Final exam pickup

Hello professor Taylor,

Is there anyway I could pick up my final exam, I would like to see which ones I got wrong. 

Thank you.

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

No.  In fact I'm required to keep them for seven years.  But you can look at it, and if you like we can make a copy for you.

Final Score Tally

% Total figured as

0.5*(Extra Q+Extra credit+3 midterm scores)/3+0.25*FinalExam + 0.15*%HW + 0.1*%Quiz

Office Hours This Week

I don't know if I'll be able to make it to office hours today, but I thought I heard you mention having office hours on Tuesday as well? If so, what time?

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

Let's do 12:30-1:30 both today (Monday) and tomorrow.

13.7#6

Hi, Dr. Taylor. I keep completing this problem and I get the same answer
(-8pi) regardless of the method I use to solve for the flux. I also tried
using the divergence theorem, but I receive a different (incorrect) answer.
Any thoughts on what I'm doing wrong?

Thanks.

*********************
well, you haven't told me what you have done, but I surmise a couple of mistakes: the first is that your normal vector ∂r/∂t x ∂r/∂s is in the opposite of the positive y-direction, while for the second you either made an algebraic mistake computing the cross product or you made an integration error equivalent to ∫ xdx = 2x^2 + C
(the divergence theorem doesn't apply to this problem)

Clearing Up the Confusion (Extra Credit Assignment)

Dr. Taylor,

I have attached my electronic copy of the extra credit to this email, however, I also have questions about the physical copy. Your initial blog post says we should put it in your mailbox at the School of Mathematics building, but in person you told us the PSA building. Then, on your blog you said to just submit it electronically. Which of these 3 answers should we follow? Thanks for the help.

Hi professor, I am ******* from ur cal 3 MWF :8:35 AM class. I just want to confirm the mail box place. I understand ur office is at GWC 356, but i am not sure where can i find the mail box for that.
PS : Is it necessary turn in the hard copy? it is kind of a pain XD.

Thank you!

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

Electronic Copy: a .txt file by email attachment to my email.  Due by Midnight on 12/7/16.

Hard Copy: To my mail box in Wexler Hall (WXLR) Room 216  Due on 12/7/16 by 5pm or when the math office closes, whichever comes first.

(both soft copy and hard copy if you want credit)

extra credit electronic submission

Hi Professor,

You said you wanted the extra credit paper turned in online and to your mailbox, right? Where is your mailbox in the school of mathematics? And do you want the electronic copy submitted through Blackboard?

Thanks,

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

Just email it to me, please.  Do this: from whatever software you wrote it in, under the "File" menu, click "Save as",  then click "Text". Then attach your text file to your email

Lecture Notes Dump (more on the way)

October 12, 2016

October 14, 2016

October 17, 2016

October 24, 2016

October 26, 2016

October 28, 2016

October 31, 2016

November 2, 2016

November 9, 2016

November 14, 2016

November 21, 2016

November 28, 2016

November 30, 2016

How to learn math

People are asking me how to learn math again. I'm not an expert, but here are a couple of links that might be useful:

Stanford Math 53: How to study math.

Barbara Oakley's Learning How to Learn

All about the final exam

Hello Dr. Taylor, I just wanted to confirm where we are going to be taking the final for this class. I am in your MWF class. I assume we take the final exam in the lecture classroom?

Thank you, 

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

Nope. Per the syllabus (remember that thing?), the final exam will take place in DISCVRY 250 on December 6, from 7:10PM-9:00PM. Plan to be well studied, well rested, wide awake and sober.

Class on the 23rd: Yes

Hi Professor, just wondering if you are holding class on Wednesday the 23rd of November. Thanks!

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

Yes.  The great State of Arizona is paying me to the teach on the 23rd, and so teach I must.

A common point of confusion...

When a problems says to integrate ∫ F.dr along a path, or any path, from point A=(a, b, c) to point
B=(d, e, f),  that means that the direction is taken FROM A TO B.  For some reason lots of people want to do this in reverse, which means they get the negative of the right answer.  In the particular if
F=▽f, lots of people will write ∫ F.dr=f(a, b, c) -f(d, e, f), instead of the correct answer which is
f(d, e, f) -f(a, b, c)

13.3#11

Hello Professor! I'm a bit stumped here. I am not really sure where to go from this point. My understanding of a conservative vector field is that it is path (in!)dependent, and should turn up the same value for work for any path between two points.  (true!)

However, I don't see what I am doing wrong. I have been drawing out the fields and attempting to estimate paths, but I cannot get past these answers. Could you please enlighten me as to what
paths I am thinking of incorrectly?

