x=2sint,y=cost,z=4t,t∈[-π,π]的MATLAB程序怎么编

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x=2sint,y=cost,z=4t,t∈[-π,π]的MATLAB程序怎么编
将空间曲线的参数方程x=3sint,y=4sint,z=5cost化为一般方程

x²+y²=25sin²tz²=25cos²t所以x²+y²+z²=25

已知x=exp(t)sint ,y=exp(t)cost,证明下列方程

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`)`

x=(e^t)sint y=(e^t)cost 求d^2y/dx^2

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

设x=1+t^2、y=cost 求 dy/dx 和d^2y/dx^2 sint-tcost/4t^3 和 sint-tc

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

如何用matlab画x=2sint,y=cost,z=4t,t=-pi:pi图象

t=-pi:0.01:pi;%设定变量区间和绘图步长x=2*sin(t);y=cos(t);plot(t,x,t,y);%分别画出t-x和t-y的曲线gridon;%开网格注:plot函数还可以有其它

∫cost/(sint^2) dt =∫dsint/sint^2 =-1/sint + C

中间那步不用那样的.因为d(sint)=costdt,先把cost换到d里面就是:原式=∫【1/(sint^2)】dsint设sint=x化为∫(1/x^2)dx=-1/x+C再把x换回sint

x=sint+cost y=sintcost 化为普通方程.

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

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=

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)/

把曲线的参数方程化为一般方程:x=3sint,y=4sint,z=5cost (0小于等于t小于2pai)

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

设x=cost,y=sint则(dy)/(dx)=

(costdt)/(-sintdt)=-cott再答:或-1/tant

x(t)=t-sint y(t)=1-cost,想建立x与y的方程,

t=arccos(1-y)x=arccos(1-y)-sin[arccos(1-y)]【sin(arccosx)=√(1-x²)】=arccos(1-y)-√[1-(1-y)²]=

参数方程x=cost+sint,y=sint*cost*(t为参数)的普通方程是多少

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

设x=t^2+cost,y=1-sint,求dy/dx

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

x=sint-cost y=sint+cost 求它得普通方程

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)^

已知﹛x=7(t-sint),y=7(1-cost),则dy/dx=

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

x=a(cost+tsint) y=a(sint—tcost) 求导dy/dx

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

The graph in the xy-plane represented by x=3sint and y=2cost

选B椭圆x=3sinty=2cost是典型的椭圆的参数方程

设x=sint,y=cost则dy/dx=

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

参数方程x=4+5cost,y=5+5sint怎么消去参数

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

设x=cost y=sint-tcost 求dy/dx

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