Prediction of Final Exam 2013-2014 Semester I

Module:                 MA 1505 Mathematics I

Time:                      2 hours ( 120 minutes )

Questions:             8 questions, each question contains two questions. i.e. 16 questions.

Average speed:     7.5 minutes per question.

Scores:                  20% mid-term exam, 80% final exam. i.e. Each question in the final              exam is 5%.

Remark:                 Another Possibility: 5 Chapters, each chapter contains 1 big question, and each question contains three small questions, i.e. 15 questions. 8 minutes per question.

The contents in high school:

Trigonometric functions, some basic inequalities and identities.

The contents before mid-term exam: Please review the details of them.

Chapter 2: Differentiation

Derivatives of one variable functions, derivatives of parameter functions, Chain rule of derivatives, the tangent line of the curve, L.Hospital Rule, critical points of one variable, local maximum and local minimum of one variable function.

Chapter 3: Integration

Integration by parts, Newton-Leibniz Formula, the area of the domain in the plane, the volume of the solid which is generated by a curve rotated with an axis.

Chapter 4: Series

Taylor Series and Power Series, radius of convergence of power series, the sum of geometric series and arithmetic series.

Chapter 5: Three Dimensional Spaces

Cross Product and Dot Product of vectors, projection of vectors, the equation of the plane and the line in 3-dimensional space, Distance from a point to a plane, Distance from a point to a line, the distance between two lines in two or three dimensional spaces, the distance between two parallel planes. Intersection points of two different curves.

The contents after mid-term exam: Must prepare them.

By the way, 2-3 questions means at least 2 questions, at most 3 questions. 0-1 question means 0 question or 1 question.

Geometric Graphs in Three Dimensional Space:

http://www.wolframalpha.com

z=x^{2}+y^{2}             infinite paraboloid

z=x^{2}-y^{2}             hyperbolic paraboloid

(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=R^{2}  sphere with radius R and center (x_{0},y_{0},z_{0})

x^{2}+y^{2}=R^{2}          cylinder

ax+by+cz=d, \text{ where } a,b,c,d \in \mathbb{R}             Plane

y=x^{2}+c \text{ and } x=y^{2}+c, \text{ where } c\in \mathbb{R}             Parabola

Chapter 6: Fourier Series:

Fourier series, Parseval’s identity: 2-3 questions. ( Integration by parts, calculate the sum of Fourier coefficients, period 2L functions ( where L is a positive real number), calculate the value of some special series from Fourier series, cosine expansion and sine expansion of function on the half domain).

Chapter 7: Multiple Variable Functions

Directional derivatives, partial derivatives, gradient of functions with two or three variables, Chain Rule of partial derivatives: 1-2 questions. (Pay attention to whether the vector is a unit vector or not. If it is not a unit vector, you should change it to a unit vector first, and then calculate the directional derivatives).

Critical points of two variable functions ( saddle point, local maximum, local minimum): 0-1 question. ( Calculate the partial derivatives first, then evaluate the critical points, so we can decide the property of the critical points from some rules).

Lagrange’s method: 0-1 question. ( Calculate the maximum value of functions under some special conditions. Construct the function first, evaluate partial derivatives secondly, and calculate the critical points of the new functions. In addition, if you use  inequality “arithmetic mean” is greater than “geometric mean”, then the question will become easier.)

Chapter 8: Multiple Integration

Double integration, polar coordinate: 1 question. ( The formula of polar coordinate in the plane).

Reverse the order of integration of double integration: 1 question. ( Draw the picture of domain R and reverse the order of dx and dy).

Volume of the solid: 1 question. ( Double integrals).

Area of the surface: 1 question. ( Partial Derivatives of two variable functions, Polar Coordinate).

Chapter 9: Line Integrals

Length of the curve: 0-1 question. ( Parameter equation of the curves).

Line integrals of scalar fields: 1 question. ( The equation of line segment, the equation of the circle with radius R, the length of vectors). Geometric meaning: the area of the wall along the curve.

Line integrals of vector fields: 1 question. ( The equation of line segments, the equation of the circle with radius R, Dot product of vectors). Physical meaning: Work done.

Conservative vector fields and Newton-Leibniz formula of gradient vector fields: 0-1 question. ( Definition of conservative vector field and its equivalent condition).

Green’s Theorem: 1 question. ( Two cases: the boundary is open; the boundary is closed. If the curve is open, you should close it by yourself.) Pay attention to the orientation, i.e. anticlockwise.

Chapter 10: Surface Integrals

Tangent plain of a surface: 0-1 question. ( Partial derivatives, Cross product of two vectors, Normal vector of a plane)

Surface integrals of scalar fields: 1 question. ( The equation of surface, Cross product of vectors, the length of vectors).

Surface integrals of vector fields: 1 question. ( The equation of surface, Cross product and Dot product of vectors).

Stokes’ Theorem: 1 question. ( Pay attention to the orientation).

Divergence Theorem: 0-1 question. ( Triple integrals).

