高等数学第二题的3题怎么做的，如何分析 10
展开全部
x = 2te^t + 1, y = t^3 - 3t
dx/dt = 2e^t + 2te^t = 2(1+t)e^t
dy/dt = 3t^2 - 3 = 3(t+1)(t-1)
dy/dx = (dy/dt)/(dx/dt) = 3(t-1)/(2e^t)
d^2y/dx^2 = d(dy/dx)/dx = [d(dy/dx)/dt](dt/dx)
= [d(dy/dx)/dt] / (dx/dt)
= (3/2)[(2-t)/e^t] / [2(1+t)e^t] = (3/4)(2-t)/[(1+t)e^t]
t = 1 时， d^2y/dx^2 = (3/4)/(2e) = 3/(8e)
dx/dt = 2e^t + 2te^t = 2(1+t)e^t
dy/dt = 3t^2 - 3 = 3(t+1)(t-1)
dy/dx = (dy/dt)/(dx/dt) = 3(t-1)/(2e^t)
d^2y/dx^2 = d(dy/dx)/dx = [d(dy/dx)/dt](dt/dx)
= [d(dy/dx)/dt] / (dx/dt)
= (3/2)[(2-t)/e^t] / [2(1+t)e^t] = (3/4)(2-t)/[(1+t)e^t]
t = 1 时， d^2y/dx^2 = (3/4)/(2e) = 3/(8e)
本回答由网友推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
|
广告 您可能关注的内容 |
