score update (after corrections)

Apologies & correction

I went home yesterday afternoon after classes, seminars and meetings and got pukey-sick, which is hopefully the only cause of my stupidity on Friday going over the quiz, for which I apologize and which I'm going to correct here.  If you still are confused on this issue, I'll be in my office tomorrow 11:30-12:30pm and Tuesday 12:30-1:30pm to discuss with you

The second problem on the quiz was to sketch out the regions of positive and negative divergence, which are as follows (blue is positive divergence and red is negative divergence)--I have imposed a coordinate system on the graph to give a language to describe parts of the graph:

Here's the correct reasoning. Calling the vector field F=Pi +Qj , the divergence is ∂P/∂x+∂Q/∂y; we can see that the Q is positive above the line y=1 and also below the line y=-1, and increasing as |y|>1 is more positive; you can see this in the way the vectors point upward-ish and are increasingly upward on each vertical line for x big enough or small enough.   On the other hand Q is negative between y=1 and y=-1, as you can see from the fact that the vectors point slightly downward.  Likewise, for P is positive for x bigger than 1 and also for x less than -1, and is more positive the more |x| is greater than 1 and is negative while x is  between x=-1 and x=1; you can see this from the way that the vectors point somewhat to the left when |x|<1 and to the right when |x|>1, and point increasingly to the right along horizontal lines as |x| is large enough.  

Now, since div(F)=∂P/∂x+∂Q/∂y, div(F) will be positive if the sum of ∂P/∂x and ∂Q/∂y is positive, that is if the rate of increase of P in the positive x-direction is greater than the rate of decrease of Q in the positive y-direction OR vice versa.  In the first quadrant, we can see that both ∂P/∂x and ∂Q/∂y are positive, in that they both P and Q increase from negative to positive values going from left to right (P) and from down to up (Q): this means div(F) must be positive.  The reverse is true in the third quadrant, so  div(F) must be negative.  In the half of the second quadrant above the line y=-x, Q increases a lot with increasing y while P decreases slightly with increasing x so ∂P/∂x is less negative than ∂Q/∂y is positive so div(F) must be positive; the opposite is true in the second quadrant below the line.  In the same manner, in the fourth quadrant above the line P is increasing quite a bit moving from left to right, while Q decreases not so much--it follows that ∂P/∂x is more positive than ∂Q/∂y is negative so div(F) must be positive, and again in the same manner the opposite is true in the fourth quadrant below the line.

Office Hours Monday and Tuesday

I announced this in class, but here it is anyway:
M 11:30AM-12:30PM
T 12:30-1:30PM

Finals Cram and Finals Workshop

From Brian England: I have decided to hold another problem session outside of the normal time frames next week.  I get out of my own final exam <<on Tuesday>> at 2pm and will head to the engineering tutoring center and be there by 2:30.  They have agreed to keep the place open until 5pm.  As always, your students are welcome to come.

Also: There will be a MAT267 Finals workshop Tuesday May 3 at from Noon to 1:00pm in BA201.  "Finals Workshops requires the student to bring class notes, practice material and questions to the workshop. Students should be prepared to collaborate with peers and tutors. This learning environment should foster recall and understanding of information for an upcoming exam."

  See <https://tutoring.asu.edu/calendar/>, Sign up: http://bit.ly/MAT267Final

13.7no7

I don't know why its coming out wrong!

*******************
A few reasons.  First the vector element of area dA here is what our book calls dS, and not what the book calls dA--that is dA=(r_u x r_v )dudv, or since your parameters are called  r, θ  
dA=(r_r x r_θ )drdθ.  Second,  r, θ here are just parameters, not polar coordinates: you only have drdθ for your area element in the parameter plane, not rdrdθ (in fact I guess that the whole point of this problem is to fool you into making the mistake you did make, so that you can learn not to make it).  Finally, in your integral ∫∫_S F.dA it looks like you dropped the 9 from your F and also did the dθ integral incorrectly: the answer is not 2π because the F . (r_r x r_θ ) actually not independent of θ.

