Question

求曲线在x=sint ,y=cos2t t=π/6的切线方程及法线方程？

Answer 1

t=π/6时x=y=1/2,
dx=costdt,dy=-2sin2tdt,
所以dy/dx=-4sint=-2,
所以所求切线方程是2x+y-3/2=0,
法线方程是x-2y+1/2=0.

Answer 2

x=sint
x(π/6) =1/2
dx/dt = cost
y=cos2t
y(π/6) = 1/2
dy/dt =-2sin2t
dy/dx = (dy/dt)/(dx/dt) = -4sint
dy/dx | t=π/6 = -4sin(π/6) = -2
切线方程 t=π/6
y-y(π/6) = (dy/dx |t=π/6 ). [x-x(π/6) ]
y -1/2 = -2 ( x- 1/2)
2y -1 = -2(2x -1)
4x+2y -3 =0
法线方程 t=π/6
y -1/2 = (1/2) ( x- 1/2)
2y - 1 = x- 1/2
4y -2 = 2x- 1
2x-4y+1 =0