主题:【求教数学问题】如何判别这个函数在原点附近的性质? -- 晨枫
这是女儿的微积分题,高三水平。这还是第一次被难道,汗颜中
已知dy/dx=(6*x^2+2*x-5*x^4)/(4*y^3-6*y^2+2*y)
1、根据dy/dx,可以看出函数在原点附近是怎样的(原文是behaviorof the curve at the origin)?
2、嘉定在原点的斜率既非零也非无穷大,如何用极限求出函数在原点的斜率?考虑从四个象限接近原点。
3、描述函数在中心区域的大体形状
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初步设想:
把x和y从分子分母提出来,那剩下的分子就是6*x+2-5*^3,分母就是4*y^2-6*y+2
假设分子的根为x1、x2、x3,分母的根为y1、y2,那函数在y=y1或y2的水平线上但x不等于x1、x2、x3的所有点上,函数的斜率为无穷大;同理,函数在x=x1或x2或x3的垂直向上弹y不等于y1或y2的所有点上,函数的斜率为零。然后呢?
一般高中,大学的微积分不要求这些。
分区间讨论轨迹。
相当于大作业。
分区间讨论轨迹怎么入手呢?
不过at the origin, 应该是(0,0)的情况,而不是附近.
把两边的分子分母对角乘,然后积分,
左边=(y^2)(y-1)^2
右边=(x^2)(1+2x-x^3)
那么x=0,y=0是一个解.(0,0)在曲线上.
原点的斜率,按x,y都无穷小,取极限,就用原题中的dy/dx,按不同象限接近,最后结果应该是+/-1.
那么在原点附近,曲线是一条过原点,斜率为+/-1的两直线.
水平太高啦。。。
什么地方的中学呀?
右边应该是
(-X^3)*(X-1)^2
school level math, but you must noticed that at (0,0) it is a special point for the first derivative. At this point, the value is different when you approach from different quadrates. When you approach from 1st quadrate, x>0 and y> 0 so the limit is 1. From second quadrate, x>0 and y< 0 so it is –1. and so on.
To calculate the above, simply use the 'delta' approach so only the lowest powered items (2x and 2y) are count.
This is obviously a saddle like surface. Not exactly sure what else they what to know.
First consider x!=0 and y!=0
For (x,y) in a neighborhood near (0,0), we can ignore the higher order terms, and the numerator becomes 2x, denominator becomes 2y, therefore, the ordinary differential equation can be approximated as
dy/dx = x/y [1]
This gives us sign of dy/dx in four quadrants, they are positive in the first and third quadrants and negative in the second and fourth quadrants. So we know how the curves of y goes.
integrate [1] gives
y^2 = x^2 +c [2]
where c is a constant. Hence, the functions we sought is a cluster functions, we can then discuss their properties in the four quadrants.
For x=0 and y!=0, dy/dx=0, so the curve is horizontal across y axis.
For x!=0 and y=0, dy/dx is not defined, i.e. the curve is vertical across x axis, indeed, this agrees with the sign changes of dy/dx discussed above.
积分后没有x^4项啊
不知道她是不是已经学过L'Hopital's rule。如果是这样,因为当(x,y)\to(0,0)的时候分子、分母都趋于零,我们有
dy/dx |_{(x,y)=(0,0)}
=\lim_{x\to 0}(6*x^2+2*x-5*x^4)/(4*y^3-6*y^2+2*y)
=\lim_{x\to 0}((6*x^2+2*x-5*x^4)')/((4*y^3-6*y^2+2*y)')
=\lim_{x\to 0}(12*x+2-20*x)/(12*y^2*y'-12*y*y'+2*y'),
所以
(dy/dx |_{(x,y)=(0,0)})^2=\lim_{x\to 0}(12*x+2-20*x)/(12*y^2-12*y+2)=1.
因此在原点处的斜率是1或者-1,在原点附近的图象是近似为一个X形(y^2=x^2)。
不用L'Hopital's rule的话,利用你注意到的因式分解,我们可以注意到(x,y)\to(0,0)的时候近似地我们有
dy/dx=x/y
所以x*dx=y*dy,即d(x^2)=d(y^2),所以x^2=y^2+C,其中C是一个常数。特别的,因为图象过原点,C=0。于是我们再次看到在原点附近的图象近似于两条垂直相交的直线y=x和y=-x。
嗯,大概是这样吧。希望有所帮助。Good luck!
多元函数不存在 L'Hopital's rule, 呵呵
大概是有点隐函数的意思吧......
积分。。。
只要用极限本身的定义就可以了。不过你前面的那个,C其实没有必要。因为题目中已经给出了要求在原点处的behavior,所以Cluster就不存在了。毕竟是高三的题,也没那么复杂的。
x^5当然有。
(6*x^2+2*x-5*x^4)dx 积分,能积出x^4这一项么?
你的 -(X^3)*(X-1)^2 展开后有个 2x^4哦。x^3的系数也不对。你再算算