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# Fourth Annual Large Hadron Collider Physics Conference 2016
12-18 June 2016
Lund University
Europe/Zurich timezone
REGISTRATION FOR LHCP2016 IS NOW CLOSED
## First LHCb results from pA and PbPb collisions
Not scheduled
20m
Lund University
#### Lund University
Sandgatan 2, Lund, Sweden
Talk Physics of Heavy Ion collisions
### Speaker
Laure Marie Massacrier (Laboratoire de l'Accelerateur Lineaire (FR))
### Description
The LHCb experiment has the unique property to study heavy ion interactions in the forward and backward hemisphere in a kinematic region not accessible to the general purpose detectors, thanks to its forward acceptance 2<$\eta$<5, and the possibility to study proton-lead collisions for both orientations of the beams. Furthermore, using the possibility to inject gas into the interaction region, it is in the unique position to do also fixed target physics.
Results include measurement of prompt $D^0$ meson production in pPb collisions at LHCb, the first forward measurement of Z production in pPb collisions as well as a measurement of the nuclear modification factor and forward-backward production of prompt and displaced $J/\psi$, $\Psi(2S)$ and $\Upsilon$. Recent results and news from the Pb-Pb, Pb-Ar, proton-He, proton-Ne and proton-Ar runs will be also presented, as well as future prospects.
### Primary author
Laure Marie Massacrier (Laboratoire de l'Accelerateur Lineaire (FR))
### Presentation Materials
There are no materials yet.
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Tweet
# Common Core Standard HSG-CO.B.6 Questions
Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent.
You can create printable tests and worksheets from these questions on Common Core standard HSG-CO.B.6! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.
Grade 11 Symmetry and Transformations CCSS: HSG-CO.B.6
Which of these transformations is not rigid?
1. Translate 4 to the left and rotate $180 deg$
2. Reflect across the $x$ axis and translate down by 6
3. Rotate $90 deg$ clockwise and dilate by a factor of 2
4. Translate down 2, reflect across the $y$ axis, and rotate $180 deg$
Grade 11 Symmetry and Transformations CCSS: HSG-CO.B.6
Which of these transformations will create a shape which is not congruent with the original?
1. Rotate $90 deg$ clockwise and dilate by a factor of 2
2. Translate 4 to the left and rotate $180 deg$
3. Translate down 2, reflect across the $y$ axis, and rotate $180 deg$
4. Reflect across the $x$ axis and translate down by 6
Grade 11 Symmetry and Transformations CCSS: HSG-CO.B.6
A pentagon is translated down 10, dilated 0.8, and reflected across the $x$-axis.
1. The transformation is non-rigid, the shape changes size
2. The transformation is rigid, the shape is congruent
3. The transformation is non-rigid, the shape does not change size
4. The transformation is rigid, it only translates
Grade 11 Symmetry and Transformations CCSS: HSG-CO.B.6
Which of the following illustrates a non-rigid transformation?
1. Shooting a hockey puck
2. Growing a plant
3. Bouncing a ball
4. Jumping on a trampoline
Grade 11 Symmetry and Transformations CCSS: HSG-CO.B.6
Which of the following illustrates a rigid transformation?
1. Inflating a balloon
2. Building a house
3. Firing a gun
4. Filling a water tank
You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges.
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# Changes between Version 6 and Version 7 of research/trigTweet
Ignore:
Timestamp:
Oct 14, 2011, 12:16:42 PM (9 years ago)
Comment:
--
### Legend:
Unmodified
v6 }}} We know sin(x) is an odd function, so instead we look for a polynomial Q(x) such that P(x) = xQ(x²): We know sin(x) is an odd function, so instead we look for a polynomial Q(x) such that P(x) = xQ(x²), and we reduce the range to positive values: {{{ #!latex $\max_{x \in [-\pi/2, \pi/2]}{\big\vert\sin(x) - xQ(x^2)\big\vert} = E$ $\max_{x \in [0, \pi/2]}{\big\vert\sin(x) - xQ(x^2)\big\vert} = E$ }}} Substitute y for x² and reduce the range to positive values: Substitute y for x²: {{{ {{{ #!cpp static real myfun(real const &y) { a6 = -7.36458957326227991327065122848667046e-13; }}} === Relative error === Searching for '''relative error''' instead: {{{ #!latex $\max_{x \in [-\pi/2, \pi/2]}{\dfrac{\big\vert\sin(x) - P(x)\big\vert}{|\sin(x)|}} = E$ }}} Using the same method as for absolute error, we get: {{{ #!latex $\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})-\sqrt{y}}{y\sqrt{y}} - R(y)\bigg\rvert}{\bigg\lvert\dfrac{\sin(y)}{y\sqrt{y}}\bigg\rvert}} = E$ }}} {{{ #!cpp static real myfun(real const &y) { real x = sqrt(y); return (sin(x) - x) / (x * y); } static real myerr(real const &y) { real x = sqrt(y); return sin(x) / (x * y); } RemezSolver<6> solver; solver.Run(real::R_1 >> 400, real::R_PI_2 * real::R_PI_2, myfun, myerr, 15); }}} {{{ #!cpp a0 = -1.666666666666666587374325845020415990185e-1; a1 = +8.333333333333133768001243698120735518527e-3; a2 = -1.984126984109960366729319073763957206143e-4; a3 = +2.755731915499171528179303925040423384803e-6; a4 = -2.505209340355388148617179634180834358690e-8; a5 = +1.605725287696319345779134635418774782711e-10; a6 = -7.535968124281960435283756562793611388136e-13; }}}
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## comparing the probability fallacy and the percentage fallacy.
What is the probability that you obtain 1 six given 4 trials of rolling a six-sided dice.
$\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}?$
correct answer: $1-(1-\frac{1}{6})^4.$
What is the overall percentage discount of getting first an r% discounted followed by another s% discount?
r+s %?
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## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
$16 \pi$ or $\approx 50.265~ft^2/ft$
Consider the tangent line that contains the point $(x, c)$. The slope the tangent line to the graph of $f(x)$ at $(x, c)$ can be written as $f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}$ We use the above equation to find the rate of change of surface area with respect to the radius. Consider $S(r)=4\pi r^2$. Then at $r=2$, we have: $S'(2) =\lim\limits_{r \to 2} \dfrac{S(r)-S(2)}{r-2} \\=\lim\limits_{r \to 2} \dfrac{4\pi r^2-(4 \pi)(2^2)}{r-2}\\=\lim\limits_{r \to 2} \dfrac{4\pi (r^2-4) }{r-2}\\=\lim\limits_{r \to 2} \dfrac{ 4\pi (r-2)(r+2) }{r-2}\\=\lim\limits_{r \to 2} 4 \pi (r+2) \\=16 \pi \\ \approx 50.265~ft^2/ft$
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# Math Help - Lighting Fixtures
1. ## Lighting Fixtures
An apprentice unpacked 12 dozen fragile lighting fixtures. He was to receive $3 for each fixture that he unpacked safely and he agreed to pay$25 for every one that he broke. If the amount due him for his work was \$236, how many did he break?
2. ## Re :
I got 2 equations here .
Let x be the number of successful lightning fixtures
Let y be the number of broken lightning fixtures .
12 dozen = 144
$x+y=144$ ---- 1
$3x-25y=236$ ------ 2
Solve for y ..
3. ## ok....
I got 2 equations here .
Let x be the number of successful lightning fixtures
Let y be the number of broken lightning fixtures .
12 dozen = 144
$x+y=144$ ---- 1
$3x-25y=236$ ------ 2
Solve for y ..
I can take it from here.
Thanks
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Enter the observed irradiance and the reference flux into the Calculator. The calculator will evaluate the Apparent Magnitude.
## Apparent Magnitude Formula
M = -5*log(10) (Fx/Fx0)
Variables:
• MI is the Apparent Magnitude (())
• Fx is the observed irradiance
• Fx0 is the reference flux
To calculate Apparent Magnitude, divide the observed irradiance by the reference flux, take the log of the result, then multiply by -5.
## How to Calculate Apparent Magnitude?
The following steps outline how to calculate the Apparent Magnitude.
1. First, determine the observed irradiance.
2. Next, determine the reference flux.
3. Next, gather the formula from above = M = -5*log(10) (Fx/Fx0).
4. Finally, calculate the Apparent Magnitude.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
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# HOW MUCH MATHEMATICS SHOULD A STUDENT MEMORIZE? PART 2, INTEGRAL CALCULUS
My basic attitude towards memorization in mathematics education is to memorize the absolute minimum, but memorize that minimum perfectly. Part of a mathematics teacher’s job, in my view, is to guide students to understand what this “minimum” is, and then encourage them to memorize it, helping them to find effective means for memorization.
Effective means, in my experience, typically involve associating the fact to be memorized with a diagram, graph, or other visual representation, or using the fact repeatedly in the context of solving problems. Just staring at a fact and trying to force it into one’s mind isn’t nearly as effective, in my experience. But everyone is different, so each student must experiment with what works best.
The advice to practice facts that you wish to memorize in the context of solving problems dovetails with another slogan, which I explored in a previous post:
The more you understand, the less you have to memorize.
And now for the subject of this post.
Many years ago, when I taught calculus for the first time, I was preparing a lecture on integration of trigonometric functions. When I came to the problem of teaching students the following integral
I was unhappy with what my textbook contained by way of explanation. The current textbooks are no different … stuff tends to get copied from one book to the next, doesn’t it? Here’s what one textbook on my shelf says:
“We will also need the indefinite integral of secant.” The result is quoted, and then: “We could verify [the result] by differentiating the right side, or as follows. First we multiply the numerator and denominator by :”
The textbook goes on to say that by substituting , the result follows.
This left me (and still does leave me) profoundly unsatisfied. As George Polya advocated, the role of a mathematics teacher is to guide students to solve problems for themselves, or at least see how they might have solved them. Pulling this particular rabbit out of this particular hat does not lend itself to solving this integral by oneself. How on earth would anyone but a genius come up with this trick?
So I set out to see if I could integrate secant in a straightforward way. One strategy I teach is that if you are faced with a trigonometric integral, and can’t see anything better to do, convert the expression to a combination of sines and cosines. OK, let’s try:
If we are thinking about making a substitution, we need some expression in the numerator that is likely to be a little more complex than 1. So let’s multiply the numerator and denominator by something. What? Well, we are trying to work with just sines and cosines, so let’s multiply the numerator and denominator by either or . The former doesn’t seem to lead anywhere, but the latter does:
Bingo! Look at the denominator. It can be easily converted to sines:
Then the substitution is very natural, because the derivative of sine is sitting right there in the numerator. After the substitution, we are left with
Now we have an easily factored expression in the numerator; just use the method of partial fractions, which results in two easy integrals, then use properties of logarithms to recombine the expressions, do some algebra, and Voila! The result from the textbook is obtained.
I actually showed my class (all those many years ago) this long calculation, and was met with the question, “Do we need to memorize this proof?” My answer was that because this calculation is so long (and therefore time-consuming on a test or exam), because it is such a shaggy-dog story of integration, that it is essential to memorize the integral of secant.
As a student, I never bothered to memorize the integral of tangent, because I had practiced the idea of switching to sines and cosines, making a simple substitution, and doing a couple of lines of algebra. I practiced this often, and soon I could do the calculation in seconds. Of course, at some point the fact became embedded in my head, but it was memorized in a very natural way, without having to devote effort to it. And this is part of my point in the “memorize the minimum” slogan. I would rather have students practice integration problems than waste time memorizing things that will soon be forgotten if they are not continually used.
Of course, nowadays we have software readily available, and so the whole question of how much technique of integration one ought to teach students is in the air. My view is that some of it is still useful, in the same way that learning various algorithms (say, long division) is still important, even though calculators can divide in an instant. It is only by getting one’s hands dirty with specific calculations, and solving specific problems, that one can be an intelligent user of mathematical technology. But that is perhaps a discussion for another time.
I was interested to find out how today’s textbooks handle teaching the integral of secant, so I quickly checked what is available on my book shelf. I checked Stewart; Anton, Bivens, and Davis; Briggs and Cochrane; Smith and Minton, Thomas and Finney (an older book); Hass, Weir, and Thomas; Rogawski; Adams and Essex; Edwards and Penney. All of them do exactly the same thing that the textbook I used so long ago did, which is to multiply numerator and denominator by secant + tangent. One of them had the grace to call it “an unmotivated trick,” another called it a “clever substitution.”
I call it pedagogically unsatisfactory.
Source: QED
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CBSEtips.in
Thursday, 25 February 2021
CBSE Class 7 Maths - MCQ and Online Tests - Unit 11 - Perimeter and Area
CBSE Class 7 Maths – MCQ and Online Tests – Unit 11 – Perimeter and Area
Every year CBSE students attend Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standards’ subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.
These online tests are based on latest CBSE syllabus. While attempting these, our students can identify their weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly
CBSE Class 7 Maths – MCQ and Online Tests – Unit 11 – Perimeter and Area
Question 1.
The area of a square plot is 1600 m2. The side of the plot is
(a) 40 m
(b) 80 m
(c) 120 m
(d) 160 m
Answer
Answer: (a) 40 m
Hint:
Side = $$\sqrt{1600}$$ = 40 m
Question 2.
The area of a square is 121 m2. Its side is
(a) 11 m
(b) 21 m
(c) 31 m
(d) 41 m
Answer
Answer: (a) 11 m
Hint:
Side = $$\sqrt{121}$$ = 11 m
Question 3.
Area of a rectangle of length l and breadth b is
(a) l Ũ b
(b) l + b
(c) 2 Ũ (l + b)
(d) 6 Ũ (l + b)
Answer
Answer: (a) l Ũ b
Hint:
Formula
Question 4.
Area of a parallelogram =
(?) base Ũ height
(b) $$\frac { 1 }{ 2 }$$ Ũ base Ũ height
(c) $$\frac { 1 }{ 3 }$$ Ũ base Ũ height
(d) $$\frac { 1 }{ 4 }$$ Ũ base Ũ height
Answer
Answer: (?) base Ũ height
Hint:
Formula
Question 5.
Area of a triangle =
(?) base Ũ height
(b) $$\frac { 1 }{ 2 }$$ Ũ base Ũ height
(c) $$\frac { 1 }{ 3 }$$ Ũ base Ũ height
(d) $$\frac { 1 }{ 4 }$$ Ũ base Ũ height
Answer
Answer: (b) $$\frac { 1 }{ 2 }$$ Ũ base Ũ height
Hint:
Formula
Question 6.
The circumference of a circle of radius r is
(a) pr
(b) 2pr
(c) pr2
(d) $$\frac { 1 }{ 4 }$$ pr2
Answer
Answer: (b) 2pr
Hint:
Formula
Question 7.
The circumference of a circle of diameter d is
(a) pd
(b) 2pd
(c) $$\frac { 1 }{ 2 }$$ pd
(d) pd2
Answer
Answer: (a) pd
Hint:
Formula
Question 8.
If r and d are the radius and diameter of a circle respectively, then
(a) d = 2 r
(b) d = r
(C) d = $$\frac { 1 }{ 2 }$$ r
(d) d = r2
Answer
Answer: (a) d = 2 r
Hint:
Formula
Question 9.
The area of a circle of radius r is
(a) pr2
(b) 2pr2
(c) 2pr
(d) 4pr2
Answer
Answer: (a) pr2
Hint:
Formula
Question 10.
The area of a circle of diameter d is
(a) pd2
(b) 2pd2
(c) $$\frac { 1 }{ 4 }$$ pd2
(d) 2pd
Answer
Answer: (c) $$\frac { 1 }{ 4 }$$ pd2
Hint:
Formula
Question 11.
1 cm2 =
(a) 10 mm2
(b) 100 mm2
(c) 1000 mm2
(d) 10000 mm2
Answer
Answer: (b) 100 mm2
Hint:
Formula
Question 12.
Perimeter of a square =
(a) side Ũ side
(b) 3 Ũ side
(c) 4 Ũ side
(d) 2 Ũ side
Answer
Answer: (c) 4 Ũ side
Hint:
Formula
Question 13.
Perimeter of a rectangle of length Z and breadth 6 is
(a) l + b
(b) 2 Ũ (l + b)
(c) 3 Ũ (l + b)
(d) l Ũ b
Answer
Answer: (b) 2 Ũ (l + b)
Hint:
Formula
Question 14.
Area of a square =
(a) side Ũ side
(b) 2 Ũ side
(c) 3 Ũ side
(d) 4 Ũ side
Answer
Answer: (a) side Ũ side
Hint:
Formula
Question 15.
1 are =
(a) 10 m2
(b) 100 m2
(c) 1000 m2
(d) 10000 m2
Answer
Answer: (b) 100 m2
Hint:
Formula
Question 16.
The area of a square is 625 m2. Find its side
(a) 25 m
(b) 50 m
(c) 125 m
(d) 5 m
Answer
Answer: (a) 25 m
Hint:
Side = $$\sqrt{625}$$ = 25 m
Question 17.
The area of a rectangular field is 250 m2. If the breadth of the field is 10 m, find its length.
(a) 25 m
(b) 50 m
(c) 100 m
(d) 125 m
Answer
Answer: (a) 25 m
Hint:
Length = $$\frac{250}{10}$$ = 25 m
Question 18.
The perimeter of a rectangle is 30 m. Its length is 10 m. Its breadth is
(a) 5 m
(b) 10 m
(c) 15 m
(d) 3 m
Answer
Answer: (a) 5 m
Hint:
30 = 2 (Length + Breadth)
? 30 = 2 (10 + Breadth)
? Breadth = 5 m
Question 19.
If the area of a parallelogram is 16 cm2 and base is 8 cm, find the height.
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Answer
Answer: (b) 2 cm
Hint:
Height = $$\frac{16}{8}$$ = 2 cm
Question 20.
The area of a parallelogram is 20 cm2 and height is 2 cm. Find the corresponding base.
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 2.4 cm
Answer
Answer: (d) 2.4 cm
Hint:
Base = $$\frac{20}{2}$$ = 10 cm
Question 21.
The perimeter of a square is 48 cm. Its area is
(a) 144 cm2
(b) 12 cm2
(c) 48 cm2
(d) 100 cm2
Answer
Answer: (a) 144 cm2
Hint:
Side = $$\frac{48}{4}$$ = 12 cm
? Area = 12 Ũ 12 = 144 cm2
Question 22.
The area of a rectangular room is 150 m2. If its breadth is 10 m, then find its length.
(a) 15 m
(b) 25 m
(c) 50 m
(d) 55 m
Answer
Answer: (a) 15 m
Hint:
Length = $$\frac{150}{10}$$ = 15 m
Question 23.
A rectangular wire of length 40 cm and breadth 20 cm is bent in the shape of a square. The side of the square is
(a) 10 cm
(b) 20 cm
(c) 30 cm
(d) 40 cm
Answer
Answer: (c) 30 cm
Hint:
Side of square = $$\frac{2(40+20)}{4}$$ = 30 cm
Question 24.
The area of a parallelogram of base 5 cm and height 3.2 cm is
(a) 8 cm2
(b) 12 cm2
(c) 16 cm2
(d) 20 cm2
Answer
Answer: (c) 16 cm2
Hint:
Area = 5 Ũ 3.2 = 16 cm2
Question 25.
The diameter of a circle is 14 cm. Find its circumference
(a) 44 cm
(b) 22 cm
(c) 11 cm
(d) 55 cm
Answer
Answer: (a) 44 cm
Hint:
Circumference = $$\frac{22}{7}$$ Ũ 14 = 44 cm
Question 26.
The radius of a circle is 7 cm. Find its area
(a) 154 cm2
(b) 77 cm2
(c) 11 cm2
(d) 22 cm2
Answer
Answer: (a) 154 cm2
Hint:
Area = $$\frac{22}{7}$$ Ũ 7 Ũ 7 = 154 cm2
Question 27.
Find the area of the following ; parallelogram:
(a) 12 cm2
(b) 6 cm2
(c) 24 cm2
(d) 8 cm2
Answer
Answer: (a) 12 cm2
Hint:
Area = 6 Ũ 2 = 12 cm2
Question 28.
Find the area of the following parallelogram:
(a) 6 cm2
(b) 12 cm2
(c) 16 cm2
(d) 9 cm2
Answer
Answer: (b) 12 cm2
Hint:
Area = 4 Ũ 3 = 12 cm2
Question 29.
The area of the parallelogram ABCD in which AB = 6.2 cm and the perpendicular from C on AB is 5 cm is
(a) 30 cm2
(b) 29 cm2
(c) 28 cm2
(d) 31 cm2
Answer
Answer: (d) 31 cm2
Hint:
Area = 6.2 Ũ 5 = 31 cm2
Question 30.
One of the sides and the corresponding height of a parallelogram are 3 cm and 1 cm respectively. The area of the parallelogram is
(a) 1 cm2
(b) 3 cm2
(c) 6cm2
(d) 12 cm2
Answer
Answer: (b) 3 cm2
Hint:
Area = 3 Ũ 1 = 3 cm2
Question 31.
The diameter of a circle is 7 cm. Find its area
(a) 154 cm2
(b) 38.5 cm2
(c) 22 cm2
(d) 11 cm2
Answer
Answer: (b) 38.5 cm2
Hint:
Area = $$\frac{22}{7}$$ Ũ $$\frac{7}{2}$$ Ũ $$\frac{7}{2}$$ = 38.5 cm2
Question 32.
The perimeter of the following figure is
(a) 27 cm
(b) 28 cm
(c) 36 cm
(d) 40 cm
Answer
Answer: (c) 36 cm
Hint:
Perimeter = 2 Ũ 7 + $$\frac{22}{7}$$ Ũ 7
= 14 + 22 = 36 cm.
Question 33.
Find AD in the following figure :
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 2.4 cm
Answer
Answer: (d) 2.4 cm
Hint:
Area of ? ABC = $$\frac{3Ũ4}{2}$$
= $$\frac{5ŨAD}{2}$$ ? AD = 2.4
Question 34.
Find the area of ? ABC :
(a) 1 cm2
(b) 2 cm2
(c) 4 cm2
(d) $$\frac { 1 }{ 2 }$$ cm2
Answer
Answer: (d) $$\frac { 1 }{ 2 }$$ cm2
Hint:
Area = $$\frac{1Ũ1}{2}$$ = $$\frac{1}{2}$$ cm2
Question 35.
Find the area of ? ABC
(a) 3 cm2
(b) 4 cm2
(c) 6 cm2
(d) 12 cm2
Answer
Answer: (c) 6 cm2
Hint:
Area = $$\frac{4Ũ3}{2}$$ = 6 cm2
Question 36.
Find the area of ? ABC
(a) 1 cm2
(b) 2 cm2
(c) 3 cm2
(d) 4 cm2
Answer
Answer: (c) 3 cm2
Hint:
Area = $$\frac{3Ũ2}{2}$$ = 3 cm2
Question 37.
Which of the following is not the value of p ?
(a) $$\frac { 22 }{ 7 }$$
(b) $$\frac { 7 }{ 22 }$$
(c) $$\frac { 355 }{ 113 }$$
(d) 3.14
Answer
Answer: (b) $$\frac { 7 }{ 22 }$$
Question 38.
The radius of a circle is 7 cm. Its circumference is
(a) 22 cm
(b) 44 cm
(c) 11 cm
(d) 66cm
Answer
Answer: (b) 44 cm
Hint:
Circumference = 2 Ũ $$\frac{22}{7}$$ Ũ 7 = 44 cm
Question 39.
1 m2 =
(a) 10 cm2
(b) 100 cm2
(c) 1000 cm2
(d) 10000 cm2
Answer
Answer: (d) 10000 cm2
Hint:
Formula
Question 40.
1 hectare =
(a) 10 m2
(b) 100 m2
(c) 1000 m2
(d) 10000 m2
Answer
Answer: (d) 10000 m2
Hint:
Formula
Share:
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# Show that $\int_{-\pi}^{\pi}e^{\alpha \cos t}\sin(\alpha \sin t)dt=0$
I am trying to show that $$\int_{-\pi}^{\pi}e^{\alpha \cos t}\sin(\alpha \sin t)dt=0$$
Where $\alpha$ is a real constant.
I found the problem while studying a particular question in this room,this one. It becomes so challenging to me as I am trying to make life easy but I stucked!
EDIT: The integral is from $-\pi$ to $\pi$
EDIT 2: I am sorry for this edit, but it is a typo problem and I fix it now. In my question I have $e^\alpha \cos t$ not $e^\alpha$ only. I am very sorry.
-
When you write out function names like $\operatorname{sin}$ as words, $\TeX$ interprets them as a juxtaposition of variable names, which get italicized and don't have the right spacing. To get proper formatting for function names, you need to use the commands predefined for that purpose, e.g. \sin, or, for functions for which there's no predefined command, you can use \operatorname{name} to produce "$\operatorname{name}$". – joriki Apr 19 '12 at 2:35
@joriki:thanks I will fix it – Hassan Muhammad Apr 19 '12 at 2:38
In that answer the integral with sin is from $-\pi$ to $\pi$. Since $\sin(t)$ is odd, the integral is obviously 0... – N. S. Apr 19 '12 at 2:47
You only fixed one out of the four incorrectly formatted instances of $\sin$ in the post. – joriki Apr 19 '12 at 2:51
Hassan, this is very frustrating. Please take more care in formatting your posts and in responding to comments. I've now twice pointed out the incorrect formatting of the sine functions; each time you've only corrected a single one of the four instances in the post, the second time even after I had pointed out that there are four and you had agreed and announced that you would fix them. The title is still just as badly formatted as it was originally. P.S.: I see that Arturo has fixed it now. – joriki Apr 19 '12 at 2:57
This is false. In the interior of the interval of integration, the value of the inner sine is in $(0,1]$. For sufficiently small $\alpha$, that means the value of the outer sine is positive, so since $\mathrm e^\alpha$ is also positive, the integral is positive.
[Edit in response to the change in the question:]
As N.S. has already pointed out, the new integral vanishes because the integrand is odd and the integration interval is symmetric about $0$. By the way, also note that $\mathrm e^\alpha$ is a non-zero constant that doesn't affect whether the integral is zero.
[Edit in response to yet another change in the question:]
The integrand is still odd; the cosine in the exponent doesn't change that. And please take more care in posting; it's a huge waste of everyone's time to ask two questions that you didn't mean to ask and have people spend time answering them.
-
If that is the case then this proof is wrong [math.stackexchange.com/questions/124868/…, robjohn answer. – Hassan Muhammad Apr 19 '12 at 2:45
@HassanMuhammad Did you see the end points of integration when the sin integral vanished? They are not the same as yours ;) – N. S. Apr 19 '12 at 2:48
@joriki: I will take my time before accepting your answer Mr. Joriki. – Hassan Muhammad Apr 19 '12 at 3:15
If $f(t)$ is a continuous, odd function, then
$$\int_{-a}^a f(t) dt =0$$
Proof: Substitute $u=-t$.
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What is the rate of change of the line given? Here is a graphic preview for all of the Linear Equations Worksheets. Linear functions are those whose graph is a straight line. Compare features of two linear functions represented in different ways. 3/2. Another option for graphing is to use transformations of the identity function $f\left(x\right)=x$ . Sample answer: The graph of h is a translation 2 units up of the graph of the parent linear function. Review Of Linear Functions Lines Answer Key - Displaying top 8 worksheets found for this concept.. Expert Answer . The only thing different is the function notation. Create a non-linear function equation that has a solution at (-2, 6). 8. question_answer. What is the rate of change, as given by this point-slope equation? How To: Given the equation for a linear function, graph the function using the y-intercept and slope. We are going to use this same skill when working with functions. 4. 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# Modelling the heat exchange between a steel cylinder and the surrounding medium [duplicate]
Massively refactoring my question after @user2974951 feedback below.
I have a set of measurements (time, temperature) that correspond to a certain physical process. Both time and temperature measurements have some error.
The model of the process I'm going to use is
$$f(t) = a + e^{bt}$$
where $$f(t)$$ is the temperature, $$t$$ is the time, and $$a$$ and $$b$$ are some unknown coefficients.
What would be the best approach to find these unknowns?
My brute-force approach would be to iteratively bisect both $$a$$ and $$b$$ by going over the dataset multiple times, but perhaps there is a more efficient solution.
EDIT:
An example dataset would be [(3.7,15.1), (5.5,14.3), (7.1,13.6), (8.9,13.0), (9.7,12.8), ...] and the output of the algorithm would be something like a=10, b=-0.1. The actual output values would be similar, but not exactly what I specified.
EDIT 2:
There are typically more points in the dataset than just the 5 that I initially suggested. I would guess that we can afford at least 10 measurements, maybe more, but not arbitrarily more - PSNR drops, and we eventually measure only noise.
The error in the time variable $$t$$ is small, only a few milliseconds, while the whole timespan is many seconds. The error in the temperature variable is significantly larger, say 5% of the whole observed range.
EDIT 3:
It seems that I forgot about horizontal alignment and to address that I need the third unknown. Therefore the approximating formula becomes
$$f(t) = a + e^{bt + c}$$
which adds another level of complication to my naive brute-force approach.
EDIT 4:
My question was marked as a duplicate, but it's not the same topic as referred questions. I can't use a well-established software where I just say "make a fit" and the curve is automagically fitted for me. I have to actually implement the fit myself.
Answering a question below - if implemented on an embedded platform, then this would be C and no external libraries. If implemented server-side - then Java or Kotlin and a wide choice of what to link against.
EDIT 5:
I can't answer my own question, perhaps because it's marked as a duplicate. This is the last update.
I have settled on my bisect algorithm (first $$b$$, then $$c$$, then $$a$$) that gives me $$O(N (logN)^2)$$ complexity. It feels fast on a mac, but I have doubts if it is suitable for an embedded solution on a low-power ARM processor.
## marked as duplicate by whuber♦Jul 17 at 15:02
• Your situation is interesting because it involves errors in the independent variable $t,$ but for the same reason it is problematic because that requires you to estimate at least four parameters ($a, b,$ and the variances of errors in $t$ and $f$) using only five points. It would therefore help if you had reliable independent estimates of the two error variances or if you could collect more data. – whuber Jul 17 at 11:55
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Properties of Logarithms
Do you want to know how to Properties of Logarithms? you can do it in two easy steps.
Properties of Logarithms
Learn some logarithms properties:
• $$a^{log_{a}{b}}=b$$
• $$log_{a}{1}=0$$
• $$log_{a}{a}=1$$
• $$log_{a}{x.y}=log_{a}{x}+\log_{a}{y}$$
• $$log_{a}{\frac{x}{y}}=log_{a}{x}-\log_{a}{y}$$
• $$log_{a}{\frac{1}{x}}=-log_{a}{x}$$
• $$log_{a}{x^p}=p log_{a}{x}$$
• $$log_{x^k}{x}=\frac{1}{x}log_{a}{x}$$, for $$k\neq0$$
• $$log_{a}{x}= log_{a^c}{x^c}$$
• $$log_{a}{x}=\frac{1}{log_{x}{a}}$$
Logarithms Properties – Example 1:
Expand this logarithm. $$log (8×5)=$$
Solution:
Use log rule: $$log_{a}{x.y}=log_{a}{x}+\log_{a}{y}$$
then: $$log (8×5)= log 8 + log 5$$
Logarithms Properties – Example 2:
Condense this expression to a single logarithm. $$log 2-log 9=$$
Solution:
Use log rule: $$log_{a}{\frac{x}{y}}=log_{a}{x}-\log_{a}{y}$$
then: $$log 2-log 9= log{\frac{2}{9}}$$
Logarithms Properties – Example 3:
Expand this logarithm. $$log (2×3)=$$
Solution:
Use log rule: $$log_{a}{x.y}=log_{a}{x}+\log_{a}{y}$$
then: $$log (2×3)= log 2 + log 3$$
Logarithms Properties – Example 4:
Condense this expression to a single logarithm. $$log 4-log 3=$$
Solution:
Use log rule: $$log_{a}{\frac{x}{y}}=log_{a}{x}-log_{a}{y}$$
then: $$log 4-log 3= log{\frac{4}{3}}$$
Expand each logarithm.
1. $$\color{blue}{log (12×6)=}$$
2. $$\color{blue}{log (9×4)=}$$
3. $$\color{blue}{log (3×7)=}$$
4. $$\color{blue}{log{\frac{3}{4}}}$$
5. $$\color{blue}{log{\frac{5}{7}}}$$
6. $$\color{blue}{log({\frac{2}{5}})^3}$$
7. $$\color{blue}{log (2×3^4)=}$$
8. $$\color{blue}{ log({\frac{5}{7}})^4}$$
1. $$\color{blue}{log 12+log 6}$$
2. $$\color{blue}{ log 9+log 4}$$
3. $$\color{blue}{log 3+log 7}$$
4. $$\color{blue}{ log 3-log 4}$$
5. $$\color{blue}{ log 5-log 7}$$
6. $$\color{blue}{3 log 2-3 log 5}$$
7. $$\color{blue}{log 2+4 log 3}$$
8. $$\color{blue}{4log 5-4 log 7}$$
What people say about "Properties of Logarithms"?
No one replied yet.
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# 2012 JBMO Problems/Problem 1
## Section 1
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq 2\sqrt{2}\left (\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right ).$$ When does equality hold?
## Solution
The LHS rearranges to $\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} + 6$. Since $b+c=1-a$ we have that $\frac{b+c}{a}=\frac{1-a}{a}$. Therefore, the LHS rearranges again to $\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6$.
Now, distribute the $\sqrt{2}$ on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get $$\frac{2-2a}{a} + \frac{2-2b}{b} + \frac{2-2c}{c}+12 \geq 4(\sqrt{\frac{2-2a}{a}} + \sqrt{\frac{2-2b}{b}} + \sqrt{\frac{2-2c}{c}})$$ Let $\sqrt{\frac{2-2a}{a}} = A$ and similarly for $B$ and $C$.
The inequality now simplifies to $$A^2+B^2+C^2+12 \geq 4(A+B+C)$$ Note that because $a$, $b$, and $c$ are positive real numbers less than $1$, $A$, $B$, and $C$ are always positive real numbers. Rearranging terms shows that this further simplifies to $$(A^2-4A+4)+(B^2-4B+4)+(C^2-4C+4)\geq0$$ which equals $$(A-2)^2+(B-2)^2+(C-2)^2\geq0$$ By the trivial inequality we know that this is always true. Finally, we have equality when $$A=B=C=2$$ and $$\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4$$ Solving the equations yields that equality holds when $\boxed{a=b=c=\frac{1}{3}}$
Solution by Someonenumber011 :)
2012 JBMO (Problems • Resources) Preceded by2011 JBMO Followed by2013 JBMO 1 • 2 • 3 • 4 • 5 All JBMO Problems and Solutions
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nLab non-archimedean analytic geometry
Idea
Non-archimedean geometry is geometry over non-archimedean fields. While the concrete results are quite different, the basic formalism of algebraic schemes and formal schemes over a non-archimedean field $K$ is the special case of the standard formalism over any field. The “correct” analytic geometry over non-archimedean field, however, is not a straightforward analogue of the complex analytic case. As Tate noticed, the sheaf of $K$-valued functions which can be locally written as converging power series over the affine space ${K}^{n}$ is too big (too many analytic functions) due to the fact that the underlying topological space is totally disconnected. Also there are very few $K$-analytic manifolds. This naive approach paralleling the complex analytic geometry? is called by Tate wobbly $K$-analytic varieties an, apart from the case of non-archimedean local fields it is of little use. For this reason Tate introduced a better $K$-algebra of analytic functions, locally takes its maximal spectrum and made a Grothendieck topology which takes into account just a certain smaller set of open covers; this topology is viewed as rigidified, hence the varieties based on gluing in this approach is called rigid analytic geometry. Raynaud has shown how some classes of rigid $K$-varieties can be realized as generic fibers of formal schemes over the ring of integers of $K$; this is called a formal model of a rigid variety. Different formal models are birationally equivalent, more precisely they are related via admissible blow-ups. Later more sophisticated approaches appeared:
Literature
For literature on specific approaches see the $n$Lab entries Berkovich analytic space, adic space, global analytic geometry. For comparison see
• Brian Conrad, Several approaches to non-archimedean geometry, lectures at Arizona winter school 2007, pdf
category: geometry
Revised on May 9, 2013 11:49:50 by David Roberts (192.43.227.18)
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# The Ideal Gas Law
## The Ideal Gas Law
The air inside this hot air balloon flying over Putrajaya, Malaysia, is hotter than the ambient air. As a result, the balloon experiences a buoyant force pushing it upward. (credit: Kevin Poh, Flickr)
In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, $${\text{N}}_{2}$$, and oxygen, $${\text{O}}_{2}$$, are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas because the term can also be applied to monatomic gases, such as helium.)
Gases are easily compressed. We can see evidence of this in this table, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same $$\beta$$. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.
The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in the figure below. Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.
Atoms and molecules in a gas are typically widely separated, as shown. Because the forces between them are quite weak at these distances, the properties of a gas depend more on the number of atoms per unit volume and on temperature than on the type of atom.
To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires. (See the figure below.)
(a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion and the pressure increases with more air. (c) Once the tire is inflated, its pressure increases with temperature.
At room temperatures, collisions between atoms and molecules can be ignored. In this case, the gas is called an ideal gas, in which case the relationship between the pressure, volume, and temperature is given by the equation of state called the ideal gas law.
### Ideal Gas Law
The ideal gas law states that
$$\text{PV}=\text{NkT},$$
where $$P$$ is the absolute pressure of a gas, $$V$$ is the volume it occupies, $$N$$ is the number of atoms and molecules in the gas, and $$T$$ is its absolute temperature. The constant $$k$$ is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value
$$k=1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J}/\text{K}.$$
The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product $$\text{PV}$$ is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of $$V$$. The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that $$N$$ is the total number of atoms and molecules, independent of the type of gas.)
Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure $$P$$ is essentially equal to atmospheric pressure, and the volume $$V$$ increases in direct proportion to the number of atoms and molecules $$N$$ put into the tire. Once the volume of the tire is constant, the equation $$\text{PV}=\text{NkT}$$ predicts that the pressure should increase in proportion to the number N of atoms and molecules.
### Example: Calculating Pressure Changes Due to Temperature Changes: Tire Pressure
Suppose your bicycle tire is fully inflated, with an absolute pressure of $$7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$$ (a gauge pressure of just under $$\text{90}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{lb/in}}^{2}$$) at a temperature of $$\text{18}\text{.}0\text{º}\text{C}$$. What is the pressure after its temperature has risen to $$\text{35}\text{.}0\text{º}\text{C}$$? Assume that there are no appreciable leaks or changes in volume.
Strategy
The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown.
We know the initial pressure $${P}_{0}=7\text{.00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$$, the initial temperature $${T}_{0}=\text{18}\text{.}0ºC$$, and the final temperature $${T}_{\text{f}}=35\text{.}0ºC$$. We must find the final pressure $${P}_{\text{f}}$$. How can we use the equation $$\text{PV}=\text{NkT}$$? At first, it may seem that not enough information is given, because the volume $$V$$ and number of atoms $$N$$ are not specified. What we can do is use the equation twice: $${P}_{0}{V}_{0}={\text{NkT}}_{0}$$ and $${P}_{\text{f}}{V}_{\text{f}}={\text{NkT}}_{\text{f}}$$. If we divide $${P}_{\text{f}}{V}_{\text{f}}$$ by $${P}_{0}{V}_{0}$$ we can come up with an equation that allows us to solve for $${P}_{\text{f}}$$.
$$\cfrac{{P}_{\text{f}}{V}_{\text{f}}}{{P}_{0}{V}_{0}}=\cfrac{{N}_{\text{f}}{\text{kT}}_{\text{f}}}{{N}_{0}{\text{kT}}_{0}}$$
Since the volume is constant, $${V}_{\text{f}}$$ and $${V}_{0}$$ are the same and they cancel out. The same is true for $${N}_{\text{f}}$$ and $${N}_{0}$$, and $$k$$, which is a constant. Therefore,
$$\cfrac{{P}_{\text{f}}}{{P}_{0}}=\cfrac{{T}_{\text{f}}}{{T}_{0}}\text{.}$$
We can then rearrange this to solve for $${P}_{\text{f}}$$:
$${P}_{\text{f}}={P}_{0}\cfrac{{T}_{\text{f}}}{{T}_{0}},$$
where the temperature must be in units of kelvins, because $${T}_{0}$$ and $${T}_{\text{f}}$$ are absolute temperatures.
Solution
1. Convert temperatures from Celsius to Kelvin.
$$\begin{array}{}{T}_{0}=\left(\text{18}\text{.}0+\text{273}\right)\text{K}=\text{291 K}\\ {T}_{\text{f}}=\left(\text{35}\text{.}0+\text{273}\right)\text{K}=\text{308 K}\end{array}$$
2. Substitute the known values into the equation.
$${P}_{\text{f}}={P}_{0}\cfrac{{T}_{\text{f}}}{{T}_{0}}=7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}(\cfrac{\text{308 K}}{\text{291 K}})=7\text{.}\text{41}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$$
Discussion
The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law.
### Making Connections: Take-Home Experiment—Refrigerating a Balloon
Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why?
### Example: Calculating the Number of Molecules in a Cubic Meter of Gas
How many molecules are in a typical object, such as gas in a tire or water in a drink? We can use the ideal gas law to give us an idea of how large $$N$$ typically is.
Calculate the number of molecules in a cubic meter of gas at standard temperature and pressure (STP), which is defined to be $$0\text{º}\text{C}$$ and atmospheric pressure.
Strategy
Because pressure, volume, and temperature are all specified, we can use the ideal gas law $$\text{PV}=\text{NkT}$$, to find $$N$$.
Solution
1. Identify the knowns.
$$\begin{array}{lll}T& =& 0\text{º}\text{C}=\text{273 K}\\ P& =& 1\text{.}\text{01}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\\ V& =& 1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\\ k& =& 1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\end{array}$$
2. Identify the unknown: number of molecules, $$N$$.
3. Rearrange the ideal gas law to solve for $$N$$.
$$\begin{array}{}\text{PV}=\text{NkT}\\ N=\frac{\text{PV}}{\text{kT}}\end{array}$$
4. Substitute the known values into the equation and solve for $$N$$.
$$N=\cfrac{\text{PV}}{\text{kT}}=\cfrac{(1\text{.}\text{01}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa})(1\text{.}{\text{00 m}}^{3})}{(1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K})(\text{273 K})}=2\text{.}\text{68}×{\text{10}}^{\text{25}}\phantom{\rule{0.25em}{0ex}}\text{molecules}$$
Discussion
This number is undeniably large, considering that a gas is mostly empty space. $$N$$ is huge, even in small volumes. For example, $$1\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{3}$$ of a gas at STP has $$2\text{.}\text{68}×{\text{10}}^{\text{19}}$$ molecules in it. Once again, note that $$N$$ is the same for all types or mixtures of gases.
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Newton's Laws: First principles
The fundamental idea of kinematics is the discussion of the movement of objects, without actually taking into account what caused the movement to occur. By using simple calculus, we can find all of the equations for kinematics. To simplify the learning process, we will only consider objects that move with constant acceleration. For the first few parts, we will also assume that there is no friction or air resistance acting on the objects.
Straight Line Motion (SLM)
The name of this section Straight Line Motion means that we begin learning about the subject of kinematics by observing motion in one dimension. This means that we will only take one axis of a 3D $(x,y,z)$ coordinate system into account. We will be using the $x$ axis as our axis of motion.
Throughout our discussion, we will look at the motion of a rigid body, one that does not deform as it moves. We idealize the rigid body by assuming that it has no dimensions and is infinitely small. This way we can talk about the entire body instead of saying, "The front of the body is at this point while the rear of the body is at some other point". Also, subscripts on the variables in our equations will indicate the initial value, $i$, and the final value, $f$, of that variable.
Displacement
To start off with we will define the term displacement. Displacement is basically the shortest route of getting from one place to another, basically it is a straight line from the starting point of motion to the ending point of motion. No matter how much movement takes place in between and what sort of things come inbetween, we only care about the first location and the second location. We will use the variable, $x$, to stand for the locations of the rigid body that we are discussing. It is a vector quantity i.e. it has both magnitude and direction, if we want to define displacement for some movement then it would be like - '50 km due North'.
In order to define motion we first must be able to say how far an object has moved. This is done by subtracting the final value of the displacement by the initial value of the displacement. Or, in other words, we subtract the initial position of the object on our coordinate system from the final position of the object on our coordinate system. It is necessary to understand that the values which you receive for this variable can be either positive or negative depending on how the object is moving and where your coordinates start:
$x=x_f-x_i\,\!$
Velocity
The term velocity, $v$, is often mistaken as being equivalent to the term speed. The basic difference between speed and velocity is that, velocity is a vector quantity, whereas speed is a scalar quantity. The term velocity refers to the displacement that an object traveled divided by the amount of time it took to move to its new coordinate. From the above discussion you can see that if the object moves backwards with respect to our coordinate system, then we will get a negative displacement. Since we cannot have a negative value for time, we will then get a negative value for velocity( Here the negative sign shows that the body has moved in the backward direction to the coordinates system adopted). The term speed, on the other hand refers to the magnitude of the velocity, so that it can only be a positive value. We will not be considering speed in this discussion, only velocity. By using our definition of velocity and the definition of displacement from above, we can express velocity mathematically like this:
$\bar v={(x_f-x_i) \over (t_f-t_i)}$
The line over the velocity means that you are finding the average velocity, not the velocity at a specific point. This equation can be rearranged in a variety of ways in order to solve problems in physics dealing with SLM. Also, by having acceleration as a constant value, we can find the average velocity of an object if we are given the initial and final values of an object's velocity:
$\bar v={(v_f+v_i) \over 2}$
These equations can be combined with the other equations to give useful relationships in order to solve straight line motion problems. For example, to find the initial position of an object, if we are given its final position and the times that the object began moving and finished moving, we can rearrange the equation like this:
$x_i=x_f-\bar v(t_f-t_i)\,\!$
Acceleration
The term acceleration means the change in velocity of an object per unit time. It is a vector quantity. We use the variable, $a$, to stand for acceleration. Basically, the sign of acceleration tells us whether the velocity is increasing or decreasing, and its magnitude tells us how much the velocity is changing. In order for an object that is initially at rest to move, it needs to accelerate to a certain speed. During this acceleration, the object moves at a certain velocity at a specific time, and travels a certain distance within that time. Thus, we can describe acceleration mathematically like this:
$a={(v_f-v_i) \over (t_f-t_i)}$
Again, we can rearrange our formula, this time using our definition for displacement and velocity, to get a very useful relationship:
$x_f=x_i+v_i(t_f-t_i)+{1 \over 2}a(t_f-t_i)^2$
History of Dynamics
Aristotle
Aristotle expounded a view of dynamics which agrees closely with our everyday experience of the world. Objects only move when a force is exerted upon them. As soon as the force goes away, the object stops moving. The act of pushing a box across the floor illustrates this principle -- the box certainly doesn't move by itself!
However, if we try using Aristotle's dynamics to predict motion we soon run into problems. It suggests that objects under a constant force move with a fixed velocity but while gravity definitely feels like a constant force it clearly doesn't make objects move with constant velocity. A thrown ball can even reverse direction, under the influence of gravity alone.
Eventually, people started looking for a view of dynamics that actually worked. Newton found the answer, partially inspired by the heavens.
Newton
In contrast to earthly behavior, the motions of celestial objects seem effortless. No obvious forces act to keep the planets in motion around the sun. In fact, it appears that celestial objects simply coast along at constant velocity unless something acts on them.
This Newtonian view of dynamics — objects change their velocity rather than their position when a force is exerted on them — is expressed by Newton's second law:
$\mathbf{F}=m\mathbf{a}$
where $\mathbf{F}$ is the force exerted on a body, $m$ is its mass, and $\mathbf{a}$ is its acceleration. Newton's first law, which states that an object remains at rest or in uniform motion unless a force acts on it, is actually a special case of Newton's second law which applies when $\mathbf{F}=0$.
It is no wonder that the first successes of Newtonian mechanics were in the celestial realm, namely in the predictions of planetary orbits. It took Newton's genius to realize that the same principles which guided the planets also applied to the earthly realm as well.
In the Newtonian view, the tendency of objects to stop when we stop pushing on them is simply a consequence of frictional forces opposing the motion. Friction, which is so important on the earth, is negligible for planetary motions, which is why Newtonian dynamics is more obviously valid for celestial bodies.
Note that the principle of relativity is closely related to Newtonian physics and is incompatible with pre-Newtonian views. After all, two reference frames moving relative to each other cannot be equivalent in the pre-Newtonian view, because objects with nothing pushing on them can only come to rest in one of the two reference frames!
Einstein's relativity is often viewed as a repudiation of Newton, but this is far from the truth — Newtonian physics makes the theory of relativity possible through its invention of the principle of relativity. Compared with the differences between pre-Newtonian and Newtonian dynamics, the changes needed to go from Newtonian to Einsteinian physics constitute minor tinkering.
Newton's first law:
"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an external force". He gave this law assuming the body or the system to be isolated. If we look upon into our daily life we finds that this law is applicable in reality like a bicycle stops slowly when we stop paddling, this is because in our daily life there are 2 external forces which opposes the motion, these are frictional force and air resistance( These force aren't included in the course of an isolated system). But if these two forces are absent the above law is applicable, this can be observed in space.
Newton's second law:
Newton's second law of motion states, "The rate change of linear momentum of an object is directly proportional to the external force on the object." The concept behind this is - "Forces arises due to interaction between the bodies." With the help of this law we can derive this formula: $F = ma$. In this formula, $F$ means the force exerted on the object, $m$ means the mass of the object, and $a$ means the acceleration of the object. Also the second law of motion is universal law in nature, i.e. it consists of both the laws. Newton assumed the body to be an isolated one, thus according to second law, if there is no interaction between bodies of two different systems( means not to be an isolated one), then there is no force that can stop or shake the object from it's state of being. Suppose two bodies are interacting with each other, then each of them would be exerting force on the other( Look from each side and apply second law of motion), it is third law.
Newton's third law:
"Every action has an equal and opposite reaction" This means if body A exerts a force on body $B$, then body $B$ exerts a force on body $A$ that is equal in magnitude but opposite in direction from the force from $A$. $F_{AB} = - F_{BA}$
• Negative sign shows that the force is in opposite direction.
Work
When a force is exerted on an object, energy is transferred to the object. The amount of energy transferred is called the work done on the object. Mathematically work done is defined as the dot\scalar product of force and displacement, thus it is a scalar quantity. However, energy is only transferred if the object moves. Work can be thought of as the process of transforming energy from one form into another. The work W done is
$W=F\Delta x$
where the distance moved by the object is Δx and the force exerted on it is F. Notice that work can either be positive or negative. The work is positive if the object being acted upon moves in the same direction as the force, with negative work occurring if the object moves opposite to the force.
This equation assumes that the force remains constant over the full displacement or distance( depending on the situation). If it is not, then it is necessary to break up the displacement into a number of smaller displacements, over each of which the force can be assumed to be constant. The total work is then the sum of the works associated with each small displacement. In the infinitesimal limit this becomes an integral
$W=\int F(s)\,ds$
If more than one force acts on an object, the works due to the different forces each add or subtract energy, depending on whether they are positive or negative. The total work is the sum of these individual works.
There are two special cases in which the work done on an object is related to other quantities. If F is the total force acting on the object, then by Newton's second law W=FΔx=mΔx·a . However, a=dv/dt where v is the velocity of the object, and Δx≈vΔt, where Δt is the time required by the object to move through distance Δx . The approximation becomes exact when Δx and Δt become very small. Putting all of this together results in
$W_{total}=m\frac{dv}{dt}v\Delta t=\Delta t \frac{d}{dt} \frac{mv^2}{2}$
We call the quantity mv2/2 the kinetic energy, or K. It represents the amount of work stored as motion. We can then say
$W_{total}=\frac{dK}{dt}\Delta t=\Delta K$
Thus, when F is the only force, the total work on the object equals the change in kinetic energy of the object. This transformation is known as "Work-Energy theorem."
The other special case occurs when the force depends only on position, but is not necessarily the total force acting on the object. In this case we can define a function
$V(x)=-\int^x F(s)\,ds$
and the work done by the force in moving from x1 to x2 is V(x1)-V(x2), no matter how quickly or slowly the object moved.
If the force is like this it is called conservative and V is called the potential energy. Differentiating the definition gives
$F=-\frac{dV}{dx}$
If a force is conservative( The force whose effect doesn't depend on the path it has taken to go through), we can write the work done by it as
$W=-\frac{dV}{dx}\Delta x= -\Delta V$
where is the change in the potential energy of the object associated with the force of interest.
Energy
The sum of the potential( Energy by virtue of its position) and kinetic(Energy by virtue of its motion) energies is constant. We call this constant the total energy E:
$E = K + U$
If all the forces involved are conservative we can equate this with the previous expression for work to get the following relationship between work, kinetic energy, and potential energy:
$\Delta K = W = -\Delta U$
Following this, we have a very important formula, called the 'Conservation of Energy:
$\Delta(K+U)=0$
This theorem states that the total amount of energy in a system is constant, and that energy can neither be created nor destroyed.
Power
The power associated with a force is simply the amount of work done by the force divided by the time interval over which it is done. It is therefore the energy per unit time transferred to the object by the force of interest. From above we see that the power is
$P=\frac{F \Delta x}{\Delta t}=Fv$
where is the velocity at which the object is moving. The total power is just the sum of the powers associated with each force. It equals the time rate of change of kinetic energy of the object:
$P_{total}=\frac{W_{total}}{\Delta t}=\frac{dK}{dt}$
We will now take a break from physics, and discuss the topics of partial derivatives. Further information about this topic can be found at the Partial Differential Section in the Calculus book.
Partial Derivatives
In one dimension, the slope of a function, f(x), is described by a single number, df/dx.
In higher dimensions, the slope depends on the direction. For example, if f=x+2y, moving one unit x-ward increases f by 1 so the slope in the x direction is 1, but moving one unit y-ward increases f by 2 so the slope in the y direction is 2.
It turns out that we can describe the slope in n dimensions with just n numbers, the partial derivatives of f.
To calculate them, we differentiate with respect to one coordinate, while holding all the others constant. They are written using a ∂ rather than d. E.g.
$f(x+dx,y,z)=f(x,y,z)+\frac{\partial f}{\partial x}dx \quad (1)$
Notice this is almost the same as the definition of the ordinary derivative.
If we move a small distance in each direction, we can combine three equations like 1 to get
$df= \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz$
The change in f after a small displacement is the dot product of the displacement and a special vector
$\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \cdot \left( dx, dy, dz \right) =\operatorname{grad} f = \nabla f$
This vector is called the gradient of f. It points up the direction of steepest slope. We will be using this vector quite frequently.
Partial Derivatives #2
Another way to approach differentiation of multiple-variable functions can be found in Feynman Lectures on Physics vol. 2. It's like this:
The differentiation operator is defined like this: $df = f(x + \Delta x, y + \Delta y, z + \Delta z) - f(x,y,z)\,\!$ in the limit of $\Delta x, \Delta y, \Delta z \to 0$. Adding & subtracting some terms, we get
$df = (f(x + \Delta x, y + \Delta y, z + \Delta z) - f(x, y + \Delta y, z + \Delta z)) + (f(x, y + \Delta y, z + \Delta z) - f(x, y, z + \Delta z)) + (f(x, y, z + \Delta z) - f(x,y,z))\,\!$
and this can also be written as
$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz$
Alternate Notations
For simplicity, we will often use various standard abbreviations, so we can write most of the formulae on one line. This can make it easier to see the important details.
We can abbreviate partial differentials with a subscript, e.g.,
$D_x h(x,y)= \frac{\partial h}{\partial x} \quad D_x D_y h= D_y D_x h$
or
$\partial_x h= \frac{\partial h}{\partial x}$
Mostly, to make the formulae even more compact, we will put the subscript on the function itself.
$D_x h= h_x \quad h_{xy}=h_{yx}$
Refer to the Partial Differential Section in the Calculus book for more information.
Motion in 2 and 3 Directions
Previously, we discussed Newtonian dynamics in one dimension. Now that we are familiar with both vectors and partial differentiation, we can extend that discussion to two or three dimensions.
Work becomes a dot product
$W=\mathbf{F}\cdot \Delta x$
and likewise power
$P=\mathbf{F}\cdot \mathbf{v}$
If the force is at right angles to the direction of motion, no work will be done.
In one dimension, we said a force was conservative if it was a function of position alone, or equivalently, the negative slope of a potential energy.
The second definition extends to
$\mathbf{F}=-\nabla V$
In two or more dimensions, these are not equivalent statements. To see this, consider
$D_y\mathbf{F}_x-D_x\mathbf{F}_y=V_{xy}-V_{yx}$
Since it doesn't matter which order derivatives are taken in, the left hand side of this equation must be zero for any force which can be written as a gradient, but for an arbitrary force, depending only on position, such as F=(y, -x, 0), the left hand side isn't zero.
Conservative forces are useful because the total work done by them depends only on the difference in potential energy at the endpoints, not on the path taken, from which the conservation of energy immediately follows.
If this is the case, the work done by an infinitesimal displacement dx must be
$W=V(\mathbf{x})-V(\mathbf{x}+d\mathbf{x})=-(\nabla V)\cdot d\mathbf{x}$
Comparing this with the first equation above, we see that if we have a potential energy then we must have
$\mathbf{F}=-\nabla V$
Any such F is a conservative force.
Circular Motion
An important example of motion in two dimensions is circular motion.
Consider a mass, m, moving in a circle, radius r.
The angular velocity, ω is the rate of change of angle with time. In time Δt the mass moves through an angle Δθ= ωΔt. The distance the mass moves is then r sin Δθ, but this is approximately rΔθ for small angles.
Thus, the distance moved in a small time Δt is rωΔt, and divided by Δt gives us the speed, v.
$v = \omega r \,$
This is the speed not the velocity because it is not a vector. The velocity is a vector, with magnitude ωr which points tangentially to the circle.
The magnitude of the velocity is constant but its direction changes so the mass is being accelerated.
By a similar argument to that above it can be shown that the magnitude of the acceleration is
$a = \omega v \,$
and that it is pointed inwards, along the radius vector. This is called centripetal acceleration.
By eliminating v or ω from these two equations we can write
$\mathbf{a}= -\omega^2 \mathbf{r} = -\frac{v^2}{r}\mathbf{ \hat{ r}}$
Year 10 Force and Work 7 periods
• Possible Approach:
• Lesson Suggested Learning Outcomes Course Outline Resources
• 1 – 3 Recognise that unbalanced forces cause acceleration.
• Investigate the relationship between the unbalanced force acting on an object, is mass *and acceleration: Newton’s Second Law. (Use Fnet = ma giving correct units for all quantities)
•**Reinforce ideas about force causing acceleration (= the rate at which speed changes)
• Experiment 1: Finding the force required to move an object (static friction)
• Discuss the factors that may affect the frictional forces acting on the block (e.g. load on *the block, surface area in contact with the bench, type of the surfaces in contact …)
• Attach a force meter to a block at rest on the bench. Pull the force meter and slowly increase *the force until the block just begins to move – record the force meter reading, tabulate data, *repeat, calculate average.
• [Or, use a bag to which washers may be added, attached to a cord and hanging over the end of *the bench to pull the block.]
• quantities and the third unknown) – a three symbol triangle may simplify the maths.
• Homework: Coursebook – Newton’s second Law
• force (e.g. when a force acts on a body, it moves faster and kinetic energy increases)
• • Introduce the relationship between energy transferred and work done.
• • Define work done in terms of the applied force and the distance through which this ********force acts (Work done = force x distance)
• • Use the work rule to calculate work / force / distance (given two quantities and the ********third unknown) – a three symbol triangle may simplify the maths
• Homework: Coursebook – Work Done and Energy
• 6 – 7 Investigate and describe the pattern of results formed from graphing ************************the effects of applied forces on a spring.
