Monday, December 14, 2015
Monday, December 7, 2015
this is your warning(your handwriting on the exam)
I'm just grading your quizzes, and am thereby motivated to point this out to you:
If your handwriting is so small or so illegible that I can't read what you've written, you will get no credit for your work.
If your handwriting is so small or so illegible that I can't read what you've written, you will get no credit for your work.
Sunday, December 6, 2015
The Final Exam, 2nd Notice
The final exam will take place on Tuesday, December 8, from 7:10PM to 9:00PM in BAC216
Saturday, December 5, 2015
Topics to be covered and other random angst...
A random sample of emails:
1) Hi professor, There are two questions i want to know about final exam. Firstly, i'm not sure the contents to be covered. I finished the final review, but i found it didn't cover all sections which we learned. So, I wanna know did it include all things we need to review? or we need to review more sections? In addition, could you tell me the form of this final exam? does it include both multiple choice and free response, or only free response, or something else?
2) Dear Dr. Taylor,
1) Hi professor, There are two questions i want to know about final exam. Firstly, i'm not sure the contents to be covered. I finished the final review, but i found it didn't cover all sections which we learned. So, I wanna know did it include all things we need to review? or we need to review more sections? In addition, could you tell me the form of this final exam? does it include both multiple choice and free response, or only free response, or something else?
2) Dear Dr. Taylor,
When looking through the final review that was posted here, https://math.asu.edu/first-
Thank you for your time.
3) Professor,
You said about two weeks ago that you'd
post an update on everyone's gardes insofar onto the blog. I figured
you'd eventually get around to it, but that hasn't happened. Did you
think you shouldn't since most people in the class don't bother to read
your blog?
********************
OK, first of all, the final exam is cumulative. For you, that means anything and everything we have covered is fair game. Since I spent my breath and time lecturing on it, it's also likely to be covered on the exam. Since subjects learned early in the course are used later in the course, I will sometimes kill two birds with one stone and test the early material and the later material in the same question--but this does not constitute a promise that I will do so.
My first priority is grading at the moment is getting your quizzes graded and recorded. If I have the time before the final exam I might get around to re-estimating your grades. BUT REALLY-you have everything that I have graded and at this point in the semester you should have a very good idea about where you stand.
Friday, December 4, 2015
webwork problems
A small sample of the emails I've just been getting:
1) Hello Professor,
1) Hello Professor,
I'm in your MWF 10:30-11:50 class and I was trying to get onto Webwork to finish up the homework due tonight
after *************, but I couldn't log in to myASU. I know it's
really close to finals, but is there any possibility you could extend
the deadline a day? If I'm the only one who's been having the problem, I
understand if you don't. But I could really use all the points I could
get, so if you do I would appreciate that.
Your student,
*************
2) Hello Professor Taylor,
I noticed while i was trying to
complete the last couple of problems on the 13.7 webwork that blackboard
is down and is unable to sign anyone into their my asu to log on to
webwork. if it would the possible to extend the deadline for the
webwork.
Thank You,
*************
3) Web work pausing and not allowing access. I have like three problems left. Please let us in. Or something.
**************
Webwork deadline is delayed until Sunday night due to network problems.
13.7 #7
Dear Professor Taylor,
This problem has me really stumped I don't know what it is asking for the second to last portion because I have entered r dr dtheta as the usual substitution of dA. But it wants it in vector form. Can you please clarify what WebWork wants exactly.
Thank you,
ok, you have a couple of different confusions going on here:
Thank you,
ok, you have a couple of different confusions going on here:
1) this is a good example of the diversity of notation surrounding this integral: what this problem wants to call a vector dA is what we have been calling n dS. (In support of my rant earlier today about "Flux Integrals", you might want to note that wikipedia distinguishes two kinds of "surface integrals" as I do, surface integrals of functions and surface integrals of vector fields, that the textbook calls flux integrals)
2) just to confuse the issue a little bit more, this is a parametric integral, not an integral in polar coordinates. What this means is that the element of area for polar coordinates r dr dθ is NOT the appropriate element of area for this problem. What you need is just (r_r x r_θ)dr dθ, (note the additional confusion of r being the parametric equation while r is just a parameter) and the correction term r is not used (it sort of gets absorbed into the r_r x r_θ). Since
r_r x r_θ=<cos(θ),sin(θ),1>x<-r sin(θ), r cos(θ),0>=<-r cos(θ), -r sin(θ), r>, which is pointing upward and not downward, so your correct answer to part (b) would be
- r_r x r_θ drdθ.
This is *almost* what you have, you just forgot to multiply the drdθ times the first two components of your vector.
- r_r x r_θ drdθ.
This is *almost* what you have, you just forgot to multiply the drdθ times the first two components of your vector.
Wednesday, December 2, 2015
Monday, November 30, 2015
13.7 problem 8
Hi Dr. Taylor,
I was wondering if you might help me get started on this
problem. I'm having some difficulty figuring out the parameterization as
well as the bounds.
Thank you,
****************************
First of all the cylinder x^2+y^2=16 has radius 4 and is parallel to the z axis. Then the planes x+y+z=3 and x+y+z=7 are parallel to each other, and you can solve these for z as z=3-x-y and
z=7-x-y. In particular you can get from first plane to the other by increasing z by 4. Then points on the cylinder can be specified by writing x=4 cos(u), y=4 sin(u), while the intersection of the cylinder with the first plane also specifies z=3-4 cos(u)-4 sin(u) while the intersection with second plane specifies instead that z=7-4 cos(u)-4 sin(u). You can put this all together to get a parameterization of the space on the cylinder between the two planes by saying
x=4 cos(u), y=4 sin(u), z=3+v-4 cos(u)-4 sin(u)
where 0≤u≤2π and 0≤v≤4.