***********
Trying to guess about wha their line integrals will be for various paths is *NOT* very helpful, because there are so many possible paths.  There are two keys required for these problems.  The first is remembering that being path independent/conservative is the same as (if the domain is reasonable) being a gradient.  The second is to have a gallery of examples of what gradient fields look like.   For example, the gradient of a linear function is always a constant vector field, that is at every point you have the same vector with the same magnitude pointing in the same direction. Do you have any of those?  I suggest you look at your favorite quadratic functions and sketch out their gradient fields, there are some of those in the collection above.  You should also be able to look at a vector field sketch above, and guess what are the functions F_1 and F_2, and from that understand if
∂F_1/∂y=∂F_2/∂x is  possible.

Section 13.4 due date

Hello Professor Taylor!

Quick question: Is section 13.4 meant to be due on Sunday? 

Thanks! 

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

Now it's Friday

Extra Credit Opportunity

Here's what I propose: a three page, single spaced essay on the topic of "What Multivariable Calculus is Good for in My Chosen Major".

Here are the rules:
1) No plagiarism, i.e. you can not copy paste from wikipedia, or the text book or anywhere else. It must be in your own words. Your are allowed to quote from your chosen source, but it must be attributed in a Citations section.

2) If you make an assertion you should document the source/reason why that assertion is true, again in the citation section

3) The essay should be structured with a one paragraph introduction, in which you tell me what you are going to tell me, a one paragraph Conclusions section, in which you tell me what you told me, and in between those at least two pages worth of making your case, clearly and cogently.

This essay will be worth 10 midterm exam valued points.

To receive credit Essays must be submitted in both electronic text (not ms word, not pdf) and as hardcopy to my mailbox in the school of mathematics.  
THE DUE DATE: 11:59PM, December 7th   

13.3#1

Hi Professor, for problem one on section 13.3, I have the following answers:
a. N
b. N
c. 5xsin(y)+5y^2


I know they are all correct, but WebWork keeps saying that the variable "N" isn't allowed, even though it accepts it for Part b.


Please get back to me as soon as you can.
Thanks.*******************

The Webwork software has a random number generator that inputs slightly different parameters for each user; everybody gets to work a slightly different problem.  If you click the 'email instructor' button at the bottom of the page, webwork will send me a link to your specific problem and I can give you an answer specific to that problem.  That being said, I just worked the problem from my webwork account and there was just one "N", so I guess you may have computed the partial derivatives incorrectly when you decided that a. was not conservative.

Last minute worries

Dr. Taylor,

Currently, the homework for 13.4 is due 11/20, but considering that we never got to it, can we have an extension for that section?

Thank you,

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

Yes.

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

Hello, Dr. Taylor.

I recall you telling us in class that the method we learned to determine whether or not a vector field is conservative <true!> (and how to calculate the potential function of a conservative field <not true!>) did not apply to three dimensional functions; you mentioned that we would learn this after the Friday's test.

Accordingly, how should we calculate the potential function for this 3D vector field?

Thanks.

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

In any case, you should start doing partial integrals to compute the function while enforcing constraints relating partial derivatives to the components of the vector field.   Just like in the two dimensional case, except that a partial derivative kills an arbitrary function of the two other variables. 

Practice Test3 #3

Using spherical coordinates, I have the limits 0<theta<2pi, 0<phi<pi/4, but I am not sure about rho. What would be the lower limit of rho? (I think the upper limit is 3)

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

That's correct. Since you are in the cone, every straight line leaving the origin stays in the cone, and intersects the sphere.  This means that the lower limit for ρ has to be 0.

Practice Test 3 #7

For this one I set me integral up as:  Int[0,2pi] Int[0,4] Int[0,16-r^2] rdzdrdtheta

I am not getting the right answer can you please tell me what I am doing wrong?

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

Yes, the inequality x^2+y^2≥1 means that the lower limit of integration for r  is 1. This means the domain has a hole in it, from above it looks like this:

 

13.2 #4

I am a bit confused about this question. I know that the middle one (the
one at the top, perpendicular to the vector field) has its integral, and
therefore work, equal to zero. However, the other two don't make sense. The
one traveling in the direction of the vector field should be positive,
right?

The vector field is in the same direction as the path, so that
equates to a positive line integral? If that is the case, why is it the
smallest? Am I just confused?

**************
The one traveling in the same direction of the vector field, C3, will have dr parallel to F, hence F.dr is positive, hence the integral will be positive (this is greater than the integral over C1, which is zero). On the other hand integrating over C2, dr will be in the opposite direction as F, hence F.dr is negative hence so is the integral over C2.

