MATH 280B

Mathematics 280B (Fall 2012)

Course Logistics

Course Information (Printable)
Tentative Course schedule
Grade Information
TBA --- Math Tutoring Hours (Thompson 390)

Day-by-Day Notes

Tuesday December 4

Topics: Questions on Divergence; Brief introduction to Curl; Fundamental Theorems; What's on the exam

Text: Fundamental Theorems

Tomorrow: Exam #4

Comments:

Today we covered the Divergence Theorem (one of the "Fundamental Theorems" referred to in this handout. We then began discussing the concepts outlined in the "Exam #4 objectives" handout.

Over the past few weeks, we've seen various types of integral involving vector fields and we've seen one new type of derivative for a vector field (the divergence of that vector field. Today, we looked at how how most of the the pieces we've been developing over the past few weeks fit together into the important Divergence Theorem. There is another important result which involves circulation density of a vector field. We can interpret the circulation density of a vector field as another type of derivative of that field. Similar to the Divergence Theorem which relates the double integral of the vector field over a closed surfaceS to to the triple integral of a derivative of the vector field over the solidD bounded by the surface S there is Stokes' Theorem which relates the line integral of a vector field around a closed curve to the double integral of a different derivative of the vector field over the region bounded by the curve. These two results are examples of fundamental theorems of calculus, all of which have the same basic structure: Integrating the derivative of a function over a region gives the same value as integrating the function itself over the (oriented) boundary of the region. In the case of a one-dimensional region such as a curve, the edge consists of only two points so integrating over the edge reduces to adding together two values.

Here is a handout explaining Curl and Stokes' Theorem and this handout summarizes the fundamental theorems of calculus.

Exam #4 is on Monday, December 10 from 8 to 10 am although I will allow you to stay until 11am. Here is the list of specific objectives and my office hours.

Monday December 3

Topics: Questions on integrating vector fields over a surface and 14.3; Divergence; Operator notation

Text: Divergence Handout; ~~possibly Handout on~~ ~~Curl~~

Tomorrow: Questions on Divergence; Fundamental Theorems; What's on the exam

Comments:

Today we reviewed the meaning of a density function and then applied that definition to finding the flux density of a vector field.

This process led to looking at derivatives of vector fields. In particular, we looked at the partial derivatives of each component of a vector field \(\vec{F}\). Each partial derivative tells us how a particular component of \(\vec{F}\) changes with respect to change in a particular coordinate direction. Of the nine partial derivatives (for a vector field in space), two specific combinations turn out to be most important. We looked at the first of these today.

Our starting point for defining divergence of a vector field was to look at flux density at a point. We defined this as a ratio of a flux integral to volume enclosed by the surface in a limit as the surface is shrunk to the point. This definition gives us insight on what divergence tells us about a vector field. In particular, for the fluid flow interpretation, the value of divergence at a point tells us the (percentage) rate at which fluid volume is being created/injected into the flow. At a point where the divergence is positive, fluid is being created or injected into into the flow. At a point where the divergence is negative, fluid is being destroyed or sucked out of the flow.

Computing divergence directly from the definition as flux density is feasible only for simple cases. In order to compute more generally, we developed a coordinate expression for the divergence of a vector field \(\vec{F}=P\,\hat{i}+Q\,\hat{j}+R\,\hat{k}\). That expressions turns out to be \[ \operatorname{div}\vec{F}=\frac{\partial P}{\partial x} +\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}. \]

Details on what we did in class today are on this handout. The handout also has some problems to give you practice with computing divergence.

Here is the list of objectives for Exam #4

Friday November 30

Topics: Questions on integrating a vector field over a surface; Divergence; Operator notation

Text: 14.3; 14.8

Tomorrow: Questions on integrating vector fields over a surface and 14.3; Divergence; Operator notation

Comments:

Today we focused on working examples of computing the integral of a vector field over an oriented surface. In the process we reviewed the method for finding infinitesimal area vectors \(d\vec{A} \) for a surface which in turn required we review how to find the infinitesimal displacement vectors \(d\vec{A} \) for curves.

We also briefly discussed what units are associated with such an integral as a means of gaining intuition from a physical interpretation. In particular, if \(\vec{F} \) is a vector velocity field for a fluid, then \(\iint_{S} F\cdot d\vec{A}\) has the units of \(\frac{m}{s} \times m^2 = \frac{m^3}{s}\). This leads to the physical interpretation of the integral as the (time) rate at which fluid volume flows through the surface \(S\) in the direction of the vectors \(d\vec{A}\).

Thursday, November 29

Topics: Questions on 14.3; Integrating a vector field over a surface
Text: Handout on Integrating a vector field over a surface
Tomorrow: Integrating a vector filed over a surface; Divergence; Operator notation
Comments:: Today we discussed Flux of a vector field across an oriented surface. We reviewed the geomtric meaning of the dot product and expoited it to determine \(\vec{F}\cdot d\vec{A}\), \the amount of a vector field that is affecting an infinitesimal region in the direction of the region's normal vector \(d\vec{A}\). Tomorrow we will discuss the physical interpretation of flux when the vector field is the velocity vector field for a fluid and will do some examples of how to compute the flux. We will also begin discussing the concept of the divergence of a vector field.

Tuesday November 27

Topics: Questions on 14.3, Line integrals around closed loops, Exact differential forms; Surface integrals in vector fields

Text: 14.3; Integrating a vector field over a surface

Tomorrow: Questions on 14.3; Integrating a vector field over a surface

Comments:

Today we talked about four ways of saying that a vector field is conservative. The first is that a vector field is conservatice if and only if there is a potential function for the vector fieldf. The second is that a vector field \(\vec{F}=M\,\hat\imath+N\,\hat\jmath+P\,\hat{k}\) is conservative (i.e., has a potential function) on a "nice" region if and only if \[ \frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}, \qquad \frac{\partial M}{\partial z}=\frac{\partial P}{\partial x}, \qquad\textrm{and}\qquad \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}. \] The third is that a vector field is conservatice if and only if it is independent of path. And the fourth is that a vector field is conservative if and only if the line integral round any closed curve is 0.

We did not discuss the fifth method (although it is in the text) involving exact differential forms. However, that notation is sometimes used in physical chemistry so, depending on your interests, you might wish to read that section of the text.

