Tuesday, February 28, 2017

11.6#5

I am also stuck on part 2 of this one















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You're almost right: you just made one little sign error.  You correctly realize that the maximum rate of change is the same as the magnitude of the gradient, and I *think* you correctly realize that the direction of the maximum increase is the unit vector in the direction of the gradient, but it looks like you incorrectly compute that the partial derivative ∂(y/z)/∂z = y/z^2 to get the gradient as
j + 5 k instead of the correct ∂(y/z)/∂z = - y/z^2 to get the correct gradient j - 5 k

Friday, February 24, 2017

10.3#8

Hi Dr. Taylor,

Will you remind me again what the notation means for E and F?

Please and thank you.











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for example f_{xy} means ∂/∂x ∂f/∂y or the derivative with respect to x of the derivative with respect to y of f(x,y). In other words, first take the derivative of f with respect to y, then take the derivative of that with respect to x.






Friday, February 17, 2017

11.2 #3

Hey Dr. Taylor, 

I did ok on WW 11.2 but I was wondering if you have any resources like khan academy videos that I could view to ensure I really know the material. I had a hard time with when y=mx or y=x with the limits and I am not sure what to look up to clarify it.

I'll insert a screenshot of one of the problems for reference. 

Thank you,














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I don't know of a specific Khan Academy or other online video that deals with this, but if anyone comes up with it I'll post the link.  This specific problem is actually not that bad though, once you apply the idea of substitution: for example if y=mx, then you replace y by mx in the above limit to get rid of the y-variable, from which you get:




11.2#6


Hello professor Taylor, 
I am sending you the picture of my work for problem 6 of 11.2 webwork.  I have simpliefed the expression after using polar coordinates but I don't know what next after there!  Need your help!

Thanks, 



















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What you do after is think this way:  
"Well, now I am taking the limit of a function that is the product of two factors, and some junk that depends on θ.  The factor is interesting because it just measures how far (x,y) is from (0,0), and since (x,y)-->(0,0) that means that the limit of r is just zero.  At the same time, the junk involving θ, although it doesn't really have a limit, it is staying bounded above and below: since -1 ≤ cos(θ) ≤ 1 and -1 ≤ sin(θ) ≤ 1 we get   
-10 ≤ cos^3(θ) + 9 sin^3(θ) ≤ 10.
This means that  -10 r ≤ r(cos^3(θ) + 9 sin^3(θ) )≤ 10 r.   Since both 10 r and -10 r have zero as a limit, the limit of my function must be zero too!"

President's day


On Fri, Feb 17, 2017 at 5:28 PM,  ********* wrote:
Hello Mr. Taylor,

I have noticed that Monday is President's Day and wondering if there is still class or not. The syllabus says there is, but I just want to make sure.

Sincerely,
*********

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Yes, class the same as always on Monday

Friday, February 10, 2017

Section 11.1 resched


Is it correct that the homework for section 11.1 is due today? I ask
because we have not yet lectured on it.


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Good point. I pushed it back a week.

Friday, February 3, 2017

Quiz on Monday, practice tests, extra pretest problem session

1) You should notice that there is material we covered this week that will be incorporated in the final exam next week--even though the homework on the subject isn't due until next Friday.  To encourage you to engage with this material, I am announcing that there will be a 10 minute quiz on Monday on topics that we covered this week, namely vector functions and things that can be done with them.

2) Practice exams and review questions can be accessed at this link.  After the quiz we will be having a review session, so work the practice exams and come with your questions.

3) One of the mat267 instructors will be running a problem session on Monday afternoon:
MAT267 Instructors,

Apparently there is no class scheduled in the same room prior to my Monday Wednesday class in WXLR A118.  Therefore, I chose to switch 2 hours of office hours for 2 hours of problem session work immediately before their class from 3:30 to 4:30, (our class starts at 4:35).

Feel free to let your students know.  I will be there simply to work problems on the board.  Usually these problem sessions only consist of a few students and thus fire code issues due to occupancy are minimal.  If it becomes excessive we can take action at that point.