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)

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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.

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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,

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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'



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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.

Practice tests...

I have a lot of requests asking for my extra practice test.  What I don't have is anyone asking me questions about the practice tests already available to through the syllabus.  It seems improbable that people are actually working on that material and *not* coming up with questions, so that I'm questioning whether you people are actually doing that  If you're not, my extra practice test won't help you and there is no use in me pulling it together for you.  Just in case you lost the syllabus link, here is the link to the practice exams:  https://math.asu.edu/mat267

10.7#6

So I am confused on how to tackle this problem. Since the function starts with x = t I put the Z function with respect to x by substituting y for 5x^2 and then replaced x with t. I don't know what I'm doing wrong.

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You're making it too hard. First of all, as they tell you, r(t)=< t, some stuff, some other stuff >.  So right away you know that x(t)=t in your parameterization.  Then since y = 5x^2, you know that y=5t^2.  You almost had the right notion since 5t^2 is what you used for z(t).  Then the paraboloid gives you z in terms of x and y and you alread have expressions for those in terms of t, so you just need to substitute those.  Where do you think you should put that? 

Calculators, exams, quizzes

1)  This is an engineering calculus class. You should plan to give a numerical answer to all quiz and exam problems unless instructed otherwise. Come prepared to do so. You should expect to get less than full credit for less than a numerical answer, unless instructed otherwise.

2)  Quiz and exam problems should be presented legibly. If I can't easily read your answers, I won't give you credit for them.

3) I grade in red ink.  If I have to get up to get a different color pen because you wrote your exam problem in red the score I will give you is 0 (zero).

Lecture Notes 9/5/18 and 9/7/18

Lecture Notes 9/5/18

Lecture Notes 9/7/18

10.9#8

I assume this attached the list of all of the answers that I've put in so far. Here's the rest of what was in the email:

For part B, I was able to solve the magnitude of the force by using rotational formulas from the physics class I took last semester. I found that to be 17.55 N.

Back to part A, looking in the textbook, I found what I believe the relevant equation to be:
F(t) = - m * w^2 * (a * cos(w * t) i + a * sin(w * t) j)

I have tried plugging in everything, leaving t in the expression, and I also tried plugging in 12 for t. WeBWork says I'm wrong for both. Here's my work for the x-component:
w = 2pi / 12 = pi/6
a = w^2 / r
F(t) = - 8 * (pi/6)^2 * ((pi/6)^2 / 8) * cos(pi/6 * t)
F(t) = - (pi/6)^2 * (pi/6)^2 * cos(pi/6 * t)
F(t) = - (pi/6)^4 * cos(pi/6 * t)
Is there something else I'm supposed to do after this point?

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(nope, it doesn't attach everything you've done, I just get to see that last thing you did)

Ok, looks like you started off sort of in the right direction, in that you start with a correct vector equation for F=-mw^2(a*cos(wt) i + a sin(wt)j),  but then a) got yourself tangled up figuring out what to put in for w and a (and it looks like you kind of got some of that stuff correct but were insecure about it and kept going when you should have stopped) and then b) you forgot that F is a vector and only kept the i component.

And if not for those confusions you could have worked it out that way.  But I propose that you take a step back and remember Newton's Law, which is covered in the textbook and says that F=ma; note that these are vector equations. And we talked in class about the fact that the acceleration vector a is the second derivative of the position r.  So the basic thing for you to know is: what does it take for you to make r(t) travel around a circle of radius 8 in 12 seconds?  (the w=2pi/12 you have is correct and plays into it).  Once you have that, take the derivative of it twice correctly, using the chain rule of differentiation, and multiply by the mass to get the force. Note that the force will have two nonzero components, and that they will both depend on time through the parameterization r(t)  you  used.

10.5#19

Dear professor,

For Question 19, it asked to "Find an equation of the plane consisting of all points that are equidistant from A(4,−5,−4) and B(0,−4,−4)."

What I tried to do was first find the mid-point between the two points, which I got (2, -9/2, -4). Then I tried to find the vector from A to B, which i got <-4,1,0>. Then I used equation for a plane <-4,1,0>dot<x-2,y+(9/2), z+4>. I got -4x +y +8 + (9/2) = 0. I move the 12.5 to the other side. -4x +y = -12.5.

Professor i don't know what i'm doing wrong. Is this not the formula to solve for this?

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Notice the minus sign that you have in your email that you don't have in the answer you entered?

10.5#17

Dr. Taylor

So i did my dot product and got -26 my magnitude that was calculated were 24 and 20 respectively in the order received from the problem. for theta = arccos (-26/sqrt(24*20)) gives me an error why is that ?

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Ok, I believe we talked about this in class today but here goes.  Your computation is correct--sort of--but also sort of incorrect.  Here's a picture of two lines intersecting:

This intersection makes some angles, I've drawn two of them, one in green, the other in blue.  Which of the angles pictured is the angle between the two lines?  Your calculation gives the other one. 

10.5#10

Hello, Dr. Taylor.
I have repeatedly crunched numbers for this problem but it has not been working out.

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1) You're not asking a question. Often the reason for this is that people don't know enough to ask a question, so I'm a little suspicious that *maybe* you haven't done your diligence.  So, this is just a hunch, but I'm guessing if you read the text book section 10.5 things will be a lot more clear.

2) When you write back to me to tell me *how* you crunched your numbers, I can get lot more helpful, and I'll know that you're prepared to learn how to do this problem.  At that point I'll update this blog post with additional information.

10.4: Problem 9 (updated)

Dr.Taylor
In class we talked about finding the volume of a paralelipiped for a couple minutes and I tried to find the scalar triple product but the answer I got was wrong and I don't know if it has to do with the ordering of the vectors or if it doesn't matter. Thank you for any help you can give.

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Well, it's hard to help without doing the problem for you, because you didn't tell me what you did, your didn't tell me what you tried to do, not to mention also that you didn't actually ask a question. However, taking a stab in the dark, you might want to pay attention to the fact that the scalar triple product is the scalar triple product, while you have four points. What do you make of that?

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I read the blog post and I realize where I went wrong with my wording. I'm taking the scalar triple product of the 3 vectors PQ, PR, and PS in that order. I've tried using different orders as well but none of them seem to work. I wanted to know if the order of the vectors matters and if it does, how do I know which order to put them in.
Sincerely,
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Ok, got it. It sounds like you mostly know what you're doing.  The order matters only in the sign of the determinant.  And since you're taking the absolute value of the scalar triple product in order to get a positive number for the volume,  remember, the order doesn't affect your answer at all.  So...a few thoughts:
1) if, as it sounds like, you're getting different answers for different orders you're doing one of the operations wrong, either the dot product or the cross product but you might also have made a mistake in computing the three vectors PQ, PR and PS.
2) the true answer is smaller than either of the two answers I've seen you try.
3) there are a lot of minus signs in those vectors, a common source of arithmetic mistakes is forgetting that two minuses is a plus. I'd guess that the mistake is most likely in the cross product (but I wouldn't bet even money on it).

PS: if you wrote up your computations LEGIBLY you could take a photo with your phone and email it to me