Monday November 26

Topics: Questions on 14.2 and Curve integrals in vector fields; Potential functions; Path Independence; Conservative fields; Exact differential forms

Text: 14.3

Tomorrow: Questions on 14.3, Line integrals around closed loops, Exact differential forms; Surface integrals in vector fields

Comments:

Today, we reviewed potential functions for vector fields and the Fundamental Theorem of Line Integrals. This FTC generalizes the familiar FTC that you learned earlier in your calculus career. Whereas any function of one variable that is continuous on an interval has an antiderivative defined on that interval, a vector field that is continuous on a region (of the plane or space) does not necessarily have a potential function. So, the FTC for line integrals is useful for integrating a vector field over a curve only if there is a potential function for a region containing the curve and if we can find that potential function.

A vector field that has a potential function for a given region is said to be conservative for that region. Tomorrow we will look at the various ways for determining if whether or not a vector field is conservative without directly looking for a potential function. We will find that the simplest such test only applies to functions defined on a "nice" region. In the context we are working, "nice" means a region that is open, path connected, and simply connected. You should read the relevant part of Section 14.3 for definitions of these.

We also noted that another way to characterize whether or not a vector field is conservative depends on whether or not the value of a line integral for that vector field depends on the specific curve joining the endpoints. If the value of any line integral for a vector field is path-independent for all pairs of endpoints within a given region, then the vector field is conservative.

Tomorrow we will look at three more ways of characterizing whether or not a vector field is conservative: the component test, line integrals around closed loops , and exact differential forms.

Those of you who are or will be taking physical chemistry might want to look at the subsection "Exact Differential Forms" since this language is sometimes used in that course. A differential \(M\,dx+N\,dy+P\,dz\) is exact if it is the result of "d-ing" a function \(V\). In other words, a differential \(M\,dx+N\,dy+P\,dz\) is exact if there is a function \(V\) so that \(dV=M\,dx+N\,dy+P\,dz\). The question of whether or not a differential \(M\,dx+N\,dy+P\,dz\) is exact is equivalent to the question of whether or not the vector field \(\vec{F}=M\,\hat\imath+N\,\hat\jmath+P\,\hat{k}\) is conservative.

Tuesday November 20

Topics: Questions on curve integrals; Circulation; Flux; Potential functions; Fundamental Theorem of Line Integrals
Text: 14.2; 14.3
Tomorrow: Questions on 14.2 and Curve integrals in vector fields; Potential functions; Path Independence; Conservative fields; Exact differential forms
Comments:: Today we reviewed how to compute line integrals of a vector field along a curve C. We also stated and outlined a proof of the Fundamental Theorem of Calculus for Line Integrals and noted that this implied that when our vector field is a gradient field, then the value of the line integral \( \int_{C} \vec{\nabla} f \cdot d\vec{r}\) only depends on the values of f at the beginning and ending points of the curve C and not on how the curve gets from one to the other. In such a situation, we say the scalar function \(f\) is a potential function for the vector field \(\vec\nabla f=\vec{F}\) and that \( F \) is a conservative vector field.

The language of potential function and conservative vector field comes from physics. If we wanted to use language that parallels antiderivative, we might say that \(f\) is an antigradient of the vector field \(\vec{F}\) if \(\vec\nabla f=\vec{F}\). However, the word antigradient is not in standard use.

In our next class, we will revisit the fundamental theorem and go into more detail on potential functions, conservative fields and path independence. In particular, we will work more on the the question Given a vector field \(\vec{F}\), is there a function \(f\) such that \(\vec{F}\) is the gradient of \(f\)?. In other words, is there a function \(f\) such that \(\vec{F}=\vec\nabla f\)? Today, we saw at least one example in which there is such a function for the given vector field. We also saw at least one example in which there is no such function for the give vector field \(\vec{F}\). In this case, we say that the vector field is not conservative.

Monday November 19

Topics: Integrating vector fields along a curve; Section 14.2

Text: Curve integrals in vector fields

Tomorrow: Questions on curve integrals; Circulation; Flux; Potential functions; Conservative vector fields

Comments:

In class, we discussed the idea of integrating a vector field over a curve. As with other types of integration, we can ask

What is it?
How do we compute it?
What does it tell us?

Today, we addressed the first two questions and got some practice in computing. You will need to get more practice by doing the assigned problems from the handout Curve integrals in vector fields and Section 14.2. As you work on these problems, try to visualize each vector field and curve so you can use your geometric intuition to get some sense of what value to expect for the integral.

Friday November 16

Topics: Questions on surface integration problems; Vector fields

Text: 14.2

Tomorrow: Questions on Section 14.2 problems and vector field worksheet; Integrating a vector field along a curve

Comments:

Today, we began looking at vector fields. For much of the course, we have been working with scalar fields (without using this name). A scalar field assigns a number (i.e., a scalar) to each point in a given domain (which can be part of a line, a plane, or space). A vector field assigns a vector to each point in a given domain (which can be part of a line, a plane, or space). Examples of scalar fields include temperature, presssure, and density. Examples of vector fields include velocity, force, and electric fields. In class, we looked at this surface air velocity vector field.

For the last part of the course, we will study the calculus of vector fields. Today, we got a bit of practice visualizing some basic planar vector fields on this handout. As homework, you should finish these problems and work on the assigned problems from Section 14.2. On Monday, we will look at the idea of integrating a vector field along a curve.

Thursday November 15

Topics: Computing the differential area element for surfaces.

Text: Integrating over a surface;14.5

Tomorrow: Questions on integrating over a surface; Vector fields

Comments:

In class, we looked at integrating over a surface. A key part of evaluating a surface integral is expressing the area element dA in terms of the coordinates chosen to describe the surface. In the simplest cases, we can use a geometric argument to deduce an expression for dA (we did this for a sphere earlier this week). In other cases, we need a computational approach. The basic idea is to exploit "nice" families of curves that partition the surface and to compute dA as the magnitude \(\|d\vec{r_1} \times d\vec{r_2} \| \). More details and some examples are on this handout on integrating over a surface. The handout also has the assigned problems for this material.

As with integrating over a curve, we will follow an approach that differs somewhat from the main approach used in the text. The text approaches integrating over curves and surfaces in terms of parametrizing the curve or surface. Our approach is a bit more general. You are welcome to read about the text's approach to integrating over a curve in Section 14.1 and integrating over a surface in Section 14.5.