13.7no2

Can you help me with Let S be the part of the plane 3x+2y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=1i+1j+4k across the surface S.

******************************
OK, here's an answer in general terms, so I don't do your homework problem for you.  You can write the plane as the graph z=f(x,y) of the function f(x,y)=4-3x-2y.  You can then use the usual parameterization of the graph of a function: r(u,v)=<u,v,4-3u-4v>, and then compute that
  r_u x r_v = ai+bj+ck, 
for whatever numbers a,b,c you get from computing the cross product.  Since the orientation is upward, you have to be a little careful: c has to be positive, so if   r_u x r_v  has c negative you have to use the -(r_u x r_v) instead to get the upward orientation.  The shadow of S in the xy-plane is the triangle enclosed by the x-axis, the y-axis, and the line 4-3x-2y=0, and the same triangle is the domain D of the parameterization. Then ∫∫_S F.dS=∫∫_D ∫∫_S F.(r_u x r_v) dA

13.6 no 3

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

OK, the problem here is a practical one.  The "flow outward" this problem wants is in the units kg/sec. Since the density is kg/m^3,  and since the vector field v is in m/sec, and since dS has units m^2, the product density*v.dS has units of kg/sec.  Thus the rate of flow outward will be

5∫v.dS=5x(8/3)pi

 or (40/3)Pi.

Attendance up to 4/20

(click on image to enlarge)

Exam 3 update glitch

Just at the end of my office hour (and a half) today I crashed my spreadsheet.  I was saving often, so this could only possibly affect the people who were in my office after 12:30PM.  I *think* I recovered all of the correct scores, but if you were in my office after 12:30PM today, and correcting your exam or quiz 9 scores, check the following list.  If there is a problem, bring your additional scores (the new points I gave you on your paper) to my office hour tomorrow or class on Friday and I'll fix the problem.

Course Evaluations (a.k.a. grade your instructor)

MyASU -> My Classes -> Course Evaluations

Current scores and projected final grade

(click on image to enlarge)








Notes:
1)"Estimated final score" is made from the current total scores combined according to the rule (Exam Scores/300)*75+Homework*15 +Quizzes*10
2) these scores do not include test corrections if you havent done them yet.

Extra credit projects

Here is your extra credit opportunity--20pts total.  The project will be three pages double spaced, 1 inch margins.  Projects will be delivered in pdf format to my inbox AND in my department mailbox (in Wexler Hall 231) before Wednesday May 4 at noon.  No credit will be given for projects that do not meet the format or deadline.  Each project will have statements clearly cited.  While I offer starting references in as Wikipedia pages, each project will require you to locate a minimum of two other published references; you may wish to speak to your engineering professors for suggestions.  Below I offer two topics, I'm open to adding other topics to the list if you can come up with one that has a compelling application of multivariable calculus.



Topics:

1) The use of divergence, curl and gradient in electrodynamics.
Some starting references:

A) https://en.wikipedia.org/wiki/Electromagnetism
B) https://en.wikipedia.org/wiki/Classical_electromagnetism
C) https://en.wikipedia.org/wiki/Maxwell's_equations

Specific questions to answer:  What does the divergence of an electric field tell you?  What does the curl of an electric field tell you?  What does the divergence of a magnetic field tell you?  What does the curl of a magnetic field tell you?  What does the gradient of an electric potential tell you? Suppose that there is an electron sitting at the origin. What is flux integral over the sphere of radius 1?


2) The use of divergence, curl and gradient in fluid dynamics.
Some starting references:

A) https://en.wikipedia.org/wiki/Fluid_mechanics
B) https://en.wikipedia.org/wiki/Fluid_dynamics
C) https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations

Specific questions to answer with regard to a velocity field of a fluid in motion: What does the divergence  tell you?  What does the curl tell you?  How are these related to pressure?  What are the properties of a potential flow?

