Question

(2)求曲线 x=sin2t y=cos t/2 在 t=/2 处的切线方程

Answer 1

x=sin2t，
dx=2cos2tdt
y=cost/2
dy=-1/2sint/2dt
k=dy/dx=-1/2sin1/2cos4

Answer 2

估计是求t=π/2处的切线方程。
x=sin2t
dx/dt=2cos2t
y=cost/2
dy/dt=-1/2sint/2
dy/dx=(dy/dt)/(dx/dt)
=(-1/2sint/2)/(2cos2t)
=(-sint/2)/(4cos2t)
t=π/2时
x=sin(2×π/2)=0
y=cos[(π/2)/2]=√2/2
切点(0,√2/2)
斜率：k=[-sin(π/2)/2]/[4cos(2×π/2)]
=(-√2/2)/[4×(-1)]
=√2/8
切线方程：y-√2/2=√2/8(x-0)
√2x-8y+4√2=0