MA 1505 Tutorial 11: Surface Integral, Divergence Theorem and Stokes’ Theorem

Surface Integrals of Scalar Fields: Assume f: U \subseteq \mathbb{R}^{3} \rightarrow \mathbb{R} is a function, r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a surface S. Then the surface integral is

\iint_{S} f dS= \iint_{D} f(\textbf{r}(x,y)) || \textbf{r}_{x} \times \textbf{r}_{y} || dxdy

where the left hand side is the surface integral of the scalar field and the right hand side is the multiple integration. \textbf{r}_{x} \times \textbf{r}_{y} denotes the cross product between \textbf{r}_{x} and \textbf{r}_{y} ,

|| \textbf{r}_{x} \times \textbf{r}_{y} ||  denotes the length of the vector \textbf{r}_{x} \times \textbf{r}_{y}.

Remark.  If f(x,y,z)=1 for all (x,y,z) \in \mathbb{R}^{3} , and r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a surface, then

the left hand side is \iint_{S} dS = \text{ the surface area of } S.

the right hand side is \iint_{D} \sqrt{1+(f_{x})^{2}+ (f_{y})^{2} } dxdy , since \textbf{r}(x,y)=(x,y,f(x,y)), \text{ where } (x,y) \in D, \textbf{r}_{x}=(1,0,f_{x}) and \textbf{r}_{y}=(0,1,f_{y}), the cross product \textbf{r}_{x} \times \textbf{r}_{y}= (-f_{x}, -f_{y},1).

That means:

\text{ the surface area of } S= \iint_{D} \sqrt{1+(f_{x})^{2}+(f_{y})^{2} }dxdy.

Surface Integrals of Vector Fields:

Imagine that we have a fluid flowing through S, such that \bold{F}(x) determines the velocity of the fluid at \bold{x}. The flux is defined as the quantity of fluid flowing through S per unit time.

This illustration implies that if the vector field is tangent to S at each point, then the flux is zero, because the fluid just flows in parallel to S, and neither in nor out. This also implies that if \bold{F} does not just flow along S, that is, if F has both a tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of \bold{F} with the unit normal vector to S at each point, which will give us a scalar field, and integrate the obtained field as above.

1280px-Surface_vectors

Assume \textbf{F} : U \subseteq \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} is a vector field, r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a surface S. Then the surface integrals of the vector field F is

\iint_{S} \textbf{F} \cdot d \textbf{S} = \iint_{S} \textbf{F} \cdot \textbf{n} dS

The left hand side is the surface integral of vector field and the right hand side is the surface integral of scalar function, since \textbf{F} \cdot \textbf{n} is a scalar function. That means,

\iint_{S} \textbf{F} \cdot d \textbf{S} = \iint_{S} \textbf{F} \cdot \textbf{n} dS = \iint_{D} \textbf{F}( \textbf{r}(x,y)) \cdot ( \textbf{r}_{x} \times \textbf{r}_{y}) dxdy

Divergence Theorem (Gauss’s theorem or Ostrogradsky’s theorem)

This theorem is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface. More precisely, the divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that the sum of all sources minus the sum of all sinks gives the net flow out of a region.

\iint_{S} \textbf{F} \cdot d \textbf{S} = \iiint_{V} \nabla \cdot \textbf{F} dV = \iiint_{V} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) dxdydz

where V \subseteq \mathbb{R}^{3} is a bounded domain and \partial V=S, \textbf{F}=(P,Q,R) is a vector field.

800px-Divergence_theorem.svg

Stokes’ Theorem

\int_{\partial \Sigma} \textbf{F} \cdot d\textbf{r} = \iint_{\Sigma} ( \textbf{curl F} ) \cdot d \textbf{S}

where \textbf{curl F}= (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) \textbf{i} + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) \textbf{j} + ( \frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}) \textbf{k} is a vector field. \Sigma is a compact surface and \partial \Sigma  is the boundary of \Sigma. The curve \partial\Sigma has the positive orientation, that means following the right hand rule.

429px-Stokes'_Theorem.svg

[转载] 给梅加强老师的一封信

作 者: zr9558

标 题: [转载] [合集]给梅加强老师的一封信

时 间: Thu Mar 11 22:37:21 2010

点 击: 111

【 以下文字转载自 D_Maths 讨论区 】

【 原文由 zhaoying 所发表 】

gfzhang (gf) 于Thu Mar 11 20:04:23 2010) 提到:

加强:

您好!