Lecture Notes 4/18/16 through 4/22/16

Lecture Notes 4/18/16

Lecture Notes 4/20/16

Lecture Notes 4/22/16

13.5 number 2

I'm not sure what I am doing wrong? why do i keep getting it wrong, i've
reworked it a few times already.

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

Well, what are you trying?  Let's see....you've done the divergences correctly, so that's not the problem.   Uh...wiki has this formula for the curl:

which is the same as ours...which for part A) gives curl(F)=-xi + (2y-(-y))j+(-2z)k, so it looks like you made an sign error leading to an arithmetic error in the j component.  In part B) you've done the curl correctly, so there was only the one error

13.5 number 9 (updated)

I do not know how to approach this at all.  Please give me some insight!  I
work the majority of the weekend, so an extension until Sunday would be
heaven-sent too.  If not, I'll do what I can before midnight.  Just
throwing that in there! ;-)

*******************************
First of all, the book talks about this in some detail. I recommend that everybody read it.  The *general* notion is that the curl measures both the amount and direction of circulation around an axis--for which counter-clockwise is positive--as well as the direction of the axis as a vector.  The important examples to keep in mind are the ones we worked in class last Friday: the vector fields

F=xi+yj+zk for which curl(F)=0

and the vector field F=yi-xj+0k, which looks like this:

for which curl(F)=-2k (and the negative is because the vector field is circulating clockwise around the z-axis).

The Final Exam

TUESDAY, May 3: 7:10-9:00pm in COOR 184

Homework Due This Friday

Hello Professor Taylor, 

I was hoping that we could extend the 13.6 homework to Wednesday or Friday next week considering we finished the notes today. Also it would be extremely hard for me to make it into the tutoring center before Friday and spend a decent amount of time on it, so any extension into next week would help. Thank you!

Regards,

AND

Hello Dr. Taylor,

I was planning to begin HW for section 13.6 this evening
(since I'm fairly certain it was open since Friday), but after walking away
from my computer for a bit and refreshing my browser I no longer have
access to 13.6.  Were we supposed to have access to it originally?  Or was
this recent change an error?

Thank you for your time.

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

Oops, right you are.  OK, 13.6 is now due on the 29th along with 13.7. (which is the last day of classes, mind you.)  It's still open though, so you can work on it, even though it's not due until next week.

Lecture Notes 4/15/16

Lecture Notes 4/15/16

About negative volumes...

There aren't any.  If you get a negative answer for a volume you did something (very) wrong.  You should figure out what it is.  Then fix it.
(Of course, you could hope that you just made a sign mistake at the last minute and that the true answer is just the negative of your negative answer.  From what I've seen on a lot of exams, though, odds are that you made a *much* bigger mistake than that)

RE: taking notes

Article on NPR

<<Not that it's easy to take laptop notes in math class.>>

13.4 no 5

Don't know what's wrong on this one either...

did green's theorem for the
entire figure as if it was closed, then subtracted the AD portion with a
simple line intergral and got 18 for the area of the trapezoid, and got
81/2 for the AD integral.

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

As you didn't say but I suppose you know ∂Q/∂x-∂P/∂y=9-8=1, so the integral counterclockwise around the closed path is equal to the area. This means, as you know, that the integral ABCD plus the integral DA is equal to the area.  In other words it's the area minus the integral DA is equal to the integral of ABCD, which is the same as the area PLUS the integral AD:  the correct answer is
 18+(81/2).

13.4 no 4

I'm trying to answer 13.4 #4 but its saying its wrong.

Int (dQ/dx-dP/dy) dA

Qx=4
Py=0

Int (4) * dA = 4A(Area of paralellogram) which is Area*Base

x0=6, y0=6, therefore A = 6*6 =36

=4*36 = 144

Is there anything I'm doing
wrong?