Tuesday November 13

Topics: 12.7, 13.1-13.5, 13.7, non-uniform density (linear, area, volume), and integrating over curves.
Text: Exam 3
Tomorrow: Integrating over a surface
Comments:

Monday November 12

Topics: Exam Review
Text: 12.7, 13.1-13.5, 13.7, non-uniform density (linear, area, volume), and integrating over curves.
Tomorrow: Exam 3
Comments:

Friday November 9

Topics: Questions on line integrals (integration over curves)

Text: 10.4, Surface Integrals

Tomorrow: Review for Exam 3

Comments:

We started class with a first example of integrating over a surface. As part of this, we needed to describe the surface in terms of a chosen coordinate system and then work out an expression for the area element of the surface in that coordinate system. For the first example, the surface was a sphere and we were able to use a geometric argument to deduce an expression for the area element in spherical coordinates. We will develop a more general approach for more general situations. As part of that, we will use a new tool called the cross product of two vectors.

In the latter part of class, we defined the cross product of two vectors. The cross product of \(\vec{u}\) and \(\vec{v}\) is a new vector denoted \(\vec{u}\times\vec{v}\) and defined geometrically in relationship to the parallelogram that has \(\vec{u}\) and \(\vec{v}\) as its edges. Specfically, \(\vec{u}\times\vec{v}\) is defined geometrically by these two properties:

The direction of \(\vec{u}\times\vec{v}\) is perpendicular to the \(\vec{u}\vec{v}\)-parallelogram so that \(\{\vec{u},\vec{v},\vec{u}\times\vec{v}\}\) has a right-hand orientation.
The magnitude of \(\vec{u}\times\vec{v}\) is the area of the \(\vec{u}\vec{v}\)-parallelogram.

From this definition, it is straightforward to work out the cross product for each pair of the unit coordinate vectors \(\hat\imath\), \(\hat\jmath\), \(\hat k\). For example, \(\hat k\times\hat\imath=\hat\jmath\). Algebraic properties of the cross product include

anticommutative property:	\(\vec{u}\times\vec{v}=-\vec{v}\times\vec{u}\)
distributive property:	\(\vec{u}\times(\vec{v}+\vec{w}) =\vec{u}\times\vec{v}+\vec{u}\times\vec{w}\)
scalar factor property:	\(\alpha(\vec{u}\times\vec{v})=(\alpha\vec{u})\times\vec{v} =\vec{u}\times(\alpha\vec{v})\)

Using these properties and results for crossing pairs of unit coordinate vectors, we can compute the cross product of any two vectors in terms of their cartesian components.

In Section 10.4 of the text, the authors show another way of computing cross products that uses determinants. You can ignore that method if you wish (particularly if you have not seen previously seen determinants.)

Exam 3 is scheduled for Tuesday November 13. As usual, you may use the entire 80 minute block of time and not just the 50 minutes we normally meet on Tuesday. The exam will cover the material in sections 12.7, 13.1-13.5, 13.7, the handout on integrating over curves, and the various handouts on non-uniform density (linear, area, volume), and integrating over curves. Note that neither the material on Lagrange multipliers (section 12.8) nor the material on cross products (section 10.4) is included on this exam. Here is the Exam Objectives handout.

Thursday November 8

Topics: Integration Over Curves; Integration over surfaces.

Text: Integration Over Curves

Tomorrow: Questions on line integrals (integration over curves); Cross Product (10.4)

Comments:

Today, we looked at the idea of integrating over a curve. Here, we think about a curve in the plane or in space that we (conceptually) break into small pieces each of which has a length ds. In some cases, we will add up these small contributions to get the total length of the curve. In other cases, we will have a length density λ defined at each point on the curve and we will add up small contributions of the form λds to get a total (of some quantity such as charge or mass). This handout has some details and the assigned problems.

will not pursue that part of the technique.

Exam 3 is scheduled for Tuesday November 13. As usual, you may use the entire 80 minute block of time and not just the 50 minutes we normally meet on Tuesday. The exam will cover the material in sections 12.7, 13.1-13.5, 13.7, the handout on integrating over curves, and the various handouts on non-uniform density (linear, area, volume), and integrating over curves. Note that the material on Lagrange multipliers (section 12.8) is not included on this exam. Here is the Exam Objectives handout.

Tuesday November 6

Topics: Length of curves; Totals from density function defined along curves.

Text: 13.7 ~~Integration Over Curves~~

Tomorrow: Integration Over Curves; Integration over surfaces.

Comments:

We spent today doing examples where we set up triple integrals for computing totals from volume density functions. Our focus on the set up was to use spherical and/or cylindrical coordinates.

At this point, given a density function, we know:

if it is a linear density function, we can compute the total over a straight line segment.
if it is an area density function, we can compute the total over various "nice" regions in the plane (rectangles, disks, interiors of lemniscapes, etc.)
if it is a volume density function , we can compute the total over various "nice" solids in three-dimensional space (rectangular solids, solid spheres, between spheres and paraboloids, etc.)

Beginning on Thursday we will develop methods for using linear or area density functions to compute the corresponding totals over, respectively, one-dimensional curves that are not straight line segments or two dimensional surfaces that are not planar. The method extends to computing the total associated with a volume density function over a three-dimensional object that lives in four-dimensional space although we will not pursue that part of the technique.

Exam 3 is scheduled for Tuesday November 13. As usual, you may use the entire 80 minute block of time and not just the 50 minutes we normally meet on Tuesday. The exam will cover the material in sections 12.7, 13.1-13.5, 13.7, the handout on integrating over curves, and the various handouts on non-uniform density (linear, area, volume), and integrating over curves. Note that the material on Lagrange multipliers (section 12.8) is not included on this exam. Here is the Exam Objectives handout.

Monday November 5

Topics: Questions on 13.5, 13.7

Text:13.7; ~~Integration Over Curves~~

Tomorrow: Integration Over Curves

Comments:

In class, did more examples of iterating triple integrals using cylindrical coordinates and introduced spherical coordinates.

Tomorrow we will work through the Total from volume density handout (and maybe the Total from area density handout) to give more examples of iterating triple integrals using either cylindrical or spherical coordinates and will begin the discussion of how to integrate over curves.

Exam 3 is scheduled for Tuesday November 13. As usual, you may use the entire 80 minute block of time and not just the 50 minutes we normally meet on Tuesday.

Friday November 2

Topics: Questions on triple integrals in Cartesian coordinates; Introduction to cylindrical and spherical coordinates.
Text: 13.7
Tomorrow: Questions on 13.5, 13.7; Introduction to integration over curves
Comments:

Thursday November 1

Topics: Triple integrals and iterated integrals in Cartesian coordinates; Introduction to cylindrical and spherical coordinates.

