Final Scores

* If you took a make up final exam the score here is the larger of the two. (click on image to enlarge)

Review Session in one hour

Hi all, sorry for the late notice but I've just received confirmation from classroom scheduling that we can have a review session 11:00-Noon in SS105.

Review session in waiting

Dear Dr. Taylor,

  You mentioned on Friday that you may have a review session tomorrow. I would be interested in attending such a review session. Will you be holding one? If so, please post it on the blog. 

Thanks,

and

Dear Professor Taylor,

Where and when will the review session be on Monday? I would attend if you organized one.

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

Well, the plan is to have one 11:00-12:00 tomorrow Dec 4.  HOWEVER, right after class last Friday I contacted one person whose responsibility is scheduling rooms, who got back to me at 4:00PM saying I had to contact the other person who scheduled classrooms, who by that point had gone home, and who (unlike some people) obviously doesn't deal with work email after hours.  So stay tuned on this channel, I'll post a notice right here as soon as I have a classroom. 

typo practice final solutions

Doctor Taylor,

I attached a screenshot from the solutions from the final exam review that is on the math department website.

When looking at where it says "System" it goes from 2x+2y-8=0 to x+y=8. Is that a mistake?

Thank you,



















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

Yes, a big mistake.  It should read x + y = 4, and from this you get -2y = -2 ==> y = 1 & x = 3.

Practice Finals

Official mat267 practice final, follow the link here.

My old final

Reminder:  I'm happy to help those who are helping themselves, so if you are asking me for help on a problem, please tell me what first what you already did.

extra credit paper

The rules:
1) Research how vector calculus applies in your chosen discipline of engineering. 
This means actually research it. This is not meant to be easy or something you can just imagine.
2) Write *about* what you discovered in your own words. Two pages double spaced,
2) Cite your references, with a number referring to the references section, in your writing.  This insures that you actually did research something.
3) Document your references, page number and/or section, in a separate reference section.  There should be at least four *distinct* references. At most one should be wikipedia.
4) No plagiarism.  I.e. no copy-pasta, no using other peoples words or ideas without crediting them.

The value to you: UP TO ten points of midterm exam level credit.  You will  have to earn your points, though.

Final Exam Protocol

Just because there will be so many people taking the final exam all in the same room as well as multiple proctors, we're going to have to change the exam protocol a little bit.  Your ID WILL BE REQUIRED for you to receive an exam AND for you to turn it in.  This is to say that if you do not have your ID you will not be able to take your final exam and will receive a score of zero for the final.

Exam 3 Scores, updated quiz scores and estimated final grade as of right now.

click of image to enlarge.

HW & Final Exam Notices

1)  I postponed the homework due date till tomorrow
2) From the Syllabus:

December 5 (Tuesday)

Final Exam 7:10-9:00 PMPSH 150

Comprehensive!(just because some excitable people will insist on misreading the final exam schedule and get flustered, please go to this link and search on the keyword "267"

(a couple years ago some people missed the final exam because they were sure it wasn't when I said it was)

3) Evaluation forms for this course are available to you at myasu

Exam 3 Review

Hello Dr. Taylor,

           I was just wanted to see if you would be willing to post the answers to the review for Exam III just so I can check my answers and make sure I am solving the problems correctly.

Thank You,

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

No. As usual.  As I've explained before, when I do that, people will look at the problems and the solutions together and convince themselves that they've studied for the test.  This doesn't work.  What does work is doing the problems and getting confused and then understanding why you were confused and how to fix it.   

I am happy to answer here any specific questions you have about specific problems of this type: I did this and I got that and is it right or why isnt it right, provided you tell or show me what you did.  A photo of your work is a good way to accomplish this.   

 (i.e. show me your diligence and I'll help you out)

Greens Theorem clarification

Hi Doctor Taylor,

When using Green's Theorem, if we have the problem set up with the double integral of the partial of F2 - partial of F1. If the orientation is counter clockwise we leave the double integral positive, and if it is clockwise we make the double integral negative. Is that right?

Thank you,

******************
yes. specifically when the boundary curve ∂E is oriented counter clockwise and blah-blah-blah*



but when ∂E is oriented clockwise



*blah-blah-blah means E is a simply connected domain and its boundary ∂E is piecewise differentiable closed curve that doesn't intersect itself

What will be on the exam?

