Question

设曲线x=x(t),y=y(t)由方程组x=te^t e^t+e^y=2e 确定,求该曲线在t=1处的曲率k.答案是k=

Answer 1

x对t求导：x'(t)=e^t+te^t=e^t(t+1)

y对t求导：e^t+e^y *y'(t)=0,得：y'(t)=-e^(t-y)

故y'(x)=y'(t)/x'(t)=-e^(-y)/(t+1)

记A(t)=d(y'(x))/dt=-[-y'(t)e^(-y)(t+1)-e^(-y)]/(t+1)^2=e^(-y)[-e^(t-y)(t+1)+1]/(t+1)^2

t=1时,x(1)=e,y(1)=1,xt'(1)=2e,y'x(1)=-e^(-1)/2,A(1)=-e^(-1)/4

y"(x)=d(dy/dx)/dt /(dx/dt)

y"(1)=A(1)/x'(1)=-e^(-1)/(8e)=-1/(8e^2)

k=|y"|/(1+y'^2)^(3/2)=1/[(8e^2)*(1+e^(-2)/4)^(3/2)]=e(1+4e^2)^(-3/2)

书上的答案是对的.注意在算二阶导数的时候容易出错.