神马作文网微积分和微分方程问题Suppose f(x) is a continuous function with positiv
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微积分和微分方程问题
Suppose f(x) is a continuous function with positive values and
f(0)=1.If for any x>0,the length of the curve y=f(x) over the interval [0,x] is always equal to the area of the region below this curve and above the x-axis,find the equation of this curve.
设f（x）是一个连续函数并且恒正的函数
f（0）= 1.如果对任意的x>0,曲线的Y长度=函数F（x）在间隔[0,x]是始终等于下面的这条曲线的面积及 以上的X轴,求这个方程曲线.
Suppose f(x) is a continuous function with positive values and
f(0)=1.If for any x>0,the length of the curve y=f(x) over the interval [0,x] is always equal to the area of the region below this curve and above the x-axis,find the equation of this curve.
设f（x）是一个连续函数并且恒正的函数
f（0）= 1.如果对任意的x>0,曲线的Y长度=函数F（x）在间隔[0,x]是始终等于下面的这条曲线的面积及 以上的X轴,求这个方程曲线.
翻译都有问题.
由题意∫(0,x)f(x)dx=∫(0,x)√(1+f'(x)^2)dx
求导得:f=√(1+f'^2),f'=√(f^2-1)
df/√(f^2-1)=dx.两边积分即可.最后利用f(0)=1可解出常数C
再问： √(1+f'(x)^2)是如何来的？
再答： the length of the curve y=f(x) over the interval [0,x] (曲线y=f(x)在区间[0,x]上的长度）=∫(0,x)√(1+f'(x)^2)dx 这是弧长公式
由题意∫(0,x)f(x)dx=∫(0,x)√(1+f'(x)^2)dx
求导得:f=√(1+f'^2),f'=√(f^2-1)
df/√(f^2-1)=dx.两边积分即可.最后利用f(0)=1可解出常数C
再问： √(1+f'(x)^2)是如何来的？
再答： the length of the curve y=f(x) over the interval [0,x] (曲线y=f(x)在区间[0,x]上的长度）=∫(0,x)√(1+f'(x)^2)dx 这是弧长公式
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