Text: 13.5, 13.7

Tomorrow:

Comments:

Today we discussed another example of using polar coordinates to evaluate a double integral.

We also looked at a simple example of a triple integral and the corresponding iterated integrals in three variables. A triple integral involves adding up infinitely many infinitesimal contributions to a total over a region of space. To describe this type of region, we need a three-dimensional coordinate system so we end up with an iterated integral in three variables (that is, the three coordinate variables). For the example we looked at today, we used cartestian coordinates. In some other example, we might find it convenient to use some other coordinate system. Tomorrow we will look at more examples using Cartesian coordinates and introduce two other coordinate systems for three dimensions: cylindrical coordinates and spherical coordinates.

Here are handouts for using double and triple integrals to compute totals from area and volume density functions. Area density problems and volume density problems.

Exam #3 will be on Tuesday November 13.

Tuesday October 30

Topics: Questions on 13.4; Triple integrals and iterated integrals in Cartesian coordinates

Text: ~~13.5~~; 9.2, 13.4

Tomorrow: Triple integrals and iterated integrals in Cartesian coordinates; Introduction to cylindrical and spherical coordinates.

Comments:

Today we did not cover any new material. Instead we did more examples of graphing equations given in polar coordinates and revisited using polar coordinates to evaluate double integrals.

The writing exercise is due on Thursday November 1. You should refer to the writing exercise handout and the notes on writing in mathematics for general directions.

I have posted new homework problems.

Monday October 29

Topics: Questions on polar coordinates and graphing; Double integrals in polar coordinates

Text: 13.4

Tomorrow: Questions on 13.4; Triple integrals and iterated integrals in Cartesian coordinates

Comments:

Today, we looked another example of evaluating a double integral using an iterated integral in polar coordinates. To set up an interated integral in polar coordinates (r,θ), we need to do three things:

Describe the region with appropriate bounds on r and θ (with at least one of the two having constant bounds).
Find an expression for the integrand in terms of r and θ.
Use \(dA=r dr d\theta\) or \( dA=r d\theta dr \) with the choice dictated by which of the variables has constant bounds.
Compute the resulting iterated double integral using our standard techniques.

The writing exercise is due on Thursday November 1. You should refer to the writing exercise handout and the notes on writing in mathematics for general directions.

Friday October 26

Topics: Questions on double integrals and polar coordinates; Area density problems;

Text: 9.1, 9.2

Tomorrow: Questions on polar coordinates and graphing; Double integrals in polar coordinates

Comments:

Today, we discussed the basics of polar coordinates and plotting polar curves. The animation below shows the curve r=cos(2θ) being traced out as θ increases. Note that half of the points have negative \(r\) values and hence are being plotted on the ray opposite to the ray specified by θ. Monday, we'll address questions from Section 9.1 and 9.2 problems. If this does not provide you with enough comfort using polar coordinates, we can arrange a time to talk outside of class.

< figure class=center> polar curve animation

Many graphing calculators have a polar graphing feature. On a TI-8X, you can get to this by going to the MODE menu and choosing the option Pol from the list Func Par Pol Seq. If you then go to the Y= menu, you will see r1= where you can enter a formula for a polar curve. On WolframAlpha, you can just type in the polar relation.

The writing exercise is due on Thursday November 1. You should refer to the writing exercise handout and the notes on writing in mathematics for general directions. Also review the paragraph on projects in the course information handout.

Thursday October 25

Topics: Questions on double integrals; Double integrals over general regions (13.2); Polar coordinates and graphing using polar coordinates (9.1, 9.2)

Text: 13.2, Area density problems, 9.1, 9.2

Tomorrow: Questions on double integrals and polar coordinates; Area density problems; Double integrals in polar coordinates

Comments:

Today, we looked at double integrals over non-rectangular regions. To set up an equivalent iterated integral, we need to describe the region with bounds on the cartesian coordinates x and y. If the region is rectangular, we will have constant lower and upper bounds on both x and y. If the region is not rectangular, we will have constant lower and upper bounds on either x or y and at least one nonconstant bound on the other one. The variable with constant bounds must be the outer variable in the iterated integral. In some cases, it is best to split the original region into smaller pieces and to then describe each piece with appropriate bounds on x and y.

Tuesday October 23

Topics: Questions on non-uniform density, Double integrals over rectangles and general regions

Text: 13.1, 13.2

Tomorrow: Questions on double integrals; Double integrals over general regions (13.2); Polar coordinates and graphing using polar coordinates (9.1, 9.2)

Comments:

Today, we began looking at multivariable integration. In particular, we looked at double integrals. A double integral involves adding up infinitely many infinitesimal contributions over a two-dimensional region. A double integral of the function f over the region R is denoted \[ \iint\limits_{R}f\,dA. \] To evaluate a double integral, we choose a coordinate system and set up an equivalent iterated integral. In cartesian coordinates, we will have either \[ \int_a^b\int_c^d f(x,y)\,dydx \qquad\textrm{or}\qquad \int_c^d\int_a^b f(x,y)\,dxdy. \] Fubini's Theorem states that if f is continuous throughout R (except, possibly, on a finite set of curves with zero area), then \[ \iint\limits_{R}f\,dA=\int_a^b\int_c^d f(x,y)\,dydx = \int_c^d\int_a^b f(x,y)\,dxdy. \] So, we can evaluate a double integral of a continuous function by setting up and evaluating either of the corresponding iterated integrals. Each of the iterated integrals represents a particular way or organizing the "adding up" represented by the double integral. Typically, we will evaluate an iterated integral by successively applying the Fundamental Theorem of Calculus.

In Section 13.1, the authors approach double and iterated integrals within the context of computing volume for a solid region bounded by the graph of a function of two variables. This generalizes the idea of computing area for a planar region bounded by the graph of a function of one variable and supplies valuable insight into why double integrals do what we claim they do. In class, we will put more focus on the "total from density" interpretation/application because this context is relevant in other settings -- particularly settings that involve higher dimensions. The "area under a curve" or "volume under a surface" interpretation/application is less readily generalized to high dimensions.

For reference, here is the handout on the Greek alphabet I passed out on Monday.

Monday October 22

Topics: Questions on Lagrange multipliers; Non-uniform density; Double Integrals

Text: Non Uniform Density; 13.1

Tomorrow: Questions on non-uniform density, Double integrals over rectangles and general regions (13.1, 13.2)

Comments:

After translating an inscribed cylinder in a sphere homework problem into the language of mathematics, we started talking about nonuniform density and linking it to integration.

