Attendance Totals to Date

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Lecture Notes 2/19/16 through 2/26/16 (Updated!)

Lecture Notes 2/19/16

Lecture Notes 2/22/16

Lecture Notes 2/24/16

Lecture Notes 2/26/16

the numbers are in...

Just for funzies...for all of my MAT267 classes for the last couple of years, over 200 students,  I plotted everybody's total homework score vs their final grade.  The data look like this:


Clearly there is a strong trend that relates performance on the homework to overall performance in the course.  The red line is the linear least squares fit; hopefully you've heard of it in one of your lab classes, it's an important tool for data analysis--Engineers use it a lot, so you should feel comfortable with its use.   The precise linear least squares description is this:

%Final Score=8.90 + 0.80 %HW + "Noise"

In this model "Noise" indicates a random effect with a Standard Deviation of 10 points in the final score and an average of zero--this is to say that homework scores don't absolutely determine the final score in this model, but they nevertheless have a strong effect on the outcome.  Of course by "random" we mean really mean all the effects that I don't have data on, for instance how much of the homework was actually done by each students in question.

Note however, that the homework is only 15% of the final, while the exams are 75% of the final score. If the only effect of the homework on the final score was from the homework itself, we would expect a slope of the least squares line to be at most 0.15.  This is to say that the homework has an effect on your grade much larger than just the numerical contribution from the homework scores.

This course has an average homework score of 73% with a standard deviation of 25%.  Clearly there is a lot of room for a lot of people to do a lot better.

Moral of the story: work hard on your homework, it will help you do better in this course.

section 11.1#10

For problem 10 i do not understand how to complete it. I plugged (0,0) into
the equation and I got the first answer 140. after that I attempted to plug
in (1,1) which did not work, (.4,0) with the same result, and then trying
to subtract 10 from 140 since it said it was spaced by 10 but still no
answer.

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This is interesting....I spoke to another student about this problem earlier gave the wrong answer.  OK,  on the line at y=0, the function f(x,0)=140e^x which is increasing with x so since they are equally spaced multiples of 10,  f(.4,0)=140e^.4=140*1.49 which is approximately 210--which should be =B. The gap between the contours thus should be 70, so that C=280 and D=350

on the unit normal vector....

“When I use a word,’ Humpty Dumpty said in rather a scornful tone, ‘it means just what I choose it to mean — neither more nor less.’

’The question is,’ said Alice, ‘whether you can make words mean so many different things.’ Through the Looking Glass, by Lewis Carroll, page 367

So let me try to correct the record.  It is true that for a parametric curve r(t), the velocity vector v(t)=r'(t) is simply related to the unit tangent vector T(t)=r'(t)/||r'(t)||=v(t)/||v(t)||.  And it's also true that the acceleration vector is a(t)=v'(t)=r"(t).  While grading the exam I saw about fifty wishful but incorrect statements to the effect that there is a simple relationship between the unit normal vector and the acceleration:  N(t)=a(t)/||a(t)|| BUT THIS IS INCORRECT.

Like Humpty Dumpty, we can define things to mean whatever we want, and a(t)/||a(t)|| is a perfectly good unit vector. The people who thought of the unit normal vector as something useful were looking for more. They wanted N(t) to point in the direction that v(t) is turning toward AND to be orthogonal to v(t).  Unfortunately for the sake of simplicity a(t) can't do this job.  This is easy to see: sometimes a(t) points in the same direction that v does, for instance when v is increasing in magnitude but doesn't change direction.  In order to get what we need, we need to define a different unit vector N(t)=T'(t)/||T'(t)||, which is always perpendicular to the direction of motion.

Your scores and extrapolated final grade

These are your scores as of Wednesday 2/17/16.  The extrapolated score is based on the course syllabus, which gives 10% weight to quizzes, 15% weight to homework, 50% weight to midterms and 25% weight to the final exam.  Since we haven't hit the final exam, I gave 75% weight to the midterm, 10% to quizzes and 15% to homework.  The extrapolated score represents what you might get if you continue to proceed as you have been--you could go up or down depending on how hard you work of course.  As of this time, everyone is capable of pulling their grade up to a passing or better grade.



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11.1 Problem 6

Could we go over contour plots in class? They were not in the notes and it
came up in the homework which confused me greatly.

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Sure, you only need to ask me in class.  But since the deadline on the HW is creeping up, let's address it a little bit now.  First of all, I recommend that you consult the textbook on this subject, you paid a lot of money for it and it covers this subject pretty well.    Second, we have used contour plots in class, though usually written on the whiteboard and as you point out not in the lecture notes.  Thirdly, this link takes you to a topo map of the downtown Tempe and ASU area, including Tempe Butte.  I recommend that you print the map and go climb A-Mountain and compare the features of the map to the topography of the butte.

The general notion is that each contour represents a trace of the graph of a function z=f(x,y) for some value of z, and the contour map is a whole family of contours for a variety of different values of z.  Usually the z-values are chosen to be evenly spaced, for example it looks like the contours in the topo map of Tempe are spaced every 20 feet in height.  Sometimes the contours are closer together,  this happens when the graph of the function steeper, and sometimes they are further apart, which happens when the graph is less steep.  In some regions where you don't see any contours at all the graph of the function is so flat that the height doesn't change enough to take you to the next contour. As a specific example, let's look at your contour plot b)

Notice that the contour lines are all straight lines and all parallel to the line y=x--this means that f(x,y)  depends only on the value of y-x.  The graph is pretty flat around the line y=x but on both sides the contours are getting closer and closer as you move away from y=x, hence the graph of the function is getting larger and steeper.  If you look at what function could do this, you focus in on b) (x-y)^2 which has the requisite properties.