Java...java....java...

Hello Professor, problems numbered 13 and 15 have java plugins that were
not supported by both my PC and Mac. Just wanted to let you know, I have
looked for a solution and will continue to.

As of September 1st, Google Chrome no longer supports Java due to HTML5's
restrictions. Will this be fixed for future problems? 

Hi for question 13 I cannot load the four images.  It shows up with a
message that says plug in not supported. I ended up just guessing to try
and get partial credit. If i cannot get any credit for this i understand,
just though i would let you know about the possible problem other people
might be having too. It may just be because I am running linux but i am not
sure.

Thanks,

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Ok, here is the deal.  There is no plugin, there is only a lonely Java applet.  You need to use a browser that supports Java.  Note that the Microsoft Edge browser also does not support Java, but Internet Explorer does, and you can access it directly from Edge.  Safari supports Java.  The Firefox browser also does support Java and is available for pretty much any operating system; I suggest you download it.   You also need to enable Java on your machine and on your browser.   Here is a link to describe how to do that for Firefox.  You may need to install Java.  Here is a link describing how to do that for OS X.  Note that the reason you're having problems is that lots of people wanted to stop Java from automatic implementation for *security reasons*.  I suggest once you are done with these problems you disable Java again.

11.1 Problem9

I arrived at the right answer however I have no idea why those are the
right answers. Can you go through the steps to solve this?

Each level curve is the solution of an equation  f(x,y)=A,B,C,D.  The curve for f(x,y)=A is the easiest, because it *looks* like A must be equal f(0,0)=140e^0-70*0^2=140, and this is a multiple of 10.  For f(x,y)=B the problem is a little more complicated--it looks like f(x,0)=B for an x close to 0.4--and we need B to be an integer multiple of 10. Since y=0 on the x-axis and increasing values of x mean increasing values of f, it looks like B must be equal to 140e^0.4, but this is equal to 208.86 which is not a multiple of 10.  The nearest multiple of 10, though is 210, which means that x=ln(210/140)=0.405, which is close enough to what we see

10.9 prob 8

Hi,
I had a question about chapter 10.9 number 8. I don't know how to find
the components of centripetal force.I'm hoping you could guide me on how
to
solve this problem.

OK....

The first thing you need to know is the definition of centripetal force--otherwise you will have no idea how to do this problem. This is actually in the textbook.  Read the part of 10.9 that covers this--you can look for the term in the index of the book to get the exact page number. Or you could look at online resources like Wikipedia (which is accurate much more often than it's not)

Second, you should know that centripetal force IS a force.  But the Newton's Law F=ma relates the force to the acceleration. This means that to get the force you're going to need to get the acceleration, which means you need to use the information they give you about the size of the circle and the speed of rotation to get a parameterization of the motion that represents the position of the path vs time accurately.  And then from that you can compute the acceleration.

Let me know how it goes.  

webwork 10.9 no 7

Hello Professor Taylor, can you help explain this problem to me I cannot
get the correct answer. I keep getting aprox. 15.7m/s

This is almost exactly the same as problem 6, which is discussed in a previous blog post.  The only difference here is that the angle is 45˚ instead of 30˚, and they give you that y(t)=0 when x(t)=25, and ask you to find v_0 instead of the other way around.  Try the problem with perspective in mind, and let me know how it goes.

your direction (grades, etc)

Below are: your midterm 1 score, your homework % score, and the grade I would assign right now, if I had to assign a grade based on course policies and your performance to date.  If you keep going the way you are you're likely to end up with approximately this grade.  I haven't included +'s or -'s, because there is not enough information yet to justify these

webwork 10.9 question 6

I had a question about chapter 10.9 number 6. Ive tried multiple tries
myself and I have even gone to tutoring centers and they say the work is
right but webwork marks it as wrong. I am hoping you could guide me on how
to solve this problem.

The acceleration due to gravity is a=<0,-9.8>.  The Sin(30˚)=1/2, Cos(30˚)=√3/2, so the initial velocity vector is v_0=250<√3/2,1/2>=<125√3,125>. Since a(t)=v'(t), and a(t)=a is constant, it follows that

v(t)=a*t+v_0=<125√3,125-9.8*t>.  

Since v(t)=r'(t), it follows that

r(t)=a*t^2/2+v_0*t+r_0.  

But since the projectile is fired from the ground, and it's a reasonable thing to make our coordinate system so that the origin is at (0,0), we can assume that r_0=0, hence

  r(t)=a*t^2/2+v_0*t=<125√3*t,125*t-9.8*t^2/2>.

Writing r(t)=<x(t),y(t)>, the projectile will be at ground level when y(t)=0, i.e. when 

0=125*t-9.8*t^2/2=t(125-9.8*t/2).

This equation has two solutions: t=0, before the projectile is fired, and t=250/9.8=25.51 when the particle hits the ground again.  At that time x(t)=125√3*25.51=5523.1 meters, so the range is 5523.1. At the time the particle hits the ground, the speed will be the magnitude of the vector 

||v(25.51)||=||<125√3,125-9.8*t>||=||<125√3,125-9.8*25.51>||=||<125√3,-125>||=250.

The maximum height will be attained with the vertical component of the velocity is zero, i.e. when 

0=125-9.8*t  or t=125/9.8=12.76.

At this time the height is y(12.76)=125*12.76-9.8*(12.76^2)/2=796.6 meters. 

