The scores that made up your grade

this is your warning(your handwriting on the exam)

I'm just grading your quizzes, and am thereby motivated to point this out to you:
If your handwriting is so small or so illegible that I can't read what you've written, you will get no credit for your work.

Office Hours Finals Week

I will be holding my normal office hours before your final exam.

Final Exam, 2nd Notice

The final exam will take place on Tuesday December 8 from 7:10PM to 9:00PM in BAC216

Topics to be covered and other random angst...

A random sample of emails: 
1) Hi professor, There are two questions i want to know about final exam. Firstly, i'm not sure the contents to be covered. I finished the final review, but i found it didn't cover all sections which we learned. So, I wanna know did it include all things we need to review? or we need to review more sections?  In addition, could you tell me the form of this final exam? does it include both multiple choice and free response, or only free response, or something else?

2)  Dear Dr. Taylor,

When looking through the final review that was posted here, https://math.asu.edu/first-year-math/mat-267-calculus-engineers-iii , I noticed that it does not cover all of the sections we learned; for example, it skips from 10.5 to 10.9. Does this accurately reflect what is going to be on the final, or should we study for all of the topics we covered in class? 

Thank you for your time.

3) Professor,

You said about two weeks ago that you'd post an update on everyone's gardes insofar onto the blog. I figured you'd eventually get around to it, but that hasn't happened. Did you think you shouldn't since most people in the class don't bother to read your blog?

********************

OK, first of all, the final exam is cumulative.  For you, that means anything and everything we have covered is fair game. Since I spent my breath and time lecturing on it, it's also likely to be covered on the exam. Since subjects learned early in the course are used later in the course, I will sometimes kill two birds with one stone and test the early material and the later material in the same question--but this does not constitute a promise that I will do so.   

My first priority is grading at the moment is getting your quizzes graded and recorded.  If I have the time before the final exam I might get around to re-estimating your grades.  BUT REALLY-you have everything that I have graded and at this point in the semester you should have a very good idea about where you stand.  

webwork problems

A small sample of the emails I've just been getting:

1) Hello Professor,

I'm in your MWF 10:30-11:50 class and I was trying to get onto Webwork to finish up the homework due tonight after *************, but I couldn't log in to myASU. I know it's really close to finals, but is there any possibility you could extend the deadline a day? If I'm the only one who's been having the problem, I understand if you don't. But I could really use all the points I could get, so if you do I would appreciate that.

Your student,

*************

2) Hello Professor Taylor,

I noticed while i was trying to complete the last couple of problems on the 13.7 webwork that blackboard is down and is unable to sign anyone into their my asu to log on to webwork. if it would the possible to extend the deadline for the webwork.

Thank You,

*************

3)   Web work pausing and not allowing access. I have like three problems left. Please let us in. Or something.

**************

Webwork deadline is delayed until Sunday night due to network problems. 

13.7 #7

Dear Professor Taylor, This problem has me really stumped I don't know what it is asking for the second to last portion because I have entered r dr dtheta as the usual substitution of dA. But it wants it in vector form. Can you please clarify what WebWork wants exactly.

Thank you,
























ok, you have a couple of different confusions going on here:

1) this is a good example of the diversity of notation surrounding this integral: what this problem wants to call a vector dA is what we have been calling n dS. (In support of my rant earlier today about "Flux Integrals",  you might want to note that wikipedia distinguishes two kinds of "surface integrals" as I do, surface integrals of functions and surface integrals of vector fields, that the textbook calls flux integrals)

2) just to confuse the issue a little bit more, this is a parametric integral, not an integral in polar coordinates.  What this means is that the element of area for polar coordinates r dr dθ is NOT the appropriate element of area for this problem.  What you need is just (r_r x r_θ)dr dθ, (note the additional confusion of r being the parametric equation while r is just a parameter) and the correction term r is not used (it sort of gets absorbed into the  r_r x r_θ). Since

  r_r x r_θ=<cos(θ),sin(θ),1>x<-r sin(θ), r cos(θ),0>=<-r cos(θ), -r sin(θ), r>, which is pointing upward and not downward, so your correct answer to part (b) would be
- r_r x r_θ drdθ.  
This is *almost* what you have, you just forgot to multiply the drdθ times the first two components of your vector.

announcing...

Extra credit quiz covering section 13.7 on FRIDAY

13.7 problem 8

Hi Dr. Taylor,
I was wondering if you might help me get started on this
problem. I'm having some difficulty figuring out the parameterization as
well as the bounds.
Thank you,
****************************


First of all the cylinder x^2+y^2=16 has radius 4 and is parallel to the z axis.  Then the planes x+y+z=3 and x+y+z=7  are parallel to each other, and you can solve these for z  as z=3-x-y and
z=7-x-y.  In particular you can get from first plane to the other by increasing z by 4.  Then points on the cylinder can be specified by writing x=4 cos(u), y=4 sin(u), while the intersection of the cylinder with the first plane also specifies z=3-4 cos(u)-4 sin(u) while the intersection with second plane specifies instead that z=7-4 cos(u)-4 sin(u).  You can put this all together to get a parameterization of the space on the cylinder between the two planes by saying
x=4 cos(u), y=4 sin(u), z=3+v-4 cos(u)-4 sin(u)
where 0≤u≤2π and 0≤v≤4.

