曲线y＝sinx(0＜x＜π)上哪一点的曲率半径最小,并求出该点处的曲率半径

Question

曲线y＝sinx(0＜x＜π)上哪一点的曲率半径最小,并求出该点处的曲率半径

Answer 1

y'=1/x(x>0)y''=-1/x^2(x>0)ρ=1/K,曲率半径ρ越小,曲率K越大K=|y''/(1+y'^2)^(3/2)|=|-1/x^2/(1+1/x^2)^(3/2)|=x/(x^2+1)^(3/2),x>0令dK/dx=[1*(x^2+1)^(3/2)-x*(3/2)*(x^2+1)^(1/2)*2x]/(x^2+1)^3=[(x^2+1)-3x^2]/(x^2+1)^(5/2)=(1-2x^2)/(x^2+1)^(5/2)=0,得1-2x^2=0,x^2=1/2,x=1/√2在x=1/√2附近,x0；x>1/√2时,dK/dx