HW 13.2

Would it be alright to request a homework extension to section 13.2? Just to the weekend, so we can have more time to study up and work on other exams. And of course, go over the material in detail. Just to compensate for the interrupted schedule math has taken lately. 

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

OK, 13.2 is now due on Sunday night.

How to remember formulas

This link talks about scientific results on how to study. Especially note the section on how to study formulas.

12.7n#13

I am very confused about this problem. Not only does it not state what form the variables should take for theta and rho, but the problem randomly resets upon different answers.

***********************
I'm not sure what was going on for you but it looks like you did it correctly.

Current Estimated Grades 11/3/16

12,3#10

I've tried almost two times to figure out what exactly this problem is
asking for in the incorrect box. I've tried a ton of different expressions,
and the generic f(r,theta), f(r,t), etc. I am just very confused as to what
exactly it wants as an answer.

I can handle picky problems, but this is
one silly demand here.

Any idea what is wrong with it?


Thanks!

******
Sometimes webwork does picky, silly things, but this isn't one of those times. You  are given a function of x&y, f(x,y) You just decide to replace x by r and y by t, but that's not the way it works. Remember : x=r cos(t) and y = r sin(t), and you need to substitue these in for x&y.

Some things not to do (updated 10/27/16)

OK. I'm grading the exam and not pulling my hair out, but only because I don't have enough hair to do that. So, purely for your edification, I thought I would make a rogue's gallery of things not to do.

1) DO NOT REVERSE THE ORDER OF AN ITERATED INTEGRAL BY SIMPLY REWRITING THE INTEGRALS IN THE REVERSE ORDER.  (Limits of integration that are functions can never, ever,wind up in the outside integral.  

2) Do not simply change replace the variables in the limits of integration for the inside integral without changing the functional form.  This will never work either.

3) The antiderivative of a function is NOT the same as the derivative.  For instance 

∫ 1/(1+x^5)dx≠5x^4(1+x^5)^-2

4) 1/(a+b)≠1/a + 1/b.  NEVER, EVER.

5) Do not write a negative answer for an integral when the integrand is strictly non-negative on the domain of integration. <---NOTE: THIS IS A CHEAP WAY TO DETECT IF YOU DID YOUR CALCULATION CORRECTLY

6) When you are computing the value of an iterated integral, be sure you compute the antiderivative of the inside integral with respect to the inside variable,  for instance
∫_a^b∫_f1(x)^f2(x) x dy dx = ∫_a^b x y |_y=f1(x)^y=f2(x) dx = ∫_a^b x f2(x)-x f1(x) dx

but
∫_a^b∫_f1(x)^f2(x) x dy dx ≠ ∫_a^b 1/2 x^2 |_y=f1(x)^y=f2(x) dx = ∫_a^b 1/2(f2(x)^2-f1(x)^2) dx

Section 12.3 HW not due until next friday

Currently the homework 12.3 is due this Saturday and I was wondering if that was a mistake or not considering we have not gone over it yet and it's not on the test? Thank you for the clarification.

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

Thanks for bringing this to my attention. I've moved the due date for section 12.3 back to Friday the 28th.

MAT Test 2 review materials

Excuse me Professor Taylor, but do you have any practice materials up for the second midterm?

Thanks!

*************** 

Same place (as noted on the syllabus): <https://math.asu.edu/resources/math-courses/mat267>

Estimated Grades as of 10/17/16

If I had to give you a grade today, these are the grades I would give you.  Here's how I figured it:  Since your final grade depends 50% on your three midterms, 25% on the final exam, 15% on the homework and 10% on quizzes and other activities, I gave a 75% weight to the one midterm you've taken 15% weight to your homework percentage and 10% to your quiz percentage.  Then I assigned grades on the principle <90%=A, <80%=B, <70%=C, <60%=D.  I did not bother to put +/- grades at this time.

11.6 #12

Hi Dr. Taylor!

I am not sure on how to take the partial derivative of
this function with x, y, and z variables.

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

This question is designed to fool you.   :-)

Yes! If you try to treat this equation as the graph of a function of two variables, you'll wind up with a z on both sides of the equation, which will make you crazy.  Instead, just treat the surface as the 2-level surface {(x,y,z): f(x,y,z)=2} where f(x,y,z)=z(1-xe^y cos(z)).  Then you take partials with respect to z the same way you take any partial derivative. The equations for the tangent plane of a level surface of a function of three variables is covered n the textbook (remember that?) and in the lecture notes for 10/3/16.

11.6 #14

I have a question about the first part of this problem.

I have tried getting the gradient vector, composed of the partial derivative of Z with respect to x and Z with respect to y. I then made it a unit vector by dividing the vector evaluated at (1,1) by the magnitude. I then got this answer, which is incorrect.