Dear Professor Taylor,

On the syllabus, it says that 13.4 will be covered on the exam this Friday. However, the 13.4 homework isn't due until 11/26. Will 13.4 be covered on the exam this Friday? Also, can you please post the links to the practice exams on the blog?

Thank you,

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

See the syllabus: exactly what it says. As we discussed in class, yes 13.4 will be in there. 

practice for exam 3

Here--the old one.

Here--a newer one.

13.2#6

I just had a mini question about this one, since I tried to evaluate it as
integral from 0 to 3/2pi with the integrand sin(t)cos(t) + 2t, I figured
the step before was the integrand of -3sin(t)cos(t)+4sin(t)cos(t) +2t. Are
you not allowed to perform algebra and reduce the coefficients the answers
were very close to each other.

Wrong answer (22.399..)
Correct answer (22.7066)

**********************
I only get the correct answer, and the wrong ones I can think of give me answers more wrong than yours, so it's hard for me to guess. How did you get the incorrect answer, maybe numerical integration on your calculator?  Btw, you can eyeball the vector field and figure out which function it's the gradient of.

Homework due dates.

Dear Professor Taylor,

My name is ***** and I’m in your ********** class. Are 13.1-13.3 all three webworks due tomorrow night at midnight? Or did you forget to change them or something?

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

Let's see.  What did we cover last week?
Vector fields:√.
Line integrals: √
Fundamental theorem....well, sort-of kinda not quite really. 

OK, then. 13.1 and 13.2 are due, but tomorrow is a holiday so lets make it due on Sunday.

And let's make 13.3 due next Friday.

Lecture Notes 11/6/17 through 11/8/17

Lecture Notes 11/6/17

Lecture Notes 11/8/17

Attendance

Click to enlarge

12.7#9

Good evening Professor Taylor,
I was working on this problem and I
understand the boundaries for rho, and theta, when I see the vertex at the
origin is 90* I want to set my boundaries from 0 to pi/2. I'm not positive
why it's required to set from 0 to 45* unless it's because the picture
show's only the 45* region shaded in? Thank you again for extending the
homework assignment :)
Thank you,

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

This question is meant to confuse you, which it did.  Remember that φ is measured from the z-axis, which is poking out of the middle of ice cream on top of the cone.  The confusion arises because the problem invites you to pay attention to the angle from one side of the cone to the opposite side of the cone--that is 90*--and oh by the way don't pay any attention to the fact that those two opposite sides have values of θ that are 180* apart.  Instead of paying attention to that you should be paying attention to the angle between the z-axis and some radial line passing through the origin and going along the side of the cone somewhere--this will be at some fixed  θ because you only go to one side of the cone--and this angle can only be 45*=90*/2

12.7#3 (edit)

I'm not sure why the system is saying that it can't take the arccos(1.61644771824097) when I plug that in for A) as the answer to phi. I divided the z/p, the argument for the arccos(z/p), which was  (9sqrt(3)/2) /  (sqrt(93)/2).

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

Well,  the short answer is that it's saying that because you can't take the arccos(1.61644...).  A slightly longer answer is in fact you can't take the arccos of any number bigger than 1 or less than -1, because the domain of arccos is the closed interval [-1,1], because the range of cos is [-1,1].

This information is incredibly important: it means that you made a mistake, in fact at least two mistakes. There is no shame in this, by itself, because everybody makes mistakes.  It is critical thought that you should not waste the opportunity to identify and correct the mistakes. 

In your case the first instance is that you've miscalculated ρ, it's quite a bit larger than that and simpler. Another mistake is conceptual: you don't have a clear picture of cos(x) and it's relationship with arccos(x).  This will hinder in the future you unless you correct it

HW Due date changed.

The new deadline is Tomorrow at 11:59pm.

Lecture Notes 10/23/17 through 11/3/17 (complete!)