Saturday, November 28, 2015
announcing a quiz for monday 11/30
Yah, that last one wasn't so good. It has been pointed out to me that all of your webwork on Green's theorem dealt with using a double integral to find the value of a line integral, whereas the quiz did the reverse: it asked you to find an area by computing a line integral.
OK.
So I'm going to give another quiz on exactly the same topic as the last: finding the area of a region by computing a line integral around the boundary. I'll give you a parameterization of the boundary, just like before. ALL YOU NEED TO KNOW HOW TO DO IS TO COMPUTE THE AREA BY USING A LINE INTEGRAL.
OK.
So I'm going to give another quiz on exactly the same topic as the last: finding the area of a region by computing a line integral around the boundary. I'll give you a parameterization of the boundary, just like before. ALL YOU NEED TO KNOW HOW TO DO IS TO COMPUTE THE AREA BY USING A LINE INTEGRAL.
ok....integration of a vector F field along a curve...(exam fall-out)
means that you have to integrate the F along the curve. That means that you need to find a parameterization of the curve. Usually there will be one of three situations: 1) we will give you the parameterization--use it, or 2) there is a natural parameterization of the curve, i.e. the curve is a circle or the curve is the graph of a function g(x)--use <cos(t),sin(t)> for the circle or use <t,g(t)> for the function, or 3) the curve doesn't matter anyway because the vector field is conservative--in which case you need to find the function f(x,y) so that F=∇f and evaluate f at the endpoints of the curve.
You don't get to chose some random path that doesn't fit the curve, and don't need a path at all in the case that F is conservative.
You don't get to chose some random path that doesn't fit the curve, and don't need a path at all in the case that F is conservative.
Saturday, November 21, 2015
grading the exam: problems with setting up integrals
Aside from the usual difficulty in recognizing upper and lower limits of integration in theta, one common confusion that people have is confusing the values of coordinates at the boundary of the region of integration--which need to become input into the selection of limits of integration--with the form of the integrand. For instance, if a cylinder has radius R, and the integrand has x in it, in the integrand x is written as R cos(theta) instead of r cos(theta) as it should be.
Friday, November 20, 2015
Thursday, November 19, 2015
Review Questions
Hi Dr. Taylor,
I was wondering if you might help me out with one of the questions on the review. I'm having a lot of difficulty getting the right bounds for problem 2 in 12.6. If you could help me out I would be extremely grateful!
*******
******************************************************************
Sure, here are my notes from reviewing this problem and a couple others on 11-18
Review 12.6#2
Review 12.7#'s 3 and 4
I was wondering if you might help me out with one of the questions on the review. I'm having a lot of difficulty getting the right bounds for problem 2 in 12.6. If you could help me out I would be extremely grateful!
*******
******************************************************************
Sure, here are my notes from reviewing this problem and a couple others on 11-18
Review 12.6#2
Review 12.7#'s 3 and 4
Wednesday, November 18, 2015
13.4#4 (updated)
Dear Professor Taylor,
I am working on this problem and I can't quite figure it out. For P I get x^3 because that's multiplied by dx. And for Q I get 5x. So dq/dx-dp/dy = 5.
Plugging that in I get double integral over D of 5 dA. Next I realize that there to do dydx as my dA I have to split the parallelogram into two pieces. So I'm integrating two separate triangles. My guess is I'm doing that wrong but I can't figure out why. I decided to do the line integral by finding four vectors going all the way around the shape by that just got me a completely different answer.
What am I doing wrong?
Thank you,
Hello sir. I keep trying to do this problem as you described on your blog,
but I keep reaching the same answer I got last time. Does this mean I'm
still doing it wrong or that the webwork is wrong??
**************************************************
It's hard to see what you did wrong without having the details of your calculation. First of all note that the orientation of the contour is clockwise, while Green's theorem is framed for counterclockwise line integrals. So: just go ahead and use Green's theorem as normal and then multiply your answer by -1. Incidentally, I would do the double integral as an iterated integral with dx first, since that way I can do only one piece. The leftmost sloping line segment runs from (0,0) to (x_0,y_0) so that line can be written as y=y_0/x_0 x or (as you will need it) x=x_0/y_0 y. The rightmost sloping line segment is displaced from the first by an amount x_0 so has equation x=x_0/y_0 y + x_0, so the iterated integral is
which is pretty easy to do. (Updated)Since here x_0=2 and y_0=2, the inner integral is 5((2/2)y+2-(2/2)y)=10. Then you get the integral of 10 with respect to dy from 0 to 2, which is positive 20. The negative of that is -20, which is the answer you got so I believe webwork is failing to properly check your answer.
I am working on this problem and I can't quite figure it out. For P I get x^3 because that's multiplied by dx. And for Q I get 5x. So dq/dx-dp/dy = 5.
Plugging that in I get double integral over D of 5 dA. Next I realize that there to do dydx as my dA I have to split the parallelogram into two pieces. So I'm integrating two separate triangles. My guess is I'm doing that wrong but I can't figure out why. I decided to do the line integral by finding four vectors going all the way around the shape by that just got me a completely different answer.
What am I doing wrong?
Thank you,
Hello sir. I keep trying to do this problem as you described on your blog,
but I keep reaching the same answer I got last time. Does this mean I'm
still doing it wrong or that the webwork is wrong??