Could I know what I am doing wrong?
Thank
you very much!

********************
The only thing you've done wrong is that you're giving the vector going up the mountain, while you need to go down.

11.6 #18

Dr. Taylor,

Hello,
I have a few problems I am stuck on. And they are always one of three things.
The first is when it asks me to find a normal line I struggle. I can find the tangent plane equation easily. I understand the normal line is the given point dot product t times grad f. But, the point has an x, y, and a z even though the equation is only in two variables(like x=... in this case). I have tried putting in everything for the partial derivative for x in the normal line equation and yet I invariably get it wrong. What key concept am I missing?
Second, how do I do the problems that say find the angle above horizontal? I understand I would have to use <0,1> for a 2d plane and <0,0,1> for a 3d plane to start the problem, but that is all I know.
Third, what is the value of maximal increase? Like on one I found the vector of direction of maximal increase (easy) and then it asked, what is the value of maximal increase and I took the magnitude of the vector I had and that got me the wrong answer. I think my issue with that is just more conceptual.
If you could clear up any of those, that would be greatly appreciated! Thank you in advance.

*********************************'
That's a lot of questions.
First, there are some different x,y,z's that can get confused here.  The tangent plane is the tangent plane at a point.  That means you have a specific gradient vector at the point, and a specific tangent plane that is perpendicular to that specific gradient vector. So before you can get those, the x,y,z that are in the definition of the graph z=f(x,y) of the function have disappeared and been replaced by some specific numbers (a,b,c) that satisfy c=f(a,b).  This means that the point (a,b,c) is on the graph of the function. It's also on the tangent plane of the graph z=f(x,y), which passes through that point. In addition a normal vector to that tangent plane is -del(f)(a,b)+k; this means that the equation of the tangent plane is -∂f/∂x(a,b)(x-a)-∂f/∂y(a,b)(y-b)+(z-c)=0.  There are a lot of normal lines to that plane; one of them passes through the point (a,b,c). A parametric equation to this normal line is <a,b,c>+t<-∂f/∂x(a,b),-∂f/∂y(a,b), 1>. Hope this clarifies.

Second, the tangent of an angle in a triangle is the opposite divided by the adjacent, which you could rephrase as the rise divided by the run. But the rise divided by the run has another name: it's called the slope.  When you have a motion along a curve in the plane this means that the tangent of the angle above the horizontal is dz/ds, where s is arclength, or equivalently (dz/dt)/(ds/dt). The numerator you can get from the chain rule, and the denominator is the speed.

Third when you find the direction of maximal increase--that's given by a unit vector--it just tells you direction to go in, not how fast you go up when you go in that direction.  How fast you go up is given by the directional derivative. The directional derivative in the direction of the greatest increase has a special form, which is discussed in the lecture notes on directional derivatives, *and* in the text book.

Attendance through 10/12/16

Lecture Notes 10/3/16-10/7/16

Lecture Notes 10/3/16

Lecture Notes 10/5/16

Lecture Notes 10/7/16

Lecture Notes 9/26/16-9/30/16

Lecture Notes 9/26/16

Lecture Notes 9/28/16

Lecture Notes 9/30/16

Attendance to 9/23

Some people  already have over the limit of unexcused absences, and even more are getting close...

Lecture Notes 9/19/16 through 9/23/16

Lecture Notes 9/19/16

Lecture Notes 9/21/16

Lecture Notes 9/23/16

10.9#3

Shouldn't the 'j' component of the velocity for this be 3cos(3t)+0 since  v(0)=i+0j+k?

*********************
No, because v(t)=∫_0^t a(s)ds+v(0), and ∫_0^t -9sin(3s)ds=3cos(3t)-3cos(3x0)

Exam&Quiz Scores 9/21/16 (Edited: some typo's fixed)

10.7#6 Hint

Dr. Taylor,

How do I go about finding this?  The example in the book isn't very helpful here because they have a graph of the two surfaces and can tell what x and y are just by looking at the graph.  What is a good
analytical way to arrive at the answer every time?

**********************************
Well the short answer has three steps 1) substituting one equation into the other to get a single equation in two variables,  2) remember what you learned in Calc2 about parameterizing plane curves, then 3) re-substitute your parameterization of a planar curve back into 3-D to get your intersection

Try that and let me know how it goes and what you try, I'll post your answer here.

10.8#3 (edited)

I can not figure out how to solve this problem. I find the derivative of
the vector then set my limits from t=1 to t=2 and then use the formula:

Integral from 1 to 2 sqrt((4^2)+(4x)+(2/x)) and this does not work. What am
I doing wrong?