Lecture Notes 10/23/17

Lecture Notes 10/25/17

Lecture Notes 10/27/17

Lecture Notes 10/30/17

Lecture Notes 11/1/17

Lecture Notes 11/3/17

Scores and Estimate Grade

Please refer to the scoring described in the course syllabus.  The score listed is computed from that description with the difference that the estimated score assigns 75% value to the two midterms instead of 50% to three midterms and 25% to the final exam--because you haven't taken the third midterm or final yet.
Note that you have two estimated grades, the first referring to the linear grade scale, the second referring to the curved grade scale, your estimated grade is the better of these two. (please click on the image to enlarge)

linearization and the equation of the tangent plane

Dr, Taylor

What is the difference between linearization and finding the eq of the tangent plane at a specific point. I find them to be practically the same.

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

They are almost the same thing.  The linearization is just a rearrangement of the equation of the tangent plane; from the equation of the tangent plane 

∂f/∂x(x_0,y_0)(x-x_0) + ∂f/∂y(x_0,y_0)(y-y_0) - (z-z_0) = 0 

you can rearrange to get the linearization

                              z = ∂f/∂x(x_0,y_0)(x-x_0) + ∂f/∂y(x_0,y_0)(y-y_0) + z_0

Lecture Notes 10/16/17 and 10/18/17 (to be completed)

Lecture Notes 10/16/17


Lecture Notes 10/18/17 (to be completed)

Review Session and Practice for Friday's Exam

1) On Wednesday we'll have a review session for the exam on Friday. Please come prepared with questions.

2)  The generic mat267 practice exams are available here.

3) An old practice exam written by me can be accessed here.

Lecture Notes 10/11/17 through 10/13/17

Lecture Notes 10/11/17

Lecture Notes 10/13/17

11.7#10

Dr. Taylor,Could you help me with this problem. My Fy is slightly different than the one from the 

solution. But I think I got it right. It is underlined on Red.

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

Congratulations, you found an error in the webwork computation of f_y; your computation is the correct one.  It looks like they were thinking about the answer you got, because they got the correct critical point, as I guess you did too.

11.7#13

Hey Dr. Taylor,

How am I suppose to get this? I thought when looking at contour plots for local extrema that we look for concentric circles that come around a point. This diagram only shows parts of it, and I've been
guessing the points and level curves because the problem's picture only shows part of the curves, and not even the parts that would show the extrema (concentric circles around a point, the point representing a max or min).
I know we aren't supposed to guess on answers, but I don't know how else to solve this?? I literally guessed the first coordinate by estimating where the circle's center would be by looking at the section of the circle that was given in this section of the contour diagram... But I can't estimate the second one's coordinates.
Thank you!

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

you did just fine. as you point out, guessing is the only thing you can do here. that happens a lot when you have incomplete information as you do here.

11.7#6

Hello. I found the derivative with respect to x which is [ y-4xy-5y^2] and the derivative with respect to y which is [ x-10xy-2x^2]. I then used those equations to find the critical points. I got that x=0,x=1/2 and y=0,y=1/5. So using a combination of these I got three out of the four critical points, but I can't get the fourth from what I have.

**************
Well, the derivative with respect to x is (1 - 4 x - 5 y) y and the derivative with respect to y is
x(1-2x-10y) so you have the equatoins (1 - 4 x - 5 y) y=0 and x(1-2x-10y)=0. The first equation can be solved if either y=0 or (1 - 4 x - 5 y)=0 and the second equation if  either x=0 or (1-2x-10y) =0. One of each pair must be true at the same point (x,y), so that the options are (x=0,y=0)-->(0,0), or
(x=0, 1-4x-5y=0)-->(0,1/5), or (1-2x-10y=0, y=0)-->(1/2,0) or
(1-2x-10y=0, 1-4x-5y=0)-->(1/6, 1/15).

10.6#9

I'm not sure what I'm doing wrong with this, so I'm probably doing it in
the wrong way altogether

***************************
OK, first of all, you didn't tell me anything that you did to get to that point like you are supposed to.  Second, notice that you are engaging in a little psychological warfare against yourself? Stop that, it's harmful to yourself.  Webwork even told you got the first answer right, so why say that you're altogether wrong? Given that how far wrong could you be about the second question?

