Friday, March 24, 2017

The funzies quiz today


Please tell me that was not a real quiz today. **********************

That was not a real quiz today.  He was just a little confused about what he was supposed to do.  I'll pass that back to you ungraded on Monday.

Webwork section 12.3 due date

Section 12.3 is now due Sunday night @midnight

12.8#8 again

Hi Dr. Taylor, I can't seem to get this one, I've tried entering   f(rcos(t),rsin(t) as well.





































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OK, you and a lot of people are making this harder than it has to be.  Still.  First of all, the only answer is doesn't like is the one about dA. It's not talking about f at all, so you don't have to worry about that.  There is a formula for dA in polar coordinates, I talk about it in my lecture notes, I talked about it about ten times in my video lecture last week that is still online, and the book talks about it in section 12.3. It is the single most important thing for you to know about polar integration. This formula DOES NOT DEPEND ON the limits of integration--that's one nice thing about it. WHAT IS THIS FORMULA?

Tuesday, March 21, 2017

Q's and Peruse

Greetings, 
You sent out a prep exam that is suppose to look like a real test. Are the answers to that anywhere? I am in dire need of another killer test grade. Also, where can I find the other tests with answers. I seem to have misplaced them. 
Is there any way to look at our current grade or will you possibly update us after the test? That would be great. Thank you for your time. 

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1) Why yes, the answer is in your hands. Other than that, the way I roll is that you demonstrate your due diligence by telling me what you did, and why you think what you did is right or wrong. Then I will likely be motivated by your studiousness to ease any remaining confusions that you have.

2) Other practice tests are in an old blog post-what? you didn't memorize them? Diligence...

oh what the heck, look here.

3) I *will* update you after the test.

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Professor Taylor,

Sorry to bother you, I just had a quick question.
I'm in your MAT 267 class and was recently working on the Test from Spring 2014 that you posted on the blog. I wanted to see if you happen to have the answer key as well that you could easily post online. If you don't already have one made then don't worry about it, I don't want it to be a hassle for you. But if you already have one made I would really appreciate it if you could post that as well.


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Yeah, this was an actual test that I gave in 2014, I wrote an answer key back then, but it's gone now.  And yeah, it would have been great to ask questions about this at the review session on Monday.

Practice Exam Question5 Typo

Hey Dr. Taylor, I've noticed when taking the practice exam, question 5 has integral of sqrt(4+y^2) dydx, the outer bounds for x are from 2 to 0, but the inner bounds are from 2 to y?? Shouldn't the inner bounds be from 2 to x instead? I thought you had to have the opposite term in the first limits of integration. 








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Yes, the lower limit of integration should be x.

(BTW, this would have been a perfect question to ask the substitute instructor yesterday)



extra credit assignment

Hi, I'd like to offer an extra credit assignment. In general terms it would involve a properly formatted & properly cited, non-plagiarized 1000 word essay on what impact multivariable calculus has on your particular chosen career. Details to follow--but you could google the words in this post that you don't already understand.

Monday, March 20, 2017

Sunday, March 19, 2017

12.3#7

I don't know if i missed this somewhere but how exactly do you know what your function is in scenarios like this? How do you know what your f(x, y) is? Would it be 1/(x^2+y^2)^13?



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Yes, that's exactly right--the graphs mentioned are the  xy-plane and the graph of the function.  Of course what might be the interesting (or tricky) part is limits of integration.

Friday, March 17, 2017

12.2#3

Hi professor,
I believe that there is a problem with two of these Webwork answers. I may, of course, just be an idiot. In which case please let me know.





































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Everybody gets to be an idiot sometimes, but I don't call anybody that except myself.  
So here's the deal: that figure is *not* a rectangle.  This means in each case your inner limit of integration that corresponds to being on the diagonal line *can't* be a constant--it has to be a function of the outer variable of integration.  The question you need to answer then is which function.

12.3#8

The region goes from pi/2 to 3pi/2 with radius' of 3 and 5 so wouldn't
those be my limits of integration? I am unsure of what to do since this
didnt work.
























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There's nothing wrong with your limits of integration. The thing you did wrong here is putting the (rcos(t), rsin(t)) into the dA, instead of leaving it with the f where it belongs.

12.1#10

I don't know what I am doing wrong with this one.
I tried using iterated integrals, I tried doing the point thing. Neither worked. 














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You do need to use iterated integrals--since the region is a rectangle the limits of integration will be constants--which makes your life easier. It would also make your life a little easier if you would do  the integral with respect to on inside--if you remember that x is a constant for that y-integral you can use substitution. I can't tell exactly what you did wrong--your answer is only about 30% wrong though.

Tuesday, March 14, 2017

12.1#8




Hi Dr. Taylor,

For this problem, does it matter if one chooses to integrate with respect to y or x first?




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No, it only matters that you have the right limits of integration for the right variable

Monday, March 13, 2017

Another practice test2

Some people were complaining that the practice tests online look nothing like the currents tests. Here is another practice test that you *might* think (no promises) be more closely related.