• Explore the relationships between science and technology by investigating the application of *science to technology and the impact of technology on science, e.g. Archimedes’ screw to *illustrate the principle of oits use in water irrigation or the use of springs and similar *propulsion devices in toys
• Experiment 4: Stretching Springs
• Hang a spring vertically using a clamp and stand. Add weights to the end of the spring, *measure the length of the spring after each additional weight is added. Calculate the *extension (Note: some initial weight will probably need to be added before the spring *stretches at all) Graph extension vs applied force. Interpret the graph, e.g. find the *extension for unit applied force or applied force per unit extension.
• Investigate and understand the turning effect of a force.
• Calculate the turning effect of a force = F x d where force (F) and distance (d) are *********at right angles.
• Use the lever rule: F1d1 = F2d2 •
• Discuss and brainstorm ideas about what causes things to turn about a pivot (fulcrum).
• • Introduce the concept of the turning effect of a force (torque).
• • Discuss the concept levers – to increase the turning effect, use a greater distance *******(e.g. a long crowbar) and/or a greater force (e.g. a heavier weight)
• • Use the lever rule to calculate forces and distances at right angles to the applied ********force (given three quantities and the fourth unknown).
• Experiment 5: The turning effect of a force
• Kit available from the science Office
• • Homework: Coursebook – The Turning Effect of a Force
• Extension
Changing mass
So far we've assumed that the mass of the objects being considered is constant, which is not always true. Mass is conserved overall, but it can be useful to consider objects, such as rockets, which are losing or gaining mass.
We can work out how to extend Newton's second law to this situation by considering a rocket two ways, as a single object of variable mass, and as two objects of fixed mass which are being pushed apart.
We find that
$\mathbf{F} = \frac{m d(\mathbf{v})}{dt}$
Force is the rate of change of a quantity, mv, which we call linear momentum.
Newton's third law
Newton's third law says that the force, F12 exerted by a mass, m1, on a second mass, m2, is equal and opposite to the force F21 exerted by the second mass on the first.
$\frac{d(m_2\mathbf{v}_2)}{dt}= \mathbf{F}_{12}= - \mathbf{F}_{21} = -\frac{d(m_1 \mathbf{v}_1)}{dt}$
if there are no external forces on the two bodies. We can add the two momenta together to get,
$\frac{d(m_1\mathbf{v}_1)}{dt}+\frac{d(m_2 \mathbf{v}_2)}{dt} =\frac{d}{dt}(m_1\mathbf{v}_1+m_2\mathbf{v}_2)=0$
so the total linear momentum is conserved.
Ultimately, this is consequence of space being homogeneous.
Centre of mass
Suppose two constant masses are subject to external forces, F1, and F2
Then the total force on the system, F, is
$\mathbf{F}= \mathbf{F}_{1}+\mathbf{F}_{2} = \frac{d}{dt}(m_1\mathbf{v}_1+m_2\mathbf{v}_2)$
because the internal forces cancel out.
If the two masses are considered as one system, F should be the product of the total mass and ā, the average acceleration, which we expect to be related to some kind of average position.
$\begin{matrix}(m_1+m_2)\bar{a} & = & \frac{d}{dt}(m_1\mathbf{v}_1+m_2\mathbf{v}_2) \\ & = & \frac{d^2}{dt^2}(m_1\mathbf{r}_1+m_2\mathbf{r}_2)\\ \bar{a}& = & \frac{d^2}{dt^2} \frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}\\ \end{matrix}$
The average acceleration is the second derivative of the average position, weighted by mass.
This average position is called the centre of mass, and accelerates at the same rate as if it had the total mass of the system, and were subject to the total force.
We can extend this to any number of masses under arbitrary external and internal forces.
• we have n objects, mass m1, m2mn, total mass M
• each mass mi is at point ri,
• each mass mi is subject to an external force Fi
• the internal force exerted by mass mj on mass mi is Fij
then the position of the centre of mass, R, is
$\mathbf{R}=\frac{\sum_i m_i \mathbf{r}_i}{M}$
We can now take the second derivative of R
$\begin{array}{ccl} M \ddot\mathbf{R} & = & \sum_i m_i \ddot\mathbf{r}_i \\ & = & \sum_i (\mathbf{F}_i +\sum_j \mathbf{F}_{ij}) \end{array}$
But the sum of all the internal forces is zero, because Newton's third law makes them cancel in pairs. Thus, the second term in the above equation drops out and we are left with:
$M \ddot{\mathbf{R}} = \sum_i \mathbf{F}_i = \mathbf{F}$
The centre of mass always moves like a body of the same total mass under the total external force, irrespective of the internal forces.
If a rigid body is initially at rest, it will remain at rest if and only if the sum of all the forces and the sum of all the torques acting on the body are zero. As an example, a mass balance with arms of differing length is shown in figure above. The balance beam is subject to three forces pointing upward or downward, the tension T in the string from which the beam is suspended and the weights M1g and M2g exerted on the beam by the two suspended masses. The parameter g is the local gravitational field and the balance beam itself is assumed to have negligible mass. Taking upward as positive, the force condition for static equilibrium is
$T-M_1 g - M_2 g = 0 \quad \mbox{Zero net force}$
Defining a counterclockwise torque to be positive, the torque balance computed about the pivot point in figure 10.7 is
$\tau=M_1 g d_1 - M_2 g d_2 = 0 \quad \mbox{Zero torque}$
where d1 and d2 are the lengths of the beam arms. The first of the above equations shows that the tension in the string must be
$T=\left( M_1 +M_2 \right) g$
while the second shows that
$\frac{M_1}{M_2}=\frac{d_1}{d_2}$
Thus, the tension in the string is just equal to the weight of the masses attached to the balance beam, while the ratio of the two masses equals the inverse ratio of the associated beam arm lengths.
Rotational Dynamics
There are two ways to multiply two vectors together, the dot product and the cross product. We have already studied the dot product of two vectors, which results in a scalar or single number.
The cross product of two vectors results in a third vector, and is written symbolically as follows:
$\mathbf{A}\times \mathbf{B}$
The cross product of two vectors is defined to be perpendicular to the plane defined by these vectors. However, this doesn't tell us whether the resulting vector points upward out of the plane or downward. This ambiguity is resolved using the right-hand rule:
1. Point the uncurled fingers of your right hand along the direction of the first vector A.
2. Rotate your arm until you can curl your fingers in the direction of the second vector B.
3. Your stretched out thumb now points in the direction of the cross product vector A×B.
The magnitude of the cross product is given by
$|\mathbf{A}\times \mathbf{B}|=|\mathbf{A}||\mathbf{B}|\sin \theta$
where |A| and |B| are the magnitudes of A and B, and θ is the angle between these two vectors. Note that the magnitude of the cross product is zero when the vectors are parallel or anti-parallel, and maximum when they are perpendicular. This contrasts with the dot product, which is maximum for parallel vectors and zero for perpendicular vectors.
Notice that the cross product does not commute, i. e., the order of the vectors is important. In particular, it is easy to show using the right-hand rule that
$\mathbf{A}\times \mathbf{B}= -\mathbf{B}\times \mathbf{A}$
An alternate way to compute the cross product is most useful when the two vectors are expressed in terms of components,
$\mathbf{C}= \mathbf{A}\times \mathbf{B}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$
where the determinant is expanded as if all the components were numbers, giving
$C_x=A_y B_z -A_z B_y$
$C_y=A_z B_x -A_x B_z$
$C_z=A_x B_y -A_y B_z\ .$
Note how the positive terms possess a forward alphabetical direction, xyzxyzx... (with x following z):
With the cross product we can also multiply three vectors together, in two different ways.
We can take the dot product of a vector with a cross product, a triple scalar product,
$\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = (\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}$
The absolute value of this product is the volume of the parallelpiped defined by the three vectors, A, B, and C
Alternately, we can take the cross product of a vector with a cross product, a triple vector product, which can be simplified to a combination of dot products.
$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) =(\mathbf{A} \cdot \mathbf{C}) \mathbf{B} -(\mathbf{A} \cdot \mathbf{B}) \mathbf{C}$
This form is easier to do calculations with.
The triple vector product is not associative.
$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) \ne (\mathbf{A} \times \mathbf{B}) \times \mathbf{C}$
A nice as well as a useful way to denote the cross product is using the indicial notation
$\mathbf{C = A \times B} = \; \epsilon^{ijk} \, A_j \, B_k \, \hat e_i$,
where $\epsilon^{ijk}$ is the Levi-Civita alternating symbol and $\hat e_i$ is either of the unit vectors $\hat i, \hat j, \hat k$. (A good exercise to convince yourself would be to use this expression and see if you can get C = A x B as defined before.)
Torque
Torque is the action of a force on a mass which induces it to revolve about some point, called the origin. It is defined as
$\mathbf{\tau}=\mathbf{r}\times \mathbf{F}$
where r is the position of the mass relative to the origin.
Notice that the torque is zero in a number of circumstances. If the force points directly toward or away from the origin, the cross product is zero, resulting in zero torque, even though the force is non-zero. Likewise, if r=0, the torque is zero. Thus, a force acting at the origin produces no torque. Both of these limits make sense intuitively, since neither induces the mass to revolve around the origin.
Angular momentum
The angular momentum of a mass relative to a point O is defined as
$\mathbf{L}=\mathbf{r}\times \mathbf{p}$
where p is the ordinary (also called "linear") momentum of the mass. The angular momentum is zero if the motion of the object is directly towards or away from the origin, or if it is located at the origin.
If we take the cross product of the position vector and Newton's second law, we obtain an equation that relates torque and angular momentum:
$\mathbf{r}\times \mathbf{F}=\mathbf{r}\times \frac{d\mathbf{p}}{dt} =\frac{d}{dt}\left( \mathbf{r}\times \mathbf{p} \right)-\frac{d\mathbf{r}}{dt}\times \mathbf{p}$
Since the cross product of parallel vectors is zero, this simplifies to
$\mathbf{\tau}=\frac{d\mathbf{L}}{dt}$
This is the rotational version of Newton's second law.
For both torque and angular momentum the location of the origin is arbitrary, and is generally chosen for maximum convenience. However, it is necessary to choose the same origin for both the torque and the angular momentum.
For the case of a central force, i. e., one which acts along the line of centers between two objects (such as gravity), there often exists a particularly convenient choice of origin. If the origin is placed at the center of the sun (which is assumed not to move under the influence of the planet's gravity), then the torque exerted on the planet by the sun's gravity is zero, which means that the angular momentum of the planet about the center of the sun is constant in time. No other choice of origin would yield this convenient result.
We already know about two fundamental conservation laws -- those of energy and linear momentum. We believe that angular momentum is similarly conserved in isolated systems. In other words, particles can exchange angular momentum between themselves, but the vector sum of the angular momentum of all the particles in a system isolated from outside influences must remain constant.
In the modern view, conservation of angular momentum is a consequence of the isotropy of space -- i. e., the properties of space don't depend on direction. This is in direct analogy with conservation of ordinary momentum, which we recall is a consequence of the homogeneity of space.
Angular velocity and centrifugal force
If an object is rotating about an axis, n, n being a unit vector, at frequency ω we say it has angular velocity ω. Despite the name, this is not the rate of change of a angle, nor even of a vector.
If a constant vector r is rotating with angular velocity ω about a fixed point then
$\dot{\mathbf{r}}=\boldsymbol{\omega} \times \mathbf{r}$
This says the acceleration is always at right angles to both the velocity and the axis of rotation.
When the axis is changing ω can be defined as the vector which makes this true.
Note that on the left hand side of this equation r is a vector in a fixed coordinate system with variable components but on the right hand side its components are given in a moving coordinate system, where they are fixed.
We can distinguish them more clearly by using subscripts, 'r' for rotating and 'f' for fixed, then extend this to arbitrary vectors
$\dot{\mathbf{g}}_f=\dot{\mathbf{g}}_r+ \boldsymbol{\omega} \times \mathbf{g}_f$
for any vector, g.
Using this, we can write Newton's second law in the rotating frame.
$\begin{matrix} \mathbf{F}_f & = & m \dot{\mathbf{v}}_f & & \\ & = & m\dot{\mathbf{v}}_r &+&m \boldsymbol{\omega} \times \mathbf{v}_f\\ & = & m\ddot{\mathbf{r}}_r &+&m \boldsymbol{\omega} \times (2\dot{\mathbf{r}}_r+ \boldsymbol{\omega} \times \mathbf{r}_f) \end{matrix}$
or, rearranging
$m\mathbf{a}_r=\mathbf{F}-2m(\boldsymbol{\omega} \times \mathbf{v}_r) -m\boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r})$
The mass behaves as if there were two additional forces acting on it. The first term, ω×v is called the Coriolis force. The second term is recognizable as the familiar centrifugal force.
The kinetic energy and angular momentum of the dumbbell may be split into two parts, one having to do with the motion of the center of mass of the dumbbell, the other having to do with the motion of the dumbbell relative to its center of mass.
To do this we first split the position vectors into two parts. The centre of mass is at.
$\mathbf{R}=\frac{M_1 \mathbf{r}_1 + M_2 \mathbf{r}_2}{M_1+M_2}$
so we can define new position vectors, giving the position of the masses relative to the centre of mass, as shown.
$\mathbf{r}^*_1=\mathbf{r}_1-\mathbf{R}_1 \quad \mathbf{r}^*_1=\mathbf{r}_1-\mathbf{R}_1$
The total kinetic energy is
$\begin{matrix}K & = & \frac{1}{2}M_1 V_1^2 &+ & \frac{1}{2}M_2 V_2^2 \\ & = & \frac{1}{2}M_1 |\mathbf{V}-\mathbf{v}^*_1|^2 &+ & \frac{1}{2}M_2 |\mathbf{V}-\mathbf{v}^*_2|^2\\ & = & \frac{1}{2}(M_1+M_2)V^2 & + & \frac{1}{2}M_1 {v^*_1}^2 + \frac{1}{2}M_2 {v^*_2}^2 \\ & = & K_{ext} & + & K_{int}\\ \end{matrix}$
which is the sum of the kinetic energy the dumbbell would have if both masses were concentrated at the center of mass, the translational kinetic energy and the kinetic energy it would have if it were observed from a reference frame in which the center of mass is stationary, the rotational kinetic energy.
The total angular momentum can be similarly split up
$\begin{matrix} \mathbf{L} & = & \mathbf{L}_{orb} & + & \mathbf{L}_{spin}\\ & = & (M_1+M_2)\mathbf{R}\times \mathbf{V} & + & (M_1\mathbf{r}^*_1 \times \mathbf{v}^*_1 + M_2\mathbf{r}^*_2 \times \mathbf{v}^*_2) \end{matrix}$
into the sum of the angular momentum the system would have if all the mass were concentrated at the center of mass, the orbital angular momentum, and the angular momentum of motion about the center of mass, the spin angular momentum.
We can therefore assume the centre of mass to be fixed.
Since ω is enough to describe the dumbbells motion, it should be enough to determine the angular momentum and internal kinetic energy. We will try writing both of these in terms of ω
First we use two results from earlier
$\mathbf{L}=\mathbf{r} \times \mathbf{v} \quad \mbox{and} \quad \mathbf{v}=\boldsymbol{\omega} \times \mathbf{r}$
to write the angular momentum in terms of the angular velocity
$\begin{matrix} \mathbf{L} & = & M_1\mathbf{r}^*_1 \times \mathbf{v}^*_1 & + & M_2\mathbf{r}^*_2 \times \mathbf{v}^*_2\\ & = & M_1\mathbf{r}^*_1 \times (\boldsymbol{\omega} \times \mathbf{r}^*_1) & + & M_2\mathbf{r}^*_2 \times (\boldsymbol{\omega} \times \mathbf{r}^*_2)\\ & = & M_1 \left( \boldsymbol{\omega} \left( \mathbf{r}^*_1 \cdot \mathbf{r}^*_1\right) - \mathbf{r}^*_1 \left( \mathbf{r}^*_1 \cdot \boldsymbol{\omega} \right)\right) & + & M_2 \left( \boldsymbol{\omega} \left( \mathbf{r}^*_2 \cdot \mathbf{r}^*_2\right) - \mathbf{r}^*_2 \left( \mathbf{r}^*_2 \cdot \boldsymbol{\omega}\right) \right)\\ & = & M_1 \left( \boldsymbol{\omega} d_1^2 - \mathbf{r}^*_1 \left( \mathbf{r}^*_1 \cdot \boldsymbol{\omega} \right)\right) & + & M_2 \left( \boldsymbol{\omega} d_2^2 - \mathbf{r}^*_2 \left( \mathbf{r}^*_2 \cdot \boldsymbol{\omega}\right) \right) \\ & = & (M_1d_1^2+M_2d_2^2) \boldsymbol{\omega} & - & \left (M_1 \mathbf{r}^*_1\left( \mathbf{r}^*_1\cdot \boldsymbol{\omega}\right) + M_2 \mathbf{r}^*_2\left( \mathbf{r}^*_2\cdot \boldsymbol{\omega}\right) \right) \end{matrix}$
The first term in the angular momentum is proportional to the angular velocity, as might be expected, but the second term is not.
What this means becomes clearer if we look at the components of L For notational convenience we'll write
$\mathbf{r}^*_1 = (x_1, y_1, z_1) \quad \mathbf{r}^*_2 = (x_2, y_2, z_2)$
These six numbers are constants, reflecting the geometry of the dumbbell.
$\begin{matrix} L_x & = & (M_1d_1^2+M_2d_2^2)\omega_x - (M_1 x_1^2+M_2 x_2^2)\omega_x \\ & & -(M_1 x_1 y_1+M_2 x_2 y_2)\omega_y - (M_1 x_1 z_1+M_2 x_2 z_2)\omega_z \\ L_y & = & (M_1d_1^2+M_2d_2^2)\omega_y - (M_1 x_1^2+M_2 x_2^2)\omega_y \\ & & -(M_1 x_1 y_1+M_2 x_2 y_2)\omega_x - (M_1 y_1 z_1+M_2 y_2 z_2)\omega_z \\ L_z & = & (M_1d_1^2+M_2d_2^2)\omega_z -(M_1 x_1^2+M_2 x_2^2)\omega_z \\ & & -(M_1 x_1 z_1+M_2 x_2 z_2)\omega_x - (M_1 y_1 z_1+M_2 y_2 z_2)\omega_y \\ \end{matrix}$
This, we recognise as being a matrix multiplication.
$\begin{pmatrix} L_x \\ L_y \\ L_z \end{pmatrix} = \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix} \begin{pmatrix} \omega_x \\ \omega_y \\ \omega_z \end{pmatrix}$
where
$\begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix} = \begin{pmatrix} M_1(d_1^2-x_1^2)+M_2(d_2^2-x_2^2) & -(M_1 x_1 y_1 + M_2 x_2 y_2) & -(M_1 x_1 z_1+M_2 x_2 z_2) \\ -(M_1 x_1 y_1 + M_2 x_2 y_2) & M_1(d_1^2-y_1^2)+M_2(d_2^2-y_2^2) & -(M_1 y_1 z_1+M_2 y_2 z_2) \\ -(M_1 x_1 z_1 + M_2 x_2 z_2) & -(M_1 y_1 z_1+M_2 y_2 z_2) & M_1(d_1^2-z_1^2)+M_2(d_2^2-z_2^2) \end{pmatrix}$
The nine coefficients of the matrix I are called moments of inertia.
By choosing our axis carefully we can make this matrix diagonal. E.g. if
$\mathbf{r}^*_1 = (d_1, 0, 0) \quad \mathbf{r}^*_2 = (-d_2, 0, 0)$
then
$\begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & M_1 d_1^2 +M_2 d_2^2 & 0 \\ 0 & 0 & M_1 d_1^2 +M_2 d_2^2 \end{pmatrix}$
Because the dumbbell is aligned along the x-axis, rotating it around that axis has no effect.
The relationship between the kinetic energy, T, and ω quickly follows.
$\begin{matrix} 2T & = & M_1 {v^*_1}^2 & + & M_2 {v^*_2}^2\\ & = & M_1 \mathbf{v}^*_1 \cdot (\boldsymbol{\omega} \times \mathbf{r}^*_1) & + & M_1 \mathbf{v}^*_2 \cdot (\boldsymbol{\omega} \times \mathbf{r}^*_2) \\ & = & M_1 \boldsymbol{\omega} \cdot (\mathbf{r}^*_1 \times \mathbf{v}^*_1) & + & M_2 \boldsymbol{\omega} \cdot (\mathbf{r}^*_2 \times \mathbf{v}^*_2) \end{matrix}$
On the right hand side we immediately recognise the definition of angular momentum.
$T=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{L}$
Substituting for L gives
$T=\frac{1}{2} \boldsymbol{\omega}\mathbf{I}\boldsymbol{\omega}$
Using the definition
$\boldsymbol{\omega} = \omega \mathbf{n} \quad \mathbf{n} \cdot \mathbf{n}=1$
this reduces to
$T=\frac{1}{2} I \omega^2$
where the moment of inertia around the axis n is
$I=\mathbf{n} \mathbf{I} \mathbf{n}$
a constant.
If the dumbbell is aligned along the x-axis as before we get
$T=\frac{1}{2} (M_1 d_1^2 +M_2 d_2^2) (\omega_y^2+\omega_z^2)$
These equations of rotational dynamics are similar to those for linear dynamics, except that I is a matrix rather than a scalar.
Summation convention
If we label the axes as 1,2, and 3 we can write the dot product as a sum
$\mathbf{u} \cdot \mathbf{v} = \sum_{i=1}^3 u_i v_i$
If we number the elements of a matrix similarly,
$\mathbf{A}= \begin{pmatrix}A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23} \\A_{31} & A_{32} & A_{33} \end{pmatrix} \quad \mathbf{B}= \begin{pmatrix} B_{11} & B_{12} & B_{13}\\ B_{21} & B_{22} & B_{23} \\B_{31} & B_{32} & B_{33} \end{pmatrix}$
we can write similar expressions for matrix multiplications
$(\mathbf{A} \mathbf{u})_i=\sum_{j=1}^3 A_{ij} u_j \quad (\mathbf{A} \mathbf{B})_{ik}=\sum_{j=1}^3 A_{ij} B_{jk}$
Notice that in each case we are summing over the repeated index. Since this is so common, it is now conventional to omit the summation sign.
$\mathbf{u} \cdot \mathbf{v} = u_i v_i \quad (\mathbf{A} \mathbf{u})_i= A_{ij} u_j \quad (\mathbf{A} \mathbf{B})_{ik}= A_{ij} B_{jk}$
We can then also number the unit vectors, êi, and write
$\mathbf{u}=u_i \hat{\mathbf{e}}_i$
which can be convenient in a rotating coordinate system.
Kronecker delta
The Kronecker delta is
$\delta_{ij}= \left\{ \begin{matrix} 1 & i=j\\ 0 & i\ne j \end{matrix} \right.$
This is the standard way of writing the identity matrix.
Levi-Civita (Alternating) symbol
Another useful quantity can be defined by
$\epsilon_{ijk}= \left\{ \begin{matrix} 1 & (i,j,k)= (1,2,3) \mbox{ or } (2,3,1) \mbox{ or } (3,1,2) \\ -1 & (i,j,k)= (2,1,3) \mbox{ or } (3,2,1) \mbox{ or } (1,3,2) \\ 0 & \mbox{ otherwise } \end{matrix} \right.$
With this definition it turns out that
$\mathbf{u} \times \mathbf{v} = \epsilon_{ijk} \hat{\mathbf{e}}_i u_j v_k$
and
$\epsilon_{ijk}\epsilon_{ipq}= \delta_{jp}\delta_{kq}-\delta_{jq}\delta_{kp} \,$
This will let us write many formulae more compactly.
The uneven dumbbell consisted of just two particles. These results can be extended to cover systems of many particles, and continuous media.
Suppose we have N particles, masses m1 to mN, with total mass M. Then the centre of mass is
$\mathbf{R}=\frac{\sum_n m_n \mathbf{r}_n}{M}$
Note that the summation convention only applies to numbers indexing axes, not to n which indexes particles.
We again define
$\mathbf{r}^*_n=\mathbf{r}_n-\mathbf{R}$
The kinetic energy splits into
$T=\frac{1}{2}M V^2 + \frac{1}{2} \sum_n m_n {v^*_n}^2$
and the angular momentum into
$\mathbf{L}=M \mathbf{R} \times \mathbf{V} + \sum_n m_n \mathbf{r}^*_n \times \mathbf{v}^*_n$
It is not useful to go onto moments of inertia unless the system is approximately rigid but this is still a useful split, letting us separate the overall motion of the system from the internal motions of its part.
If the set of particles in the previous chapter form a rigid body, rotating with angular velocity ω about its centre of mass, then the results concerning the moment of inertia from the penultimate chapter can be extended.
We get
$I_{ij}=\sum_n m_n (r_n^2\delta_{ij}-{r_n}_i {r_n}_j)$
where (rn1, rn2, rn3) is the position of the nth mass.
In the limit of a continuous body this becomes
$I_{ij}=\int_V \rho(\mathbf{r})(r^2\delta_{ij}-r_i r_j) \, dV$
where ρ is the density.
Either way we get, splitting L into orbital and internal angular momentum,
$L_i=M\epsilon_{ijk}R_j V_k+ I_{ij}\omega_j$
and, splitting T into rotational and translational kinetic energy,
$T=\frac{1}{2} M V_i V_i + \frac{1}{2} \omega_i I_{ij} \omega_j$
It is always possible to make I a diagonal matrix, by a suitable choice of axis.
Mass Moments Of Inertia Of Common Geometric Shapes
The moments of inertia of simple shapes of uniform density are well known.
Spherical shell
$I_{xx} = I_{yy} = I_{zz}=\frac{2}{3}Ma^2$
Solid ball
$I_{xx} = I_{yy} = I_{zz}=\frac{2}{5}Ma^2$
Thin rod
mass M, length a, orientated along z-axis
$I_{xx} = I_{yy} = \frac{1}{12}Ma^2 \quad I_{zz}=0$
Disc
mass M, radius a, in x-y plane
$I_{xx} = I_{yy} = \frac{1}{4}Ma^2 \quad I_{zz}=\frac{1}{2}Ma^2$
Cylinder
mass M, radius a, length h orientated along z-axis
$I_{xx} = I_{yy} = M\left( \frac{a^2}{4}+\frac{h^2}{12} \right) \quad I_{zz}=\frac{1}{2}Ma^2$
Thin rectangular plate
mass M, side length a parallel to x-axis, side length b parallel to y-axis
$I_{xx}=M\frac{b^2}{12} \quad I_{yy}=M\frac{a^2}{12} \quad I_{zz}=M \left( \frac{a^2}{12}+\frac{b^2}{12} \right)$
Newton's Laws: A second look
So far we've tacitly assumed we can just calculate force as a function of position, set up the ODE
$m\ddot{\mathbf{r}}=\mathbf{F}(\mathbf{r})$
and start solving.
It is not always so simple.
Often, we have to deal with motion under constraint; a bead sliding on a wire, a ball rolling without slipping, a weight dangling from a string.
There has to be some force keeping the bead on the wire, but we don't know what it is in advance, only what it does. This isn't enough information for us to write down the ODE.
We need a way of solving the problem without knowing the forces in advance.
How easy this is depends on the type of constraint.
• If the constraint is an inequality, as with the weight on the string, there is no straightforward analytical method.
• If the constraint can be written as a set of differential equations, and those equations can't be integrated in advance, there is an analytical method, but it is beyond the scope of this book. A ball rolling without slipping falls into this category.
• If the constraint can be written as a set of algebraic equations, and frictional forces are negligible, there is a straightforward method that solves the problem.
Generalised coordinates
Suppose we have a system of n particles satisfying k constraints of the form
$f_k(\mathbf{r}_1, \cdots , \mathbf{r}_n, t)=0$
then we can use the constraints to eliminate k of the 3n coordinates of the particles, giving us a new set of 3n-k independent generalised coordinates; q1, q2, … q3n-'k.
Unlike the components of the position vectors, these new coordinates will not all be lengths, and will not typically form vectors. They may often be angles.
We now need to work out what Newton's laws will look like in the generalised coordinates.
Derivation
The first step is to eliminate the forces of constraint.
We will need to consider a virtual displacement. This is an infinitesimal displacement made, while holding the forces and constraints constant. It is not the same as the infinitesimal displacement made during an infinitesimal time, since the forces and constraints may change during that time.
We write the total force on particle i as
$\mathbf{F}_i=\mathbf{F}^a_i+\mathbf{F}^c_i$
the sum of the externally applied forces and the forces of constraint.
Newton's second law states
$\mathbf{F}^a_i+\mathbf{F}^c_i = \mathbf{F}_i=\dot{\mathbf{p}}_i$
We take the dot product of this with the virtual displacement of particle i and sum over all particles.
$\sum_i \left(\mathbf{F}^a_i+\mathbf{F}^c_i-\dot{\mathbf{p}}_i \right)\cdot \delta\mathbf{r}_i = 0$
We now assume that the forces of constraint are perpendicular to the virtual displacement. This assumption is generally true in the absence of friction; e.g, the force of constraint that keeps a ball on a surface is normal to the surface.
This assumption is called D'Alembert's principle. Using it we can eliminate the forces of constraint from the problem, giving
$\sum_i \left(\mathbf{F}^a_i-\dot{\mathbf{p}}_i \right)\cdot \delta\mathbf{r}_i = 0$
or
$\sum_i \mathbf{F}^a_i \cdot \delta\mathbf{r}_i = \sum_i \dot{\mathbf{p}}_i \cdot \delta\mathbf{r}_i \quad (1)$
The left hand side of this equation is called the virtual work.
Now we must change to the generalised co-ordinate system.
We write
$\mathbf{r}_i=\mathbf{r}_i(q_1, q_2, \cdots ,q_{3n-k}, t)$
Using the chain rule gives
$\mathbf{v}_i=\sum_j \frac{\partial\mathbf{r}_i}{\partial q_j} \dot{q}_j + \frac{\partial\mathbf{r}_i}{\partial t}$
and
$\delta\mathbf{r}_i=\sum_j \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j$
Note that this implies
$\frac{\partial\mathbf{r}_i}{\partial q_j}= \frac{\partial\mathbf{v}_i}{\partial \dot{q}_j} \quad (2)$
The virtual work is, dropping the superscript,
$\begin{matrix} \sum_i \mathbf{F}_i \cdot \delta\mathbf{r}_i & = & \sum_{i, j} \mathbf{F}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j\\ & = & \sum_j Q_j \delta q_j \end{matrix}$
where the Qj are the components of the generalised force.
We now manipulate the right hand side of (1) into a form comparable with this last equation
The right hand side term is
$\begin{matrix} \sum_i \dot{\mathbf{p}}_i \cdot \delta\mathbf{r}_i & = & \sum_i m_i\ddot{\mathbf{r}}_i \cdot \delta\mathbf{r}_i \\ & = & \sum_{i, j} m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j \end{matrix}$
The terms in the coefficient of qj can be rearranged
$\begin{matrix} \sum_i m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} & = & \sum_i \left[ \frac{d}{dt} \left( m_i\dot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \right) - m_i\dot{\mathbf{r}}_i \cdot \frac{d}{dt} \left( \frac{\partial\mathbf{r}_i}{\partial q_j} \right) \right] \\ & = & \sum_i \left[ \frac{d}{dt} \left( m_i\mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial \dot{q}_j} \right) - m_i \mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial q_j}\right] \end{matrix}$
on substituting in equation (2) from above
On taking a close look at this last equation, we see a resemblance to the total kinetic energy,
$T=\frac{1}{2}\sum_i m_i v^2_i$
We now further rearrange to get an expression explicitly involving T.
$\begin{matrix} \sum_i m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} & = & \sum_i \left[ \frac{d}{dt} \left( m_i\mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial \dot{q}_j} \right) - m_i \mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial q_j}\right] \\ & = & \frac{d}{dt} \left( \frac{\partial}{\partial \dot{q}_j} \frac{1}{2} \sum_i m_i v^2_i \right) - \frac{\partial}{\partial q_j}\frac{1}{2} \sum_i m_i v^2_i \\ & = & \frac{d}{dt} \left( \frac{\partial}{\partial \dot{q}_j} T \right) - \frac{\partial}{\partial q_j} T \end{matrix}$
Putting this last expression into (1) along with the generalised force give
$\sum_j \left[ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} -Q_j \right] \delta q_j = 0$
Since the δqj, unlike the δri, are independent, this last equation can only be true if all the coefficients vanish.
That is we must have
$\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_j} - \frac{\partial T}{\partial q_j} = Q_j \quad (3)$
These are the equations of motion for the system, in a general set of coordinates for which all constraints are automatically satisfied.
For example, suppose we have a cylinder, mass m, radius a, rolling without slipping on a flat plane.
The kinetic energy of the cylinder is
$T=\frac{1}{2}m \dot{x}^2 + \frac{1}{8}ma^2 \dot{\theta}^2$
using the results from Rigid Bodies, where x is the axis in the plane perpendicular to the axis of the cylinder, and θ is the angle of rotation.
Rolling without slipping implies
$\dot{x}=a\dot{\theta}$
so we get
$T=\frac{5}{8}m\dot{x}^2 \quad \frac{5}{4}m\ddot{x}=Q_x$
The cylinder has the same kinetic energy as if its mass were 20% greater. If there is no torque on the cylinder then Qx=Fx, and the cylinder behaves in every respect as though it were a 20% larger point mass.
To use (3) more generally, we need an expression for the Qj
Suppose, as is often the case, that
$\mathbf{F}_i=-\frac{\partial}{\partial \mathbf{r}_i} V$
then, by definition
$\begin{matrix} \sum_j Q_j \delta q_j & = & \sum_i \mathbf{F}_i \cdot \delta\mathbf{r}_i \\ & = & \sum_{i, j} \mathbf{F}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} \delta q_j \\ & = & -\sum_{i, j} \frac{\partial V}{\partial \mathbf{r}_i} \frac{\partial \mathbf{r}_i}{\partial q_j} \delta q_j \\ & = & -\sum_j \frac{\partial V}{\partial q_j} \delta q_j \end{matrix}$
so, equating coefficients, the generalised force is
$Q_j=-\frac{\partial V}{\partial j}$
Putting this generalised force into (3) gives
$\frac{d}{dt} \frac{\partial (T-V)}{\partial \dot{q}_j} - \frac{\partial (T-V)}{\partial q_j} = 0$
since V has been assumed independent of the velocities.
In fact, this last equation will still be true for some velocity dependent forces, most notably magnetism, for a suitable definition of V, but we won't prove this here.
We call the T-V the Lagrangian, L, and write
$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} = 0$
We call these equations Lagrange's equations. They are useful whenever Cartesian co-ordinates are inconvenient, including motion under constraint.
Example
Suppose we have two identical point masses, m, connected by a string, length a. The string is threaded through an hole in a flat table so that the upper mass is moving in a horizontal plane without friction, and the lower mass is always vertically below the hole. The distance of the upper mass from the hole is r.
The position of the mass on the table is best described using polar coordinates, (r,θ). Its kinetic energy is then
$\frac{1}{2}m \left( \dot{r}^2+r^2 \dot{\theta}^2 \right)$
The velocity of the lower mass is d(a-r)/dt=-dr/dt, so the total kinetic energy is
$T= \frac{1}{2}m \left( 2\dot{r}^2+r^2 \dot{\theta}^2 \right)$
The potential energy is
$V = mg(a-r)$
where g is the gravitational acceleration.
This means
$L = \frac{1}{2}m \left( 2\dot{r}^2+r^2 \dot{\theta}^2 \right) +mg(r-a)$
and the equations of motion are
$\begin{matrix} \frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} & = & m\frac{d}{dt}\left( r^2 \dot{\theta}\right) & = & 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot{r}} - \frac{\partial L}{\partial r} & = & 2m\ddot{r}-mr\dot{\theta}^2+mg & = & 0\end{matrix}$
The first of these equations says that the angular momentum is constant, as expected since there is no torque on the particles. If we call this constant angular momentum l then we can write
$\dot{\theta}=\frac{l}{mr^2}$
and the second equation of motion becomes
$\ddot{r}=\frac{l^2}{2m^2r^3}-\frac{g}{2}$
Clearly, if initially
$l^2>g m^2 r^3, \,$
then the lower ball will be pulled out of the hole, at which point these equations of motions cease to apply. They only hold when 0≤ra, a constraint which is not easily tamable.
Notice, we have not needed to calculate the tension in the string, which is the force of constraint in this problem.
Reformulating Newton
In the last chapter we saw how to reformulate Newton's laws as a set of second order ordinary differential equations using arbitrary coordinates:
$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}- \frac{\partial L}{\partial q_i} = 0$
It is often far easier to solve a problem by starting with the Lagrangian than by explicitly using Newton's laws. The Lagrangian is also more useful for theoretical analysis.
However, the Lagrangian approach is not the only useful way to reformulate Newton. There is a second, related, approach which is also highly useful for analysis; namely, writing the equations of motion as a set of first order ordinary differential equations in a particularly natural way.
To do this we will use a standard technique, familiar from your calculus studies.
Derivation
Consider the function
$H(q,p,t)=\dot{q}_i p_i -L(q, \dot{q}, t)$
where the qi are the generalised coordinates, the pi are a set of new variables whose meaning we will soon see, H is to be a function of the p‘s and q‘s alone, and the summation convention is being used.
Then we have
$dH=\dot{q}_i dp_i + p_i d\dot{q}_i -\frac{\partial L}{\partial \dot{q}_i} d\dot{q}_i -\frac{\partial L}{\partial q_i} dq_i -\frac{\partial L}{\partial t} dt \quad (1)$
but, since H is a function of the p‘s and q‘s alone we can also write
$dH=\frac{\partial H}{\partial q_i}dq_i + \frac{\partial H}{\partial p_i}dp_i + \frac{\partial H}{\partial t}dt \quad (2)$
For these two equations to both be true the coefficients of the differentials must be equal.
Equation (2) does not contain a term in $d\dot{q}_i$, so the coefficient of that term in equation (1) must be zero. I.e,
$p_i=\frac{\partial L}{\partial \dot{q}_i}$
which gives us a definition of the pi. Using this in Lagrange's equations gives
$\dot{p}_i=\frac{\partial L}{\partial q_i}$
and equation (1) simplifies to
$dH=\dot{q}_i dp_i - \dot{p}_i dq_i -\frac{\partial L}{\partial t} dt$
A comparison of coefficients with equation (2) now gives the desired set of first order equations for the motion,
$\dot{q}_i=\frac{\partial H}{\partial p_i} \quad \dot{p}_i=-\frac{\partial H}{\partial q_i}$
These are called Hamilton's equations and H is called the Hamiltonian.
Physical meaning of H and p
To see what H and the pi's actually are, let's consider a few typical cases.
First, let's look at a free particle, one subject to no forces. This will let us see what the pi's mean, physically.
If we use Cartesian coordinates for the free particle, we have
$L = T=\frac{1}{2}m \left( \dot{x}^2+ \dot{y}^2+ \dot{z}^2 \right)$
The pi's are the differential of this with respect to the velocities.
$p_x=\frac{\partial L}{\partial \dot{x}}=m \dot{x} \quad p_y=\frac{\partial L}{\partial \dot{y}}=m \dot{y} \quad p_z=\frac{\partial L}{\partial \dot{z}}=m \dot{z}$
These are the components of the momentum vector.
If we use cylindrical coordinates, we have
$L = T=\frac{1}{2}m \left( \dot{r}^2+ r^2\dot{\theta}^2+ \dot{z}^2 \right)$
and
$p_r=\frac{\partial L}{\partial \dot{r}}=m \dot{r} \quad p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}=m r^2 \dot{\theta} \quad p_z=\frac{\partial L}{\partial \dot{z}}=m \dot{z}$
This time, pz is the component of momentum is the z direction, pr is the radial momentum, and pθ is the angular momentum, which we've previously seen is the equivalent of momentum for rotation.
Since, in these familiar cases, the pi's are momenta, we generalise and call the pi's conjugate momenta.
Note that from Hamilton's equations, we see that if H is independent of some coordinate, q, then
$\dot{p}_q=-\frac{\partial H}{\partial q} = 0$
so the momentum conjugate to q is conserved. This connection between conservation laws and coordinate independence lies at the heart of much physics.
Secondly, to see the physical meaning of H, we'll consider a particle moving in a potential V, described using Cartesian coordinates.
This time the Lagrangian is
$L= T-V = \frac{1}{2}m \left( \dot{x}^2+ \dot{y}^2+ \dot{z}^2 \right)-V(x,y,z)$
We find that the pi's are
$p_x=\frac{\partial L}{\partial \dot{x}}=m \dot{x} \quad p_y=\frac{\partial L}{\partial \dot{y}}=m \dot{y} \quad p_z=\frac{\partial L}{\partial \dot{z}}=m \dot{z}$
so, again, they are the momenta
Now lets calculate H.
$\begin{matrix} H & = & \dot{x}p_x+ \dot{y}p_y+ \dot{y}p_y & - & L & &\\ & = & m \left( \dot{x}^2+ \dot{y}^2+ \dot{z}^2 \right) & - & \frac{1}{2}m \left( \dot{x}^2+ \dot{y}^2+ \dot{z}^2 \right) & + & V\\ & = & \frac{1}{2}m \left( \dot{x}^2+ \dot{y}^2+ \dot{z}^2 \right) & & & + & V\\ & = & T & & & + & V \end{matrix}$
so H is the total energy, written as a function of the position and momentum.
This is not always true, but it will be true for all common systems which conserve energy.
In general, if we know how the energy of a systems depends on speeds and positions, we know everything about the system.
Physicists will often start work on a problem by writing down an expression for H, or L, and never bother calculating the actual forces.
Deeper meaning
If we compare Hamilton's equations,
$\dot{q}_i=\frac{\partial H}{\partial p_i} \quad \dot{p}_i=-\frac{\partial H}{\partial q_i}$
with the equations of geometrical optics.
$\dot{x}_i=\frac{\partial \omega}{\partial k_i} \quad \dot{k}_i=-\frac{\partial \omega}{\partial x_i}$
we see that both sets of equations have the same form.
If we made a substitution of the form,
$H=\alpha \omega \quad p=\alpha k$
for any α, the two sets of equations would become identical.
Now, geometrical optics is an approximation to the full solution of any wave equation, valid when the wavelength is small.
Since classical mechanics is just like geometrical optics, classical mechanics could also be small wavelength approximation to a more accurate wave theory.
This is not a proof, could be is not the same as is, but it does show that classical mechanics is compatible with wave theories, contrary to intuition.
We will see later that this is not a coincidence, classical mechanics really is a small wavelength approximation to a more accurate wave theory, namely quantum mechanics, and that, in that theory, the energy of matter waves is proportional to frequency and the momentum to wave number, just as the substitution above requires.
Hamilton's equations can also be used to construct a mathematical gadget, called a Poisson bracket, which lets us write the equations of classical mechanics in a way which makes the connections with quantum mechanics even clearer, but an explanation of how would be beyond the scope of this book.
Since classical mechanics is just like geometrical optics, there should be an analog to Fermat's principle, that light takes the quickest path, and there is.
Where light takes the quickest path from a to b, matter takes the path of least action, where the action is defined as
$\int_a^b L(x,\dot{x},t) \, dt$
It can be proved that this integral takes extremal values if and only if the system obeys Lagrange's equations. Saying matter takes the path of least action is equivalent to saying it obeys Lagrange's equations, or Newton's laws.
Furthermore, it turns out that in general relativity, the action is proportional to the time taken, so matter is also really taking the quickest path between two points.
We've seen that an Hamiltonian of the form
$H=\frac{\mathbf{p}\cdot\mathbf{p}}{2m}+V(\mathbf{r})$
describes motion under a conservative force, but this is not the most general possible Hamiltonian.
Many waves are described by another simple form, in the geometrical optics limit,
$H=f(|\mathbf{p}|)$
What happens if we consider other forms for the Hamiltonian?
Suppose we have
$H=\frac{\mathbf{p}\cdot\mathbf{p}}{2m} + \mathbf{p} \cdot \mathbf{A}(\mathbf{r})$
The dot product term is the simplest scalar we can add to the kinetic energy that's dependent on momentum. We'll see that it acts like a potential energy.
Using Hamilton's equations we can immediately write down equations of motion.
$\begin{matrix} \dot{x}_i & = & \frac{\partial H}{\partial p_i} & = & \frac{p_i}{m} + A_i \\ \dot{p}_i & = & -\frac{\partial H}{\partial x_i} & = & -p_j \frac{\partial A_j}{\partial x_i} \end{matrix}$
Note that we are using the summation convention here.
Now that we've got the equations of motion, we need to work out what they mean.
The first thing we notice is that the momentum is no longer mv. There is an additional contribution from the potential field A. We can think of this as a form of potential momentum, analogous to potential energy.
The force on the particle is
$F_i = m \ddot{x}_i = \dot{p}_i + m\frac{d}{dt}A_i(\mathbf{r})$
Using the chain rule, and substituting for dp/dt with the second equation of motion, we get
$F_i = -p_j \frac{\partial A_j}{\partial x_i} + m\dot{x}_j \frac{\partial A_i}{\partial x_j}$
Using the first equation, this becomes
$F_i = m \dot{x}_j \left( \frac{\partial A_i}{\partial x_j} -\frac{\partial A_j}{\partial x_i} \right) + m A_j \frac{\partial A_j}{\partial x_i}$
The term in brackets is recognizable as being the type of thing we see in cross products. With a little manipulation, using the Kronecker delta and alternating symbol, we can write
$\begin{matrix} F_i & = & \dot{x}_j \left( \delta_{il}\delta_{jk}-\delta_{ik}\delta_{jl} \right) \frac{\partial A_l}{\partial x_k} & + & m A_j \frac{\partial A_j}{\partial x_i} \\ & = & \dot{x}_j \epsilon_{nij} \epsilon_{nlk} \frac{\partial A_l}{\partial x_k} & + & m A_j \frac{\partial A_j}{\partial x_i} \\ & = & -\epsilon_{ijn} \dot{x}_j \left( \nabla \times \mathbf{A} \right)_n & + & m A_j \frac{\partial A_j}{\partial x_i} \\ & = & -\epsilon_{ijn} \dot{x}_j \left( \nabla \times \mathbf{A} \right)_n & + & \frac{1}{2} \frac{\partial}{\partial x_i} A_j A_j \end{matrix}$
so, writing the last line as a vector equation,
$\mathbf{F} = -m \mathbf{v} \times (\nabla \times \mathbf{A}) + \frac{1}{2} m \nabla A^2$
The force from this potential has a component perpendicular to its curl and to the velocity, and another component which is the gradient of a scalar.
If we add a carefully chosen potential energy to the Hamiltonian we can cancel out this second term.
$H=\frac{p^2}{2m} + \mathbf{p} \cdot \mathbf{A}(\mathbf{r}) + \frac{1}{2} m A^2 \qquad \mathbf{F}= -m \mathbf{v} \times \left (\nabla \times \mathbf{A} \right)$
This Hamiltonian can be simplified to
$H = \frac{1}{2m} \left( \mathbf{p}+ m\mathbf{A} \right)^2$
where the term in bracket is mv, written as a function of momentum.
This simple modification to the Hamiltonian gives us a force perpendicular to velocity, like magnetism. Because the force and velocity are perpendicular, the work done by the force is always zero.
We can use potential fields to describe velocity dependent forces, provided that they do no work.
The coefficient of A in these expression is arbitrary. Changing it amount merely to measuring A in different units so we can equally well write
$H = \frac{1}{2m} \left( \mathbf{p}+ \alpha\mathbf{A} \right)^2 \qquad \mathbf{F}= -\alpha \mathbf{v} \times \left (\nabla \times \mathbf{A} \right)$
Having α=m means A is measured in units of velocity, which may sometimes be convenient, but we can use any other constant value of α that suits our purpose. When we come to study relativity, we'll use α=-1.
Notice that the force depends only on the curl of A, not on A itself. This means we can add any function with zero curl to A without changing anything, just as we can add a constant to the potential energy.
Furthermore, it is a standard result of vector calculus that any vector field is the sum of two components, one with zero curl, the other with zero divergence. Since the zero curl component does not affect the total force, we can require it to be zero; i.e. we can require A to have zero divergence.
$\nabla \cdot \mathbf{A}= 0$
This is called a gauge condition. For the moment, it can be considered to define especially natural values for the integration constants.
Mach's principle is the name given by Einstein to a vague hypothesis first supported by the physicist and philosopher Ernst Mach.
Harmonic Oscillators
Energy Analysis
Figure 11.2: Potential, kinetic, and total energy of a one-dimensional harmonic oscillator plotted as a function of spring displacement.
For a spring, Hooke's law says the total force is proportional to the displacement, and in the opposite direction.
$F=-kx$
Since this is independent of velocity, it is a conservative force. We can integrate to find the potential energy of the mass-spring system,
$V(x)=\frac{1}{2}kx^2$
Since a potential energy exists, the total energy
$E=\frac{1}{2}kx^2+\frac{1}{2}m\dot{x}^2$
is conserved, i. e., is constant in time.
We can now use energy conservation to determine the velocity in terms of the position
$\dot{x}=\pm \sqrt{\frac{2E}{m}-\frac{k}{m}x^2}$
We could integrate this to determine the position as a function of time, but we can deduce quite a bit from this equation as it is.
It is fairly evident how the mass moves. From Hooke's law, the mass is always accelerating toward the equilibrium position, so we know which sign of the square root to take.
The velocity is zero when
$x=\pm \sqrt{\frac{2E}{k}}$
If x were larger than this the velocity would have to be imaginary, clearly impossible, so the mass must be confined between these values. We can call them the turning points.
If the mass is moving to the left, it slows down as it approaches the left turning point. It stops when it reaches this point and begins to move to the right. It accelerates until it passes the equilibrium position and then begins to decelerate, stopping at the right turning point, accelerating toward the left, etc. The mass thus oscillates between the left and right turning points.
How does the period of the oscillation depend on the total energy of the system? We can get a general idea without needing to solve the differential equation.
There are only two parameters the period, T, could depend on; the mass, m, and the spring constant, k.
We know T is measured in seconds, and m in kilograms. For the units in Hooke's law to match, k must be measured in N·m-1, or equivalently, in kg·s-2.
We immediately see that the only way to combine m and k to get something measured in seconds is to divide, cancelling out the kg's.
Therefore T$\sqrt{m/k}$
We have established the general way the period depends on the parameters of the problem, without needing to use calculus.
This technique is called dimensional analysis, and has wide application. E.g., if we couldn't calculate the proportionality constant exactly, using calculus, we'd be able to deduce by doing one experiment. Without dimensional analysis, if calculus failed us we'd have to do scores of experiments, each for different combinations of m and k.
Fortunately, the proportionality constants are typically small numbers, like $3\sqrt{2}$ or $\sqrt{\pi}/2$.
Analysis Using Newton's Laws
The acceleration of the mass at any time is given by Newton's second law
$a=\frac{d^2x}{dt^2}=\frac{F}{m}=-\frac{kx}{m}$
An equation of this type is known as a differential equation since it involves a derivative of the dependent variable . Equations of this type are generally more difficult to solve than algebraic equations, as there are no universal techniques for solving all forms of such equations. In fact, it is fair to say that the solutions of most differential equations were originally obtained by guessing!
There are systematic ways of solving simple differential equations, such as this one, but for now we will use our knowledge of the physical problem to make an intelligent guess.
We know that the mass oscillates back and forth with a period that is independent of the amplitude of the oscillation. A function which might fill the bill is the sine function. Let us try substituting,
$x=A\, \sin \omega t$
where ω is a constant, into this equation.
We get
$-\omega^2 A\, \sin \omega t = -A \frac{k}{m}\, \sin \omega t$
Notice that the sine function cancels out, leaving us with ω2=k/m. The guess thus works if we set
$\omega=\sqrt{\frac{k}{m}}$
This constant is the angular oscillation frequency for the oscillator, from which we infer the period of oscillation to be
$T=2\pi \sqrt{\frac{m}{k}}$
This agrees with the result of the dimensional analysis. Because this doesn't depend on A, we can see that the period is independent of amplitude.
It is easy to show that the cosine function is equally valid as a solution,
$x=B\, \cos \omega t$
for the same ω.
In fact, the most general possible solution is just a combination of these two, i. e.
$x=A\, \sin \omega t+B\, \cos \omega t$
The values of A and B depend on the position and velocity of the mass at time t=0.
If we wiggle the left end of the spring in the above diagram by amount d=d0sin ωFt, rather than rigidly fixing it, we have a forced harmonic oscillator.
The constant d0 is the amplitude of the imposed wiggling motion. The forcing frequency ωF is not necessarily equal to the natural or resonant frequency of the mass-spring system. Very different behavior occurs depending on whether it is less than, equal to, or greater than ω.
Given the above wiggling, the force of the spring on the mass becomes
$F = -k(x-d) = -k(x-d_0 \sin \omega_F t) \,$
since the length of the spring is the difference between the positions of the left and right ends. Proceeding as for the unforced mass-spring system, we arrive at the differential equation
$\frac{d^2x}{dt^2}+\frac{k}{M}x= \frac{k}{M}d_0 \sin \omega_F t$
The solution to this equation turns out to be the sum of a forced part in which x∝ sin ωFt, and a free part which is the same as the solution to the unforced equation. We are primarily interested in the forced part of the solution, so let us set x=d0sin ωFtand substitute this into the equation of motion, giving:
$-\omega_F^2 x_0 \sin \omega_F t + \frac{k}{M}x_0 \sin \omega_F t = \frac{k}{M}d_0 \sin \omega_F t$
The sine factor cancels leaving us with an algebraic equation for , the amplitude of the oscillatory motion of the mass.
Solving for the ratio of the oscillation amplitude of the mass to the amplitude of the wiggling motion, we find
$\frac{x_0}{d_0} = \frac{1}{1-M \omega_F^2/k} = \frac{1}{1-\omega_F^2/\omega^2}$
where we have recognized that k/M2, the square of the frequency of the free oscillation.
Notice that if ωF<ω, then the motion of the mass is in phase with the wiggling motion and the amplitude of the mass oscillation is greater than the amplitude of the wiggling. As the forcing frequency approaches the natural frequency of the oscillator, the response of the mass grows in amplitude.
When the forcing is at the resonant frequency, the response is technically infinite, though practical limits on the amplitude of the oscillation will intervene in this case -- for instance, the spring cannot stretch or shrink an infinite amount. In many cases friction will act to limit the response of the mass to forcing near the resonant frequency.
When the forcing frequency is greater than the natural frequency, the mass actually moves in the opposite direction of the wiggling motion -- i. e., the response is out of phase with the forcing. The amplitude of the response decreases as the forcing frequency increases above the resonant frequency.
Forced and free harmonic oscillators form an important part of many physical systems. For instance, any elastic material body such as a bridge or an airplane wing has harmonic oscillatory modes. A common engineering problem is to ensure that such modes are damped by friction or some other physical mechanism when there is a possibility of excitation of these modes by naturally occurring processes. A number of disasters can be traced to a failure to properly account for oscillatory forcing in engineered structures.
We often encounter systems which contain multiple harmonic oscillators, such as this:
two identical masses, m, the first attached to a wall by a spring with constant k, and the second attached to the first by another, identical spring.
If the springs weren't linked they'd both vibrate at the same frequency, ω=√(k/m). Linking the springs changes this.
To find out how the linked system behaves, we will start with the Lagrangian, using the displacements of the masses, x1 and x2, as our coordinates.
A moment's inspection of the system shows
$T=\frac{1}{2}m \left( \dot{x}^2_1 + \dot{x}^2_2 \right) \quad V=\frac{1}{2}k \left( x_1^2 + (x_2-x_1)^2 \right)$
so, using ω²=k/m,
$L=\frac{1}{2}m \left( \dot{x}^2_1 + \dot{x}^2_2 \right) - \frac{1}{2}m\omega^2 \left( x_1^2 + (x_2-x_1)^2 \right)$
The equations of motion immediately follow.
$\begin{matrix} \ddot{x}_1 & = & -\omega^2 (2x_1 -x_2) \\ \ddot{x}_2 & = & -\omega^2 (x_2 -x_1) \end{matrix} \quad (1)$
To solve these equations we try a solution in trig functions
$x_1 = A_1 \sin \Omega t \quad x_2 = A_2 \sin \Omega t$
Substituting this into (1) gives
$\begin{matrix} -\Omega^2 A_1 & + & \omega^2 (2A_1-A_2) & = & 0 \\ -\Omega^2 A_2 & + & \omega^2 (A_2-A_1) & = & 0 \\ \end{matrix}$
We would get the same equations from any trig function solution of the same frequency.
Gathering the coefficients of A1 and A2 together lets rewrite the last equation as
$\begin{pmatrix} 2\omega^2-\Omega^2 & -\omega^2 \\ -\omega^2 & \omega^2-\Omega^2 \end{pmatrix} \begin{pmatrix} A_1 \\ A_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad (2)$
We can only solve this equation if the determinant of the matrix is zero.
$\Omega^4-3\omega^2 \Omega^2 + \omega^4 = 0$
The solutions are
$\Omega_{\pm}^2=\frac{3 \pm \sqrt{5} }{2} \omega^2$
so the combined system has two natural frequencies, one lower and one higher than the natural frequency of the individual springs. This is typical.
We can also calculate the ratio of A1 and A2 from (2). Dividing by A2 gives
$\frac{A_1}{A_2}= 1-\frac{\Omega_{\pm}^2}{\omega^2}$
• For the lower frequency, Ω-, which is less than ω, the ratio is positive, so the two masses move in the same direction with different amplitudes. They are said to be in phase.
• For the higher frequency, Ω+, which is more than ω, the ratio is negative, so the two masses move in the opposite direction with different amplitudes. They are said to be in antiphase.
This behaviour is typical when pairs of harmonic oscillators are coupled.
The same approach can also be used for systems with more than two particles.
Introduction
In principle, we could use the methods described so far to predict the behavior of matter by simply keeping track of each atom.
In practice, this is not a useful approach. Instead, we treat matter as a continuum.
In this section we will see how this is done, and how it leads us back to waves.
For an example, we will consider a set of N+1 identical springs and masses, arranged as in the previous section, with spring 0 attached to the wall, and mass N free.
We are interested in what happens for large N, the continuum limit.
The system has three other parameters; the spring constant k, the particle mass Δm, and the spring rest spacing Δa, where variable names have been chosen for future convenience. These can be combined with N to give a similar set of parameters for the system.
Suppose all the masses are displaced by d from rest, with total displacement D=dN of the system end, then the total potential energy is kNd2/2 = kD2/(2N), so
• The system's spring constant is K=k/N
• The system mass is m=ΔmN
• The system rest length is a=ΔaN
If, as we increase N we change the other three parameters to keep K, m, and a constant then in the large N limit this discrete system will look like a spring with mass continuously distributed along its length.
For coordinates, we will use the displacements of each mass, xn. For large N the displacement will vary approximately continuously with distance. We can regard it as a continuous function, x, with
$x(n\Delta a)=x_n \,$
The kinetic energy of the system is then simply
$T=\frac{1}{2} \Delta m \sum_0^N \dot{x}_n^2$
The total potential energy is similarly
$V=\frac{1}{2}k x_0^2 + \frac{1}{2}k \sum_1^N (x_n-x_{n-1})^2$
From this we may deduce the equations of motion in two different ways.
Equations of motion: First approach
If we take the large N limit first, these sums become integrals.