Using the arctan is the correct thing to do.  The argument of the arctan is not the ratio of the y and x directions though, since what this is talking about is the angle above the horizontal.  The horizontal in this case is any direction in the xy-plane.  You have to get the slope of z in the direction of the unit vector you use in the first part

Lecture Notes 9/25/17 through 9/29/17 (complete)

Lecture Notes 9/25/17


Lecture Notes 9/27/17


Lecture Notes 9/29/17

11.3#9

Dear Dr. Taylor,
I have been working on this problem and I am a little confused why the answers are positive instead of negative. When I took the derivative of the function I got (dz/dx)=-3cos(-3x-5y+z) and (dz/dy)=-5cos(-3x-5y+z). From here I plugged in the point (0,0,0) and received the answer -3,-5.

Thanks

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

well, I'm not sure but I guess you're thinking of the equation z = sin(-3x-5y) instead of the equation that's really there. If you think about the real one, you have one equation and three variables, so you can think of one of the variables as a function of the other two--in this case you should think of z = z(x,y).   Then you can find the partial derivatives of both sides of the equation with respect to x and y; the partials of the right hand side are both zero, while the partials of the left side can be  computed using the chain rule--and the part of the chain rule that involves multiplying by the derivatives of the argument will include multiplying by ∂z/∂x or by ∂z/∂y.  (BTW, one follow on effect of this is that it will change the sign of your derivatives.)

11.1#10

Hey Dr. Taylor so I emailed you once about this already but you didn't respond so I'm nagging you. I only got this question correct because I google'd it, I honestly have no idea how to do it other than the (0,0) point, which I got by plugging in 0 and 0 for x and y.

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

You *googled it*?? How did you do that?  Good job on the nagging btw.
So: yes, as you surmise, plugging in (0,0) will give the the value of 140 on contour "A".  My eye tells me that "B" hits the x-axis at approximately 0.4.  If I plug in x=0.4 and y = 0 into f(x,y) I get f(0.4, 0) = 208.85, to which the closest multiple of 10 is  210,  so that must be the value of "B". This gives me the gap between the contours and so I'm basically done.

11.2#10

Hi Dr. Taylor,

I solved this question with a math tutor, and we weren't positive why the answer was choice B, I feel choice E should be the answer since one could write f(x,y) as ((inf-n, inf - n - 1)) such that the values
can be massive or equal to each other and allowed in the domain?

Any feedback regarding a proof for B being the answer would be appreciated.

Thanks,

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

Well, the point is that you need the arguments of both square roots to be non-negative. For the first square root this means that x + y ≥ 0 or equivalently y ≥ -x. For the second square root this means that x - y ≥ 0  or x ≥ y.  If you just put these together you get x ≥ y ≥ -x.   Notice that this implies that
x ≥ -x; this means that x ≥ 0 (because if x < 0 , -x is positive so x ≥ -x is not true ), so the domain is the wedge between the lines y = x and y = -x with x ≥ 0:

The grade scale

I've added this section to the course syllabus

Grade Scale: This class will be graded on a qualified curve, backed up by an absolute grade scale. This means that the higher of the two grades specified by tables below, where µ denotes the class average and ß denotes the class standard deviation:

Grade	Score		Grade	Class Rank
A+	97%+		A+	r >µ+1.3ß, historically the top 4.8%
A	93-96.99%		A	µ+1.3ß ≥ r > µ + 1.1ß, historically the top 10.2%
A-	90-92.99%		A-	µ+1.1ß ≥ r > µ + 0.95 ß, historically the top 16%
B+	87-89.99%		B+	µ+0.95 ß ≥ r > µ + 0.8 ß, historically the top 21%
B	83-86.99		B	µ+0.8 ß ≥ r > µ + 0.6ß, historically the top 29%
B-	80-82.99		B-	µ+0.6 ß ≥ r > µ + 0.45ß, historically the top 36%
C+	77-79.99		C+	µ+0.45 ß ≥ r > µ + 0.3 ß, historically the top 43%
C	70-76.99		C	µ+0.3 ß ≥ r > µ - 0.2 ß, historically the top 66%
D	60-69.99%		D	µ-0.2 ß ≥ r > µ - 0.55 ß, historically the top 77%
E	<60%		E	µ-0.55 ß ≥ r  historically bottom 23% (usually primarily people who did not complete the class)

Lecture Notes 9/18/17 through 9/22/17

Lecture Notes 9/18/17


Lecture Notes 9/20/17


Lecture Notes 9/22/17

Scores to 9/22/17

Click on the image to enlarge.

