x=2sin(t),y=cos(t+pie),求dy/dx,和d^2y/dx^2,在点（0,-1）

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/12 17:21:15

x=2sin(t),y=cos(t+pie),求dy/dx,和d^2y/dx^2,在点（0,-1）
我求出了dy/dx=sin(t)/2cos(t),在点（0,-1）时,dy/dt=0
哪个高手帮我求一下d^2y/dx^2,

x=2sin(t),y=cos(t+pie
x=2sint
y=-cost
x/2)^2+y^2=1
x^2+4y^2-4=0
则：2x+8ydy/dx=0
dy/dx=-x/4y=-2sint/(-4cost)=sint/(2cost)
2x+8ydy/dx=0
1+4[(dy/dx*dy/dx)+yd^2y/dx^2]=0
代入dy/dx=-x/4y
1+4[（-x/4y)^2+yd^2y/dx^2]=0
1+4(x^2/(16y^2)+ 4yd^2y/dx^2=0
4y^2+x^2+16y^3 d^2y/dx^2=0
d^2ydx^2=-(4y^2+x^2)/(16y^3)
代入x=2sint
y=-cost
d^2ydx^2=-(4(cost)^2+4(sint)^2)/(-16(cost)^3)=((cost)^2+(sint)^2)/(4(cost)^3)=1/(4(cost)^3)

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