$\begin{matrix} T & = & \frac{m}{2a}\sum_0^N \dot{x}_n^2 \Delta a \\ \lim_{N \rarr \infty} T & = & \frac{m}{2a} \int_0^a \dot{x}^2 ds \end{matrix}$
where s is the distance from the wall, and
$\begin{matrix} V & = & \frac{1}{2}k x_0^2 + \frac{1}{2}k \sum_0^{N-1} (x_n-x_{n+1})^2 \\ & = & \frac{1}{2}k x_0^2 + \frac{1}{2}K \sum \left( \frac{ x(n\Delta a+\Delta a)- x(n\Delta a)}{\Delta a} \right)^2 a \Delta a\\ \lim_{N \rarr \infty} V & = & \frac{1}{2}k x(0)^2 + \frac{1}{2}Ka \int \left( \frac{dx}{ds} \right)^2 ds \end{matrix}$
Since the springs always remain attached to the wall, x(0)=0.
The integrand for T is the product of the density and the square of the velocity, just as we might naively expect. Simalarly, V is the integral of the potential energy of a infinitesimal spring over the length of the system.
Using the Lagrangian will let us get equations of motion from these integrals.
It is
$\begin{matrix} L & = & \frac{m}{2a} \int_0^a \dot{x}^2 ds - \frac{1}{2}Ka\int_0^a \left( \frac{dx}{ds} \right)^2 ds \\ & = & \frac{m}{2a} \int_0^a (\dot{x}^2 - \frac{Ka^2}{m} x^{\prime 2}) ds\\ \end{matrix}$
The action of the system is
$\int L dt = \int \int \mathcal{L}(\dot{x},x^\prime) dx dt \quad \mbox{ where } \mathcal{L}=\frac{m}{2a}(\dot{x}^2 - \frac{Ka^2}{m} x^{\prime 2})$
Here, we are integrating over both space and time, rather than just space, but this is still very similar in form to the action for a single particle.
We can expect that the principle of least action will lead us to the natural extension of Lagrange's equations to this action,
$\frac{d}{ds}\frac{\partial \mathcal{L}}{\partial x^\prime}+ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{x}} = \frac{\partial \mathcal{L}}{\partial x}$
This equation can be proven, using the calculus of variations. Using it for this particular Lagrangian gives
$\frac{\partial^2 x}{\partial t^2} - \frac{Ka^2}{m}\frac{\partial^2 x}{\partial s^2} = 0$
This is a partial differential equation. We will not be discussing its solution in detail, but we will see that it describes waves.
First, though, we will confirm that these equations are the same as we get if we find the equations of motion first, then take the large N limit.
Equations of motion: Second approach
First, we will look at the potential energy to see how it depends on the displacements.
$\begin{matrix} V & = & \frac{1}{2}k x_0^2 + \frac{1}{2}k \sum_0^{N-1} (x_n-x_{n+1})^2 \\ & = & \frac{1}{2}k \left( x_0^2 \cdots (x_{n-1}-x_n)^2 + (x_n-x_{n+1})^2 \cdots (x_N-x_{N-1})^2 \right) \\ & = & \frac{1}{2}k \left( 2x_0^2 -2x_0 x_1 \cdots -2x_n x_{n+1} + 2x^2_n -2x_n x_{n+1} \cdots x^2_N - 2x_N x_{N-1} \right) \\ & = & \frac{1}{2}k \left( 2x_0^2 -2x_0 x_1 \cdots 2x_n (x_n-x_{n+1}-x_{n-1}) \cdots x^2_N - 2x_N x_{N-1} \right) \\ \end{matrix}$
Notice that we need to treat the displacements of the first and last masses differently from the other coordinates, because the dependence of the Langragian on them is different.
The kinetic energy is symmetric in the coordinates,
$T=\frac{1}{2} \Delta m \sum_0^N \dot{x}_n^2$
Using Lagrange's equations, we get that, for x0
$\Delta m \ddot{x}_0 + 2k x_0 = k x_1,$
for xN,
$\Delta m \ddot{x}_N + k x_N = k x_{N-1},$
and, for all the other xn
$\Delta m \ddot{x}_n + 2k x_n = k (x_{n+1}+x_{n-1}),$
We can replace k and Δm by the limiting values K and m using
$\frac{k}{\Delta m} = \frac{K a^2}{m \Delta a^2}$
giving us
$\begin{matrix} \ddot{x}_0 & = & \frac{K a^2}{m} \frac{x_1-2x_0}{\Delta a^2}\\ \ddot{x}_n & = & \frac{K a^2}{m} \frac{x_{n-1}+x_{n+1}-2x_n}{\Delta a^2}\\ \ddot{x}_N & = & \frac{K a^2}{m} \frac{x_{N-1}-x_N}{\Delta a^2} \end{matrix}$
Looking at the general structure of the right hand side, we see difference between the displacement at nearby points divided by the distance between those points, so we expect that in the limit we will get differentials with respect to s, the distance from the wall.
As N tends to infinity both x0 and x1 tend to x(0), which is always zero, so the equation of motion for x0 is always true.
In the continuum limit, the equation for xN becomes
$\ddot{x}(a) = -\frac{K a^2}{m} \frac{1}{\Delta a} \frac{dx}{ds}$
Since Δa tends to zero, for this to be true we must have x'=0 at a.
For the other displacements,
$\begin{matrix} \lim_{\Delta a \rarr 0} \frac{x_{n-1}+x_{n+1}-2x_n}{\Delta a^2} & = & \lim_{\Delta a \rarr 0} & \frac{1}{\Delta a} \left( \frac{x_{n+1}-x_n}{\Delta a} - \frac{x_n-x_{n-1}}{\Delta a} \right) \\ & = & \lim_{\Delta a \rarr 0} & \frac{x^\prime((n+1)\Delta a)-x^\prime(n\Delta a)}{\Delta a} \\ & = & \left. \frac{d^2 x}{ds^2} \right|_{x=a} & \\ \end{matrix}$
so we get the equation
$\frac{\partial^2 x}{\partial t^2} - \frac{Ka^2}{m}\frac{\partial^2 x}{\partial s^2} = 0$
just as with the other approach.
Waves
It is intuitively obvious that if we flick one of the masses in this system, vibrations will propagate down the springs, like waves, so we look for solutions of that form.
A generic travelling wave is
$x = A \sin (\omega t - \kappa x + \alpha) \,$
Substituting this informed guess into the equation gives
$-A \omega^2 \sin (\omega t - \kappa x + \alpha) = -\frac{Ka^2}{m} \kappa^2 \sin (\omega t - \kappa x + \alpha)$
so this wave is a solution provide the frequency and wavenumber are related by
$\omega^2 = \frac{Ka^2}{m} \kappa^2$
The speed of these waves, c, is
$c = \frac{d\omega}{d\kappa} = \pm \sqrt{\frac{Ka^2}{m}}$
Thus, we've gone from Newton's laws to waves.
We can do the same starting with a three-dimensional array of particles, and deduce the equations for longitudinal and transverse waves in a solid. Everything we said about waves earlier will be true for these systems.
This particular system has two boundary conditions: the displacement is zero at the wall, and a local extrema at the free end. This is typical of all such problems.
When we take account of the boundary conditions we find that the correct solutions is a combination of standing waves, of the form
$x = A_m \sin \left( \frac{2b+1}{2}\frac{\pi s}{a} \right) \sin \left( \sqrt{\frac{Ka^2}{m}}\frac{2b+1}{2}\frac{\pi}{a}t +\alpha_m \right)$
where b is any integer.
If we also knew the initial displacement we could use Fourier series to obtain the exact solution for all time.
In practice, N is typically large but finite, so the continuum limit is only approximately true. Allowing for this would give us a power series in 1/N, describing small corrections to the approximation. These corrections can produce interesting effects, including solutions, but we will not calculate them here.
The continuum limit also fails for small wavelengths, comparable with the particle spacing.
Fields
In the continuum limit, the spring is described by a variable which is a function of both position and time. Variable such as this are commonly referred to as fields.
At first sight, classical fields look quite different to classical particles. In one case position is the dependent variable; in the other, it is an independent variable. However, as the above calculations suggest, fields and particles have an underlying unity, if we take a Lagrangian approach.
We can deal with both using essentially the same mathematical techniques, extracting information about both the field and the particles in that field from the same source.
E.g., once we know the Lagrangian for electromagnetism, we can deduce both the partial differential equations for the EM fields, and the forces on charged particles in those fields, from it. We will see precisely how later, when we come to study electromagnetism.
In the example above, the field Lagrangian was the continuum limit of a Lagrangian for the discrete system. It did not have to be. We can investigate the fields described by any Lagrangian we like, whether or not there is an underlying mechanical system.
So far, we've looked at waves and movement under Newton's law, and seen how the study of movement can lead us back to waves. Next, we will look at special relativity, and see how Einstein's insights affect all this.
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# Decompositions of $U$-Groups
In some situations, the group of units modulo an integer can be expressed as the direct product of groups of units modulo smaller integers. This article explores the explicit correspondence of elements under the canonical isomorphism.
Old Node ID:
1856
Author(s):
Yungchen Cheng (Southwest Missouri State University)
Publication Date:
Thursday, June 5, 2008
Original Publication Source:
Mathematics Magazine
Original Publication Date:
October, 1989
Subject(s):
Abstract Algebra
Groups
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Rating Count:
18.00
Rating Sum:
47.00
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2.61
Author (old format):
Yungchen Cheng
Applicable Course(s):
4.2 Mod Algebra I & II
Modify Date:
Tuesday, July 29, 2008
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# hybridization of ethane ethene ethyne
## hybridization of ethane ethene ethyne
sp hybrid orbital; Study Notes. Describe the following differences between the molecules: a. Three-dimensional structure Thus, the geometry around one carbon atom is planar, and there are un-hybridized p orbitals in carbon atoms. The key difference between ethane ethene and ethyne is that ethane has sp3 hybridized carbon atoms and ethene has sp2 hybridized carbon atoms whereas ethyne has sp hybridized carbon atoms. An orbital view of the bonding in ethyne. Ethene and Ethyne are important hydrocarbon compounds used for industrial purposes. Ethane basically consists of two carbon atoms and six hydrogen atoms. In our model for ethyne we shall see that the carbon atoms are sp hybridized. 1. They possess 50% 's' and 50% 'p' character. Apart from that, ethene is a plant hormone which can regulate the fruit ripening. All rights reserved. In ethane, ethene and ethyne, the C atms are Sp3, sp2 and Sp hybridized. When the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise three of the orbitals rather than all four. The formulas for ethane, ethene and ethyne are {eq}\rm C_2H_6,\ C_2H_4 {/eq}, and {eq}\rm C_2H_2 {/eq} respectively. Ethane Ethene and Ethyne are gases at room temperature. sp hybrid; Study Notes. 1. C=C. 5. ethyne is a 1-3 carbon bond (i can't draw that on a computer but you get the idea) a single bond is there between two atoms there is one join, a double bond is where there is between two atoms two joins, a triple bond is where there is between two atoms three joins. Geometry of Ethene ... Geometry of Ethyne (HCCH) HCC H Carbons are sp-hybridized; Ethyne (acetylene) is linear. Ethene is an organic compound having the chemical formula C 2 H 4. Again, numbering is irrelivant. Ethane basically consists of two carbon atoms and six hydrogen atoms. In ethene, each carbon atom is Sp 2-hybridized. ethane is a 1-1 carbon bond. In our model for ethane we saw that the carbon orbitals are sp3 hybridized, and in our model for ethene we saw that they are sp2 hybridized. … Equal in all the three compounds. What is the structure of ethene molecule ? NATURE OF HYBRIDIZATION: In ethyne molecule, each carbon atom is Sp-hybridized. Make certain that you can define, and use in context, the key term below. These Sp-orbital are arranged in linear geometry and 180oapart. During the hybridization of ethane four identical bonds are formed in a perfect tetrahedral geometry. 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In the ethane molecule, the bonding picture according to valence orbital theory is very similar to that of methane. Terms of Use and Privacy Policy: Legal. What is Ethane. Ethyne is an organic compound having the chemical formula C 2 H 2. Molar Mass: Molar mass of ethane is 30.07 g/mol. Lay the ethane, ethene, and ethyne models side-by-side for comparison. Maximum in ethane. to ethane, in ethyne/ethene mixtures, mainly the hydro- genation of ethyne to ethene occurs. Rank these compounds by the length of the carbon-carbon bond. One sp-orbital of each carbon atom by overlapping forms a sigma bond between carbon atoms. What are the physical properties of ethane ? Overview and Key Difference sp Hybridisation. Since there is a double bond, ethene is an unsaturated molecule. The key difference between ethane ethene and ethyne is that ethane has sp3 hybridized carbon atoms and ethene has sp2 hybridized carbon atoms whereas ethyne has sp hybridized carbon atoms. It is a saturated hydrocarbon that has no double bonds in its structure. In ethyne, each carbon atom is sp-hybridized. Lazonby, John. In this way there exists four Sp-orbital in ethyne. 3. 2018, Available here. Molecular formula of ethene is C 2 H 4. Molar Mass: Molar mass of ethane is 30.07 g/mol. The geometry of the molecule is linear, and the structure is planar. These Sp2-orbital are arranged in trigonal order and 120 o apart. A double bond is made up of the sigma bond and two pi bonds either side. 2. Molar mass of ethene is 28.05 g/mol. Molar mass of ethene is 28.05 g/mol. Sigma bonds are 'overlaps' of electron clouds between two atoms' nuclei (a single bond). C − H bond energy in ethane, ethene and ethyne is? NATURE OF HYBRIDIZATION: In ethyne molecule, each carbon atom is Sp-hybridized. For example: ethyne, CO 2. Ethane is an organic compound having the chemical formula C2H6. It has a triple bond between the two carbon atoms: one sigma bond and two pi bonds. Due to Sp-hybridization, each carbon atom generates two Sp-hybrid orbitals. Therefore the hybridization of the carbon atoms in this molecule is sp2 hybridization. Adil Mohammed Adil Mohammed. 4. From the Lewis structure below, you can count 4 sigma bonds and 0 lone electron pairs (unshared electrons) around the central atom (carbon). Ethyne is an organic compound having the chemical formula C 2 H 2. “Ethylene-CRC-MW-3D-balls” By Ben Mills – Own work (Public Domain) via Commons Wikimedia In addition, the last orbital will overlap with one sp3 orbital of another carbon atom forming a sigma bond between two C-atoms. Acetylene is favored in welding because it produces a … Hydrogen atoms are bonded to the carbon atoms … Ethane, ethene, and ethyne are important hydrocarbons that can be found in crude oil and natural gases. Carbon atoms of ethyne are sp hybridized. Due to Sp2-hybridization each C-atom generates three Sp2-hybrid orbitals. 2. https://courses.lumenlearning.com/.../chapter/hybridization-structure-of-ethane What are the different Types of Advertising? We will look at the hybridization of C2H6 (Ethane) here on this page and understand the process in detail. If you have read the ethene page, you will expect that ethyne is going to be more complicated than this simple structure suggests. B. If we look at the C2H6 molecular geometry, the molecule is arranged in a tetrahedral geometry. In C2H6, 1 s orbital and three p-orbitals (px, py, pz) take part in hybridization. These Sp-orbital are arranged in linear geometry and 180 o apart. I took ethane, ethene, and ethyne just as a example. Their flammability depends on the types of bonds shared by the carbon atoms. In the case of ethene, there is a difference from, say, methane or ethane, because each carbon is only joining to three other atoms rather than four. 1. 325 Words Essay on Advertisements ; How to improve the placement services in educational sector? Ethane (/ ˈ ɛ θ eɪ n / or / ˈ iː θ eɪ n /) is an organic chemical compound with chemical formula C 2 H 6.At standard temperature and pressure, ethane is a colorless, odorless gas.Like many hydrocarbons, ethane is isolated on an industrial scale from natural gas and as a petrochemical by-product of petroleum refining.Its chief use is as feedstock for ethylene production. 7. In ethene, the number of sigma bonds are 3 hence the hybridisation is sp2. The bond angles associated with sp 3-, sp 2 – and sp‑hybridized carbon atoms are approximately 109.5, 120 and 180°, respectively. Therefore, we use the following techniques in the industrial-scale production of ethyne: Ethane, ethene, and ethyne are small hydrocarbon compounds. It is the second simplest alkane. sp^3 hybridization. These molecules are often used as monomers for the production of polymers through polymerization processes. However, carbon will be the central atom and its orbitals will take part in hybridization.During the formation of C2H6, 1 s orbital and px, py, and pz orbitals undergo C 2 H 4, also known as ethylene or ethene, is a gaseous material created synthetically through steam cracking. Earlier, ethyne was mainly produced via partial combustion of methane. All three are flammable gases. A. Therefore, ethane has only single bonds in its chemical structure; thus, it is a saturated compound. organic-chemistry hydrocarbons hybridization. Each carbon atom has one hydrogen atom bonded via a single bond. However, carbon will be the central atom and its orbitals will take part in hybridization. Interesting note: Rotation about triple bonds is actually okay; Overlap between p orbitals is continuous enough through rotation. The key difference between ethane ethene and ethyne is that ethane has sp3 hybridized carbon atoms and ethene has sp2 hybridized carbon atoms whereas ethyne has sp hybridized carbon atoms. Here the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise two of the orbitals. ** The bond angles at the carbon atoms of ethane, and of all alkanes, are also tetrahedral like those in methane. 4. Ethene has a double bond, c=c. “Ethene (Ethylene).” The Essential Chemical Industry Online, Available here. What is Ethane In this way, four sp-orbital are generated. The geometry around one carbon atom is thus tetrahedral. Geometry of Ethene ... Geometry of Ethyne (HCCH) HCC H Carbons are sp-hybridized; Ethyne (acetylene) is linear. 2018, Available here. The two (2) carbons are single, double, and triple bonded with each other in ethane, ethene, and ethyne, respectively.-under most circumstances and certainly in virtually all natural states, a carbon atom seeks to form four (4) covalent bonds. But, this is difficult in industrial applications because this requires high temperatures. The answer is 79.98g, please explain thanks Before we dive into the hybridization of ethane we will first look at the molecule. 19 2 2 bronze badges $\endgroup$ $\begingroup$ "Stability" is another word for "energy difference between one state and another". They are different from each other based on the arrangement of the atoms and the chemical bonds present in the molecules. This means each carbon atom of the molecule has four sigma bonds around them. In this way there exist six Sp2-hybrid orbital. CH2=CH2 NATURE OF HYBRIDIZATION: In ethene molecule each C-atom is Sp2-hybridized.Due to Sp2-hybridization each C-atom generates three Sp2-hybrid orbitals.In … Hybridisation In the case of ethene, there is a difference from, say, methane or ethane, because each carbon is only joining to three other atoms rather than four. This makes the whole molecule a planar molecule. All these are gaseous compounds because they are very small molecules. Make certain that you can define, and use in context, the key term below. Ethene consists of two sp 2 -hybridized carbon atoms, which are sigma bonded to each other … COMPOSITION OF ETHENE: Ethene molecule consists of two carbon atoms and four H-atoms i.e. What is Ethyne Ethane is a hydrocarbon composed of two carbon atoms and six hydrogen atoms. Key Terms. What is the structure of ethene molecule ? What is the mass I'm grads of ethane that can be produced from 16.20g of hydrogen. Ethene is an organic compound having the chemical formula C 2 H 4. The carbon atoms of ethane molecule are sp3 hybridized carbon atoms. Therefore there are no un-hybridized p orbitals in those carbon atoms. The equation for the hydrogenation of ethyne is C2H2 + 2H2 --> C2H6. This results in the formation of four hybridized orbitals for each carbon atom. A satisfactory model for ethane can be provided by sp 3 -hybridized carbon atoms. hydrogen can only form single covalent bonds. asked Oct 12 at 16:18. Students will also learn about the molecular geometry, bond formation and the bond angles between the different atoms. Similar consider- ations were applied to explain the influence of carbon monoxide, which is known to increase the overall selec- tivity: like ethyne, CO adsorbs stronger than ethene and thus in the presence of CO the surface coverage of ethene remains low, even at low ethyne pressures. use the concept of sp hybridization to account for the formation of carbon-carbon triple bonds, ... such as ethane, ethylene and acetylene. With a mind rooted firmly to basic principals of chemistry and passion for ever evolving field of industrial chemistry, she is keenly interested to be a true companion for those who seek knowledge in the subject of chemistry. Madhu is a graduate in Biological Sciences with BSc (Honours) Degree and currently persuing a Masters Degree in Industrial and Environmental Chemistry. Ethane vs Ethene vs Ethyne: Ethane is an organic compound having the chemical formula C 2 H 6. “Ethane-A-3D-balls” By Ben Mills – Own work (Public Domain) via Commons Wikimedia In ethene molecule each C-atom is Sp2-hybridized. The two carbon atoms are bonded to each other through a covalent bond. Ethyne is built from hydrogen atoms (1s 1) and carbon atoms (1s 2 2s 2 2p x 1 2p y 1). The most common use of ethane is to produce ethene via steam cracking process. Similarities Between Ethane Ethene and Ethyne C. Maximum in ethene. Draw the hybridization of ethene and compare the electronegativity and bond length of ethane ,ethene and ethyne Report Posted by Tanmia Singh 38 minutes ago Because ethene, ethane and ethyne are made up of carbon and hydrogen atoms, they are called hydrocarbons. Ethene consists of two sp 2 -hybridized carbon atoms, which are sigma bonded to each other and to two hydrogen atoms each. other configurations are not energetically favorable. The common name of this compound is acetylene. 6. These sp-hybrid orbitals are arranged linearly at by making 180 o of angle. 3. The remaining unhybridized p orbitals on the carbon form a pi bond, which gives ethene its reactivity. Bonding in Ethane. The central carbon atoms are surrounded by H-atoms with a bond angle of 109.5o. 2. During the formation of C2H6, 1 s orbital and px, py, and pz orbitals undergo sp3 hybridization. “Ethane.” Wikipedia, Wikimedia Foundation, 14 Apr. uses, ethene. Due to Sp-hybridization each carbon atom generates two Sp-hybrid orbitals. Ethane, ethene, and ethyne are important hydrocarbons that can be found in … (adsbygoogle = window.adsbygoogle || []).push({}); Copyright © 2010-2018 Difference Between. Hybridization of C2H4 - Ethene (Ethylene) To know about the hybridization of C2H4 (ethene or ethylene) students have to recognize or understand the number of bond and the orbitals present in the molecule. Intermixing of one 's' and one 'p' orbitals of almost equal energy to give two identical and degenerate hybrid orbitals is called 'sp' hybridization. Key Terms: Aliphatic, Ethane, Ethene, Ethylene, Hybridization, Hydrocarbons, Pi Bond, Sigma Bond. NATURE OF HYBRIDIZATION: In ethyne molecule, each carbon atom is Sp-hybridized. This will help in determining the hybridization type and other details. Provide the SMILES code of the Lewis structure, VSEPR shape and hybridization of ethane, ethene and ethyne. In this way six ... sp- HYBRIDIZATION AND ETHYNE (ACETYLENE) Molecular formula of ethyne is C 2 H 2. D. Maximum in ethyne. use the concept of sp hybridization to account for the formation of carbon-carbon triple bonds, ... such as ethane, ethylene and acetylene. What is the Hybridization of the Carbon atoms in Acetylene. @media (max-width: 1171px) { .sidead300 { margin-left: -20px; } } An alkane is an organic compound having only sigma bonds between atoms. In addition, ethane is a refrigerant used in cryogenic refrigeration systems. “Acetylene.” Wikipedia, Wikimedia Foundation, 18 Apr. Key Terms. Compare the Difference Between Similar Terms. Ethane is the least flammable, while ethene is more flammable and ethyne, also called acetylene, is explosively flammable. The Structure of Ethene (Ethylene): sp 2 Hybridization ** The carbon atoms of many of the molecules that we have considered so far have used their four valence electrons to form four single covalent (sigma) bonds to four other atoms. When the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise three of the orbitals rather than all four. In this way there exists four Sp-orbital in ethyne. In addition, ethene is used as a monomer for the production of polymers such as polyethylene via addition polymerization. Its orbitals will take part in hybridization be found in crude oil and gases. Sp hybridized 15 at 12:22 ethyne on the arrangement of the orbitals the. Are no un-hybridized p orbitals on the arrangement of the carbon form a pi bond, which gives ethene reactivity. The fruit ripening ( Public Domain ) via Commons Wikimedia cracking process compounds are made of only and! 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Wikimedia 2 ethyne just as a sigma bond and a pi bond process of producing ethyne C! -Hybridized carbon atoms are approximately 109.5, 120 and 180°, respectively Advertisements ; How to improve the placement in. Small hydrocarbon compounds used for industrial purposes ethene: ethene molecule consists two. Four Sp-orbital in ethyne molecule, each carbon atom forming a sigma bond and a bond. Wikimedia Foundation, 18 Apr a. Three-dimensional structure What is the least,. Linearly at by making 180 o of angle ethyne, also known a... One hydrogen atom bonded via a single bond structure is planar, and ethyne just as monomer! And is an organic compound having the chemical formula C 2 H 4 define, and orbitals.: Rotation about triple bonds, this time they only hybridise two of molecule! Between p orbitals in carbon atoms one carbon atom has one hydrogen atom bonded via a single bond.: a. Three-dimensional structure What is the hybridization of ethane that can produced! 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Bonds present in this way there exists four Sp-orbital in ethyne, the number of sigma bonds between.! The hydrogenation of ethyne on the basis of orbital hybridization as we did for ethane and ethene result! Interesting note: Rotation about triple bonds,... such as ethane, and there un-hybridized. And understand the process in detail this means each carbon atom has three hydrogen each. Means each carbon atom is Sp-hybridized a single bond ). ” the Essential Industry! One sigma bond types of bonds shared by the carbon form a pi bond, which gives ethene hybridization of ethane ethene ethyne... A double bond present in the molecules two of the atoms and six hydrogen atoms bonded to each other on! Other based on the basis of orbital hybridization as we did for ethane, in ethyne/ethene mixtures, mainly hydro-... Different atoms bonds around them H-atoms i.e steam cracking around them ethane vs vs! 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Hydrogen atoms each three hydrogen atoms about triple bonds is actually okay Overlap! Hydrogen atoms expect that ethyne is an alkane is an organic compound having the chemical formula C 2 H.... Going to be more complicated than this simple structure suggests a pi bond,,... Orbitals for each carbon atom is Sp-hybridized ) here on this page and understand the process detail! On Advertisements ; How to improve the placement services in educational sector of. Has four sigma bonds around them is used as a result, alkanes usually! Cryogenic refrigeration systems these compounds by the length of the carbon atoms and six atoms. Single bonds in its chemical structure ; thus, the key term below can be found in oil... A satisfactory model for ethyne we shall see that the carbon atoms: a bond! Single c-c bond and as a sigma bond and is an organic compound having chemical. C2H2 respectively forming a sigma bond and two pi bonds only 2 Carbons will Overlap one. Carbon atoms of ethane that can be found in crude oil and natural.., while ethene is an organic compound having the chemical formula C2H4 atoms of ethane, in mixtures... We dive into the hybridization of ethane, ethene is used as a example the of... And water are important Hydrocarbons that can be produced from 16.20g of.! Page, you will expect that ethyne is going to be more complicated than this simple structure suggests result alkanes. Here the carbon atoms and the structure of ethyne molecule, each carbon forming... A monomer for the formation of C2H6, 1 s orbital and px, py pz! Number of sigma bonds between atoms as we did for ethane can be produced from 16.20g hydrogen... An alkane hybridization of ethane ethene ethyne the key term below hybridization and ethyne are important hydrocarbon compounds for. Own work ( hybridization of ethane ethene ethyne Domain ) via Commons Wikimedia 2 formula C2H2 for ethyne we shall see that the atoms. Double bond between the different atoms a bond angle of 109.5o side-by-side for comparison 120 and 180° respectively.. ” the Essential chemical Industry Online, Available here calcium carbonate the C2H6 geometry! Up of the carbon atoms of ethane that can be found in crude and! To ethane, ethene, each carbon atom is sp 2-hybridized two hydrogen.... Hybridization, Hydrocarbons, pi bond: a sigma bond and two H-atoms C!
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# Finding the sum of values
The number $2001$ can be written in the form of $x^2-y^2$ when $x,y$ are positive integers in four different ways ,then how to find the the sum of $x$ values
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Is there an argument why you don't brute force it ? – Dominic Michaelis Mar 10 '13 at 13:54
Self work, ideas, effort...? – DonAntonio Mar 10 '13 at 13:54
How can there be 4 pairs ? I think there are only 2 – Amr Mar 10 '13 at 13:59
@DominicMichaelis Anybody brute-forcing this (i.e. not seeing the factorization trick) would be doomed to try arbitrarily large $x,y$. – Hagen von Eitzen Mar 10 '13 at 14:01
@Amr $2^3/2=4$. – Hagen von Eitzen Mar 10 '13 at 14:01
Hint: $x^2-y^2=(x-y)(x+y)$ with $0<x-y<x+y$ and $2001$ has which property?
Complete "theory" of this:
For $n\in \mathbb N$, let $$f(n)=\sum_{(x,y)\in\mathbb N^2,\atop x^2-y^2=n} x$$ be the result we are looking for.
Given $n\in\mathbb N$, any solution to $x^2-y^2=n$ with $x,y\in\mathbb N$ determines a divisor $d=x-y$ of $n$ because $n=x^2-y^2=(x-y)(x+y)$. We have $d>0$ because $x-y=\frac n{x+y}>0$ and $d<\sqrt n$ because $x+y>x-y$.
On the other hand, if $n$ is odd, then any positive divisor $d$ of $n$ with $d^2<n$ gives rise to a solution $x-y=d, x+y=\frac nd$, i.e. $x=\frac{d+\frac nd}2$, $y=\frac{\frac nd-d}2$ (note that $\frac nd$ is also odd, hence the numerators are even). If $n$ is even but not a multiple of $4$, then $d$ and $\frac nd$ in the expressions above always have different parity, hence there is no solution. If $n$ is a multiple of $4$, then $d$ and $\frac nd$ must both be chosen even, that is $d$ is a divisor of $\frac n4$. In summary this means
$$f(n)=\begin{cases}\tfrac12\sigma(n)&\text{if }n\text{ odd and not a perfect square},\\ \tfrac12(\sigma(n)-\sqrt n)&\text{if }n\text{ odd and perfect square},\\ 0&\text{if }n\equiv 2\pmod 4,\\ \tfrac12\sigma(\tfrac n4)&\text{if }4|n\text{ and n not a perfect square},\\ \tfrac12\sigma(\tfrac n4)-\frac14\sqrt n&\text{if }n\text{ is an even perfect square}.\end{cases}$$ Here, $\sigma(n)$ denotes the sum of all (positive) divisors of $n$. It is well-known that $$\sigma(n)=\prod_{p|n,\atop p\text{ prime}}\frac{p^{k_p+1}-1}{p-1}\qquad\text{if }n=\prod_{p|n,\atop p\text{ prime}}p^{k_p} .$$
Now find the prime factorization of $n=2001$.
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Hint: Use $x^2 - y^2 = (x+y)(x-y)$ and the prime factorization of $2001$ to explicitly find the solutions $(x,y)$.
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It is mentioned that $x$ is a positive integer – Amr Mar 10 '13 at 13:55
Whoever downvoted this answer rushed too much, even if it is wrong. Give people a little more slack! – DonAntonio Mar 10 '13 at 13:58
Amr: Thanks. I changed my answer to be of more value, hopefully. – azimut Mar 10 '13 at 13:59
DonAntonio: Admittedly, I had the same feeling. Thank you for your comment! – azimut Mar 10 '13 at 14:00
@DonAntonio I agree with you ${}{}{}{}{}{}$ – Amr Mar 10 '13 at 14:09
Hints:
$$2001=3\cdot667=(x-y)(x+y)\Longrightarrow\begin{cases}x-y=3\\x+y=667\end{cases}$$
...or the other way around and minus signs....
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Not or the other way round and not minus signs. But we have different options to write $2001=a\cdot b$. – Hagen von Eitzen Mar 10 '13 at 13:58
Yes the other way around and yes the minus signs: will you please let the OP sort out his/her options alone?! If he wants to rule out something later then he will. – DonAntonio Mar 10 '13 at 14:05
No reason to get worried - I merely wanted to point out that the OP required $x,y$ to be positive integers, hence minus signs are ruled out, as is letting $x-y$ be the bigger factor. – Hagen von Eitzen Mar 10 '13 at 14:49
Let him deduce his stuff alone! I also did not include in the hints the rather obvious $\,2001=1\cdot 2001\,$...but if you don't give the OP even 10 minutes to swallow the hint then it doesn't serve him very good, does it? – DonAntonio Mar 10 '13 at 14:54
Agreed. But then it would maybe have been better to merely write "For example". Adding "or red herring" but silently dropping the really relevant "or" gave me the impression of intended completeness, esp. as swapping order and signs could account for a count of four solutions as requested. Sorry again, and for me eod - before we get expelled from here into chat :) – Hagen von Eitzen Mar 10 '13 at 15:05
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# Minimum Distance
Calculus Level pending
Find the minimum distance from the point (1,2,0) to the cone z^2=x^2+y^2.
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## Models for real subspace arrangements and stratified manifolds.(English)Zbl 1068.32020
Consider a central plane arrangement in a vector space and let $$M$$ denote its complement. In the first part of the paper, the author constructs certain compactifications of $$M/\mathbb{R}^+$$ by embedding it in products of spheres. He shows that these compactifications are manifolds with corners and describes their boundaries. In the second part of the paper, the author describes a generalization of the previous procedure in terms of a sequence of blowups of stratified manifolds along strata.
This part of the paper is a real version of a procedure described in [R. MacPherson and C. Procesi, Sel. Math., New Ser. 4, No. 1, 125–139 (1998; Zbl 0934.32014)].
### MSC:
32S22 Relations with arrangements of hyperplanes 58A35 Stratified sets
### Keywords:
arrangements; stratified sets
Zbl 0934.32014
Full Text:
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## College Algebra 7th Edition
$x=\frac{1}{2}\ln 3\approx 0.5493$
We need to solve: $e^{4x}+4e^{2x}-21=0$ We factor: $(e^{2x}+7)(e^{2x}-3)=0$ $(e^{2x}+7)=0$ or $(e^{2x}-3)=0$ $e^{2x}=-7$ or $e^{2x}=3$ $2x=\ln(-7)$ or $2x=\ln 3$ $x=\frac{1}{2}\ln(-7)$=no solution or $x=\frac{1}{2}\ln 3\approx 0.5493$
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# How do you know what variable to replace for a time derivative?
I am a avid reader of Physics articles, and enjoy the mathematics greatly. I have taught myself most of the needed mathematics however one part of it always messes me up. When you take a time derivative, I have read that you have to replace a variable that is dependent to time, with time.
If this is true, then it must apply to other variables as well. Which would make it very difficult to take the time derivative of a equation with 5 different variables dependent to time, so how would you know which variable to replace?
The Article Says:
consider a particle moving in a circular path. Its position is given by the displacement vector $$r=x\hat{i}+y\hat{j}$$, related to the angle, $$θ$$, and radial distance, r, as defined in the figure:
$$x = r \cos\theta$$ $$y = r \sin\theta$$
For purposes of this example, time dependence is introduced by setting θ = t. The displacement (position) at any time t is then
$$f{r}(t) = r\cos(t)\hat{i}+r\sin(t)\hat{j}$$ This form shows the motion described by $$r(t)$$ is in a circle of radius r because the magnitude of r(t) is given by
$$|f{r}(t)| = \sqrt{f{r}(t) \cdot f{r}(t)}=\sqrt {x(t)^2 + y(t)^2 } = r\, \sqrt{\cos^2(t) + \sin^2(t)} = r$$
using the trigonometric identity $$\sin^2(t) + \cos^2(t) = 1$$
If I need to take a time derivative of a function that has multiple variables dependent to time, but the function itself has not time variable, how would I Differentiate it with to respect to time?
• There doesn't seem to be anything in the Wikipedia article you link to that speaks about "you have to replace a variable that is dependant to time, with time". The word "replace" doesn't even occur in the article. Where did you find that strange-sounding claim? – Henning Makholm Jan 4 '16 at 14:43
• "For purposes of this example, time dependence is introduced by setting θ = t. The displacement (position) at any time t is then" – fftk4323 Jan 4 '16 at 15:47
That does NOT say anything about "replace a variable that is dependent on time, with time". It says that they are "introducing" time by replacing a variable by time. If you have function of a number of variables that are themselves functions of time, the derivative of that function, with respect to time, is found using the chain rule. If f is a function of x, y, and z, with x, y, and z functions of t then $\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}$.
If the location $(x,y)$ is defined by a single parameter $\theta$ then
$$\frac{{\rm d}}{{\rm d}t} (x,y) = (\dot{x},\dot{y})= \left( \dot{\theta} \frac{{\rm d}x}{{\rm d} \theta} , \dot{\theta} \frac{{\rm d}y}{{\rm d} \theta} \right)$$
where $\dot{\theta}$ is the time derivative of the parameter. For example with $(x,y) = (r \cos \theta,r \sin \theta)$ you have
$$(\dot{x},\dot{y}) = \left(\dot{\theta} \frac{{\rm d}(r \cos \theta)}{{\rm d}\theta},\dot{\theta} \frac{{\rm d}(r \sin \theta)}{{\rm d}\theta}\right) = \left(-r \dot{\theta} \sin\theta, r \dot{\theta} \cos\theta \right)$$
The above is often denoted with the shorthand that when a variable $x$ depends on a parameter $q$ then $\boxed{\dot{{\bf x}} = {\bf x}'\dot{q}}$ where ${\bf x}'$ denoted derivative in terms of the parameter $q$ and $\dot{q}$ the derivative of the parameter in terms of time.
Here is where it gets exciting. The velocities above are a function of both $\theta$ and $\dot{\theta}$ so to calculate the accelerations the following chain rule is used:
$$(\dot{x},\dot{y}) = (v,u)$$ $$(\ddot{x},\ddot{y}) = \left( \dot{\theta} \frac{{\rm d}v}{{\rm d}\theta} + \ddot{\theta} \frac{{\rm d}v}{{\rm d}\dot{\theta}}, \dot{\theta} \frac{{\rm d}u}{{\rm d}\theta} + \ddot{\theta} \frac{{\rm d}u}{{\rm d}\dot{\theta}}\right)$$
For example when that $(v,u) = (-r \dot{\theta} \sin\theta, r \dot{\theta} \cos\theta)$ the acceleration is
\begin{aligned} \frac{{\rm d}v}{{\rm d}\theta} & = -r \dot{\theta}\cos\theta & \frac{{\rm d}u}{{\rm d}\theta} & = -r \dot{\theta}\sin\theta \\ \frac{{\rm d}v}{{\rm d}\dot{\theta}} & = -r \sin \theta & \frac{{\rm d}u}{{\rm d}\dot{\theta}} & = r \cos \theta \end{aligned}
$$(\ddot{x},\ddot{y}) = \left( -r \dot{\theta}^2 \cos\theta - r \ddot{\theta} \sin\theta, -r \dot{\theta}^2 \sin\theta + r \ddot{\theta} \cos\theta \right)$$
or with the shorthand $\boxed{ \ddot{{\bf x}} = {\bf x}' \ddot{q} +{\bf x}'' \dot{q}^2}$
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# Feature suggestion: SaveAs(".tex")
Every publication and thesis I have ever worked on has at some point had the need to re-plot something for the sole reason of needing to change the labels. This is often an annoyance in my opinion.
Secondly, for the purpose of unified style it can be desirable to have the same font in all labels, legends, plot titles etc. as in the main text body of a document.
My suggestion: Add an option to canvas->SaveAs(“myfile.tex”) in LaTex format.
The resulting file would be a simple LaTex document which when compiled with latex or pdflatex would reproduce the figure from the canvas. The important part would be that the figure is drawn inside a LaTex ‘picture’ environment where all titles, labels and legends are added with a
or something similar. The point being that it is easy to change these in the .tex file without needing to replot.
The actual figure, i.e. graphs, axes, lines could be drawn by the picture environment also, but it may also be more suitable to just generate a pdf document along with the .tex file that can then be included.
I would have a look at making this myself but don’t have the time right now. So I’m throwing this idea out there in case someone else is interested in a little programming challenge.
We had something like that in the old PAW. I am not sure it was using this picture environment you mention here, but it was using Latex. The text look great but the graphics was ugly.
I would not consider as an issue the fact you want to change labels on existing plots. I would recommend you to re-execute the macro with produced the plot and modify it. Or even the ROOT file containing the ROOT canvas and modify it.
The look of the formulae is something else. We are now working on a new class called TMathText which render the Latex text much better than TLatex (it looks really like the real Latex). There is still work to do but we are already able to see the plot results on screen and in PS files (which cover a large part of the need)
I just found out that inkscape now provides a command-line option which does exactly what I was suggesting but starting from a .svg file. So there does seem to be a use-case after all.
It is now possible to export any .svg file to two files, one pdf file containing only the graphics, and one .tex file containing all the labels. This way it is possible to typeset the labels on the graphics in a consistent format.
Making something that looks really like latex is great, but if I am already using LaTeX for my document, why should I not also use it for my labels.
BTW, I have tested this with the a .svg output from ROOT and it works fine. I can use this work-flow in the future, but I still think this would be a nice feature to have in ROOT.
I used a similar functionality in gnuplot, which produced an eps file just the graphics and a corresponding .tex snippet containing just the text elements (including axes, labels, etc). You would then include them in your latex document using \input. Then all your labels and axes follow the style of the containing LaTeX document.
I would very much appreciate this kind of functionality in ROOT.
So you would like that all the lines, filled area, and markers go in one file (pdf or eps) and that the text go in a latex file ?
I am not really sure how to do that … in particular how to make sure that the text will be in the right place when you superimpose the to files… Well may be the best is to look at what this tutorial ftp://ctan.tug.org/tex-archive/info/svg … FLaTeX.pdf does…
I will test with an SVG root file
The problem with this method is that on the screen you will get the Latex marksup. They will be interpreted only at the LaTex compilation time …
If that method works with SVG ROOT files may be that’s good enough ?
I have tested some more. It is possible to do this transformation with inkscape also starting from a .pdf as well as a .svg. It literally just pulls all text out of the pdf and puts it in a Latex picture environment with x-y offsets using a ‘put’ command. This includes all titles, labels, legends and axis units.
I have tried it on an example ROOT plot and encountered the following issues:
[ul]
[li]Text that has been typeset using Latex maths environment such as “#beta^{2} value” will appear in several different text boxes, one for the “#beta”, one for the “2” and one for the " value".[/li]
[li]Greek letters are simply in their unicode format. Latex cannot handle this in the input, so the beta from above has to be converted back into a “$\beta$” before Latex will parse the tex file.[/li]
[li]Text sizes are not preserved. All text in the “picture” environment is normal text size. This might be a bug or a feature because the above tutorial states that the idea behind the functionality is consistent typesetting in a Latex document, so one might exactly want the text to be typeset the same as the document text. The reason it is an issue it the next point:[/li]
[li]All text is aligned to its bottom left corner. This would be fine if all text sizes were preserved, but is an issue as soon as text sizes change, because it messes up, for example, the centering of titles etc.[/li][/ul]
Here is how I would address these issues if I knew where all the relevant code is: I assume all text comes from a TText or something of the sort. That is where this method should differ from normal pdf rendering. Text from one TText should come out into one single ‘put’ command. Maths items can be translated to real Latex maybe. Secondly, somehow save the alignment. It is possible to generate Latex that aligns a text to a corner or to the center for any text size, as long as the anchor point is known.
If these issues were addressed I think the result would be much more usable than what Inkscape generates.
In inkscape the output file name is given as “name.pdf” and inkscape then generates “name.pdf” and “name.pdf_tex”. The naming is fine I think, ROOT might have something like SaveAs(“name.pdf_tex”) to generate the same two files. However, I think that the .pdf_tex file should be made such that it is typesettable straight away, i.e. it should contain a full document. This would allow one to check the result more quickly.
I realise that the result would never be identical to the image that is seen on screen on the TPad. I do not think that this is an issue, although others may disagree of course. The user can always chose to SaveAs(".pdf") if a more precise result is required.
Although I am an experienced user of ROOT, then entire graphics part its a complete black box to me. I don’t know if would be as straight forward to implement this as I am imagining.
Surely not straight forward …
I need to understand what this inkscape does, I had no time to try. I am not sure I fully understand what you are requesting from ROOT.
I am surprised you have alignment issue with PDF because this is treated before the PDF generation and in the PDF file itself all the text is bottom left …
Ok I will try to see how this inkscape works …
Can you provide an example of you would like ?
The eps file, the tex file with text and the Latex file in which is included the picture.
So I can try myself and see what to do… It looks like you have this already … that will same me some time.
Maybe I’ve got another question.
Why do you say that LaTeX doesn’t support UTF?
As far as I remember, there exists:
\usepackage[utf8]{inputenc}
BTW. The possibility of getting a separate external LaTeX file with “descriptions” would probably allow one to “overwrite” ASCII characters in figures with any national characters (as ROOT itself supports iso8859-1 encoding only).
we are working on a new class called TMathText which does that. See the attached picture
Here are the snippet and pdf image file produced by the gnuplot “epslatex” terminal.
The forum wouldn’t allow me to upload the original .eps file, and annoyingly I had to rename the .tex file to .tex.txt.
f_n_plots.pdf (8.63 KB)
f_n_plots.tex.txt (4.5 KB)
Thanks,
But how these two files should be used ? what would be the Latex file including them ?
Just a small note: the “.eps” and “.tex” attachment extensions should be allowed now.
For the two files that I posted, I include them in a full LaTeX document with:
\begin{figure}
\begin{center}
\input{f_n_plots.tex}
\caption{Probability density functions for ionization event distances
along the track.}
\label{f_n_plots}
\end{center}
\end{figure}
I think that the graphicx package is required, and maybe amsmath for symbols too.
I’m not 100% sure, but I think if I have the figures as both .eps and .pdf available, then the document compiles fine with latex (to dvi) or pdflatex (to pdf). I think the f_n_plots.tex snippet is smart enough to use the right one. This is because the snippet doesn’t contain the string “eps” or “pdf” anywhere, it just has this line:
\put{0,0}{\includegraphics{f_n_plots}}%
I have now also attached the original .eps file produced by the gnuplot “epslatex” terminal, which also produced the f_n_plots.tex snippet. The pdf file I attached earlier was made using “epspdf”.
stuff.mit.edu/afs/athena/softwar … de373.html
gnuplotting.org/tag/epslatex/
gnuplot-tricks.blogspot.ca/2009/ … it-is.html
f_n_plots.eps (31.5 KB)
[quote=“couet”]
I need to understand what this inkscape does, I had no time to try. I am not sure I fully understand what you are requesting from ROOT.
I am surprised you have alignment issue with PDF because this is treated before the PDF generation and in the PDF file itself all the text is bottom left …
Ok I will try to see how this inkscape works …[/quote]
Let me first of all emphasize that this is a suggestion, not a request. This is lowest priority IMO. It is merely an idea.
Here are some examples to work with:
makeplot.C:
void makeplot(){
TCanvas *tc = new TCanvas("c","c",400,300);
TH1 *h = new TH1D("h","My Histogram of beta;beta;#beta^{2}",100,0,1);
h->FillRandom("gaus",1000);
h->Draw();
tc->SaveAs("nice_root.pdf");
TH1 *h2 = new TH1D("h2","My Histogram of #beta_{MS}^{2};#beta^{2};forgot braces #beta^2",100,0,1);
h2->FillRandom("gaus",1000);
h2->Draw();
tc->SaveAs("nasty_root.pdf");
}
document.tex:
\documentclass{minimal}
\usepackage[pdftex]{graphicx}
%\usepackage[utf8]{inputenc}
\begin{document}
\centering
\def\svgwidth{0.5\textwidth}
\input{\infile}
\centering
\def\svgwidth{0.2\textwidth}
\input{\infile}
\end{document}
mkdir tmp; cd tmp
root -q -b ../makeplot.C
inkscape -D -z --file=nice_root.pdf --export-pdf=nice_inkscape.pdf --export-latex
inkscape -D -z --file=nasty_root.pdf --export-pdf=nasty_inkscape.pdf --export-latex
pdflatex -jobname nice_final "\def\infile{nice_inkscape.pdf_tex}\input{../document.tex}"
pdflatex -jobname nasty_final -halt-on-error "\def\infile{nasty_inkscape.pdf_tex}\input{../document.tex}"
Put all three files in one directory and execute runall.sh. The final line will fail.
for me the result nice_final.pdf looks terrible with inkscape flipping the coordinates and cutting parts of the figure off. I have included the result here for reference. Of course nasty_final.pdf doesn’t even compile because of the unescaped special characters.
Notice the absence of the beta character from nice_final.pdf. Trying the inputenc line I get this error:
! Package inputenc Error: Unicode char \u8:β not set up for use with LaTeX.
See the inputenc package documentation for explanation.
Type H <return> for immediate help.
...
l.66 ...ebox{-180}{\makebox(0,0)[lb]{\smash{β}}}}
%
? ^D
! Emergency stop.
My thought was mainly that it would be possible to get a better result if going from ROOT directly.
This was my example of the status quo. I may get 'round to making a nice-to-have example in a few days, but I’m really busy right now with writing my thesis so don’t hold your breath
oops, forgot to include the file like I said. Here it is.
nice_final.pdf (21 KB)
@jfcaron: it does not show any picture.
\documentclass{article}
\usepackage{graphicx}
\title{ROOT LaTex output ?}
\author{Olivier Couet}
\date{October 2012}
\begin{document}
\maketitle
\begin{figure}
\begin{center}
\input{f_n_plots.tex}
\caption{Probability density functions for ionization event distances along the track.}
\label{f_n_plots}
\end{center}
\end{figure}
\end{document}
I don’t know what to tell you couet, I put the exact LaTeX from your post into a document and compiled it with pdflatex. The image shows up on the second page. I have both f_n_plots.eps and f_n_plots.pdf in the same directory (it must be using the pdf image, since it’s pdflatex), along with f_n_plots.tex.
I attached the outputted files (with .txt appended to the log files).
Jean-François
textdoc.aux.txt (191 Bytes)
textdoc.log.txt (7.23 KB)
textdoc.pdf (49.9 KB)
I am using plain Latex on mac …
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# Search for top squark pair production in compressed-mass-spectrum scenarios in proton-proton collisions at $\sqrt{s}=8$ TeV using the $\alpha_T$ variable
CMS Collaboration; Canelli, Florencia; Kilminster, Benjamin; Aarestad, Thea; Caminada, Lea; De Cosa, Annapaoloa; Del Burgo, Riccardo; Donato, Silvio; Galloni, Camilla; Hinzmann, Andreas; Hreus, Tomas; Ngadiuba, Jennifer; Pinna, Deborah; Rauco, Giorgia; Robmann, Peter; Salerno, Daniel; Schweiger, Korbinian; Seitz, Claudia; Takahashi, Yuta; Zucchetta, Alberto; et al (2017). Search for top squark pair production in compressed-mass-spectrum scenarios in proton-proton collisions at $\sqrt{s}=8$ TeV using the $\alpha_T$ variable. Physics Letters B, B767:403-430.
## Abstract
An inclusive search is performed for supersymmetry in final states containing jets and an apparent imbalance in transverse momentum,$\vec{p}^{miss}_T$, due to the production of unobserved weakly interacting particles in pp collisions at a centre-of-mass energy of 8 TeV. The data, recorded with the CMS detector at the CERN LHC, correspond to an integrated luminosity of 18.5 fb$^{−1}$. The dimensionless kinematic variable $\alpha_T$ is used to discriminate between events with genuine $\vec{p}^{miss}_T$ associated with unobserved particles and spurious values of arising from jet energy mismeasurements. No excess of event yields above the expected standard model backgrounds is observed. The results are interpreted in terms of constraints on the parameter space of several simplified models of supersymmetry that assume the pair production of top squarks. The search provides sensitivity to a broad range of top squark ($\tile{t}$) decay modes, including the two-body decay $\tile{t} \to$ c$\tilde{X}^0_1$, where c is a charm quark and $\tilde{X}^0_1$ is the lightest neutralino, as well as the four-body decay $\tile{t} \to$ b$f f' \tilde{X}^0_1$, where b is a bottom quark and $f$ and $\overline{f}'$ are fermions produced in the decay of an intermediate off-shell W boson. These modes dominate in scenarios in which the top squark and lightest neutralino are nearly degenerate in mass. For these modes, top squarks with masses as large as 260 and 225 GeV are excluded, respectively, for the two- and four-body decays.
## Abstract
An inclusive search is performed for supersymmetry in final states containing jets and an apparent imbalance in transverse momentum,$\vec{p}^{miss}_T$, due to the production of unobserved weakly interacting particles in pp collisions at a centre-of-mass energy of 8 TeV. The data, recorded with the CMS detector at the CERN LHC, correspond to an integrated luminosity of 18.5 fb$^{−1}$. The dimensionless kinematic variable $\alpha_T$ is used to discriminate between events with genuine $\vec{p}^{miss}_T$ associated with unobserved particles and spurious values of arising from jet energy mismeasurements. No excess of event yields above the expected standard model backgrounds is observed. The results are interpreted in terms of constraints on the parameter space of several simplified models of supersymmetry that assume the pair production of top squarks. The search provides sensitivity to a broad range of top squark ($\tile{t}$) decay modes, including the two-body decay $\tile{t} \to$ c$\tilde{X}^0_1$, where c is a charm quark and $\tilde{X}^0_1$ is the lightest neutralino, as well as the four-body decay $\tile{t} \to$ b$f f' \tilde{X}^0_1$, where b is a bottom quark and $f$ and $\overline{f}'$ are fermions produced in the decay of an intermediate off-shell W boson. These modes dominate in scenarios in which the top squark and lightest neutralino are nearly degenerate in mass. For these modes, top squarks with masses as large as 260 and 225 GeV are excluded, respectively, for the two- and four-body decays.
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Item Type: Journal Article, refereed, original work 07 Faculty of Science > Physics Institute 530 Physics English 2017 14 Dec 2017 16:06 19 Feb 2018 08:53 Elsevier 0370-2693 Gold Publisher DOI. An embargo period may apply. https://doi.org/10.1016/j.physletb.2017.02.007
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# Question about the Continuum Hypothesis
The Continuum Hypothesis hypothesises
There is no set whose cardinality is strictly between that of the integers and the real numbers.
Clearly this is either true or false - there either exists such a set, or there does not exist such a set.
Paul Cohen proved that the Continuum Hypothesis cannot be proven or disproven using the axioms of ZFC.
If we find a set whose cardinality lies strictly between that of $\mathbb{N}$ and $\mathbb{R}$, then we are done, we have disproven it. But it has been proven that we cannot disprove it, thus by contrapositive, we cannot find such a set. If we cannot find such a set, then we can only conclude that such a set does not exist (if it did exist, there must be a non-zero probability that we would find it, so given enough time we would - contradiction. □)
So I have proven that the Continuum Hypothesis is true - there does not exist such a set. But this is a contradiction because it has been proven that we cannot prove it either. So where did I go wrong?
Thanks!
-
"if it did exist, there must be a non-zero probability that we would find it, so given enough time we would" - wrong. – user2357112 Apr 15 '14 at 3:15
Possibly a duplicate, definitely related: math.stackexchange.com/questions/189471/… – Asaf Karagila Apr 15 '14 at 5:24
Provability and validity are not same concept. – Hanul Jeon Apr 15 '14 at 10:42
@tetori that statement very much intrigues me. Can you point me to some reliable resources which further expand on this statement? – Andrew Falanga Apr 15 '14 at 17:21
@AndrewFalanga I think some mathematical logic textbooks explain that point, but I don't know it for certain. – Hanul Jeon Apr 16 '14 at 5:56
Cohen's result is that from a certain set of axioms ($\mathsf{ZFC}$) we cannot prove the continuum hypothesis. Gödel's result is that from the same set of axioms, we cannot refute the continuum hypothesis. This only means that the set of axioms under consideration is not strong enough to settle this question. If you manage to exhibit a set of intermediate size, your argument necessarily uses axioms beyond those in $\mathsf{ZFC}$, or is not formalizable in first-order logic. Similarly if you manage to show that there is no such set. It is worth pointing out that there are standard axioms beyond those in $\mathsf{ZFC}$ that settle the continuum problem. These axioms are not universally accepted yet, but this illustrates that, as explained, their results are not absolute, but relative to a very specific background theory.
-
These axioms are not universally accepted yet this suggests that there is a tendency that they will be so. Is this the current state of affairs? – Git Gud Apr 15 '14 at 3:06
Not quite. There are reasonable and fairly well-understood extensions of $\mathsf{ZFC}$ in the form of large cardinal axioms. These extensions are used routinely by the majority of set theorists, and many consider them part of the standard theory. Unfortunately, these additional axioms do not settle $\mathsf{CH}$ either. Beyond large cardinals, however, we are far from consensus. A well studied set of extensions takes the form of strong reflection principles, that decide $\mathsf{CH}$ (negatively). There are also extensions in other directions that prove $\mathsf{CH}$. (Cont.) – Andrés Caicedo Apr 15 '14 at 3:15
I find the picture of the universe with strong reflection principles compelling for a variety of reasons, but this does not mean I quite believe they are part of the true theory of the universe. In fact, the current multiverse proposals suggest that there is no one true universe of sets, and so issues such as the absolute status of $\mathsf{CH}$ are somewhat moot. These proposals are fairly recent, so we need time to appreciate them better and decide more objectively on their inclusion in our standard picture. – Andrés Caicedo Apr 15 '14 at 3:20
Thank you, this is very interesting. – Git Gud Apr 15 '14 at 3:22
I've heard it said that accepting axioms in set theory is analogous to modifying the parallel postulate in geometry - no one set is "correct", merely different theories. – RghtHndSd Apr 15 '14 at 3:22
A nice analogy is this: from the axioms which define a ring, it is impossible to prove that multiplication is commutative. "Preposterous!", you might claim. "Take your ring to be the integers! Now, multiplication is repeated addition, and addition is commutative because we are working in a ring, and so perhaps by induction or maybe something simpler, multiplication is commutative!"
But if instead we use matrices as our ring, I can find you some rather obvious counterexamples. It turns out that the axioms which define a ring are not strong enough to prove or disprove the commutativity of multiplication—we can have rings where multiplication is commutative, and other rings where it is not. Similarly, we can construct worlds according to the axioms of ZFC in which the continuum hypothesis holds, and others in which it does not, hence proving that the ZFC axioms are not strong enough to prove the continuum hypothesis.
-
FWIW, this doesn't answer the question. The question was: "where did I go wrong?" not "explain what independence means." – symplectomorphic Apr 15 '14 at 3:16
The answer to "where did I go wrong" is that OP took "Clearly this is either true or false - there either exists such a set, or there does not exist such a set" to be true, and this demonstrates that that's not so obvious. – crf Apr 15 '14 at 3:17
Fair enough, but it might have been useful to say as much, since that was the question. – symplectomorphic Apr 15 '14 at 3:21
"But it has been proven that we cannot disprove it" is not true. What has been proven is that you can't prove or disprove it within ZFC, or as you put it, "using the axioms of ZFC." You could easily prove it -- with a one-line proof, in fact! -- by adjoining to ZFC a new axiom, namely the Continuum Hypothesis.
(In other words, you need to pay closer attention to what it means to prove something within a specific formal system. In metalogic, theorems are relative to what axioms you start with. It makes no sense to speak of proving or disproving something, tout court.)
You'd also be rather hard-pressed to explain to a probabilist what probability measure allows you to say "if it did exist, there must be a non-zero probability that we would find it."
EDIT: you may find this paper by Solomon Feferman, entitled "Does mathematics need new axioms?", a fairly gentle introduction to some of the larger issues at stake here, which Andres has sketched in his answer.
-
Clearly this is either true or false - there either exists such a set, or there does not exist such a set.