*************************
You have the limits of integration correct, but it looks like you forgot to square the derivatives of the y and z-components of r.

*********************************
Hello Professor. I've got the steps down to do this problem. I found the upper and lower T values needed, and I have set up the equation. The problem is, I have no clue how to integrate the mess after it is all set up. A clue or step in the right direction would be greatly appreciated!

************************
Well, a good thing to ask yourself might be something like: "Huh. There are very few functions that I've been taught how to integrate. The arc-length formula has the square root of some stuff for an integrand.  What stuff could  be inside that square root that I might reasonably be expected to come up with an anti-derivative for, after taking the square root?"

In fact there are very few things that could be inside the square root that would allow you to do that integral, and if you've done the sum of squares of the components of the derivative correctly when you look at that you see that it turns out to be one of those very few things.

10.7#1

I am having a bit of an issue with problem 1 on the 10.7 WebWork homework. I believe that WebWork is not grading my work correctly (it is marking a correct answer as incorrect).  The problem is to find the domain of the vector function r(t)=<e^(-4t),t/sqrt(t^2-49),t^(1/3)>.

I calculated the domain of the first component as (-INF,INF).

I calculated the domain of the second component as (-INF,7) U (7,INF).

I calculated the domain of the third component as [0,INF).

Therefore, all of the components should be defined only on the interval (7,INF). I have used Wolfram|Alpha to back me up on this:

However, WebWork marks this answer as incorrect:

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

No, it's grading you correctly (this time, although it does make mistakes sometimes).   It's interesting that Wolfram alpha is incorrect, isn't it?So, you made a combination of two mistakes: the domain of the second component is
(-INF,-7) U (7,INF) and the domain of the third component is (-INF,INF)--the cube of a negative number is negative, so likewise for the cube root.  Hence the domain is (-INF,-7) U (7,INF) .
 

10.7#16

 I can't seem to solve question 16 on 10.7 and need help. I got part A, but my method doesn't work for parts B or C, so I think I just got lucky on part A. Thanks.

*******************
One parameterization of the line will be
r(t)=u+tv,
where u=<1,-1,2> and v=<4,1,-4>, but v can be replaced by α<4,1,-4> for any nonzero constant α and u can be replaced by any other vector with time on the line,  i.e. u can be replace by
<1,-1,2> +β<4,1,-4> for any real number β.  Since u+0v = <1,-1,2> and u+1v =<5,0,-2>, all we have to do to get the answer to (a) is replace v by v*=v/5, so now we have the equations
u+0v* = <1,-1,2> and u+5v* =<5,0,-2>, and your answer <1,-1,2>+t<4/5,1/5,-4/5>  is the correct one.  In more generality, you would need to look at something like
r(t)=(<1,-1,2> +β<4,1,-4>)+tα<4,1,-4> 
and choose α,β to fit your parameters.  For example in case (b) r(5)= (<1,-1,2> +β<4,1,-4>)+5α<4,1,-4>=<1,-1,2>  and
r(7)= (<1,-1,2> +β<4,1,-4>)+7α<4,1,-4>=<5,0,2>  so we get equations
5α+β=0 and 7α+β=1 from which 2α=1=>α=1/2, and β=-5/2,
r(t)=(<1,-1,2> -5/2<4,1,-4>)+t/2<4,1,-4>
    = <-9,-7/2,12> +t<2,1/2,-2>

Exact Answers? (edited)

Hello, Professor, this is ********** in your MAT 267 MWF class. In question 1 of section 10.8 on the "Test 1 Review," it asks us to compute the length of a curve. To solve, I set up an integral and then plugged the integral into my calculator's integral solver to get a numerical answer (28.73). However, the answer key lists the answer as (1/27)(85^(3/2)-8), which computes to approximately 28.73. My question is, on the exam if I had skipped the step of doing a u-substitution to solve the integral by hand and instead set up the initial integral for arc length, plugged it in to a calculator, and given a numerical answer, would I be deducted points?

Sincerely,

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

It depends on the question.  If the question asks for an exact answer, you would not get full credit.  If the question just asks for an answer you can use your calculator.

******************
Pardon me, reading your new post I found that under "arc length" you specify we need an exact answer (no calculator answer).

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

No, I said it depends on the way the question is worded. 

Test 1 Important Concepts

Test 1 Concepts

Exam Q

Professor,

I was wondering if section 10.8 will be included on our exam this Friday?

***************
10.8 is fair game 

Lecture Notes 9/12/16&9/14/16 (updated)

Lecture Notes 9/12/16

Lecture Notes 9/14/16

Q on 10.8#2

Hello, Dr. Taylor.