Lecture Notes 11/23/15,11/25/15

Lecture Notes 11/23/15

Lecture Notes 11/25/15

announcing a quiz for monday 11/30

Yah, that last one wasn't so good.  It has been pointed out to me that all of your webwork on Green's theorem dealt with using a double integral to find the value of a line integral, whereas the quiz did the reverse: it asked you to find an area by computing a line integral.

                    OK. 

So I'm going to give another quiz on exactly the same topic as the last: finding the area of a region by computing a line integral around the boundary.  I'll give you a parameterization of the boundary, just like before.  ALL YOU NEED TO KNOW HOW TO DO IS TO COMPUTE THE AREA BY USING A LINE INTEGRAL.  

ok....integration of a vector F field along a curve...(exam fall-out)

means that you have to integrate the F along the curve.  That means that you need to find a parameterization of the curve. Usually there will be one of three situations: 1) we will give you the parameterization--use it, or 2) there is a natural parameterization of the curve, i.e. the curve is a circle or the curve is the graph of a function g(x)--use <cos(t),sin(t)> for the circle or use <t,g(t)> for the function, or 3) the curve doesn't matter anyway because the vector field is conservative--in which case you need to find the function f(x,y) so that F=∇f and evaluate f at the endpoints of the curve.

You don't get to chose some random path that doesn't fit the curve, and don't need a path at all in the case that F is conservative.

grading the exam: problems with setting up integrals

Aside from the usual difficulty in recognizing upper and lower limits of integration in theta,  one common confusion that people have is confusing the values of coordinates at the boundary of the region of integration--which need to become input into the selection of limits of integration--with the form of the integrand. For instance, if a cylinder has radius R, and the integrand has x in it, in the integrand x is written as R cos(theta) instead of r cos(theta) as it should be.

Final Exam, 1st Notice

The Final Exam will take place from 7:10-9:00pm in BAC 216

Lecture Notes week of 11-16 & Quiz

Quiz 11/18/15

Lecture Notes 11/16/15

Review Questions

Hi Dr. Taylor,
I was wondering if you might help me out with one of the questions on the review. I'm having a lot of difficulty getting the right bounds for problem 2 in 12.6. If you could help me out I would be extremely grateful!
*******

******************************************************************
Sure, here are my notes from reviewing this problem and a couple others on 11-18

Review 12.6#2

Review 12.7#'s 4 and 4

13.4#4 (updated)

Dear Professor Taylor,

I am working on this problem and I can't quite figure it out. For P I get x^3 because that's multiplied by dx. And for Q I get 5x. So dq/dx-dp/dy = 5.

Plugging that in I get double integral over D of 5 dA. Next I realize that there to do dydx as my dA I have to split the parallelogram into two pieces. So I'm integrating two separate triangles. My guess is I'm doing that wrong but I can't figure out why. I decided to do the line integral by finding four vectors going all the way around the shape by that just got me a completely different answer.

What am I doing wrong?
Thank you,

Hello sir. I keep trying to do this problem as you described on your blog,
but I keep reaching the same answer I got last time. Does this mean I'm
still doing it wrong or that the webwork is wrong??
**************************************************

It's hard to see what you did wrong without having the details of your calculation.  First of all note that the orientation of the contour is clockwise, while Green's theorem is framed for counterclockwise line integrals. So:  just go ahead and use Green's theorem as normal and then multiply your answer by -1.   Incidentally, I would do the double integral as an iterated integral with dx first, since that way I can do only one piece. The leftmost sloping line segment runs from (0,0) to (x_0,y_0) so that line can be written as y=y_0/x_0 x or (as you will need it) x=x_0/y_0 y. The rightmost sloping line segment is displaced from the first by an amount x_0 so has equation  x=x_0/y_0 y + x_0,  so the iterated integral is

which is pretty easy to do.  (Updated)Since here x_0=2 and y_0=2, the inner integral is 5((2/2)y+2-(2/2)y)=10.  Then you get the integral of 10 with respect to dy from 0 to 2, which is positive 20. The negative of that is -20, which is the answer you got so I believe webwork is failing to properly check your answer.

13.2 #14

Can you tell me what I'm doing wrong for part B?

Well, it's hard for me to tell how you got that answer, because it's pretty far off. You need to understand the definition of line integral: since you are given a particular vector field, you need to do the line integral of that vector field, along the curve you have, which looks like the following:

Next you have to use the parameterization you already wrote down x(t)=t, y(t)=t^2, which will give you x'(t)=1 and y'(t)=2t, hence you have:

 
which gives you

comment on Reviews for Test 3

On Sun, Nov 15, 2015 at 6:54 PM, Unknown <noreply-comment@blogger.com> wrote:

Unknown has left a new comment on your post "Reviews for Test 3":

Is there any changes to the chapters on the exam?( Syllabus says 12.5-13.4)


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No.

Reviews for Test 3

Monday November 16th at 7:15pm in Noble Library Tutoring Center.