※ 引述《kkgfdsaa (Jared)》之銘言：
: Find a differentable function y=f(x) such that f(0)=0
: and
: the arc length of the graph of f between (0,0) and (x,y) is e^(x)+y-1.
: 懇請各位大大幫忙一下
: 謝謝~
x
∫ sqrt( 1 + (f'(t))^2 ) dt = exp(x) + f(x) - 1
0
1 - exp(2x)
=> f'(x) = -----------------
2exp(x)
=> f(x) = -cosh(x) + c
代入 f(0) = 0 得到 f(x) = 1 - cosh(x)
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