Question

数学微积分求解

Find an equation of the tangent line to the parametric curve x=2sin(2t) y=2sin(t) at point (sqrt(3),1) , where does this curve have horizontal or vertical tangent ?

参数方程x=2sin(2t) y=2sin(t),求在坐标 （根号下3，1）上的切线

Answer 1

x= 2sin(2t)
dx/dt = 4cos2t
y=2sint
dy/dt = 2cost
dy/dx = dy/dt.dt/dx = cost/(2cos2t)
√3 = 2sin2t (1)
1= 2sint (2)
(1)/(2)
2cost =√3
t = π/6
dy/dx| t=π/6 = (√3/2)

tangent of the equation (√3,1)
y-1 = (√3/2)(x-√3)

horizontal tangent ie dy/dx = 0
cost =0
t = π/2 or 3π/2
ie
(0, 2) or (0,-2)

vertal tangent , 1/dy/dx = 0
cos2t =0
t = π/4 or 3π/4 or 5π/4 or 7π/4
ie
(2, √2) or (-2,√2) or (2,-√2) or (-2,-√2)

追问

dy/dx| t=π/6 = (√3/2)

等于根号3/3吧？

追答

dy/dx = cost/(2cos2t)
dy/dx| (t=π/6) = cos(π/6)/(2cos(π/3))
= √3/2

Answer 2