**************************************************
It's hard to see what you did wrong without having the details of your calculation. First of all note that the orientation of the contour is clockwise, while Green's theorem is framed for counterclockwise line integrals. So: just go ahead and use Green's theorem as normal and then multiply your answer by -1. Incidentally, I would do the double integral as an iterated integral with dx first, since that way I can do only one piece. The leftmost sloping line segment runs from (0,0) to (x_0,y_0) so that line can be written as y=y_0/x_0 x or (as you will need it) x=x_0/y_0 y. The rightmost sloping line segment is displaced from the first by an amount x_0 so has equation x=x_0/y_0 y + x_0, so the iterated integral is
which is pretty easy to do. (Updated)Since here x_0=2 and y_0=2, the inner integral is 5((2/2)y+2-(2/2)y)=10. Then you get the integral of 10 with respect to dy from 0 to 2, which is positive 20. The negative of that is -20, which is the answer you got so I believe webwork is failing to properly check your answer.
Tuesday, November 17, 2015
13.2#14
Well, it's hard for me to tell how you got that answer, because it's pretty far off. You need to understand the definition of line integral: since you are given a particular vector field, you need to do the line integral of that vector field, along the curve you have, which looks like the following:
Next you have to use the parameterization you already wrote down x(t)=t, y(t)=t^2, which will give you x'(t)=1 and y'(t)=2t, hence you have:
which gives you
Sunday, November 15, 2015
comment on Reviews for Test 3
On Sun, Nov 15, 2015 at 6:54 PM, Unknown <noreply-comment@blogger.com> wrote:
Unknown has left a new comment on your post "Reviews for Test 3":
Is there any changes to the chapters on the exam?( Syllabus says 12.5-13.4)
**************************
No.
Friday, November 13, 2015
ok, the quiz today.
Well. That was instructive. The median score was 4/10. This should be a big wake-up call for the exam next week.
1) You need to know what to do with a parameterized path. Really. You need to know what one looks like, where it goes into a line integral and how it goes into the line integral (two different ways--into the functions as x&y, and into the dr as dr/dt dt). You need to know the difference between the vector field and the path.
2) you need to know the difference between the line integral of a scalar function along a curve (which are not very important) and the line integral of a vector field along a curve (which are very important). In particular if you're doing section 13.2 stuff (like |dr/dt|) on a section 13.3 problem (where you need a dot product with dr/dt), you won't be happy.
3) You need to be able to use the gift given to you when a vector field is conservative. Otherwise something that's very nice, becomes something that's very mean.
4) btw, when you have an line integral
5) Polar coordinates won't help you.
1) You need to know what to do with a parameterized path. Really. You need to know what one looks like, where it goes into a line integral and how it goes into the line integral (two different ways--into the functions as x&y, and into the dr as dr/dt dt). You need to know the difference between the vector field and the path.
2) you need to know the difference between the line integral of a scalar function along a curve (which are not very important) and the line integral of a vector field along a curve (which are very important). In particular if you're doing section 13.2 stuff (like |dr/dt|) on a section 13.3 problem (where you need a dot product with dr/dt), you won't be happy.
3) You need to be able to use the gift given to you when a vector field is conservative. Otherwise something that's very nice, becomes something that's very mean.
4) btw, when you have an line integral
this is NOT the same thing as an integral with respect to a radial coordinate r. Your clue in this is the dot product.
6) oh, and *please*, nobody ever tell me again that the following formulas are valid. (prepare to tell me why on Monday)
Announcing a For-Credit Quiz Next Wednesday.
There will be a 10 point, 15 minute long quiz covering section 13.4 on Wednesday November 18.
(if you are feeling the need for privacy, make sure you know your posting ID)
By the way, your actual student ID is identity theft material, don't write it on your quiz.
(if you are feeling the need for privacy, make sure you know your posting ID)
By the way, your actual student ID is identity theft material, don't write it on your quiz.
Problems with webwork 13.1-13.3
*********************************
Hello Professor Taylor, I noticed recently that assignments 13.1 through 13.3 are due tonight November 13th. Yesterday, I checked webwork and for my account, these assignments were assigned. One of my fellow classmates hasn't even been assigned the assignments for tonight. Is there a problem with webwork? Thank you very much!
Yes, there was a random problem with the webwork in which some people did not have the homework assigned and some other people did--I'm not sure how it happened since at the beginning of the semester everyone signed up for the course was supposed to be assigned all of the problems by the webwork admin. I started hearing about the problem yesterday, and now I've gone through the whole assignment file and I *think* I fixed the problem. If anyone is still not seeing their problems assigned on webwork please contact me and I'll take a look.
Sections 13.1-13.3
Hey Professor Taylor, a lot of people just found out that homework 3.1-3.3 are due tonight when before it stated that they weren't going to be due. Is there anyway you can extend their due dates for at least a few days because that is not really fair to make 3 sections of homework due all at the same time without any prior knowledge that they were even due at all. Please and Thank you.
***************************
I'm not sure what a lot of people could have been thinking, given that there is a test covering that material coming up next Friday, but OK. Sections 13.1-13.3 are now due on Tuesday. NOTE that SECTION 13.4, WHICH WILL ALSO BE COVERED ON THE EXAM NEXT FRIDAY is due as well as being tested next Friday, so you might as well get started.
Wednesday, November 11, 2015
[WWfeedback] course:Taylor_MAT_267_Fall_2015 set:Section_13.3 prob:7
***** The feedback message: *****
Dear Professor Taylor,
I am having trouble apply open and connected to a
set. I'm looking back into the book and the lecture notes and I see how it
is applied to the path of a line integral but I don't see how it applies to
a set.
Can you explain these concepts again?
Thank you,
*****************************
"Connected" means that between any two points in the set there is a path lying completely in the set that connects those two points. Here are three connected regions and one disconnected one, crossed out.
Open means that every point inside the set is some nonzero distance away from points outside the set--think of the open unit interval {x:0<x<1}: even points really close to 0, or 1 are some nonzero distance from points outside the set, e.g. 0.9999 is in the set but is a distance at least 0.00001 from points outside the set.