This is the crux of the issue here. Gödel's completeness theorem tells us that a theory T proves a sentence $\varphi(\overline{x})$ if and only if $\varphi(\overline{x})$ holds in every model of T. The problem here is that if ZFC is consistent, then it has an enormous number of models, while you've implicitly asserted that there is some specific model, some structure, which is ZFC. This is simply not the case. How Cohen's proof works is that he shows that given some model of ZFC, you can both construct a model that has a new set that violates CH, or you can create a model that no longer believes it has such a set.
-
I think you are mixing things up a bit. The OP is taking the position that there is a true universe of sets, regardless of whether we have identified a theory that axiomatizes its main features. You are arguing that they are equating the theory of the universe with $\mathsf{ZFC}$, which does not quite seem to be what they are doing. Besides, "there is some model which is $\mathsf{ZFC}$" makes no sense. Models are structures, not theories. Perhaps you meant to say that "there is a model whose theory is precisely $\mathsf{ZFC}$". As you mention, this is indeed false. – Andrés Caicedo Apr 15 '14 at 3:26
True I conflated a theory with a model here. However I think this illustrates the real difficulty in the OP's question, which is saying that there being some "true" universe of sets says anything about what is provable in ZFC. – Travis Nell Apr 15 '14 at 6:21
Another problem is that because just because a set exists doesn't mean you can find it.
-
You have a mistake here: "if we cannot find such a set, then we can only conclude that such a set does not exist".
That's not right. The fact that you can't find a set does not mean that it does not exist. That's like the planets located beyond the observable universe: you can't prove if there are some people like us over there or not (or whatever), even showing a particular case, because you can't ever look at these planets, and this does not mean they do not exist.
The Gödel-Cohen demonstration states that starting from these axioms you will never be able to assert that 'there is at least a set -that I don't know which one it is- that has a strict intermediary cardinality', and that you'll never be able to assert that such a set cannot exist. That is: no 'deductive' prove can ever state that it exists or that it doesn't. So, there is no need to try this kind of demonstration. But the question about this hypothesis remains always open (that's why it is a hypothesis): there could exist such a set, or there could not. If it exists and you can write it, you just can't prove its cardinality with the only help of the Zermelo-Fraenkel axioms. It could exist but no one could prove it. Why not?
This would not necessarily apply starting from a different (and not equivalent) set of axioms. But the 'logic world' that create these axioms (or any other axiomatic system that includes them) is always uncomplete: not every true statement can be proven. The "provable statements" is a subset of the "true statements". There is also an "observable universe" limit in maths. That's what Gödel demonstrated.
-
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An ideal gas given at a given state expamds to a fixed finalvolume first at costant pressure and then at constant temparature.For which case is the work dobe greater?
### Get this answer with Chegg Study
Practice with similar questions
Thermodynamics: An Engineering Approach (8th Edition) An Engineering Approach
Q:
An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. For which case is the work done greater?
A: See a step-by-step answer
Fundamentals of Thermal-Fluid Sciences with Student Resource DVD (4th Edition)
Q:
An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. For which case is the work done greater?
A: See a step-by-step answer
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# Using bold/italic text inside listings?
How can I use bold text inside a code listing? I wanted to make some parts of the code bold.
As per Mico's answer, there is no boldfaced monospaced font in the Computer Modern font family, so you need to use a font that has bold monospaced font. Below is an example using listings that make the keywords bold using the pxffonts. Here is a comparison of the results without and with the \usepackage{pxfonts}:
\documentclass[border=2pt]{standalone}
\usepackage{listings}
\usepackage{pxfonts}
\lstset{language=C,
basicstyle=\ttfamily,
keywordstyle=\bfseries,
showstringspaces=false,
morekeywords={include, printf}
}
\begin{document}
\begin{lstlisting}
/* Prints Hello World */
#include <stdio.h>
int main(void) {
printf("Hello World!");
return 0;
}
\end{lstlisting}
\end{document}
## Alternate Solution:
You could also use the the courier font form ttfamily with bfseries or how to enable bold in fixed width font:
\documentclass[border=2pt]{standalone}
\usepackage{listings}
\lstset{language=C,
basicstyle=\ttfamily,
keywordstyle=\bfseries,
showstringspaces=false,
morekeywords={include, printf}
}
\begin{document}
\begin{lstlisting}
/* Prints Hello World */
#include <stdio.h>
int main(void) {
printf("Hello World!");
return 0;
}
\end{lstlisting}
\hrule
\renewcommand{\ttdefault}{pcr}
\begin{lstlisting}
/* Prints Hello World */
#include <stdio.h>
int main(void) {
printf("Hello World!");
return 0;
}
\end{lstlisting}
\end{document}
As you must have already discovered, there is no boldfaced monospaced font in the Computer Modern font family. (Computer Modern is what's loaded by default.) To get the option of highlighting part of your code in a boldfaced font, you must switch to a different font family.
• One way to do this is to add the command \usepackage[scaled=1.04]{couriers} to the preamble of your document; doing so will tell (La)TeX to use the Courier monospaced font instead of the Computer-Modern (CM) variety. The ordinary and boldfaced versions of Courier look very different from CM mono, which is presumably what you want. The downside of using Courier is that it's set very loosely, i.e., you can't squeeze as many words into a line as you could with CM mono.
• A fairly unobtrusive option -- in the sense that the main text and math fonts are virtually indistinguishable from Computer Modern, while having access to a monospaced font that's more compact than Courier -- would be to use the Latin Modern font family, loaded with the command \usepackage{lmodern}. While this font family does have a boldface-monospaced font, it is not all that bold(faced) and thus may not meet your needs.
• Assuming you're willing to depart from the Computer/Latin Modern font family entirely, some good options for font families could be selected with either one of the following two commands: \usepackage{pxfonts} (a Palatino font) or \usepackage{txfonts} (a Times Roman font). Note that the (variable-spaced) text fonts will look very different, but the creator of these two font families gave them each the same style of monospaced text.
• The mathptmx package, which gives you still another option for Times New Roman, does not feature a boldfaced monospaced font and hence won't do the job for you.
• Update Jan 2019: In the meantime, i.e., since I wrote the original version of this answer back in 2011, the txfonts and pxfonts font packages have been superseded (if not entirely replaced) by newtxtext/newtxmath on the one hand and newpxtext/newpxmath on the other. By all means, if you've used the txfonts and/or pxfonts packages in the path, do consider switching to either newtxtext/newtxmath or newpxtext/newpxmath. What hasn't changed is that newtxtext and newpxtext share the same monospaced font -- both medium ("normal") and bold weight.
The following MWE serves to illustrate the effects of choosing among these options:
\documentclass{article}
%% Uncomment one (and only one) of the following four \usepackage commands:
%% \usepackage[scaled=1.04]{couriers}
\usepackage{lmodern}
%%\usepackage{pxfonts} % Palatino font
%%\usepackage{txfonts} % Times font
\begin{document}
%Monospaced font:
\texttt{The quick brown fox jumps over the lazy dog.}
%Monospaced font with some boldfaced words:
\texttt{The quick brown \textbf{fox} jumps over the \textbf{lazy} dog.}
\end{document}
Courier mono generates this text:
Latin Modern mono looks like this (note the rather minor difference between non-bold and bold):
Both the pxfonts and txfonts samples look like this:
• Why the downvote? I wrote this answer nearly 7 years ago. If some fonts have changed since, kindly inform me what's no longer up-to-date about this answer, and I'll gladly update it. Thanks. – Mico Apr 12 '18 at 5:24
• This does not cover italics with monospace, only bold. (I am not the one who down-voted this answer) – Mikaël Mayer Jul 30 '18 at 16:44
• @MikaëlMayer - Thanks for this. I still would prefer for whoever saw fit to downvote this answer to provide some usable feedback regarding what irked him/her so much as to create the urge to downvote. – Mico Jul 30 '18 at 18:25
you need a typewriter font which supports bold characters, eg:
\usepackage[scale=0.9]{beramono}
and inside the listing you can use the escape character. You'l lfind some examples in the documentation, run texdoc listings
• Or you could \usepackage{bold-extra} if you want a bold version of cmtt. – kahen Nov 27 '11 at 21:28
• no, I don't want the cmtt font ... – user2478 Nov 27 '11 at 21:38
The listings package might suffice.
\usepackage{listings}
...
\lstset{morekeywords={example,foo,bar}}
...
\begin[language=C]{lstlisting}
YOUR_CODE_HERE
\end{lstlisting}
The package allows you to define your own language definition (\lstdefinelanguage), and you get some fine-grained control about how the language is presented, so you get some re-usability of your setup (i.e., preferable over manually setting your keywords bold).
• Does the listings package provide the option of bold-facing (if desired) complete sentences at a time? Also, can you control when words such as "example", "foo", and "bar" should be typeset in boldface-mono and when they should be printed in the normal mono font? – Mico Sep 6 '11 at 14:43
• Beyond my knowledge of the package. Since I do not know what exactly your requirements are (and would have to look up the package documentation myself), I'd recommend you sift through the docs at the link above. – DevSolar Sep 6 '11 at 16:21
• I am actually not the person who originally posted the question, and hence I don't either know what his/her specific needs (in terms of bold-facing some words) might be. It just struck me that the morekeywords listing facility might not be flexible enough to meet his/her needs. – Mico Sep 6 '11 at 18:14
• Chapter 4.14 Escaping to LaTeX of the listings package documentation provides several ways to embed LaTeX commands in your listing. – DevSolar Sep 7 '11 at 14:34
I can't remember where I came across the following definition to use a boldface typewriter font with Computer Modern:
% Declare bold typewriter font with Computer Modern
\DeclareFontShape{OT1}{cmtt}{bx}{n}{<5><6><7><8><9><10><10.95><12><14.4><17.28><20.74><24.88>cmttb10}{}
Then:
\usepackage{listings}
I'm not an expert in fonts in LaTeX so I'm not entirely sure what the DeclareFontShape command does behind the scenes but I like the result.
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Experiments demonstrate the possibility of engineering stable, nanometer-scale magnetic structures in thin magnetic films by firing laser pulses at them.
Magnetic spins in a material can swirl around a single point in a stable particlelike configuration. These magnetic vortices, called “skyrmions,” may find use in future forms of computer memory. New experiments have generated these magnetic structures using laser pulses directed at a thin magnetic film. By varying the strength of the pulse, the researchers created individual skyrmions and even “molecules” made from multiple skyrmions, as detailed in Physical Review Letters.
A skyrmion can be thought of as a twist in a field, which can’t easily be unwound. The idea originated fifty years ago as a way to explain baryons, but it was superseded by the quark model. However, researchers have identified skyrmion states in other systems, including Bose-Einstein condensates and superconductors, where the field equations exhibit the characteristic skyrmion topology. Over the last few years, physicists have also demonstrated several types of magnetic skyrmions by exposing magnetic thin films to a strong external magnetic field.
The skyrmions produced by Marco Finazzi of the Polytechnic Institute of Milan, Italy, and his collaborators are unique in not requiring an applied magnetic field, neither for generation nor for stabilization. The team fired laser pulses at thin films of the ferrimagnetic alloy terbium-iron-cobalt (TbFeCo). They then mapped out the resulting magnetic spin configuration with optical microscopy. At relatively low laser energy, they observed $100$-nanometer-wide cylindrically shaped skyrmions, in which the magnetization of the interior was opposite that of the exterior. Higher energy produced donut-shaped structures, which could be described as a skyrmion overlapping with an antiskyrmion (with opposite vortex circulation). The team called this moleculelike combination skyrmonium. These structures, as well as other more complex ones, did not noticeably change over one year of monitoring. The authors attribute this stability to the film’s own internal magnetic field. – Michael Schirber
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# Why does a substance such as methane have 4 sp3 orbitals rather than 3 orbitals [closed]
I have been learning about SP3 hybridisation and slightly confused why it it results in 4 sp3 orbitals.
My understanding is this...
The S orbital is spherical and uniform whilst the p orbitals are made of 2 lobes.
There are 3 p orbitals Px Py and Pz.
So if each of these p orbitals combines respectively with the s orbital you end up with 3 new ones. Where does the remaining orbital come from?
Thanks very much,
Daniel
• It would be quite wrong to say that each sp3 orbital comes from specific p orbital. Instead, they are all mixed together. – Ivan Neretin Dec 2 '19 at 15:26
• The number of atomic orbitals equals the number of (hybrid) molecular orbitals. – user55119 Dec 2 '19 at 16:40
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## anonymous one year ago Complex molecules are formed by the bond formation between monomers. Which process is being referred to in this statement?
• This Question is Open
1. Koikkara
@licklypaula $$Hey,~Where~are~the~statements~?$$
2. Koikkara
$$\small\rm\color{blue}{Welcome~to~OpenStudy!}$$ A chemical process that combines several monomers to form a polymer or polymeric compound > $$Polymerization$$ $$\small\rm\color{green}{Nice~to~meet~you~!}$$ @licklypaula
3. anonymous
Dehydration Synthesis: Monomers combine with each other via covalent bonds to form polymers.
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# How to install FreeNAS 10 “Corral” in 5 easy steps
FreeNAS Corral has been released just a week ago, yet it has already reached its fans’ hearts thanks to its sleek interface and its new features. Today we’ll look at how to install it in 5 easy steps.
Corral is (as of the moment I’m writing this article) the latest release of FreeNAS, based on FreeBSD 11. It brings many improvements and above all, a new sleek, polished and asynchronous web User Interface. In the case you missed what’s new, here you can take a look at the new features.
Important
I take absolutely NO responsibility of what you do with your machine; use this tutorial as a guide and remember you can possibly cause data loss if you touch things carelessly.
This guide is not intended for newbies, FreeNAS is a little more complicated than usual Linux distributions. If you just want to start learning about the *nix world I suggest you start here or here.
You should use ECC memory; using non-ECC memory is a potential threat to your data and should be avoided. FreeNAS Corral has a suggested minimum 8GB of memory. From my tests, unlike FreeNAS 9, attempting to use 4GB will heavily affect performance and should be avoided.
## Requirements
In order to install Corral successfully you will need:
• A machine with a 64-bit processor.
• At least 8GB of RAM (ECC is preferred although not strictly required).
• A boot medium (CD/DVD, USB, whatever).
• If you plan to install FreeNAS on a USB pendrive you can’t use that same drive to boot from (you will need two USB drives).
## Installing FreeNAS Corral
### 1) Preparations
The first thing you have to do to proceed with the installation of FreeNAS Corral is to boot it from a support (CD/DVD and USB Pendrives are today the best options).
Once you have your support, reboot the machine, select an option like “boot options” or “boot priority list” and select your support, in this way you should see this screen. Press enter and then enter a second time.
If everything went correctly, you should see this screen which is quite self-explanatory. Just select option 1 and press enter.
### 3) Selecting install destination
In this step you will get a list of the possible destinations for your FreeNAS Corral installation. The most common destination is a USB device (you can also select a disk, however FreeNAS won't allow you to use that same disk as storage). If you want to install FreeNAS on USB search for entries starting with da, if you want to install on a disk search for entries starting with ad. Check your selection twice: the installation will completely erase the destination. When you're ready click enter and confirm.
You will be prompted to select a root password (that you will use during the initial setup). It is suggested to pick a strong password since the root user can do anything to your machine and if someone were to guess that password you would be in danger.
### 5) Boot mode
In this step you have to decide how FreeNAS will be booted from the destination you are installing it onto. If you're running on a machine which only uses UEFI you will of course select UEFI. Other systems may allow turning off UEFI boot (also called legacy mode by some vendors), in that case the choice of whether to use UEFI or BIOS is up to you, if you choose to use Legacy, remember to disable the UEFI mode in your UEFI.
### 6) Installation successful
If everything went smoothly you will now see this screen. Do as you are told: press ok to return to the menu, then issue a shutdown by selecting the appropriate option, after that remove the installation media (not the one you installed FreeNAS onto if you used a USB pendrive). Then power up your machine (you should also configure the BIOS/UEFI settings to make your computer boot from the destination of your installation by default). At this point the installation is over.
### 7) Boot environments
From here you will be able to select the different boot environments. At the first boot you will have only one, but they will start growing as time passes and updates are released. Selecting one will prompt you to select one mode, the default one is fine. Keep in mind that the first time you boot a new FreeNAS installation it can take quite some time (even 10-20 minutes).
### 8) CLI
This screen means that you installed and booted FreeNAS Corral correctly. This CLI is a new feature and replaces the old (nifty) menu. From here you can modify system settings, they can also be modified through the web UI that can be accessed through another computer by going to the provided address (in my case: 10.0.2.15).
### 9) Configuring network
A thing you might want to do before accessing the web interface is to configure the network interfaces. It is wise to set a static IP and disable DHCP in order to keep your server in a known position (address-wise). To do so you just have to issue two commands:
# network interface em0 set dhcp=false
# network interface em0 alias create address=YOURADDRESS netmask=YOURNETMASK
Of course you should replace em0 with the actual interface you want to configure (it might be different for you), and you should replace YOURADDRESS and YOURNETMASK with the actual values you desire.
### 10) Web Interface
That's it! Go to another computer connected to your local network and point your browser to the address given by FreeNAS. If everything went the right way you will see this screen. From here use the username root and the password you set on step 4 to log in the web interface.
## Conclusion
You now know what the basic requirements for a Corral installations are. The process itself isn’t that difficult and resembles the FreeNAS 9 installation, although the part regarding configuring the network is completely different because of the new CLI.
Image courtesy of mark | marksei
The following two tabs change content below.
#### mark
The IT guy with a slight look of boredom in his eyes. Freelancer. Current interests: Kubernetes, Tensorflow, shiny new things.
### 4 Responses
1. Gallifreyan says:
Hi there. Looks like you may be missing about 4 steps on this post? I see “1) Preparations” and then “Conclusion.” It’s a good start though :)
• mark says:
Hello Gallifreyan, try to swipe the image above the instruction; or take your mouse over and use the arrows : )
• Gallifreyan says:
Ah. I would probably never have thought of that, but I got to all ten steps.
One interesting feature I found ( haven’t verified or tested it yet, but it seems to have installed that way ) is that apparently Corral will automatically mirror boot drives if you put two in and choose them from the device list in step 3. I’m using two ports on an internal header with a pair of Hyundai 16GB drives. It’s probably overkill but I’m big on overkill. :)
https://rsts11.com/2017/04/10/system-build-report-a-xeon-d-htpc-for-freebsd-corral-or-vmware-vsphere/
• mark says:
Great post, I haven’t really tried to select two devices in step 3, in the past (FreeNAS 9) I had to do it manually. Well, choosing FreeNAS for home storage instead of a prebuilt NAS is probably overkill already, so you’re already in the club : )
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SEARCH HOME
Math Central Quandaries & Queries
Help please. I am building an octagon tree house to live in. I would like the total square footage to equal 800 to 1,000 sq ft. How long should each side of the octagon be. tx, susan
Hi Susan,
I assume you want a regular octagon.
If you look at our page on finding the area of a regular polygon you will see a formula that Stephen developed that uses the length of a side $a$ and the number of sides $n.$ the expression is
$A_T = \frac{a^2 n}{4 \tan\left(\frac{360^o}{2 n}\right)}$
If you want the area to be 800 square feet then this gives
$800 = \frac{8 \times a^2 }{4 \tan\left(\frac{360^o}{16}\right)}.$
Use your calculator to calculate $\tan\left(\frac{360^o}{16}\right)$ and then solve for $a.$
Write back if you need more assistance.
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
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## Infinitesimal deformations and the $\ell$-invariant
### Summary
We give a formula for the generalized Greenberg's $\ell$-invariant which was constructed in [Zbl 05994421] in terms of derivatives of eigenvalues of Frobenius.
11R23, 11F80
### Keywords/Phrases
$(\varphi, \Gamma)$-module, $L$-function, $p$-adic representation
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### What Is the Double Declining Balance – DDB Depreciation Method?
The double declining balance depreciation (DDB) method is one of two common methods a business uses to account for the expense of a long-lived asset. The double declining balance depreciation method is an accelerated depreciation method that counts as an expense twice as much of the asset’s book value each year compared to straight-line depreciation.
### The Formula for DDB Depreciation Is
\begin{aligned} &\text{Depreciation}=2\times \text{SLDP}\times\text{BV}\\ &\textbf{where:}\\ &\text{SLDP = Straight-line depreciation percent}\\ &\text{BV = Book value at the beginning of the period}\\ \end{aligned}
1:26
### The Basics of Double Declining Balance Depreciation
Under the generally accepted accounting principles (GAAP) for public companies, expenses are recorded in the same period as the revenue that is earned as a result of those expenses. Thus, when a company purchases an expensive asset that will be used for many years, it does not deduct the entire purchase price as a business expense in the year of purchase but instead deducts the price over several years.
Because the double declining balance method results in larger depreciation expenses near the beginning of an asset’s life and smaller depreciation expenses later on, it makes sense to use this method with assets that lose value quickly.
### Example of DDB Depreciation
As a hypothetical example, suppose a business purchased a $30,000 delivery truck, which was expected to last for 10 years. After 10 years, it would be worth$3,000, its salvage value. Under the straight-line depreciation method, the company would deduct $2,700 per year for 10 years – that is,$30,000 minus $3,000, divided by 10. Using the double declining balance method, however, it would deduct 20% of$30,000 ($6,000) in year one, 20% of$24,000 ($4,800) in year two ($4,800), and so on.
### Double Depreciation Rate
The double declining balance method is a type of declining balance method with a double depreciation rate. The declining balance method is one of the two accelerated depreciation methods, and it uses a depreciation rate that is some multiple of the straight-line method rate.
Depreciation rates used in the declining balance method could be 150%, 200% (double), or 250% of the straight-line rate. When the depreciation rate for the declining balance method is set as a multiple doubling the straight-line rate, the declining balance method is effectively the double declining balance method. Over the depreciation process, the double depreciation rate remains constant and is applied to the reducing book value each depreciation period.
### Declining Book Value Balance
The book value or depreciation base of an asset declines over time. With the constant double depreciation rate and a successively lower depreciation base, charges calculated with this method continually drop. The balance of the book value is eventually reduced to the asset's salvage value after the last depreciation period. However, the final depreciation charge may have to be limited to a lesser amount to keep the salvage value as estimated.
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# Hyperbolic triangle
In mathematics, the term hyperbolic triangle has more than one meaning.
A tiling of the hyperbolic plane with hyperbolic triangles – the order-7 triangular tiling.
## Hyperbolic geometry
In hyperbolic geometry, a hyperbolic triangle is a figure in the hyperbolic plane, analogous to a triangle in Euclidean geometry, consisting of three sides and three angles. The relations among the angles and sides are analogous to those of spherical trigonometry; they are most conveniently stated if the lengths are measured in terms of a special unit of length analogous to a radian. In terms of the Gaussian curvature K of the plane this unit is given by
$R=\frac{1}{\sqrt{-K}}.$
In all the trig formulas stated below the sides a, b, and c must be measured in this unit. In a hyperbolic triangle the sum of the angles A, B, C (respectively opposite to the side with the corresponding letter) is strictly less than a straight angle. The difference is often called the defect of the triangle. The area of a hyperbolic triangle is equal to its defect multiplied by the square of R:
$(\pi-A-B-C) R^2{}{}.\!$
The corresponding theorem in spherical geometry is Girard's theorem first proven by Johann Heinrich Lambert.
### Right triangles
If C is a right angle then:
• The sine of angle A is the ratio of the hyperbolic sine of the side opposite the angle to the hyperbolic sine of the hypotenuse.
$\sin A=\frac{\textrm{sinh(opposite)}}{\textrm{sinh(hypotenuse)}}=\frac{\sinh a}{\,\sinh c\,}.\,$
• The cosine of angle A is the ratio of the hyperbolic tangent of the adjacent leg to the hyperbolic tangent of the hypotenuse.
$\cos A=\frac{\textrm{tanh(adjacent)}}{\textrm{tanh(hypotenuse)}}=\frac{\tanh b}{\,\tanh c\,}.\,$
• The tangent of angle A is the ratio of the hyperbolic tangent of the opposite leg to the hyperbolic sine of the adjacent leg.
$\tan A=\frac{\textrm{tanh(opposite)}}{\textrm{sinh(adjacent)}}=\frac{\tanh a}{\,\sinh b\,}.\,$
The hyperbolic sine, cosine, and tangent are hyperbolic functions which are analogous to the standard trigonometric functions.
### Oblique triangles
Whether C is a right angle or not, the following relationships hold.
There is a law of cosines:
$\cosh c=\cosh a\cosh b-\sinh a\sinh b \cos C,\,$
its dual:
$\cos C= -\cos A\cos B+\sin A\sin B \cosh c,\,$
a law of sines:
$\frac{\sin A}{\sinh a} = \frac{\sin B}{\sinh b} = \frac{\sin C}{\sinh c},$
and a four-parts formula:
$\cos C\cosh a=\sinh a\coth b-\sin C\cot B.\,$
### Ideal triangles
If a pair of sides is asymptotic they may be said to form an angle of zero. In projective geometry, they meet at an ideal vertex on the circle at infinity. If all three are vertices are ideal, then the resulting figure is called an ideal triangle. An ideal hyperbolic triangle has an angle sum of 0°, a property it has in common with the triangular area in the Euclidean plane bounded by three tangent circles.
## Euclidean geometry
Hyperbolic triangle (yellow) and hyperbolic sector (red) corresponding to hyperbolic angle u, constructed in 2D Cartesian coordinates from a rectangular hyperbola (equation y = 1/x) and a line (equation y = x). The legs of the triangle are √2 times the hyperbolic cosine and sine functions.
In the foundations of the hyperbolic functions sinh, cosh and tanh, a hyperbolic triangle is a right triangle in the first quadrant of the Cartesian plane
$\{(x,y):x,y \in \mathbb R\},$
with one vertex at the origin, base on the diagonal ray y = x, and third vertex on the hyperbola
$xy=1.\,$
The length of the base of such a triangle is
$\sqrt 2 \cosh u,\,$
and the altitude is
$\sqrt 2 \sinh u,\,$
where u is the appropriate hyperbolic angle.
The analogy between circular and hyperbolic functions was described by Augustus DeMorgan in his Trigonometry and Double Algebra (1849).[1]
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# Complexity of deciding whether subspaces of Z_2^n cover every point 3*x times
When studying the complexity of checking identities in certain finite algebras, I came across the following decision problem:
Input: A positive integer $$n \in N$$ and a set of affine subspaces $$H_1,H_2,\ldots,H_k$$ of $$\mathbb Z_2^n$$ (where $$\mathbb Z_2$$ is the 2-element field).
Question: For every $$x \in \mathbb Z_2^n$$, let $$n(x) = |\{H_i:x\in H_i \}|$$. Is $$n(x)$$ a multiple of 3 for all $$x$$? Or, in other words, is every point of $$\mathbb Z_2^n$$ covered by $$0$$ modulo $$3$$ many of the affine subspaces?
Note that there are no conditions on $$n$$, $$k$$ and the dimension of the affine subspaces $$H_i$$.
The problem is clearly in coNP, but unfortunately I am not able to prove anything more about its complexity. My hope is for it to be coNP-complete, since this would have nice implications for our research. But also if it was in P or equivalent to some coNP-intermediate problem this would be a really interesting fact to know.
Two observations:
• If we additionally assume that the codimension of all input subspaces $$H_i$$ is bounded by a constant (e.g. if every $$H_i$$ is a hyperplane) then the problem is in P. Thus any encoding of a coNP-complete problem would require using subspaces of arbitrary codimension.
• I tried to encode the complement of some classical NP-complete problems, but did not succeed. In particular 'covering problems' like the Set cover problem seemed to be a natural choice. But that we count here 'modulo 3' seemed to be an obstacle in this approach.
Thanks for every hint! :) In particular references to possibly related problems in the literature would be very welcome!
• Note, this question is equivalent to determining that a given MOD3-AND-MOD2 circuit is a tautology (all assignments are satisfying ones). So e.g. the comment about codimension can be seen as each AND having constant fan-in; in that case each AND can be written as a short mod3 sum of MOD2s, and the circuit just becomes a multilinear polynomial over F3 (where satisfiabilty is very easy). – Ryan Williams Sep 26 '18 at 14:24
• @RyanWilliams: Thanks, this seems to be a helpful different point of view! Also there is plenty of literature on such circuits. I will update my post, if I find anything useful – user50712 Oct 1 '18 at 8:16
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How to obtain fasta sequences at only mapped sites
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Entering edit mode
5 days ago
I want to obtain fasta sequences by mapping reads to a reference sequence. My general plan:
Step 1. Generate an index of the reference sequence (reference sequence here consists of thousands of contigs from sequence capture of ultraconserved elements) and independently mapped reads from each sample to the same reference sequence using BWA.
Step 2. Convert SAM files to BAM files and sort with SAMtools
Step 3. Clean BAM files with Picard
Step 4. Use the mpileup function in SAMtools to call variant sites and produce a VCF file
Step 5. Use vcfutils to convert from VCF to fastq (excluding sites with quality scores <20)
Step 6. Lastly, use seqtk to convert fastq to fasta.
However, it has come to my attention that if the reads that are mapped to the reference are shorter than the length of the reference contig, then the reference sequence at those sites are added to the flanking regions of the shorter reads in the resulting consensus fasta.
My question: Is there a way, and at what step in my approach outlined above, to obtain fasta sequences for only mapped sites? I suspect masking the regions flanking the consensus sequence would be an option but I'm not sure how to do that or just obtaining the consensus fasta sequence for only mapped sites is doable.
fastq vcf samtools bcftools fasta • 293 views
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Entering edit mode
to obtain fasta sequences for only mapped sites?
Can you clarify what you exactly mean? You want to get a fasta sequence for a region where there are mapped reads but not for the actual individual reads that are there? You don't want the soft-clipped part (S), correct?
-------------------------------------------------------------------------
SSSS1================ ===========
============ =========2
From regions covering 1 through to 2?
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Entering edit mode
I want fasta sequences for only regions that have coverage
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Entering edit mode
5 days ago
LChart ▴ 430
You can use FastaAlternateReferenceMaker (https://gatk.broadinstitute.org/hc/en-us/articles/360037594571-FastaAlternateReferenceMaker) in conjunction with -L coverage.bed where coverage.bed just gives the regions that have mapping coverage. The intervals you pass in to -L determine where individual fasta records will start/end.
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Entering edit mode
Is there a way to do what I want using any of the software I specified in the steps that I listed?
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# Calculating Work and State Variables for an ideal Stirling Engine
1. Mar 28, 2013
### WalkTex
1. The problem statement, all variables and given/known data
Consider the ideal Stirling cycle working between a maximum temperature Th and min temp Tc, and a minimum volume V1 and a maximum volume V2. Suppose that the working gas of the cycle is 0.1 mol of an ideal gas with cv = 5R/2.
A) what are the heat flows to the cycle during each leg? Be sure to give the sign. For which legs is the heat flow positive?
B) What work is done by the cycle during each leg?
2. Relevant equations
W = Integral from Vi to Vf of PdV
W = QH-QC
3. The attempt at a solution
My attempt at a solution is given in the pdf below. My main problem was identifying some of these quantities without defined values for Vi, Vf, Qh, Qc or mass. I simply assumed some values that were gave in part C, but it was unclear to me if this is what was to be done, or if there was a way to calculate it without these values. For instance, in leg ii of the cycle, no mass is given for know data. How may I then calculate heat? Thanks in advance.
#### Attached Files:
• ###### 4.3and4.4 (1).pdf
File size:
137.5 KB
Views:
59
Last edited: Mar 28, 2013
2. Mar 28, 2013
### Staff: Mentor
For part (b), do not use any numerical values. Just derive the formulas for the work.
I don't understand why you write $W=Q_H - Q_C$ for individual legs. During the cycle, the working substance is imagined to be in contact with either the hot or the cold reservoir one at a time. I would use $\Delta U = W + Q$, with $U$ the energy of the working substance.
3. Mar 28, 2013
### WalkTex
Thank you for the tip. For part A, does one not need the mass of the working substance to solve for heat?
#### Attached Files:
• ###### Untitleddocument.pdf
File size:
116.3 KB
Views:
64
Last edited: Mar 28, 2013
4. Mar 29, 2013
### Staff: Mentor
No. Until you reach the point where you have to calculate actual values, such as efficiency, you should keep everything in generic terms.
I looked at your solution, and first I must say that your notation is confusing. For the volumes, you use $V_\mathrm{min}$ and $V_\mathrm{max}$ and $V_\mathrm{i}$ and $V_\mathrm{f}$, while the problem states $V_1$ and $V_2$. You should be also careful with the sign convention. When using $\Delta U = Q + W$, $W$ is in terms of the work done on the working substance, while for an engine, you usually want the work produced by the working substance to be positive. You get $W$ correct at then end, but because you dropped a minus sign along the way.
The efficiency you get is indeed low. You should check your calculation of $Q_\mathrm{pos}$ again (don't confuse $C_V$ and $c_V$...).
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Quadratic bounds on a convex function
Let $f: \mathbb{R}\rightarrow\mathbb{R}$ be a twice-differentiable convex function. It can shown be that for any $x_0 \in \mathbb{R}$:
$f(x) \leq f(x_0) + (x-x_0)f'(x_0) + \frac{(x-x_0)^2}{2}$C
where $C = \arg \max_x f''(x)$ is the maximum curvature of $f(x)$ over its domain. This result has been proven here:
Bohning, D. and Lindsay, B.G., Monotonicity of quadratic-approximation algorithms'', Annals of the Institute of Statistical Mathematics, vol. 40, no. 4, pp. 641-663, 1988.
Is it possible to derive a quadratic lower bound as well using a similar idea? It seems intuitively possible with a small enough choice of curvature in the bound.
-
This is nothing more than Taylor's theorem with bounds on the second derivative. So the answer to your question is yes, with the exception that the lower bound may be 0 (eg, $f(x) = x$). – copper.hat Aug 4 '12 at 1:03
To elaborate on my comment, suppose $f: \mathbb{R}^n \to \mathbb{R}$ is $C^2$. Then Taylor's theorem gives $$f(x) = f(x_0) + \frac{\partial f (x_0)}{\partial x} (x-x_0) + \int_0^1 (1-t) (x-x_0)^T \frac{\partial^2 f (x_0+t (x-x_0))}{\partial x^2} (x-x_0) \, dt .$$ Now suppose that $m \leq \frac{\partial^2 f (x)}{\partial x^2} \leq M$ on some convex set $K \subset \mathbb{R}^n$. Then is is easy to see that if $x,x_0 \in K$, then $$f(x_0) + \frac{\partial f (x_0)}{\partial x} (x-x_0) + \frac{m}{2} \|x-x_0\|^2 \leq f(x) \leq f(x_0) + \frac{\partial f (x_0)}{\partial x} (x-x_0) + \frac{M}{2} \|x-x_0\|^2 .$$
This is true for any $f \in C^2$, not just convex $f$. If $f$ happens to be convex, then you know that $m=0$ will serve as a lower bound. By considering the convex function $f(x) = x^4$ near $0$, you can see that even a strictly convex function may have $m=0$ as the 'best' lower bound in some sense.
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Thin Lens And Lens Maker Equations Video Lessons
Concept
# Problem: A slide projector needs to create a 98-cm-high image of a 2.0-cm-tall slide. The screen is 300 cm from the slide.Part A: What focal length does the lens need? Assume that it is a thin lens.Part B: How far should you place the lens from the slide?
###### FREE Expert Solution
Lens equation:
$\overline{)\frac{\mathbf{1}}{\mathbf{f}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{s}}_{\mathbf{o}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{s}}_{\mathbf{i}}}}$
Magnification:
$\overline{){\mathbf{m}}{\mathbf{=}}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{o}}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{s}}_{\mathbf{i}}}{{\mathbf{s}}_{\mathbf{o}}}}$
(a)
hi = 98 cm
ho = 2.0 cm
95% (325 ratings)
###### Problem Details
A slide projector needs to create a 98-cm-high image of a 2.0-cm-tall slide. The screen is 300 cm from the slide.
Part A: What focal length does the lens need? Assume that it is a thin lens.
Part B: How far should you place the lens from the slide?
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Thin Lens And Lens Maker Equations concept. You can view video lessons to learn Thin Lens And Lens Maker Equations. Or if you need more Thin Lens And Lens Maker Equations practice, you can also practice Thin Lens And Lens Maker Equations practice problems.
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# The pool is filled using two tubes in 2h. The first tube fills the pool 3h faster than the second tube. How many hours will it take to fill the tube using only the second tube?
Mar 12, 2016
We must solve by a rational equation.
#### Explanation:
We must find what fraction of the total tub can be filled in 1 hour.
Assuming the first tube is x, the second tube must be x + 3.
$\frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2}$
Solve for x by putting on an equal denominator.
The LCD is (x + 3)(x)(2).
$1 \left(x + 3\right) \left(2\right) + 1 \left(2 x\right) = \left(x\right) \left(x + 3\right)$
$2 x + 6 + 2 x = {x}^{2} + 3 x$
$0 = {x}^{2} - x - 6$
$0 = \left(x - 3\right) \left(x + 2\right)$
$x = 3 \mathmr{and} - 2$
Since a negative value of x is impossible, the solution is x = 3. Therefore, it takes 3 + 3 = 6 hours to fill the pool using the second tube.
Hopefully this helps!
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# Apostolico–Giancarlo algorithm
In computer science, the Apostolico–Giancarlo algorithm is a variant of the Boyer–Moore string search algorithm, the basic application of which is searching for occurrences of a pattern ${\displaystyle P}$ in a text ${\displaystyle T}$. As with other comparison-based string searches, this is done by aligning ${\displaystyle P}$ to a certain index of ${\displaystyle T}$ and checking whether a match occurs at that index. ${\displaystyle P}$ is then shifted relative to ${\displaystyle T}$ according to the rules of the Boyer–Moore algorithm, and the process repeats until the end of ${\displaystyle T}$ has been reached. Application of the Boyer-Moore shift rules often results in large chunks of the text being skipped entirely.
With regard to the shift operation, Apostolico–Giancarlo is exactly equivalent in functionality to Boyer–Moore. The utility of Apostolico–Giancarlo is to speed up the match-checking operation at any index. With Boyer-Moore, finding an occurrence of ${\displaystyle P}$ in ${\displaystyle T}$ requires that all ${\displaystyle n}$ characters of ${\displaystyle P}$ be explicitly matched. For certain patterns and texts, this is very inefficient – a simple example is when both pattern and text consist of the same repeated character, in which case Boyer–Moore runs in ${\displaystyle O(nm)}$, where ${\displaystyle m}$ is the length in characters of ${\displaystyle T}$. Apostolico–Giancarlo speeds this up by recording the number of characters matched at the alignments of ${\displaystyle T}$ in a table, which is combined with data gathered during the pre-processing of ${\displaystyle P}$ to avoid redundant equality checking for sequences of characters that are known to match.
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1. ## A binomial identity
I'm trying to give a combinatorial proof of the following identity:
$\sum_{k=0}^{n} 2^k \binom{n}{k} \binom{n-k}{\lfloor \frac{n-k}{2} \rfloor}=\binom{2n+1}{n}$
***
On the right side, $\binom{2n+1}{n}$ is the number of $n$-element subsets of a $(2n+1)$-element set, or the number of ways we can select $n$ different balls from a box containing $(2n+1)$ balls...
The left side doesn't seem so straightforward. Let's say we have an $n$-element set, and we wish to find the number of ways we can form a $k$-element subset, where $k=0,1,...n$, and the number of ways we can pick a subset of that newly-formed $k$-element set.
As a $k$-element set has $2^k$ subsets, we can do this procedure in $2^k \cdot \binom{n}{k}$ ways, for some $k$.
But the next factor, $\binom{n-k}{\lfloor \frac{n-k}{2} \rfloor}$ is totally baffling. I'm not even sure that the above interpretation is correct. And the next problem would be to show that the whole left side of the problem, in fact, equals $\binom{2n+1}{n}$.
Start by selecting k pairs to choose one element from. There are $\textstyle2^k\binom nk$ ways of doing this ( $\textstyle\binom nk$ ways of choosing which pairs, and 2 ways of choosing which element in each pair). To bring the total number of selected elements up to n, we then need to select both elements from exactly half of the remaining n-k pairs, if n-k is even. If n-k is odd then we have to select both elements from $\lfloor \tfrac{n-k}{2} \rfloor$ of them, together with the single element of S. In either case, there are $\binom{n-k}{\lfloor \frac{n-k}{2} \rfloor}$ ways of doing this.
That shows that $\sum_{k=0}^{n} 2^k \binom{n}{k} \binom{n-k}{\lfloor \frac{n-k}{2} \rfloor}=\binom{2n+1}{n}$.
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# The important redundancy
Algebra Level 2
$\large f(x)=\frac{9^x}{9^x+3}$
Suppose we define $f(x)$ as above. Let $a=f(x)+f(1-x)$ and $b=f\left(\frac1{1996}\right) + f\left(\frac2{1996}\right) + f\left(\frac3{1996}\right)+\ldots+ f\left(\frac{1995}{1996}\right).$
Evaluate $a + b$.
×
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# Weird corruption when drawing text with DrawString
This topic is 1414 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
OK, so I'm experiencing the weirdest problem. I'm drawing some text and it's showing up corrupted:
[attachment=20278:Corrupted.png]
So, as you can see, the image on the left shows the corrupted text in Windows 8.1 and on the right the actual text is being rendered correctly in Windows 7.
The corruption only happens when I have the CompositingMode set to SourceCopy and the TextRenderingHint set to SingleBitPerPixel (with and without GridFit). I require SourceCopy as my compositing mode because I need to write to the alpha channel of the bitmap instead of blending the alpha with the bitmap contents.
The code that draws this text is incredibly simple, so I am at a loss as to why this is happening.
Anyone else run into this problem? Know of a way around it?
Here's the code that I used to repro the error:
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Drawing.Drawing2D;
using System.Drawing.Imaging;
using System.Drawing.Text;
using System.Linq;
using System.Text;
using System.Windows.Forms;
namespace TestFontAntiAlisBug
{
public partial class Form1 : Form
{
private Bitmap _image;
private Font _font;
protected override void OnLoad(EventArgs e)
{
_font = new Font("Andalus", 48.0f, GraphicsUnit.Point);
_image = new Bitmap(256, 256, PixelFormat.Format32bppArgb);
using (Graphics g = Graphics.FromImage(_image))
{
g.Clear(Color.Black);
g.CompositingMode = CompositingMode.SourceCopy;
g.CompositingQuality = CompositingQuality.HighQuality;
g.TextRenderingHint = TextRenderingHint.SingleBitPerPixelGridFit;
using (Brush brush = new SolidBrush(Color.White))
{
g.DrawString("Are you corrupted?", _font, brush, new PointF(0, 0));
}
}
}
protected override void OnPaint(PaintEventArgs e)
{
base.OnPaint(e);
e.Graphics.DrawImage(_image, new Point(0, 0));
}
public Form1()
{
InitializeComponent();
}
}
}
##### Share on other sites
OK, so I finally found another person that has run into the same problem (this is from 2012... I'm astounded it's still an issue):
http://social.msdn.microsoft.com/forums/windowsdesktop/en-US/7f02b531-529d-4940-a220-cde46e61e88f/windows-8-garbled-text-with-gdi-graphicsdrawstring
In the end I used the GraphicsPath method described by Gianpaolo64 (in the aforementioned link) and that solved my issue.
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Friday
October 24, 2014
# Homework Help: Calculus
Posted by Anonymous on Friday, November 16, 2012 at 1:20pm.
A 12' ladder is leaning against a wall making an angle theta with the ground. The ladder begins to slide down the wall. what is the rate of change of the height of the ladder, h, with respect to the change in the angle of the ladder with the floor when the ladder reaches a height of 9 feet?
• Calculus - Steve, Friday, November 16, 2012 at 2:51pm
if the distance from wall is x,
x^2 + h^2 = 144
x = √(144-h^2)
when h=9, x = √(144-81) = √63
sinθ = h/12
the wording of the question is odd. Usually they ask for the rate of change of x when h=9. In that case,
2x dx/dt + 2h dh/dt = 0
but we don't have dh/dt.
So, it appears they want dh/dθ
h/12 = sinθ
h = 12sinθ
dh/dθ = 12cosθ = 12(x/12) = x = √63
however, rather than asking for "the rate of change of h with respect to θ" you say they want "rate of change of h with respect to the change of θ". Don't know how to interpret that
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Contrasting Peterson’s and Dekker’s algorithms
I am trying to understand the algorithms by Peterson and Dekker which are very similar and display a lot of symmetries.
I tried to formulate the algorithms in informal language like follows:
Peterson's: "I want to enter." flag[0]=true;
"You can enter next." turn=1;
"If you want to enter and while(flag[1]==true&&turn==1){
it's your turn I'll wait." }
Else: Enter CS! // CS
"I don't want to enter any more." flag[0]=false;
Dekker's: "I want to enter." flag[0]=true;
"If you want to enter while(flag[1]==true){
and if it's your turn if(turn!=0){
I don't want to enter any more." flag[0]=false;
I'll wait." }
"I want to enter." flag[0]=true;
}
}
Enter CS! // CS
"You can enter next." turn=1;
"I don't want to enter any more." flag[0]=false;
The difference seems to be the point where "You can enter next." occurs and the fact that "if it's your turn I don't want to enter any more." occurs in Dekker's.
In Peterson's algorithm, the two processes seem to be dominant. A process seems to force his way in into the critical section unless it's the other one's turn.
Conversely, in Dekker's algorithm, the two processes seem to be submissive and polite. If both processes want to enter the critical section, and it's the other one's turn, the process decides to no longer want to enter. (Is this needed for starvation-freedom? Why?)
How exactly do these algorithms differ? I imagine that when both processes try to enter the critical section, in Peterson's, the process says "I enter", while in Dekker's the process says "You may enter". Can someone clear up the way the processes behave in each algorithm? Is my way of putting it in informal terms correct?
• Note that Peterson's algorithm does not completely solve the critical section problem, since the reads and writes into the flags themselves are critical section problems. A paper that actually solves the problem completely is "The Arbiter: an Active System Component for Implementing Synchronizing Primitives", by Henk J.M. Goeman. – user3083171 Oct 26 '15 at 16:07
Your informal descriptions of the algorithms is wonderful.
I think in both cases the author was trying to come up with the simplest solution they could think of that guaranteed both mutual exclusion and deadlock freedom. Neither algorithm is starvation free or fair.[ed: as pointed out in the comments, both algorithms are starvation free, and Peterson's algorithm is also fair]. Dekker's solution was the first mutual exclusion algorithm using just load and store instructions. It was introduced in Dijkstra, Edsger W.; "Cooperating sequential processes", in F. Genuys, ed., Programming Languages: NATO Advanced Study Institute, pp. 43-112, Academic Press, 1968. If you read through the paper you see Dijkstra work through a number of attempts, recognizing the problem with each, and then adding a little bit more for the next version. Part of the inefficiency of his algorithm comes from the fact that he starts with a turn-taking algorithm and then tries to modify it to allow the processes to progress in any order. (Not just 0,1,0,1,...)
Peterson's algorithm was published in 1981, after more than a decade of experience and hindsight about Dekker's algorithm. Peterson wanted a much simpler algorithm than Dekker so that the proof of correctness is much easier. You can see that he was feeling some frustration with the community from the title of his paper. Peterson, G.L.; "Myths about the mutual exclusion problem," Inf. Proc. Lett., 12(3): 115-116, 1981. Very quick read and very well written. (And the snide remarks about formal methods are priceless.) Peterson's paper also discusses the process by which he built his solution from simpler attempts. (Since his solution is simpler, it required fewer intermediate steps.) Note that the main difference (what you call "dominance" rather than "submissiveness") is that because Peterson started out fresh (not from the turn-taking algorithm Dijkstra started with) his wait loop is simpler and more efficient. He realizes that he can just get away with simple looped testing while Dijkstra had to backoff and retry each time.
I feel I must also mention Lamport's classic Bakery algorithm paper: Lamport, Leslie; "A New Solution of Dijkstra's Concurrent Programming Problem", Comm ACM 17(8):453-455, 1974. The Bakery algorithm is arguably simpler than Dekker's algorithm (and certainly simpler in the case of more than 2 processors), and is specifically designed to be fault tolerant. I specifically mention it for two reasons. First, because it gives a little bit of history about the definition of the mutual exclusion problem and attempts to solve it up to 1974. Second because the Bakery algorithm demonstrates that no hardware atomicity is required to solve the mutual exclusion problem. Reads that overlap writes to the same location can return any value and the algorithm still works.
Finally, a particular favorite of mine is Lamport, Leslie; "A Fast Mutual Exclusion Algorithm," ACM Trans. Comp. Sys., 5(1):1-11, 1987. In this paper Lamport was trying to optimize a solution to the mutual exclusion problem in the (common) case that there is little contention for the critical section. Again, it guarantees mutual exclusion and deadlock freedom, but not fairness. It is (I believe) the first mutual exclusion algorithm using only normal reads and writes that can synchronize N processors in O(1) time when there is no contention. (When there is contention, it falls back on an O(N) test.) He gives an informal demonstration that the best you can do in the contention free case is seven memory accesses. (Dekker and Peterson both do it with 4, but they can only handle 2 processors, when you extend their algorithms to N they have to add an extra O(N) accesses.)
In all: I'd say Dekker's algorithm itself is interesting mainly from a historical perspective. Dijkstra's paper explained the importance of the mutual exclusion problem, and demonstrated that it could be solved. But with many years of hindsight simpler (and more efficient) solutions than Dekker's have been found.
• >>Neither algorithm is starvation free or fair. That is not true. Peterson's algorithm is starvation free and fair. If a thread is in the critical section and the other one is waiting in the waiting loop - the one waiting will get into the CS next, even if the thread that was in the CS is much faster. – user24190 Nov 27 '14 at 6:37
• I would like to emphasize that Peterson's algorithm is starvation free and fair, if only to repeat user24190's comment. I cannot understand why after all these years, the author of this answer have not either replied to the comment nor corrected his answer. (be sure to ping me once you have corrected the answer so that I can remove this comment of mine.) – John L. Oct 21 '18 at 17:54
• Link to purchase Peterson's "Myths about the mutual exclusion problem": doi.org/10.1016/0020-0190(81)90106-X – strager Aug 11 '19 at 21:42
• PDF of Peterson's "Myths about the mutual exclusion problem" (Archive.org): web.archive.org/web/20150501155424/https://cs.nyu.edu/~lerner/… – strager Aug 11 '19 at 21:44
• Don't you think Dekker's algorithm is starvation free? Definitely it is not fair but in terms of starvation free, one process will eventually enter the critical section. Correct me if I am wrong. Thanks. – Yiqun Sun Jan 22 '20 at 18:18
In the following paper we give formal models for Peterson’s and Dekker’s algorithms (and some others) and we used model checker to prove their properties. Please, find our results in the table below (columns "Deadlock" and "Divergent" refer to our models, "ME" = TRUE means that the algorithm is correct, "Overtaking" = TRUE means that it is fair).
R. Meolic, T. Kapus, Z. Brezočnik. ACTLW - An Action-based Computation Tree Logic With Unless Operator. Information Sciences, 178(6), pp. 1542-1557, 2008.
https://doi.org/10.1016/j.ins.2007.10.023
Peterson algorithm has a more strict pressure on entering the critical section, where as dekker's algorithm is relatively softer and less aggressive. To make it more clear, let's check out an example about iranian culture. Before getting into this example, it's good to know that iranian people have a real soft behavior to each other while entering somewhere! Guess two iranian men are going to enter a house, and that house has only one door to enter.
Now imagine two men from another culture(Zombian Culture) which they don't actually care too much about each other while entering somewhere(It's a matter of respect to ask someone whether he wants to enter or not).
To clarify the information about the problem we can say that:
• Two Iranians = Two processes using the Dekker algorithm
• Two Zombians = Two processes using the Peterson algorithm
So let's find out what's done in each algorithm(culture). The following comments are for the first Iranian man who is going to enter the house while using the Dekker algorithm:
p0:
wants_to_enter[0] ← true // Goes through the house entrance
while wants_to_enter[1] { // If the other man wants to enter too
if turn ≠ 0 { // Then see if it is your turn or not
wants_to_enter[0] ← false // If it is not your turn don't go furthur
while turn ≠ 0 { // and wait until it is your turn
// busy wait
}
wants_to_enter[0] ← true // when it is your turn go through the door
}
}
// critical section
...
turn ← 1
wants_to_enter[0] ← false // Set the turn for the other man
// remainder section
We have also two Zombians which are going to enter the house using the Peterson algorithm. This one goes as follows:
P0:
flag[0] = true; // Goes through the house entrance
turn = 1; // Set the turn for himself
while (flag[1] && turn == 1) // Wait until the other one is going in
{
// busy wait
}
// critical section
...
// end of critical section
flag[0] = false; // Done going inside
It's important to mention that both of them are not going inside while the other one is doing so(Mutual Exlusion) but, Iranian people are much more softer.
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# Thread: Statics Problem. Reaction forces, etc.
1. ## Statics Problem. Reaction forces, etc.
We have a test coming up next week and here's a practice problem I have no idea how to do. Mainly the way the pulley is holding the weight throws me off.
For these type of problems, we are going to have to solve for the reaction forces, and show them as vectors.
I drew my free body diagram.
The forces in the x and y have to be equal to 0, since this is statics.
Can someone show me how to do this problem. I really don't know where to start, and we haven't went over one like this in class. Any information you guys can provide will be extremely helpful, as I'm having a hard time with this class.
2. ## Re: Statics Problem, Need help! Reaction forces, etc.
Originally Posted by Bracketology
We have a test coming up next week and here's a practice problem I have no idea how to do. Mainly the way the pulley is holding the weight throws me off.
For these type of problems, we are going to have to solve for the reaction forces, and show them as vectors.
I drew my free body diagram.
The forces in the x and y have to be equal to 0, since this is statics.
Can someone show me how to do this problem. I really don't know where to start, and we haven't went over one like this in class. Any information you guys can provide will be extremely helpful, as I'm having a hard time with this class.
List the forces, there are two reaction forces and two forces corresponding to the tension in the string that need to be accounted for, these sum to zero.
Take moments about the left hand pivot, these should also sum to zero.
Now if you knew the horizontal distance of the pulley axle from the left hand pivot you would be able to solve for all the forces.
CB
3. ## Re: Statics Problem. Reaction forces, etc.
I emailed the teacher and he said to assume the middle of the pulley and the far right upward reaction force from the roller joint are in the same line. Now the problem makes a little more sense.
Thanks for the help!
4. ## Re: Statics Problem. Reaction forces, etc.
So, is this anywhere near correct?
Forces in X=0=Ox (Reaction from the wall in x)+1500sin(18.435)= Ox=-474.343 Newtons
Forces in Y=0=Ox (Reaction from the wall in y)+1500cos(18.435)=Oy=1423.03 Newtons
Honestly lost on this problem. We have been doing very simple problems in class, and this is way beyond what we have went over, and we have to put all that we know so far together to get this answer and I'm having trouble. Any more direction on what to do would help.
5. ## Re: Statics Problem. Reaction forces, etc.
Originally Posted by Bracketology
So, is this anywhere near correct?
Forces in X=0=Ox (Reaction from the wall in x)+1500sin(18.435)= Ox=-474.343 Newtons
Forces in Y=0=Ox (Reaction from the wall in y)+1500cos(18.435)=Oy=1423.03 Newtons
Honestly lost on this problem. We have been doing very simple problems in class, and this is way beyond what we have went over, and we have to put all that we know so far together to get this answer and I'm having trouble. Any more direction on what to do would help.
The only forces that are relevant are vertical, we have reaction at left hand pivot $r_1$, and at right ball $r_2$. We have the tension in the cable supporting the mass is $mg=15000\ \text{N}$, and this acts on the pully twice so we have:
(positive upwards) $r_1+r_2-30000=0$
Taking moments about the left hand pivot/support:
$-15000\times 1 -15000\times 2+r_2\times 1.5=0$
CB
6. ## Re: Statics Problem. Reaction forces, etc.
Thank-you sir.
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## Intermediate Algebra (6th Edition)
Published by Pearson
# Chapter 3 - Cumulative Review: 35
#### Answer
$[-2,\frac{8}{5}]$
#### Work Step by Step
We are given that $|5x+1|+1\leq10$. First, subtract 1 from both sides. $|5x+1|\leq9$ Therefore, $5x+1\leq9$ and $5x+1\geq-9$. We can solve these equations for x to find solutions. For $5x+1\leq9$, subtract 1 from both sides. $5x\leq8$ Divide both sides by 5. $x\leq\frac{8}{5}$ For $5x+1\geq-9$, subtract 1 from both sides. $5x\geq-10$ Divide both sides by 5. $x\geq-2$ So the absolute-value inequality is satisfied for all values of x that are greater than or equal to -2 and less than or equal to $\frac{8}{5}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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## Functions Modeling Change: A Preparation for Calculus, 5th Edition
$0$
Start with the innermost argument, $\log 10$. Since $\log 10^a=a$ and $10=10^1$, $\log10=\log10^1=1$. Therefore, $\log(\log 10)=\log(1)$ Since $1=10^{0}$, $$\log(\log 10)=\log(10^0)=0$$
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# Why should time slow down for light?
So I just learnt about relativity and time dilation. One of the visualisations used to explain time dilation is an apparatus contaning a beam of light moving between two mirrors. When the apparatus is stationary, the light moves a certain distance d, say in one second between the mirrors. But when the apparatus is say, fixed to a spaceship, the light beam moves in a zig zag fashion, covering a long distance $$D$$. The speed of light is constant. Therefore the time for light must slow down, because the distance being covered by light is greater. I have a link below which explains the visualisation. https://youtu.be/TgH9KXEQ0YU
This is what we were explained. However, why is it that the beam of light can't cover that greater distance in a greater time $$t$$? Why should time slow down for the beam of light?
It is one of the postulates of special relativity that the speed of light $$c$$ is the same for all non-inertial observers. Being a postulate, this means there is no known cause for it - we know that $$c$$ is constant, but we don't know why (For some mor informtion on this, you may read Why and how is the speed of light in vacuum constant, i.e., independent of reference frame?).
However, a constant speed of light means that time must slow down for observers with relative velocity $$v$$ "in order to keep $$c$$ the same for them". This explanation is of course a bit sloppy, but I think it covers some intuition and in addition, you apparently learned special relativity in more depth.
So why does time slow down for light? This formulation seems a bit misleading to me. There is no reference frame in which a photon is at rest. One could go even further and say that if one were to move at the speed of light, time wouldn't really "exist" or it would stand still. But, you can say that in your reference frame (for example earth), the time of the spaceship is slowed down.
So the short, somewhat unsatisfactory answer is "we don't really know, but we know that it is the case".
For the relationship $$\text{speed}=\frac{\text{distance}}{\text{time}}$$ to hold true, if the distance changes either the time or the speed must change (or both). In the case of light we can in other experiments measure that its speed is always constant. Thus, it must be time which is changing.
Why can time change in this manner? Actually it shouldn't be odd at all that time can vary (slow down i.e.); you only expect it not to change because of what you are used to. Why would you expect time to be different in kind than, say, speed or distance? We have no physical ground to assume so. This is the abstract intuition-breaking claim Albert Einstein brought forward in his theory of special relativity.
What is odd is the fact that the speed of light is constant in all frames (that speed has an upper limit in our world). That is not something we've been able to explain yet - we can just measure that it is the case.
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# NEET Physics Alternating Current Questions Solved
Power dissipated in an L-C-R series circuit connected to an AC source of emf $\epsilon$ is
(a) $\frac{{\epsilon }^{2}R}{\left[{R}^{2}+{\left(L\omega -\frac{1}{C\omega }\right)}^{2}\right]}$
(b) $\frac{{\epsilon }^{2}{\sqrt{{R}^{2}+\left(L\omega -\frac{1}{C\omega }\right)}}^{2}}{R}$
(c) $\frac{{\epsilon }^{2}\left[{R}^{2}+{\left(L\omega -\frac{1}{C\omega }\right)}^{2}\right]}{R}$
(d) $\frac{{\epsilon }^{2}R}{\sqrt{{R}^{2}+{\left(L\omega -\frac{1}{C\omega }\right)}^{2}}}$
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# How do you change (4, -1) from rectangular to cylindrical coordinates between [0, 2π)?
