11.3#11 and #12

#11  Dear Professor Taylor,
I have no idea if it is just me, but for some reason I was completely unable to solve this problem even after consulting the book and lecture notes. I don't believe this concept was yet covered in class or know if it will be. Please help me escape the land of not knowing.
Thanks,

AND

#12  Dear Professor Taylor,
Similar to the last problem, for some reason I was completely unable to solve this problem as well and for the same reasons. I assume that this problem is related to the last as it asks for some sign in relation to some partial derivative. Again, I would like some help in understanding this concept.
Thanks,

(lets talk about #11 because #12 uses the same thinking)

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

OK, so you are supposed to be familiar with these concepts from Calculus 1 :

1) derivatives are approximated by difference quotients, 

2) difference quotients are a ratio [f(x_2)-f(x_1)]/[x_2-x_1]

3) you can read the values x_1, f(x_1), x_2, f(x_2) from the graph of the function.

If you are not familiar with or don't remember these concepts you need to pick them up, either reviewing on your own or in the tutoring center or in my office hours. 

The new concepts you should have picked up in lecture, where we did indeed talk about them, are that 

1) partial derivatives of two variable functions are approximated by directional difference quotients; the x-direction for f_x = ∂f/∂x and the y-direction for f_y = ∂f∂y.

2) that contours of a function in two dimensions, aka the level curves of the function, describe values taken by the function at different places 

While you can eyeball the value of the function anywhere in the square 0≤x≤10, 0≤y≤10 from the above contours the easiest place to get the value of the function is right on the contour.  So looking at the point (3,5), we see that the contour of 14 passes right through that point, so we know that f(3,5)=14.  Now we need to find a point (x,y) that still has y=5 but x≠3,  for which preferably it's easy to figure out the value of f(x,5).   And sure enough, looking on the line y=5 we see that the contour for 16 passes almost right through the point (6,5)! 
Therefore, I can tell you that ∂f/∂x is approximately equal to 

[f(6,5)-f(3,5)]/[6-3] = [16-14]/[6-3] = 2/3

By the same token if you had wanted to compute the ∂f/∂y at (3,5) you would have wanted to look for values of f(x,y) on the line x=3 to compute your difference quotient.

Announcement: Kaust Undergraduate Poster Competition

Dear Colleagues,      

King Abdullah University of Science and Technology (KAUST), Saudi Arabia, is pleased to announce the launch of the Eighth Annual Undergraduate Poster Competition.

This event is open to undergraduate students from around the world and is being held as part of the KAUST Winter Enrichment Program (WEP) for January 2019. WEP is a 12-days innovative interlude program created to inspire and broaden the student and community experience at KAUST with keynotes, workshops, lectures, and events that go beyond the traditional academic curriculum.

 

A selection of past keynote speakers includes Dirk Ahlborn, CEO at Hyperloop; Carolyn Porco, Imaging Science Team Leader of the NASA Cassini Mission, artist Theo Jansen, and Kip Thorne, Nobel Laureate in Physics, Caltech.

KAUST invites your students to enter this competition by submitting an abstract of their original work before November 10th 2018. For more details, please refer to submission guidelines at:

http://studentopportunities.kaust.edu.sa/WEP/

Authors of the best 50 abstracts will be invited to spend a week at KAUST to present their research and participate in WEP from 20-24 January 2019. The best presentations will be recognized in a final awards ceremony with prizes.

Kindly promote our event to your undergraduates. For any queries, please email: weppostersession@kaust.edu.sa

To stay tuned, visit the Enrichment Office’s website and make sure you follow them on Facebook, Twitter, and Instagram.

We look forward to receiving inspiring, creative research abstracts from your students.

Valerio Orlando
Professor, Bioscience
Biological and Environmental Science and Engineering Division
King Abdullah University of Science and Technology

Thuwal, Saudi Arabia

Marie-Laure Boulot, Enrichment Programs Director: marielaure.boulot@kaust.edu.sa

Lecture Notes 9/17/18 through 9/21/18

Lecture Notes 9/17/18

Lecture Notes 9/19/18

Lecture Notes 9/21/18

Exam Scores

Average=68.6875, Standard deviation= 20.016, 9 A's, 7 B's, 9 C's, 6 D's and 17 E's

The relationship between homework score and exam score

PracticeTest1 problem6

I have no clue how to approach problem 6. i got sqrt(21)/sqrt(5) but using the formula |axb|/|a|
but I'm not sure if that's right.

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

Do you mean this problem?

You might take a look at Problem 45 on page 565 of the textbook, which validates your approach if you're willing to believe the result of the problem without actually doing it.  Or, for P=(1,1,1), Q=(2,1,3),  R=(4,2,3), you could compute it as we've discussed extensively in class  (for example see the lecture notes for 8/31/18) as

||PR -  (PR∙PQ)/(PQ∙PQ) PQ|| 

Test 1 Review Question MAT 267

Dr. Taylor,

The question asks to find the length of the curve r(t) = <2+3t, 1-4t, -4+3t> from (5,-3,-3) to (20,-23,14). I took the derivative of r(t) and then found its magnitude. I then took the integral of ||r'(t)|| and got sqrt(34)t. What point do i plug in for t? In the back of the packet are the answers and it says the answer is 5sqrt(34). Do you always just plug in the first point given?

Thank you,

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

Well, you know you need to compute an integral, but it has to be a definite integral ∫_a^b ||r'(t)|| dt, and you have discovered that real (and only) trick that is required is to figure out the limits of integration a and b.  You can figure this out from the clause "from (5,-3,-3) to (20,-23,14)"; this means that at the start the curve is at (5,-3,-3), which says that a is the solution of  

<5,-3,-3>= <2+3t, 1-4t, -4+3t>

which gives you that t=1.   Then at the end the curve is at the point (20, -23, 14), which gives the equation <20,-23,14>= <2+3t, 1-4t, -4+3t> has the solution t=6, so your integral is from 1 to 6.

10.7#20

Dear Dr. Taylor,

The problem asked to find the solution r(t) of the differential equation with the given initial condition: r′(t)=⟨sin6t,sin4t,3t⟩,r(0)=⟨3,8,8⟩

so I took the antiderivatives of each component and got
r(t)=⟨3-1/6cos(6t),8-1/4cos(4t),3/2t^2+8⟩ but it says that's wrong. when I take the derivate of that answer I get the r'(t) because the derivative of -cos(u) is sin(u)*u'



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

It looks like you figured it out in the meantime, and clearly you figured out that r(t)=∫r'(t)dt + constant vector.  The problem with your original answer, as you now know, is that it didn't account for the nonzero value of  ∫r'(t)dt at t=0, so you needed to correct for that in the constant vector you added.