At a fundamental level, integration is adding up infinitely many infinitesimal contributions to a total. In your first look at integration (in first year calculus), you probably focused on one main application, namely computing area under a curve. You can think about this as computing the area for a "rectangle with variable height". You might also have used integration to accumulate a total of some quantity from a variable rate of accumulation (a rate of change). In this course, we will use a third context as our primary application: using integration to compute a total amount of stuff from a density for that stuff.

Your first introduction to density (a long time ago) likely came as something like "density is mass divided by volume". Turning this around, we can say "mass is density times volume". Getting a total mass from a density by multiplication works for situations with uniform density. For nonuniform density, we will get a total from a density using integration.

In addition to generalizing to nonuniform density, Our use of density will be more general than your initial view in two other ways:

other dimensions: length, area, or volume
other quantities: mass, number, charge, cost, probability,...

To denote a length density, we will typically use \(\lambda\) (the Greek letter "lambda"). For area density, we will generally use \(\sigma\) (the Greek letter "sigma"). For the more familiar volume density, we will use either \(\rho\) (the Greek letter "rho") or \(\delta\) (the Greek letter "delta"). If the stuff (mass, number, charge,...) is spread out uniformly (that is, the density is constant), then we can get the total amount of stuff by multiplication. If the stuff is not spread out uniformly, we need integration to compute the total amount of stuff. In class, we looked at doing this with stuff spread out on a line segment. This handout on nonuniform density has related examples and problems.

Here is a Greek alphabet handout with information about how those letters are used in mathematics.

Friday October 19

Topics: Questions on Lagrange multipliers; Nonuniform density; Integration
Text: Lagrange Multipliers, Non Uniform Density
Tomorrow: Questions on nonuniform density problems; Integration in more than one variable (13.1)
Comments:: We spent more time looking understanding why the method of Lagrange multipliers and looking at examples of how to use it to solve constrained optimization problems. Here is a highly detailed write-up of the example we did in class.
The homework on Lagrange Multipliers (12.8) is due on Tuesday.

Thursday October 18

Topics: Questions on global extrema; Lagrange multipliers; Constrained optimization (12.8);

Text: 12.8; A pertinent slide-show

Tomorrow: Questions on Lagrange multipliers; Nonuniform density; Integration

Comments:

Here is a handout with hints for exam solutions.

We started today by reviewing how to locate and classify local extrema as well as how to translate applied optimization problems from English into mathematics.

We also began looking at constrained optimization problems using the method of Lagrange multipliers. This method is motivated geometrically by looking for points at which a level curve/surface of the objective function is tangent to the constraint curve/surface. This is equivalent to points at which the objective function gradient vector is aligned with the constraint function gradient vector. Here is a "slide-show" illustrating both this Lagrange multiplier approach and the earlier "solve for a variable and substitute approach.

I have added a problem to be turned in from Section 12.8 (Lagrange multipliers) on the Homework page. It is due on Tuesday October 23.

Friday October 12

Topics: Global Extrema; Constrained optimization
Text: 12.7
Tomorrow: Lagrange multipliers; Constrained optimization (12.8); A pertinent slide-show
Comments:: Today, we worked on finding global (absolute) maxima and minima when the domain is constrained. In constrained optimization problems we are given a function to maximize (or minimize) called the objective function and we are also given an equation, called the constraint, that specifies the domain we are interested in. We focused on the first of two methods that can be used when working on a constrained optimization problem. This involves solving the constraint equation for one of the variables and then substituting the result into the objective function. This reduces the number of variables in the objective function and also leaves us with a simpler domain to consider. As part of setting up the function to be optimized, you should also set up any domain restrictions on the independent variables. We also focused on the method to employ when the domain is both closed and bounded. In this case we are guaranteed the existence of both a global maximum and a global minimum. Since we also know the only places an extremum can occur are where the gradient is the zero vector, or where the gradient is undefined, or at points on the boundary of the domain, we noted we could build a complete list of all points where a maximum or minimum could occur and that by evaluating the objective function on those points we can identify the global maximizers and minimizers. This handout contains some additional "applied" optimization problems as homework.

Thursday October 11

Topics: Exam #2
Text: Sections 10.2, 10.3, the second planes handout, 11.1, 11.3, 12.3, 12.5, and 12.6
Tomorrow: Global Extrema; Constrained optimization
Comments:

Tuesday October 9

Topics: Exam overview

Text: Sections 10.2, 10.3, the second planes handout, 11.1, 11.3, 12.3, 12.5, and 12.6

Tomorrow: Exam #2

Comments:

Exam #2 will be on Thursday October 11. We will use the 80-minute period from 9:30 to 10:50. The exam will cover material from Sections 10.2, 10.3, the second planes handout, 11.1, 11.3, 12.3, 12.5, and 12.6. This handout has a list of specific objectives for the exam. There are also a number of handouts pertinent to this material at this link.

You might also want to look at questions from old exams. Since we have covered the material in a different order than previous course offerings, here is a list of appropriate problems from the previous exams that are on my website.

Exam 1: #4,5,6,7,8
Exam 2: #B2, B3, C4
Exam 3: #1,5,6
Exam 4: #1,2
Exam 5: #2

I will be available for office hour today (Tuesday) from 3:00 to 4:30. On Wednesday, I will have time in the late morning and after 3:30. If you have questions, email or call to set up a time to talk. I can address questions sent by email. I have a request in to reserve a room on the third floor of Thompson starting at 7 pm on Wednesday October 11 for an informal Math 280 study session in TH 395. ~~I will post the room number as soon as I hear back from the room scheduler~~. If you are looking for others to study with, come to the classroom after 7 pm and find a small group working on something of interest.

Monday October 8

Topics: More on local extrema; Global extrema; Applied problems

Text: 12.7

Tomorrow: Questions on 12.7; Review for examination; Sections 10.2, 10.3, the second planes handout, 11.1, 11.3, 12.3, 12.5, and 12.6

Comments:

Today, we first discussed the second-derivative test. To better understand why the second-derivative test works, we put together two pieces:

the Taylor series expansion of f based at the critical point \((x_0,y_0)\)
functions of the form \(Ax^2+2Bxy+Cy^2\) for various choices of the constants A, B, and C.

. We then turned our attention to global extreme values. As with local extreme values, we want to distinguish between what it is and how to find it. "What it is" for global extreme values is given by the following definition.