***********************
Your contour is clockwise, instead of counter clockwise, i.e. the opposite of the direction that Green's theorem is about.  The result is that true answer is -144.

final exam quibbles

I read the syllabus and it says the final is May 3rd a Tuesday at 7:10 pm to 9:00 pm. According to the ASU finals schedule it says our final is Monday May 2nd at 9:00 am. Which day and time is the final at?
*********************

I wish. To be fair, I've been fooled a few times too, and I should know better.  However, MAT267 has a common final, and is listed here in the Common Finals on May 3rd in the evening.

Lecture Videos MAT267

<https://math.la.asu.edu/~surgent/video/mat267_exp.html>

13.2no5

Professor,

We got value 9031 and have no idea where the sqrt(5) came from???? 

The other image attached is the work we have done for this problem please steer us in the right direction.

Thanks, 

Team Night TRAIN

(click to enlarge image)







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

you multiplied by ||<1,2>||^2=5 instead of multiplying by ||<1,2>||=√5.

mat267 reviews

T/TH - COOR L1-88    6:00pm - 8:00pm

M/W - CPCOM 213  6:00pm - 8:00pm

review 12.6 no4

Having trouble setting up the bounds for theta in this situation z is bounded by 0 to 1 an r is bounded 0 to sqrt(2) but we realized that because the of the bounded region it is not a simple 0 to 2pi. Theta is also cut down because of the Z being only from 0 to 1 which cuts off part of the radius. We know it is something less the pi/2 but do not know what, so what do we do to find the region in which theta is bounded? 

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

I guess you mean this problem?

Between y=0 and y=sqrt(2-x^2) is the half circle in the upper half plane of radius sqrt(2)--that makes theta run from 0 to pi. z between 0 and 1 has nothing to do with it.

webwork typo

Hello Professor,
I am confused about the notation used to describe the gradient vector fields.

I see that there is a problem which has the function f(x,y,z)=4x+4y+2z and we are told to take the gradient of the function  <grad>f(x,y)=_i+_j+_k

The answer that was considered correct was based on a gradient which created a R^3 vector field.

Please help me understand, I don't know what I just did.

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

I see what you are saying:  problem D has a function f(x,y,z) of three variables but <grad>f(x,y) has just two variables.  That's a typo in the code, and I think I fixed it. Reload the webwork and let me know if it changed.

BTW, thanks for getting started on this early.

Lecture Notes 4/4/16-4/8/16

Lecture Notes 4/4/16

Lecture Notes 4/6/16

Lecture Notes 4/8/16

12.6 number 6

Do i have the boundaries of z correct?  Also how exactly do i calculate boundaries for x and y?  Replace something w zero or set it equal to zero or..?

Thank You,

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

Yes.  To calculate the boundaries for x&y you have to understand what is the shadow of the figure on the xy-plane.  Since E is part of three dimensional space below z=4-4(x^2+y^2)  and above the xy plane, and since the xy plane is give by the equation z=0, the intersection is given by the equation 0=4-4(x^2+y^2) or x^2+y^2=1--this is the unit circle of course. Depending on the order of your integration you have to use this formula to get the limits above and below in the middle integration, while the outside integration has the limits which allow the full extent of the circle. 

12.6 Number 4

***** The feedback message: *****

Why am I not getting theta correct?  I used inverse tan(-3/5) and got
-.5404 and it is saying it is incorrect.

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

Well, tan^{-1}(-3/5) is indeed -0.5404 as you suggest, which is the same as 2.601 as you wrote. *BUT* (-5,3) is in the second quadrant, while the formula θ=tan^{-1}(y/x) only works for (x,y) in the 1st and 4th quadrants.  The correct formula for the second and third quadrants is instead
θ=π + tan^{-1}(y/x).

Lecture Notes 3/28/16-4/1/16

Lecture Notes 3/28/16

Lecture Notes 3/30/16

Lecture Notes 4/1/16

HW section 12.5 reopened.

Hello professor, i am here at work half panicking and trying to do my homework without getting in-trouble. Can you please extend the homework until tomorrow please. I won't get out of here until 11pm. Thank you so much! 