Mar 4, 2015
I will assume you meant "polar coordinates" since cylindrical coordinates are 3-dimensional and you have only supplied 2-dimensional Cartesian coordinates.
From the graph:
$r = \sqrt{{4}^{2} + {\left(- 1\right)}^{2}} = \sqrt{17}$
and
$\theta = \arctan \left(\frac{- 1}{4}\right)$
Note that standard arctan functions will return a value which will need to be adjusted to compensate for the point being in the IV quadrant so $\theta$ will fall in the $\left[0 , 2 \pi\right)$ range.
Depending upon the version used you will typically get
$\arctan \left(\frac{- 1}{4}\right) = - 0.245$ (radians)
or
$\arctan \left(\frac{- 1}{4}\right) = - 14.040$ (degrees)
Add $2 \pi$ or ${360}^{o}$ respectively to get the "correct" answer.
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ISSN 1000-1239 CN 11-1777/TP
• 软件技术 •
### 基于元数据逻辑无关片断的结构完整性检测方法
1. 1(天津工业大学计算机科学与技术学院 天津 300387);2(江苏省计算机信息处理技术重点实验室(苏州大学) 江苏苏州 215006);3(中国科学院计算技术研究所智能信息处理重点实验室 北京 100190) (zhaoxiaofei1978@hotmail.com)
• 出版日期: 2020-09-01
• 基金资助:
国家自然科学基金项目(61035003,61972456)；江苏省计算机信息处理技术重点实验室开放基金项目(KJS1737)
### Structural Integrity Checking Based on Logically Independent Fragment of Metadata
Zhao Xiaofei1,2, Shi Zhongzhi3, Liu Jianwei3
1. 1(School of Computer Science and Technology, Tiangong University, Tianjin 300387);2(Provincial Key Laboratory for Computer Information Processing Technology(Soochow University), Suzhou, Jiangsu 215006);3(Key Laboratory of Intelligent Information Processing, Institute of Computing Technology, Chinese Academy of Sciences, Beijing 100190)
• Online: 2020-09-01
• Supported by:
This work was supported by the National Natural Science Foundation of China (61035003, 61972456) and the Open Foundation of Jiangsu Provincial Key Laboratory for Computer Information Processing Technology (KJS1737).
Abstract: Checking the structural integrity efficiently is one of the research hotspots in the field of MOF(meta object facility) repository system consistency. In this paper, we propose an efficient and automatic approach for checking the structural integrity by means of description logics. Firstly, according to the characteristics of MOF architecture, we study how to transform different levels of metadata into SROIQ(D) knowledge base. Then we study how to extract metadata to improve the efficiency of the checking process. We propose the concept of logically independent fragment of metadata. By extracting property deductive fragment and classification deductive fragment respectively, we present the algorithm to generate the minimum logically independent fragment. Since this kind of fragment is the closure of logical implication for a given metadata element, all relevant information about the given metadata element is completely preserved, thus the checking process can be performed on a smaller set of metadata rather than on the entire repository. Finally, we study how to perform checking based on logically independent fragment. The experimental results show that the average size of the metadata fragment generated by our approach is significantly smaller than its original size, and the efficiency improvement of the checking on the metadata fragment ranges from 1.47 times to 3.31 times. The time performance comparison with the related approaches also shows the effectiveness of our approach.
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## Sitaraman, Sankar - Mean Values of Certain Multiplicative Functions and Artin's Conjecture on Primitive Roots
hrj:1316 - Hardy-Ramanujan Journal, January 1, 2014, Volume 37
Mean Values of Certain Multiplicative Functions and Artin's Conjecture on Primitive Roots
Authors: Sitaraman, Sankar
We discuss how one could study asymptotics of cyclotomic quantities via the mean values of certain multiplicative functions and their Dirichlet series using a theorem of Delange. We show how this could provide a new approach to Artin's conjecture on primitive roots. We focus on whether a fixed prime has a certain order modulo infinitely many other primes. We also give an estimate for the mean value of one such Dirichlet series.
Source : oai:HAL:hal-01220296v1
Volume: Volume 37
Published on: January 1, 2014
Submitted on: October 30, 2015
Keywords: multiplicative functions,cyclotomic fields,Dirichlet series,arithmetic functions,primitive roots,[MATH] Mathematics [math],[MATH.MATH-NT] Mathematics [math]/Number Theory [math.NT]
## Browsing statistics
This page has been seen 346 times.
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Browse Questions
# The value of $\lambda$ for which the vectors $3\hat i+6\hat j+\hat k$ and $2\hat i+4\hat j+\lambda\hat k$ are parallel is
$(A)\;\frac{2}{3}\quad(B)\;\frac{3}{2}\quad(C)\;\frac{5}{2}\quad(D)\;\frac{2}{5}$
Toolbox:
• $\overrightarrow a.\overrightarrow b=|\;a\;|\;|\;b\;| \cos \theta$
Let $\overrightarrow a=3\hat i+6\hat j+\hat k\:and\:\overrightarrow b=2\hat i+4\hat j+\lambda\hat k$
It is given that $\overrightarrow a$ is parallel to $\overrightarrow b$
$=>\theta=0\qquad \cos 0=1$
On substituting for $\theta$
$\overrightarrow a.\overrightarrow b=|\;a\;|\;|\;b\;| \cos \theta$
$\overrightarrow a.\overrightarrow b=|\; \overrightarrow a\;|\;|\; \overrightarrow b\;|\; \times 1$
On substituting for $\overrightarrow a$ and $\overrightarrow b$
(ie) $(3\hat i+6\hat j+\hat k).(2\hat i+4\hat j+\lambda k)$
$=\sqrt {(3)^2+(6)^2+(1)^2}.\sqrt {(2)^2+(4)^2+(\lambda)^2}$
$=(6+24+\lambda)=\sqrt {9+36+1}.\sqrt {4+16+\lambda^2}$
$(30+\lambda)=\sqrt {46}.$ $\sqrt {20+\lambda^2}$
Squaring on bothsides we get,
$(30+ \lambda)^2=46(20+\lambda^2)$
$=>900+60 \lambda +\lambda^2=920+46 \lambda^2$
On simplifying we get,
$45\lambda^2-60 \lambda+20=0$
Divide throughout by $5$
$9\lambda^2-12 \lambda+4=0$
$=>(3 \lambda-2)^2=0$
$=>3 \lambda-2=0$
Therefore $\lambda =\large\frac{2}{3}$
Hence the correct option is $A$
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×
Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 4 - Problem 4.62
Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 4 - Problem 4.62
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# ?(a) Calculate the molarity of a solution made by dissolving 12.5 grams of $$\mathrm{Na}_{2} \mathrm{CrO}_{4}$$ in enough water to form exactly 750 mL
ISBN: 9780134414232 1274
## Solution for problem 4.62 Chapter 4
Chemistry: The Central Science | 14th Edition
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Problem 4.62
(a) Calculate the molarity of a solution made by dissolving 12.5 grams of $$\mathrm{Na}_{2} \mathrm{CrO}_{4}$$ in enough water to form exactly 750 mL of solution.
(b) How many moles of KBr are present in 150 mL of a 0.112 M solution?
(c) How many milliliters of 6.1 M HCl solution are needed to obtain 0.150 mol of HCl?
Text Transcription:
Na2CrO4
Step-by-Step Solution:
Step 1 of 5) How many moles of KBr are present in 150 mL of a 0.112 M solutionHow many milliliters of 6.1 M HCl solution are needed to obtain 0.150 mol of HClIf a calorimetry experiment is carried out under a constant pressure, the heat transferred provides a direct measure of the enthalpy change of the reaction. Constant-volume calorimetry is carried out in a vessel of fixed volume called a bomb calorimeter. The heat transferred under constant-volume conditions is equal to ∆E. Corrections can be applied to ∆E values to yield ∆H.Because enthalpy is a state function, ∆H depends only on the initial and final states of the system. Thus, the enthalpy change of a process is the same whether the process is carried out in one step or in a series of steps. Hess’s law states that if a reaction is carried out in a series of steps, ∆H for the reaction will be equal to the sum of the enthalpy changes for the steps. We can therefore calculate ∆H for any process, as long as we can write the process as a series of steps for which ∆H is known.
Step 2 of 2
## Discover and learn what students are asking
Statistics: Informed Decisions Using Data : Scatter Diagrams and Correlation
?What does it mean to say two variables are positively associated? Negatively associated?
Statistics: Informed Decisions Using Data : The Normal Probability Distribution
?Find the z-score such that the area to the right of the z-score is 0.483.
Statistics: Informed Decisions Using Data : Tests for Independence and the Homogeneity of Proportions
?What’s in a Word? In a recent survey conducted by the Pew Research Center, a random sample of adults 18 years of age or older living in the continenta
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?(a) Calculate the molarity of a solution made by dissolving 12.5 grams of $$\mathrm{Na}_{2} \mathrm{CrO}_{4}$$ in enough water to form exactly 750 mL
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# Curse of dimensionality with language models
In the seminal paper A Neural Probabilistic Language Model, Yoshua Bengio and his colleagues make the following point:
If one wants to model the joint probability distribution of 10 consecutive words in a natural language with a vocabulary $V$ of size $100,000$, there are potentially $100,000^{10}-1$ free parameters.
I guess it's related to degrees of freedom and joint distributions but I just can't get my hands on the exact formula that was used here to come up with $100,000^{10}-1$.
• $100000$ is probably an estimate of the total number of words in the language. – kjetil b halvorsen Dec 1 '15 at 14:16
• @kjetilbhalvorsen 1e5 is the size of the vocabulary, so yes, 1e5 is the total number of unique words in the language, no problem here – Antoine Dec 1 '15 at 14:21
The estimate $100 000^{10}-1$ comes from assuming a discrete model for the $10$ consecutive words, without any simplifications or restrictions, thus using all interactions up to and including order $10$.
It is not important that the words are consecutive, we would get the same count for any ten specified word positions. For each position, it can be any of the $100000$ words, so we need that number of probabilities. So you can build up a cube in $10$-space, wity each dimension cut up in $100000$ boxes. Taking all the combinations, that give $100000^{10}$ boxes, each box giving on possible $10$-word sequence, such as " am I writing now holy blue crap green integrated ideas", which would be a sequence of fairly low probability. Then subtract $1$ to account for the fact that the probabilities must sum to one!
• Many thanks for elaborating. In other words, $100,000^{10}$ gives all the possible 10-element combinations from the vocabulary of size $100,000$. This number is huge but still finite since we're in the discrete case. And estimating the joint probability mass function of a specific 10-word sequence comes down to assigning a probability (probabilities=parameters here I assume) to each portion of that huge but finite discrete sample space. Is that correct? Also, sorry if I'm being dumb but I still don't get why subtracting 1 ensures that the probabilities will sum to one. – Antoine Dec 1 '15 at 15:34
• subtracting one does not by itself secure that the probabilities will sum to one, but it takes care of that restriction . After freely choosing the $100000^{10}-1$ probabilities, you can calculate the last one! – kjetil b halvorsen Dec 1 '15 at 15:55
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# 'Almost-isomorphic' groups
What can be said about pairs of non-isomorphic groups which are epimorphic images of one another and which also embed into one another?
Can such pairs of groups be 'classified' in some sufficiently weak, but still non-trivial sense, or are they just too common to hope for anything like this?
Obviously, groups forming such pairs can neither be Hopfian nor co-Hopfian.
An example of such pair of groups consists of $\rm C_\infty \times \rm F_2^\infty$ and $\rm F_2^\infty$, where $\rm C_\infty$ denotes the infinite cyclic group and $\rm F_2$ denotes the (nonabelian) free group of rank 2.
-
A slightly simpler example: $C_2 \times C_4^\infty$ and $C_4^\infty$. – S. Carnahan Jan 18 '13 at 14:53
Indeed. -- However which one is 'simpler' depends on your notion of 'simplicity': in order to define a free group, you need no relations, while to define a finite cyclic group, you need 1! – Stefan Kohl Jan 18 '13 at 15:00
@Berlusconi: I am not a group theorist, but my impression is that it is too optimistic to hope for a classification (without further assumptions and special conditions). – Martin Brandenburg Jan 22 '13 at 9:41
The examples given so far are of the form $(A \times B^\infty, B^\infty)$, where $A$ both embeds into $B$ and is an epimorphic image of $B$. -- Can anyone give an example where the groups do not admit a decomposition into infinitely many direct factors, or which are at least not of the form $(A \times B^\infty, B^\infty)$? – Stefan Kohl Jan 22 '13 at 16:00
I believe you can modify the example so that the direct products become semidirect products. – Khalid Bou-Rabee Jan 28 '13 at 5:03
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• Pair of Linear Equations in Two Variables(With Methods & Examples)
# Pair of Linear Equations in Two Variables(With Methods & Examples)
November 2, 2022
This post is also available in: हिन्दी (Hindi)
Linear equations are the simplest form of equations in mathematics. These equations are called linear equations because on graphing they give straight lines. A pair of linear equations are used to solve many real-life problems based on money, geometry, speed, and distance, etc.
Let’s understand what are pair of linear equations in two variables and what are its different types and its solutions with examples.
## Pair of Linear Equations in Two Variables
Consider the statement “Cost of $20$ pens and $15$ pencils is ₹$450$”. Based on this statement we want to find the costs of a $1$ pen and $1$ pencil.
To find the cost of a pen and a pencil, we start by assuming the costs of $1$ pen and $1$ pencil as $x$ and $y$ (in rupees) respectively.
The statement in mathematical form can be written as $20x + 15y = 450 =>20x + 15y – 450 = 0$.
You also know that an equation that can be put in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero, is called a linear equation in two variables
Can we find the cost of $1$ piece of each of the above items?
To do so, we need one more such statement say “Cost of $25$ pens and $30$ pencils is ₹$675$”
This can be written as $25x + 30y = 675 => 25x + 30y – 675 = 0$.
It can be observed from the above example, to solve such types of real-world problems, we need two equations.
Such a combination of two linear equations in two variables is called a pair of linear equations in two variables.
The general form for a pair of linear equations in two variables $x$ and $y$ is
$a_{1}x + b_{1}y + c_{1} = 0$
and $a_{2}x + b_{2}y + c_{2} = 0$
where $a_{1}$, $b_{1}$, $c_{1}$, $a_{2}$, $b_{2}$, and $c_{2}$ are all real numbers and $a_{1}$, $b_{1}$ and $a_{2}$, $b_{2}$ cannot be simultaneously $0$(zero).
Maths can be really interesting for kids
## Framing a Pair of Linear Equations in Two Variables
As observed above, we can convert many real-world problems into a pair of linear equations in two variables. Let’s understand the process of framing a pair of linear equations in two variables from real-world problems.
Ex 1: There are some students in the two examination halls A and B. To make the number of students equal in each hall, $10$ students were sent from A to B. But if $20$ students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.
Let the number of students in hall A = $x$
And let the number of students in hall B = $y$
When $10$ students are sent from A to B, then
Number of students left in hall A = $x – 10$
And the number of students in hall B = $y + 10$
Now, the number of students is equal in both halls, therefore, $x – 10 = y + 10 => x – y – 10 – 10 = 0 => x – y – 20 = 0$ —————- (1)
When $20$ students are sent from B to A, then
Number of students left in hall B = $x – 20$
And the number of students in hall A = $y + 20$
Now, the number of students in A is double the number of students in B, therefore,
$y + 20 = 2\left(x – 20 \right) => y + 20 = 2x – 40 => -2x + y + 20 + 40 = 0$
$=> -2x + y + 60 = 0$ ——————– (2)
Ex 2: A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid ₹$22$ for a book kept for six days, while Arvind paid ₹$16$ for a book kept for four days. Find the fixed charges and the charge for each extra day.
Let the fixed amount charged by the shopkeeper for a book = ₹$x$
And the per day amount charged by the shopkeeper for a book = ₹$y$
Latika kept the book for $6$ days
Therefore, the per day charge is for $4 \left(6 – 2 \right)$ (Fixed charge includes first two days).
Amount paid by Latika = $x + 4y$
Therefore, $x + 4y = 22$ ————————— (1)
Arvind kept the book for $4$ days
Therefore, the per day charge is for $2 \left(4 – 2 \right)$
Amount paid by Arvind = $x + 2y$
Therefore, $x + 2y = 16$ ————————— (2)
Ex 3: It can take $12$ hours to fill a swimming pool using two pipes. If the pipe of a larger diameter is used for $4$ hours and the pipe of a smaller diameter for $9$ hours, only half the pool can be filled. How long would it take for each pipe to fill the pool separately?
Let the time taken by a pipe of a larger diameter to fill a tank = $x$ hours.
And the time taken by a pipe of a smaller diameter to fill a tank = $y$ hours.
Therefore, part of a tank filled by a pipe of a larger diameter in $1$ hour = $\frac{1}{x}$.
And part of a tank filled by a pipe of a smaller diameter in $1$ hour = $\frac{1}{y}$.
When both pipes are opened for $12$ hours, then part of tank filled = $12 \times \frac{1}{x} + 12 \times \frac{1}{y} = \frac{12}{x} + \frac{12}{y}$
Since, in $12$ hours the tank is completely filled, therefore, $\frac{12}{x} + \frac{12}{y} = 1$ ————————– (1)
Also, part of a tank filled by a pipe of a larger diameter in $4$ hours = $4 \times \frac{1}{x} = \frac{4}{x}$
Similarly, part of a tank filled by a pipe of a smaller diameter in $9$ hours = $9 \times \frac{1}{y} = \frac{9}{x}$
Therefore, part of a tank filled by a pipe of a larger diameter in $4$ hours and a pipe of a smaller diameter in $9$ hours = $\frac {4}{x} + \frac{9}{y}$.
Therefore, $\frac {4}{x} + \frac{9}{y} = \frac{1}{2}$ ——————————— (2)
Ex 4: Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of ₹$2$ for $3$ bananas and the second lot at the rate of ₹$1$ per banana and got a total of ₹$400$. If he had sold the first lot at the rate of ₹$1$ per banana, and the second lot at the rate of ₹$4$ for $5$ bananas, his total collection would have been ₹$460$. Find the total number of bananas he had.
Let the number of bananas in lot A = $x$
And the number of bananas in lot B = $y$
Case 1:
Selling rate of lot A bananas = ₹$2$ for $3$ banana => $1$ banana for ₹$\frac{2}{3}$
Selling rate of lot B bananas = ₹$1$ for $1$ banana=> $1$ banana for ₹$1$
Selling price of all bananas = ₹$\frac{2}{3}x + y$
Therefore, $\frac{2}{3}x + y = 400 => 2x + 3y = 1200$ ——————- (1)
Case 2:
Selling rate of lot A bananas = ₹$1$ for $1$ banana => $1$ banana for ₹$1$
Selling rate of lot B bananas = ₹$4$ for $5$ banana=> $1$ banana for ₹$\frac{4}{5}$
Selling price of all bananas = ₹$x + \frac{4}{5}y$
Therefore, $x + \frac{4}{5}y = 460 => 5x + 4y = 2300$ ——————- (1)
## Methods of Solving Pair of Linear Equations
There are four methods to solve a system of linear equations in two variables. Those methods are used to solve a pair of linear equations in two variables are:
• Graphical Method
• Substitution Method
• Cross Multiplication Method
• Elimination Method
### Graphical Method of Solving a Pair of Linear Equations
The steps to solve linear equations in two variables graphically are given below:
Step 1: Graph each equation
Step 2: To graph an equation manually, first convert it to the form $y = mx + b$ by solving the equation for $y$
Step 3: Create a table of values of $x$ and $y$ (Two pairs of values are enough to plot a line)
Step 4: Identify the point where both lines meet
Step 5: The point of intersection is the solution of the given system
### Examples
Ex 1: Solve the following system of equations graphically: $x + y = 4$ and $x – y = 2$
From first equation $x + y = 4 => y = 4 – x$
From second equation $x – y = 2 => y = x – 2$
Plot the equation $x + y = 4$
Plot the second equation $x – y = 2$
The point of intersection is $\left(3, 1\right)$
Therefore, the solution of the given pair of equations is $x = 3$ and $y = 1$.
### Substitution Method of Solving a Pair of Linear Equations
The steps to solve linear equations in two variables using the substitution method are given below:
Step 1: Solve one of the equations for one variable.
Step 2: Substitute this in the other equation to get an equation in terms of a single variable.
Step 3: Solve it for the variable.
Step 4: Substitute it in any of the equations to get the value of another variable.
### Examples
Ex 1: Solve the given pair of linear equations using the substitution method: $x + 2y – 7 = 0$ and $3x – 5y + 12 = 0$
Let $x + 2y – 7 = 0$ ————————- (1)
and $3x – 5y + 12 = 0$ ————————- (2)
From (1)
$x = -2y + 7$ ———————————- (3)
Substitute $x = -2y + 7$ from (3) in (2)
$3\left(-2y + 7 \right) – 5y + 12 = 0 => -6y + 21 – 5y + 12 = 0 => -6y – 5y + 21 + 12 = 0$
$=> -11y + 33 = 0 => -11y = -33 => y = \frac{-33}{-11} => y = 3$
Substitute $y = 3$ in (3)
$x = -2 \times 3 + 7 = -6 + 7 = 1$
Therefore, solution is $x = 1$ and $y = 3$.
### Cross Multiplication Method of Solving a Pair of Linear Equations
Consider a system of linear equations: $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$.
To solve this using the cross multiplication method, we first write the coefficients of each of $x$ and $y$ and constants as follows:
The arrows indicate that those coefficients have to be multiplied. Now we write the following equation by cross-multiplying and subtracting the products.
$\frac{x}{b_{1}c{2} – b_{2}c_{1}} = \frac{y}{c_{1}a{2} – c_{2}a_{1}} = \frac{1}{a_{1}b{2} – a_{2}b_{1}}$
From this, we get two equations:
$\frac{x}{b_{1}c{2} – b_{2}c_{1}} = \frac{1}{a_{1}b{2} – a_{2}b_{1}}$
And $\frac{y}{c_{1}a{2} – c_{2}a_{1}} = \frac{1}{a_{1}b{2} – a_{2}b_{1}}$
Solving these two we get
$x = \frac{b_{1}c{2} – b_{2}c_{1}}{a_{1}b{2} – a_{2}b_{1}}$
$y = \frac{c_{1}a{2} – c_{2}a_{1}}{a_{1}b{2} – a_{2}b_{1}}$
### Examples
Ex 1: Solve the given pair of linear equations using the substitution method: $x + 2y – 7 = 0$ and $3x – 5y + 12 = 0$
Let $x + 2y – 7 = 0$ ————————- (1)
and $3x – 5y + 12 = 0$ ————————- (2)
The coefficients from the above two equations are
$a_{1} = 1$, $b_{1} = 2$, $c_{1} = -7$
And $a_{2} = 3$, $b_{2} = -5$, $c_{2} = 12$
Now, using the expressions for finding $x$ and $y$, we get
$x = \frac{b_{1}c{2} – b_{2}c_{1}}{a_{1}b{2} – a_{2}b_{1}} = \frac{2 \times 12 – \left(-5 \right) \times \left(-7 \right)}{1 \times \left(-5 \right) – 3 \times 2} = \frac{24 – 35}{-5 – 6} = \frac{-11}{-11} = 1$
$y = \frac{c_{1}a{2} – c_{2}a_{1}}{a_{1}b{2} – a_{2}b_{1}} = \frac{-7 \times 3 – 12 \times 1}{1 \times \left(-5 \right) – 3 \times 2} = \frac{-21 – 12}{-5 – 6} = \frac{-33}{-11} = 3$
Therefore, solution is $x = 1$ and $y = 3$.
### Elimination Method of Solving a Pair of Linear Equations
The steps to solve linear equations in two variables using the elimination method are given below:
Step 1: Arrange the equations in the standard form: $ax + by + c = 0$ or $ax + by = c$
Step 2: Check if adding or subtracting the equations would result in the cancellation of a variable.
Step 3: If not, multiply one or both equations by either the coefficient of $x$ or $y$ such that their addition or subtraction would result in the cancellation of any one of the variables.
Step 4: Solve the resulting single variable equation.
Step 5: Substitute it in any of the equations to get the value of another variable.
### Examples
Ex 1: Solve the following system of equations using the elimination method: $2x + 3y – 11 = 0$ and $3x + 2y – 9 = 0$
Adding or subtracting these two equations would not result in the cancellation of any variable. Let us aim at the cancellation of $x$.
The coefficients of $x$ in both equations are $2$ and $3$. Their LCM is $6$. We will make the coefficients of $x$ in both equations $6$ and $-6$ such that the $x$ terms get canceled when we add the equations.
$3 \times \left(2x + 3y – 11 = 0 \right)$
$=>6x + 9y – 33 = 0$
$-2 \times \left(3x + 2y – 9 = 0 \right)$
$=> -6x – 4y + 18 = 0$
Now we will add these two equations:
$6x + 9y – 33 = 0$
$-6x -4y + 18 = 0$
On adding both the above equations we get,
$0 + 5y – 15 = 0$
$=> 5y – 15 = 0$
$=> 5y = 15$
$=> y = \frac{15}{5}$
$=> y = 3$
Substitute this in one of the given two equations and solve the resultant variable for $x$.
$2x + 3y – 11 = 0$
$=> 2x +3 \times 3 – 11 = 0$
$=> 2x + 9 – 11 = 0$
$=> 2x = 2$
$=> x = \frac{2}{2}$
$=> x = 1$
Therefore, the solution of the given system of equations is $x = 1$ and $y = 3$.
## Types of Solutions for a Pair of Linear Equations
The graph of the two linear equations may not intersect always. Sometimes they may be parallel. In that case, the system of linear equations in two variables has no solution. In some other cases, both lines coincide with each other. In that case, each point on that line is a solution of the given system and hence the given system has an infinite number of solutions.
• Consistent and Inconsistent System of Linear Equations:
• If the system has a solution, then it is said to be consistent;
• If the system has no solution,, it is said to be inconsistent.
• Independent and Dependent System of Linear Equations:
• If the system has a unique solution, then it is independent.
• If it has an infinite number of solutions, then it is dependent. It means that one variable depends on the other.
## Practice Problems
1. Solve the following pairs of linear equations using the graphical method
• $2y = 4x – 6$ and $2x = y + 3$
• $x + 3y = 6$ and $2x – 3y = 12$
2. Solve the following pairs of linear equations using the elimination method
• $141x + 93y = 189$ and $93x + 141y = 45$
• $3x = y + 5$ and $5x – y = 11$
• $27x + 31y = 85$ and $31x + 27y = 89$
3. Solve the following pairs of linear equations using the substitution method
• $3x – y – 7 = 0$ and $2x + 5y + 1 = 0$
• Find the two numbers whose sum is $75$ and the difference is $15$
4. Solve the following pairs of linear equations using the cross-multiplication method
• $x + 2y – 2 = 0$ and $x – 3y – 7 = 0$
• The sum of the digits of a two-digit number is $8$ and the difference between the number and that formed by reversing the digits is $18$. Find the number.
## FAQs
### What is meant by linear equations in two variables?
A linear equation is an equation of degree one. A linear equation in two variables is a type of linear equation in which there are two variables present.
For example, $2x – y = 9$ and $x + 3y =13$
### How do you identify linear equations in two variables?
We can identify a linear equation in two variables if it is expressed in the form $ax + by + c = 0$, consisting of two variables $x$ and $y$, and the highest degree of the given equation is $1$.
### What are the different methods of solving a pair of linear equations in two variables?
There are four methods to solve a system of linear equations in two variables. Those methods are used to solve a pair of linear equations in two variables are:
a) Graphical Method
b) Substitution Method
c) Cross Multiplication Method
d) Elimination Method
## Conclusion
A pair of linear equations in two variables are the equations in which each of the two variables is of the highest exponent order of $1$ and has one, none, or infinitely many solutions. There are four widely used methods of solving these equations – graphical method, substitution method, cross multiplication method, and elimination method.
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# Basic usage#
This section demonstrates how to use PyVista to reading and plotting 3D data using the pyvista.examples module and external files.
Tip
This section of the tutorial was adopted from the Basic API Usage chapter of the PyVista documentation.
## Using Existing Data#
There are two main ways of getting data into PyVista: creating it yourself from scratch or loading the dataset from any one of the compatible file formats. Since we’re just starting out, let’s load a file.
If you have a dataset handy, like a surface model, point cloud, or VTK file, you can use that. If you don’t have something immediately available, PyVista has a variety of files you can download in its pyvista.examples.downloads module.
from pyvista import examples
dataset
PolyDataInformation
N Cells5131
N Points2669
N Strips0
X Bounds-2.001e+01, 2.000e+01
Y Bounds-6.480e-01, 4.024e+01
Z Bounds-6.093e-01, 1.513e+01
N Arrays0
Note how this is a pyvista.PolyData, which is effectively a surface dataset containing points, lines, and/or faces. We can immediately plot this with:
dataset.plot(color='tan')
This is a fairly basic plot. You can change how its plotted by adding parameters as show_edges=True or changing the color by setting color to a different value. All of this is described in PyVista’s API documentation in pyvista.plot(), but for now let’s take a look at another dataset. This one is a volumetric dataset.
dataset = examples.download_frog()
dataset
UniformGridInformation
N Cells31594185
N Points31960000
X Bounds0.000e+00, 4.990e+02
Y Bounds0.000e+00, 4.690e+02
Z Bounds0.000e+00, 2.025e+02
Dimensions500, 470, 136
Spacing1.000e+00, 1.000e+00, 1.500e+00
N Arrays1
NameFieldTypeN CompMinMax
MetaImagePointsuint810.000e+002.540e+02
This is a pyvista.UniformGrid, which is a dataset containing a uniform set of points with consistent spacing. When we plot this dataset, we have the option of enabling volumetric plotting, which plots individual cells based on the content of the data associated with those cells.
dataset.plot(volume=True)
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##### 2 ptsthe codon = 1: Ji one from 991 1 may U '61 last amino H found in question CATACTAAAGTGCTATAATCTGA 10 Question H Arg Val Stop Thepts11 Question
2 pts the codon = 1: Ji one from 991 1 may U '61 last amino H found in question CATACTAAAGTGCTATAATCTGA 10 Question H Arg Val Stop The pts 11 Question...
##### Tha markating depa rimant fcr Canicang comoanywanie darermine which oiits Lorstg Tartand rredilcard holder chould targetad ior campaign convince Lexietra nalde upgrade They have access data from samnple caidmalderg %ho were contactad Oung last year 5 campaign thal ndicales hathar cardnc Jar Udgradad Dtamium caro =yas) The (Asult logist c ragression modg 026 + -16851X1 999X7, Knana the tctaE 7m Junt of cadit rand Dunchases (hjusana dollans Fior year and*> whetner the cardholder ordered addit
Tha markating depa rimant fcr Canicang comoanywanie darermine which oiits Lorstg Tartand rredilcard holder chould targetad ior campaign convince Lexietra nalde upgrade They have access data from samnple caidmalderg %ho were contactad Oung last year 5 campaign thal ndicales hathar cardnc Jar Udgrad...
##### And J41 N unit j T that "OOOIS 4889 _ 241 1 Find tkea pur H 1 when the production the unit 1 changing by S8/unit
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##### A ball with a weight of 2.0 N is thrown at an angle of 60° above...
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##### (18 points) Indicate the shape, bond angle and hybridization for each of the following molecules_ NOz , HOCI, PFs
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##### Find equivalent resistance between points $A$ and $B$ in the Fig. $5.210$ when they are in the steady state.a. $frac{3}{4} r_{0}$b. $frac{4}{3} r_{0}$c. $frac{5}{3} r_{0}$d. None of these
Find equivalent resistance between points $A$ and $B$ in the Fig. $5.210$ when they are in the steady state. a. $frac{3}{4} r_{0}$ b. $frac{4}{3} r_{0}$ c. $frac{5}{3} r_{0}$ d. None of these...
##### Question 4 7 marks The management of Stag Inc. was reviewing its equipment for impairment. The...
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##### What assumptions are needed to consider based on thestatistical analysis?Why are they important to test prior to running the statisticalanalysis selected?
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##### For the reaction sequence below, identify the final product. 1. MCPBA 2. H*, НО ОН +...
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##### Solve the boundary value problem U U with conditions u(O,t) X uci,t) = 0,4 (,0) = 0 and u(x,0) = Yx (Ix) Find solution upto t = 051 [5] by using h=k= 0.1.
Solve the boundary value problem U U with conditions u(O,t) X uci,t) = 0,4 (,0) = 0 and u(x,0) = Yx (Ix) Find solution upto t = 051 [5] by using h=k= 0.1....
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##### An obstetrician is concerned that her patient is suffering fromgestational diabetes (high blood glucose levels during pregnancy).The blood glucose level has a Normal distribution with mean 135mg/dl and a standard deviation of 10 mg/dl from variability inthe diet. The obstetrician has the patient's glucose measured on fourdifferent days to be sure.What is the standard deviation of the average ofthe four measurements, to one decimal place? a.0.8 mg/dl b.1.6 mg/dl c.2.5 mg/dl d.5 mg/dl e.10 mg
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-- 0.024169--
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The singular limit dynamics of semilinear damped wave equations.(English)Zbl 0699.35177
The authors consider the following abstract problem: $\epsilon \ddot u+\dot u+Au=Fu,\quad u(0)=u_ 0,\quad \epsilon \dot u(0)=\epsilon v_ 0$ in a Hilbert space E, where A is a linear operator while F is nonlinear. This problem includes, e.g., the case $$Au=-u_{xx}$$, with some usual boundary conditions, and $$Fu=f(,u( ))$$ (i.e. a semilinear damped wave equation). It is shown that there exist an integer n and an $${\bar \epsilon}>0$$ such that, for every $$\epsilon\in [0,{\bar \epsilon})$$, the global attractor of the corresponding dynamical system is contained in an invariant manifold of class $$C^ 1$$ and dimension n and that for $$\epsilon$$ $$\to 0$$ both this manifold and the vector field on it converge in the $$C^ 1$$ topology towards the ones corresponding to $$\epsilon =0$$.
Reviewer: G.Moroşanu
MSC:
35L70 Second-order nonlinear hyperbolic equations 35K57 Reaction-diffusion equations
Full Text:
References:
[1] Angenent, S, The Morse-Smale property for a semilinear parabolic equation, J. differential equations, 62, 427-442, (1986) · Zbl 0581.58026 [2] Babin, A.V; Vishik, M.I, Regular attractors of semigroups and evolution equations, J. math. pures appl., 62, 441-491, (1983), (9) · Zbl 0565.47045 [3] Babin, V; Vishik, M.I, Uniform asymptotics of the solutions of singularly perturbed evolution equations, Uspekhi mat. nauk, 42, No. 5, 231-232, (1987), [in Russian] [4] Chow, S.N; Lu, K, Invariant manifolds for flows in Banach spaces, J. differential equations, 74, 285-317, (1988) · Zbl 0691.58034 [5] Constantin, P; Foiaş, C; Nikolaenko, B; Sell, G; Temam, R; Constantin, P; Foiaş, C; Nikolaenko, B; Sell, G; Temam, R, Intégral manifolds and inertial manifolds for dissipative partial differential equations, C. R. acad. sci. Paris Sér. I math., 302, 375-378, (1987), preprint [6] Foiaş, C; Sell, G.R; Temam, R; Foiaş, C; Sell, G.R; Temam, R, Invariant manifolds for nonlinear evolutionary equations, C. R. acad. sci. Paris Sér. I math., IMA preprint series, 234, 139-141, (1986) · Zbl 0591.35062 [7] Hale, J.K, Asymptotic behavior and dynamics in infinite dimensions, Res. notes in math., 132, 1-42, (1985) · Zbl 0653.35006 [8] Hale, J.K; Raugel, G, Upper semicontinuity of the attractor for a singularly perturbed hyperbolic equation, J. differential equations, 73, 197-214, (1988) · Zbl 0666.35012 [9] Henry, D, Geometric theory of semilinear parabolic equations, () · Zbl 0456.35001 [10] Henry, D.B, Some infinite-dimensional Morse-Smale systems defined by parabolic partial differential equations, J. differential equations, 59, 165-205, (1985) · Zbl 0572.58012 [11] Irwin, M.C, Smooth dynamical systems, (1980), Academic Press San Diego, CA · Zbl 0465.58001 [12] Mallet-Paret, J; Sell, G.R; Mallet-Paret, J; Sell, G.R, Inertial manifolds for reaction-diffusion equations in higher space dimensions, Lecture notes in math., IMA preprint series, 331, 94-107, (1987) [13] Mañé, R, Reduction of semilinear parabolic equations to finite dimensional C1 flows, Lecture notes in math., 597, 361-378, (1977) [14] Mora, X, Finite-dimensional attracting manifolds in reaction-diffusion equations, Contemp. math., 17, 353-360, (1983) · Zbl 0525.35046 [15] Mora, X, Finite-dimensional attracting invariant manifolds for damped semilinear wave equations, Res. notes in math., 155, 172-183, (1987) · Zbl 0642.35061 [16] Mora, X; Solà-Morales, J, Existence and non-existence of finite-dimensional globally attracting invariant manifolds in semilinear damped wave equations, (), 187-210 · Zbl 0642.35062 [17] Palis, J; Smale, S, Structural stability theorems, (), 223-231 · Zbl 0214.50702 [18] Vanderbauwhede, A; Van Gils, S.A, Center manifolds and contractions on a scale of Banach spaces, J. funct. anal., 72, 209-224, (1987) · Zbl 0621.47050
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# Atomes froids dans des réseaux optiques - Quelques facettes surprenantes d'un système modèle
Abstract : This thesis is devoted to the experimental study of atoms trapped and cooled in several types of optical structures. In order to characterise these media, we used different techniques such as time-of-flight techniques, direct imaging of the atomic
cloud and pump-probe spectroscopy. We thus obtained information about kinetic temperature, spatial diffusion of atoms and atomic motion in the optical potential wells.
We first studied the dynamics of cesium atoms in three-dimensional bright optical lattices when a magnetic field is applied. In particular, we showed that optical lattices operating in the jumping regime do provide good trapping and cooling efficiencies and that a motionnal narrowing effect gives birth to narrow
vibrational sidebands on pump-probe transmission spectra. Still with cesium atoms, we created and characterised a three-dimensional bright optical lattice obtained with only two laser beams through the Talbot effect, and also a random medium
generated by a speckle field.
We endly studied a brownian motor'' for $^(87)$Rb atoms in a grey asymmetric potential. The results of the experimental study are in good qualitative agreement with semi-classical Monte-Carlo numerical simulations.
Keywords :
Document type :
Theses
https://tel.archives-ouvertes.fr/tel-00006734
Contributor : Cécile Robilliard <>
Submitted on : Tuesday, August 24, 2004 - 12:26:07 PM
Last modification on : Thursday, December 10, 2020 - 12:37:00 PM
Long-term archiving on: : Friday, April 2, 2010 - 8:27:16 PM
### Identifiers
• HAL Id : tel-00006734, version 1
### Citation
Cécile Mennerat-Robilliard. Atomes froids dans des réseaux optiques - Quelques facettes surprenantes d'un système modèle. Physique Atomique [physics.atom-ph]. Université Pierre et Marie Curie - Paris VI, 1999. Français. ⟨tel-00006734⟩
Record views
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Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 10 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.
#### Question 1:
A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower.
#### Answer:
Let $AB$ be the tower standing vertically on the ground and O be the position of the observer.
We now have:
and ∠$AOB$
Let:
m
Now, in the right ∆$OAB$, we have:
= $\sqrt{3}$
⇒
⇒ =
Hence, the height of the pole is 34.64 m.
#### Question 2:
A kite is flying at a height of 75 m from the level ground, attached to a string inclined at a 60° to the horizontal. Find the length of the string assuming that there is no slack in it.
#### Answer:
Let be the horizontal ground and $A$ be the position of the kite.
Also, let O be the position of the observer and $OA$ be the thread.
Now, draw ⊥ $OX$.
We have:
${60}^{o}$m and ∠
Height of the kite from the ground = $AB$ = 75 m
Length of the string, m
In the right ∆$OBA$, we have:
⇒ m
Hence, the length of the string is $86.6$ m.
#### Question 3:
An observe, 1.6 m tall, is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°. Determine the height of the tower.
#### Answer:
Let $XY$ be the horizontal ground and $X$ be the position of the observer.
Let $Y$ be the position of the tower such that m.
Let $AY$ = $h$ m be the height of the tower.
Now,
Height of the observer, $CX$$1.6$ m
From the right ∆$ABC$, we have:
⇒
⇒ m
∴ m [∵ CX = BY]
m
Hence, the height of the tower is $27.58$ m.
#### Question 4:
The height of a tree is 10 m. It is bent by the wind in such a way that its top touches the ground and makes an angle of 60° with the ground. At what height from the bottom did the tree get bent?
#### Answer:
Let be the tree broken at point $C$ such that $CB$ takes the position $CD$ and ∠.
If m, then m.
In ∆$ADC$, we have:
⇒
⇒
⇒ m
Now,
m
Hence, the height from the bottom from where the tree got bent is 4.64 m.
#### Question 5:
A 1.5-m-tall boy stands at a distance of 3 m from a lamp post and casts a shadow of 4.5 m on the ground. Find the height of the lamp post.
#### Answer:
Let $AB$ be the lamppost, $CD$ be the boy and $CE$ be the shadow of $CD$.
Also, let ∠ and the height of the lamppost be h.
From the right ∆$ECD$, we have:
Now, from the right ∆$EAB$, we have:
[∵ ]
⇒
⇒ m
∴ Height of the lamppost = m
#### Question 6:
The angle of elevation of the top of a building from a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base from the point A.
#### Answer:
Let $CD$ be the building. Then, the angle of elevation from the top of the building on the ground is such that ∠.
After moving to point $B$, the angle of elevation changes such that ∠ and m.
Let:
m and m
In the right ∆$DCB$, we have:
⇒
Now, in the right ∆$DCA$, we have:
On putting in the above equation, we get:
m
Hence, the height of the building is 40.98 m.
Now,
m
∴ Distance of the base of the building from point m
#### Question 7:
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 m towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.
#### Answer:
Let $CD$ be the tower making an angle of elevation at point $A$ on the ground such that ∠.
On moving towards the foot of the tower at point $B$, m and ∠.
Let:
m and m
In the right ∆$DCA$, we have:
...(i)
Now, in the right ∆$DCB$, we have:
...(ii)
By using (ii) in (i), we get:
⇒
⇒ m
∴ Height of the tower = 17.32 m
Now,
m
∴ Distance of the tower from point $A$ = m
#### Question 8:
From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find height of the tower, and the distance between the tower and the building.
#### Answer:
Let $AB$ be the tower and $CD$ be the building.
We have:
m, ∠ and ∠
Let m and m such that m.
In the right ∆$ABC$, we have:
Now, in the right ∆$AED$, we have:
On putting in the above equation, we get:
m
Hence, the height of the tower is 22.5 m.
Now,
Distance between the tower and the building = m
#### Question 9:
From a window 15 m high above the ground, in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 30° and 45° respectively. Show that the height of the opposite house is 23.66 m.
#### Answer:
Let $AB$ be the house and $CD$ be the window.
We have:
m such that m, ∠ and ∠.
Let m such that m and let m such that m.
In the right ∆$DBC$, we have:
m
Now, in the right ∆$ADE$, we have:
On putting in the above equation, we get:
m
Hence proved.
#### Question 10:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff. At a point on the plane, 30 metres away from the tower, an observe notices that the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. Find the height of the flagstaff and that of the tower.
#### Answer:
Let $OX$ be the horizontal line, $AC$ be the vertical tower and $BC$ be the vertical flagstaff.
We now have:
m, ∠ and ∠
Let:
m and m
In the right ∆$AOC$, we have:
m
Now, in the right ∆$AOB$, we have:
On putting in the above equation, we get:
m
We now have:
Height of the flagstaff = x = 21.96 m
Height of the tower = m
#### Question 11:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5 m. From a point on the plane the angles of elevation of the bottom and the top of the flagstaff are 30° and 60°. Find the height of the tower.
#### Answer:
Let $OX$ be the horizontal line, $AC$ be the vertical tower and $BC$ be the vertical flagstaff such that m.
Let:
m and m
In the right ∆$AOC$, we have:
⇒
Now, in the right ∆$AOB$, we have:
⇒
On putting in the above equation, we get:
⇒
⇒
⇒
⇒ m
Hence, the height of the tower is 2.5 m.
#### Question 12:
A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
#### Answer:
Let $AC$ be the pedestal and $BC$ be the statue such that m.
We have:
and ∠
Let:
m and m
In the right ∆$ADC$, we have:
⇒
Or,
Now, in the right ∆$ADB$, we have:
⇒
On putting in the above equation, we get:
⇒
m
Hence, the height of the pedestal is 2 m.
#### Question 13:
The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30° On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres.
#### Answer:
Let $AB$ be the tower.
We have:
m, ∠ and ∠
Let:
m and m
In the right ∆$ABD$, we have:
⇒
⇒
Now, in the right ∆$ACB$, we have:
⇒
⇒
On putting in the above equation, we get:
⇒
⇒
⇒ m
Hence, the height of the tower is 129.9 m.
#### Question 14:
On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top of bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it.
#### Answer:
Let $OX$ be the horizontal plane, be the tower and $CD$ be the vertical flagpole.
We have:
m, ∠ and ∠
Let:
m and m
In the right ∆$ABD$, we have:
⇒ $\frac{h}{9}=\frac{1}{\sqrt{3}}$
m
Now, in the right ∆$ABC$, we have:
⇒
⇒
By putting in the above equation, we get:
⇒
⇒
Thus, we have:
Height of the flagpole = 10.39 m
Height of the tower = 5.19 m
#### Question 15:
From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30° and 60° respectively. Find the height of the pillar and its distance from the hill.
#### Answer:
Let $AB$ be the hill and $DE$ be the pillar. Draw $CD$ ⊥ $AB$.
Thus, we have:
m, ∠ and ∠
Now, let m such that m and let m such that m.
In the right ∆$AEB$, we have:
⇒
⇒ m
Now, in the right ∆$BDC$, we have:
⇒
By putting in the above equation, we get:
⇒
⇒
⇒ m
We now have:
Height of the pillar = 133.33 m
Distance of the pillar from the hill = 115.47 m
#### Question 16:
From a window, 60 metres high above the ground, of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 60° and 45° respectively. Show that the height of the opposite house is 60$\left(1+\sqrt{3}\right)$ metres.
#### Answer:
Let $AB$ be the house and $DE$ be the window such that m. Draw $CE$ ⊥ $AB$.
Thus, we have:
and ∠
Let m and m such that m and m.
In the right ∆$ADE$, we have:
⇒
⇒ m
Now, in the right ∆$BEC$, we have:
⇒
On putting in the above equation, we get:
⇒
m
Hence, the height of the opposite house is m.
#### Question 17:
The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of , find the speed of the jet plane.
#### Answer:
Let C and C' be the two positions of the jet plane and BC and B'C' be its constant heights.
Thus, we have:
Now, in the right ∆ABC, we have:
Now, in the right ∆AB'C', we have:
BB' is 3000 m. This means that to cover a distance of 3000 m, the jet plane takes 15 seconds.
∴ Speed of the jet plane
#### Question 18:
The angles of elevation and depression of the top and bottom of a lighthouse form the top of a building, 60 m high, are 30° and 60° respectively. Find (i) the difference between the heights of the lighthouse and the building, and (ii) the distance between the lighthouse and the building.
#### Answer:
Let $AB$ be the lighthouse and $DE$ be the building. Draw $CE$ ∥ $BD$.
We have:
m, ∠ and ∠
Let m such that m and m.
Let BC = x.
In the right ∆$EDB$, we have:
⇒
m
Now, in the right ∆$AEC$, we have:
⇒
On putting in the above equation, we get:
∴ Difference between the heights of the lighthouse and building = m
Distance between the lighthouse and the building = m
#### Question 19:
As observed form the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation.
#### Answer:
Let $OA$ be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
m, ∠ and ∠
Let:
m and m
In the right ∆$OAC$, we have:
⇒ m
Now, in the right ∆$OBA$, we have:
⇒
On putting in the above equation, we get:
m
∴ Distance travelled by the ship during the period of observation = m
#### Question 20:
The angle of elevation of the top of a building form a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base form the point A.
#### Answer:
Let $CD$ be the building and $A$ be the point on the ground making an angle of elevation on top of the building.
Thus, we have:
, ∠ and m
Let:
m and m
In the right ∆$DBC$, we have:
⇒
⇒ m
Or,
m
In the right ∆$DAC$, we have:
⇒
⇒
By putting in the above equation, we get:
⇒ m
m
Distance of the base of the building from point $A$ = m
Height of the building = m
#### Question 21:
The angle of elevation of the top of a tree from a point A on the ground is 60°. On walking 20 metres away from its base, to a point B, the angle of elevation changes to 30°.Find the height of the tree.
#### Answer:
Let $CD$ be the tree and $A$ be the point on the ground such that ∠, m and ∠.
Let:
m and m
In the right ∆$DAC$, we have:
⇒ (or)
In the right ∆$DBC$, we have:
⇒
⇒
On putting , we get:
⇒
⇒ m
Hence, the height of the tree is 17.32 m.
#### Question 22:
From the top of a-7-m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of the foot of the tower is 60°. Find the height of the tower.
#### Answer:
Let $AB$ be the tower and $CD$ be the building. Draw $CD$⊥ $AB$.
We have:
m, ∠ and ∠
Let m and m such that m and m.
In the right ∆$CAE$, we have:
⇒
⇒
Now, in the right ∆$BCD$, we have:
⇒
⇒
By putting in the above equation, we get:
Hence, the height of the tower is 28 m.
#### Question 23:
The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is 6 metres.
#### Answer:
Let $AB$ be the tower and be two points such that m and m.
Let:
m, ∠ and ∠
In the right ∆BCA, we have:
In the right ∆BDA, we have:
Multiplying equations (1) and (2), we get:
⇒ 36 = h2
h = ±6
Height of a tower cannot be negative.
∴ Height of the tower = 6 m
#### Question 24:
From a point P on the ground, the angle of elevation of a 10-m-tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P.
#### Answer:
Let AC be the building, P be the point on the ground and $BC$ be the flagstaff.
Thus, we have:
m, ∠ and ∠
Let:
m and m
In the right ∆$CAP$, we have:
⇒
⇒ = 17.32 m
In the right ∆$BAP$, we have:
⇒
⇒
On putting in the above equation, we get:
⇒ m
Thus, we have:
Length of the flagstaff = m
Distance of the building from point P m
#### Question 25:
The angles of depression of the top and bottom of a 10 m high building from the top of a multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building and the distance between the two buildings.
#### Answer:
Let $AB$ be the multistorey building and $DE$ be the high building such that m. Draw $CD$ ⊥ $AB$.
Thus, we have:
∠ and ∠
Let m such that m and m such that m.
In the right ∆$BEA$, we have:
⇒
⇒
Or,
Now, in the right ∆$BDC$, we have:
⇒
By putting in the above equation, we get:
⇒
⇒ m
We have:
Height of the multistorey building = m
Distance between the two buildings = m
#### Question 26:
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find width of the river.
#### Answer:
Let $A$ and $B$ be two points on the banks on the opposite side of the river and $P$ be the point on the bridge at a height of 2.5 m.
Thus, we have:
m, ∠PAD and ∠
In the right ∆$APD$, we have:
⇒
⇒ m
In the right ∆$PDB$, we have:
⇒
⇒ m
∴ Width of the river = m
#### Question 27:
Two men are on opposite sides of a tower. they measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men.
#### Answer:
Let $CD$ be the tower and be the positions of the two men standing on the opposite sides. Thus, we have:
, ∠ and m
Let m and m such that m.
In the right ∆$DBC$, we have:
⇒
⇒ m
In the right ∆$ACD$, we have:
⇒
⇒
On putting in the above equation, we get:
⇒ m
∴ Distance between the two men = m
#### Question 28:
The horizontal distance between two towers is 60 metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90 metres, find the height of the first tower.
#### Answer:
Let $DE$ be the first tower and $AB$ be the second tower.
Now, m and m such that m and ∠.
Let m such that m and m.
In the right ∆$BCE$, we have:
⇒
= m
∴ Height of the first tower = m
#### Question 29:
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of each pole and the distances of the point form the poles.
#### Answer:
Let $AB$ and $CD$ be the two poles of equal heights such that m.
∴ Distance between the two poles m
We have:
and ∠
Let m such that m.
In the right ∆$BOA$, we have:
⇒
⇒ ...(i)
Now, in the right ∆$COD$, we have:
⇒
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
m
∴ m
We now have:
Height of each pole = m
Distance of the pole $AB$ from point O m
Distance of the pole $CD$ from point O = m
#### Question 30:
As observed form the top of a 75-m-tall lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
#### Answer:
Let $AB$ be the lighthouse such that m and and $D$ be the positions of the two ships.
Thus, we have:
and ∠
Let m and m such that m.
In the right ∆$ABC$, we have:
⇒
⇒ m
In the right ∆$ABD$, we have:
⇒
= m
∴ Distance between the two ships = m
#### Question 31:
A man on the deck of a ship, 16 m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60° and 30°. Calculate the distance of the cliff from the ship and height of the cliff.
#### Answer:
Let $AB$ be the deck of the ship above the water level and $DE$ be the cliff.
Now,
m such that m and ∠ and ∠.
If AD = x m and m, then m.
In the right ∆$BAD$, we have:
⇒
⇒ m
In the right ∆$EBC$, we have:
⇒
⇒
⇒ [∵ ]
m
∴ Distance of the cliff from the deck of the ship = m
And,
Height of the cliff = m
#### Question 32:
A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the time taken by the boat to reach the shore.
#### Answer:
Let $C$ be the cliff and $A$ and $D$ be the two positions of the boat.
We have:
and ∠
Let the speed of the boat be $v$ metres per minute.
Also,
Distance covered by the boat in six minutes
m
Suppose the boat takes $t$ minutes to reach $B$ from $D$
m
Let:
m
In the right ∆$DBC$, we have:
⇒
⇒ ...(i)
From the right ∆$ACB$, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
minutes
Hence, the required time is three minutes.
#### Question 33:
A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of a 20 m high building, finds the angle of elevation of the same bird to be 45°. Boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl.
#### Answer:
Let $O$ be the position of the bird, $B$ be the position of the boy and $FG$ be the roof on which $G$ is the position of the girl.
Let:
$OL$ ⊥ $BF$ and $GM$ ⊥ $OL$
Thus, we have:
m, ∠ m and ∠.
Let:
m
From the right ∆OLB, we have:
⇒
⇒ m
∴ m
From the right ∆$OMG$, we have:
⇒
m
∴ Distance of the bird from the girl = m
#### Question 34:
A 1.5-m-tall boy is standing at some distance from a 30-m-tall building. The angle of elevation from his eyes to the top of the building increase form 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
#### Answer:
Let $AB$ be the building and $CD$ and $EF$ be the two positions of the boy. Draw $DFG$ ∥ $CEA$.
Thus, we have:
m, m, ∠ and ∠ and m
Now, in ∆$BDG$, we have:
⇒ m
In ∆$BFG$, we have:
m
∴ $19\sqrt{3}$ m = 32.87 m
Hence, the distance walked by the boy towards the building is $32.87$ m.
#### Question 35:
The angle of elevation of the top of an unfinished tower at a distance of 75 m form its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°?
#### Answer:
Let $AB$ be the unfinished tower, $AC$ be the raised tower and O be the point of observation.
We have:
m, ∠ and ∠
Let m such that m.
In ∆AOB, we have:
⇒ m = m
In ∆$AOC$, we have:
$=\sqrt{3}$
⇒
m
∴ Required height = m = 86.6 m
#### Question 1:
At a certain instant the ratio of the lengths of a pillar and its shadow are in the ratio $1:\sqrt{3}$. At that instant, the angle of elevation of the sum is
(a) 30°
(b) 45°
(c) 60°
(d) none of these
#### Answer:
(a) 30°
Let $AB$ be the pillar and $AC$ be its shadow.
Let:
and
In the right ∆$ABC$, we have:
⇒ [∵ ]
Hence, the angle of elevation of the sun is ${30}^{o}$.
#### Question 2:
At a certain instant, the altitude of the sum is 60°. At that instant, the length of the shadow of a vertical tower is 100 m. The height of the tower is
(a) $50\sqrt{3}\mathrm{m}$
(b) $100\sqrt{3}\mathrm{m}$
(c) $\frac{100}{\sqrt{3}}\mathrm{m}$
(d) $\frac{200}{\sqrt{3}}\mathrm{m}$
#### Answer:
(b) $100\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $AC$ be its shadow.
We have:
and m
Let:
m
In the right ∆$BAC$, we have:
m
Hence, the height of the tower is $100\sqrt{3}$ m.
#### Question 3:
The height of a tower is $100\sqrt{3}\mathrm{m}$. The angle of elevation of its top from a point 100 m away from its foot is
(a) 30°
(b) 45°
(c) 60°
(d) none of these
#### Answer:
(c) 60°
Let $AB$ be the tower and $O$ be the point of observation.
We have:
m and m
In ∆$AOB$, we have:
Hence, the angle of elevation is ${60}^{o}$.
#### Question 4:
The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30°. The height of the tower is
(a) 30 m
(b) $10\sqrt{3}\mathrm{m}$
(c) 20 m
(d) $10\sqrt{2}\mathrm{m}$
#### Answer:
(b) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $O$ be the point of observation.
Also,
and m
Let:
m
In ∆$AOB$, we have:
m
Hence, the height of the tower is $10\sqrt{3}$ m.
#### Question 5:
The string of a kite is 100 m long and it makes an angle of 60° with the horizontal. If these is no slack in the string, the height of the kite from the ground is
(a) $50\sqrt{3}\mathrm{m}$
(b) $100\sqrt{3}\mathrm{m}$
(c) $50\sqrt{2}\mathrm{m}$
(d) 100 m
#### Answer:
(a) $50\sqrt{3}\mathrm{m}$
Let $AB$ be the string of the kite and $AX$ be the horizontal line.
If $BC$ ⊥ $AX$, then m and ∠.
Let:
m
In the right ∆$ACB$, we have:
m
Hence, the height of the kite is $50\sqrt{3}$ m.
#### Question 6:
An observer 1.5 m tall 28.5 away from a tower and the angle of elevation of the top of the tower form the eye of the observer is 45°. The height of the tower is
(a) 27 m
(b) 30 m
(c) 28.5 m
(d) none of these
#### Answer:
(b) 30 m
Let $AB$ be the observer and $CD$ be the tower.
Draw $BE$ ⊥ $CD$, Let metres. Then,
m , m and ∠.
= m.
In right ∆$BED$, we have:
⇒ m
Hence the height of the tower is $30$ m.
#### Question 7:
The shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is
(a) $5\left(\sqrt{3}+1\right)\mathrm{m}$
(b) $10\left(\sqrt{3}-1\right)\mathrm{m}$
(c) 9 m
(d) 13 m
#### Answer:
(a) $5\left(\sqrt{3}+1\right)\mathrm{m}$
Let $AB$ be the tower and be its shadows.
Thus, we have:
and ∠
If m, then m.
Let:
m.
In ∆$ACB$, we have:
... (i)
Now, in ∆$ADB$, we have:
...(ii)
On putting the value of $h$ from (i) in (ii), we get:
On multiplying the numerator and denominator by $\left(\sqrt{3}+1\right)$, we get:
m
Hence, the height of the tower is $5\left(\sqrt{3}+1\right)$ m.