The way these assumptions are used in section 13.3is to have a region where a function or vector field is reasonably nice and to have contours strictly inside that region--because once you let a contour get on a boundary things infinitely close might mess things up.
Dear Professor Taylor,
I am having trouble apply open and connected to a
set. I'm looking back into the book and the lecture notes and I see how it
is applied to the path of a line integral but I don't see how it applies to
a set.
Can you explain these concepts again?
Thank you,
*****************************
"Connected" means that between any two points in the set there is a path lying completely in the set that connects those two points. Here are three connected regions and one disconnected one, crossed out.
Open means that every point inside the set is some nonzero distance away from points outside the set--think of the open unit interval {x:0<x<1}: even points really close to 0, or 1 are some nonzero distance from points outside the set, e.g. 0.9999 is in the set but is a distance at least 0.00001 from points outside the set.
The way these assumptions are used in section 13.3is to have a region where a function or vector field is reasonably nice and to have contours strictly inside that region--because once you let a contour get on a boundary things infinitely close might mess things up.
[WWfeedback] course:Taylor_MAT_267_Fall_2015 set:Section_11.1 prob:14
On Wed, Nov 11, 2015 at 7:27 PM, ************** wrote:
The two that I am missing are 11.1-14 and 11.3-12. I used all my attempts when they are due, and now I am no able to submit any new answers.
-------- Original message --------
From: Thomas Taylor <tom.taylor@asu.edu>
Date: 11/11/2015 5:51 PM (GMT-07:00)
To: *******************
Subject:[WWfeedback] course:Taylor_MAT_267_Fall_2015 set:Section_11.1 prob:14 Well, which ones? If you looked at the blog you noticed that I fixed one of those, are there more?
***** The feedback message: *****
Hello Professor,
In regards to this "Homework Marathon", there are some
problems like this one that I could not complete because I ran out of
attempts. I just wanted to know if these kind of problems will be opened up
and be able to complete or will they not be able to complete and we just do
not receive any partial credits for them.
Thank you very much,
**************
*******************************************************
Fixed
Calc 3 Wednesday Homework Marathon
On Wed, Nov 11, 2015 at 11:02 AM, ********************* wrote:
Dear Professor Taylor,I am looking at the homework marathon and there is only one question I failed to answer correctly previously, 10.7 Q-15. I failed to answer it correctly in the three attempts so I failed the question. I thought that during the homework marathon I would be able to re-answer that question but it appears that since I used up all of my opportunities that I can't try the credit for it.I was wondering if you were intending to reset the problem opportunities as well or not? If that was not what you were meaning to do then oh well I missed out on one point. But if you were intending to reset those problems I still wanted the opportunity to get that half point back.Thank you,
********************************
Fixed
Sunday, November 8, 2015
COMING UP THIS WEEK (11/9/15-11/13/15)
Wednesday November 11:
Marathon make up all your webwork up until 12.5 (for up to 50% of the webwork credit you missed)
Friday November 13:
For-credit quiz covering sections 13.1-13.3 (10pts)
Marathon make up all your webwork up until 12.5 (for up to 50% of the webwork credit you missed)
Friday November 13:
For-credit quiz covering sections 13.1-13.3 (10pts)
Friday, November 6, 2015
12.7 problem 3
Dear Professor Taylor,
I have tried many times on number 3. The the last
two are counting my second coordinate as wrong, even though I believe they
are correct.
My Cartesian coordinates are ((5/4)sqrt6, -(5/4)sqrt2,
-(5/2)sqrt2)
My answer is (5, -pi/6, 2.356)
(-(5/4)sqrt6, -(5/4)sqrt2,
-(5/2)sqrt2)
My answer- (5, -2.6179938,2.256)
Thank you,
****************************
The only problem that you are having here is that you are thinking of negative angles, while this webwork problem is set up to only consider positive angles measured counter-clockwise from the positive x-axis.
I have tried many times on number 3. The the last
two are counting my second coordinate as wrong, even though I believe they
are correct.
My Cartesian coordinates are ((5/4)sqrt6, -(5/4)sqrt2,
-(5/2)sqrt2)
My answer is (5, -pi/6, 2.356)
(-(5/4)sqrt6, -(5/4)sqrt2,
-(5/2)sqrt2)
My answer- (5, -2.6179938,2.256)
Thank you,
****************************
The only problem that you are having here is that you are thinking of negative angles, while this webwork problem is set up to only consider positive angles measured counter-clockwise from the positive x-axis.
12.7 Problem 11
Good morning professor,
My question in regard to this problem is that I do
not know where to begin. If you can help me out with it, I'll appreciate
it.
Thank you,
******************************
Well, 12.7 so we're talking spherical coordinates. Inside x^2+y^2+z^2=64 and above the xy-plane says were in the northern hemisphere of a sphere of radius 8. Outside the cone says you have to poke a hole in the top of the cone: the cone is pretty sharp, the radius of the cone is one seventh times the height. This means that the angle phi satisfies tan(phi)=1/7 or phi =~0.14 radians. Thus the limits of integration on theta are 0 to 2pi, on phi from 0.14=arctan(1/7) and on rho from 0 to 8,
My question in regard to this problem is that I do
not know where to begin. If you can help me out with it, I'll appreciate
it.
Thank you,
******************************
Well, 12.7 so we're talking spherical coordinates. Inside x^2+y^2+z^2=64 and above the xy-plane says were in the northern hemisphere of a sphere of radius 8. Outside the cone says you have to poke a hole in the top of the cone: the cone is pretty sharp, the radius of the cone is one seventh times the height. This means that the angle phi satisfies tan(phi)=1/7 or phi =~0.14 radians. Thus the limits of integration on theta are 0 to 2pi, on phi from 0.14=arctan(1/7) and on rho from 0 to 8,
12.6 problem 5
Can you please explain what I am doing wrong with the two answers I'm not
getting correct? Is something strange happening with the coding or am I
just doing something wrong?