Your Posting ID. Your Posting ID will be used to identify your scores. You should not share your Posting ID or do anything to compromise it's security. To quote from this link:=

Posting ID
Your Posting ID is a seven-digit number composed of the last four digits of your ASU ID number plus the last three digits of your Campus ID number, separated by a hyphen. Your Posting ID is printed on the class rosters and grade rosters your professors work with. You can also view your Posting ID on the My Profile tab in My ASU.

10.9#6

Hi, I've been attempting this problem for some time and can't seem to get
it. I've been following the book's example to the letter. I first created a
parametric model for the projectile, where x=200cos(30)t, which simplifies
to 100sqrt(3)t. My y model was 200sin30 - (1/2)(9.8)t^2, which simplifies
to 100 - 4.9t^2. Setting this to 0 and solving gives the answer
sqrt(100/4.9) = t. I plugged this value into the x equation to get the
range, and the answer was marked incorrect. Can you explain what I did
wrong?

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

OK, so your first mistake: you were trying to do too many things in your head. You should write out the steps. These involve a lot of integration: your starting ingredients should be a(t) = -9.8j, and v(0)=<200cos(30), 200sin(30)> and r(0) = <0,0>. Then you need to integrate twice using the fundamental theorem of calculus each time and substituting in v(0) and r(0) as appropriate.  From your answer it looks like you forgot to include the effect of v(0) in integrating out to r(t). 

yes, 11.1 hw due date changed

Have we covered 11.1 yet? The homework on web work for that section is due tomorrow. 

**********

Not any more.

Practice Exam Q5 (Fall 2015)

Dear Professor Taylor,

I am having trouble finishing this problem. This is what I have done so far.

I think I have the y and z components correct for the vector, but I'm having trouble finding the x component.

Thank you,

(please click on image to enlarge)

Thanks for showing me what you did, it really helps I think

10.9#1

Hi Professor,

I am struggling to find the speed of the particle. As you
can see, I have attempted this problem multiple times. Any help would be
great! Thanks.

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

1) OK, first of all, let's go through this again.  How much you are struggling or how many times you attempted the problem, while it may be a sign that you are working hard, is also a sign that you are not working well. When you attempt a problem and do not get the right answer,  that is a sign that you should try something different. Just trying the same thing over again will not get you a different answer, it just serves to tire you out.

2) Second, I am much better able to help you learn if I know specifically what you already tried, so I can point out exactly where you made your mistake instead of guessing based on your wrong answer. Please tell me what you tried when you ask the question.

3)  Finally, as it looks like you understand, the speed is the square root of the sum of the squares of the components of the velocity.  You have the z-component squared correctly, but it looks like you did FOIL incorrectly with the sum of the squares of the x- and y-components: in particular you'll get another t^2 term and some t cos(t) sin(t) terms that don't add up the way you have them

10.8#3

Hello Mr. Taylor!
I am doing the integral from 0 to 3 of the magnitude of
the derivative, but I'm not sure how to do it due to 0 being in the
denominator for one of the parts. What do I need to do different?
Thanks!




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

Ehr....OK, I guess you're deducing the wrong limits of integration from those two endpoints of the curve, (12,6,0) and (36,54,6 ln(3)).  You see, t=0 is not what you need to plug into r(t) to get (12,6,0) because that says that 6 ln(0) = 0, when really it is not even defined. What *is* true is that ln(1)=0.  So the real limits of integration should be from t=1 to t=3.

Practice Test Links

My old test#1 from 2015

ASU exam review page

Lecture Notes 9/6/17 and 9/8/17

Lecture Notes 9/6/17

Lecture Notes 9/8/17

10.7#1

Hi,

Just wondering why in letter C) the answer is not (-inf,-5)U(-5,5)U(5,inf) but only (-inf,-5)U(5,inf). Thanks

******************
Well, because for -5 < t < 5, t^2-25 is negative so √(t^2-25) is not defined (no imaginary numbers are allowed in this class)

10.5#17

I know that to find the angle I have to use the formula a DOT b =
|a||b|cos(theta). For my two normal vectors I get <5,-5,1> and <2,3,-4>.
This leaves theta = cos^-1(-9/sqrt(1479)). I get 103.53 degrees or 1.807
rad. What did I do wrong?