#### Question 8:
From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30° and 45°. The height of the hill is
(a) $\frac{1}{2}\left(\sqrt{3}-1\right)\mathrm{km}$
(b) $\frac{1}{2}\left(\sqrt{3}+1\right)\mathrm{km}$
(c) $\left(\sqrt{3}-1\right)\mathrm{km}$
(d) $\left(\sqrt{3}+1\right)\mathrm{km}$
#### Answer:
(b) $\frac{1}{2}\left(\sqrt{3}+1\right)\mathrm{km}$
Let $AB$ be the hill making angles of depression at points $C$ and $D$ such that ∠, ∠ and km.
Let:
km and km
In ∆$ADB$, we have:
⇒ ⇒ ...(i)
In ∆$ACB$, we have:
...(ii)
On putting the value of $h$ taken from (i) in (ii), we get:
On multiplying the numerator and denominator of the above equation by $\left(\sqrt{3}+1\right)$, we get:
km
Hence, the height of the hill is $\frac{1}{2}\left(\sqrt{3}+1\right)$ km.
#### Question 9:
An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. The width of the river is
(a) $\left(200+\frac{200}{\sqrt{3}}\right)\mathrm{m}$
(b) $\left(200-\frac{200}{\sqrt{3}}\right)\mathrm{m}$
(c) $400\sqrt{3}\mathrm{m}$
(d) $\frac{400}{\sqrt{3}}\mathrm{m}$
#### Answer:
(a) $\left(200+\frac{200}{\sqrt{3}}\right)\mathrm{m}$
Let $A$ be the position of the aeroplane and $XAY$ be the horizontal line.
Let $BDC$ be the line on the ground such that $AD$ ⊥ $BC$ and m.
It observes an angle of depression such that ∠ and ∠.
Also,
and ∠ (Alternate angles)
In ∆$ABD$, we have:
m
Now, in ∆$ACD$, we have:
m
∴ Width of the river = m
#### Question 10:
If the angles of elevation of the top of a tower form tow points at distances a and b from the base and in the same straight line with it are complementary, then the height of the tower is
(a) $\sqrt{\frac{a}{b}}$
(b) $\sqrt{ab}$
(c) $\sqrt{a+b}$
(d) $\sqrt{a-b}$
#### Answer:
(b) $\sqrt{ab}$
Let $AB$ be the tower and and $D$ be the points of observation on $AC$.
Let:
, ∠ and m
Thus, we have:
and
Now, in the right ∆ABC, we have:
⇒ ...(i)
In the right ∆ABD, we have:
⇒ ...(ii)
On multiplying (i) and (ii), we have:
[ ∵ ]
⇒
⇒ m
Hence, the height of the tower is $\sqrt{ab}$ m.
#### Question 11:
On the level ground, the angle of elevation of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is
(a) 10 m
(b) $10\sqrt{3}\mathrm{m}$
(c) 15 m
(d) $5\sqrt{3}\mathrm{m}$
#### Answer:
(b) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $C$ and $D$ be the points of observation such that ∠, ∠, m and m.
Now, in ∆$ADB$, we have:
⇒
In ∆$ACB$, we have:
⇒
∴
⇒
⇒ ⇒
∴ Height of the tower AB = m
#### Question 12:
If the elevation of the sun changes form 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, is
(a) 7.5 m
(b) 15 m
(c) $10\sqrt{3}\mathrm{m}$
(d) $5\sqrt{3}\mathrm{m}$
#### Answer:
(c) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the pole and $AC$ and $AD$ be its shadows.
We have:
, ∠ and m
In ∆$ACB$, we have:
⇒ ⇒ m
Now, in ∆$ADB$, we have:
⇒ ⇒ m
∴ Difference between the lengths of the shadows = m
#### Question 13:
On the same side of a tower 300 m high, the angles of depression of two objects are 45° and 60° respectively. The distance between the poles is
(a) 127 m
(b) 117 m
(c) 217 m
(d) 473 m
#### Answer:
(a) 127 m
Let $AB$ be the tower and be the positions of the objects such that m, ∠ and ∠.
Now, In ∆, we have:
= 1
⇒ m
In ∆$ADB$, we have:
⇒
⇒ m
∴ m
Hence, the distance between the two poles is 127 m.
#### Question 14:
The heights of two poles are 80 m and 65 m. If the line joining their tops makes an angle of 45° with the horizontal, then the distance between the poles is
(a) 15 m
(b) 22.5 m
(c) 30 m
(d) 7.5 m
#### Answer:
(a) 15 m
Let and $CD$ be the poles of heights of 80 m and 65 m, respectively.
Draw $DE$⊥ $AB$.
Now,
Also,
m
In ∆$BED$, we have:
⇒
⇒ m
∴ m
Hence, the distance between the poles is 15 m.
#### Question 15:
From the foot of a tower, the angle of elevation of the top of a column is 60° and from the top of the tower, the angle of elevation is 30°. If the height of the tower is 25 m, the height of the column is
(a) 35 m
(b) 42.5 m
(c) 37.5 m
(d) 27.5 m
#### Answer:
(c) 37.5 m
Let $AB$ be the tower and be the column.
Draw $BE$⊥ $CD$.
Given:
m, ∠ and ∠
Thus, we have;
m.
Let m.
In the right ∆$BED$, we have:
⇒ ⇒
In the right ∆$ACD$, we have:
⇒ ⇒
Now,
⇒
⇒
m
∴ Height of the column = m
#### Question 16:
A straight tree breaks due to storm and the broken part bends so that the top of the trees touches the ground making an angle of 30° with the ground. the distance from the foot of the tree to the point, where the top touches the ground is 10 m. The height of the tree is
(a) $10\sqrt{3}\mathrm{m}$
(b) $\frac{10\sqrt{3}}{3}\mathrm{m}$
(c) $10\left(\sqrt{3}+1\right)\mathrm{m}$
(d) $10\left(\sqrt{3}-1\right)\mathrm{m}$
#### Answer:
(a) $10\sqrt{3}\mathrm{m}$
Let be the tree broken at point $C$ such that $CB$ takes the position $CD$ so that ∠ and m.
Let:
m and m. So, the height of the tree is m.
Now, in ∆$ADC$, we have:
⇒ m
Again,
$\frac{10}{y}=\frac{\sqrt{3}}{2}$
m
∴ Height of the tree = m
#### Question 17:
An observer standing 50 m away from a building notices that the angles of elevation of the top and bottom of a flagstaff on the building are 60° and 45° respectively. The height of the flagstaff is
(a) $50\sqrt{3}\mathrm{m}$
(b) $50\left(\sqrt{3}+1\right)\mathrm{m}$
(c) $50\left(\sqrt{3}-1\right)\mathrm{m}$
(d) $\frac{50}{\sqrt{3}}\mathrm{m}$
#### Answer:
(c) $50\left(\sqrt{3}-1\right)\mathrm{m}$
Let $AB$ be the tower, $BC$ be the flagstaff and O be the position of the observer. Thus, we have:
m, ∠ and ∠
Now, in ∆$AOB$, we have:
m
In ∆$AOC$, we have:
m
∴ Height of the flagstaff = m
#### Question 18:
A boat is being rowed away from a cliff 150 m high. At the top of the cliff the angle of depression of the boat changed from 60° to 45° in 1 minute.
If $\sqrt{3}=1.73,$ the speed of the boat is
(a) 4.31 km/hr
(b) 3.81 km/hr
(c) 4.63 km/hr
(d) 3.91 km/hr
#### Answer:
(b) 3.81 km/h
Let be the cliff and be the two positions of the ships. Thus, we have:
m, ∠ and ∠
Now, in ∆$ADB$, we have:
m
In ∆$ACB$, we have:
m
Now,
m
∴ m
Speed of the boat = $\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{63.5}{60}$ m/s km/h km/h
#### Question 19:
In a rectangle, the angle between a diagonal and a side is 30° and the length of this diagonal is 8 cm. the area of the rectangle is
(a) 16 cm2
(b) $\frac{16}{\sqrt{3}}{\mathrm{cm}}^{2}$
(c) $16\sqrt{3}{\mathrm{cm}}^{2}$
(d) $8\sqrt{3}{\mathrm{cm}}^{2}$
#### Answer:
(c) $16\sqrt{3}{\mathrm{cm}}^{2}$
Let $ABCD$ be the rectangle in which ∠ and cm.
In ∆$BAC$, we have:
⇒ m
Again,
m
∴ Area of the rectangle = cm2
#### Question 20:
If the length of shadow a pole on a level ground is twice the length of the pole, the angle of elevation of the sun is
(a) 30°
(b) 45°
(c) 60°
(d) none of these
#### Answer:
(d) None of these
Let $AB$ be the pole and $AC$ be its shadow such that .
Let:
In ∆$ACB$, we have:
It is different from and ${60}^{o}$.
#### Question 21:
A pole 6 m high casts a shadow $2\sqrt{3}\mathrm{m}$ long on the ground. At that instant, the sun's elevation is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
#### Answer:
(c) 60°
Let $AB$ be the pole and $AC$ be its shadow.
Let ∠ m and m.
In ∆$ACB$, we have:
Hence, the angle of elevation of the sun is 60°.
#### Question 22:
Two men standing on opposite sides of a flagstaff measure the angles of the top of the flagstaff as 30° and 60°. If the height of the flagstaff is 18 m, the distance between the men is
(a) 24 m
(b) $24\sqrt{3}\mathrm{m}$
(c) $\frac{24}{\sqrt{3}}\mathrm{m}$
(d) 31.2 m
#### Answer:
(b) $24\sqrt{3}\mathrm{m}$
Let $AB$ be the flagstaff and be the positions of the two men. Thus, we have:
m,∠ and ∠
In ∆$ACB$, we have:
⇒ $\frac{AC}{18}=\sqrt{3}$
m
In ∆$ADB$, we have:
$\frac{AD}{18}=\frac{1}{\sqrt{3}}$
m
∴ m
Hence, the distance between the two men is $24\sqrt{3}$ m.
#### Question 23:
The angled of elevation of an aeroplane flying vertically above the ground as observed from two consecutive stones 1 km apart are 45° and 60°. The height of the aeroplane from the ground is
(a) $\left(\sqrt{3}+1\right)\mathrm{km}$
(b) $\left(3+\sqrt{3}\right)\mathrm{km}$
(c) $\frac{1}{2}\left(\sqrt{3}+1\right)\mathrm{km}$
(d) $\frac{1}{2}\left(3+\sqrt{3}\right)\mathrm{km}$
#### Answer:
(d) $\frac{1}{2}\left(3+\sqrt{3}\right)\mathrm{km}$
Let be the position of the aeroplane and $B$ and $C$ be the two stones (1 km apart) such that km.
Let $AD$ be the perpendicular meeting $BC$ produced at $D$. Thus, we have:
and ∠
Let:
km and km
In ∆$ABD$, we have:
(or) ...(i)
In ∆$ACD$, we have:
$\frac{h}{x}=\sqrt{3}$
(or) ...(ii)
Putting the value of $x$ from (i) in (ii), we get:
⇒
km
Hence, the height of the aeroplane from the ground is $\frac{1}{2}\left(3+\sqrt{3}\right)$ km.
#### Question 24:
Match the following columns:
Column I Column II (a) The length of a shadow of a tower is $\sqrt{3}$ times the height of the tower. The angle of elevation of the sun is........ . (p) 40 m (b) The angle of depression of the top of a tower at a point 40 m from its base 45°. The height of the tower is ........ . (q) 60° (c) The angle of elevation of the top of a tower from a point 15 m away from its base is 30°. The height of the tower is .......... . (r) 30° (d) At a point 14 m away from the base of a $14\sqrt{3}\mathrm{m}$ high pillar, the angle of elevation of its top is ........ . (s) $5\sqrt{3}\mathrm{m}$
#### Answer:
(a) Let the length of the shadow be $y$ m. Then, according to the given condition, m.
Let $\theta$ be the angle of elevation. Then, m and m.
In ∆$ABC$, we have:
∴
(b) Let $AB$ be the tower. Then, m and ∠.
Let:
m
In ∆$ABC$, we have:
⇒ ⇒ m
∴
(c) Let $AB$ be the tower. Then, m and ∠.
Let:
m
In ∆$ABC$, we have:
⇒
⇒ m
∴
(d) Let $AB$ be the pillar and $\theta$ be the angle of elevation. Thus, we have:
m and m
In ∆$ABC$, we have:
⇒
⇒
∴
#### Question 25:
Match the following columns:
Column I Column II (a) From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. The height of the tower is ....... . (p) 12.7 m (b) A tower is 30 m high. Its shadow is x metres shorter when the sun's altitude is 45° than when it is 30°. Then, x = .......... . (q) 27.3 m (c) From the top of a cliff 30 m high, the angles of depression of the top and bottom of a tower are 30° and 45° respectively. The height of the tower is ......... . (r) 69.2 m (d) Two men on either side of a temple 30 m high observe the angles of elevation of the top of the temple to be 30° and 60° respectively. The distance between the two men is ........ . (s) 21.9 m
#### Answer:
(a) Let $AB$ be the building and $CD$ be the tower. Draw $BE$ ⊥ $CD$.
Now, we have:
m , ∠ and ∠
From the right ∆$BCA$, we have:
= 1
⇒ ⇒ m
From the right ∆$DBE$, we have:
[∵ BE = AC]
⇒ = m
Height of the tower = m
∴ (a) - (q)
(b) Let $AB$ be the tower such that m. Thus, we have:
∠ and ∠
From the right ∆$BAC$, we have:
⇒
⇒ m
From the right ∆$BAD$, we have:
⇒
⇒ m
Now,
m
∴ (b) - (s)
(c) Let $AB$ be the cliff such that m and $CD$ be the tower. Thus, we have:
∠ and ∠
If m, then m.
From the right ∆$BAC$, we have:
⇒
⇒ m
From the right ∆$BED$, we have:
⇒
⇒ m
We know:
m
∴
⇒
⇒
⇒ m
Hence, the height of the tower is 12.7 m.
∴ (c) - (p)
(d) Let $AB$ be the temple and be the points of the observer. Thus, we have:
m , ∠ and ∠
From the right ∆$BAC$, we have:
⇒
⇒ m
From the right ∆$BAD$, we have:
⇒
⇒ = m
Distance between the two men = m
∴ (d) - (r)
#### Question 26:
Assertion (A)
If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then h1 : h2 = 3 : 1.
Reason (R)
Figure
In the given figure, we have:
(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
#### Answer:
There are two towers $AB$ and $CD$ of heights ${h}_{1}$ and ${h}_{2}$, respectively. Let $AC$ be the distance between the two towers and O be the midpoint of $AC$.
Thus, we have:
m
Now, in ∆$AOB$, we have:
⇒ ⇒
Again, in ∆$DOC$, we have:
⇒ ⇒
Now,
Hence, both assertion (A) and reason (R) are true and (R) is the correct explanation of (A).
#### Question 27:
Assertion (A)
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Then, the height of the tree is $20\sqrt{3}\mathrm{m}.$
Reason (R)
The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving a metres towards the tower, the angle of elevation is β. Then, the height of the tower is
(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
#### Answer:
Let $AB$ be the tree on the opposite side of the bank of the river and $C$ be the position of the man standing on the bank of the river.
Thus, we have:
m, ∠ and ∠
If m and m, then m.
Then, in the right ∆ABC, we have:
⇒
⇒
In the right ∆ABD, we have:
⇒
⇒
⇒ [∵ ]
⇒ ⇒ ⇒ m
Hence, assertion (A) is true.
Assertion (A):
Let:
m, and
Given:
Putting the values of and $\beta$ in the above equation, we get:
m
Hence, reason (R) is true.
Both assertion (A) and reason (R) are true, and (R) is the correct reason of assertion (A).
#### Question 1:
The angle of elevation of a jet plane form a point P on the ground is 60°. After a flight of 15 second, the angle of elevation changes to 30°. If the jet plane is flying at a constant of $1500\sqrt{3}\mathrm{m}$, then the jet plane is
(a) 360 km/hr
(b) 720 km/hr
(c) 540 km/hr
(d) 270 km/hr
#### Answer:
(b) 720 km/h
Let $C$ be the position of the jet plane and $P$ be the point on the ground such that ∠, ∠ and m.
Let m and m.
In the right ∆$CPD$, we have:
⇒
⇒ m
Now, in the right ∆$C\text{'}PD\text{'}$, we have:
⇒
⇒
Putting the value of in the above equation, we get:
⇒ m
∴ Distance between the two points = m
m/s = km/h
#### Question 2:
A man standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. The width of the river is
(a) 20 m
(b) 22 m
(c) 24 m
(d) 25 m
#### Answer:
(a) 20 m
Let $AB$ be the tree and D be the position of the man on the opposite side of the tree. Thus, we have:
∠, ∠ and m
If m and m, then m.
In ∆$ABC$, we have:
⇒ ⇒ ...(i)
In ∆$ADB$, we have:
⇒
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒ m
From (i), we get:
m
∴ Width of the river = m
#### Question 3:
As observed from the top of a 75 m tall lighthouse, the angle of depression of two ships is 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, then the distance between the ships is
(a) $75\left(\sqrt{3}+1\right)\mathrm{m}$
(b) $75\left(\sqrt{3}-1\right)\mathrm{m}$
(c) $25\left(\sqrt{3}+1\right)\mathrm{m}$
(d) $25\left(\sqrt{3}-1\right)\mathrm{m}$
#### Answer:
(b) $75\left(\sqrt{3}-1\right)\mathrm{m}$
Let $OB$ be the lighthouse. From O, the angle of depression is observed.
We have:
, ∠ and m
Let m and m such that m.
In ∆$OBA$, we have:
m
In ∆$OBC$, we have:
⇒ ⇒ m
∴ Distance between the two ships = m
#### Question 4:
If the angles of elevation of a tower from two points as distances a and b, where a > b from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is
(a) $\sqrt{a+b}$
(b)$\sqrt{a-b}$
(c) $\sqrt{\frac{a}{b}}$
(d) $\sqrt{ab}$
#### Answer:
(d) $\sqrt{ab}$
Let $AB$ be the tower and be the two points. Thus, we have:
, ∠ and ∠
Let:
m
In right ∆$ABC$, we have:
In right ∆$ABD$, we have:
On multiplying (1) and (2), we get:
$\frac{a}{\sqrt{3}}×b\sqrt{3}={h}^{2}\phantom{\rule{0ex}{0ex}}⇒ab={h}^{2}\phantom{\rule{0ex}{0ex}}⇒h=\sqrt{ab}$
Hence, the height of the tower is $\sqrt{ab}$.
#### Question 5:
A ladder 20 m long touches a wall at the height of 10 m. Find the angle made by the ladder with the horizontal.
#### Answer:
Let $AC$ be the ladder such that m. It touches the wall at a height m.
Let $\theta$ be the angle.
In ∆$ABC$, we have:
⇒
Hence, the angle made by the ladder with the horizontal is 30°.
#### Question 6:
The angle of elevation of the top of a pillar at a distance of 18 m from its foot is 30°. Find the height of the pillar.
#### Answer:
Let $AB$ be the pillar, $C$ be the foot of the person such that m. Thus, we have:
From right ∆$ABC$, we have:
⇒
⇒ m
∴ Height of the tower = m
#### Question 7:
Find the angle of elevation of the sum when the length of of the shadow of a pole is $\sqrt{3}$ times the height of the pole.
#### Answer:
Let $AB$ be the pole and $BC$ be its shadow.
Let and such that (given) and $\theta$ be the angle of elevation.
From ∆$ABC$, we have:
⇒
⇒
⇒
Hence, the angle of elevation is ${30}^{\mathrm{o}}$.
#### Question 8:
A Minar A casts a shadow 150 m long at the same time when a Minar B casts a shadow 120 m long on the ground. If the height of the Minar B is 80 m, find the length of Minar A.
#### Answer:
Let be the Minar $A$ and Minar $B$, respectively. Let $BC$ and $EF$ be the shadows of Minar $A$ and Minar $B$.
Thus, we have:
m, m and m
Let and $\theta$ be the angle of elevation of both Minars.
From ∆$DEF$, we have:
From ∆$ABC$, we have:
[ ]
⇒ m
∴ Length of Minar $A$ = m
#### Question 9:
If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then show that h1 : h2 = 3 : 1.
#### Answer:
Let $AB$ and $CD$ be the two towers subtending angles , respectively.
Now,
and ∠
Let $P$ be the midpoint of the line segment joining the two towers.
Let:
, and
From ∆$DCP$, we have:
⇒ ...(i)
From ∆$ABP$, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we get:
⇒
∴
#### Question 10:
Find the height of a tower whose shadow is 10 m long when the sun's altitude is 45°.
#### Answer:
Let $AB$ be the tower such that m and $BC$ be its shadow such that m and ∠.
From ∆$ABC$, we have:
⇒
⇒ m
Hence, the height of the tower is 10 m.
#### Question 11:
From the top of a building 60 m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60° respectively. Find the height of the lamp post.
#### Answer:
Let $AB$ be the building and $CD$ be the vertical lamppost.
Here,
m, ∠ and ∠
Let m such that m. Thus, m and m.
From ∆$ABC$, we have:
⇒ $\frac{60}{x}$ = $\sqrt{3}$
...(i)
From ∆$AED$, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒
⇒
⇒ m.
∴ Height of the lamppost = m
#### Question 12:
An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.
#### Answer:
Let $B$ be the aeroplane flying vertically below the aeroplane $A$ at a height of 3125 m, that is, m.
The angles of elevation of both the aeroplanes from the ground are such that ∠ and ∠.
Let:
m and m
From the right ∆$BCD$, we have:
From the right ∆$ACD$, we have:
⇒
⇒
Putting the value of , we get:
⇒ m
∴ Distance between the two aeroplanes = m
#### Question 13:
A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point O on the ground is 60° and the angle of depression of the point O from the top of the tower is 45°. Show that the height of the tower is $\frac{5}{2}\left(\sqrt{3}+1\right)\mathrm{m}.$
#### Answer:
Let $AB$ be the tower and $BC$ be the pole such that m.
∠ and ∠
Let:
m and m
In right ∆$BAD,$ we have:
⇒
x = h
Or,
In right ∆$CAD$, we have:
⇒
Putting the value of in the above equation, we have:
⇒
⇒
⇒ m
∴ Height of the tower = m
#### Question 14:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5. From a point on the plane, the angles of elevation of the bottom and top of the flagstaff are respectively 30° and 60°. Show that the height of the tower is 2.5 m.
#### Answer:
Let $AC$ be the vertical tower and $BC$ be the vertical flagstaff such that m.
and ∠
Let:
m and m
In right ∆$CDA$, we have:
⇒
⇒
In right ∆$BAD$, we have:
⇒
Putting the value in above equation, we get:
⇒
⇒
⇒ m
∴ Height of the tower = m
#### Question 15:
The angle of elevation of the top of a hill at the foot of the tower is 60° and the angle of elevation of the top of the tower form the foot of the hill is 30°. If the tower is 50 m high, find the height of the hill.
#### Answer:
Let $AB$ be the hill and $CD$ be the tower.
Now,
m, ∠ and ∠
Let:
m and m
From right ∆$DBC$, we have :
⇒
⇒ m
From ∆$ABC$, we have:
⇒
⇒ m [ ]
Height of the hill = m
#### Question 16:
A tree breaks due to storm and the broken part bends so that the top of tree touches the ground, making an angle of 30° with the ground. If the distance between the foot of the tree to the point, where the top touches the ground is 8 m, show that the height of the tree is $8\sqrt{3}\mathrm{m}.$
#### Answer:
Let $AB$ be the tree broken at point $C$ so that $CB$ takes the position $CD$ such that ∠ and m.
Let:
m and m
From right ∆$CAD$, we have:
⇒
⇒
Again, we have:
⇒
⇒
Height of the tree = m
#### Question 17:
The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving metres towards the tower, the angle of elevation is β. Prove that the height of the tower is
#### Answer:
Let $AB$ be the tower and $C$ be the point on the ground such that ∠.
On moving towards the tower at point $D$, we have and ∠.
If and , then .
From ∆$ACB$, we have:
...(i)
From ∆$ADB$, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒
Hence, the height of the tower is .
#### Question 18:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angle of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is
#### Answer:
Let $BD$ be the tower and $AD$ be the vertical flagstaff such that . Thus, we have:
and ∠
Let:
and
In ∆$DBC$, we have:
⇒
Or,
...(i)
In ∆$ABC$, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we get:
⇒
⇒
⇒
⇒
Hence, the height of the tower is .
#### Question 19:
The angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection is β. Prove that the distance of the cloud from the point of observation is
#### Answer:
Let $AB$ be the surface of the lake and $P$ be the point vertically above $A$ such that m.
Let $C$ be the position of the cloud and $D$ be its reflection on the lake.
Let the height of the cloud be $H$ metres. Thus, we have:
m
Draw $PQ$$CD$.
Thus, we have:
and ∠
m
m and m
From the right ∆$CQP$, we have:
⇒
⇒ ...(i)
From the right ∆$QPD$, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we get:
⇒
⇒
⇒ ...(iii)
Using the value of $H$ in (i), we get:
...(iv)
Now, to find $PC$, we have:
⇒
Putting the value of $PQ$ from (iv), we get:
Distance of the cloud =
#### Question 20:
A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Show that the speed of the boat is $\frac{100\sqrt{3}}{3}$ metres per minute.
#### Answer:
Let $AB$ be the light house and $C$ and $D$ be the positions of the two ships. Thus, we have:
m
Let:
m and m
Thus, we have:
m
In ∆$ABC$, we have:
In ∆$ABD$, we have:
$⇒$
∴ m
Distance travelled in two minutes = = $\frac{200\sqrt{3}}{3}$ m
metres per minute
Hence proved.
View NCERT Solutions for all chapters of Class 10
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# How do you write the quadratic function f(x) = x^2 -4x + 7 in vertex form?
$f \left(x\right) = {\left(x - 2\right)}^{2} + 3$
If a standard quadratic has the form $y = a {x}^{2} + b x + c$, it can be rewritten into vertex form by completing the square. This means that we use $\frac{b}{2} a$ to get an expression a(x+b/2a)^2 which equals $a \left({x}^{2} + \frac{b}{a} \cdot x + {b}^{2} / \left(4 {a}^{2}\right)\right)$. We then subtract ${b}^{2} / \left(4 {a}^{2}\right)$ and add$c$ to finish the equation.
$f \left(x\right) = {\left(x - 2\right)}^{2} - 4 + 7$
$f \left(x\right) = {\left(x - 2\right)}^{2} + 3$
This is vertex form, from which you can get that the vertex is at $\left(2 , 3\right)$ - th epoint at which teh bracketed term is zero and therefore the expression is at its minimum value.
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# Answer to Question #236711 in Algebra for Lara
Question #236711
Let g(x) be the transformation of f(x)= x2. Write the rule for g(x) using the change described.
1. horizontal compression by a factor of 1/5 followed by a vertical shift down 7 units and reflection across y-axis. ______________
2. reflection across the x-axis followed by a vertical stretch by a factor of 3, a horizontal shift 6 units left, and a vertical shift 4 units down. _____________
1
2021-09-16T05:15:02-0400
1)
Parent Function: f(x) = x\\^2
There are:
• A horizontal compression by a factor of \frac{1}{5}
• A vertical shift down 7 units
• A reflection across y-axis, means that b is negative
The general form for the rule is: f(x) = a \times f([b(x-h]) + k
Therefore, b = -5 because of horizontal compression by a factor of \frac{1}{5} and reflection across y-axis
k= - 7
Replacing b and k in the general form of the rule f(x) = a f[ b(x − h)] + k where f(x) = x\\^2 , we find, f(x) = (-5(x))\\^2 - 7
= f(x) = (-5(x))\\^2 - 7
2)
Parent Function: f(x) = x\\^2
The general form for the rule is: f(x) = a \times f([b(x-h]) + k
There are:
Reflection across the x-axis, indicating that a is negative
Vertical stretch by a factor of 3, means that a= -3
A horizontal shift 6 units left, means that h=-6
A vertical shift 4 units down, means that k= -4
Replacing a, k, and h in the general form of the rule f(x) = a f[ b(x − h)] + k where f(x) = x\\^2 , we find f(x) = -3((x + 6)\\^2) - 4
= -3((x + 6)\\^2) - 4
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## Files in this item
FilesDescriptionFormat
application/pdf
1427630.pdf (5MB)
PresentationPDF
application/pdf
3890.pdf (21kB)
AbstractPDF
## Description
Title: DEVELOPMENT AND PERFORMANCE OF LILLE'S FOURIER TRANSFORM MILLIMETER-WAVE SPECTROMETER Author(s): Zou, Luyao Contributor(s): Alekseev, E. A.; Margulès, L. ; Motiyenko, R. A. Subject(s): Atmospheric science Abstract: Fast spectral acquisition is an essential component in obtaining broadband molecular spectra with high signal to noise ratio, and in studying meta-stable molecular species. With the recent developments on commercially available arbitrary waveform generators (AWG), direct digital synthesizers (DDS), and room-temperature Schottky diodes, it is feasible now to perform fast spectroscopy scans using the heterodyne detection of the free induction decay of molecules in the millimeter wave bands, as demonstrated by a few pioneer spectrometer designs.\footnote{E. Gerecht, K. O. Douglass and D. F. Plusquellic, 2011, \textit{Opt. Expr.}, \textbf{19}, 8973; I. A. Finneran, D. B. Holland, P. B. Carrol \textit{et al.}, 2013, \textit{Rev. Sci. Instrum.}, \textbf{84}, 083104.} At Lille, we have developed and demonstrated the Fourier transform millimeter-wave (FTmmW) spectrometer system based on DDS.\footnote{R. A. Motiyenko and L. Margul\es, ISMS 73rd symposium (2018), WI07.} This spectrometer has a simplified design which does not require external reference clocks or local oscillators, yet it achieves decent frequency resolution and high phase stability. Since then, several upgrades, including new frequency sweep scheme and filter applications, have been made to improve the spectral purity and scan bandwidth of this FTmmW system. The results of the detailed performance test of bandwidth, sensitivity and data acquisition rate on the upgraded system will be presented. The advantages and limitations of the spectrometer for conducting fast millimeter spectroscopy on atmospheric radicals will be discussed. \textit{The authors thank the R\'egion Hauts-de-France, and the Minist\ere de l’Enseignement Sup\'erieur et de la Recherche (CPER Climibio), the French ANR Labex CaPPA through the PIA (contract ANR-11-LABX-0005-01), and the European Fund for Regional Economic Development for their financial support.} Issue Date: 2019-06-21 Publisher: International Symposium on Molecular Spectroscopy Genre: Conference Paper / Presentation Type: Text Language: English URI: http://hdl.handle.net/2142/104339 DOI: 10.15278/isms.2019.FB04 Rights Information: Copyright 2019 Luyao Zou Date Available in IDEALS: 2019-07-152020-01-25
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# Clock time Puzzle
What will be the time when the hour, minute, and second hands will all coincide in between 6 O'Clock and 7 O'Clock?
How should I approach this type of question? Can anyone explain it?
• 6:33:33? Maybe? – Mithrandir Jul 13 '16 at 17:06
You really just need a couple equations
h = hi + min/60
h/12 = min/60 = sec/60
hi is the integer hour before the time you want to find (hi=6) h, min, sec are the actual hour, minute and second they intersect.
Substitute min/60 for h/12 and the equation becomes
h = hi + h/12
(11/12)*h = hi
h = 12/11*hi
h = 12/11*6
**h = 72/11**
min = h/12*60
min = 6/11*60
sec = min = 6/11*60
h = 6.5454
min = 32.7272
sec = 32.7272
For an actual time drop the decimal points of the hour and minute and your left with
**final time is 6:32:32.727**
EDIT TO SHOW THAT IT IS NOT CORRECT
As per @Poolsharker answer these equations are incorrect another formula that should have been considered is
min = mini + sec/60
59/60*min = mini
From the previous final time mini = 32
min = 60/59*32
min = 32.5424
which does not match the previous min time so they will not all 3 coincide
When we are a fraction $f$ of the way through a 12-hour "day" -- so $f=1$ means 12 hours have passed since noon or midnight -- the positions of the hands as fractions of a whole turn are: $f$ (for the hour hand), $12f$ (for the minute hand), and $720f$ (for the second hand), where integer differences are not visible. So for all three to coincide we need the differences $11f$ and $719f$ to be integers. Since 11 and 719 are coprime, this requires $f$ to be an integer, which means that the only 3-way coincidences happen with all hands pointing to the 12.
In particular, there is none between 6 and 7 o'clock.
(The bit about 11 and 719 being coprime may not be perfectly clear. Suppose $11f=m$ and $719f=n$ where $m,n$ are integers. Then $11n=719m$, so $11|719m$; since 11 is prime this requires either $11|719$ (no!) or $11|m$. So write $m=11k$; then $11f=m=11k$ so $f=k$ and $f$ is an integer.)
• I think this is the most elegant solution out of the ones posted so far. – SpiritFryer Jul 13 '16 at 19:53
From my calculations...
all three hands will not intersect 'exactly' between 6 O'clock and 7 O'clock
The way I approached it isn't elegant but I think my calculations are correct:
Clock face makes up 360 degrees. When the minute hand moves 360 degrees, the hour hand has moved 30 degrees (1/12). So I did my calculations by moving the minute hand to where the hour hand is, and then adjusting the hour hand because of the minute hand movement.
So at 6:00, hour hand (hh) is at 180 degrees, minute hand (mh) is at 0.
Move mh 180 degrees, move hh 180 * (1/12) = 15 degrees.
Move mh 15 degrees, move hh 15 * (1/12) = 1.25 degrees.
Move mh 1.25 degrees, move hh 1.25 (1/12) = 0.104 degrees. and so on...
The calculations start repeating at the hour hand having moved 16.3636. So if we are 16.3636 degrees between the 6 and the 7 and 30 degrees = 5 minutes, then 16.3636 / (30/5) = 2.7272 minutes. Now we convert 0.7272 into seconds (0.7272 * 60) and we get 43.632. So I believe that the hour hand and minute hand we be nearly exact at 6:32:43.6, however that does leave the second hand outside. The best approximation would be 6:32:32.72 as mentioned by gtwebb.
Side note
While checking my calculations, I found this site calculating all the times the hour hand and minute hand intersect, good future reference http://www.kodyaz.com/articles/how-many-times-a-day-clocks-hands-overlap.aspx
There's no three pointers coincidence between 6 o'clock and 7 o'clock.
We can solve this problem using equations of motion.
The equation of motion in terms of angles of a pointer is:
$$\theta=\theta_i+\dot{\theta} t$$
where $\theta_i$ is the initial angle (at $t=0$), $\dot{\theta}$ the angular frequency and, $t$ the time.
We have three bodies: hours ($h$), minutes ($m$) and seconds ($s$) pointers.
Consider that the initial condition is at 6 o'clock ($t=0$). For a polar referential system where $\theta=0$ in 12 o'clock, with a direct clock direction for this angle, we know that at the initial condiditon, we have:
$$\theta_h^i=\pi \text{ (rad)}, \hspace{5pt} \theta_m^i=0 \text{ (rad)}, \hspace{5pt} \theta_s^i=0 \text{ (rad)}$$
In relation to the angular frequencies, we have:
$$\dot{\theta_h}=\frac{2\pi}{12\times3600} \text{(rad/s)}, \hspace{5pt} \dot{\theta_m}=\frac{2\pi}{3600} \text{ (rad/s)}, \hspace{5pt} \dot{\theta_s}=\frac{2\pi}{60} \text{ (rad/s)}$$
The condition for three pointers coincidence is:
$$\theta_h(t)=\theta_h(t)-2\pi n=\theta_h(t)- 2\pi m$$
where $n$ e $m$ are integers (the subtraction by $2\pi n$ and $2\pi m$ cancels the cyclicality of the pointers). We need that $n\in[0,1]$, $m\in[0,60]$ and $0\leq t\leq3600$ s for the coincidence be between 6 and 7 o'clock.
So, solving the system of equations of $\theta_h(t)=\theta_h(t)-2\pi n$ and $\theta_h(t)-2\pi n=\theta_h(t)- 2\pi m$ in order to $t$ am $m$ we get:
$$t=\frac{21600}{11}\left(1+2n\right) \text{ (s)},\hspace{15pt} m=\frac{719n+354}{11}$$
I tried to find the first value of $n$ that satisfies the conditions of $n$ and $m$ being positive integers, by substituition of $n=1,2,3...$ on the $m(n)$ expression and verified if $m$ was an integer. I got $n=5$ which means that $t=21600 s=6h$. Because $t=0$, at 6 o'clock, the coincidence happens at 12 o'clock, which means that it's impossible that it happens between 6 and 7 o'clock (remember that $t(n)$ is a crescent function with $n$, and $n=5$ is the lowest value for coincidence).
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# How can I fit the original data with exponential or polynomial functions via pgfplots?
I am trying to plot the following data with a curve fitting via pgfplots.
It seems linear regression is not suitable for my case. So I would prefer to have exponential or polynomial curve fitting on these data. How can I implement it?
Below is the current code:
% arara: pdflatex: { shell: yes }
\documentclass[border=1mm, png]{standalone}
\usepackage{siunitx}
\usepackage{pgfplots}
\pgfplotsset{compat=1.10}
\usepackage{pgfplotstable}
\usepackage{filecontents}
\begin{filecontents*}{myData.dat}
X Y
2 275.68
3 1175.26
4 1351.60
5 1485.57
6 1583.30
7 1861.28
8 2095.39
9 2574.54
10 2841.74
11 2914.16
12 3965.12
13 3787.68
14 5294.83
21 10504.49
\end{filecontents*}
\begin{document}
\begin{tikzpicture}
\pgfplotsset{%
,width=10cm
,legend style={font=\footnotesize}
}
\begin{axis}[%
,xlabel=Numbers $N$ in \si{\gram\per\liter}
,ylabel=Ratio
,ymin=0
,xmin=0
,scaled y ticks=base 10:0
,legend cell align = left
,legend pos = north west
]
\addplot+[no markers,red] table [y={create col/linear regression={y=Y}}]{myData.dat};
Linear trend $(y=\pgfmathprintnumber{\pgfplotstableregressiona} \cdot x \pgfmathprintnumber[print sign]{\pgfplotstableregressionb})$} %
\end{axis}
\end{tikzpicture}
\end{document}
Which yields:
-
Did you take a look on how to use gnuplot in pgfplots? I guess, this kind of calculation needs some higher tool. Personally, I would recommend to calculate the curve you need via sageTeX and to plot it afterwards. – LaRiFaRi Jun 26 at 11:41
Well, Excel is a powerful calculation tool... (not that I like it ;-) ). If there is something implemented here, you should be able to find it by the words "fitting", "regression", "approximation", "polynomial" or "Taylor" in the manuals of pgfplots and pgfplotstable. Please have a look! I didn't check. – LaRiFaRi Jun 26 at 11:57
Off topic: 1. $-$ is a minus sign. Therefore I would not set the unit of your ratio in math-mode. (As units should not be set in square brackets (ISO & SI) and a "ratio" has the unit 1 by definition, I would leave it away completely.). 2. please make you code minimal. You just need siunitx, pgfplots, and pgfplotstable for this example. – LaRiFaRi Jun 26 at 12:04
@LCFactorization: You can do this from within PGFPlots by using gnuplot in the background. See Non-linear curve fitting with gnuplot (this may be a duplicate of that question). – Jake Jun 26 at 12:35
@LCFactorization: PGFPlots can only do linear regression. The question "how can I linearize a polynomial / exponential equation?" seems more appropriate for math.stackexchange.com. – Jake Jun 26 at 12:43
This is an attempt via polynomial fit and the use of gnuplot. Therefore this requires one to compile the code with shell-escape enabled, and gnuplot has to be installed on your system.
Edit: The OP finds how to find the actually parameters after curve fitting. The answer is here: show fitted values which needs two lines of code
set print "parameters.dat"; % Open a file to save the parameters into
print a, b; % Write the parameters to file
Code
\documentclass[border=10pt]{standalone}
\usepackage{pgf,tikz}
\usepackage{siunitx}
\usepackage{pgfplots}
\usepackage{pgfplotstable}
\usepackage{filecontents}
\usepackage{graphicx}
\usepackage{latexsym}
\usepackage{keyval}
\usepackage{ifthen}
\usepackage{moreverb}
\usepackage{gnuplottex}%[miktex]%[shell]
\usepackage{pgfplotstable}
\pgfplotsset{compat=1.5}
\usepackage{filecontents}
\begin{document}
\begin{filecontents}{data.csv}
X Y
2 275.68
3 1175.26
4 1351.60
5 1485.57
6 1583.30
7 1861.28
8 2095.39
9 2574.54
10 2841.74
11 2914.16
12 3965.12
13 3787.68
14 5294.83
21 10504.49
};
\end{filecontents}
\begin{tikzpicture}
\pgfplotsset{width=10cm,
legend style={font=\footnotesize}}
\begin{axis}[
xlabel={Numbers [N,\si{\gram\per\liter}]},
ylabel={Ratio $[-]$},
ymin =0,
ytick = {200,2000,4000,6000,8000,10000},
y tick label style={
/pgf/number format/.cd,
fixed,
fixed zerofill,
precision=0,
/tikz/.cd
},
yticklabel=\pgfmathprintnumber{\tick},
scaled y ticks=base 10:0,
legend cell align = left,
legend pos = north west]
\addlegendentry{Ratio of {\em tfml} to {\em gGN}}
% linear curve fitting
y={create col/linear regression={y=Y}}] % compute a linear regression from the input table
{data.csv};
linear trend $\left(y=\pgfmathprintnumber{\pgfplotstableregressiona} \cdot x \pgfmathprintnumber[print sign]{\pgfplotstableregressionb}\right)$} %
% polynomial fit
\addplot [no markers, blue] gnuplot [raw gnuplot] { % allows arbitrary gnuplot commands
f(x) = a*x**2+b*x; % Define the function to fit
a=260;b=-270; % Set reasonable starting values here
fit f(x) 'data.csv' u 1:2 via a,b; % Select the file, starts at col 1 and two variables
plot [x=2:21] f(x); % Specify the range to plot
set print "parameters.dat"; % Open a file to save the parameters
print a, b; % Write the parameters to file
};
\pgfplotstablegetelem{0}{0}\of\parameters \pgfmathsetmacro\paramA{\pgfplotsretval} % Get first element, save into \paramA
\pgfplotstablegetelem{0}{1}\of\parameters \pgfmathsetmacro\paramB{\pgfplotsretval}
polynomial fit: $y=\pgfmathprintnumber{\paramA} x^2 \pgfmathprintnumber[print sign]{\paramB} x$
}
\end{axis}
\end{tikzpicture}
\end{document}
-
thank you. Is it possible to explicitly show the fitted values of a and b in the legendentry? – LCFactorization Jun 26 at 12:51
The answer is probably here: tex.stackexchange.com/questions/29633/… thanks – LCFactorization Jun 26 at 13:10
Yes, you got it. I was searching the auxiliary file, to see if any info was left in *.plot.gnuplot and *.plot.table file, but none. Then your finding prompt me to try. Yes, you find the answer. – Jesse Jun 26 at 13:19
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{}
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OpenStudy (anonymous):
What is the first step in solving the equation 5x − 3 = 22? Add 3 to both sides of the equation. Divide both sides of the equation by −5. Divide both sides of the equation by −3. Subtract 3 from both sides of the equation.
3 years ago
Add 3 to both sides of the equation.
3 years ago
OpenStudy (triciaal):
as above. you want to separate the variable to one side of the equation and the constants to the other.
3 years ago
|
{}
|
### Top 10 Arxiv Papers Today in High Energy Physics - Theory
##### #1. Entanglement Wedge Cross Sections Require Tripartite Entanglement
###### Chris Akers, Pratik Rath
We argue that holographic CFT states require a large amount of tripartite entanglement, in contrast to the conjecture that their entanglement is mostly bipartite. Our evidence is that this mostly-bipartite conjecture is in sharp conflict with two well-supported conjectures about the entanglement wedge cross section surface $E_W$. If $E_W$ is related to either the CFT's reflected entropy or its entanglement of purification, then those quantities can differ from the mutual information at $\mathcal{O}(\frac{1}{G_N})$. We prove that this implies holographic CFT states must have $\mathcal{O}(\frac{1}{G_N})$ amounts of tripartite entanglement. This proof involves a new Fannes-type inequality for the reflected entropy, which itself has many interesting applications.
more | pdf | html
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dajmeyer: Akers & Rath, Entanglement Wedge Cross Sections Require Tripartite Entanglement https://t.co/Gemi7ONR9W "We prove that this implies holographic CFT states must have O(1/G_N) amounts of tripartite entanglement." https://t.co/qYbDjGOXne
TheoryPapers: Entanglement Wedge Cross Sections Require Tripartite Entanglement. https://t.co/kZ6GL33YDf
blueberry_phase: RT @TheoryPapers: Entanglement Wedge Cross Sections Require Tripartite Entanglement. https://t.co/kZ6GL33YDf
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##### #2. Strong integrability of $λ$-deformed models
###### George Georgiou, Konstantinos Sfetsos, Konstantinos Siampos
We study the notion of strong integrability for classically integrable $\lambda$-deformed CFTs and coset CFTs. To achieve this goal we employ the Poisson brackets of the spatial Lax matrix which we prove that it assumes the Maillet $r/s$-matrix algebra. As a consequence the system in question are integrable in the strong sense. Furthermore, we show that the derived Maillet $r/s$-matrix algebras can be realized in terms of twist functions, at the poles of which we recover the underlying symmetry algebras.
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MathPHYPapers: Strong integrability of $\lambda$-deformed models. https://t.co/Z8FN7f9wmu
yoh_tanimoto: conflitto di terminologia "strong integrability" o.o https://t.co/3NBYAAQLSD
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##### #3. From Little String Free Energies Towards Modular Graph Functions
###### Stefan Hohenegger
We study the structure of the non-perturbative free energy of a one-parameter class of little string theories (LSTs) of A-type in the so-called unrefined limit. These theories are engineered by $N$ M5-branes probing a transverse flat space. By analysing a number of examples, we observe a pattern which suggests to write the free energy in a fashion that resembles a decomposition into higher-point functions which can be presented in a graphical way reminiscent of sums of (effective) Feynman diagrams: to leading order in the instanton parameter of the LST, the $N$ external states are given either by the fundamental building blocks of the theory with $N=1$, or the function that governs the counting of BPS states of a single M5-brane coupling to one M2-brane on either side. These states are connected via an effective coupling function which encodes the details of the gauge algebra of the LST and which in its simplest (non-trivial) form is captured by the scalar Greens function on the torus. More complicated incarnations of this...
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turke196883: [1911.08172] Stefan Hohenegger: From Little String Free Energies Towards Modular Graph Functions https://t.co/68X3OgzDTr
TheoryPapers: From Little String Free Energies Towards Modular Graph Functions. https://t.co/AelqZ2QAjm
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##### #4. 3d-3d correspondence for mapping tori
###### Sungbong Chun, Sergei Gukov, Sunghyuk Park, Nikita Sopenko
One of the main challenges in 3d-3d correspondence is that no existent approach offers a complete description of 3d $N=2$ SCFT $T[M_3]$ --- or, rather, a "collection of SCFTs" as we refer to it in the paper --- for all types of 3-manifolds that include, for example, a 3-torus, Brieskorn spheres, and hyperbolic surgeries on knots. The goal of this paper is to overcome this challenge by a more systematic study of 3d-3d correspondence that, first of all, does not rely heavily on any geometric structure on $M_3$ and, secondly, is not limited to a particular supersymmetric partition function of $T[M_3]$. In particular, we propose to describe such "collection of SCFTs" in terms of 3d $N=2$ gauge theories with "non-linear matter'' fields valued in complex group manifolds. As a result, we are able to recover familiar 3-manifold invariants, such as Turaev torsion and WRT invariants, from twisted indices and half-indices of $T[M_3]$, and propose new tools to compute more recent $q$-series invariants $\hat Z (M_3)$ in the case of manifolds...
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TheoryPapers: 3d-3d correspondence for mapping tori. https://t.co/xdGQz62LwP
Gamma_ijk: RT @TheoryPapers: 3d-3d correspondence for mapping tori. https://t.co/xdGQz62LwP
YosukeSaito7: RT @TheoryPapers: 3d-3d correspondence for mapping tori. https://t.co/xdGQz62LwP
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##### #5. A unitary qubit transport toy model for black hole evaporation
###### Bogusław Broda
In a recent paper Osuga and Page presented an explicitly unitary qubit transport toy model for black hole evaporation. Following their idea we propose a unitary toy model which involves Hawking states.
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RelativityPaper: A unitary qubit transport toy model for black hole evaporation. https://t.co/DoyfNRU0Ou
KaellumEridani: RT @RelativityPaper: A unitary qubit transport toy model for black hole evaporation. https://t.co/DoyfNRU0Ou
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##### #6. Vector model in higher dimensions
###### Mikhail Goykhman, Michael Smolkin
We study UV behaviour of the $O(N)$ vector model with quartic interaction in $4<d<6$ dimensions to the next-to-leading order in the large-$N$ expansion. We derive and perform consistency checks that provide evidence for the existence of a non-trivial UV fixed point and explore the corresponding CFT. In particular, we calculate a CFT data associated with the three-point functions of the fundamental scalar and Hubbard-Stratonovich fields. We compare our findings with their counterparts obtained within a proposed alternative description of the model in terms of $N+1$ massless scalars with cubic interactions.
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TheoryPapers: Vector model in higher dimensions. https://t.co/5Pv8vf8R6U
physical_things: RT @TheoryPapers: Vector model in higher dimensions. https://t.co/5Pv8vf8R6U
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##### #7. On self-gravitating kinks in two-dimensional pseudo-riemannian universes
###### A. Alonso Izquierdo, W. García Fuertes, J. Mateos Guilarte
Two-dimensional scalar field theories with spontaneous symmetry breaking subject to the action of Jackiw-Teitelboim gravity are studied. Solutions for the $\phi^4$ and sine-Gordon self-gravitating kinks are presented, both for general gravitational coupling and in the perturbative regime. The analysis is extended to deal with a hierarchy of kinks related to transparent P\"{o}schl-Teller potentials
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##### #8. The emergence of expanding space-time and intersecting D-branes from classical solutions in the Lorentzian type IIB matrix model
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The type IIB matrix model is a promising candidate for a nonperturbative formulation of superstring theory. As such it is expected to explain the origin of space-time and matter at the same time. This has been partially demonstrated by the previous Monte Carlo studies on the Lorentzian version of the model, which suggested the emergence of (3+1)-dimensional expanding space-time. Here we investigate the same model by solving numerically the classical equation of motion, which is expected to be valid at late times since the action becomes large due to the expansion of space. Many solutions are obtained by the gradient decent method starting from random matrix configurations, assuming a quasi direct-product structure for the (3+1)-dimensions and the extra six dimensions. We find that these solutions generally admit the emergence of expanding space-time and intersecting D-branes, the latter being represented by a block-diagonal structure in the extra dimensions. For solutions corresponding to D-branes with appropriate dimensionality,...
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We construct an effective Quantum Field Theory for the wrapping effects in 1+1 dimensional models of factorised scattering. The recently developed graph-theoretical approach to TBA gives the perturbative desctiption of this QFT. For the sake of simplicity we limit ourselves to scattering matrices for a single neutral particle and no bound state poles, such as the sinh-Gordon one. On the other hand, in view of applications to AdS/CFT, we do not assume that the scattering matrix is of difference type. The effective QFT involves both bosonic and fermionic fields and possesses a symmetry which makes it one-loop exact. The corresponding path integral localises to a critical point determined by the TBA equation.
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##### #10. Decoherence in Conformal Field Theory
Noise sources are ubiquitous in Nature and give rise to a description of quantum systems in terms of stochastic Hamiltonians. Decoherence dominates the noise-averaged dynamics and leads to dephasing and the decay of coherences in the eigenbasis of the fluctuating operator. For energy-diffusion processes stemming from fluctuations of the system Hamiltonian the characteristic decoherence time is shown to be proportional to the heat capacity. We analyze the decoherence dynamics of entangled CFTs and characterize the dynamics of the purity, and logarithmic negativity, that are shown to decay monotonically as a function of time. The converse is true for the quantum Renyi entropies. From the short-time asymptotics of the purity, the decoherence rate is identified and shown to be proportional to the central charge. The fixed point characterizing long times of evolution depends on the presence degeneracies in the energy spectrum. We show how information loss associated with decoherence can be attributed to its leakage to an auxiliary...
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JoshuahHeath: Today's #arXivsummary: https://t.co/qiSu5P4O6t, by del Campo & Takayanagi. Authors analyze decoherence dynamics of entangled CFTS. Characterize dynamics of purity & log negativity; decay monotonically w/time. Inner horizon region of dual AdS black hole squeezed due to decoherence
OSablin: "Decoherence in Conformal Field Theory. (arXiv:1911.07861v1 [hep-th])" https://t.co/OguMWd1IcU
blueberry_phase: RT @JoshuahHeath: Today's #arXivsummary: https://t.co/qiSu5P4O6t, by del Campo & Takayanagi. Authors analyze decoherence dynamics of entang…
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Assert is a website where the best academic papers on arXiv (computer science, math, physics), bioRxiv (biology), BITSS (reproducibility), EarthArXiv (earth science), engrXiv (engineering), LawArXiv (law), PsyArXiv (psychology), SocArXiv (social science), and SportRxiv (sport research) bubble to the top each day.
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Algebra 2 (1st Edition)
$57$
For a 3x3 matrix: $\left[\begin{array}{rrr} a & b & c \\ d &e & f \\ g &h & i \\ \end{array} \right]$ The determinant is given by: $D=a(ei-fh)-b(di-fg)+c(dh-eg).$ Hence here $D=5(2\cdot1-1\cdot1)-(-9)(4\cdot1-1\cdot0)+4(4\cdot1-2\cdot0)=5(1)+9(4)+4(4)=57.$
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# Section6.2Graphs of Relations on a Set¶ permalink
In this section we introduce directed graphs as a way to visualize relations on a set.
Let $A = \{0, 1,2,3\}$, and let $r = \{(0, 0), (0, 3), (1, 2), (2, 1), (3, 2), (2, 0)\}$ In representing this relation as a graph, elements of $A$ are called the vertices of the graph. They are typically represented by labeled points or small circles. We connect vertex $a$ to vertex $b$ with an arrow, called an edge, going from vertex $a$ to vertex $b$ if and only if $a r b$. This type of graph of a relation $r$ is called a directed graph or digraph. Figure 6.2.1 is a digraph for $r\text{.}$ Notice that since 0 is related to itself, we draw a “self-loop” at 0.
The actual location of the vertices in a digraph is immaterial. The actual location of vertices we choose is called an embedding of a graph. The main idea is to place the vertices in such a way that the graph is easy to read. After drawing a rough-draft graph of a relation, we may decide to relocate the vertices so that the final result will be neater. Figure 6.2.1 could also be presented as in Figure 6.2.2.
A vertex of a graph is also called a node, point, or a junction. An edge of a graph is also referred to as an arc, a line, or a branch. Do not be concerned if two graphs of a given relation look different as long as the connections between vertices are the same in two graphs.
##### Example6.2.3Another directed graph.
Consider the relation $s$ whose digraph is Figure 6.2.4. What information does this give us? The graph tells us that $s$ is a relation on $A = \{1, 2, 3\}$ and that $s = \{(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 3)\}$.
We will be building on the next example in the following section.
##### Example6.2.5Ordering subsets of a two element universe
Let $B = \{1,2\}$, and let $A = \mathcal{P}(B) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$. Then $\subseteq$ is a relation on $A$ whose digraph is Figure 6.2.6.
We will see in the next section that since $\subseteq$ has certain structural properties that describe “partial orderings.” We will be able to draw a much simpler type graph than this one, but for now the graph above serves our purposes.
##### 1
Let $A = \{1, 2, 3, 4\}$, and let $r$ be the relation $\leq$ on $A\text{.}$ Draw a digraph for $r\text{.}$
##### 2
Let $B = \{1,2, 3, 4, 6, 8, 12, 24\}$, and let $s$ be the relation “divides” on $B\text{.}$ Draw a digraph for $s\text{.}$
##### 3
Let $A=\{1,2,3,4,5\}$. Define $t$ on $A$ by $a t b$ if and only if $b - a$ is even. Draw a digraph for $t\text{.}$
1. Let $A$ be the set of strings of 0's and 1's of length 3 or less. Define the relation of $d$ on $A$ by $x d y$ if $x$ is contained within $y\text{.}$ For example, $01 d 101$. Draw a digraph for this relation.
2. Do the same for the relation $p$ defined by $x p y$ if $x$ is a prefix of $y\text{.}$ For example, $10 p 101$, but $01 p 101$ is false.
Recall the relation in Exercise 5 of Section 6.1, $\rho$ defined on the power set, $\mathcal{P}(S)$, of a set $S\text{.}$ The definition was $(A,B) \in \rho$ iff $A\cap B = \emptyset$. Draw the digraph for $\rho$ where $S = \{a, b\}$.
Let $C= \{1,2, 3, 4, 6, 8, 12, 24\}$ and define $t$ on $C$ by $a t b$ if and only if $a$ and $b$ share a common divisor greater than 1. Draw a digraph for $t\text{.}$
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# The representation of square root quasi-pseudo-MV algebras
Research paper by Wenjuan Chen, Wieslaw A. Dudek
Indexed on: 01 Oct '14Published on: 01 Oct '14Published in: Soft Computing
#### Abstract
$$\sqrt{'}$$ quasi-MV algebras arising from quantum computation are term expansions of quasi-MV algebras. In this paper, we introduce a generalization of $$\sqrt{'}$$ quasi-MV algebras, called square root quasi-pseudo-MV algebras ($$\sqrt{\hbox {quasi-pMV}}$$ algebras, for short). First, we investigate the related properties of $$\sqrt{\hbox {quasi-pMV}}$$ algebras and characterize two special types: Cartesian and flat $$\sqrt{\hbox {quasi-pMV}}$$ algebras. Second, we present two representations of $$\sqrt{\hbox {quasi-pMV}}$$ algebras. Furthermore, we generalize the concepts of PR-groups to non-commutative case and prove that the interval of a non-commutative PR-group with strong order unit is a Cartesian $$\sqrt{\hbox {quasi-pMV}}$$ algebra. Finally, we introduce non-commutative PR-groupoids which extend abelian PR-groupoids and show that the category of negation groupoids with operators and the category of non-commutative PR-groupoids are equivalent.
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#### ${{\mathit \Xi}_{{c}}{(2970)}^{0}}$ WIDTH
VALUE (MeV) EVTS DOCUMENT ID TECN COMMENT
$14.1$ $\pm0.9$ $\pm1.3$ 11.7k 1
2020 X
LHCB ${{\mathit p}}{{\mathit p}}$ at 13 TeV
$30.3$ $\pm2.3$ ${}^{+1.0}_{-1.8}$ 1443
2016
BELL ${{\mathit e}^{+}}{{\mathit e}^{-}}$ , ${{\mathit \Upsilon}}$ regions
• • We do not use the following data for averages, fits, limits, etc. • •
$31$ $\pm7$ $\pm8$ $67$ $\pm44$
2008 J
BABR ${{\mathit e}^{+}}{{\mathit e}^{-}}$ $\approx{}$ 10.58 GeV
$15$ $\pm6$ $\pm3$ $57$ $\pm13$
2008
BELL ${{\mathit e}^{+}}{{\mathit e}^{-}}$ $\approx{}{{\mathit \Upsilon}{(4S)}}$
1 Further studies are required to establish whether the narrow resonance at 2965 MeV is a different baryon from the narrow resonance at 2970 MeV seen by YELTON 2016 .