OK, the critical thing for you to understand is that tan is continuous between -pi/2 and pi/2, so when you take the arctan of something it will only give you a value in this range. SO, when you're looking for an angle, you need to be aware of what quadrant your vector (x,y) is in: for example when (x,y)=(-3/2 , 3√3/2) , you are in the second quadrant. Arctan will give you an angle in the fourth quadrant though corresponding to the coordinates (x,y)=(3/2,-3√3/2) instead, which will be -pi/3. The angle *you* need though is pi-pi/3=2pi/3. Likewise when (x,y)=(-3/2 , -3√3/2), you are in the third quadrant, arctan will give you an angle corresponding to (x,y)=(3/2 , 3√3/2) or pi/3, while you need pi+pi/3=4pi/3
getting correct? Is something strange happening with the coding or am I
just doing something wrong?
OK, the critical thing for you to understand is that tan is continuous between -pi/2 and pi/2, so when you take the arctan of something it will only give you a value in this range. SO, when you're looking for an angle, you need to be aware of what quadrant your vector (x,y) is in: for example when (x,y)=(-3/2 , 3√3/2) , you are in the second quadrant. Arctan will give you an angle in the fourth quadrant though corresponding to the coordinates (x,y)=(3/2,-3√3/2) instead, which will be -pi/3. The angle *you* need though is pi-pi/3=2pi/3. Likewise when (x,y)=(-3/2 , -3√3/2), you are in the third quadrant, arctan will give you an angle corresponding to (x,y)=(3/2 , 3√3/2) or pi/3, while you need pi+pi/3=4pi/3
Tuesday, November 3, 2015
Monday, November 2, 2015
Thursday, October 22, 2015
Tomorrow's test! (a request)
Hello Dr. Taylor,
I had a question about tomorrows test. I was wondering if I could bring some blank graph paper in, to use as scratch paper. It will help me draw pictures to scale (for when switching limits of integration), and sometimes I like to write out the solution as it helps me figure out step by step what I need to do to solve the problem. Do you think this is a possibility? I will be okay with or without it. It just helps me draw my pictures which can be key to the answer. Let me know what you think. Thanks!
******************************
I'd prefer to supply the scratch paper; the functions in the limits of integration will be very simple and any switching of limits of integration will depend on an understanding of these functions and their inverse functions and not on any precise rendering.
Partial homework extension
I have extended the due date for section 12.5 to next Monday at midnight.
No cheat sheets allowed on the exam
Would you be so kind as to let us have a HAND WRITTEN cheat sheet
with general formulas for ourselves? I think if you allowed us to have
example problems it would literally be cheating. But, perhaps if we could
have formulas like for linear approximation, and what information the
determinate tells us about our critical points. We could turn it in with
our tests so you know what was on there and that students don't share one
copy. I hope you take this into consideration as something allowed for this
test.
with general formulas for ourselves? I think if you allowed us to have
example problems it would literally be cheating. But, perhaps if we could
have formulas like for linear approximation, and what information the
determinate tells us about our critical points. We could turn it in with
our tests so you know what was on there and that students don't share one
copy. I hope you take this into consideration as something allowed for this
test.
Sorry, I asked the course coordinator and the rest of the instructors for this course, the consensus is that there we cannot allow cheat sheets.
Tuesday, October 20, 2015
Friday, October 16, 2015
12.2 no 3
I'm having difficulty understanding how to solve this problem. Where
should I start?
Well, the first step is to read section 12.2 in the textbook and the notes. The basic notion is that you are trying to write the double integral as an iterated integral. For part (a), this means that you are integrating first with respect to y and then with respect to x, this means that you need to integrate from the horizontal line at the bottom of the triangle--which gives you the lower limit of integration of the inside integral, to the slanting line at the top of the triangle. In order to do this, you *must* find the equation of the line relating from the graph of the line; this means that you write a linear equation. for y as a function of x. This function of x becomes your upper limit of integration.
For part (b) you must turn this around--the vertical line on the right gives you the lower limit of the inside integral, and the upper limit becomes the equation for x in terms of y that you get by solving the equation of the slanting line for x.
should I start?
Well, the first step is to read section 12.2 in the textbook and the notes. The basic notion is that you are trying to write the double integral as an iterated integral. For part (a), this means that you are integrating first with respect to y and then with respect to x, this means that you need to integrate from the horizontal line at the bottom of the triangle--which gives you the lower limit of integration of the inside integral, to the slanting line at the top of the triangle. In order to do this, you *must* find the equation of the line relating from the graph of the line; this means that you write a linear equation. for y as a function of x. This function of x becomes your upper limit of integration.
For part (b) you must turn this around--the vertical line on the right gives you the lower limit of the inside integral, and the upper limit becomes the equation for x in terms of y that you get by solving the equation of the slanting line for x.
12.1 no 1
Can you tell me what I'm doing wrong here?