本周三在班车上我们聊的很愉快。其间您告诉我去年尤老师被评为最受学生欢迎的老师 。 您说对于这件事情我们两人要好好反思一下:我们都是给大一上数学分析课的,可为什么我们两人都没有获此殊荣。回来之后我仔细的思考了这一问题。现把我自己的想法总结如下。仅供您参考。

就对教材的熟练程度上,以及解题能力上,我想尤老师都无法与你我相比。尤老师自己也说,他在大学时成绩很差,只是到了大四成绩才一下子好了起来。可我们都知道,本科大学的第四年除了一些音乐,美术之类的基本上就没什么主课了。而数学分析课是大一大 二开的。可想在三十年以前尤老师就没有学好这门课程。而你我则不同。如同我们的名字 一样,你我都属于那种志向高远,精益求精的人。讲数学分析这种基础课程,我俩自然都是驾轻就熟,游刃有余。

说到这里,您自然会问,既然如此,那为何尤老师讲的课最受学生的欢迎呢?这其实正是问题的关键所在。 尤老师深知自己对这门课程掌握的不够火候,因此他讲课时只讲那些 自己有相当把握的。 比如凡是证明过程超过十行的,尤老师都留给习题课上讲。凡是计算 中间需要技巧的,尤老师就循循善诱的告诉大家,他在课上主要讲思想,这些技巧上的东西就留给习题课上讲吧。如此一来,尤老师的课不仅讲的轻松,学生听的也轻松。就连工科毕业的辅导员杨靖在听完尤老师的课后,都感慨的说:原来数学系的数分课比她们当初 学的大学数学容易多了。当然尤老师有时也会小试牛刀一把。比如有的同学会在课间休息 时请教课后的习题。在确信自己有十足的把握拿下该题时,尤老师总会毅然走上讲台,奋 笔疾书。每当这种时刻,重修的同学都会背起书包,悄悄离开了教室。因为他们知道在接下来的一节课,尤老师会一直专注在这道题上。直到下课铃声响起。

然而让尤老师在教学上取得极大成功的却正是他的这种教学风格。几乎所有同学都觉得尤老师讲课思路清晰,浅显易懂。就连那些入学时被调剂到数学系的学生都对数学产生了极大的兴趣。特别是当他们目睹了头顶无数光环的尤老师在一个并不很难的问题上久攻不克时,更是信心倍增。尤老师的课与其说是一堂数学课,倒不如说是一堂励志课。大家都凝 视着尤老师,对自己的未来充满了梦想。

然而在大一的第二学期所有这一切都变了。你教一班,我教二班。你我讲课行云流水,面面俱到。大多数同学都不能当堂消化所讲内容。而你不时在黑板的角落处写下的思考题, 好多人学期结束时也没做出来。上学期大家都觉得自己是天才,这学期似乎还不如普通人 。前后对比,尤老师与我俩形成了鲜明的对比。在埋怨你我的同时,大家更加想念那个能让他们自信满满的尤老师。

而恰好此时,学校开展了评比活动。结果可想而知。据杨靖告知,在抽查的120人中,有116人把票投给了尤老师。有3个人投票给我。他们都是我河北老乡。只有一人投了你一票。 但那个人不是你的湖北老乡。因为你从来没告诉大家你是湖北人。那个人是辅导员杨靖。 她知道你一向工作很认真,怕你一下子接受不了。

好了,加强,就说这么多吧。我们都还年轻。只要我们好好努力,未来还有好多机会。希望就在前方。

此致

礼!

高飞

[转载] 尤老师的乒乓球之路

作 者: zr9558

标 题: [转载] 尤老师的乒乓球之路

时 间: Fri Jan 8 20:51:28 2010

点 击: 129

【 以下文字转载自 D_Maths 讨论区 】

【 原文由 gfzhang 所发表 】

我们系里和所里各有一张乒乓球台子。水平低一些的就在系里玩,高一些的就在所里。就像CBA与NBA,分得很清楚。比如,何老师和孙老师就总在系里。汪老师,钟老师和武海军他们一般都在所里。在系里打得好的,比如老纪,有时就去一下所里。而状态不能保持的 ,比如朱老师,现在就只能在系里打一打。 当然也有例外,比如尤老师一直在所里打。那只是因为他的办公室在所里。

尤老师不仅自己对乒乓球保持着浓厚的兴趣,而且也热心关注我们系的乒乓球事业。比如在最近的一场师生对抗赛中,尤老师带病参加了比赛。虽然以零比三惜败给我系一个女同 学,可是他那种战斗到底永不言输的精神给我们留下了深刻的印象。

尤老师除了训练的很刻苦之外,还请人买了很贵的球拍,同时从网上下载了教学录像来纠 正自己的动作。尤老师的心中有一个梦想:就是希望有一天能够得到大家的认可,从而名正言顺的加入到我们系的高手俱乐部。因为尤老师是主任,所以汪老师,朱老师他们和尤 老师比赛时总打假球。有时故意回一个很高的球,等尤老师把球扣过来时再大声喊“好球 ”!有一次刚把球回过去,就开始喊“好球”。结果尤老师一拍子把球抡在了网子上。我在旁边实在看不下去了。就说:“尤老师,他们是说这颗球是刚买的,是个好球”。

尽管如此,在过去的一年里有尤老师的乒乓球技术还是取得了可喜的进步。比如他学会了拉下旋球,学会了搓球,并且不再吃我发的不转球。有时我甚至想或许真的有那么一天, 即使我不让球也输给了尤老师。 那该是怎样的一个世界啊。

— ※ 来源:.南京大学小百合站 http://bbs.nju.edu.cn [FROM: 202.119.34.246] —

※ 来源:.南京大学小百合站 http://bbs.nju.edu.cn [FROM: 123.114.37.153]