Friday, January 29, 2016

Gandalf...

It has come to my attention that I am not very blog-savvy.  On 10.3 #9, I
am having trouble setting up how to do the problem.  To get from point
(3,2) to (4,7) starting with a vector 5i+2j and turning once at 90 degrees
to get to that specific endpoint.  I know that in order to be orthogonal
the dot products of the 2 vectors needs to be zero....but I don't know what
2 vectors to use and then once I have the two vectors to use....do I just
subtract or add one of those vectors to the start or end point...?  Please
help!










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OK the direction of the Iron Hills from the forest of Mirkwood is <1,5>=<4,7>-<3,2>.  Good old Gandalf doesn't go that way though (because that would be too simple).  Instead he goes in the direction of the vector <5,2> for awhile (which is *not* the same direction of <1,5>), and after awhile takes a single right turn and heads in that direction and ends up at <1,5> (I emphasize the words "right turn" because that is the crucial bit of information that tells us what we need to do)

So the picture looks like this:

If you put it in this way, it should gradually become clear that this is a problem about orthogonal projection. The right pointing red arrow is the orthogonal projection of <1,5> on <5,2>, i.e.
=(<1,5>.<5,2>)/(<5,2>.<5,2>)<5,2>=15/29<5,2>=<75/29,30/29>
This is the point where he makes his turn.
The upward pointing red arrow is the difference or residual <1,5>-<75/29,30/29>=<-46/29,115/29>.

Wednesday, January 27, 2016

Monday, January 25, 2016

Homework due this week, plus wondering...

Section 10.3

Section 10.4


Should I give a quiz on Friday? The ideas in these sections will be used all semester long...

Friday, January 22, 2016

Problem 10.2#17

 Hello Dr. Taylor,
I am can't figure out what to multiply the first part of
number 17 by to get half its length and it in the opposite direction. I got
the unit vector but can't figure out the proper scalar please help.








(click on image to enlarge)
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Well, for part 1 you're doing good things with dividing by the length to find a unit vector, but you're making it harder than it needs to be.  To get a vector half the length of u, you just have to multiply it by 1/2.  And to get it going in the opposite direction you just have to multiply it by -1. You *don't* need to divide by the length of u.

For part 2, though, you need to get a vector of exactly length 3 pointing in the direction of v, so you're doing exactly the right thing by computing the unit vector v/|v| and then multiplying by 3

Tuesday, January 19, 2016

Friday, January 15, 2016

Office Hours Posted

Course office hours are posted on the online syllabus and here. They will be:

Tues 12:30-1:30PM
Wed 11:30-12:30PM
Thurs 11:30-12:30PM

Announcement: Homework and Quiz

The homework for this week is now open and will be due a week from today at midnight.
Section 10.1
Section 10.2

Also, there will be a quiz next Friday, 1/22/16.

Monday, January 11, 2016

Welcome & First Homework Assignement

Hi! Welcome.
1) The syllabus has been updated--the tentative schedule is correct, but the actual sequence and timing of material may still be subject to small variations.
2) The first homework assignment will cover sections 10.1 and 10.2.  The webwork site for this course is still not online, however this is not a problem because the homework won't open until January 15 at Noon and will be due January 22 at midnight.