Lecture Notes 9/25/17 through 9/29/17 (complete)

Lecture Notes 9/25/17


Lecture Notes 9/27/17


Lecture Notes 9/29/17

11.3#9

Dear Dr. Taylor,
I have been working on this problem and I am a little confused why the answers are positive instead of negative. When I took the derivative of the function I got (dz/dx)=-3cos(-3x-5y+z) and (dz/dy)=-5cos(-3x-5y+z). From here I plugged in the point (0,0,0) and received the answer -3,-5.

Thanks

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well, I'm not sure but I guess you're thinking of the equation z = sin(-3x-5y) instead of the equation that's really there. If you think about the real one, you have one equation and three variables, so you can think of one of the variables as a function of the other two--in this case you should think of z = z(x,y).   Then you can find the partial derivatives of both sides of the equation with respect to x and y; the partials of the right hand side are both zero, while the partials of the left side can be  computed using the chain rule--and the part of the chain rule that involves multiplying by the derivatives of the argument will include multiplying by ∂z/∂x or by ∂z/∂y.  (BTW, one follow on effect of this is that it will change the sign of your derivatives.)

11.1#10

Hey Dr. Taylor so I emailed you once about this already but you didn't respond so I'm nagging you. I only got this question correct because I google'd it, I honestly have no idea how to do it other than the (0,0) point, which I got by plugging in 0 and 0 for x and y.

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You *googled it*?? How did you do that?  Good job on the nagging btw.
So: yes, as you surmise, plugging in (0,0) will give the the value of 140 on contour "A".  My eye tells me that "B" hits the x-axis at approximately 0.4.  If I plug in x=0.4 and y = 0 into f(x,y) I get f(0.4, 0) = 208.85, to which the closest multiple of 10 is  210,  so that must be the value of "B". This gives me the gap between the contours and so I'm basically done.

11.2#10

Hi Dr. Taylor,

I solved this question with a math tutor, and we weren't positive why the answer was choice B, I feel choice E should be the answer since one could write f(x,y) as ((inf-n, inf - n - 1)) such that the values
can be massive or equal to each other and allowed in the domain?

Any feedback regarding a proof for B being the answer would be appreciated.

Thanks,

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Well, the point is that you need the arguments of both square roots to be non-negative. For the first square root this means that x + y ≥ 0 or equivalently y ≥ -x. For the second square root this means that x - y ≥ 0  or x ≥ y.  If you just put these together you get x ≥ y ≥ -x.   Notice that this implies that
x ≥ -x; this means that x ≥ 0 (because if x < 0 , -x is positive so x ≥ -x is not true ), so the domain is the wedge between the lines y = x and y = -x with x ≥ 0:

The grade scale

I've added this section to the course syllabus

Grade Scale: This class will be graded on a qualified curve, backed up by an absolute grade scale. This means that the higher of the two grades specified by tables below, where µ denotes the class average and ß denotes the class standard deviation:

Grade	Score		Grade	Class Rank
A+	97%+		A+	r >µ+1.3ß, historically the top 4.8%
A	93-96.99%		A	µ+1.3ß ≥ r > µ + 1.1ß, historically the top 10.2%
A-	90-92.99%		A-	µ+1.1ß ≥ r > µ + 0.95 ß, historically the top 16%
B+	87-89.99%		B+	µ+0.95 ß ≥ r > µ + 0.8 ß, historically the top 21%
B	83-86.99		B	µ+0.8 ß ≥ r > µ + 0.6ß, historically the top 29%
B-	80-82.99		B-	µ+0.6 ß ≥ r > µ + 0.45ß, historically the top 36%
C+	77-79.99		C+	µ+0.45 ß ≥ r > µ + 0.3 ß, historically the top 43%
C	70-76.99		C	µ+0.3 ß ≥ r > µ - 0.2 ß, historically the top 66%
D	60-69.99%		D	µ-0.2 ß ≥ r > µ - 0.55 ß, historically the top 77%
E	<60%		E	µ-0.55 ß ≥ r  historically bottom 23% (usually primarily people who did not complete the class)

Lecture Notes 9/18/17-9/22/17

Lecture Notes 9/18/17


Lecture Notes 9/20/17


Lecture Notes 9/22/17

Scores to 9/22/17

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10.9#6

Hi, I've been attempting this problem for some time and can't seem to get
it. I've been following the book's example to the letter. I first created a
parametric model for the projectile, where x=200cos(30)t, which simplifies
to 100sqrt(3)t. My y model was 200sin30 - (1/2)(9.8)t^2, which simplifies
to 100 - 4.9t^2. Setting this to 0 and solving gives the answer
sqrt(100/4.9) = t. I plugged this value into the x equation to get the
range, and the answer was marked incorrect. Can you explain what I did
wrong?