Definition: Given a function \(f:\rightarrow \mathbb{R}\) we say an input \((x_0,y_0)\) is a global maximizer and the corresponding output \(f(x_0,y_0) \) is a global maximum for a given region R if \(f(x_0,y_0) \leq f(x,y)\) for all \((x,y)\) in R. Global minimizer and global minimum are defined similarly with the inequality reversed.

We will do an example of "How to find it" for global extreme values after the examination.

Friday October 5

Topics: Questions on Linearization; Extreme Values

Text: 12.7

Tomorrow: Extreme Values; Absolute extrema

Comments:

Today, we developed the terminology for extrema. Definition: Given a function \( f:\mathbb{R}\to\mathbb{R}\) we say an input \((x_0,y_0)\) is a local maximizer and the corresponding output \(f(x_0,y_0)\) is a local maximum for a given region \(R\) if \(f(x_0,y_0)\geq f(x,y)\) for all \((x,y)\) in some open disk contaied in the domain \(R\). Local minimizer and local minimum are defined similarly with the inequality reversed.

We also set up the algebra for showing that a purely quadratic function on two variables \(z=Ax^2 +2Bxy +cy^2 \) can be rewritten in a form recognizable as related to elliptic paraboloids \( z= \frac{1}{A} \left[ (Ax+By)^2 +(AC-B^2)y^2 \right]\). In this form we can see that depending on the signs of \( A\) (or C) and \( AC-B^2 \), the quadratic is the equation of a paraboloid vertexed at \( (0,0 \) (opening up or down depending on the sign of \(A\) or a parabolic hyperboloid (with saddle point at \((0,0) \).

We also reviewed a bit about Taylor polynomials and how they are used to verify the second derivative test for functions on one variable.

Perhaps the most important thing we did was use Theorem 10 in the text book to note that the only points that can be local mazimizers or minimizers are those that are in one of the following places

on the boundary of the domain of the function,
a point where at least one of the partial derivatives fails to exist, or
an interior point of the domain at which all partial derivatives equal zero.

Exam #2 will be on Thursday October 11. We will use the 80-minute period from 8:00 to 9:20.

Thursday October 4

Topics: Questions on Tangent planes and linearization; Differentials; Extreme values of functions

Text: 12.6; 12.7

Tomorrow: Questions on Linearization; Extreme Values; Second derivative test

Comments:

Today we summarized our recent studies on gradients, directional derivatives, differentiability, and linearization by reviewing our geometric intuition of these topics and the analytical formulas for working with them (which rely on the algebraic manipulations of actual solution procedures). We also looked at how to use differentials to relate (infinitesimally) small changes among variables. Generally, we start with a nonlinear relation among various variables and we then compute a linear relation among the differentials for those variables. Differentials can be thought of as coordinates in the "zoomed-in world". Differentials are always related linearly. Ratios of differentials give rates of change. No limit is needed since the limit process has already been taken care of in "zooming in" process.

Working with differentials complements working with the linearization function L. Differentials are useful when we want to focus on change and rate of change. Linearizations are useful when we want to focus on approximating specific output values.

Exam #2 is scheduled for Thursday October 11. We will use the full 80 minute period the room is available. Here are Exam #2 objectives.

Tuesday October 2

Topics: Questions on 12.5; Tangent planes and linearization

Text: 12.6

Tomorrow: 12.6, 12.7; Differentials; Extreme Values

Comments:

In class, we reviewed tangent lines for functions of one variable and then considered tangent planes for functions of two variables. We then recast these ideas in terms of linearization. The linearization of a function f is a linear function L built using information about f at a specific point P [P=(x0,y0) if f is a funtion on two variables]. If the function is differentiable at a point, then the linearization based at that point is the best linear approximation. There are many contexts in which one trades in the full accuracy of a function for the simplicity of the linearization. In making this trade, it is often essential to have some handle on how much error is introduced by trading in for the linearization. We looked at an error bound for the linearization of a function of one variable. There is a similar error bound for the linearization of a function of two variables. In class, we did not do an example of getting an upper bound on the error in a linearization for a function of two variables. I've written up the details of an example in this handout. You might find this useful in looking at Problem 33 from Section 12.6.

The text's approach to tangent planes and linearization for functions of two variables differs from what we did in class. The text starts with the more general idea of a tangent plane to a surface at a point where that surface is not necessarily the graph of a function z=f(x,y). In reading Section 12.6, you can focus on

Example 2 on page 749 and the box just above that example
The subsection "How to linearize a function of two variables" on pages 751-752
The subsection "Functions of more than two variables"

For reference, here's the applet we looked at in class that allows you to look at tangent planes for the graph of a function of two variables. Tomorrow, we will talk about differentials. This will provide us with a clean and powerful way to look at the linear relations that underlie linearization.

Exam Two is scheduled for Thursday October 11.

Monday October 1

Topics: Questions on Section 12.5 problems; Vector fields, Differentiability, Tangent planes and linearization; Handout on vector fields.

Text: 12.5; 12.6

Tomorrow: Questions on 12.5 and 12.6; Tangent planes and linearization (12.6)

Comments:

Many of the problems on gradient from the text involve looking at a single gradient vector. Last week, to emphasize that there is a gradient vector at each point in the domain, we drew a gradient vector field for the first problem on this handout. As homework, you should complete the two problems on the handout.

At the end of class, we addressed the idea of a directional derivative. For a function f, there are many other directions besides those parallel to the coordinate axes and the direction of largest rate of change. In particular, we denote the directional derivative at a point and in the direction of any specific unit vector \(\hat{u}\) by \(df/ds\) where df represents an infinitesimal rise and ds represents an infinitesimal run. We then compute a directional derivative by finding the component of the gradient vector along the direction of interest. Since the unit vector \(\hat{u}\) gives the direction of interest, the directional derivative is given by \[ \frac{df}{ds}=\vec{\nabla} f\cdot\hat{u}. \] An alternate notation for direction derivative is \(D_{\hat{u}}f\). With this, we can write the result as \[ D_{\hat{u}}f=\vec{\nabla} f\cdot\hat{u}. \] Note again that we can think of a directional derivative as the component of the gradient vector in the direction \(\hat{u}\).

Many of you did not have/take time to look deeply at the Section 12.5 problems before class today. So, we'll take a few minutes at the beginning of class on Tuesday to address more questions and I've pushed back the due date for the problems to be submitted until Thursday.