I am not sure what integration technique I can use to integrate the magnitude of r'(t). 
Should I be approaching this problem
differently?

Thanks.

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

Well, r'(t)=<10√2, 10e^(10t), -10e^(-10t)>, so

thus

at which point you can look up the answer in your table of integrals.
Remark: I bet someone thought that you would never need to know hyperbolic trig functions

Test 1 Review Question

Hello Professor,

I was wondering if there was an answer key to the "Test 1 Review" so I can make sure I am doing my work correctly. 

Thank you

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

There is an answer key to Practice Test 1, how did that one go?  For the review, what specific questions do you have, and what did you do?

Test 1 Question

Dr. Taylor,

While looking at the practice exam I noticed that there was a link to "Test 1" in addition to the review and practice. Is this how we will be taking the actual test on Friday, or are we doing it in class? Thanks.

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

The link to test 1 takes you to another practice exam. We'll be taking the exam in class.

10.5#17 again

Hello, Dr. Taylor.

I am lost on this question - if I need to find the angle of intersection between the planes, I would assume I can just find the angle between their two normal vectors using the dot product formula 
A dot B = ||A||||B||cos(theta), where A and B are the two normal vectors to the planes. I plotted both planes in Wolfram and visually it seems like that should give the right answer. I also found an answer on MathExchange that uses this method to solve a similar problem (and defines the angle of intersection between two planes as being the angle between the normal vectors). However, no matter the angle I find (obtuse or acute) with this method, the angle is reported as being incorrect

How should I go about doing this if this is not the correct method?

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

You aren't doing anything wrong, except you let webwork fool you into thinking that the normal vector of the first plane is <1,-4,1>. It's not.  Notice that the coefficient of x is -4, while that of y is 1?  The coefficient of x goes in the x-position and of y goes in the y-position.  This means that the normal vector is <-4,1,1>.

10.5#8 hints

Hello! I have a question about this problem. I feel that I am missing
something here. I don't know where to begin, and I am unsure of the
process. Any help would be greatly appreciated!

***********
OK, first of all, go back to the lecture notes, we covered this.  The line of intersection consists of points that are in both planes. The parametric form of a line is given by data that consist of 1) a vector (which we called u in the lecture notes) with endpoint at some single point in the line (which must consist a point in both planes) and 2) a vector parallel to the line (which we called v in the lectures notes and which must therefore of a vector in both planes, hence perpendicular to both normal vectors--hint, hint, hint).  The problem helps you choose both of these by constraining u so that its z-component is zero and constraining  v so that its x-component is 12.

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(You can click on the image below to enlarge)

Lecture Notes 9/7/16-9/916

Lecture Notes 9/7/16

Lecture Notes 9/9/16

10.5#10

On problem 10 of 10.5, two parametric equations are given. To solve for the intersection we solve for two variables, usually t and s in the equations. Unfortunately these equations have only t’s. The problem can be solved by assuming the equations for L2 use a different variable, independent from t.

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No typo, really, but you make a useful point.  It's easy to make the mistake of thinking that the t in the parametric equation of a line refers to anything real.  It doesn't, it's just a dummy variable.  It's just a way of labeling points on the line. For example L1: x=2+12t,   y=6+8t,  z=4+2t is another labeling that would do just as well, and the dummy variable associated t associated with this labeling would be different.

It's true that if you expect t to be a real thing and look for an intersection by setting 14+6t=-18+7t, 14+4t=-10+6t, 6+t=-5+4t, there will almost always be no such t.  And yes, the correct way to look for an intersection would be to allow give two different names to the parameters for the two different lines. The names t and s would work as well as any other two names.

(to misquote Wild Bill Shakespeare, a variable by any other name would be just as good, or smell as sweet.  Wild Bill had much to say on this subject, btw:

JULIET

'Tis but thy name that is my enemy;
Thou art thyself, though not a Montague.
What's Montague? it is nor hand, nor foot,
Nor arm, nor face, nor any other part
Belonging to a man. O, be some other name!
What's in a name? that which we call a rose
By any other name would smell as sweet;
So Romeo would, were he not Romeo call'd,
Retain that dear perfection which he owes
Without that title. Romeo, doff thy name,
And for that name which is no part of thee
Take all myself.

)

10.5#17

Hi Dr. Taylor,

I've attempted this problem 8 times and my answers are
still wrong. I used the equation for the angle between planes. Am I doing
something wrong?

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Well, 142 radians is a lot of radians and 2.49 degrees is very few degrees. That should make you suspicious about which one is which.   In fact, 2.49 radians divided by π x 180 degrees = 142 degrees, whereas 142 radians divided by π x 180 degrees amounts to 8136 degrees. So: at the very least you have degrees and radians reversed.

