Question

大一学渣问下数学大神，这道题怎么做

Answer 1

x=e^t.sint
dx/dt = (sint+cost)e^t

y=e^t.cost
dy/dt= (cost -sint) e^t

dy/dx
= dy/dt/ dx/dt
=( cost - sint)/(cost +sint)
= (1 - tant)/(1 +tant)
= 1 - 2tant/(1+tant)

d/dt (dy/dx)
=-2[ (1+tant)(sect)^2 - tant.(sect)^2 ]/(1+tant)^2
=-2(sect)^2/(1+tant)^2

d^y/dx^2
=d/dt (dy/dx) / (dx/dt)
=[-2(sect)^2/(1+tant)^2] /[(sint+cost)e^t]
=-2(sect)^2/[ (1+tant)^2.(sint+cost).e^t]

Answer 2

dy/dt=e^t*(cost-sint)
dx/dt=e^t*(cost+sint)
所以dy/dx=(dy/dt)/(dx/dt)
=(cost-sint)/(cost+sint)
=ctg(t+π/4)
d(dy/dx)/dt=[csc(t+π/4)]^2

所以d^2y/dx^2=[d(dy/dx)/dt]/(dx/dt)
=[csc(t+π/4)]^2/[e^t*(cost+sint)]
=[csc(t+π/4)]^3/(√2*e^t)

Answer 3

x=e^t·sint
y=e^t·cost
x'=e^t·sint+e^t·cost=e^t(sint+cost)
y'=e^t·cost-e^t·sint=e^t(cost-sint)
x''=e^t(sint+cost)+e^t(cost-sint)=2e^t·cost
y''=e^t(cost-sint)+e^t(-cost-sint)=-2e^t·sint
dy/dx=y'/x'
d²y/dx²=d(dy/dx)/dx
=(y''x'-y'x'')/(x')²
=[-2e^t·sint·e^t(sint+cost)-e^t(cost-sint)2e^t·cost]/[e^t(sint+cost)]²
=[-2e^2t·sin²t-e^2tsin2t-2e^2t·cos²t+e^2tsin2t]/[e^t(sint+cost)]²
=-2/(1+sin2t)