高数设参数方程x=2cost,y=sint-搜答案-智慧作业帮

x²+y²=25sin²tz²=25cos²t所以x²+y²+z²=25

由x=2(sec^2α-1)(-90`

-1

dy/dx=y'(t)/x'(t)=(sint+tcost)/(1-cost+tsint)再问：要过程谢谢再答：dy=y'(t)dt.dx=x'(t)dt=>dy/dx=y'(t)/x'(t)

x=e^t*sinty=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)/

（I）曲线C1的参数方程式x＝4＋5costy＝5＋5sint（t为参数）,得（x-4)^2+（y-5）^2=25即为圆C1的普通方程,即x^2+y^2-8x-10y+16=0．将x=ρcosθ,y=

x^2=9sin^ty^2=16sin^tz^2=25cos^t三式相加可得一般方程x^2+y^2+z^2=25

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

先化为直角坐标方程：（x-4)/5=cost、(y-5)/5=sint=>(x-4)^2/5^2=cos^2t、(y-5)^2/5^2=sin^2t=>（x-4)^2/5^2+(y-5)^2/5^2=

1.x=4+3ty=2+t3y=6+3t相减x-3y=-2x-3y+2=02.x=cos^2ty=sint平方,相加x+y^2=13.x=a/costcost=a/xy=b*tanty*coxt=b*

x=t+t^2,y=cost所以dx/dt=1+2t,dy/dt=-sint于是dy/dx=(dy/dt)/(dx/dt)=-sint/(1+2t)而d^2y/dx^2=(dy/dx)/dt*dt/d

由∫ydx把y=a(2sint-sin2t),dx=a(-2sint+2sin2t)dt代入计算就行了代入时要注意对称性,只对y>0部分求积分

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

sint=t-x/acost=1-y/asint^2+cost^2=1所以(at-x)^2+(a-y)^2=a^2

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=a(t-sint)∴dx=d[a(t-sint)]=(a-cost)dt∴y=a(1-cost)∴dy=d[a(1-cost)]=asintdt∴dy/dx=(asint)/(a-cost)再问

二阶导数再导一次就好了

没错啊,dx/dt=cost/sint楼主可以把题目拍下来吗?再问：您看下红笔写的是标准答案黑色是我写的再答：答案是不是这个

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