Now let's see, you didn't show me how you computed your angles but you should have had:
cos(θ)=u.v/(|u| |v|)
          =<4,-4,-1>.<4,-4,-3>/(√(4^2+(-4)^2+(-1)^2)√(4^2+(-4)^2+(-3)^2))
          =(16+16+3)/(√(16+16+1)√(16+16+9))
          =39/(√37√45)
          =0.956
which is a number very close to 1, hence we expect that θ is close to zero.  In fact computing the arccos what we arrive at is θ=0.30 radians, which is about 17.1 degrees.

10.5#11

Following the book, I believe that I have correctly solved for the plane's
equation. The normal vector is expressed as <-2,4,1> and following the book
my specific points should lead to -2(x--4)+4(y--3)+1(z-2)=0. Am I correct
in this assumption or am I missing something?

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

Nope, not correct. A parametric equation for a line looks like r(t)=u+tv, where the vector u just gives a point on the line and the vector v determines the direction of the line: the direction of the line is read from the coefficients of t.   Instead you've used the u.

Practice Exams

MAT267 Page

question regarding section 10.5

Professor,

I’m wondering about an equation, equation 4, given on page 569 of our textbook. Do these statements show the right relationship?

    

__________________________________

The first equation is fine, it describes the line segment starting at the vector (point) r_0  and passing through to the endpoint vector (point) r_1, because r(0)=r_0 and r(1)=r_1.  Note that this equation can be rewritten as 

r(t) = (1-t)r_0 + t r_1 

      = r_0 - t r_0 + t r_1 

      = r_0 + t (r_1 - r_0)   

      = r_0 + t v   

where v= (r_1 - r_0), which is just the form of a line as we described today in lecture, and allows all real values of t, i.e. does not require 0≤t≤1.

I'm not sure what the second equation means, since I'm not sure what you mean by r_a and r_b, but it would give some line segment not necessarily beginning or ending at  r_a and r_b/

(Edit the morning after: after thinking a bit more about your question I think I understand where you were coming from: you wanted to know if 0 and 1 are special or if any numbers would do.  The answer is no, any numbers will not do and 0 and 1 are special.  The whole point is that

(1-t)u+t v 

interpolates between u and v as t goes from 0 to 1, no matter what vectors u and v are.)

10.4#5

I need help on this problem on how to find the orthogonal vectors
of 'a' and 'b'. I've tried it many times! I computed the cross product of
'a' and 'b' and then used the dot product using each vectors. Please
explain to me what I'm doing wrong.

(you can click on the image to enlarge it)

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Well, you're doing a number of things wrong, starting with your interpretation of the problem.
First of all, your supposed to find two unit vectors u^ and v^, each of which is orthogonal to both of the vectors a and b.   One thing you choose to do is to compute the cross product a x b, because this will give you a vector orthogonal to both a and b.  It looks like you made a simple arithmetic error in this computation however, because for example
 a x b = (-3i-5j+4k)x (-2i+4j+4k)
         = -12( i x j)+10( j x i)-12( i x k)-8( k x i)-20( j x k)+16( k x j)
         = -22( i x j) - 4( i x k) - 36( j x k)
         = -22 k + 4 j - 36 i
         = - 36 i + 4 j - 22 k

While this vector is orthogonal to both a and b,  it is not a unit vector.  Since the norm of this vector is ||a x b||=√(36^2+4^2+22^2)=√1796=2√449, the unit vector pointing in the direction of  a x b is
 a x b/(2√449). Of course another unit vector perpendicular to both  a and b is b x a/(2√449), which points in the opposite direction 

10.4#8

Hello, Dr. Taylor.

I solved this problem by projecting vector QP onto vector QR, finding the angle measure between vectors QP and QR with the arccos function, and then calculating sqrt(14)sin(theta) to find the
distance d.  However, I am also aware that using cross products rather than projections is the focus of this lesson - how can this problem be solved using cross products? Using the formula ||U x V|| = ||U|| ||V|| sin(theta), I would end up with 14sin(theta) instead of sqrt(14), since both QP and QR
have magnitudes of sqrt(14).

Thanks for your assistance.











************
OK, some short answers are 1) you started on one possible right track using projections, but then you got off the trail and started bushwhacking through the weeds, and 2) that nice formula
||U x V|| = ||U|| ||V|| sin(theta)
you use comes in three pieces, it's up to you to figure out how to take the pieces you want.