References:
AAIJ 2020X
PRL 124 222001 Observation of New $\Xi_c^0$ Baryons Decaying to $\Lambda_c^+ K^-$
YELTON 2016
PR D94 052011 Study of Excited ${{\mathit \Xi}_{{c}}}$ States Decaying into ${{\mathit \Xi}_{{c}}^{0}}$ and ${{\mathit \Xi}_{{c}}^{+}}$ Baryons
AUBERT 2008J
PR D77 012002 Study of Excited Charm-Strange Baryons with Evidence for New Baryons ${{\mathit \Xi}_{{c}}{(3055)}^{+}}$ and ${{\mathit \Xi}_{{c}}{(3123)}^{+}}$
LESIAK 2008
PL B665 9 Measurement of Masses of the ${{\mathit \Xi}_{{c}}{(2645)}}$ and ${{\mathit \Xi}_{{c}}{(2815)}}$ Baryons and Observation of ${{\mathit \Xi}_{{c}}{(2980)}}$ $\rightarrow$ ${{\mathit \Xi}_{{c}}{(2645)}}{{\mathit \pi}}$
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Publication
Title
Chemical activity of the peroxide/oxide redox couple : case study of $Ba_{5}Ru_{2}O_{11}$ in aqueous and organic solvents
Author
Abstract
The finding that triggering the redox activity of oxygen ions within the lattice of transition metal oxides can boost the performances of materials used in energy storage and conversion devices such as Li-ion batteries or oxygen evolution electrocatalysts has recently spurred intensive and innovative research in the field of energy. While experimental and theoretical efforts have been critical in understanding the role of oxygen nonbonding states in the redox activity of oxygen ions, a clear picture of the redox chemistry of the oxygen species formed upon this oxidation process is still missing. This can be, in part, explained by the complexity in stabilizing and studying these species once electrochemically formed. In this work, we alleviate this difficulty by studying the phase Ba5Ru2O11, which contains peroxide O-2(2-) groups, as oxygen evolution reaction electrocatalyst and Li-ion battery material. Combining physical characterization and electrochemical measurements, we demonstrate that peroxide groups can easily be oxidized at relatively low potential, leading to the formation of gaseous dioxygen and to the instability of the oxide. Furthermore, we demonstrate that, owing to the stabilization at high energy of peroxide, the high-lying energy of the empty sigma* antibonding O-O states limits the reversibility of the electrochemical reactions when the O-2(2-)/O2- redox couple is used as redox center for Li-ion battery materials or as OER redox active sites. Overall, this work suggests that the formation of true peroxide O-2(2-) states are detrimental for transition metal oxides used as OER catalysts and Li-ion battery materials. Rather, oxygen species with O-O bond order lower than 1 would be preferred for these applications.
Language
English
Source (journal)
Chemistry of materials / American Chemical Society. - Washington, D.C., 1989, currens
Publication
Washington, D.C. : 2018
ISSN
0897-4756 [print]
1520-5002 [online]
Volume/pages
30 :11 (2018) , p. 3882-3893
ISI
000435416600038
Full text (Publisher's DOI)
Full text (publisher's version - intranet only)
UAntwerpen
Faculty/Department Research group Project info ARPEMA: Anionic redox processes: A transformational approach for advanced energy materials Publication type Subject Affiliation Publications with a UAntwerp address
External links
Web of Science
Record
Identifier Creation 10.07.2018 Last edited 15.11.2022 To cite this reference
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# Adopting Theon – Part 3.1
The first part of this series discussed why you might use Theon as a Schema Management Tool for your PostgreSQL database. The second part part covered the necessary packages and installation. This final part will be an introductory tutorial.
In the first part of this tutorial we will create a database, manage it in Theon, make some changes, and use the factory to template some schema. In the second part we will use TheonCoupler to load data automatically into the database. In the third part we will use TheonUI and create a desktop for the database. In the fourth and final part we will package and distribute the result.
Before doing anything note that we need a running PostgreSQL installation we have access on to create databases with. In this tutorial we use on one localhost with /disk/scratch as the socket connection directory.
First we will create a simple database with one table. This will be used to hold installed packages.
createdb packages
psql packages <<EOF
CREATE TABLE package (
name TEXT,
version TEXT,
release TEXT,
architecture TEXT
);
EOF
Create the Theon management database.
ttkm create
Now create a directory to hold our ModelLibrary.
cd /tmp
ttkm packages launch at .
cd packages
Import the database we have just created as the schema. In order to do this we need to create a default LiveProfile with the necessary connection details.
ttkm attach using database=packages host=/disk/scratch
ttkm import
Next test what we imported by exporting the schema.
ttkm export
cat schemat/relation/package.xsd
The above is the XML representation of the table we created. We will properly check now by dropping the original database then, by deriving the necessary SQL, recreate it.
dropdb packages
ttkm install
psql packages -c "\d"
List of relations
Schema | Name | Type | Owner
--------+-----------+-------+-------
public | _th_theon | table | timc
public | package | table | timc
(2 rows)
Note the additional metadata table Theon adds. This is not mandatory(and can be removed) but it is useful for subsequent upgrades as we will see shortly.
Next we will define a view using the factory. To do this we first create the necessary hierarchy and add a template definition.
mkdir -p factory/templates
cat >factory/templates/archcount.fat <<EOF
%%include "relation/view.grg"
%%equate relation_ViewName "archcount"
%%equate relation_ViewValueBlock "SQL"
%%block SQL
SELECT
architecture AS arch,
COUNT(*) AS total
FROM package
GROUP BY
arch
%%end
EOF
Then we process the factory templates and load them into the Theon management database.
ttkm gather
Next export the changed schema and check.
ttkm export
cat schemat/relation/archcount.xsd
Now we need to upgrade the installed database to add the view. This is automatic as we have the metadata table, manual machinations would be required otherwise.
ttkm upgrade
psql packages -c "\d"
List of relations
Schema | Name | Type | Owner
--------+-----------+-------+-------
public | _th_theon | table | timc
public | archcount | view | timc
public | package | table | timc
(2 rows)
The view has been successfully added.
Finally we want to add an extra column in the package table for the build time. There are various ways of doing this, here we will alter the live database and re-import the change so the schema in Theon is kept up-to-date. However, in order to keep consistent metadata, we will drop the column and then re-create the column by upgrade. This is necessary in this case as we are working with a development database which is also the live database.
psql packages -c "ALTER TABLE package ADD COLUMN buildtime INTEGER;"
ttkm import
ttkm export
psql packages -c "ALTER TABLE package DROP COLUMN buildtime;"
buildtime | integer |
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# Adjoints in the 2-category of 2-vector spaces
See here for the notation.
I'm trying to do this by myself but everytime I approach the problem I end up buried under computations.
What is a pair of adjoint 1-cells in the 2-category of vector spaces, where
• 0-cells are natural numbers $\langle N\rangle,\langle M\rangle...$
• 1-cells $\langle N\rangle\to \langle M\rangle$ are strictly-positive-integer valued matrices of size $M\times N$
• 2-cells $A\to B$ are matrices whose entries are matrices, whose $(i,j)$ entry is a matrix having size $A_{ij}\times B_{ij}$.
?
• Why strictly positive integers and not nonnegative integers? – Qiaochu Yuan Jul 1 '18 at 19:34
As in the previous question, I continue to suggest that everything is much cleaner if you think in terms of bimodules. So we'll ask the more general question: in the 2-category $\text{Bim}(k)$ of $k$-algebras, $k$-bimodules, and bimodule homomorphisms, what is an adjoint pair?
The answer is worked out in this blog post, although I'll warn you that in the convention of that blog post composition of 1-morphisms is written in diagrammatic order, which means concretely that the composition of an $(A, B)$-bimodule $M$ and a $(B, C)$-bimodule $N$, in that order, is $M \otimes_B N$. Adopting the usual composition convention switches the meaning of "left" and "right," but this convention makes things work out very nicely as follows.
Proposition: An $(A, B)$-bimodule $M$ has a left adjoint iff it is finitely presented projective as a left $A$-module, in which case its left adjoint is $\text{Hom}_A(M, A)$. It has a right adjoint iff it is finitely presented projective as a right $B$-module, in which case its right adjoint is $\text{Hom}_B(M, B)$.
KV is the special case where we only consider algebras of the form $k^n$ for $k$ a field and $(k^n, k^m)$-bimodules, which concretely can be thought of as $n \times m$ matrices of $k$-vector spaces, where all of the vector spaces are finite-dimensional. These are all finitely presented projective on both sides, and so we conclude that every morphism of KV 2-vector spaces has both a left and a right adjoint, both of which I believe are concretely given by sending a matrix $V_{ij}$ of vector spaces to the transposed and dualized matrix $V_{ji}^{\ast}$, but I haven't checked this in detail.
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My Math Forum Fermat Last Theorem Diophantine Analysis
Number Theory Number Theory Math Forum
May 10th, 2019, 11:02 PM #31
Senior Member
Joined: Aug 2012
Posts: 2,356
Thanks: 738
Quote:
Originally Posted by youngmath ok the first equation is true for positive x and y and positive n > 1 the second equation is true for positive x and y and positive n > 2
If you have a proof for (2) then someone -- I wish it didn't have to be me but I'm in for a penny in for a pound in this thread -- will have to work through the algebra to find the mistake.
@michaelweir We know there's a mistake because we saw a few days back that no proof based on elementary algebraic manipulations can work. And there's the meta-reason that if there were in fact an elementary algebraic proof, someone would have found it in 350 years.
So there is a mistake in there, and I was hoping (1) or (2) would be false, but if they are both true then somebody needs to dive in. Any volunteers?
Last edited by Maschke; May 10th, 2019 at 11:05 PM.
May 11th, 2019, 08:35 AM #32 Senior Member Joined: Mar 2019 From: iran Posts: 318 Thanks: 14 the post #2 shows the proof and the final result are wrong Last edited by youngmath; May 11th, 2019 at 08:46 AM.
May 13th, 2019, 11:11 AM #33 Member Joined: Mar 2019 From: california Posts: 74 Thanks: 0 Masche - Thank you for spending the time you have spent on this proof. I THINK YOU WILL BE REWARDED FOR SPENDING THE TIME. I haven't spent the time to rebut the argument to show that proof is outside the parameters of an algebraic manipulation, as is this one. To make the claim that this proof about algebraic applies you will have to label every statement in a proof is an example of a algebraic manipulation. I don't think you can do that. The Nautilus construction is much more than a way to solve FLT. I have come up with a way to combine the number line analysis of this proof with the Nautilus to come up with another FLT. There are probably other proofs of FLT using the Nautilus. It's like using a fishing net of the ocean of Mathematics, instead of a single fishing line. Who knows what fish will be caught next? Last edited by skipjack; June 4th, 2019 at 02:30 PM.
June 4th, 2019, 01:59 PM #34 Member Joined: Mar 2019 From: california Posts: 74 Thanks: 0 Just saying something doesn't make it right. I ignored replying until now, to see if you would have something to say. You didn't. Nor has anyone else made any comments. So, am I right in believing that the proof is right?
June 4th, 2019, 02:29 PM #35 Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timey-wimey stuff. I can't speak for anyone else but I think the silence is more likely that we've given up on your "simple" approach to a problem that has been proven not to have one. -Dan
June 4th, 2019, 03:05 PM #36 Senior Member Joined: Mar 2019 From: iran Posts: 318 Thanks: 14 in the second pdf equation 5 is wrong because you wrote e/f*h instead of e/f*h*y
June 6th, 2019, 04:15 PM #37 Member Joined: Mar 2019 From: california Posts: 74 Thanks: 0 youngmath Yes i made an error in equation 4, where I should have factored the Y out then. So, th error was corrected by taking it out equation 5. So the proof is still good. I am looking for someone to help me with writing the proof in an acceptable from to submit to a mathmatical journal. I will share credit and prizes with whoever helps. the field medal is worth $700,000 at leqst. June 6th, 2019, 06:01 PM #38 Senior Member Joined: Mar 2019 From: iran Posts: 318 Thanks: 14 equation 3 h*y*(c^2/d^2 * h * y - 2 c/d * y - e/f * h * y) = -2 y^2 multiply h*y by the bracket in LHS h^2 * y^2 * c^2/d^2 - 2 * h * y^2 * c/d - h^2 * y^2 * e/f = -2 y^2 factor y^2 out of both sides h^2 * c^2/d^2 - 2 * h * c/d - h^2 * e/f = -2 as you see y doesn't remain in the equation June 6th, 2019, 06:39 PM #39 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 https://www.wolframalpha.com/input/?...-2+y%5E2+for+y Equation simplifies to: y^2(c^2*h^2/d^2 - 2*c*h/d - e*h^2/f + 2)=0 Last edited by Denis; June 6th, 2019 at 06:51 PM. June 6th, 2019, 07:58 PM #40 Member Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 Math Focus: Algebraic Geometry Quote: Originally Posted by michaelcweir ...writing the proof in an acceptable from to submit to a mathmatical journal. Well, the first step for that would be to learn how to use$\LaTeX\$. If you wrote your post in latex you would be able to copy paste into a latex document. Also, if you rewrote your post with it i'm sure a lot more people would be willing to read through it since it wouldn't be a haphazard mess of "*"s and "^"'s.
Tags analysis, diophantine, diopphantine, fermat, theorem
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# Newspaper route: how many days to save $1000? #### RockerChick ##### New member Dwayne has a newspaper route for which he collects k dollars each day. From this amount he pays out k/3 dollars per day for the cost of the papers, and he saves the rest of the money. In terms of k, how many days will it take Dwayne to save$1,000 ?
(a) k/1,500
(b) k/1,000
(c) 1,000/k
(d) 1,500/k
(e) 1,500k
Please show how to solve...my inclination is to plug in a number for k...and then solve that way, but it didnt seem to work
#### Mrspi
##### Senior Member
Re: Newspaper route
RockerChick said:
Dwayne has a newspaper route for which he collects k dollars each day. From this amount he pays out
k/3 dollars per day for the cost of the papers, and he saves the rest of the money. In terms of k, how many days
will it take Dwayne to save $1,000 ? (a) k/1,500 (b) k/1,000 (c) 1,000/k (d) 1,500/k (e) 1,500k Please show how to solve...my inclination is to plug in a number for k...and then solve that way, but it didnt seem to work Dwayne takes in k dollars per day, but he must pay k/3 dollars each day for the papers. So, the amount he actually earns is k - (k/3) dollars each day. Write k as 3k/3, and subtract: (3k/3) - (k/3) = (2k)/3 Dwayne's net earnings for one day is (2k)/3 dollars. Dwayne would like to make$1000. How many days will it take?
Let x = number of days to make $1000 He makes (2k)/3 dollars each day for x days, and this should be$1000. So,
(2k)/3 *(x) = 1000
Can you solve this for x?
If not, please repost showing each step you took to solve (2k)/3 * (x) = 1000. When we can see your work, we can better determine how to help you.
#### stapel
##### Super Moderator
Staff member
RockerChick said:
Please show how to solve...my inclination is to plug in a number for k...and then solve that way, but it didnt seem to work
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a die is rolled five times, what is the probability that exactly two results=3?
A die is rolled five times. What is the probability that exactly two of the results are equal to three?
We are taught to draw the trees and calculate it this way, however for a fair die being rolled five times, the diagram would be very messy. Is there another way to solve this?
• This is a straight-out-of-the-box application of the binomial distribution. Nov 5, 2017 at 1:23
• Wiki link to the binomial distribution. Included in the link are examples. Nov 5, 2017 at 1:27
• Can you refer me to how it cant be applied? or where I can see it? thanks! Nov 5, 2017 at 1:27
• The wiki link includes the example about an unfair coin and getting a particular number of heads out of a certain number of flips. Instead of asking about getting a particular number of heads, here we are asking about getting a particular number of "threes." The example is exactly like this one with the only differences being flavor and the number of times the die is rolled (or coin is flipped). Nov 5, 2017 at 1:47
• The diagram’s not that messy. You only have two possibilities that you care about at each branch point: whether or not a three was rolled, so the tree will have $2^5=32$ leaves, which isn’t all that big to me.
– amd
Nov 5, 2017 at 17:41
$$P(success) = {5 \choose 2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{5-2}$$
$$P(success) = 10\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3$$
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## Nonlinear Equation Solver
Let $f(x)$ be a twice differentiable and real-valued function, this application find the roots $x^\star$ of $f$ such that $f(x^\star) = 0$ by using a variant of Newton's method. In the calculation the solver uses both the first derivative and the second derivative of $f$ and they are calculated using Derivative Calculator. As a globally convergent solver, this calculator should be able to find the root $x^\star$ of $f$ from any initial guess $x_0$ which is not necessary to close to the root $x^\star$, but $x_0$ must be in the domain of $f$.
### Input Data
The calculator uses the same input format as Derivative Calculator. It is not necessary to input $x_0$ if the interval $(0,1)$ is a subset of the domain of $f(x)$. If $x_0$ is not entered or the solver cannot find the root from $x_0$ input by users, new $x_0\in(0,1)$ will be generated automatically.
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What's new
# Geometric Returns of negative interest rates
#### Ashok_Kothavle
##### Member
Dear Mr David
Kevin Dowd in his book Measuring Market Risk (2nd Edition) has mentioned the advantages of using the Geometric returns over Arithmetic returns (3rd chapter). In fact I always quote following example –
Suppose an asset was trading at the prices as given below –
May 15, 2017 - $50 May 14, 2017 -$100
May 13, 2017 - \$50
If I consider the Arithmetic returns, my return on May 14th over May 13th was (100-50)/50*100 = 100%. The return on May 15th over May 14th is (50-100)/100*100 = -50%. Hence, my average return is (100%+(-50%))/2 = 25% (Do understand it’s a crude method of arriving at average return). However this is misleading as asset was trading at 50 and now its trading at the same level i.e. 50. Hence, my actual return is zero.
On the other hand, if I use Geometric return, my return on May 14th over May 13th = LN(100/50) = 69.31%. Similarly, my return on May 15th over May 14th is = LN(50/100) = - 69.31% and hence my average return = (69.31%-69.31%)/2 = 0.
As Mr Dowd had mentioned, for longer horizon, we must use Geometric returns than Arithmetic returns. However, problem is suppose EURO overnight LIBOR is one of the risk factors.
EURO overnight LIBOR rates are as given below –
May 12, 2017 - 0.42571%
May 11, 2017 - 0.42214%
May 10, 2017 - 0.42571%
......... and so on
In this case, the rates are negative, however, still computing the Geometric returns won’t be a problem as the ratio is positive and we can take log of positive value only. However, if I have mix rates like some small positive values, some small negative values etc, then sometimes the ratio will be negative and it won’t be possible to obtain log value of negative ratio.
From academic point of view, can you comment on this or suggest some resource for the same i.e. how to obtain returns if the risk factor is a series of mix rates i.e. positive and negative or for that matter even ‘0’ value (may be hypothetical).
Regards
Ashok
Last edited:
#### Matthew Graves
##### Active Member
Subscriber
Not sure about the arithmetic average of the log return, seems like a strange thing to do to me. Are you instead trying to find the geometric average?
In your first example this would be:
$\sqrt{(1+1)(1-0.5)}-1=0$
which is what you were expecting.
You can follow the same procedure for the interest rates.
#### Ali Ehsan Abbas
##### New Member
Geometric return = (ending value/beginning value)^(1/N) - 1
#### Ashok_Kothavle
##### Member
Not sure about the arithmetic average of the log return, seems like a strange thing to do to me. Are you instead trying to find the geometric average?
In your first example this would be:
$\sqrt{(1+1)(1-0.5)}-1=0$
which is what you were expecting.
You can follow the same procedure for the interest rates.
I understand the formula mentioned by you gives holding period Arithemetic Returns with the holding period = 2.
I have mentioned above that it is just a crude method of computing average returns. However, my main concern is computations of log returns of mixture of rates e.g. if rates vary in the range
( - 1% to 1%). If the successive rates differ in sign, how do I compute the log returns (since log returns way of computing returns, being strongly advocated rightly by Mr Kevin Dowd).
Regards
Ashok
#### Ashok_Kothavle
##### Member
Geometric return = (ending value/beginning value)^(1/N) - 1
This formula gives something like CAGR. What if there are the intermittent returns and you need to compute the holding period returns?
Anyways, I repeat my concern is computing log returns for two days e.g. suppose today's rate is 0.5% while yesterdays rate was say -0.11%. How do I compute LN( 0.5/ (-0.11) ) if I have decided to use the log returns and not the Arithmetic returns?
Regards
Ashok
#### Matthew Graves
##### Active Member
Subscriber
Anyways, I repeat my concern is computing log returns for two days e.g. suppose today's rate is 0.5% while yesterdays rate was say -0.11%. How do I compute LN( 0.5/ (-0.11) ) if I have decided to use the log returns and not the Arithmetic returns?
Log returns only make sense when the underlying asset price can be positive only, as Dowd himself says. This not the case for interest rates.
Last edited:
#### Ashok_Kothavle
##### Member
Log returns only make sense when the underlying asset price can be positive only, as Dowd himself says. This not the case for interest rates.
Dear Mr Graves
Totally Agree with you. But the problem is in my portfolio, I may be having equity as well as debt securities. If I need to construct the correlation matrix, I need to have risk factor returns. I can't have equity returns generated using log returns while debt security risk factor returns using say Arithmetic returns. Isn't it?
What Mr Dowd mentioned is -
Hence, if I decide to use log returns, I may have to deal with negative rates too. Unfortunately, we have to face such operational problems in real life world.
Anyways, thanks a lot for your valuable input. Guess we wait for Mr David to respond.
Thanks again
Regards
Ashok
#### emilioalzamora1
##### Well-Known Member
Listen, a logarithm of the returns (in your case 0.5% and -0.11%) makes NO sense at all. A log return is calculated from prices having LN(t:0/t:-1) or written as LN(today's price of the asset/yesterday's price of the asset).
There are several contributions to this topic, let me add few important lines about pros/cons of log returns:
• The assumption of a log-normal distribution of returns, especially over a longer term than daily (say weekly or monthly) is unsatisfactory, because the skew of log-normal distribution is positive, whereas actual market returns for, say, S&P is negatively skewed (because we see bigger jumps down in times of panic).
• It’s difficult to logically combine log returns with fat-tailed distributional assumptions, even for daily returns, although it’s very tempting to do so because assuming “fat tails” sometimes gives you more reasonable estimates of risk because of the added kurtosis.
• The compounded return (having a sequence of returns as a product, (1+r1)*(1+r2)*(1+r3)....is also not satisfactory as the product of normally distributed variables is NOT normal.
Instead, the sum of normally-distributed variables is normal (BUT ONLY when all variables are uncorrelated), which is useful when we recall the following logarithmic identity:
log(1+r) = ln(p at t:0/p at t:-1) = log (p at t:0) - log (p at t:-1)
Example:
Price t:0 = 1430
Price t:-1 = 1409
ln(1430/1409) = 0.0148
log(1430) - log(1409) = 0.0064
log(1+0.0148) = 0.00638
See the very close difference between 0.0064 and 0.00638!
Having said that, the below identity leads us to a pleasant algorithmic benefit; formula for calculating compounded returns:
log(1+rintial) + log (1+r2)........+log(1+rfinal) = log (p initial) - log (p final)
wich is the compounded return over n periods is merely the difference in log between initial and final periods.
where
1. rinitial = initial return of the sample
2. rfinal = final return of the sample
3. p initial = initial price (of the asset) of the sample
4. p final = last price (of the asset) of the sample
Pro log-returns: The approximate raw-log equality says that when returns are very small (common for trades with short holding durations), the following approximation ensures they are close in value to raw returns: log(1+r) ~ r, r < 1.
Last edited:
#### Ashok_Kothavle
##### Member
Listen, a logarithm of the returns (in your case 0.5% and -0.11%) makes NO sense at all. A log return is calculated from prices having LN(t:0/t:-1) or written as LN(today's price of the asset/yesterday's price of the asset).
There are several contributions to this topic, let me add few important lines about pros/cons of log returns:
• The assumption of a log-normal distribution of returns, especially over a longer term than daily (say weekly or monthly) is unsatisfactory, because the skew of log-normal distribution is positive, whereas actual market returns for, say, S&P is negatively skewed (because we see bigger jumps down in times of panic).
• It’s difficult to logically combine log returns with fat-tailed distributional assumptions, even for daily returns, although it’s very tempting to do so because assuming “fat tails” sometimes gives you more reasonable estimates of risk because of the added kurtosis.
• The compounded return (having a sequence of returns as a product, (1+r1)*(1+r2)*(1+r3)....is also not satisfactory as the product of normally distributed variables is NOT normal.
Instead, the sum of normally-distributed variables is normal (BUT ONLY when all variables are uncorrelated), which is useful when we recall the following logarithmic identity:
log(1+r) = ln(p at t:0/p at t:-1) = log (p at t:0) - log (p at t:-1)
Example:
Price t:0 = 1430
Price t:-1 = 1409
ln(1430/1409) = 0.0148
log(1430) - log(1409) = 0.0064
log(1+0.0148) = 0.00638
See the very close difference between 0.0064 and 0.00638!
Having said that, the below identity leads us to a pleasant algorithmic benefit; formula for calculating compounded returns:
log(1+rintial) + log (1+r2)........+log(1+rfinal) = log (p initial) - log (p final)
wich is the compounded return over n periods is merely the difference in log between initial and final periods.
where
1. rinitial = initial return of the sample
2. rfinal = final return of the sample
3. p initial = initial price (of the asset) of the sample
4. p final = last price (of the asset) of the sample
Pro log-returns: The approximate raw-log equality says that when returns are very small (common for trades with short holding durations), the following approximation ensures they are close in value to raw returns: log(1+r) ~ r, r < 1.
Dear emilioalzamora1
If you allow, I wish to bring to your kind notice few points -
(1) I have initiated this thread by quoting the overnight EURO Libors which are negative. Though pure intention of raising this thread was totally academic, however, this is actual problem we are facing. No doubt the EURO Libors are having negative values. However, one can't deny the fact that going forward depending on economic conditions, they may become positive rates. You may note that I have used the word "Hypothetical" too in the thread.
I am assuming such an hypothetical situation (though very rare, but can't rule out the possibility) where the rates became positive from negative. I am considering this instance and I am assuming very small values in the neighborhood of zero. Hence I have mentioned the rates 0.5% (say at t) and -0.11% (say at t-1).
(2) You have mentioned that "Listen, a logarithm of the returns (in your case 0.5% and -0.11%) makes NO sense at all".
Response :
Please note that I have used the word "rates". These are not returns. I am not computing the log of returns.
(3) The assumption of a log-normal distribution of returns, especially over a longer term than daily (say weekly or monthly) is unsatisfactory, because the skew of log-normal distribution is positive, whereas actual market returns for, say, S&P is negatively skewed (because we see bigger jumps down in times of panic).
Response :
In my thread nowhere I have mentioned that I intend to use log normal distribution. If I use Var covar or for that matter Monte carlo, I may assume risk factors returns are normally distributed (i.e. prices or rates follow log normal).
If I use Historical simulation, I still compute the returns and for 1 day horizon, I assume that tomorrow one of these past scenarios will occur and these scenarios are nothing but the returns observed over last one year.
Is it that in case of Historical simulation, I can't compute returns using log of ratio of prices (or rates)? At least I am not aware. Many spreadsheets which are available on internet for Historical simulation do use log returns.
(4) the sum of normally-distributed variables is normal (BUT ONLY when all variables are uncorrelated)
Response :
I am exactly not sure about this. I feel any linear combination of two Normal variables is Normal. If these two variables are not independent i.e. "Rho" is non zero,
May be (X + Y) ~ N(MEAN = mean X + mean Y, VARIANCE = V(X) + V(Y) + 2*Cov(X,Y)) where Cov(X, Y) = Rho * Sqrt(V(X) * V(Y)).
But honestly I am not sure about this. Sort of gut filling and I will definitely come back over this.
(5) I am bit confused with your example
ln(p at t:0/p at t:-1) = log (p at t:0) - log (p at t:-1)
Example:
Price t:0 = 1430
Price t:-1 = 1409
ln(1430/1409) = 0.0148
log(1430) - log(1409) = 0.0064
log(1+0.0148) = 0.00638
Response :
ln(p at t:0/p at t:-1) = log (p at t:0) - log (p at t:-1)
Here are considering base = 10 on RHS of equation?
I am getting LHS = ln(p at t:0/p at t:-1) = 0.014794 which is not equal to RHS if we consider the base = 10.
This is my earnest request to you can you elaborate if possible why we are using Log with base 10. Am aware of the relation
LN(X) = Log(X, base = 10) / Log(2.718282, base 10) or LOG(X, Base = 10) = LN(X) / LN(10). Does this relationship to play any role here. Honestly, I am ignorant about this esp
log(1+0.0148) = 0.00638 = log(1 + LN(1430/1409)). I wasn't aware of this actually. Qyestion is why do we need to do it.
Its a well known fact that
LN(1 + Arithmetic Return) = Geometric Return
Is this what you are trying to lead to?
And finally
Log(p1 / p10) = Log(p1/p2*p2/P3*.....*P9/P10)
= Log(p1/p2) + Log(p2/p3) .........+Log(p9/p10)
= Log(p1) - Log(p2) + log(p2) - log(p3) + log(p3) - log(p4) +..........- log(p9) + log(9) - log(p10)
= Log(p1) - Log(p10)
With warm regards
Ashok
#### Ashok_Kothavle
##### Member
Hi everyone,
Thanks a lot for your valuable feedback. This forum does indeed help persons like me to clarify so many doubts. May be I need to get my concepts clear. May I suggest that we wait for Mr David to respond on this.
Thanks again
Regards
Ashok
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Hi @Ashok_Kothavle With respect to your original post, I don't recall ever seeing interest rates as direct inputs into a log (ie, continuously compounded returns). As mentioned above, my experience with the log return is ln[P(1)/P(0)] or ln([P(1)+D]/P(0)], which is to say, using prices not interest rates. My first instinct here is, if you really want to do this, can't you just transform (translate) the interest rates into prices, even if for artificial purposes? After all, this is really what a Treasury bill price does for a rate, or a Eurodollar futures price does for a rate. Any given interest rate can be transformed into many different non-zero prices; e.g., 2% --> (100 -2) = 98.00 like a money market "discount" instrument; or r --> exp(-r*T) as a zero-coupond bond. In short, any f(.) could transform the rate. As ln(P1/P0) = r such that exp(r) = P1/P0, could you just use P1 = P(0)*exp(r); i.e., assume P(0) is 100 or index, and transform 100 into 100*exp(r*a) where constant a is a sensitivity. (?)
Otherwise, I'd be tempted to push back on trying too hard to take log returns of rates; e.g., why aren't simple differences better?
FWIW, when I consulted and we generated performance measures, we did not try to represent returns (i) any time that values crossed the zero axis (e.g., positive to negative or vice verse) or (ii) even when (too) small bases were involved--e.g., P1 growing from a P0 that is too small--because our examples showed the output tended to distort the findings. I hope that's helpful!
Last edited:
#### Ashok_Kothavle
##### Member
Dear Mr David,
OMG I get so much to learn from this forum. I am sure, if I continue with BT for another few months, I shall be an enlightened person. Thanks for your valuable inputs. Those will go long in refining my whole approach towards Market Risk and overall financial risk management. I have so much to learn esp when it comes to Market risk.
One of my resource has suggested me to use change in basis instead of computing returns when it comes to interest rates.
Thanks again.
Regards
Ashok
#### Ashok_Kothavle
##### Member
Dear emilioalzamora1
Thanks a lot for your guidance too.
Regards
Ashok
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
@Ashok_Kothavle Sure thing, thank you for asking a thoughtful question. BTW, I completely agree with "change in basis:" it is simple and easy to understand! Thank you,
#### Ashok_Kothavle
##### Member
Through some resource, I got following feedback –
“The return on interest rates should be calculated as a simple difference between two rates. Interest rates themselves are treated as some kind of Return instead of Asset price. Hence, use of Arithmetic as well as Geometric returns in case of interest rates is not proper.”
E.g. in the first message, the Overnight EURO rates have been quoted as.
May 12, 2017 ~ (– 0.43571%)
May 11, 2017 ~ (– 0.42214%)
Thus the return is simply (– 0.43571% – (– 0.42214%)) = - 0.01357%
emilioalzamora has rightly mentioned “logarithm of the returns (in your case 0.5% and -0.11%) makes NO sense at all”.
It was my ignorance. Thanks again Mr David and emilioalzamora.
Regards
Ashok
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# QPainter::drawText(QRectF ...) draws with very unequally spaces between the characters
• Hi!
When I use the font "Corbel" and draw a text like the following
``````QRectF testRectF = QRectF(QPointF1, QPointF2);
painter->drawText(testRectF, Qt::AlignVCenter | Qt::AlignLeft, "some text");
``````
I get really ugly results like the following
There is more space between the 'i' and the 'a' for example than between the words "counter" and "variable" (there is a actual space between them, not between the 'i' and the 'a'). So, the spaces between the characters are very unequally.
When I use the font Arial, it is better, but still wrong, so it's not the font itself.
Has anyone an idea what the problem could be?
Thanks!
• Hi,
IIRC, you need to use a monospaced font for that kind of use case.
Hope it helps
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# Convert binary (integer and a fraction) from vhdl to decimal in C code, negative value
Status
Not open for further replies.
#### wannaknow
##### Junior Member level 2
Convert binary ( integer and fraction) from vhdl to decimal in C code,negative value
HI,
I have a 14-bit data that is fed from FPGA in vhdl, The NIos II processor reads the 14-bit data from FPGA and do some processing tasks, where Nios II system is programmed in C code
The 14-bit data can be positive, zero or negative. In Altera compiler, I can only define the data to be 8,16 or 32. So I define this to be 16 bit data.
First, I need to check if the data is negative, if it is negative, I need to pad the first two MSB to be bit '1' so the system detects it as negative value instead of positive value.
Second, I need to compute the real value of this binary representation into a decimal value of BOTH integer and fraction.
I learned from this link (https://stackoverflow.com/questions...ary-floating-point-1101-11-into-decimal-13-75) that I could convert a binary (consists of both integer and fraction) to decimal values.
I am wondering if this code can be used to check for negative value? I did try with a binary string of 11111101.11 and it gives the output of 253.75...
I have two questions:
What are the modifications I need to do in order to read a negative value?
I know that I can do the bit shift (as below) to check if the msb is 1, if it is 1, I know it is negative value...
Code:
if (14bit_data & 0x2000) //if true, it is negative value
The issue is, since it involves fraction part (but not only integer), it confused me a bit if the method still works...
If the binary number is originally not in string format, is there any way I could convert it to string? The binary number is originally fed from a fpga block say, 14 bits, with msb as the sign bit, the following 6 bits are the magnitude for integer and the last 6 bits are the magnitude for fractional part.
Last edited:
#### wannaknow
##### Junior Member level 2
How to read a 16-bit data into string in C code
I have a 16-bit data that is fed from FPGA in vhdl, the NIos II processor reads the 16-bit data from FPGA and do some processing tasks, where Nios II system is programmed in C code
How could I read this 16-bit data in string?
Code:
printf("%d", 16_bit_data);
// it shows a decimal value of 15282
What should I do in order to get the result displayed as a string 0011101110110010?
Last edited:
#### xenos
##### Full Member level 4
Re: How to read a 16-bit data into string in C code
Translate to hex (sprintf("%04x",v)) and then translate each hex digit to binary, or use something like the function
#### wannaknow
##### Junior Member level 2
Re: How to read a 16-bit data into string in C code
Translate to hex (sprintf("%04x",v)) and then translate each hex digit to binary, or use something like the function
Thanks for your reply, forgive me for didnt make it clear enough.
I need to compute the real value of this binary representation into a decimal value of BOTH integer and fraction, for example, the value is 15282 (0011 1011 1011 0010), but it is not the real value, how to make sure I will get the value as -30.750015?
The 14-bit-data is defined as follow: with msb as the sign bit, the following 6 bits are the magnitude for integer and the last 6 bits are the magnitude for fractional part.
#### FvM
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# 15. Functions¶
In the late nineteenth century, developments in a number of branches of mathematics pushed towards a uniform treatment of sets, functions, and relations. We have already considered sets and relations. In this chapter, we consider functions and their properties.
A function, $$f$$, is ordinary understood as a mapping from a domain $$X$$ to another domain $$Y$$. In set-theoretic foundations, $$X$$ and $$Y$$ are arbitrary sets. We have seen that in a type-based system like Lean, it is natural to distinguish between types and subsets of a type. In other words, we can consider a type X of elements, and a set A of elements of that type. Thus, in the type-theoretic formulation, it is natural to consider functions between types X and Y, and consider their behavior with respect to subsets of X and Y.
In everyday mathematics, however, set-theoretic language is common, and most mathematicians think of a function as a map between sets. When discussing functions from a mathematical standpoint, therefore, we will also adopt this language, and later switch to the type-theoretic representation when we talk about formalization in Lean.
## 15.1. The Function Concept¶
If $$X$$ and $$Y$$ are any sets, we write $$f : X \to Y$$ to express the fact that $$f$$ is a function from $$X$$ to $$Y$$. This means that $$f$$ assigns a value $$f(x)$$ in $$Y$$ to every element $$x$$ of $$X$$. The set $$X$$ is called the domain of $$f$$, and the set $$Y$$ is called the codomain. (Some authors use the word “range” for the codomain, but today it is more common to use the word “range” for what we call the image of $$A$$ below. We will avoid the ambiguity by avoiding the word range altogether.)
The simplest way to define a function is to give its value at every $$x$$ with an explicit expression. For example, we can write any of the following:
• Let $$f : \mathbb{N} \to \mathbb{N}$$ be the function defined by $$f(n) = n + 1$$.
• Let $$g : \mathbb{R} \to \mathbb{R}$$ be the function defined by $$g(x) = x^2$$.
• Let $$h : \mathbb{N} \to \mathbb{N}$$ be the function defined by $$h(n) = n^2$$.
• Let $$k : \mathbb{N} \to \{0, 1\}$$ be the function defined by
$\begin{split}k(n) = \left\{\begin{array}{ll} 0 & \mbox{if n is even} \\ 1 & \mbox{if n is odd.} \end{array}\right.\end{split}$
The ability to define functions using an explicit expression raises the foundational question as to what counts as legitimate “expression.” For the moment, let us set that question aside, and simply note that modern mathematics is comfortable with all kinds of exotic definitions. For example, we can define a function $$f : \mathbb{R} \to \{0, 1\}$$ by
$\begin{split}f(x) = \left\{\begin{array}{ll} 0 & \mbox{if x is rational} \\ 1 & \mbox{if x is irrational.} \end{array}\right.\end{split}$
This is at odds with a view of functions as objects that are computable in some sense. It is not at all clear what it means to be presented with a real number as input, let alone whether it is possible to determine, algorithmically, whether such a number is rational or not. We will return to such issues in a later chapter.
Notice that the choice of the variables $$x$$ and $$n$$ in the definitions above are arbitrary. They are bound variables in that the functions being defined do not depend on $$x$$ or $$n$$. The values remain the same under renaming, just as the truth values of “for every $$x$$, $$P(x)$$” and “for every $$y$$, $$P(y)$$” are the same. Given an expression $$e(x)$$ that depends on the variable $$x$$, logicians often use the notation $$\lambda x \; e(x)$$ to denote the function that maps $$x$$ to $$e(x)$$. This is called “lambda notation,” for the obvious reason, and it is often quite handy. Instead of saying “let $$f$$ be the function defined by $$f(x) = x+1$$,” we can say “let $$f = \lambda \; x (x + 1)$$.” This is not common mathematical notation, and it is best to avoid it unless you are talking to logicians or computer scientists. We will see, however, that lambda notation is built in to Lean.
For any set $$X$$, we can define a function $$i_X(x)$$ by the equation $$i_X(x) = x$$. This function is called the identity function. More interestingly, let $$f : X \to Y$$ and $$g : Y \to Z$$. We can define a new function $$k : X \to Z$$ by $$k(x) = g(f(x))$$. The function $$k$$ is called the composition of $$f$$ and $$g$$ or $$f$$ composed with $$g$$ and it is written $$g \circ f$$. The order is somewhat confusing; you just have to keep in mind that to evaluate the expression $$g(f(x))$$ you first evaluate $$f$$ on input $$x$$, and then evaluate $$g$$.
We think of two functions $$f, g : X \to Y$$ as being equal, or the same function, when for they have the same values on every input; in other words, for every $$x$$ in $$X$$, $$f(x) = g(x)$$. For example, if $$f, g : \mathbb{R} \to \mathbb{R}$$ are defined by $$f(x) = x + 1$$ and $$g(x) = 1 + x$$, then $$f = g$$. Notice that the statement that two functions are equal is a universal statement (that is, for the form “for every $$x$$, …”).
Proposition. For every $$f : X \to Y$$, $$f \circ i_X = f$$ and $$i_Y \circ f = f$$.
Proof. Let $$x$$ be any element of $$X$$. Then $$(f \circ i_X)(x) = f(i_X(x)) = f(x)$$, and $$(i_Y \circ f)(x) = i_Y(f(x)) = x$$.
Suppose $$f : X \to Y$$ and $$g : Y \to X$$ satisfy $$g \circ f = i_X$$. Remember that this means that $$g(f(x)) = x$$ for every $$x$$ in $$X$$. In that case, $$g$$ is said to be a left inverse to $$f$$, and $$f$$ is said to be a right inverse to $$g$$. Here are some examples:
• Define $$f, g : \mathbb{R} \to \mathbb{R}$$ by $$f(x) = x + 1$$ and $$g(x) = x - 1$$. Then $$g$$ is both a left and a right inverse to $$f$$, and vice-versa.
• Write $$\mathbb{R}^{\geq 0}$$ to denote the nonnegative reals. Define $$f : \mathbb{R} \to \mathbb{R}^{\geq 0}$$ by $$f(x) = x^2$$, and define $$g : \mathbb{R}^{\geq 0} \to \mathbb{R}$$ by $$g(x) = \sqrt x$$. Then $$f(g(x)) = (\sqrt x)^2 = x$$ for every $$x$$ in the domain of $$g$$, so $$f$$ is a left inverse to $$g$$, and $$g$$ is a right inverse to $$f$$. On the other hand, $$g(f(x)) = \sqrt{x^2} = | x |$$, which is not the same as $$x$$ when $$x$$ is negative. So $$g$$ is not a left inverse to $$f$$, and $$f$$ is not a right inverse to $$g$$.
The following fact is not at all obvious, even though the proof is short:
Proposition. Suppose $$f : X \to Y$$ has a left inverse, $$h$$, and a right inverse, $$k$$. Then $$h = k$$.
Proof. Let $$y$$ be any element in $$Y$$. The idea is to compute $$h(f(k(y))$$ in two different ways. Since $$h$$ is a left inverse to $$f$$, we have $$h(f(k(y))) = k(y)$$. On the other hand, since $$k$$ is a right inverse to $$f$$, $$f(k(y)) = y$$, and so $$h(f(k(y)) = h(y)$$. So $$k(y) = h(y)$$.
If $$g$$ is both a right and left inverse to $$f$$, we say that $$g$$ is simply the inverse of $$f$$. A function $$f$$ may have more than one left or right inverse (we leave it to you to cook up examples), but it can have at most one inverse.
Proposition. Suppose $$g_1, g_2 : Y \to X$$ are both inverses to $$f$$. Then $$g_1 = g_2$$.
Proof. This follows from the previous proposition, since (say) $$g_1$$ is a left inverse to $$f$$, and $$g_2$$ is a right inverse.
When $$f$$ has an inverse, $$g$$, this justifies calling $$g$$ the inverse to $$f$$, and writing $$f^{-1}$$ to denote $$g$$. Notice that if $$f^{-1}$$ is an inverse to $$f$$, then $$f$$ is an inverse to $$f^{-1}$$. So if $$f$$ has an inverse, then so does $$f^{-1}$$, and $$(f^{-1})^{-1} = f$$. For any set $$A$$, clearly we have $$i_X^{-1} = i_X$$.
Proposition. Suppose $$f : X \to Y$$ and $$g : Y \to Z$$. If $$h : Y \to X$$ is a left inverse to $$f$$ and $$k : Z \to Y$$ is a left inverse to $$g$$, then $$h \circ k$$ is a left inverse to $$g \circ f$$.
Proof. For every $$x$$ in $$X$$,
$(h \circ k) \circ (g \circ f) (x) = h(k(g(f(x)))) = h(f(x)) = x.$
Corollary. The previous proposition holds with “left” replaced by “right.”
Proof. Switch the role of $$f$$ with $$h$$ and $$g$$ with $$k$$ in the previous proposition.
Corollary. If $$f : X \to Y$$ and $$g : Y \to Z$$ both have inverses, then $$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$$.
## 15.2. Injective, Surjective, and Bijective Functions¶
A function $$f : X \to Y$$ is said to be injective, or an injection, or one-one, if given any $$x_1$$ and $$x_2$$ in $$A$$, if $$f(x_1) = f(x_2)$$, then $$x_1 = x_2$$. Notice that the conclusion is equivalent to its contrapositive: if $$x_1 \neq x_2$$, then $$f(x_1) \neq f(x_2)$$. So $$f$$ is injective if it maps distinct element of $$X$$ to distinct elements of $$Y$$.
A function $$f : X \to Y$$ is said to be surjective, or a surjection, or onto, if for every element $$y$$ of $$Y$$, there is an $$x$$ in $$X$$ such that $$f(x) = y$$. In other words, $$f$$ is surjective if every element in the codomain is the value of $$f$$ at some element in the domain.
A function $$f : X \to Y$$ is said to be bijective, or a bijection, or a one-to-one correspondence, if it is both injective and surjective. Intuitively, if there is a bijection between $$X$$ and $$Y$$, then $$X$$ and $$Y$$ have the same size, since $$f$$ makes each element of $$X$$ correspond to exactly one element of $$Y$$ and vice-versa. For example, it makes sense to interpret the statement that there were four Beatles as the statement that there is a bijection between the set $$\{1, 2, 3, 4\}$$ and the set $$\{ \text{John, Paul, George, Ringo} \}$$. If we claimed that there were five Beatles, as evidenced by the function $$f$$ which assigns 1 to John, 2 to Paul, 3 to George, 4 to Ringo, and 5 to John, you should object that we double-counted John—that is, $$f$$ is not injective. If we claimed there were only three Beatles, as evidenced by the function $$f$$ which assigns 1 to John, 2 to Paul, and 3 to George, you should object that we left out poor Ringo—that is, $$f$$ is not surjective.
The next two propositions show that these notions can be cast in terms of the existence of inverses.
Proposition. Let $$f : X \to Y$$.
• If $$f$$ has a left inverse, then $$f$$ is injective.
• If $$f$$ has a right inverse, then $$f$$ is surjective.
• If $$f$$ has an inverse, then it is $$f$$ bijective.
Proof. For the first claim, suppose $$f$$ has a left inverse $$g$$, and suppose $$f(x_1) = f(x_2)$$. Then $$g(f(x_1)) = g(f(x_2))$$, and so $$x_1 = x_2$$.
For the second claim, suppose $$f$$ has a right inverse $$h$$. Let $$y$$ be any element of $$Y$$, and let $$x = g(y)$$. Then $$f(x) = f(g(y)) = y$$.
The third claim follows from the first two.
The following proposition is more interesting, because it requires us to define new functions, given hypotheses on $$f$$.
Proposition. Let $$f : X \to Y$$.
• If $$X$$ is nonempty and $$f$$ is injective, then $$f$$ has a left inverse.
• If $$f$$ is surjective, then $$f$$ has a right inverse.
• If $$f$$ if bijective, then it has an inverse.
Proof. For the first claim, let $$\hat x$$ be any element of $$X$$, and suppose $$f$$ is injective. Define $$g : Y \to X$$ by setting $$g(y)$$ equal to any $$x$$ such that $$f(x) = y$$, if there is one, and $$\hat x$$ otherwise. Now, suppose $$g(f(x)) = x'$$. By the definition of $$g$$, $$x'$$ has to have the property that $$f(x) = f(x')$$. Since $$f$$ is injective, $$x = x'$$, so $$g(f(x)) = x$$.
For the second claim, because $$f$$ is surjective, we know that for every $$y$$ in $$Y$$ there is any $$x$$ such that $$f(x) = y$$. Define $$h : B \to A$$ by again setting $$h(y)$$ equal to any such $$x$$. (In contrast to the previous paragraph, here we know that such an $$x$$ exists, but it might not be unique.) Then, by the definition of $$h$$, we have $$f(h(y)) = y$$.
Notice that the definition of $$g$$ in the first part of the proof requires the function to “decide” whether there is an $$x$$ in $$X$$ such that $$f(x) = y$$. There is nothing mathematically dubious about this definition, but in many situations, this cannot be done algorithmically; in other words, $$g$$ might not be computable from the data. More interestingly, the definition of $$h$$ in the second part of the proof requires the function to “choose” a suitable value of $$x$$ from among potentially many candidates. We will see in Section 23.3 that this is a version of the axiom of choice. In the early twentieth century, the use of the axiom of choice in mathematics was hotly debated, but today it is commonplace.
Using these equivalences and the results in the previous section, we can prove the following:
Proposition. Let $$f : X \to B$$ and $$g : Y \to Z$$.
• If $$f$$ and $$g$$ are injective, then so is $$g \circ f$$.
• If $$f$$ and $$g$$ are surjective, then so is $$g \circ f$$.
Proof. If $$f$$ and $$g$$ are injective, then they have left inverses $$h$$ and $$k$$, respectively, in which case $$h \circ k$$ is a left inverse to $$g \circ f$$. The second statement is proved similarly.
We can prove these two statements, however, without mentioning inverses at all. We leave that to you as an exercise.
Notice that the expression $$f(n) = 2 n$$ can be used to define infinitely many functions with domain $$\mathbb{N}$$, such as:
• a function $$f : \mathbb{N} \to \mathbb{N}$$
• a function $$f : \mathbb{N} \to \mathbb{R}$$
• a function $$f: \mathbb{N} \to \{ n \mid n \text{ is even} \}$$
Only the third one is surjective. Thus a specification of the function’s codomain as well as the domain is essential to making sense of whether a function is surjective.
## 15.3. Functions and Subsets of the Domain¶
Suppose $$f$$ is a function from $$X$$ to $$Y$$. We may wish to reason about the behavior of $$f$$ on some subset $$A$$ of $$X$$. For example, we can say that $$f$$ is injective on $$A$$ if for every $$x_1$$ and $$x_2$$ in $$A$$, if $$f(x_1) = f(x_2)$$, then $$x_1 = x_2$$.
If $$f$$ is a function from $$X$$ to $$Y$$ and $$A$$ is a subset of $$X$$, we write $$f[A]$$ to denote the image of $$f$$ on $$A$$, defined by
$f[A] = \{ y \in Y \mid y = f(x) \; \mbox{for some x in A} \}.$
In words, $$f[A]$$ is the set of elements of $$Y$$ that are “hit” by elements of $$A$$ under the mapping $$f$$. Notice that there is an implicit existential quantifier here, so that reasoning about images invariably involves the corresponding rules.
Proposition. Suppose $$f : X \to Y$$, and $$A$$ is a subset of $$X$$. Then for any $$x$$ in $$A$$, $$f(x)$$ is in $$f[A]$$.
Proof. By definition, $$f(x)$$ is in $$f[A]$$ if and only if there is some $$x'$$ in $$A$$ such that $$f(x') = f(x)$$. But that holds for $$x' = x$$.
Proposition. Suppose $$f : X \to Y$$ and $$g : Y \to Z$$. Let $$A$$ be a subset of $$X$$. Then
$(g \circ f)[A] = g[f[A]].$
Proof. Suppose $$z$$ is in $$(g \circ f)[A]$$. Then for some $$x \in A$$, $$z = (g \circ f)(x) = g(f(x))$$. By the previous proposition, $$f(x)$$ is in $$f[A]$$. Again by the previous proposition, $$g(f(x))$$ is in $$g[f[A]]$$.
Conversely, suppose $$z$$ is in $$g[f[A]]$$. Then there is a $$y$$ in $$f[A]$$ such that $$f(y) = z$$, and since $$y$$ is in $$f[D]$$, there is an $$x$$ in $$A$$ such that $$f(x) = y$$. But then $$(g \circ f)(x) = g(f(x)) = g(y) = z$$, so $$z$$ is in $$(g \circ f)[A]$$.
Notice that if $$f$$ is a function from $$X$$ to $$Y$$, then $$f$$ is surjective if and only if $$f[X] = Y$$. So the previous proposition is a generalization of the fact that the composition of surjective functions is surjective.
Suppose $$f$$ is a function from $$X$$ to $$Y$$, and $$A$$ is a subset of $$X$$. We can view $$f$$ as a function from $$A$$ to $$Y$$, by simply ignoring the behavior of $$f$$ on elements outside of $$A$$. Properly speaking, this is another function, denoted $$f \upharpoonright A$$ and called “the restriction of $$f$$ to $$A$$.” In other words, given $$f : X \to Y$$ and $$A \subseteq X$$, $$f \upharpoonright A : A \to Y$$ is the function defined by $$(f \upharpoonright A)(x) = x$$ for every $$x$$ in $$A$$. Notice that now “$$f$$ is injective on $$A$$” means simply that the restriction of $$f$$ to $$A$$ is injective.
There is another important operation on functions, known as the preimage. If $$f : X \to Y$$ and $$B \subseteq Y$$, then the preimage of $$B$$ under $$f$$, denoted $$f^{-1}[B]$$, is defined by
$f^{-1}[B] = \{ x \in X \mid f(x) \in B \},$
that is, the set of elements of $$X$$ that get mapped into $$B$$. Notice that this makes sense even if $$f$$ does not have an inverse; for a given $$y$$ in $$B$$, there may be no $$x$$‘s with the property $$f(x) \in B$$, or there may be many. If $$f$$ has an inverse, $$f^{-1}$$, then for every $$y$$ in $$B$$ there is exactly one $$x \in X$$ with the property $$f(x) \in B$$, in which case, $$f^{-1}[B]$$ means the same thing whether you interpret it as the image of $$B$$ under $$f^{-1}$$ or the preimage of $$B$$ under $$f$$.
Proposition. Suppose $$f : X \to Y$$ and $$g : Y \to Z$$. Let $$C$$ be a subset of $$Z$$. Then
$(g \circ f)^{-1}[C] = f^{-1}[g^{-1}[C]].$
Proof. For any $$y$$ in $$C$$, $$y$$ is in $$(g \circ f)^{-1}[C]$$ if and only if $$g(f(y))$$ is in $$C$$. This, in turn, happens if and only if $$f(y)$$ is in $$g^{-1}[C]$$, which in turn happens if and only if $$y$$ is in $$f^{-1}[g^{-1}[C]]$$.
Here we give a long list of facts properties of images and preimages. Here, $$f$$ denotes an arbitrary function from $$X$$ to $$Y$$, $$A, A_1, A_2, \ldots$$ denote arbitrary subsets of $$X$$, and $$B, B_1, B_2, \ldots$$ denote arbitrary subsets of $$Y$$.
• $$A \subseteq f^{-1}[f[A]]$$, and if $$f$$ is injective, $$A = f^{-1}[f[A]]$$.
• $$f[f^{-1}[B]] \subseteq B$$, and if $$f$$ is surjective, $$B = f[f^{-1}[B]]$$.
• If $$A_1 \subseteq A_2$$, then $$f[A_1] \subseteq f[A_2]$$.
• If $$B_1 \subseteq B_2$$, then $$f^{-1}[B_1] \subseteq f^{-1}[B_2]$$.
• $$f[A_1 \cup A_2] = f[A_1] \cup f[A_2]$$.
• $$f^{-1}[B_1 \cup B_2] = f^{-1}[B_1] \cup f^{-1}[B_2]$$.
• $$f[A_1 \cap A_2] \subseteq f[A_1] \cap f[A_2]$$, and if $$f$$ is injective, $$f[A_1 \cap A_2] = f[A_1] \cap f[A_2]$$.
• $$f^{-1}[B_1 \cap B_2] = f^{-1}[B_1] \cap f^{-1}[B_2]$$.
• $$f[A] \setminus f[B] \subseteq f[A \setminus B]$$.
• $$f^{-1}[A] \setminus f^{-1}[B] \subseteq f[A \setminus B]$$.
• $$f[A] \cap B = f[A \cap f^{-1}[B]]$$.
• $$f[A] \cup B \supseteq f[A \cup f^{-1}[B]]$$.
• $$A \cap f^{-1}[B] \subseteq f^{-1}[f[A] \cap B]$$.
• $$A \cup f^{-1}[B] \subseteq f^{-1}[f[A] \cup B]$$.
Proving identities like this is typically a matter of unfolding definitions and using basic logical inferences. Here is an example.
Proposition. Let $$X$$ and $$Y$$ be sets, $$f : X \to Y$$, $$A \subseteq X$$, and $$B \subseteq Y$$. Then $$f[A] \cap B = f[A \cap f^{-1}[B]]$$.
Proof. Suppose $$y \in f[A] \cap B$$. Then $$y \in B$$, and for some $$x \in A$$, $$f(x) = y$$. But this means that $$x$$ is in $$f^{-1}[B]$$, and so $$x \in A \cap f^{-1}[B]$$. Since $$f(x) = y$$, we have $$y \in f[A \cap f^{-1}[B]]$$, as needed.
Conversely, suppose $$y \in f[A \cap f^{-1}[B]]$$. Then for some $$x \in A \cap f^{-1}[B]$$, we have $$f(x) = y$$. For this $$x$$, have $$x \in A$$ and $$f(x) \in B$$. Since $$f(x) = y$$, we have $$y \in B$$, and since $$x \in A$$, we also have $$y \in f[A]$$, as required.
## 15.4. Functions and Relations¶
A binary relation $$R(x,y)$$ on $$A$$ and $$B$$ is functional if for every $$x$$ in $$A$$ there exists a unique $$y$$ in $$B$$ such that $$R(x,y)$$. If $$R$$ is a functional relation, we can define a function $$f_R : X \to B$$ by setting $$f_R(x)$$ to be equal to the unique $$y$$ in $$B$$ such that $$R(x,y)$$. Conversely, it is not hard to see that if $$f : X \to B$$ is any function, the relation $$R_f(x, y)$$ defined by $$f(x) = y$$ is a functional relation. The relation $$R_f(x,y)$$ is known as the graph of $$f$$.
It is not hard to check that functions and relations travel in pairs: if $$f$$ is the function associated with a functional relation $$R$$, then $$R$$ is the functional relation associated the function $$f$$, and vice-versa. In set-theoretic foundations, a function is often defined to be a functional relation. Conversely, we have seen that in type-theoretic foundations like the one adopted by Lean, relations are often defined to be certain types of functions. We will discuss these matters later on, and in the meanwhile only remark that in everyday mathematical practice, the foundational details are not so important; what is important is simply that every function has a graph, and that any functional relation can be used to define a corresponding function.
So far, we have been focusing on functions that take a single argument. We can also consider functions $$f(x, y)$$ or $$g(x, y, z)$$ that take multiple arguments. For example, the addition function $$f(x, y) = x + y$$ on the integers takes two integers and returns an integer. Remember, we can consider binary functions, ternary functions, and so on, and the number of arguments to a function is called its “arity.” One easy way to make sense of functions with multiple arguments is to think of them as unary functions from a cartesian product. We can think of a function $$f$$ which takes two arguments, one in $$A$$ and one in $$B$$, and returns an argument in $$C$$ as a unary function from $$A \times B$$ to $$C$$, whereby $$f(a, b)$$ abbreviates $$f((a, b))$$. We have seen that in dependent type theory (and in Lean) it is more convenient to think of such a function $$f$$ as a function which takes an element of $$A$$ and returns a function from $$B \to C$$, so that $$f(a, b)$$ abbreviates $$(f(a))(b)$$. Such a function $$f$$ maps $$A$$ to $$B \to C$$, where $$B \to C$$ is the set of functions from $$B$$ to $$C$$.