Friday September 28

Topics: Questions on 12.5; Computing greatest rate of change -- directional derivatives and gradient vectors

Text: 12.5

Tomorrow: More questions on Section 12.5 problems; Vector fields, Differentiability, Tangent planes and linearization; Handout on vector fields.

Comments:

Homework Change: Although I mentioned in class that homework on Section 12.5 would be due on Tuesday, I am changing the due date to Thursday October 4.

In class, we continued discussing greatest rate of change. In particular, we

reviewed the definition of gradient vector that we set up on Tuesday, and
determined how to compute the components of a gradient vector (in Cartesian coordinates)

We used infinitesimals to provide the notation and intuition for the reasoning we used to connect these two things. Although initially challenging, the take-away messages are simple:

The gradient at each point is a vector \(\vec\nabla f\)
- the gradient vector points in the direction of greatest rate of change, and
- has magnitude equal to that greatest rate of change.
In cartesian coordinates, components of gradient vectors are computed as \(\displaystyle \vec\nabla f=\frac{\partial f}{\partial x}\hat{\imath}+ \frac{\partial f}{\partial y}\hat{\jmath}. \)

Here's a handout outlining the reasoning we discussed in class. A key part of this reasoning is that infinitesimal changes \(df\) in outputs are related to infinitesimal displacements \(d\vec{r}\) by \( df=\vec\nabla f\cdot d\vec{r}\).

This last relation is also a starting point for thinking about directional derivatives. For a function f, there are many other directions besides those parallel to the coordinate axes and the direction of largest rate of change. In particular, we denote the directional derivative at a point and in the direction of any specific unit vector \(\hat{u}\) by \(df/ds\) where df represents an infinitesimal rise and ds represents an infinitesimal run. We then compute a directional derivative by finding the component of the gradient vector along the direction of interest. Since the unit vector \(\hat{u}\) gives the direction of interest, the directional derivative is given by \[ \frac{df}{ds}=\vec{\nabla} f\cdot\hat{u}. \] An alternate notation for direction derivative is \(D_{\hat{u}}f\). With this, we can write the result as \[ D_{\hat{u}}f=\vec{\nabla} f\cdot\hat{u}. \] Note again that we can think of a directional derivative as the component of the gradient vector in the direction \(\hat{u}\).

Thursday September 27

Topics: Questions on 11.3 problems; Greatest rate of change with handout

Text: 12.5

Tomorrow: Computing greatest rate of change -- directional derivatives and gradient vectors

Comments:

In class, we discussed the idea of greatest rate of change for a function \(f\) of two or more variables. We started with this handout on estimating greatest rate of change for a function of two variables. Since greatest rate of change involves both direction and magnitude, we represent the greatest rate of change as a vector at each point in the domain of the function. These are called gradient vectors and denoted \(\vec\nabla f\). So, at each point in the domain of the function, a gradient vector

points in the direction of greatest rate of change
has magnitude equal to that greatest rate of change.

On Friday, we'll talk about how to compute the gradient of a function if we are given a formula for the function.

In class, I will occasionally use Sage to make pictures and do calculations. If you are interested in Sage, you can access it by visiting the website sage.pugetsound.edu, signing in, and using the "workbook" supplied. Sage is an "open license" (and hence free) suite of computer packages used by mathematicians in their research. Just as you need to use a special syntax when entering functions to graph on your calculator, there is a special syntax to use for Sage. There is a nice tutorial at the above website.

Sage's capabilities are similar to those of Mathematica, Matlab, Scientific Notebook, and other commercial symbolic computing programs. Although it is expensive to buy your own copy, Mathematica is on on many computers on campus or you can access it through vDesk from your own computer. vDesk is a virtual desktop system that the university launched over the summer. After you log on to vDesk, go to the Academic Applications folder and then to the Computer Science and Math Applications folder. Launching an application through vDesk takes a bit of time but the application generally runs at a reasonable pace once it has launched. You can also get to some of Mathematica's capabilities online through the WolframAlpha web site. (Wolfram is the company that produces Mathematica. WolframAlpha is a web-based service to provide information and do computations.) At the WolframAlpha site, you can enter Mathematica commands or just try natural language. Mathematica commands will be interpreted without ambiguity whereas natural language input generally has some ambiguity that might be interpreted in a way other than what you have in mind. I occasionally use Wolframs Alpha but will use Sage instead of Mathematica in my own work.

Tuesday September 25

Topics: Questions on 11.1 problems; a bit more on describing curves parametrically; greatest rate of change

Text: 11.3

Tomorrow: Questions on 11.3 problems; Greatest rate of change (12.5)

Comments:

In class, we looked at the idea of infinitesimal displacement vectors. We denote an infinitesimal vector as \(d\vec{r}\) and we express it in terms of components as \[ d\vec{r}=dx\,\hat\imath+dy\,\hat\jmath \qquad\textrm{or}\qquad d\vec{r}=dx\,\hat\imath+dy\,\hat\jmath+dz\,\hat k \] depending on whether we are working in two or three dimension. Along a curve, the components \(dx\), \(dy\), and \(dz\) are related to each other. If the curve is described parametrically in terms of a parameter \(t\), then the components \(dx\), \(dy\), and \(dz\) are related to \(dt\). In this case, we can express the general relationship as \[ d\vec{r}=\vec{r}^\prime(t)\,dt. \] Thinking in terms of time \(t\), position \(\vec{r}(t)\), and velocity \(\vec{r}^\prime(t)\), this says that an infinitesimal displacement along a curve is a velocity times an infinitesimal increment in time.

If we add up infinitesimal displacements along a curve, we get the total displacement. This is nothing more than the displacement from the start position to the end position. A more interesting problem is to calculate the length of a curve by adding up magnitudes of infinitesimal displacements along the curve. A common notation for the magnitude of an infinitesimal displacement is \(ds=\|d\vec{r}\|\). The magnitude \(ds\) represents an infinitesimal length. To get a total length, we need to add up infinitely many infinitesimal lengths; in other words, we need to integrate.

The text discusses length of curve from a point of view that does not emphasize infinitesimal displacement vectors. In class we'll make use of infinitesimal displacment vectors in a variety of contexts so I encourage you to think through Section 11.3 problems using infinitesimal displacement vectors rather than substituting into a formula such as the one on page 678 of the text.

Monday September 24

Topics: Questions on planes problems; Describing curves in space

Text: 11.1

Tomorrow: Questions on 11.1 problems; A bit more on describing curves parametrically (11.3) ; Greatest rate of change

Comments:

Read these notes on the Section 10.3 Homework.

