弧长s = int (1+k‘)^1/2 dx
ds/dx = (1+k‘)^1/2 = d[(e^x)-1 ]/dx = e^x
1+k'=e^2x
k=1/2* e^2x-x+c
x>0,k=1/2* e^2x-x+c>0; c>x- 1/2* e^2x
弧长s = int (1+k‘)^1/2 dx
ds/dx = (1+k‘)^1/2 = d[(e^x)-1 ]/dx = e^x
1+k'=e^2x
k=1/2* e^2x-x+c
x>0,k=1/2* e^2x-x+c>0; c>x- 1/2* e^2x