***********

OK, it's true that the angle between two planes is the angle between their normal vectors, the question is which normal vector of the plane do you use, because there are two opposite directions that are normal to a given plane.  Here is a representation of the two planes, all arrows are perpendicular to a plane, the black arrow is perpendicular to one plane and both the red and blue arrows are perpendicular to the other plane. You can see that the blue arrow has an obtuse angle with the black arrow, i.e. more than 90degrees, which has a negative cosine, while the red arrow has an acute angle with the black arrow hence the cosine is positive.  This is reflected in the fact that the two planes have two different angles between them, one less than 90degrees and the other greater than. BY DEFINITION though, the angle between the two planes is taken to be the smaller.  SO: to compute the angle between the two planes, you get to use the normal vector that makes the dot product positive.

10.2#12

Dr. Taylor,
How can I figure out the speed that the bird must go if I only have one known speed value?













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

Well, this is a problem of vector addition.  You have two known directions, "from the west" (= toward the ???) and "head northwest", and you have one magnitude "at 70mph".   You are looking for the magnitude of the other vector in such a way as to fly directly north, that is to obtain the specific direction north for the vector sum.

absolute values and magnitudes

Hi Dr. Taylor, I'm just wondering if on the last portion of the problem if
|w| is really supposed to be ||w|| since vectors don't technically have
absolute values

************
We discussed this in class: technically yes, you're a little bit right.  It's a common usage to write |w| instead of ||w|| for the very reason you mention: there's no such thing as the absolute value of a vector, so what else could  |w| mean but ||w||.  There's really no room for confusion so no harm in this usage.

10.3#5

this:
I am not sure where i have made my mistake on this problem, i took the
cross product of the vectors given <1,1,0>x<0,1,-5>  which gave me the new
orthogonal vector <-6,5,1> but since this problem needs the orthogonal
vector to have an i component that is equal to 1, i divided the 5 and 1 by
-6 getting the resultant vector <1,-5/6,-1/6>. Web work is saying that this
answer is incorrect. Is my set up wrong? Is the method for finding the
answer different that what i am attempting?
and this:
in response to the previous email, my set up was right, i ended up making a
small computational error while doing the cross product, i should have
equalled -5 not -6, throwing my answer off. Ended up getting the right
answer.

*************
Glad you figured it out on your own, this is the best possible outcome, especially because you learn to catch your own algebraic and arithmetic errors.

10.4#8

Hello Dr. Taylor!

I got the answer correct by finding the cross product of
the two vectors, then finding the ((magnitude of the cross
product)/(distance between Q and R)). Could you explain why this works
though? I would've thought it would have something to do with projection.

Thanks!

*******************
Oh. Yeah, I never thought about that, but you can do that.  Here's a picture of the points Q,R,P:

There is a dotted line passing through the two points Q, R, and a displacement vector from Q to P.  Notice that I've drawn a right triangle, one corner of which is at Q, the other corner at P, and the right angle on the line passing through Q and R.  I suppose that when you say "the two vectors" you mean the displacement vectors u=R-Q and v=P-Q.  Note that the hypotenuse of the triangle is ||v||. Then the part of v along the direction of  u  is (u.v)/(u.u) u, this is the displacement vector that goes from Q to the place where the blue line intersects the line QR, and has magnitude ||v|| Cos(θ) where θ is the angle RQP.  Of course a little trigonometry tells you that the opposite leg of the right triangle must have magnitude ||v||Sin(θ) = ||uxv||/||u||

10.3#11

Dear Dr. Taylor,
I am a little confused about what I am doing wrong.  I
have been using the equation W=|F||D|cos(Theta) and received the answer
(31sqrt(3))/2.

Thanks,

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

You meant 3*1*sqrt(3)/2.  That kind of mistake will kill you on the exam though, so be careful.

10.2#10

Hi, I'm not sure if you can see my attempts for this problem but I've done
it a bunch of times and I can't figure out where I am going wrong with the
second part to this question. If you could give me a little guidance I
would really appreciate it thank you!

