绘制x=sint,y=cost.z=t.在[0,6*pi]的图形
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x²+y²=25sin²tz²=25cos²t所以x²+y²+z²=25
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
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
∵(sint+cost)²=sin²t+2sintcost+cos²t=1+2sintcost∴x²=1+2y∴y=x²/2-1/2
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
(costdt)/(-sintdt)=-cott再答:或-1/tant
t=arccos(1-y)x=arccos(1-y)-sin[arccos(1-y)]【sin(arccosx)=√(1-x²)】=arccos(1-y)-√[1-(1-y)²]=
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=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)^
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)/
t=0:0.01:27;x=sin(t);y=cos(t);z=t;plot3(x,y,z)见图
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
直接求导,根据导数也就是微商的定义y'=dy/dx=(dy/dt)/(dx/dt)=-sint/cost=-tgt当t=Pi/4时,y'=-tgt=-1,并且曲线过点(sqrt2/2,sqrt2/2)
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