Question

计算y^2=2px在(0，0)到(p/2，P)上的一段弧长

Answer 1

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