I think so. When dividing the square [0,2]x[0,2] into four equal squares you get squares [0,1]x[0,1], [0,1]x[1,2], [1,2]x[0,1], [1,2]x[1,2]. It looks like you misinterpreted the questions--for example in part (A) you have supplied the value of the integral
I think so. When dividing the square [0,2]x[0,2] into four equal squares you get squares [0,1]x[0,1], [0,1]x[1,2], [1,2]x[0,1], [1,2]x[1,2]. It looks like you misinterpreted the questions--for example in part (A) you have supplied the value of the integral
But you should instead give the value of the Riemann sum approximation for the integral
with that partition of the square with values at the points (0,0),(1,0),(0,1),(1,1)
(25-0^2-0^2)ΔA_{11}+(25-1^2-0^2)ΔA_{21}+(25-0^2-1^2)ΔΑ_{12}+(25-1^2-1^2)ΔΑ_{22}
and since all of the ΔA's take the value 1, the correct answer for part (A) should be
25+24+24+23=96
Thursday, October 8, 2015
11.6 problem 20 (updated)
Dear Professor Taylor,
I'm not really sure how to do this. I'm fairly
certain the grad f vector point towards the upper right of the graph at
every point. But I think I just happened to guess correctly to get the
first two answers. Can you explain what how to do this?
Thank you,
At the point P=(1,4) in the direction v=(i+j)/√2, the directional derivative is approximately ______
The first step is to know where you are; in this case at the intersection of the vertical line x=1 and the horizontal line y=4. If you go up from the intersection of the two red lines, the contour changes from about 4.0 to half way between 4.0 and 6.0--i.e. 5.0. while the y-value changes from 4 to 5--hence the rise over the run is 1/1=1. Likewise the x-value changes from 1 to 2.5 while the contour changes from 4.0 and 10.0--the intersection of the red horizontal line with the yellow vertical line--which gives you another rise and a run to compute a difference quotient. For the direction vector v you add the two difference quotients and divide by √2.Btw, grad(f) points to the upper right at a few points on the graph, but not at most points: toward the top of the graph near the y-axis it points mostly to the right, on the x-axis except near x=0 it points mostly up, on the diagonal line y=x it points to the upper right corner.
Hello Professor Taylor,
I did what you said to do for this problem.
I get
that
df/dx= (10-4)/(3-2)=4i
and
df/dy=(5-4)/(5-4)=1j
so the answer along
the vector v I get
(4i+1j)/sqrt(2) I kept getting 5/sqrt(2) but I changed
it to six/sqrt(2) and that happened to be the right. What am I doing wrong?
you've got a little problem using the dot product correctly, since you kept the i and j after you did the dot product, and they're supposed go away. basically you did this correctly though, the answer 5/√2 is correct and 6/√2 is incorrect. The problem was in the software and I fixed it, it should work now
Section 11.7 Problem 8
Hey Professor Taylor,
I am very much stuck on problem 8 from section 11.7. I was hoping you would be able to help me out. Thank you very much. The problem states the following:
(Yep, this one is a bit of a detail-challenge)
I am very much stuck on problem 8 from section 11.7. I was hoping you would be able to help me out. Thank you very much. The problem states the following:
Find the absolute maximum and absolute minimum of the function f(x,y)=2x^3+y^4 on the region {(x,y)|x^2+y^2≤81}. It also asks for the x and y coordinates at which the maximum and minimum values are obtained at.
Sincerely,
Sincerely,
**********
(Yep, this one is a bit of a detail-challenge)
OK, the domain is the closed disk of radius 9 centered at the origin. The first thing you need to do is to compute the gradient of f:
It follows that the critical points inside the disk are when 6x^2=0 and 4y^3=0; i.e. only at the point (0,0). Since f(0,0)=0 and f can take both positive and negative values on the x-axis inside the disk, for example f(1,0)=2 and f(-1,0)=-2, it follows that the absolute min and absolute max are on the boundary of the domain, i.e. must be on the circle of radius 9 centered at (0,0).
We find the absolute max and min on the unit circle by parameterizing the circle of radius 9 as
x(t)=9 Cos(t), y(t)=9 Sin(t). Then f(x(t),y(t))=9^3(6 Cos^3(t)+9 Sin^4(t)), which is a differentiable function of t, and so the absolute max and min will be at critical points of this function of one variable, i.e. satisfy the equation
so the critical points are where Sin(t)=0 AND where Cos(t)=0 AND where - Cos(t)+6 Sin^2(t)=0. Thus we get t=0,π/2,π, 3π/2 and solutions of the equation Cos(t) - 6 + 6Cos^2(t)=0 (where we used the identity Sin^2(t)=1-Cos^2(t)). BUT notice that the equation is quadratic in Cos(t). Using the quadratic formula we get that Cos(t)=(-1±√(1^2-4(1)(-6))/12=(-1±√25)/12=1/3, -1/2. Then, since Sin(t)=±√(1-Cos^2(t)), we need to check the points (-1/2,±√3/2) and (1/3,±2√2/3), hence we have eight points to check: plugging in the numbers we get the values (9,0,1458), (0,9,6561), (-9,0,-1458),(0,-9, 6561), (-4.5, 9√3/2,14580), (-4.5, -9√3/2,14580), (3, 6√2,5238), (3, -6√2, 5238).
Thus the absolute max value is 14580 which takes place at (-4.5, ±9√3/2) and the absolute min value is -1458 which takes place at (-9,0).
Friday, October 2, 2015
Section 11.3 problem 11
I'm stumped on how exactly to get this answer. I just can't see the
relation between Fy and the contour lines.
relation between Fy and the contour lines.
The notion is that the partial derivative is the limit of the difference quotient
this means that when h is small, the partial derivative is approximately equal to the finite difference quotient:
This is is an important and useful fact for an engineer, because a lot of times the only thing you know about the function is some contours or some data points. For this problem you can start at the point (1,3), where you're right on the contour for 10, so you can get f(x,y)=10, and if you increment y by about h=2.6 you get the f(x,y+h)=8
--that's all the information you need to get the approximation
Thursday, October 1, 2015
Friday, September 25, 2015
Java...java....java...
