求x=t×e的t次方与lny+y=t方的参数方程(参数方程x=e的t次方*sint。y=e的t次方*cosx。求二阶导数y“。求详解)
- 参数方程x=e的t次方*sint。y=e的t次方*cosx。求二阶导数y“。求详解
- 参数方程x=e的t次方*cost。y=e的t次方*sint求其导数及二阶导数
- 求参数方程x=e^t,y=ln根号(1+t)确定的函数y=f(x)的一阶导数和二阶导数
- 验证参数方程{x=e^t*sint y=e^t*cost 所确定的函数满足关系式(d^2y/dx^2)*(x+y)^2=2(x*dy/dx-y)
参数方程x=e的t次方*sint。y=e的t次方*cosx。求二阶导数y“。求详解
应该是cost而不是cosx吧
x=(e^t)sint, y=(e^t)cost
∴dx/dt=(e^t)sint+(e^t)cost=(e^t)(sint+cost)
dy/dt=(e^t)cost-(e^t)sint=(e^t)(cost-sint)
∴y'=dy/dx=(dy/dt)/(dx/dt)
=[(e^t)(cost-sint)]/[(e^t)(sint+cost)]
=(cost-sint)/(sint+cost)
y''=dy'/dx=(dy'/dt)/(dx/dt)
=[(-sint-cost)(sint+cost)-(cost-sint)(cost-sint)]/(sint+cost)²/[(e^t)(sint+cost)]
=[e^(-t)][-(sint+cost)²-(cost-sint)²]/(sint+cost)³
=-2[e^(-t)]/(sint+cost)³
参数方程x=e的t次方*cost。y=e的t次方*sint求其导数及二阶导数
因为
dx/dt=e^t * (cost - sint)
dy/dt=e^t * (sint + cost)
所以根据公式
dy/dx=(dy/dt)/(dx/dt)
=(sint + cost)/(cost - sint)
=(tant+1)/(1-tant)
=tan(t+π/4)
另外有
d(dy/dx)/dt=(sec(t + π/4))^2
所以d2y/dx2=(d(dy/dx)/dt) / (dx/dt)
=(sec(t + π/4))^2 / [e^t * (cost - sint)]
=(1/2) * (√2 ) * e^(-t) * (sec(t + π/4))^3
求参数方程x=e^t,y=ln根号(1+t)确定的函数y=f(x)的一阶导数和二阶导数
x=e^t
y=ln√(1+t)
dy/dt=1/[2(1+t)]
dx/dt=e^t
利用参数方程求导的方法
dy/dx=(dy/dt)÷(dx/dt)
=1/[2e^(t)*(1+t)]
d²y/dx²=[d（dy/dx）/dt]÷(dx/dt)
=-0.5e^(-2t)[(2+t)/(1+t)²]
不明白可以追问，如果有帮助，请选为满意回答！
验证参数方程{x=e^t*sint y=e^t*cost 所确定的函数满足关系式(d^2y/dx^2)*(x+y)^2=2(x*dy/dx-y)
x=e^t*sint y=e^t*cost
所以dx/dt=e^t*(sint +cost) ，dy/dt=e^t*(cost-sint)
故dy/dx=(dy/dt) / (dx/dt)= (cost-sint) / (sint +cost)
而
d^2y/dx^2
=d(dy/dx) /dt * dt/dx
=d[(cost-sint) / (sint +cost)] /dt * dt/dx
= [(-sint-cost)*(sint+cost) -(cost-sint)*(cost-sint)] /(sint+cost)^2 * 1/[e^t*(sint +cost)]
= (-1-2sint*cost -1+2sint*cost)/[e^t*(sint+cost)^3]
= -2 / [e^t*(sint+cost)^3]
所以
(d^2y/dx^2)*(x+y)^2
= -2 / [e^t*(sint+cost)^3] * (e^t*sint +e^t*cost)^2
= -2e^t /(sint+cost)
而
2(x*dy/dx-y)
=2[e^t*sint * (cost-sint) / (sint +cost) - e^t*cost]
=2e^t *[sint * (cost-sint) -cost*(sint +cost)] /(sint +cost)
=2e^t *[sint*cost -(sint)^2 -cost*sint -(cost)^2] / (sint +cost)
= -2e^t /(sint+cost)
故
(d^2y/dx^2)*(x+y)^2 =2(x*dy/dx-y) = -2e^t /(sint+cost)
所以这两个式子是相等的