I'm getting frustrated with how to set up 12.5.  Is there any way you can extend the assignment until Monday when I can get some face to face time with you to figure it out?  Or with a tutor since I'm sure you'll be busy with extra credit people?

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

How about Sunday night?  It has been reopened.

Lecture Notes 3/25/16

Lecture Notes 3/25/16

12.3 no8

How is it not -pi/2 to pi/2? I am positive that is what the limits of theta
are. Correct me if I'm wrong.

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

Well, it is -pi/2 to pi/2.  It looks like you're entering 0 to 2pi though.

Review section 11.3 number 4.

Professor Taylor,

Number 4  in 11.3 there is a substitution made to go from term t to term x giving the answer as cos(x^8) please help and explain this substitution.

Thanks 

*******

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

OK, the function f(x,y) is defined as the integral from x to y, of a function of one variable. The f_x that you are supposed to compute is the derivative of the integral with respect to the upper limit of integration.  NOW, something that many people wish they didn't have to remember but always *do* have to remember is the fundamental theorem of calculus, which describes the way that integration is the opposite of differentiation.  What the fundamental theorem gives you is the differentiation with respect to the upper limit gives you the integrand evaluated at the upper limit.   In your case this means that you just get cos(x^8).

For the next hour, calculus review. Nevermind, I'm tired of waiting.

Here is the link.  You should be logged into your ASU google account

Exam Review

It will be Tuesday WXLR 203 at 6pm.  
"My class ends at 5:45 that day so I may be a bit late getting there depending on questions after class."

Lecture Notes 3/14/16-3/18/16

Lecture Notes 3/14/16

Lecture Notes 3/16/16

Answers Q8 "Funzies"

Answers Q7

Extra credit proposition....

So here is my proposition: I will give you up to 50% of the points you missed on each quiz or each exam problem IF you come to my office hours and work through the quiz or problem on my white board within 1 week of the quiz or exam being returned.

Please let me know if this of interest to you.

Webwork Section 12.3 is now open

For your viewing &exam preparation pleasure, Webwork Section 12.3 is now open.

MAT 267: WebWork- NAG

On Mon, Mar 14, 2016 at 7:47 PM, ************** wrote:

Hello Professor Taylor,

Would you please post the WebWork for this week?

Thank you,

***********

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

Done 

Lecture Notes 2/29/16-3/4/16

Lecture Notes 2/29/16

Lecture Notes 3/2/16

Lecture Notes 3/4/16

Section 11.7 HW

Can you extended the homework due date for tomorrow ? Please. This is me nagging
******


***********************
yes, done.

Attendance Totals to Date

(Click on table to enlarge)

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.

**************
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.



(Click on the image to enlarge)

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.

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

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.

Lecture Notes 2/5/16 through 2/17/16

Lecture Notes 2/17/2016

Lecture Notes 2/15/2016

Lecture Notes 2/12/2016

Lecture Notes 2/8/2016

Lecture Notes 2/5/2016

and the classics should not be neglected either...

the ever hopeful insistently tell me this on exams:

√(a+b)=√a + √b, 

for instance as in √(1+4t^2)= 1+2t

IT'S NOT TRUE!

another thing not to do...

Suppose you are given two planes, say:

5x-2y+z-3=0  and -x+y+2z-1=0

and you want the intersection of these two planes, which is a line, and 0=0 so you should set these equations equal each other

5x-2y+z-3=-x+y+2z-1, 

which gives you 

6x-3y-z-2=0,

right?  

DO NOT DO THIS.

Do not do this because your are trying to find the points <x,y,z> that make both equations true. Yeah, both equations are equal to zero and so 5x-2y+z-3=-x+y+2z-1 when x,y,z satisfy both equations, but 5x-2y+z-3=-x+y+2z-1 for a lot of triples  <x,y,z> that don't make either equation true.  You need to figure out how to solve both equations at once and this won't help you do this.