Hello Professor, problems numbered 13 and 15 have java plugins that were
not supported by both my PC and Mac. Just wanted to let you know, I have
looked for a solution and will continue to.
As of September 1st, Google Chrome no longer supports Java due to HTML5's
restrictions. Will this be fixed for future problems?
Hi for question 13 I cannot load the four images. It shows up with a
message that says plug in not supported. I ended up just guessing to try
and get partial credit. If i cannot get any credit for this i understand,
just though i would let you know about the possible problem other people
might be having too. It may just be because I am running linux but i am not
sure.
Thanks,
*******************************************
Ok, here is the deal. There is no plugin, there is only a lonely Java applet. You need to use a browser that supports Java. Note that the Microsoft Edge browser also does not support Java, but Internet Explorer does, and you can access it directly from Edge. Safari supports Java. The Firefox browser also does support Java and is available for pretty much any operating system; I suggest you download it. You also need to enable Java on your machine and on your browser. Here is a link to describe how to do that for Firefox. You may need to install Java. Here is a link describing how to do that for OS X. Note that the reason you're having problems is that lots of people wanted to stop Java from automatic implementation for *security reasons*. I suggest once you are done with these problems you disable Java again.
11.1 Problem9
I arrived at the right answer however I have no idea why those are the
right answers. Can you go through the steps to solve this?
Each level curve is the solution of an equation f(x,y)=A,B,C,D. The curve for f(x,y)=A is the easiest, because it *looks* like A must be equal f(0,0)=140e^0-70*0^2=140, and this is a multiple of 10. For f(x,y)=B the problem is a little more complicated--it looks like f(x,0)=B for an x close to 0.4--and we need B to be an integer multiple of 10. Since y=0 on the x-axis and increasing values of x mean increasing values of f, it looks like B must be equal to 140e^0.4, but this is equal to 208.86 which is not a multiple of 10. The nearest multiple of 10, though is 210, which means that x=ln(210/140)=0.405, which is close enough to what we see
right answers. Can you go through the steps to solve this?
Each level curve is the solution of an equation f(x,y)=A,B,C,D. The curve for f(x,y)=A is the easiest, because it *looks* like A must be equal f(0,0)=140e^0-70*0^2=140, and this is a multiple of 10. For f(x,y)=B the problem is a little more complicated--it looks like f(x,0)=B for an x close to 0.4--and we need B to be an integer multiple of 10. Since y=0 on the x-axis and increasing values of x mean increasing values of f, it looks like B must be equal to 140e^0.4, but this is equal to 208.86 which is not a multiple of 10. The nearest multiple of 10, though is 210, which means that x=ln(210/140)=0.405, which is close enough to what we see
Thursday, September 24, 2015
10.9 prob 8
Hi,
I had a question about chapter 10.9 number 8. I don't know how to find
the components of centripetal force.I'm hoping you could guide me on how
to
solve this problem.
OK....
The first thing you need to know is the definition of centripetal force--otherwise you will have no idea how to do this problem. This is actually in the textbook. Read the part of 10.9 that covers this--you can look for the term in the index of the book to get the exact page number. Or you could look at online resources like Wikipedia (which is accurate much more often than it's not)
Second, you should know that centripetal force IS a force. But the Newton's Law F=ma relates the force to the acceleration. This means that to get the force you're going to need to get the acceleration, which means you need to use the information they give you about the size of the circle and the speed of rotation to get a parameterization of the motion that represents the position of the path vs time accurately. And then from that you can compute the acceleration.
Let me know how it goes.
webwork 10.9 no 7
Hello Professor Taylor, can you help explain this problem to me I cannot
get the correct answer. I keep getting aprox. 15.7m/s
This is almost exactly the same as problem 6, which is discussed in a previous blog post. The only difference here is that the angle is 45˚ instead of 30˚, and they give you that y(t)=0 when x(t)=25, and ask you to find v_0 instead of the other way around. Try the problem with perspective in mind, and let me know how it goes.
get the correct answer. I keep getting aprox. 15.7m/s
This is almost exactly the same as problem 6, which is discussed in a previous blog post. The only difference here is that the angle is 45˚ instead of 30˚, and they give you that y(t)=0 when x(t)=25, and ask you to find v_0 instead of the other way around. Try the problem with perspective in mind, and let me know how it goes.
your direction (grades, etc)
Below are: your midterm 1 score, your homework % score, and the grade I would assign right now, if I had to assign a grade based on course policies and your performance to date. If you keep going the way you are you're likely to end up with approximately this grade. I haven't included +'s or -'s, because there is not enough information yet to justify these
webwork 10.9 question 6
I had a question about chapter 10.9 number 6. Ive tried multiple tries
myself and I have even gone to tutoring centers and they say the work is
right but webwork marks it as wrong. I am hoping you could guide me on how
to solve this problem.
The acceleration due to gravity is a=<0,-9.8>. The Sin(30˚)=1/2, Cos(30˚)=√3/2, so the initial velocity vector is v_0=250<√3/2,1/2>=<125√3,125>. Since a(t)=v'(t), and a(t)=a is constant, it follows that
Since v(t)=r'(t), it follows that
But since the projectile is fired from the ground, and it's a reasonable thing to make our coordinate system so that the origin is at (0,0), we can assume that r_0=0, hence
myself and I have even gone to tutoring centers and they say the work is
right but webwork marks it as wrong. I am hoping you could guide me on how
to solve this problem.
The acceleration due to gravity is a=<0,-9.8>. The Sin(30˚)=1/2, Cos(30˚)=√3/2, so the initial velocity vector is v_0=250<√3/2,1/2>=<125√3,125>. Since a(t)=v'(t), and a(t)=a is constant, it follows that
v(t)=a*t+v_0=<125√3,125-9.8*t>.
r(t)=a*t^2/2+v_0*t+r_0.
r(t)=a*t^2/2+v_0*t=<125√3*t,125*t-9.8*t^2/2>.
