x=cost,y=tcost^2--搜答案-搜搜考试网

证一：为了方便,记x`=dx/dt,y`=dy/dt.则d²y/dx²=d(dy/dx)/dx=d(y`/x`)/dx=[d(y`/x`)/dt]/(dx/dt)=(y`/x`)`

dx/dt=(e^t)sint+(e^t)cost=(e^t)(sint+cost)dy/dt=(e^t)cost-(e^t)sint=(e^t)(cost-sint)dy/dx=(dy/dt)/(d

∵x=1+t²,y=cost==>dx/dt=2t,dy/dt=-sint∴d²y/dx²=d(dy/dx)/dx=(d((dy/dt)/(dx/dt))/dt)/(dx

dx/dt=-e^(-t)sint+e^(-t)cost=e^(-t)(cost-sint)dy/dt=e^tcost+e^t(-sint)=e^t(cost-sint)dy/dx=(dy/dt)/(

=(1+e^t)/(2-sint)不通,看书.

∵(sint+cost)²=sin²t+2sintcost+cos²t=1+2sintcost∴x²=1+2y∴y=x²/2-1/2

用到的知识点：两项乘积的导数（uv）'=u'v+uv'

（costdt）/(-sintdt)=-cott再答：或-1/tant

需要注意的是有个隐藏条件：(sint)^2+(cost)^2=1即(sint+cost)^2-2sint*cost=1将x=cost+sint,y=sint*cost代入得x^2-2y=1,即y=(x

解dy/dx=(1-sint)'/(t²+cost)'=(-cost)/(2t-sint)

x=a(cost+tsint),y=a(sint-tcost)L=∫√(dx²+dy²)dx=atcostdtdy=atsintdt=∫at√((cos²t+sin&su

x=sint-costy=sint+cost则：x+y=2sintx-y=-2cost所以：(x+y)^2+(x-y)^2=2再问：这个不像圆的方程啊再答：这个是圆的方程。(x+y)^2+(x-y)^

先求dx=(cost-tsint)dt,dy=(sint+tcost)dt然后dy/dx=(sint+tcost)/(cost-tsint)根据x=tcost;y=tsint;y/x=tant所以dy

dx=(7-7cost)dtdy=(7sint)dtdy/dx=(7sint)/(7-7cost)再问：有两个答案耶，哪个是对的呀再答：我的应该是对的，当然公因子7可以约掉

解析x=acost+atsinty=asint-atcostdx=-asint+asint+atcostdy=acost-acost+atsint∴dy/dx=(acost-acost+asint)/

dy=lnt+1dx=1-sintdy/dx=(lnt+1)/(1-sint)

dy/dt=-sintdx/dt=cost∴dy/dx=-sint/cost=-tant

x-4=5cost,y-5=5sint(x-4)^2=25cos^2t,(y-5)^2=25sin^2t(x-4)^2+(y-5)^2=25(cos^2t+sin^2t)(x-4)^2+(y-5)^2

x,y随t增减趋势,大致画出图像是从A(1,0) 沿着逆时针到B(1,-2π)的一段曲线..设原题目中P=y+ye^x,Q=x+e^x因为Q'x=P'y,所以原积分与路径无关

dy/dx=y'/x'=tsint/(-sint)=-t再问：在详细一点呗再答：dy/dx=(dy/dt)/(dx/dt)=(cost-cost+tsint)/(-sint)=-t