# 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.

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.

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.

# [转载] 给梅加强老师的一封信

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# [转载] 尤老师的乒乓球之路

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