Writing r(t)=<x(t),y(t)>, the projectile will be at ground level when y(t)=0, i.e. when
0=125*t-9.8*t^2/2=t(125-9.8*t/2).
This equation has two solutions: t=0, before the projectile is fired, and t=250/9.8=25.51 when the particle hits the ground again. At that time x(t)=125√3*25.51=5523.1 meters, so the range is 5523.1. At the time the particle hits the ground, the speed will be the magnitude of the vector
||v(25.51)||=||<125√3,125-9.8*t>||=||<125√3,125-9.8*25.51>||=||<125√3,-125>||=250.
The maximum height will be attained with the vertical component of the velocity is zero, i.e. when
0=125-9.8*t or t=125/9.8=12.76.
At this time the height is y(12.76)=125*12.76-9.8*(12.76^2)/2=796.6 meters.
Tuesday, September 22, 2015
Monday, September 21, 2015
Friday, September 18, 2015
Tuesday, September 15, 2015
Monday, September 14, 2015
Friday, September 11, 2015
Thursday, September 10, 2015
Exam next week! (and related matters) Part I
Friday 9/18. Be there.
**********************
Hello,
I was hoping there would be some kind of review session or study
guide or anything of the sort before our tests in this class? I am
struggling to get through the homeworks and I often don't even understand
what I'm doing. I do my best to pay attention to lectures, but talking or
theory with hypothetical numbers doesn't help me personally. I was just
hoping there would be something more to help me prepare for the upcoming
tests.
*******
**********************
1) "I was hoping there would be some kind of review session or study guide..."
Yes, there are. There are a number of reviews and study guides, which you can access at this webpage (this link has also been available on the syllabus since the first day of class, btw). I suggest that you give yourself a timed 50 minutes to take the practice test, and then afterward review the answers there. There is also an old exam there. (I shouldn't need to give this disclaimer but I do: the practice exam and old exam are representative of the sort of problems you may seen on the actual exam, but in no way are meant to imply that you will see exactly those problems on the exam. In fact very suredly you won't see exactly those problems.)
I believe that next week there will be an evening review session hosted by the Engineering Tutoring Centers for the exam also, but haven't been notified about the particulars yet--I'll post the information here on the blog as soon as I get it. Also, and as always, I'm available by office hours and by appointment.
(I've got more to say on this, but later)
**********************
Hello,
I was hoping there would be some kind of review session or study
guide or anything of the sort before our tests in this class? I am
struggling to get through the homeworks and I often don't even understand
what I'm doing. I do my best to pay attention to lectures, but talking or
theory with hypothetical numbers doesn't help me personally. I was just
hoping there would be something more to help me prepare for the upcoming
tests.
*******
**********************
1) "I was hoping there would be some kind of review session or study guide..."
Yes, there are. There are a number of reviews and study guides, which you can access at this webpage (this link has also been available on the syllabus since the first day of class, btw). I suggest that you give yourself a timed 50 minutes to take the practice test, and then afterward review the answers there. There is also an old exam there. (I shouldn't need to give this disclaimer but I do: the practice exam and old exam are representative of the sort of problems you may seen on the actual exam, but in no way are meant to imply that you will see exactly those problems on the exam. In fact very suredly you won't see exactly those problems.)
I believe that next week there will be an evening review session hosted by the Engineering Tutoring Centers for the exam also, but haven't been notified about the particulars yet--I'll post the information here on the blog as soon as I get it. Also, and as always, I'm available by office hours and by appointment.
(I've got more to say on this, but later)
Tuesday, September 8, 2015
Monday, September 7, 2015
[WWfeedback]:Taylor_MAT_267_Fall_2015:Section_10.5 prob:21
Hi Professor Taylor,
I am attempting problem number 21 and have come across trouble on part C. To solve the problem, I combined the vectors from part A and B to get the vector for part C and I got the right answer for the second coordinate but not for the other two coordinates so I'm not sure what I'm doing wrong. Thank you in advance for your help!
****************
Consider the planes 5x+5 y+5 z=1 and 5 x+5 z=0.
A)Well, P is on the y-axis so x=z=0, which means that P is in the second plane(why??), and in the first plane means that 5y=1 so y=1/5--it sounds like you figured this out.
B) A unit vector parallel to both planes will be perpendicular to both normals (what *are* the normal vectors to each plane, and how do you use them?), which turns out to be something parallel to
C) there are a lot of vector equations that will do the job for this part, but the simplest one for you to write would be r(t)=<0,1/5,0> + t<1/√2,0,-1/√2>=(t/√2i+1/5j-t/√2k).
I am attempting problem number 21 and have come across trouble on part C. To solve the problem, I combined the vectors from part A and B to get the vector for part C and I got the right answer for the second coordinate but not for the other two coordinates so I'm not sure what I'm doing wrong. Thank you in advance for your help!
****************
A)Well, P is on the y-axis so x=z=0, which means that P is in the second plane(why??), and in the first plane means that 5y=1 so y=1/5--it sounds like you figured this out.
B) A unit vector parallel to both planes will be perpendicular to both normals (what *are* the normal vectors to each plane, and how do you use them?), which turns out to be something parallel to
±(i+0j-k),
so a unit vector that's has positive first coordinate would be (i+0j-k)/√2.--it sounds like you got this part too.C) there are a lot of vector equations that will do the job for this part, but the simplest one for you to write would be r(t)=<0,1/5,0> + t<1/√2,0,-1/√2>=(t/√2i+1/5j-t/√2k).
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