=2px取导数得 2yy′=2p，故y′=p/y=p/√(2px)
于是弧长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)，当x=0时，u=+∞；当x=p/2时，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)/2

追问

y′=p/y=p/√(2px)
于是弧长S=[0，p/2]∫[√(1+y′²)]dx=[0，p/2]∫√[1+(p/2x)]dx
y′=p/√(2px)
应该是s=[0，p/2]∫√[1+(p^2/2x)]dx吧，不过我现在会了，所以仍然很谢谢你，嘻嘻！