New Course Policy

From now on:  ONLY your writing on the actual exam paper in the actual space provided for the problem will be considered in assessing your score.  What you wrote on scratch paper will not be considered.

something not to do....

Here's a logic I'm seeing on the test, in fact I've seen it on a lot of tests for a lot of years:

r(t)=<t,t^2,t^3/3>

so

r'(t)=<1,2t,t^2>   evaluated at t=0 gives

1+0+0 so

r'(0)=1,

right?

WRONG!

<strong emphasis>DO NOT DO THIS</strong emphasis>

evaluating a vector function at a given time does NOT mean that you get to add all the components. The correct answer is r'(0)=<1,0,0>.  You have to substitute t=0 in every component, the fact that evaluating the component gives that the component is equal to 0 does not mean that the component disappears.  It is still there being zero.

homework extension

Hello Dr. Taylor,
   I don't seem to get these 2 sections at all and am getting so frustrated with them.   Is there any possible way I can get an extension on these two sections? Maybe until I can talk to someone and go through my issues with them or at least another day so I can research more online on how to do it... Sorry for the inconvenience, but I just don't seem to get it.

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

OK, given that you just took the exam, you all can have until Sunday night

webwork 10.9

Hi Dr Taylor,

I'm stuck on calculating the speed in this problem from 10.9 homework, I feel like I am close but not quite there. I'm unsure what I am doing wrong will appreciate any help.

Thank you

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

OK, you're going to need to compute the root sum of squares of the components of v; it looks like instead you've used the components of a., or maybe you forgot the t multiplying the trig functions.  The right answer should be made from

 √((7sin(t)+7tcos(t))^2+(7cos(t)-7tsin(t))^2+14^2t^2)

which you can and should simplify considerably, but you need at least another t^2 in there somewhere.

Problem 10.8#4

Hey Prof. Taylor,

I don't understand why I'm getting the wrong answer for

Tanget Unit Vector. I was able to get the Normal Unit Vector fine which

requires TUV in the first place. To find it, I did r' / |r'|.

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

ehr...it looks like you've computed |r'|=17...but r'=4sin(-4t)i-4cos(-4t)j -4k, so  |r'|=√(16sin^2(-4t)+16cos^2(-4t)+16)=√32=5.66.

MAT267 Review Questions

 I'm having troubles finding the calc 3 review from the math dept site... do you happen to have a link for it?  I know one of the fellow classmates mentioned they had a review posted on that site.  If not, do I just study webwork and the book?  And which sections is it up to again?

**************************************
The test tomorrow is on everything we covered from Chapter 10.  The webwork and book are useful indications too, especially study the questions you didn't get right the first time, if any. There's a link to the review question on the syllabus, but here's the link again.

ASU Online MAT267 Lectures

at this link.

Pretest Review Session Tomorrow Night!

There will be a review session on Tuesday night for MAT 267. It will be at Wexler 116 at 8:15pm. 

Service Announcement from a fellow student

U of A is coming to Swim Asu on Saturday the 6th at 1:00 PM.  During the breaks Michael Phelps and the other professionals will be giving an olympic preview (exhibition swims) and after the met there will be a meet and great with the same.  The meet will be held at the Mono Plummer aquatic center.

Biweekly Review Session

There will be a biweekly problem review session in WXLR A 107 on Mondays and Wednesdays from 2:30 - 4:00 pm.

attendance to 2-4-16 (&your posting ID)

Your Posting ID. To quote

"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. Many faculty use Posting IDs to post incremental grades like mid-term exam scores, lab results, etc., in public places such as the departmental bulletin board, or the wall in the classroom. The Posting ID is used to permit your professors to get your incremental grades out to you quickly and efficiently while at the same time protecting your privacy."

Lecture Notes 2/1/16-2/5/16(to be updated)

Lecture Notes 2/1/16

Lecture Notes 2/3/16

Lecture Notes 2/5/16

Gandalf...