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

Well, you got the speed right.  The angle is a mistake of course.   It's interesting though because, although it's close to the correct answer in an absolute sense, the relative error is about 10%.  I went through the usual list of mistakes people make in computing angles and they all gave smaller relative errors than that, so I really can't imagine how you got that answer, except maybe bad arithmetic or a typo. For instance using 13 instead of 14 would give about the answer you have. 

 

To reiterate a comment I made in an earlier post, when you submit a question like this you need to tell me what you tried so I can catch your mistake. 

Yes I can see how many times you tried to do the problem. So as far as guidance, just flailing and trying lots of guesses isn't going to help you. Stop doing that.  Stepping back and making sure you understand the math, and making sure you did the steps correctly will work better than that 

Lecture Notes 2/28/17 through 9/1/17 (complete!)

Lecture Notes 8/28/17

Lecture Notes 8/30/17

Lecture Notes 9/1/17

10.3#14

Mr. Taylor,

I just don't understand this question at all, and as you can

probably see I've tried it a ton of different times

Thank you for any help

**************************Well you clearly do understand something since you got the a) correct.  It *looks like maybe* you tried to do part b) correctly too, except you made a wee little mistake subtracting the one vector from the other.  I can't tell though, because *you didn't tell me what you did*.  May I suggest that you do that when send me a question?--it usually is a better teaching moment if I can tell you exactly where you made a mistake.   BTW, we used exactly the terminology this vector parallel and that vector perpendicular in the lecture notes from last Wednesday and/or Friday, and those lecture notes are already posted here in the blog, and this stuff is also discussed in section 10.3 in the textbook.  I didn't lecture on the stuff about work, but is also discussed in the textbook in glorious detail.  The upshot is that the work is defined  as the Force vector dot product with the displacement vector.

10.3#19

I've been working on a problem, specifically number 19. I keep getting
sqrt(78),sqrt(26). But the problem says I'm getting it wrong. I also had a
friend attempt the problem and he got the same thing as me, yet still
wrong. It is asking for the length of the diagonals using u=<-3,2> and
v=<7,4>. Any help on why it says it's wrong?

Thanks,

****************
Well the answers you have now are different and correct so it looks like you figured it out.

10.2#13

Dear Mr. Taylor,
I just wanted to clarify how many tries we are supposed
to get when completing homework.  While working through the homework, this
problem seems to have given me a limit to the number of tries before
marking it wrong.

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

OK, I changed the problem to allow unlimited number of attempts.  But before you go, let me point out that the reason 10.2#13 had limited number of attempts is because it was a lot of True-False problems, and it is not too difficult for *someone* without even understanding the problem to just go on guessing until they guess all correct answers and then walk away giving themselves a pat on the head for that.  This would be a *bad mistake* because they would have failed to understand what the problem was about, which would mean that that basic understanding would not be available, come all those difficult and scary exams that will come.  The fact that you bumped into the attempt limit means that you were trying to do the problems without understanding the material, when you should make sure that you understand the material before you try the problem, and certainly before you've tried the problem three or four times.  I suggest that you should attend my office hours or otherwise get tutoring before you get to that point.

10.3#9 frustration

Hi Professor,

I have been working on 10.3 problem 9 for quite some time now and am almost 100% positive the answer I have placed for y is correct. Please let me know if there is something wrong with the grading or if I am doing it incorrectly?

Thank you,

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

First of all, when you have a question about the webwork, could you *please* ask it by clicking the "Email instructor" link at the bottom of the page--you can send exactly the same message but the software will send me the same image as your screenshot above, but will also give me access to the problem as you worked it and in addition computer code that is behind your webwork problem so that I can check it for bugs.

Your answers above are correct to two decimal places.  My *hypothesis* is that webwork is being finicky about wanting three decimal places--that's often the case when this happens, but occasionally there is a bug in the code instead.  If you send do the email instructor button I'll check it out and update this post. 

Lecture Notes 8/21/17 through 8/25/17(complete!)

Lecture Notes 8/21/17

Lecture Notes 8/23/17

Lecture Notes 8/25/17

No homework Friday 8/25/17 and yes homework 9/1/17

Hi Dr. Taylor,

Will the 10.3 and 10.4 webwork assignments be due next Friday as originally posted? I know 10.1 and 10.2 were moved to that date.

