Question

一道大学高数导数题，谢谢？

Answer 1

(0,1)
y=e^t .cost
1=e^t.cost
=> t=0
x=e^t. sin(2t)

dx/dt = (2cos2t + sin2t ) .e^t
y= e^t.cost
dy/dt = ( -sint +cost) e^t
dy/dx = (dy/dt)/(dx/dt) = ( -sint +cost)/(2cos2t + sin2t )
dy/dx |t=0 = ( -0+1)/(2+0) = 1/2
切线方程 (0,1)
y-1 = (1/2)(x-0)
x-2y +2 =0
法线方程 (0,1)
y-1 = -2(x-0)
2x+y-1=0