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lim (x→0,y→0)(xy%sin(xy))/(xy%xyCos(xy))=1/3,...

0/0型等价交换公式,xy-sinxy等价于(xy)³/6, 1-cosxy等价于(xy)²/2,所以原式就等于见图吧 等价公式很重要呀,要记得

令t=xy, x->0,y->0即得t->0 Taylor展开:sin(t)=t-t^3/6+o(t^3),cos(t)=1-t^2/2+o(t^2) 所以原式=lim(t->0)(t-sin(t))/(t-tcos(t))=lim(t^3/6+o(t^3))/(t^3/2+o(t^3))【o(t^3)是无穷锌 =1/3+o(1)=1/3 编辑的格式不规范,但应该不影响理解

由圆的方程得到圆心O(0,0),半径r=5,∵圆心O到直线l的距离d=1cos2θ+sin2θ=1<5,且r-d=5-1>1,∴圆O上到直线l的距离等于1的点的个数为4,即k=4.故答案为:4

对x求导是ycos(xy).对y求导是xcos(xy)

这涉及到大学里偏导数的概念,对x求导,可以把y看做常数,这样就很好理解了吧

cosxy=x -sinxy*(y+xy')=1 y+xy'=-cscxy y'=-(cscxy+y)/x. y=cos(x+y) y'=-sin(x+y)*(1+y') y'[1+sin(x+y)]=-sin(x+y) y'=-sin(x+y)/[1+sin(x+y)].

sin(xy)+3x-y=1 y=y(x) cos(xy) (y+xy') + 3-y' =0 ycos(xy) + xy'cos(xy) - y' = -3 -ycos(xy) y'[1-xcos(xy)] = [3+ycos(xy)] y' = [3+ycos(xy)] / [1-xcos(xy)]

dz=cos(xy)xdy+cos(xy)ydx 由z/x=ycos(xy),z/y=xcos(xy)→dz=ycos(xy)dx + xcos(xy)dy. cos(x-y)dx-cos(x-y)dy

u=xysin(x+y) du=[ysin(x+y)+xycos(x+y)]dx+[xsin(x+y)+xycos(x+y)]dy

xy的坐标可不可以用sincos表示 可以, 把直角坐标系的原点作为极点,x轴的正半轴作为极轴,并在两种坐标系中取相同的长度单位,设M是平面内任意一点,它的直角坐标是(x,y), 极坐标是(ρ,θ),从点M作MN⊥Ox,由三角函数定义,可以得出x,y与ρ,θ之间的...

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