Grade estimate based on straight grade scale. The curved scale is likely different. Click on the image for enlarged view.
MAT267 Fall 2019
Wednesday, December 4, 2019
Tuesday, November 19, 2019
Scores and Estimate Grades
Friday, November 15, 2019
13.4#3
I can't seem to setup my bounds properly for this problem. Honestly,
setting up my bounds for all
of these Green's Theorem problems has been a struggle. How should I go about setup up the bounds?

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Well, ok, you've got a piece of pie of radius 4. It's 1/8 of the whole pie. *If* you were going to actually calculate the circulation, that would be the path integral along the edge of the piece of pie. Since it has three pieces, you'd have to break up the path into three pieces: the part going outward from the origin along the x-axis, which can be parameterized as r(t) = <4t,0> for 0≤t≤1, the counterclockwise circular part of radius 4 moving from the x-axis upward and to the left to the point (2√2,2√2), which can be parameterized as r(t) = <4 cos(t), 4 sin(t)> for 0≤t≤π/4, and then the radial part moving inward that can be parameterized as r(t) = (1-t)<2√2, 2√2>. You'd have to substitute x(t) and y(t) for each part and then integrate with respect to t and add them all up.
*BUT* you're not going to do it that way, instead you're going to compute
∂F_2/∂x - ∂F_1/∂y = 5-(-4) = 9,
then use ∫∫_D ∂F_2/∂x - ∂F_1/∂y dA = 9 ∫∫_D dA = 9(Area of circle)/8 = 9(π 4^2)/8
=9*π*16/8 = 18π = 56.55
By the way, I changed the due date to tomorrow night.
of these Green's Theorem problems has been a struggle. How should I go about setup up the bounds?

******************************
Well, ok, you've got a piece of pie of radius 4. It's 1/8 of the whole pie. *If* you were going to actually calculate the circulation, that would be the path integral along the edge of the piece of pie. Since it has three pieces, you'd have to break up the path into three pieces: the part going outward from the origin along the x-axis, which can be parameterized as r(t) = <4t,0> for 0≤t≤1, the counterclockwise circular part of radius 4 moving from the x-axis upward and to the left to the point (2√2,2√2), which can be parameterized as r(t) = <4 cos(t), 4 sin(t)> for 0≤t≤π/4, and then the radial part moving inward that can be parameterized as r(t) = (1-t)<2√2, 2√2>. You'd have to substitute x(t) and y(t) for each part and then integrate with respect to t and add them all up.
*BUT* you're not going to do it that way, instead you're going to compute
∂F_2/∂x - ∂F_1/∂y = 5-(-4) = 9,
then use ∫∫_D ∂F_2/∂x - ∂F_1/∂y dA = 9 ∫∫_D dA = 9(Area of circle)/8 = 9(π 4^2)/8
=9*π*16/8 = 18π = 56.55
By the way, I changed the due date to tomorrow night.
Thursday, November 7, 2019
12.7#11
I'm struggling with what to do for this problem. My phi bounds are pi/3 to pi/2. My theta bounds are 0 to 2pi. My rho bounds are 0 to 2. My integrand is rho^2 times sin(phi). Can you please give me a clue as to where I'm going wrong? I took this problem to the tutoring center and they couldn't figure it out either.
************************************
Let's see: z=3r means ρ*cos(φ)= 3ρ*sin(φ) so φ = tan^{-1}(1/3) =0.322 radians or 18.43 degrees. Your only mistake was jumping to the conclusion that webwork would default to a clean and easy lower bound.
Friday, November 1, 2019
12.6#4
Theta can be found with the following formula, right?
arctan(y/x).
The coordinate for this problem is (-3, 4, -1), and none of the follow are accepted:
arctan(4/-3) = -0.927295218001612 = -53.1301023542
The previous problem didn't have this issue, and the first entry for this problem accepts the following:
-3/(cos(arctan(4/-3))
Is there a way to fix this, or do I need to do something differently?


Tuesday, October 29, 2019
switching order of integration
....do you remember when I said *NEVER* have variables in the outside limits of integration, and that if you should find yourself with variables in the outside limits go back because you did it wrong?
sums and vectors
an amazing number of people STILL make the illogical jump equating sums and vectors: that 3+5 means the same thing <3,5>
Tuesday, October 22, 2019
Practice Exam 2 Answers
Professor Taylor,
<<<<<<<<<<<<<snip>>>>>>>>>>>> In class on Monday you passed out a
practice exam and said that we could come ask you one question if we
showed our work. Does this mean that you are not posting an answer key
for us to check against? I have checked the blog and the syllabus
practice exams to see if the answer key is available and I have not
found it. As I have already completed the review, an answer key to check
my work against would be very informative. I apologize for the late
hour of this email, however I thought if I was having this problem then
it is likely that I am not the only one.
Thank you for your time,
************************
You heard correctly. You do maintain the option to go to the syllabus and look at the practice test linked there, which does come with answers, as well as the review linked there. You may also take the practice test I passed out and go to one of the review sessions hosted by the University Academic Success program and ask them how to do the problems. As I've stated several times by now, passively getting someone else to tell you the answer fosters the delusion that you know how to do the problem, when really much of how to do the problem is learning what mistakes you're likely to make. This requires actually doing the problem and making the mistakes before you get them corrected though. The practice test I passed out is for this purpose.
Exam 2 reviews
The University Academic Success Program will be holding some test review sessions (see table below for times/locations), in case you want to announce this to your students.
Date Course Topic Time Location Monday, October 21 MAT 267 Exam 2 6:00 PM PSH 151 Tuesday, October 22 MAT 267 Exam 2 7:00 PM PSH 153
Monday, October 21, 2019
Friday, October 4, 2019
Thursday, October 3, 2019
scores and extrapolated grade
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