Today, we looked at describing curves parametrically. The idea is to give the coordinates for points on the curve in terms of an independent variable, often called the parameter and often denoted t. For a curve in the plane, this means giving x and y in terms of t. For a curve in space, this means giving x, y, and z in terms of t. A curve is a one-dimensional object sitting in two or three dimensions. The parameter t labels each point on the curve with a single value.

We can repackage a parametric description of a curve as a vector-valued function in the form \[ \vec{r}(t)=x(t)\,\hat\imath+y(t)\,\hat\jmath\] for curves in the plane or \[ \vec{r}(t)=x(t)\,\hat\imath+y(t)\,\hat\jmath+z(t)\,\hat k \] for curves in space. Each input of \( \vec{r} \) is a single real number \( t \) and each output is a vector \( x(t)\,\hat\imath+y(t)\,\hat\jmath+z(t)\,\hat k. \) Alternatively, we can think of the output as an ordered triple \( (x,y,z) \).

In terms of calculus, we can compute the derivative \( \vec{r}'(t)\) of a vector-valued function by differentiating each component with respect to t. If we think of \( \vec{r}(t) \) as giving the position of an object moving in time \( t \), then the derivative \( \vec{r}'(t) \) is the velocity of the object. The second derivative \( \vec{r}''(t) \) is the acceleration of the object.

Friday September 21

Topics: Questions on Section 10.3 problems; Point-normal form for the equation of a plane
Text: Planes Handout #2 and (optionally) Section 10.5
Tomorrow: Questions on Planes handout; Descriping curves in space
Comments:: The main idea we discussed in class is what we will call the point-normal form for the equation of a line. We derived this form of equation by specifying a plane using a point on the plane and a vector perpendicular to the plane. Such a vector is called a normal vector and tells us how a plane is "tilted" in 3-dimensions in a fashion analogous to slope telling us how a line is "tilted" in 2-dimensions. If the given point on the plane is at the tip of a position vector \( \vec{x_0} \) and a variable point on the plane is at the tip of the variable position vector \( \vec{x} \), then the point-normal equation of the plane is \( \vec{n} \cdot ( \vec{x} - \vec{x_0}=0) \)There are details on this in Section 10.5 but these are mixed in with several other ideas so I have pulled out the main idea we want on this handout. The handout includes the assigned homework problems.
We will skip over the ideas in Section 10.4 (on what is called the cross product) for now. We'll come back to these ideas later in course when we need them.

Thursday September 20

Topics: Writing exercise assignment; More on dot product
Text: Section 10.3
Tomorrow: Questions on Section 10.3 problems; Point-normal form for the equation of a plane; Planes Handout #2
Comments:

Tuesday September 18

Topics: Questions on Section 10.2 problems; Angle between two vectors and the dot product
Text: Section 10.3
Tomorrow: More on dot product
Comments:: Today we showed how to use vectors to determine the coordinates of the midpoint of a line segment in two or three dimensional space. We discussed how to extend the process to determine the coordinates of the point on the line segment between points P and Q that is any fraction of the distance from P to Q.
Using the total cost of buying books we saw one reason for the concept of the dot product of two vectors. \[< a,b,c>\cdot < x,y,z>= ax+by+cz\].; I will pass out copies of the first Project (Writing) on Thursday. It is due Friday, September 28.

Monday September 17

Topics: Questions on Section 12.4 problems; Vectors; Direction of greatest rate of change
Text: Section 10.2
Tomorrow: Questions on Section 10.2 problems; Angle between two vectors and the dot product
Comments:: Today, we began talking about vectors geometrically as quantities having both direction and magnitude which can be recorded in a picture using an arrow. We gave geometric definitions of what it means to add and scale vectors and then introduced a cartesian coordinate system so that we could add and scale vectors using components. You should work on learning to think about vectors both geometrically and in terms of components. In many situations, we will think geometrically to set up the relevant vector operations and we will then carry out those operations using components. Once we have carried out the operations, we will again think geometrically to interpret the result.

Friday September 14

Topics: Chain rule for multivariate functions
Text: 12.4
Tomorrow: Vectors; Direction of greatest rate of change
Comments: In class, we looked at an example of a chain rule involving partial derivatives. Chain rules are relevant when differentiating functions that are compositions of other functions. In the context of functions of more than one variable, there are many ways to build compositions so there are many chain rules. Rather than trying to memorize a chain rule for each specific type of composition, you should work to understand how the pieces of a chain rule fit together in general. The text shows how to use tree diagrams as one way of doing this.

Thursday September 13

Topics: Sections 9.4, 10.1, 10.6, planes handout, 12.1-12.3
Text: Exam #1
Tomorrow: Chain rules with partial derivatives (12.4)
Comments:

Tuesday September 11

Topics: Exam Overview; Questions on 12.3

Text: Sections 9.4, 10.1, 10.6, planes handout, 12.1-12.3

Tomorrow: Exam #1

Comments:

Exam #1 will be on Thursday, September 13. We will use the 80 minute period from 9:30 to 10:50 ~~8:00 to 9:20~~. The exam will cover material from Sections 9.4, 10.1, 10.6, 12.1-12.3, and the handout on planes. I posted a handout with specific exam objectives on my website. You might also want to look at questions from old exams although we are covering material in a different order than the 2008 classes.

I will be available for office hours Tuesday from 8:00 to 8:30, 10:30-11:00 and 3:00-4:30. On Wednesday, I will also have time in the late morning and early afternoon. If you have questions, email or call to set up a time to talk. Also remember that tutors are available in the Center for Writing, Learning, and Teaching and in TH 390.

I have reserved Thompson 395 from 7:00 to 9:00 PM on Wednesday September 12 for an informal study session. If you are looking for others to form a study group, come to Thompson 395 after 7:00PM and find a group working on something that interests you.

Monday September 10

Topics: Rate of change for functions of two or more variables
Text: 12.3
Tomorrow: Questions on 12.3; Exam overview
Comments:: In class we looked at rates of change of functions on two or three variables in the special directions that are parallel to the coordinate axes. To estimate a rate of change in the x direction at the point \((x_0,y_0)\) we use a specific value of \(h\) and compute \[\frac{f(x_0 +h,y_0)-f(x_0,y_0)}{h}\] To compute the exact value of these rates of change we use partial differentiation. Specifically, a partial derivative gives the rate of change of the output variable when all of the other variables are held constant.

Mathematics 280B (Fall 2012)

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