But when doing these problems I think it's important to take an organized approach, and the beginning should be TO DRAW A PICTURE OF THE PROBLEM from a perspective that captures the essential features.  This three-D plot captures EVERYTHING:

but maybe it's too much information?  This simpler diagram says what we need to know:

See that right triangle? You know the size of the hypotenuse, and the piece || to QR is the projection so you know it's size.  The other piece is the answer you need.  You can get that in several ways. Hint: cos^2(x)+sin^2(x)=1. 

Lecture Notes 8/29/16-9/2/16

Lecture Notes 8/29/16

Lecture Notes 8/31/16

Lecture Notes 9/2/16

10.3#14

I am not understanding the first 2 parts of the question, can you please
rephrase it or somehow explain what it is asking for? Thank you.

********************
First of all, reading the chapter we are covering is very important.  You should read section all of section 10.3 in before you attempt the problems.  Especially you should read pages 554 and 555.

The implication of the problem is that F can be written as the sum of a part that is parallel to v and a part that is perpendicular to v.  But this comes from the notion of the projection Proj_v(F) of F on v:
 this IS parallel to v.  In addition F-Proj_v(F) is perpendicular to v, in fact the Proj_v(F)  is defined as the unique vector w parallel to v that makes  F-w perpendicular to v.  These give you the two vectors you need.

10.3#12

Professor Taylor,

I've completed all of assignment 10.3, except for #12, which I don't know how to solve. Do you have any advise that might help me understand the problem?

The question is:

Assume that u⋅v=7, ∥u∥=4, and ∥v∥=3.
What is the value of 7u⋅(2u−2v)?

Thanks for the help.
**********************
OK,  one critical piece of information that you might have missed is that we usually write |u| instead of ∥u∥ for vectors u, but both of these mean the same thing: the magnitude of the vector. Then you just need to know basic facts about the dot produce like ∥u∥^2=u.u  and the distributive law.

10.2#11

Hi Dr. Taylor,

I have been having difficulty with problem number 11 on section 10.2 of the homework. I went to the  math tutoring center and none of the tutors were able to help me. If I could see a solution worked through it would be much appreciated! The problem is attached below. 

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Well, this isn't due for another week.  I suggest you take a look at example 7 on page 548 of the text book--notice that it uses some angles.  Now look at the statement of the problem 11--how does it tell you what the angles have to be?

10.1#5

When I type the equation of a sphere, it is saying that the "=" is an
"unexpected character" and is counting it wrong.  Is there some way Webwork
is expecting me to write the equation of a sphere without using "="?

****************
Notice how webwork has those "=0"s over at the right hand side?  So, for example, what you've written amounts to (x+7)^2+(y-5)^2+(z-8)^2=9=0, which makes no sense.  You could have written
(x+7)^2+(y-5)^2+(z-8)^2=9, if webwork would have let you, but it didn't.  Instead you need to write

(x+7)^2+(y-5)^2+(z-8)^2-9, so that (x+7)^2+(y-5)^2+(z-8)^2-9=0 makes sense.

Webworks 10.2

Hi professor Taylor,

     On the 10.2 Webworks question 3 and several other problems involving vectors will not take my answers because it says "your answer isn't a vector (it looks like a point)". So I was wondering how you input a vector into Webworks.

Thank you,

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

Basically the same way webwork gives you the vector:  as a tuple of numbers enclosed by sharp brackets, e.g.:  v=<1,2,3>.  You probably did the reasonable thing of giving it a tuple enclosed by round brackets (1,2,3), which would be fine for me, but webwork wants the other because of reasons. 

BTW,  in webwork at the bottom of each homework there is a button that says "Email instructor".  If you contact me using that button it will show me the page as you see it with your answers.

webwork 10.1#2

Hello Professor. I am wondering about this question. I know it's all fresh,
but I am a bit confused non the less. Isn't this a problem utilizing the
distance formula in three dimensions? If so, I am not getting nice answers.
Any suggestions?

*****************
Yes, you need to use the distance formula in three dimensions.  The point being that the shortest distance between two points is on a straight line, so that if the three points are on a line the sum of some two distances will equal the third distance, while if they are not on a straight line the sum of any two is larger than the third.

MAT 267 Question About WeBWork

Professor Taylor,

I didn’t find the Homework in WeBWork, the only thing I found in the website is “Taylor_MAT_267_Spring_2016”. Is the homework site open to us in this week?

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The webwork link on the syllabus or the homework assignment links on the blog take you to the correct webpages.

Webwork question

Professor Taylor,

Since I have never used the WebWork program before, I just have a quick question about submitting homework. If I completed the 10.1 assignment and each question has a status of "100%", do I need to submit the assignment as a whole? I don't see any option to do so, but I don't want to risk missing the first homework grade. Thanks for you help.

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

Each question is submitted separately using the "submit" button; e.g. see below.