It has come to my attention that I am not very blog-savvy.  On 10.3 #9, I
am having trouble setting up how to do the problem.  To get from point
(3,2) to (4,7) starting with a vector 5i+2j and turning once at 90 degrees
to get to that specific endpoint.  I know that in order to be orthogonal
the dot products of the 2 vectors needs to be zero....but I don't know what
2 vectors to use and then once I have the two vectors to use....do I just
subtract or add one of those vectors to the start or end point...?  Please
help!

****************************
OK the direction of the Iron Hills from the forest of Mirkwood is <1,5>=<4,7>-<3,2>.  Good old Gandalf doesn't go that way though (because that would be too simple).  Instead he goes in the direction of the vector <5,2> for awhile (which is *not* the same direction of <1,5>), and after awhile takes a single right turn and heads in that direction and ends up at <1,5> (I emphasize the words "right turn" because that is the crucial bit of information that tells us what we need to do)

So the picture looks like this:

If you put it in this way, it should gradually become clear that this is a problem about orthogonal projection. The right pointing red arrow is the orthogonal projection of <1,5> on <5,2>, i.e.

=(<1,5>.<5,2>)/(<5,2>.<5,2>)<5,2>=15/29<5,2>=<75/29,30/29>

This is the point where he makes his turn.

The upward pointing red arrow is the difference or residual <1,5>-<75/29,30/29>=<-46/29,115/29>.

Homework due Friday 2/5/16 (also a quiz on sections 10.5-10.7)

Section 10.5, Section 10.6 and Section 10.7

The quiz on Friday...

The same subject material as the homework: Sections 10.3 and 10.4

Lecture Notes 1/25/16 through 1/29/16 (to be updated)

Lecture Notes 1/25/16

Lecture Notes 1/27/16

Lecture Notes 1/29/16

Lecture Notes 1/20/16 and 1/22/16

Lecture Notes 1-20-16

Lecture Notes 1-22-16

Homework due this week, plus wondering...

Section 10.3

Section 10.4


Should I give a quiz on Friday? The ideas in these sections will be used all semester long...

Problem 10.2#17

 Hello Dr. Taylor,
I am can't figure out what to multiply the first part of
number 17 by to get half its length and it in the opposite direction. I got
the unit vector but can't figure out the proper scalar please help.

(click on image to enlarge)
*******************************
Well, for part 1 you're doing good things with dividing by the length to find a unit vector, but you're making it harder than it needs to be.  To get a vector half the length of u, you just have to multiply it by 1/2.  And to get it going in the opposite direction you just have to multiply it by -1. You *don't* need to divide by the length of u.

For part 2, though, you need to get a vector of exactly length 3 pointing in the direction of v, so you're doing exactly the right thing by computing the unit vector v/|v| and then multiplying by 3

Resources for learning

Tutoring:
Fulton Engineering College Tutoring

University Academic Success Programs

Math Tutoring Center


Free Online Lectures (not associated with ASU, but some of my student have found them useful):

Khan Academy, (the sections of Linear Algebra on Vectors, and the Dot and Cross Products could be useful to you now, and the lectures on Multivariable calculus will be useful starting with our chapter 11.)

Lecture Notes 1/13/16 and 1/15/16

Lecture Notes 1/13/16

Lecture Notes 1/15/16

Office Hours Posted

Course office hours are posted on the online syllabus and here. They will be:

Tues 12:30-1:30PM

Wed 11:30-12:30PM
Thurs 11:30-12:30PM

Announcement: Homework and Quiz

The homework for this week is now open and will be due a week from today at midnight.
Section 10.1
Section 10.2

Also, there will be a quiz next Friday, 1/22/16.

Welcome & First Homework Assignement

Hi! Welcome.
1) The syllabus has been updated--the tentative schedule is correct, but the actual sequence and timing of material may still be subject to small variations.
2) The first homework assignment will cover sections 10.1 and 10.2.  The webwork site for this course is still not online, however this is not a problem because the homework won't open until January 15 at Noon and will be due January 22 at midnight.