Thanks,
**************
Depends on if we do section 10.4 tomorrow.  It's looking highly likely though.

MAT 267 homework site down (Uh, no)

Dr. Taylor,

I'm assuming you won't receive this until the morning, but I was wondering if the WeBWorK site is also down on your end. I've had it bookmarked since last week but now I'm getting an error when I try to access it. 

Thanks,

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

I'm not sure if it was down when you sent this message, but the most common cause of this problem is that your login certificate to ASU is expired.  You need to have a current active login to ASU, e.g. myasu, to be able to access the webwork. If in doubt reload your myasu, and then if you have to login again reload the webwork **after** deleting everything to the left of the question mark in the URL bar

10.3#17

Good evening Dr. Taylor,

I am having trouble figuring out what the dot
product would be when looking at a graph; I realize that parallel and
orthogonal vectors would have a dot product of zero, but I can't tell when
two vectors would be positive or negative.

Thank you!
*****************************

ok, here's your hint.  the trick to this problem is exactly what we discussed in class on wednesday:

u.v = ||u|| ||v|| cos(θ)

Since u.v can be positive only if all of ||u|| ||v|| cos(θ) are positive, and can be negative only if both of  ||u||, ||v|| are positive (since they can't be negative) while cos(θ) is negative.  Looking at the vectors in the image, they are each clearly not the zero vector, hence in each case ||u||, ||v|| are positive. This means that the answer to all of these questions depends only on cos(θ).  So you can know everything you need to know by eye-balling the angles between each pair of vectors and knowing about the properties of cos(θ).  (NOTE: go study cos(θ))

Office Hours

The MAT267 Office Hours will be:
Mon 12:30-1:30
Wed 1:30-2:30
Friday 2:30-3:30

A couple of comments.

1) This schedule manages to accommodate 55 of the 61 people (which is less than half of the students taking calculus from me) who had answered the poll.

2) If you can't make my office hours, scheduling a meeting with me is always an option for you.

3) You are telling me that you will come to an office hour on Friday afternoon.  We'll see.  If attendance is very low I reserve the right to reschedule that time to one more convenient.

Office Hours Poll

Hi All, please give your office hours availability in the doodle poll at the following link:
<https://beta.doodle.com/poll/p5sgf7n34db24wf7>
Instructions: please click on every available time slot for which you could attend the office hours for at least 15 minutes out of the hour.
thanks!
--Tom Taylor

RE WebAssign

On Fri, Aug 18, 2017 at 11:55 AM, ************   wrote:

is web work different from web assign. I'm curious because the book list said to purchase web assign so I purchased the text book new. however if I do not need web assign ill return it and get a used copy. will all homework be done through web work and do I need to purchase it ?

*******************************
To quote from line 6 of the syllabus:

Text: Essential Calculus , Early Transcendentals by James Stewart, Thomson (Brooks/Cole), 2e (Or, if you are on campus and want to look in the bookstore, the book designed for ASU and has a hand on the front cover is: ACP Calculus (Custom) ASU Bundle (w/Enh WebAssign Access) We will not use WebAssign but the price of the text with it is the same as the price without it.)

Welcome to MAT267, and some useful information

Hi All, welcome to your MAT267 blog. You can look here to find assignments, posted scores & estimated grades, questions and answers. I suggest that you bookmark this page, and also subscribe to email updates to this blog in the subscription field to the right.

1) Your Posting ID. Your Posting ID will be used to identify your scores. You should not share your Posting ID or do anything to compromise it's security. To quote from this link:

Posting ID
Your Posting ID is a seven-digit number composed of the last four digits of your ASU ID number plus the last three digits of your Campus ID number, separated by a hyphen. Your Posting ID is printed on the class rosters and grade rosters your professors work with. You can also view your Posting ID on the My Profile tab in My ASU.

2) For that matter, especially don't do anything to compromise the security of your ASU or Campus ID numbers--they can be used to for identity theft or invade your privacy. For instance, DO NOT SEND ME YOUR ID'S BY EMAIL--I don't need them to interact with you and email is an inherently insecure form of communication.

3) The first homework assignment is sections 10.1 and 10.2, which due on Friday August 25 at 11:59 PM.