Tuesday December 4
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 Wednesday, December 12 from 8 to 10 am although I will allow you to stay until 11am and here is the list of specific objectives and my office hours.
Monday December 3
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
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
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
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
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
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
In class, we discussed the idea of integrating a vector field over a curve. As with other types of integration, we can ask
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
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
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
Monday November 12
Friday November 9
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:
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
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
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:
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
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
We did several examples of setting up iterated triple integrals including one using cylindrical coordinates. You can think of cylindrical coordinates as polar coordinates r and θ for a plane together with a third coordinate z measuring the distance away from that plane. I've assigned some problems from Section 13.7. I've also listed additional problems from this section that you can work on after we discuss spherical coordinates.
We will not cover the material in Section 13.6 in any detail. One of the ideas in that section is getting total mass given mass density. We have already woven that theme into our study of integration. The other ideas in Section 13.6 are center of mass and moments of inertia (aka rotational inertia). Those of you who have or are taking a first-year physics course might want to have a look at these sections to see connections between our course and your physics course.
Thursday November 1
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
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
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:
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
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> figure>
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
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.
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.
Tuesday October 23
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.
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.
Monday October 22
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:
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
The homework on Lagrange Multipliers (12.8) is due on Tuesday.
Thursday October 18
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
Thursday October 11
Tuesday October 9
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.
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
Today, we first discussed the second-derivative test. To better understand why the second-derivative test works, we put together two pieces:
. 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
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
Exam #2 will be on Thursday October 11. We will use the 80-minute period from 8:00 to 9:20.
Thursday October 4
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
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
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
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
In class, we continued discussing greatest rate of change. In particular, we
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:
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
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
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
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
Read these notes on the Section 10.3 Homework.
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
Tuesday September 18
Using the total cost of buying books we saw one reason for the concept of the dot product of two vectors. \[< a,b,c>\cdot < x,y,z>= ax+by+cz\].
Monday September 17
Friday September 14
Thursday September 13
Tuesday September 11
Exam #1 will be on Thursday, September 13. We will use the 80 minute period from 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
You will need to be proficient at taking partial derivatives for the exam on Thursday. I have assigned many homework problems that you can use for practice.
Exam #1 will be on Thursday, September 13. If possible, we will use the 80 minute period from 8:00 to 9:20. If you have not already sent me an email as to whether or not this works for your schedule, please do so as soon as possible.
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.
Friday September 7
One extremely valuable use of limits is the definition of continuity: a function is continuous at a specific point if the limit of a function at that point is equal to the output of the function at that point. This can be phrased in an intuitive fashion by saying the function value is what it ought to be at that point and more formally by the symbolic expression \[ \lim_{(x,y)\rightarrow (x_0,y_0)} f(x,y) =f(x_0,y_0)\]
Exam #1 will be on Thursday, September 13. If possible, we will use the 80 minute period from 8:00 to 9:20. If you have not already sent me an email as to whether or not this works for your schedule, please do so as soon as possible.
Thursday September 6
For functions of three variables, we can think about domain and range. The graph of a function of three variables requires thinking in four dimensions, which we will not do directly. We can, however, draw or visualize level sets. In this context, a level set is the set of points (often a surface) in the input space on which the output has a constant value.
Exam #1 will be on Thursday, September 13. If possible, we will use the 80 minute period from 8:00 to 9:20. If you have not already sent me an email as to whether or not this works for your schedule, please do so as soon as possible.
Tuesday September 4
Friday August 31
Thursday August 30
In class, we also reviewed some basics of ellipses, parabolas, and hyperbolas. In particular, we started from a purely geometric definition for each type of curve and then introduced a coordinate system to get an analytic description. In all three cases, the analytic description is a quadratic equation in two variables. Since they are in the book, we skipped many of the details of how the analytic descriptions follow from the geometric. As a small challenge, you can fill in the steps that we skipped over between the geometric definition and the most common form of an analytic descriptions.
We can also turn this around and ask about the graph of any quadratic equation in two variables. It is a fact that the graph of any quadratic equation in two variables is an ellipse, a parabola, or a hyperbola (or a degenerate case). That is, the graph of any equation of the form \[ Ax^2+2Bxy+Cy^2+Dx+Ey+F=0\qquad A,B,C\textrm{ not all zero} \] is an ellipse, a parabola, or a hyperbola. You can determine which type of curve by computing \(AC-B^2\). If this quantity is positive, the graph is an ellipse. If this quantity is zero and one of D or E is nonzero, the graph is a parabola. If this quantity is negative, the graph is a hyperbola.
Tomorrow, we move to three-dimensions so we will be dealing with quadratic equations in three variables and the corresponding graphs that are surfaces in space.
Tuesday August 28
I will often include mathematics symbols on this page and will be using MathJax to do so. Please let me know if the following looks like the quadratic formula to you. If not, please tell me which browser you are using. \[ x= -b \pm \frac{\sqrt{b^2-4ac}}{2a} \]
Sec 10.1 Homework Comments For problem 26: using geometric thinking to deduce the solution is the plane \(y=1\) is a valid answer but there are more convincing ways to verify that answer. One such method is to name an arbitrary point in the set we are to describe with variables -- for example, call such an arbitrary point \((x,y,z)\). Then the fact that this point is equidistant from \((0,0,0)\) and \((0,2,0)\) allows is to build the equation between the two distances \[ \sqrt{(x-0)^2 +(y-0)^2+(z-0)^2} = \sqrt{(x-0)^2 +(y-2)^2+(z-0)^2}\] A little algebra then simplifies this equation to \(y=1\).
Monday August 27
Topics: Course information sheet, functions of more than one variable; Points, lines, distance and spheres in space
Today we covered the basics of a 3-dimensional cartesian coordinate system and discussed the equations of spheres and planes. Equations of spheres come directly from the distance formula and equations of planes that are parallel to one of the coordinate planes (the \(xy\)-plane, \(yz\)-plane and \(xz\)-plane) have equations of the form \(z=c_1, x=c_2, y=c_3\), respectively, where \(c_1, c_2, c_3\) are constants.
[Exam 1: Spring 2008] [Exam 2: Spring 2008] [Exam 3: Spring 2008] [Exam 4: Spring 2008] [Exam 5 with Final: Spring 2008]