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