Exam 1 scores (updated)

an image from lecture today

attendance as of this week:9/14 and 9/16

Lecture Notes this week 9/14/14 & 9/16/15

Lecture Notes 9/14/15

Lecture Notes 9/16/15

REVIEW SESSION FOR EXAM TONIGHT!

In WXLR 116 (PSA116) at 8:15PM

Cumulative Attendance As Of Today, 9/11/15

 Click on the image to zoom in

Lecture Notes This Week 9/9/15 & 9/11/15

Lecture Notes 9/9/15

Lecture Notes 9/11/15

Exam next week! (and related matters) Part I

Friday 9/18.  Be there.

**********************

Hello,

I was hoping there would be some kind of review session or study
guide or anything of the sort before our tests in this class? I am
struggling to get through the homeworks and I often don't even understand
what I'm doing. I do my best to pay attention to lectures, but talking or
theory with hypothetical numbers doesn't help me personally. I was just
hoping there would be something more to help me prepare for the upcoming
tests.
*******

**********************

1)  "I was hoping there would be some kind of review session or study guide..."

Yes, there are. There are a number of reviews and study guides, which you can access at this webpage (this link has also been available on the syllabus since the first day of class, btw).  I suggest that you give yourself a timed 50 minutes to take the practice test, and then afterward review the answers there.  There is also an old exam there.  (I shouldn't need to give this disclaimer but I do: the practice exam and old exam are representative of the sort of problems you may seen on the actual exam, but in no way are meant to imply that you will see exactly those problems on the exam.  In fact very suredly you won't see exactly those problems.)

I believe that next week there will be an evening review session hosted by the Engineering Tutoring Centers for the exam also, but haven't been notified about the particulars yet--I'll post the information here on the blog as soon as I get it.  Also, and as always, I'm available by office hours and by appointment.

(I've got more to say on this, but later)

Office hour cancelled today 9/8/15

 I am canceling my office hour today due to automotive problems

[WWfeedback]:Taylor_MAT_267_Fall_2015:Section_10.5 prob:21

Hi Professor Taylor,
I am attempting problem number 21 and have come across trouble on part C. To solve the problem, I combined the vectors from part A and B to get the vector for part C and I got the right answer for the second coordinate but not for the other two coordinates so I'm not sure what I'm doing wrong. Thank you in advance for your help!
****************

Consider the planes 5x+5 y+5 z=1 and 5 x+5 z=0.


A)Well, P is on the y-axis so x=z=0, which means that P is in the second plane(why??), and in the first plane means that 5y=1 so y=1/5--it sounds like you figured this out.
B) A unit vector parallel to both planes will be perpendicular to both normals (what *are* the normal vectors to each plane, and how do you use them?), which turns out to be something parallel to

±(i+0j-k),

 so a unit vector that's has positive first coordinate would be (i+0j-k)/√2.--it sounds like you got this part too.
C) there are a lot of vector equations that will do the job for this part, but the simplest one for you to write would be r(t)=<0,1/5,0> + t<1/√2,0,-1/√2>=(t/√2i+1/5j-t/√2k).

[WWfeedback]:Taylor_MAT_267_Fall_2015:Section_10.3 prob:9

Hello Mr. Taylor:
     I have been having problems with this:
"Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (2, 0)
and arrived in the Iron Hills at the point with coordinates (3, 4). If he began walking in the direction of the vector v=5i+2j  and changes direction only once, when he turns at a right angle, what are the coordinates of the point where he makes the turn. "
     I think that it is a vector addition problem (the parallelogram rule?). However, I have no idea how to begin solving it. I know that I have one out of the two vectors, but I do not know how to start
Thank you!
     ***********

(Apologies for the late reply, I already answered this once, but it looks like I didn't save correctly.)  Sure, this is a vector addition problem, but it's also a problem about the dot product (this is section 10.3 after all).  So let's rephrase the problem:  you start at the point (2,0), then you need to move in the direction of the vector 5i+2j some unknown amount, and then there is another vector xi+yj which you don't know except that it is orthogonal to 5i+2j (the word "orthogonal" is the clue that tells you need to use the dot product somehow), and then go along that perpendicular direction for another unknown amount until you end at the point (3,4).  So here's:
RECIPE
1)Figure the total displacement vector by subtracting the initial point from the final point.
2) Use the dot product to figure out the projection of the displacement vector on the vector 5i+2j.  This will tell you how far you need to go along the direction 5i+2j.
3) subtract the projection from the displacement; this is perpendicular to 5i+2j (how would you check this?).
4) the initial point plus the projection of the displacement is the point you need (why?)

Homework New Assignment and Extension This Week (due to blackout)

1) The new homework assignments are due next Friday 9/11/15 at midnight and are available on webwork at the following links:
Section 10.4,  Section 10.5 and Section 10.6.

2) The due date for HW sections 10.2 and 10.3, which were due tomorrow at midnight has been extended to Saturday 9/5/15 at midnight (because some people who lost an evening due to blackout at residence hall feel the extra time might be important)

Lecture Notes this Week (8/31/15-9/4/15)

Lecture Notes 8/31/15

Lecture Notes 9/2/15

Lecture Notes 9/4/15

Your posting ID (and how to read your grades etc posted here.)

When I post your grades and attendance I use your Posting ID; this is not the same as your ASU ID.  Please read this page for an understanding of your various IDs.