z=uv-sint 而u=e^t
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再问:果然是大神呀。。
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
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)/(
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
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
f(t)=1/(2j)*(e^(j(w+1)t)-e^((j(w-1)t))因为查表得exp(j*2*pi*f0*t)的傅立叶变换为delta(f-f0),所以原f(x)的傅立叶变换为1/(2j)*(
du/dt=du/dx*dx/dt+du/dy*dy/dt=e^(x-2y)*cost-2e^(x-2y)*3t^2=e^(x-2y)*(cost-6t^2)αz/αx=αz/αu*du/dx+αz/
有些条件是多余的.由z-y²=u⁴,z+y²=v⁴相加得z=(u⁴+v⁴)/2≥u²v²(均值不等式).由v>u
将e^(u+v)=uv两边对u求导得: e^(u+v)*(1+v')=v+u*v' 解得v'=(v-e^(u+v))/(e^(u+v)-u) 即dv/du=(v-e^(u+v))/(e^(u+v
其实就是求z的导数,cost^2求导为2cost*(-sint),t^6求导是6t^5,cost*t^3求导是-sint*t^3+cost*3*t^2,综合起来就是2cost*(-sint)+6t^5
分别把x,y,z,t当做为之数,其余都是常数,求就行了再问:具体怎么做呢?麻烦写清楚些
dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(
dz/dx是z对x的偏导,这样把u,v都带入的话直接球偏导就好了dz/dx=y*e^(xy)*sin(x+y)+e^(xy)*cos(x+y)同理也可得到dz/dy=x*e^(xy)*sin(x+y)
①偏z/偏x=偏z/偏u偏u/偏x+偏z/偏v偏v/偏x=(2uv-v^2)siny+(2uv-v^2)cosy=(2x^2sinycosy-x^2(cosy)^2)siny+(2x^2sinycos
z=e^(x-2y)dz=e^(x-2y)(dx-2dy)(1)x=sintdx=costdt(2)y=t^2dy=2tdt(3)将(2),(3)代入(1)得dz=e^(x-2y)(cost-4t)d
t=0:0.01:27;x=sin(t);y=cos(t);z=t;plot3(x,y,z)见图
f(x)=u(x)v(x)f(x+△x)-f(x)=u(x+△x)v(x+△x)-u(x)v(x)=u(x+△x)v(x+△x)-u(x)v(x+△x)+u(x)v(x+△x)-u(x)v(x)=[u
z=u²v+3uv^4,u=e^x,v=sinx,求dz/dxdz/dx=2uu'v+u^2v'+3u'v^4+3v(4v^3)v'=2e^(2x)sinx+e^(2x)cosx+3e^x(