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OK, so your first mistake: you were trying to do too many things in your head. You should write out the steps. These involve a lot of integration: your starting ingredients should be a(t) = -9.8j, and v(0)=<200cos(30), 200sin(30)> and r(0) = <0,0>. Then you need to integrate twice using the fundamental theorem of calculus each time and substituting in v(0) and r(0) as appropriate.  From your answer it looks like you forgot to include the effect of v(0) in integrating out to r(t). 

yes, 11.1 hw due date changed

Have we covered 11.1 yet? The homework on web work for that section is due tomorrow. 

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Not any more.

Practice Exam Q5 (Fall 2015)

Dear Professor Taylor,

I am having trouble finishing this problem. This is what I have done so far.

I think I have the y and z components correct for the vector, but I'm having trouble finding the x component.

Thank you,

(please click on image to enlarge)

Thanks for showing me what you did, it really helps I think

10.9#1

Hi Professor,

I am struggling to find the speed of the particle. As you
can see, I have attempted this problem multiple times. Any help would be
great! Thanks.

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1) OK, first of all, let's go through this again.  How much you are struggling or how many times you attempted the problem, while it may be a sign that you are working hard, is also a sign that you are not working well. When you attempt a problem and do not get the right answer,  that is a sign that you should try something different. Just trying the same thing over again will not get you a different answer, it just serves to tire you out.

2) Second, I am much better able to help you learn if I know specifically what you already tried, so I can point out exactly where you made your mistake instead of guessing based on your wrong answer. Please tell me what you tried when you ask the question.

3)  Finally, as it looks like you understand, the speed is the square root of the sum of the squares of the components of the velocity.  You have the z-component squared correctly, but it looks like you did FOIL incorrectly with the sum of the squares of the x- and y-components: in particular you'll get another t^2 term and some t cos(t) sin(t) terms that don't add up the way you have them

10.8#3

Hello Mr. Taylor!
I am doing the integral from 0 to 3 of the magnitude of
the derivative, but I'm not sure how to do it due to 0 being in the
denominator for one of the parts. What do I need to do different?
Thanks!




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Ehr....OK, I guess you're deducing the wrong limits of integration from those two endpoints of the curve, (12,6,0) and (36,54,6 ln(3)).  You see, t=0 is not what you need to plug into r(t) to get (12,6,0) because that says that 6 ln(0) = 0, when really it is not even defined. What *is* true is that ln(1)=0.  So the real limits of integration should be from t=1 to t=3.

Exam Review Links

My old test#1 from 2015

ASU exam review page

Lecture Notes 9/6/17 and 9/8/17 (Complete)

Lecture Notes 9/6/17

Lecture Notes 9/8/17

10.7#1

Hi,

Just wondering why in letter C) the answer is not (-inf,-5)U(-5,5)U(5,inf) but only (-inf,-5)U(5,inf). Thanks

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Well, because for -5 < t < 5, t^2-25 is negative so √(t^2-25) is not defined (no imaginary numbers are allowed in this class)

10.5#17

I know that to find the angle I have to use the formula a DOT b =
|a||b|cos(theta). For my two normal vectors I get <5,-5,1> and <2,3,-4>.
This leaves theta = cos^-1(-9/sqrt(1479)). I get 103.53 degrees or 1.807
rad. What did I do wrong?

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OK, it's true that the angle between two planes is the angle between their normal vectors, the question is which normal vector of the plane do you use, because there are two opposite directions that are normal to a given plane.  Here is a representation of the two planes, all arrows are perpendicular to a plane, the black arrow is perpendicular to one plane and both the red and blue arrows are perpendicular to the other plane. You can see that the blue arrow has an obtuse angle with the black arrow, i.e. more than 90degrees, which has a negative cosine, while the red arrow has an acute angle with the black arrow hence the cosine is positive.  This is reflected in the fact that the two planes have two different angles between them, one less than 90degrees and the other greater than. BY DEFINITION though, the angle between the two planes is taken to be the smaller.  SO: to compute the angle between the two planes, you get to use the normal vector that makes the dot product positive.

10.2#12

Dr. Taylor,
How can I figure out the speed that the bird must go if I only have one known speed value?













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Well, this is a problem of vector addition.  You have two known directions, "from the west" (= toward the ???) and "head northwest", and you have one magnitude "at 70mph".   You are looking for the magnitude of the other vector in such a way as to fly directly north, that is to obtain the specific direction north for the vector sum.