求曲线x=3*sin(t)与y=cos(2t)在t=π/6处的切线方程和法线方程？
1个回答
展开全部
x(π/6) = 3/2, y(π/6) = 1/2
x'(π/6) = 3cos(π/6) = 3sqrt(3)/2
y'(π/6) = -2sin(π/3) = -sqrt(3)
dy/dx = -2sin(π/3)/[3cos(π/6)] = -2/3
切线方程: y = (-2/3)x + 3/2
法线方程: y = (3/2)x - 7/4
x'(π/6) = 3cos(π/6) = 3sqrt(3)/2
y'(π/6) = -2sin(π/3) = -sqrt(3)
dy/dx = -2sin(π/3)/[3cos(π/6)] = -2/3
切线方程: y = (-2/3)x + 3/2
法线方程: y = (3/2)x - 7/4
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