Tuesday, March 7, 2017

11.7#7

Mr. Taylor,
I am having trouble with this problem, the values I'm finding
for the maximum and minimum values don't seem to be correct. How do you
find the absolute max and min on this region?



















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I'd be happy to tell you.  BUT, if you want me to do that work I need you to do the work of telling me what *you* did first, and preferably also how what you did relates to what's in the textbook and my lecture notes--it's a lot better pedagogy to start with what you already do know than just to repeat the lecture I gave the other day that you didn't quite get in the first place.

Sunday, March 5, 2017

webwork problem and the rules

Hello Prof. Taylor, I have worked out this webwork problem and am doing everything right, but there only two parts of the problem which are refusing.
I thought the two part just required finding the Magnitude of the vector above them, but I am getting it wrong. Some people got theirs right, so I don't understand why my answer is rejected. I therefore need your help!
Below is the problem;
















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OK, 
1) first of all, lets talk about the RULES for getting my help on a webwork problem.  The rather than take a screenshot of the problem, click the "Email Instructor" button at the bottom of the problem--this will show me what you've done.  When you do this, you'll have a field to give me a message. In that field enter a brief description of what you have already done.  "Help, I don't know what to do" is NOT useful to you or to me. You have many resources, specifically the textbook, that covers the material, and each section of the webwork corresponds to a specific section of the book, which will tell you SPECIFICALLY what to do. 
2) The gradient 
∇f = <-160*2x/(x^2+y^2+2)^2, -160*2y/(x^2+y^2+2)^2>
   = -320/(x^2+y^2+2)^2 <x, y>
so ∇f(2,2) = -3.2 <2, 2> = <-6.4, -6.4>.  This means that the direction of maximal change, AS A UNIT VECTOR is = <-1/√2, -1/√2> .  The answer to part c) that you have is NOT A UNIT VECTOR AND NOT THE GRADIENT, but it does point in the right direction. I don't know at all why you want that vector--but I guess you are trying to normalize incorrectly.  The norm of that vector is the answer you give. What you needed, though, is the directional derivative in that direction is
D_u = u . ∇f  = <-1/√2, -1/√2> . <-6.4, -6.4> = 6.4*2/√2 √2*6.4
which is also equal to |∇f(2,2) |.  All in all I *guess* that the mistake you made was multiplying by
1/√2 while *not* taking the dot product 

Saturday, March 4, 2017

Webwork section 11.6 reopened

Due to the mass of requests, I've re-opened the section 11.6 webwork until Sunday night.

Friday, March 3, 2017

11.5#5

Hello Prof.Taylor,
This problem really needs your intervention:




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While non-interventionism has some attraction on a Friday night, OK.  This problem is really all about the chain rule (discussed in the book and in the lecture notes), for example 
∂W/∂s = ∂F/∂u ∂u/∂s + ∂F/∂v ∂v/∂s, 
and you are given all of the required components above for the values s=1 and t=0.

Thursday, March 2, 2017

The most common mistake

The most common mistake people make when working a math problem is not that they don't know how to do the math, it's that they don't know what math to use and they don't know how to find out what math to use.  The reason that they don't know what math to use is because the don't know what the words mean (and so they stare at it and ponder and nothing gets done).  The reason that they don't know what the words mean is because they didn't read the textbook, or at least they didn't read the right part of the textbook.  Now anyone could be forgiven that last, because the textbook is so thick and heavy you could use it to kill poisonous snakes just by dropping it on them, so actually reading it might be fatal.

Except. It's so easy to find the right part of the textbook. Because. Each problem set refers to just one little section in the textbook, which is by itself quite light and slender. That means that the words that you don't understand are in that one little section.  And those words will have some equations that come right after that describe what those words mean, and those equations will be the math that you need to use to solve the problem.

problem


Hi Dr. Taylor,
 I need help answering this question. I dont know how to answer without an equation. Thank you.












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OK, first of all: at the bottom of every webwork page is a little button that says "Email Instructor".  If are having trouble with a problem you push that button and it will send me all of the info from the work you have been doing.
The equation you need here is the definition of directional derivative:
D_u f = ∇f . u = <∂f/∂x, ∂f/∂y>. u  for a unit vector u.  You are given a vector <4,6> which has unit vector <4/√52, 6/√52> and a vector <3,7> that has unit vector <3/√58,7/√58>.  This means that the directional derivatives given in the problem satisfy
6/√52 =  ∇f . <4/√52, 6/√52> or 6 =  ∇f . <4, 6> and
6/√58 =  ∇f . <3/√58, 7/√58> or 6 =  ∇f . <3, 7>.
These amount to the two equations
4 ∂f/∂x + 6 ∂f/∂y = 6
3 ∂f/∂x + 7 ∂f/∂y = 6
which you can solve for the two unknowns  ∂f/∂x, ∂f/∂y