Question

高数之曲率

对数曲线y=lnx上哪一点处的曲率半径最小？求出该点处的曲率半径。需要详解！ 答案是（根号2/2,-ln2/2)处曲率半径有最小值3根号3/2. 谢谢！

Answer 1

y=lnx
y'=1/x
y''=-x^(-2)
曲率半径
公式ρ=[(1+y'^2)^(3/2)]/∣y"∣
=
(1+
(1/x)^2)^(3/2)
/
(x^(-2))
=x^2
*
(1+
x^(-2)
)^(3/2)
对它
求导
=2x*(1+x^(-2))^(3/2)
+x^2
*
3/2*(1+x^(-2))^(1/2)
*(-2)x^(-3)
=2x*(1+x^(-2))*(1+x^(-2))^(1/2)-3*x^(-1)*(1+x^(-2))^(1/2)
=(2x-1/x)*(1+x^(-2))^(1/2)
=(2x^2-1)*(1+x^2))^(1/2)/x
2x^2-1=0
x=2^(1/2)时,曲率半径最小
代入曲率半径公式ρ=2*(1+1/2)^(3/2)=2*(3/2)^(3/2)