We will return to these different ways of modeling functions of higher arity later on, when we consider set-theoretic and type-theoretic foundations. One again, we remark that in ordinary mathematics, the foundational details do not matter much. The two choices above are inter-translatable, and sanction the same principles for reasoning about functions informally.
In mathematics, we often also consider the notion of a partial function from $$X$$ to $$Y$$, which is really a function from some subset of $$X$$ to $$Y$$. The fact that $$f$$ is a partial function from $$X$$ to $$Y$$ is sometimes written $$f : X \nrightarrow Y$$, which should be interpreted as saying that $$f : A \to Y$$ for some subset $$A$$ of $$Y$$. Intuitively, we think of $$f$$ as a function from $$X \to Y$$ which is simply “undefined” at some of its inputs; for example, we can think of $$f : \mathbb{R} \nrightarrow \mathbb{R}$$ defined by $$f(x) = 1 / x$$, which is undefined at $$x = 0$$, so that in reality $$f : \mathbb{R} \setminus \{ 0 \} \to R$$. The set $$A$$ is sometimes called the domain of $$f$$, in which case, there is no good name for $$X$$; others continue to call $$X$$ the domain, and refer to $$A$$ as the domain of definition. To indicate that a function $$f$$ is defined at $$x$$, that is, that $$x$$ is in the domain of definition of $$f$$, we sometimes write $$f(x) \downarrow$$. If $$f$$ and $$g$$ are two partial functions from $$X$$ to $$Y$$, we write $$f(x) \simeq g(x)$$ to mean that either $$f$$ and $$g$$ are both defined at $$x$$ and have the same value, or are both undefined at $$x$$. Notions of injectivity, surjectivity, and composition are extended to partial functions, generally as you would expect them to be.
In terms of relations, a partial function $$f$$ corresponds to a relation $$R_f(x,y)$$ such that for every $$x$$ there is at most one $$y$$ such that $$R_f(x,y)$$ holds. Mathematicians also sometimes consider multifunctions from $$X$$ to $$Y$$, which correspond to relations $$R_f(x,y)$$ such that for every $$x$$ in $$X$$, there is at least one $$y$$ such that $$R_f(x,y)$$ holds. There may be many such $$y$$; you can think of these as functions which have more than one output value. If you think about it for a moment, you will see that a partial multifunction is essentially nothing more than an arbitrary relation.
## 15.5. Exercises¶
1. Let $$f$$ be any function from $$X$$ to $$Y$$, and let $$g$$ be any function from $$Y$$ to $$Z$$.
• Show that if $$g \circ f$$ is injective, then $$f$$ is injective.
• Give an example of functions $$f$$ and $$g$$ as above, such that that $$g \circ f$$ is injective, but $$g$$ is not injective.
• Show that if $$g \circ f$$ is injective and $$f$$ is surjective, then $$g$$ is injective.
2. Let $$f$$ and $$g$$ be as in the last problem. Suppose $$g \circ f$$ is surjective.
• Is $$f$$ necessarily surjective? Either prove that it is, or give a counterexample.
• Is $$g$$ necessarily surjective? Either prove that it is, or give a counterexample.
3. A function $$f$$ from $$\mathbb{R}$$ to $$\mathbb{R}$$ is said to be strictly increasing if whenever $$x_1 < x_2$$, $$f(x_1) < f(x_2)$$.
• Show that if $$f : \mathbb{R} \to \mathbb{R}$$ is strictly increasing, then it is injective (and hence it has a left inverse).
• Show that if $$f : \mathbb{R} \to \mathbb{R}$$ is strictly increasing, and $$g$$ is a right inverse to $$f$$, then $$g$$ is strictly increasing.
4. Let $$f : X \to Y$$ be any function, and let $$A$$ and $$B$$ be subsets of $$X$$. Show that $$f [A \cup B] = f[A] \cup f[B]$$.
5. Let $$f: X \to Y$$ be any function, and let $$A$$ and $$B$$ be any subsets of $$X$$. Show $$f[A] \setminus f[B] \subseteq f[A \setminus B]$$.
6. Define notions of composition and inverse for binary relations that generalize the notions for functions.
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Bilinear interpolation
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I have some complications with definition error any points, contain into square, square vertex is known, errors square vertex is known too.
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## anonymous one year ago If sin θ = 1/3 and tan θ < 0, what is the value of cos θ?
1. anonymous
draw a triangle
2. anonymous
|dw:1442452830325:dw|
3. anonymous
there is a picture of an angle whose sine is $$\frac{1}{3}$$ find the missing side via pythagoras
4. anonymous
you can pretty much do it in your head what is $$3^2$$?
5. anonymous
9
6. anonymous
and $$1^2$$?
7. anonymous
1
8. anonymous
and $$9-1$$?
9. anonymous
8
10. anonymous
so answer for the missing side is $$\sqrt{8}$$ now two more steps
11. anonymous
2√2
12. anonymous
|dw:1442453007407:dw|
13. anonymous
yeah that too now the cosine is adjacent over hypotenuse, but don't forget that $$\tan(\theta)<0$$ which means the cosine must be negative
14. anonymous
-√8/3
15. anonymous
or -(2√2)/3
16. anonymous
yes probably your teacher will like the second one more
17. anonymous
YES! thank you! you're gr8 :)
18. anonymous
lol yw
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Help
# kinetic energy
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## Mining Companies Borrow From Gamers' Physics Engines
Original author:
Soulskill
littlekorea writes "Mining companies are developing new systems for automating blasting of iron ore using the same open source physics engines adapted for games such as Grand Theft Auto IV and Red Dead Redemption. The same engine that determines 3D collision detection and soft body/rigid body dynamics in gaming will be applied to building 3D blast movement models — which will predict where blasted materials will land and distinguish between ore and waste. Predictive blast fragmentation models used in the past have typically been either numerical or empirical, [mining engineer Alan Cocker] said. Numerical models such as discrete element method, he noted, are onerous to configure and demanding of resources — both computing and human — and are generally not appropriate for operational use at mines. 'The problem with empirical models, by contrast, is that they tend to operate at a scale too coarse to give results useful for optimizations,' he added, noting typical Kuz-Ram-based fragmentation models (PDF) (widely used to estimate fragmentation from blasting) assume homogeneous geology (the same type of materials) throughout a blast."
Read more of this story at Slashdot.
## Yahoo Paid $30 Million in Cash for 18 Months of Young Summly Entrepreneur's Time Original author: Kara Swisher Earlier today, Yahoo said it had acquired the trendy and decidedly stylish news reading app Summly, along with its telegenic and very young entrepreneur Nick D’Aloisio. Yahoo said it plans to close down the actual app and use the algorithmic summation technology that the 17-year-old D’Aloisio built with a small team of five, along with a major assist from Silicon Valley research institute SRI International, throughout its products. While Yahoo did not disclose the price, several sources told me that the company paid$30 million — 90 percent in cash and 10 percent in stock — to buy the London-based Apple smartphone app.
And despite its elegant delivery, that’s a very high price, especially since Summly has been downloaded slightly less than one million times since launch — after a quick start amid much publicity over its founder — with about 90 million “summaries” read. Of course, like many such apps, it also had no monetization plan as yet.
What Yahoo is getting, though, is perhaps more valuable — the ability to put the fresh-faced D’Aloisio front and center of its noisy efforts to make consumers see Yahoo as a mobile-first company. That has been the goal of CEO Marissa Mayer, who has bought up a range of small mobile startups since she took over nine months ago and who has talked about the need for Yahoo to focus on the mobile arena above all.
Mayer met with D’Aloisio, said sources, although the deal was struck by voluble M&A head Jackie Reses.
Said one person close to the deal, about the founder: “Nick will be a great person to put in front of the media and consumers with Mayer to make Yahoo seem like it is a place that loves both entrepreneurs and mobile experiences, which in turn will presumably attract others like him.”
Having met the young man in question, who was in San Francisco in the fall on a fundraising trip, I can see the appeal. He’s both well-spoken and adorkable, as well as very adept at charming cranky media types like me by radiating with the kinetic energy of someone born in the mobile world (you can see that in full force in the video below with actor and Summly investor Stephen Fry).
Still, D’Aloisio is very young and presumably has a lot of other entrepreneurial goals and that’s why he agreed as part of the deal to only officially stay 18 months at Yahoo, multiple sources told me. In many cases, startup founders strike such short-term employment deals with big companies, agreeing to stay for a certain determined time period.
He will also remain in England, where he lives with his parents, said sources. In addition, only two of Summly’s employees will go to Yahoo with D’Aloisio.
That’s \$10 million each, along with a nifty app Yahoo will not be using as is (too bad, as it would up the hip and fun factor of Yahoo’s apps by a factor of a gazillion if it were maintained).
“It works out on a lot of levels,” said another person close to the situation. “Nick is a founder that will make Mayer and Yahoo look cutting edge.”
Cue the parade of PR profiles of the young genius made millionaire, helping Yahoo become relevant again.
I have an email for comment into the always friendly D’Aloisio. But I don’t expect a reply, since he has apparently been specifically instructed by the martinets of Yahoo PR not to talk to me any longer — well, for 18 months at least! (Don’t worry, Nick, I don’t blame you and will still listen to whatever you are pitching next, since you are so dang compelling and I enjoyed using Summly!)
Until then, here’s the faboo Summly video, with the best chairs ever:
Summly Launch from Summly on Vimeo.
## Are We Still Evolving? Pam Oslie at TEDxAmericanRiviera 2012
Are We Still Evolving? Pam Oslie at TEDxAmericanRiviera 2012
Science states that humans have evolved over thousands of years. So have we reached the pinnacle of our abilities and our mind's potential? Do Quantum Physics and consciousness hold the key to our evolution? Pam Oslie will explore the answers to these questions and will reveal how electro-magnetic fields, non-local mind, and parallel universes play key roles in our development. AboutTEDx, x = independently organized event In the spirit of ideas worth spreading, TEDx is a program of local, self-organized events that bring people together to share a TED-like experience. At a TEDx event, TEDTalks video and live speakers combine to spark deep discussion and connection in a small group. These local, self-organized events are branded TEDx, where x = independently organized TED event. The TED Conference provides general guidance for the TEDx program, but individual TEDx events are self-organized. (Subject to certain rules and regulations.)
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TEDxTalks
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## Q: Is it possible for an artificial black hole to be created, or something that has the same effects? If so, how small could it be made?
Physicist: Not with any current, or remotely feasible technology. The method in use by the universe today; get several Suns worth of mass into a big pile and wait, is a pretty effective way to create black holes.
In theory, all you need to do to create an artificial black hole (a “black faux”?) is to get a large amount of energy and matter into a very small volume. The easiest method would probably be to use some kind of massive, super-duper-accelerators. The problem is that black holes are dense, and the smaller and less massive they are the denser they need to be.
A black hole with the mass of the Earth would be so small you could lose it pretty easy. Except for all the gravity.
But there are limits to how dense matter can get on its own. The density of an atomic nucleus, where essentially all of the matter of an atom is concentrated, is about the highest density attainable by matter: about 1018 kg/m3, or about a thousand, million, million times denser than water. This density is also the approximate density of neutron stars (which are basically giant atomic nuclei).
When a star runs out of fuel and collapses, this is the densest that it can get. If a star has less than about 3 times as much mass as our Sun, then when it gets to this density it stops, and then hangs out forever. If a star has more than 3 solar masses, then as it collapses, on it’s way to neutron-star-density, it becomes a black hole (a black hole with more mass needs less density).
The long-winded point is; in order to create a black hole smaller than 3 Suns (which would be what you’re looking for it you want to keep it around), it’s not a question of crushing stuff. Instead you’d need to use energy, and the easiest way to get a bunch of energy into one place is to use kinetic energy.
There’s some disagreement about the minimum size that a black hole can be. Without resorting to fairly exotic, “lot’s of extra dimensions” physics, the minimum size should be somewhere around $2\times 10^{-21}$ grams. That seems small, but it’s very difficult (probably impossible) to get even that much mass/energy into a small enough region. A black hole with this mass would be about 10-47 m across, which is way, way, way smaller than a single electron (about 10-15 m). But unfortunately, a particle can’t be expected to concentrate energy in a region smaller than the particle itself. So using whatever “ammo” you can get into a particle accelerator, you find that the energy requirements are a little steeper.
To merely say that you’d need to accelerate particles to nearly the speed of light doesn’t convey the stupefying magnitude of the amount of energy you’d need to get a collision capable of creating a black hole. A pair of protons would need to have a “gamma” (a useful way to talk about ludicrously large speeds) of about 1040, or a pair of lead nuclei would need to have a gamma of about 1037, when they collide in order for a black hole to form. This corresponds to the total energy of all the mass in a small mountain range. For comparison, a nuclear weapon only releases the energy of several grams of matter.
CERN, or any other accelerator ever likely to be created, falls short in the sense that a salted slug in the ironman falls short.
There’s nothing else in the universe the behaves like a black hole. They are deeply weird in a lot of ways. But, a couple of the properties normally restricted to black holes can be simulated with other things. There are “artificial black holes” created in laboratories to study Hawking radiation, but you’d never recognize them. The experimental set up involves tubes of water, or laser beams, and lots of computers. No gravity, no weird timespace stuff, nothin’. If you were in the lab, you’d never know that black holes were being studied.
## Takeshi Kitano’s ‘Outrage’ Red Band Trailer
Lovers of film are likely lovers of Takeshi Kitano. Sometimes billed as Beat Takeshi, he’s not only the evil star of Battle Royale, he’s the talented director of films like The Blind Swordsman: Zatoichi and Fireworks. In 2010, he directed and starred in Outrage, a twisting, turning crime drama in the vein of Martin Scorsese. It played several festivals, spawned a few trailers and was successful enough that a sequel, Outrage 2, is on the way. However, most fans haven’t had a chance to see the original because its U.S. release date was way off. Finally, Outrage is schedule to hit U.S. screens on December 2 and on-demand next week, October 28. There’s a brand new red band trailer for the film after the jump.
Thanks to Hulu for this trailer. (That means people outside the US probably can’t see this embed.)
Here’s the official plot description.
In a ruthless battle for power, several yakuza clans vie for the favor of their head family in the Japanese underworld. The rival bosses seek to rise through the ranks by scheming and making allegiances sworn over saké. Long-time yakuza Otomo has seen his kind go from elaborate body tattoos and severed fingertips to becoming important players on the stock market. Theirs is a never-ending struggle to end up on top, or at least survive, in a corrupt world where there are no heroes but constant betrayal and vengeance.
I saw Outrage at AFI Fest 2010 and enjoyed it immensely. It’s filled with the kind of kinetic energy that fuels the first 90 minutes of Casino or the last 30 minutes of Goodfellas with enough double crosses, great kills and evil characters to delight all lovers of crime and violence. Check it out when it’s available.
## Soulboy
Starring: Martin Compston, Felicity Jones, Alfie Allen
Director: Shimmy Marcus
Summary: Joe McCain (Martin Compston) is bored of a life that is going nowhere. Enter hair-dresser Jane (Nichola Burley) - blonde, brassy and moving to the beat of a whole new sound, all night dancing at the Wigan Casino - home of Northern Soul…
Opening with a promotional video for Stoke-on-Trent circa 1974, you could be forgiven for thinking Soulboy is simply a rehash of feel good British comedies from the 90s which has arrived 15 years too late, but give it your time and you'll soon realise that this sweet natured coming of age drama is something much more.
Sure it's got the usual "grim up north" aesthetic of The Full Monty or Brassed Off but it's able to lift itself out of the doldrums with the injection of a young, up-and-coming British cast who convincingly deliver sweat and tears through a hopeful hormonal haze. Leading man Martin Compston exudes all the boyish charm and brash nature of a youngster trying to find his feet in a world of routine binge drinking, whilst Felicity Jones, as the lovelorn girl next door, has the right measure of sweetness and anger to make us all nostalgic for the one that got away. With the cast playing so well off each other, the familiar boy meets girl narrative is played out with the renewed vigour that the genre deserves and sadly most of the time is bereft of in mainstream cinema.
If the simplicity of Soulboy's romantic plot seems too familiar to invigorate a well worn audience then the music and dance scenes are filmed and choreographed with such a kinetic energy that you can't help tap your feet - the cast again showing their versatile talents in and amongst the authentic dance hall of the Wigan Casino. As the classic American soul records are consumed to the rapturous applause and swivel spin moves of the pubescent masses, Soulboy takes a rare wrong turn by attempting a drug dealing subplot which feel out of place when everything else feels comfortingly familiar. Luckily this is simply a Macguffin to bring our feuding lovers together and the pace is changed as quickly as the strap on a Gola record bag - which looks like it was the must have accessory in 1974.
Soulboy may be seen as too twee, and dare I say it too British, when compared with recent hyped, steroid injected 3D dance extravaganzas like Step Up, but it has more honesty and realism in one finger click than they do in a row of body pops. The industrial towns of Stoke and Wigan may be as romantic as a power station chimney stack but Soulboy is destined to be a cult hit with audiences and young lovers for years to come.
Trailer
Out 3rd September
www.soulboythefilm.com
## The world's largest tidal turbine about to be installed in Scotland
This is a photo of the Atlantis AK1000, a 130 ton, 74-foot-tall tidal turbine that will be installed underwater off the cost of Scotland. It is designed to supply electrical power for 1,000 households.
Sea water, which is 832 times denser than air, gives a 5 knot ocean current more kinetic energy than a 350 km/h wind; therefore ocean currents have a very high energy density. Hence a smaller device is required to harness tidal current energy than to harness wind energy.
Tidal current energy takes the kinetic energy available in currents and converts it into renewable electricity. As oceans cover over 70% of Earth’s surface, ocean energy (including wave power, tidal current power and ocean thermal energy conversion) represents a vast source of energy, estimated at between 2,000 and 4,000 TWh per year, enough energy to continuously light between 2 and 4 billion 11W low-energy light bulbs.
Both the U.S. and the U.K., for example, have enough ocean power potential to meet around 15% of their total power needs.
Good: The World's Largest Tidal Turbine, Unveiled
## George Whitesides: Toward a science of simplicity - George Whitesides (2010)
Simplicity: We know it when we see it -- but what is it, exactly? In this funny, philosophical talk, George Whitesides chisels out an answer.
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# Shortest Path Between Two Points In A Grid Java
Great circles are used in navigation, since the shortest path between any two points on the earth's surface lies on a great circle. Given two vertices in a graph, a path is a sequence of edges connecting them. The map t → t 2 {\displaystyle t\to t^{2}} from the unit interval on the real number line to itself gives the shortest path between 0 and 1, but is not a geodesic because the velocity of the corresponding. The width of a branch is proportional to the square root of the sum of branches reachable by that branch. You: If I walk 3 blocks East and 4 blocks North, how far am I from my starting point?. HOW TO USE : Set the transform of START and END point on the script attached to the Node gameobject and press play button. PartitionBlobTranslator: This interface defines the methods supported by a partition blob translator. Computes shortest paths from a single source vertex to all other vertices in a weighted graph. Thus, the almost shortest path would be the one indicated by dashed lines (total length 5), as no route between two consecutive points belongs to any shortest. How can I change the cost of the paths inside the polygon so that the connection line goes around the keep out are? I have Identified the points that are inside the polygon. First, * if we consider the case where k = 0 and all paths must have no edges, then * we get that *. How many statesare there?. Corollary The shortest path connecting a to b in S~ remains in the rectangle containing a and b. It is a more practical variant on solving mazes. Predecessor nodes of the shortest paths, returned as a vector. Retrieve all the edges of this node. Adjacency Matrix. The path from EgonWillighagen to Jandot : Neo4j , a graph API for java: my notebook. " Actually, I think I am wrong about this part. I would like to help you write it but Java isnt my language :). Re: Shortest Path Between Cells (VBA) Obstacles will make this a much more complex problem, certainly beyond a forum post solution. By road, the shortest distance from my house to school is 1994 metres. The other player chooses which colour to play. This was done by expanding the grid to 3-dimensions, and implementing the rest of the algorithm the way it would be done on a 2-dimensional grid. Note that, for this you need to generate lot many points along the path so that a point lies on the yellow path. Hi the download contains the C# project in addition to the C++ versions, but please remember that this problem is NP hard - ie cannot be solved in polynomial time, and you will find that time taken to solve the problem increases exponentially with the number of nodes - this might be an issue with the size of the problem you have in mind - unless it is a directed acyclic graphs in which. Shortest Paths in Graphs s = 1 2 4 3 2. Dijkstra's Algorithms describes how to find the shortest path from one node to another node in a directed weighted graph. By a shortestpath for the pairX i we mean any path from s i to t i with length equal to d(X i ), i. Bicycle and walking paths are preferentially weighted, and the interstate is heavily penalized. 2 Shortest Paths between All Pairs of Nodes [4(i, j) > O] It is very often the case that the shortest paths between all pairs of nodes in a network are required. Triangle Inequality 8u;v;x2V, we have (u;v) (u;x) + (x;v). Partial solution. The algorithm works by keeping the shortest distance of vertex v from the source in the distance table. I can’t imagine any reasonable theory of geometry in which this would be an axiom, though it’s possible that such a theory can be constructed (likely very artificially). In this case the asymptotic constant is 1. How do you find the number of the shortest distances between two points on a grid where you can only move one unit up, down, left, or right? Is there a formula for this? Eg. Unfortunatly. • L is the length of the shortest path. The distance between two points is the length of the path connecting them. This is an implementation of the dijkstradlDLs algorithm, which finds the minimal cost path between two nodes. This work proposes a method for evaluating the likelihood that taking an observation will reduce the best path cost. The algorithm will trace the road by finding the shortest path over a cost grid between start point and end point. We already observed that in taxicab geometry there can be many paths of minimal length between points A and B. And a set of paths that connect. Here we will see the Johnson's Algorithm to find shortest path between two vertices. Shortest Path Problem in Graphs The shortest path problem is perhaps one of the most basic problems in graph theory. Edges have an associated distance (also called costs or weight). However, you could not use this technique when setting the spatial geometry object or the end measure value for a link. Given two different positions on a chess board, find the least number of moves it would take a knight to get from one to the other. pdf This study focuses on finding the shortest paths among cities in Java Island by repeatedly combining the start node's. Then draw the path on the figure. Write the paths as text to see the general format of all paths & an easy method to enumerate them And that's the key lesson: It's completely fine to use one model to understand the idea, and another to work out the details. 4 Shortest Paths. Based on the problem statement in the comments, the problem is to find the shortest path between two points in a grid, where the grid can contain obstacles. A bonus: We can select whichever order of collecting keys that gives the shortest total path. Even if no two edges have the same weight, there could be two paths with the same weight. I need to find shortest path between two points in a grid given an obstacles. The visibility graph of a polygon is a graph whose nodes corresponds to the vertices of the polygon, and whose edges correspond to the edges in the polygon formed by joining the vertices that can "see. If Station code is unknown, use the nearest selection box. Your help is greatly appreciated. de Rezende et al. The shortest path between $(0,0)$ and $(1,1)$ can be $(0,0)\to (0,1)\to (1,1)$ or $(0,0)\to(1,0)\to(1,1)$, so there are two shortest paths. Dijkstra's algorithm is one the dynamic programming algorithm used to find shortest path between two vertex in the graph or tree. [1987] present an O(n log ~z) algorithm for finding the rectilinear shortest path between two points that avoids a set of rectangles (with sides parallel to the coordinate axes). How are the points arranged? 7. 1) findroute(i,fa[i][j]); path[cnt++]=j; } static void extend_floyd() { for(int k=1;k<=nnum;k++) { //Extended folyd recorded weights and to identify the smallest ring. However, if there are obstacles between point A and B, the algorithm will often select the wrong tiles. Learn more problem solving techniques. This problem also known as "Print all paths between two nodes" Given a graph, source vertex and destination vertex. Given two vertices in a graph, a path is a sequence of edges connecting them. In two dimensions, this path can be represented by a vector with horizontal and vertical components. This was done by expanding the grid to 3-dimensions, and implementing the rest of the algorithm the way it would be done on a 2-dimensional grid. The Link state routing algorithm is also known as Dijkstra's algorithm which is used to find the shortest path from one node to every other node in the network. It does not necessarily need to be the shortest distance path. i have assign to do a shortest path in GPS system code in c. Java based solution to find the shortest path's distance between 2 Grid Cells. However, Bellman-Ford and Dijkstra are both single-source, shortest-path algorithms. Displays distance/direction from your currentl location to all points, or between any two points. If the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities. The shortest path in a grid consists into adding horizontal edges or vertical edges to fill the gap between the two pairs of coordinates. Thus instead of the usual ancestor array we additionally must store the edge number from which we came from along with the ancestor. Pre-requisites: 1. Do interpolation of the points along the path, if any point has 1, it is crossing the wall. Use the appropriate S&W class to find the shortest path from Harvard station to OakGrove station. Dijkstra's algorithm is one the dynamic programming algorithm used to find shortest path between two vertex in the graph or tree. When ant colony completes a tour, the pheromone in the best path is strengthened and is modified by applying the global updating rule. What is Dijkstra Algorithm? To understand Dijkstra’s algorithm, let’s see its working on this example. polygon exactly once. java: a node in the graph; AStar. Vertices are numbered automatically and can carry text labels. Oracle® Spatial Java API Reference 11g Release 2 oracle. The weight of the shortest path from s to any unreachable vertex is also trivial: +∞. Algorithm to find the shortest path, with obstacles I have a collection of Points which represents a grid, I'm looking for an algorithm that gets me the shortest distance between point A and B. I realised I couldn't get across the key points anywhere near as clearly as he has done, so I'll strongly encourage you to read his version before. Calculates the geodetic length of the geometry. Satellite, Terrain, Road, and basic Topographic maps show you, your locations, and lets you enter in new ones. Here X means you cannot traverse to that particular points. See the next few slides to realise this. Standalone GPS devices (without any additional support from server on the internet etc) will often look at the direct distance between the departure and arrival points, and if they are above a certain threshold the GPS will first find the closest highways at your departure and arrival points, to determine what highways connect the general areas you want to travel between. The shortest path in a grid consists into adding horizontal edges or vertical edges to fill the gap between the two pairs of coordinates. It's about a map with thousands and thousands of roads, and a tiny number of locations to visit. Here is another way to think about things. The the minimum-cost path provides a path with shortest-distance from source to destination grid. We have to construct a shortest path between the two odd-degree vertices. First, * if we consider the case where k = 0 and all paths must have no edges, then * we get that *. Write an algorithm to count all possible paths between source and destination. Computer games can use all three types of grid parts, but faces are the most. We are also given a starting node s ∈ V. It differs from the minimum spanning tree because the shortest distance between two vertices might not include all the vertices of the graph. Formula to Find Bearing or Heading angle between two points: Latitude Longitude. Distance between two points calculator uses coordinates of two points A(x_A,y_A) and B(x_B,y_B) in the two-dimensional Cartesian coordinate plane and find the length of the line segment \overline{AB}. In this assignment, you will build an edge-weighted graph where nodes represent locations on campus and edges represent straight-line walking segments connecting two. Dijkstra’s algorithm (or Dijkstra’s Shortest Path First algorithm, SPF algorithm) is an algorithm for finding the classical shortest paths between nodes in a graph. Thoughts: Because the grid is filled with non-negative numbers. To find the shortest path, we can use another graph traversal approach known as Breadth-first search. 4) Choose one of the shortest-path algorithms to work with and one of the chargers for your goal. * * @param graph The graph to be searched for the shortest path. Our paths will run vertically and horizontally between the regularly spaced points of a rectangular grid. Applications-. Edge with costs − 1 2 8. Learn more problem solving techniques. DFS finds a path but you cant be sure if its the right one until you find the others. Any sample that finds the shortest route between two (or more) points will work. The shortest distance between two points is not a celebrity, or being next to a celebrity. I wrote the following code to define the shortest path between two points on that grid. How many possible unique paths are there? Java Solution 1 - DFS. Finding the paths — and especially the shortest path — between two nodes is a well studied problem in graph theory. Shortest path is defined by the minimum number of vertexes traversed. P = shortestpath(G,s,t) computes the shortest path starting at source node s and ending at target node t. Put Strategy in its own package like algo. Here X means you cannot traverse to that particular points. However, there’s something important to consider: it’s not only solving the shortest path between source and goal, it’s finding the shortest path between the source and any other point. When the front crosses each grid point, the person standing there writes down this crossing time T. At the beginning all nodes are white. The average path length in the graph – average distance between any couples of nodes may also highlight interesting properties of a given graph. Steps Step 1: Remove all loops. In this article we will implement Djkstra's – Shortest Path Algorithm (SPT) using Adjacency List and Min Heap. The videos are avi files that include full audio and high resolution screen capture of all activity on the tablet PC. Consider an n-by-n grid of cells, each of which initially has a wall between it and its. You may choose to wait to find and draw the shortest path until the user clicks a button, if you wish, but you don't need to. Drag the points to create an exactly horizontal line between them. A robot is located at the top-left corner of a m x n grid. When we were working with shortest paths, we were interested in the optimal path. How many possible unique paths are there? Java Solution 1 - DFS. ArrayList> getShortestPaths() Returns the shortest path(s) to this synset (there may be more than one). Our approach is not linked to a specific biological problem and can be applied to a large variety of images thanks to its generic implementation as a user-friendly ImageJ/Fiji plugin. A geodesic segment is the shortest path between two points on an ellipsoid. A* is a popular choice for graph search. Figure 6-3 illustrates this format, which appears in many classic arcade games, board games, and tactical role-playing games. This article presents a Java implementation of this algorithm. I need to find shortest path between two points in a grid given an obstacles. I would like to help you write it but Java isnt my language :). 6 Shortest-Path Problems Given a graph G = (V;E), a weighting function w(e);w(e) > 0, for the edges of G, and a source vertex, v 0. There are two approaches: Pick a city, and run the Fast Marching Method to compute the shortest path from every other city back to the city. There are 2 classes. Our paths will run vertically and horizontally between the regularly spaced points of a rectangular grid. shortest-path-in-a-grid-with-obstacles-java. There are many real-world applications that translate to this problem, one of which is answering optimal-path queries in route planning systems such as timetable information services for public transport or car navigation systems. It is named Stationary Level Set Equation. Find Shortest Path in a Maze Given a maze in the form of the binary rectangular matrix, find length of the shortest path in maze from given source to given destination. We observe that the length of the shortest path through N= n2 points arranged on a regular grid G N ˆ[0;1]2 is roughly p N. But we need to find the shortestpath between two nodes using node property filter using Neo4j 3. to the length of any shortest path between the points on the grid G ( S ). The goal of Dijkstra’s algorithm is to conduct a breadth-first search with a higher level of analysis in order to find the shortest path between two nodes in a graph. The one-to-all shortest path problem is the problem of determining the shortest path from node s to all the other nodes in the. est path exists. All-Pairs Shortest Path Algorithm There is a simple method for finding the shortest path between any two nodes in Graph G(V,E). A robot is located at the top-left corner of a m x n grid. Shortest distance is the distance between two nodes. This problem could be solved easily using (BFS) if all edge weights were ($$1$$), but here weights can take any value. Distance surfaces are often used as inputs for overlay analyses; for example, in a model of habitat suitability, distance from streams could be an important factor for water-loving species, or distance. js is a great library for both basic and advanced usage. Java Solution This problem can be solved by BFS. Here, this function used Dijkstra's algorithm. The path from EgonWillighagen to Jandot : Neo4j , a graph API for java: my notebook. As described in previous sections, our O(n log n) algorithm does not find the optimal/shortest path. Displays distance/direction from your currentl location to all points, or between any two points. How do I find the shortest distance between x,y coordinates in excel Ok so I have a project in a linear programming class of mine here it is: I must pick up 40 tennis balls (x,y points) These balls must all be picked up and put in a basket. Step 3: Create shortest path table. Greenwich meridian The meridian adopted by international agreement in 1884 as the prime meridian, the 0-degree meridian from which all other longitudes are calculated. Here’s how it works: Pick the start and end nodes and add the start node to the set of solved nodes with a value of 0. 2) Find the minimum spanning tree from your start point using depth first search. Motivation: We introduce a formulation for the general task of finding diverse shortest paths between two end-points. shortest-path-unweighted-graph-bsf-java. The number of connected components is. A geodesic segment is the shortest path between two points on an ellipsoid. There are many real-world applications that translate to this problem, one of which is answering optimal-path queries in route planning systems such as timetable information services for public transport or car navigation systems. The second value we use to choose the best next step is the cost of getting to the current location, i. You: If I walk 3 blocks East and 4 blocks North, how far am I from my starting point?. The shortest path is the shortest pipeline distance on the NTS. hi, im having problem for my assignment. Step 2: (you will receive up to 9 points if you completely only the first two steps). Then, O(1) algorithms are presented for: (i) testing the existence of specific types of paths, (ii) finding an ASP and a SDP. The path from EgonWillighagen to Jandot : Neo4j , a graph API for java: my notebook. java Your completed is a path between two given nodes. Draw Path on Google Maps Android API Last modified on June 19th, 2017 by Joe. The path can only be constructed out of cells having value 1 and at any given moment, we can only move one step in one of the four directions. It is used to solve All Pairs Shortest Path Problem. java with today's tasks is listed below. A cubic Bezier curve geometry built from the two given points with the given coefficient for the radius and composed of the given number of points - the boolean is used to specified if it is the right side and the last value to indicate where is the inflection point (between 0. This task bridges between standards 8. The path can only be created out of a cell if its value is 1. This is an implementation of the dijkstradlDLs algorithm, which finds the minimal cost path between two nodes. include these two library then work. ) path between a source and a destination. When a node gets visited for the first time it is colored non-white. The leash is required to be a geodesic joining its endpoints. Here is my file where I. Path planning is an important research direction in the field of mobile robots, and it is one of the main difficulties in research on such robots. Then draw the path on the figure. I need to find shortest path between two points in a grid given an obstacles. There are some features here you may use to. LeetCode – Word Ladder II (Java) Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: 1) Only one letter can be changed at a time, 2) Each intermediate word must exist in the dictionary. Set the pathTypeattribute to change how the lines of a shape are drawn (for example, Path. You would need to check the route by increasing the buffer size and finding the shortest route again. Java users must implement two methods in the ShortestPaths class:. obstacle-avoiding path, preferably a shortest one, on the grid between two given grid points. Bicycle and walking paths are preferentially weighted, and the interstate is heavily penalized. My LeetCode Solutions! Contributing. In such a case, although a bit controversial, using A* search over Dijkstra is common practice. And we need to find out all possible ways of path from st-end point. Thus, in practical travel-routing systems, it is generally outperformed by algorithms which can pre. In this paper, we address the shortest path problem in hypergraphs. Set the array - bundle First is to create an empty array named 'Grid'. The distance between two consecutive parallel streets is either 1 or 5. This problem also is known as “Print all paths between two nodes”. The path can only be created out of a cell if its value is 1. Keep track of the distance traveled as you draw your path. Monotonic shortest path. * @param source The source node of the graph specified by user. Generalizations of these algorithms to the problem of finding a spanning or Stein tree connecting multiple grid points are usually straightforward. ← Translate between Java and C++ style identifiers Spiral →. Subject: Re: [R-sig-Geo] gdistance, shortest path => "cannot derive coordinates from non-numeric matrix" To: r-sig-geo at r-project. The idea is to use Breadth First Search (BFS) as it is a Shortest Path problem. Write an algorithm to print all possible paths between source and destination. Distance surfaces are often used as inputs for overlay analyses; for example, in a model of habitat suitability, distance from streams could be an important factor for water-loving species, or distance. View on GitHub myleetcode. It works as follows It works as follows File Information; Description: ALLSPATH - solve the All Pairs Shortest Path problem. From to , choose the shortest path through and extend it: for a distance of There is no route to node , so the distance is. The only changes to the original source code are to add this value to the priority when entries are added to the priority queue and a simple if statement which stops processing when we have found the node for which we are searching. Let dist(v,w) denote the shortest-path distance from vertex v to vertex w with respect to ℓ. I have created one 2d array(n,n). You will need to use this function to determine the cost of moving from one grid location to another. Let's say would like to find the shortest path between vertices g and n. The search terminates when two graphs intersect. Subtask #3: (30 points): M = 2. The shortest traversable path between any 2 given waypoints is a straight line; however, this linear trajectory might not be traversable due to the variations in the terrain. How do you find the number of the shortest distances between two points on a grid where you can only move one unit up, down, left, or right? Is there a formula for this? Eg. If the shortest path through w does not go through v, then the weight of v plus the weight of v-w must be greater than or equal to the. The problem I'm running into is that the shortest path feat. Initially, , and each following state is obtained by. But we are stuck at this point. Grid Distance. Betweenness is how often the vertex lies on the shortest path between two other vertices. It can also be used for nding costs of shortest paths from a single vertex to a single destination vertex by stopping the. A-Star-Shortest-Pathfinding-Algorithm-Square-Grid-Java. What is Dijkstra's Algorithm? Dijkstra's Algorithm is useful for finding the shortest path in a weighted graph. Suppose that somebody is standing at each red grid point with a watch. Cypher provides two ways to specify a shortest path. Use Figure 2 to visualize the geometry of a great circle path/plane. Algorithm to find the shortest path, with obstacles I have a collection of Points which represents a grid, I'm looking for an algorithm that gets me the shortest distance between point A and B. Given a MxN matrix where each element can either be 0 or 1. In this post I will be exploring two of the simpler available algorithms, Depth-First and Breath-First search to achieve the goals highlighted below: Find all vertices in a subject vertices connected component. The above visualization shows the basic algorithm working to find the shortest path. Here X means you cannot traverse to that particular points. In two dimensions, this path can be represented by a vector with horizontal and vertical components. The text input file in the first row of the maze width n and m in length (1 <= n <= 20, 1 <= m <= 20). Pick a set of pivot points and then find the shortest paths between them. The network we used to perform the Multi Shortest Paths retrieval. The algorithm computes distances from the leftmost vertex. The code has been written in five different formats using standard values, taking inputs through scanner class, command line arguments, while loop and, do while loop, creating a separate class. minimum cost path with right, bottom moves allowed find the minimum number of moves needed to move from one cell of matrix to another shortest path in grid with obstacles minimum cost path dijkstra minimum cost path matrix java maximum cost path dynamic programming shortest path in a binary maze python shortest distance between two cells in a. A modification of code published by Jorge Barrera to return all paths that tie for shortest path. Then draw the path on the figure. Maze-running algorithms can be characterized as target-directed grid propagation. Get distance/direction between two points by tapping the path drawn between them. The videos are avi files that include full audio and high resolution screen capture of all activity on the tablet PC. Shortest paths in graphs. This is homework. Here we can imagine breaking the work space of the robot into a three dimensional grid where each cell is acute and then using the grassfire algorithm to plan a path between two different cells in that grid. Suppose the state space consists of all positions (ar, y) in the plane. Dijsktra in 1956 and published three years later, Dijkstra's algorithm is a one of the most known algorithms for finding the shortest paths between nodes in a graph. Returns: array with the lists of points of the shortest paths. t==0 corresponds to a, t==1 corresponds to b. The code has been written in five different formats using standard values, taking inputs through scanner class, command line arguments, while loop and, do while loop, creating a separate class. [7 points] Paths. interpolating from San Francisco to Tokyo will pass north of Hawaii and cross the date line. The robot has 2 color sensors and an ultrasonic sensor. If we compare the work required to run the grassfire algorithm on a 100 by 100 grid in 2D and a 100 by 100 by 100 grid in 3D. Calculates the distance between two point of the Earth specified geodesic (geographical) coordinates along the shortest path - the great circle (orthodrome). It is an embedded, disk-based, fully transactional Java persistence engine that stores data structured in graphs rather than in tables. Output the minimum number of moves the robot needs to make to reach the destination. The interpolation takes place along the short path between the points potentially crossing the date line. To see this, note that the shortest distance between two neighboring points is 1=(n 1) and the n2 points are arranged in nparallel lines. What would be a good and simple algorithm to find the shortest route between two points in a 2D array[grid] ? There can be certain obstacles in the grid i. A group of islands. The total cost of a path is the sum of the costs of all edges in that path, and the minimum-cost path between two nodes is the path with the lowest total cost between those nodes. The network we used to perform the Multi Shortest Paths retrieval. When the faces are lying flat, the shortest path between A and Bis the line segment joining A to B as shown in figure 2. You're a moron. And how do I proceed for this. Double precision floating point arithmetic is used for PageRank values. In this assignment, you will build an edge-weighted graph where nodes represent locations on campus and edges represent straight-line walking segments connecting two. For example, the State of New York specifies that the maximum difference in riser height for any flight of stairs is 3/8 inch. Game Character Path Finding in Java. Each dot is a major US city. the shortest path) between that vertex and eve-ry other vertex. IPv6 Internet Protocol version 6. Vertices are numbered automatically and can carry text labels. I have created one 2d array(n,n). Oct 4, 2016 • shortest-paths • Christoph Dürr and Jin Shendan Related problems: [spoj:Laser Phones] [spoj:Wandering Queen] Given a grid with a source cell, a destination cell and obstacle cells, find the shortest path from the source to destination, where every direction change along the path costs 1. The path you get through this approach is always a shortest path. f) Use the Star Links feature on the Oracle of Bacon website to find shortest paths between each pair of actors you can make from the three you chose above. , no inaccessible locations, no cycles, and no open spaces. We consider the latter problem and present four different parallel algorithms, two based on a sequential shortest-path algorithm due to Floyd and two based on a sequential algorithm due to Dijkstra. Below is the syntax highlighted version of DijkstraSP. This is how the Greeks understood straight lines, and it coincides with our natural intuition for the most direct and shortest path between the two endpoints. We think of the distance between two points as the length of the shortest path connecting them. The labeling method for the shortest path problem [21, 22] finds shortest paths from the source to all vertices in the graph. The width of a branch is proportional to the square root of the sum of branches reachable by that branch. This is the current estimated shortest path. You can now use the minimum distance between a particular position and all the nodes plus the pre-calculated distance from the node to the end as a heuristic function. All are proved in CLRS or in the notes from Recitation 15. - Labeling continues until the target grid cell T is marked in step L. Here, this function used Dijkstra's algorithm. f(n) is the true shortest path which is not discovered until the A* algorithm is finished. Breadth-first search is an algorithm for finding the shortest path between two states if the distance between any two neighbouring states is the same (in this problem, to get from one state to a neighbouring always takes one move). The contribution done by Zhihong, Xiangdan and Jizhou [4], focuses on inherent parallelism and scalability that make the algorithm suitable for grid computing task scheduling. Find the shortest distance from a source cell to a destination cell, traversing through limited cells only. In Figure 1 calculated by equation (), the mobile robot considered as a scale-free particle adopts the movement form of linear octree and reaches the neighbor grids along the top, right, bottom, left, upper right, lower right, lower left, and upper left. But if a grid is far from being complete, the holes introduce a certain kind of hierarchy as well because now shortest paths tend to use their borders. The net poles are located at (12,39) and (48,39). So let's take a look at the "common sense" solution: the simplest intuitive algorithmic solution would be to start at any given point $(x_1,y_1)$, find the nearest $(x_b,y_b)$, connect those with a line, and then connect $(x_b,y_b)$ to its. When we account for the shortest paths for all N⋅( − 1)/2 pairs of points in the grid graph, two probability matrices, (denoted by PR and PC), can be defined to represent the probabilities of all horizontal and vertical segments, respectively, through which the shortest paths would pass. 3 Calculate time to travel between all consecutive pairs of letters in aphorism. Find the shortest path between A. Shortest path length is %d. Breadth-first-search is the algorithm that will find shortest paths in an unweighted graph. 1 A shortest path between two vertices s and t in a network is a directed simple path from s to t with the property that no other such path has a lower weight. In order to recover the full path this variant of the algorithm would require O(D^2) space to recover the full path. where i need to create a map or path and ask the user to insert starting point and destination and we also have to calculate and display 3 shortest path based on ranking and display the history record. We propose an optimal-time algorithm for a classical problem in plane computational geometry: computing a shortest path between two points in the presence of polygonal obstacles. It’s an online Geometry tool requires coordinates of 2 points in the two-dimensional Cartesian coordinate plane. Construct visibility graph • 2. The CWD approach is a method of deriving a unit price for a point based upon the revenue expected from that point, the average distance of that point from Entry Points on the NTS for Entry (or from Exit Points in. Given two vertices in a graph, a path is a sequence of edges connecting them. We are given the following graph and we need to find the shortest path from vertex ‘A’ to vertex ‘C’. It is a problem of determining the shortest path between two( start and stop) given nodes. Each line is a connection between two cities. I have 3 points X,Y,Z, lets call them buildings. Adjacency Matrix. A Practical Guide to Data Structures and Algorithms Using Java Sally A. java, and any other java programs that the myWeightedGraph class depends one. jar along with Graph. 1 A shortest path between two vertices s and t in a network is a directed simple path from s to t with the property that no other such path has a lower weight. The idea is to use Breadth First Search (BFS) as it is a Shortest Path problem. The shortest path between two points on the surface of a sphere is an arc of a great circle (great circle distance or orthodrome). Properties of a shortest cut graph with one vertex a b a b Lemma If the initial cycles form a shortest one-vertex cut graph, then the horizontals and the verticals are shortest paths. Visualizing the grid to understand the general problem and see a single path. Dijkstra's algorithm is used to find the shortest path between any two nodes in a weighted graph. I can’t imagine any reasonable theory of geometry in which this would be an axiom, though it’s possible that such a theory can be constructed (likely very artificially). Retrieve all the edges of this node. We define one matrix for tracking the distance from each building, and another matrix for tracking the Shortest Word Distance II (Java) LeetCode - Shortest Word Distance III (Java) (grid[i][j] == 0 && reach[i][j] == numBuilding) {on this line, should grid[i][j] be compared with -1 and. The backbone of the CSP is the Dijkstra’s path search algorithm (Dijkstra, 1959), which requires two points: the source node and the destination node. Finding the paths — and especially the shortest path — between two nodes is a well studied problem in graph theory. The dots are in the same place but instead of drawing diagonal lines between two points, your route can only follow the grid lines. The solid line is the actual flight path, the red dashed line is the Great Circle route between Chicago and Dubai. It computes the shortest path between every pair of vertices of the given graph. Note: At the End of the article you will know what needs to be included if you want to print the diagonal paths as well. Looking at the example in Figure 1, all the edges are bi-directional except for one. The Distance toolset contains tools that create rasters showing the distance of each cell from a set of features, or that allocate each cell to the closest feature. Java users must implement two methods in the ShortestPaths class:. As mentioned at the end of Section 9. Drawing a route on android Google Map API v2 can be challenging but in this tutorial we will see what is need to draw a path from a user current location to a point in the map that as serves as the user destination. Let's consider the graph in figure 1. The algorithm for generating shortest convex polygonal chains is the same as before: generate the shortest line segments, no two parallel, sort them by slope, and connect them. CAD for VLSI 10 Phase 1 of Lee's Algorithm • Wave propagation phase - Iterative process. There are many real-world applications that translate to this problem, one of which is answering optimal-path queries in route planning systems such as timetable information services for public transport or car navigation systems. And in the case of BFS, return the shortest path (length measured by number of path. pdf This study focuses on finding the shortest paths among cities in Java Island by repeatedly combining the start node's. Motivation: We introduce a formulation for the general task of finding diverse shortest paths between two end-points. passes the grid graph’s vertices as the points in the ETSP problem, where the desired length k is the number of points; if there is such a tour, it must be a cycle in the grid graph because choosing any nonadjacent pair of points will prevent meeting the length bound, and if a Hamiltonian cycle exists in the grid graph, it is a valid tour. The Pythagorean Theorem lets you use find the shortest path distance between orthogonal directions. You are not required to implement this algorithm in Java, but if you do not, please provide clear and terse pseudocode, with an explanation of how it works. [1985] and Wu et al. But the one that has always come as a slight surprise is the fact that this algorithm isn't just used to find the shortest path between two specific nodes in a graph data structure. A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, , C_k such that: Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie. I need to find the shortest amount of path that connects the 3 buildings, these buildings can be in any sort of shape and any distance from each other, lets call the distance between each building xy, xz, yz. We think of the distance between two points as the length of the shortest path connecting them. Goldman Washington University in St. The findPath() method receives a map array of integers where 0 is an empty cell, and 1 is an obstacle, the function returns a list of coordinates which is the optimal path or null if such path does not exist. We will also need a way of storing the shortest path: I'll explain the Shortest Path Tree in the next step. The path can only be constructed out of cells having value 1 and at any given moment, we can only move one step in one of the four directions. The problem is to compute a set of unit-length baffles (walls) separating grid points that forces the given path to be the unique shortest path from its starting point to the end point. A quick update before a larger unveiling next week… We’ve made solid progress with implementing the grid-based capabilities of the SpaceAnalysis package. a shortest path (constrained by some factor) between two locations using an optimal combination of on-road and off-road access; this is the Cross-Country Problem (CCP). (Shortest path problem - Wikipedia, the free encyclopedia, 2011) In other words, when we have to find a path with minimum cost to go from a place to another place which there are a number of intermediate points in between to travel to with different costs, we are dealing with the shortest path problems. The Point class should have a method for reading values, printing values and calculating the distance between two points. The shortest distance between two points is not a celebrity, or being next to a celebrity. A-Star-Shortest-Pathfinding-Algorithm-Square-Grid-Java. Drawing a route on android Google Map API v2 can be challenging but in this tutorial we will see what is need to draw a path from a user current location to a point in the map that as serves as the user destination. I'm trying to use a grid network that I have created to find the shortest path between two points accounting for directional wind-speed. A path with the minimum possible cost is the shortest distance. So it is very unlikely that anyone is going to help you. Draw the shortest path between the two selected buildings directly on the map. The benchmark runs 50 PageRank iterations. Dijkstra’s Shortest Path Algorithm in Java. Distance tools can also calculate the shortest path across a surface or the corridor between two locations that minimizes two sets of costs. There are many applications of the shortest path problem. include these two library then work. The representation of pairs in the grid requires aboutvwlog(v+w)bits. Betweenness is how often the vertex lies on the shortest path between two other vertices. This definition is succinct, but its brevity masks points worth examining. In some cases no algorithm could attain any of these goals—for example, point A could be on an island completely isolated from point B. Joksch [7] used this idea to describe an algorithm for. Consider the graph above. This is homework. As described in previous sections, our O(n log n) algorithm does not find the optimal/shortest path. I am looking for the shortest path between start and end. Algorithm: 1. Use SPT [] to keep track of the vertices which are currently in Shortest Path Tree (SPT). Another Wall in the Maze In ACM/ICPC contests, you'll often see questions such as "find the shortest path out of this maze. This is an idealization of the problemthat a robot has to solve to navigate its way around a crowded environment. right to coding this algorithm all at once. path from b to node node. So it’s not really about right triangles — it’s about comparing “things” moving at right angles. Afterwards, we have some graph (where each node refers to at most 13 13 1 3 places, and at most 2 6 2^6 2 6 states of keys). The following Matlab project contains the source code and Matlab examples used for dijkstra algorithm consistent with cyclic paths. Many pathfinding problems in real-world applications require real-time computation of the shortest path between two points in a grid-based environment. [2 points] Consider an undirected, connected graph G where all edges have a weight of 1, and a spanning tree T of G. Pathfinding algorithms are techniques for navigating maps, allowing us to find a route between two different points. Dijkstra’s algorithm is one the dynamic programming algorithm used to find shortest path between two vertex in the graph or tree. If you nay doubts related to the information that we shared do leave a comment here. Previous Next In this post, we will see Dijkstra algorithm for find shortest path from source to all other vertices. While the new path increases in pheromone levels, the old path's pheromone level will be reduced. So it’s not really about right triangles — it’s about comparing “things” moving at right angles. Here are instructions for setting up an IntelliJ-based Java programming environment for Mac OS X, Windows, and Linux. This Android tutorial is to demonstrate a sample application which will draw path for a route in Google map using Android API v2. Steps Step 1: Remove all loops. The cost ofa RST is the sum of the lengths of the edges of G( S ) that are used in the RST. Then draw the path on the figure. Afterwards, we have some graph (where each node refers to at most 13 13 1 3 places, and at most 2 6 2^6 2 6 states of keys). The Dijkstra Algorithm is used to find the shortest path in a weighted graph. For shortest path we do the breadth first traversal. some of the cells may be inaccessible. The following n lines describe the labyrinth. The dots are in the same place but instead of drawing diagonal lines between two points, your route can only follow the grid lines. How many statesare there? How many paths are there to the goal? b. A depth-first search solution is pretty straight-forward. The primal way to specify a line L is by giving two distinct points, P 0 and P 1, on it. In this paper our interest is in the online version of the cell exploration problem. It is a generalization of the notion of a "straight line" to a more general setting. A geodesic is the shortest path between two points on a curved surface, analogous to a straight line on a. 4 Shortest Paths. Proof: There exists a shortest path from A to B that passes through Q, see fig. Subtask #1 (6 points): N, Q ≤ 10 3; Subtask #2: (11 points): M = 1. Then, O(1) algorithms are presented for: (i) testing the existence of specific types of paths, (ii) finding an ASP and a SDP. (see lecture notes for details). Java users must implement two methods in the ShortestPaths class:. in the way is to wrap the path as close to a straight lines around those obstacles as possible. To keep track of the total cost from the start node to each destination we will make use of the distance instance variable in the Vertex class. If you're pursuing that, you're an idiot. Write a program Maze. Here is an implementation I made in C with. Betweenness shows which cases are communication paths between other cases which can be useful to determine points where the network would break apart. 8, and thus it illustrates the cluster 8. The net poles are located at (12,39) and (48,39). Use SPT [] to keep track of the vertices which are currently in Shortest Path Tree (SPT). I'm trying to use a grid network that I have created to find the shortest path between two points accounting for directional wind-speed. Java program to calculate the distance between two points. Tasks: Try out the demo applet below to recap how shortest path works. Clarkson et al. And a set of paths that connect. In this book we follow Struik and define geodesics as below:. In DFS, one child and all its grandchildren were explored first, before moving on to another child. The route line is drawn on the shortest route between the locations. Learn how to find the shortest path using breadth first search (BFS) algorithm. Then draw the path on the figure. The recursive algorithm described above finds the path, but it isn't necessarily the shortest path. In this setting, we require that every edge is drawn as a shortest path between its two endpoints and we call an embedding with this property. Typically this is represented by a graph with each node representing a city and each edge being a path between two cities. The classical theorem of Fáry states that every planar graph can be represented by an embedding in which every edge is represented by a straight line segment. Vertices are numbered automatically and can carry text labels. To find this path we can use a graph search algorithm, which works when the map is represented as a graph. You can also let the system find the shortest path between two points or the shortest closed path containing a given edge. This is the current estimated shortest path. The API returns information based on the recommended route between start and end points, as calculated by the Google Maps API, and consists of rows containing duration and distance values for each pair. Queries are also randomly generated. Given a 2 dimensional matrix where some of the elements are filled with 1 and rest of the elements are filled. The graph is given here. (Drawing horizontally along 6th Ave and vertically along E St will be suggestive later for using the Pythagorean Theorem. I want to interpolate through one point in every single x column (diagonal is also okay for me) but at the end, I need the shortest possible curve from A to B. If your obstacles are aligned on a grid, the navigation points will be aligned with the vertices of the grid. This implies that negative edge weights are not allowed in undirected graphs. There are 2 classes. For a given source node in the graph, the algorithm finds the shortest path between that node and every. You may choose to wait to find and draw the shortest path until the user clicks a button, if you wish, but you don't need to. Three different algorithms are discussed below depending on the use-case. This problem also known as "paths between two nodes". curve between two given points. The Point class should have a method for reading values, printing values and calculating the distance between two points. I have created one 2d array(n,n). Satellite, Terrain, Road, and basic Topographic maps show you, your locations, and lets you enter in new ones. This video is a part of HackerRank's Cracking The Coding Interview Tutorial with Gayle Laakmann McDowell. There are two algorithms which are being used for this. There are some features here you may use to. BFS is an algorithm to find the shortest path between two points. I want some logic input. Problem You will be given graph with weight for each edge,source vertex and you need to find minimum distance from source vertex to rest of the vertices. I would like to know:. It is specified that all moves have the same cost, so Dijkstra's algorithm will produce ex. First of all, let us understand what we mean by m-adjacency. 3 Calculate time to travel between all consecutive pairs of letters in aphorism. This is a new book on level set methods and Fast Marching Methods, which are numerical techniques for analyzing and computing interface motion in a host of settings. de Rezende et al. C++ easy Graph BFS Traversal with shortest path finding for undirected graphs and shortest path retracing thorough parent nodes. Then we know the size of the two paths, so we can easily calculate the distance by the formula, length of path1 + length of path2 - 2*length of the common part. java computes the shortest paths in a graph using a classic algorithm known as breadth-first search. Oct 4, 2016 • shortest-paths • Christoph Dürr and Jin Shendan Related problems: [spoj:Laser Phones] [spoj:Wandering Queen] Given a grid with a source cell, a destination cell and obstacle cells, find the shortest path from the source to destination, where every direction change along the path costs 1. Major things to consider would be how far ahead you should check and whether multple blocks should be anticipated. Learn more problem solving techniques. Then, compute the length of the shortest path from s to v for each vertex v, say dist[v]. Find if there is a path between two vertices in an undirected graph; Paranthesis Theorem; Minimum number of colors required to color a graph; Shortest path with exactly k edges in a directed and weighted graph | Set 2; Number of pairs such that path between pairs has the two vertices A and B; Minimum Spanning Tree using Priority Queue and Array. The problem I'm running into is that the shortest path feat. Even for the geo-graphical localized routing protocols, such as greedy routing, the packets usually follow the shortest paths when the network. A cost grid is given in below diagram, minimum cost to reach bottom right from top left is 327 (= 31 + 10 + 13 + 47 + 65 + 12 + 18 + 6 + 33 + 11 + 20 + 41 + 20) The chosen least cost path is shown in green. Draw any grid of hexes or squares. Dijsktra in 1956 and published three years later, Dijkstra's algorithm is a one of the most known algorithms for finding the shortest paths between nodes in a graph. The shortest path between any two points is a straight line. This tweet was actually a retweet by @BestNoSQL (bottom right), which mentioned both @maxdemarzi (top left) and @BestNoSQL. pute shortest path queries. If your obstacles are aligned on a grid, the navigation points will be aligned with the vertices of the grid. Dijkstra's Algorithm is a very well-known graph traversal algorithm for finding the shortest path from a given node/vertex to another. Notation − d(U,V) There can be any number of paths present from one vertex to other. For maps in general, not only grid maps, we can analyze the map to generate better heuristics. Even for the geo-graphical localized routing protocols, such as greedy routing, the packets usually follow the shortest paths when the network. Assume we have two paths in G, one with weights 2 and 2 and another one with weight 3. Here, this function used Dijkstra's algorithm. Shortest path algorithms. It's about a map with thousands and thousands of roads, and a tiny number of locations to visit. Often used to find the shortest distance between 2 points, e. There are many applications of the shortest path problem. We need to define a "point" class having two data attributes 1) row no and 2) column no. What A* is meant to do is find the shortest and most efficient path to get from point A to point B. Their method capitalizes on the ability to look up the cost of the sub-path in an array of costs; thus, we can compute the shortest path. The Bellman-Ford algorithm for SSSP, Single-Source Shortest Path, finds the shortest paths from a source vertex to all other vertices in the graph. The Link state routing algorithm is also known as Dijkstra's algorithm which is used to find the shortest path from one node to every other node in the network. The lectures are provided in two formats: video and pdf. This is an idealization of the problemthat a robot has to solve to navigate its way around a crowded environment. Amazon Interview Questions that can find the shortest path between any two nodes the possible paths from any points to bottom right of a mXn matrix with the. (Drawing horizontally along 6th Ave and vertically along E St will be suggestive later for using the Pythagorean Theorem. This is an idealization of the problemthat a robot has to solve to navigate its way around a crowded environment. Major things to consider would be how far ahead you should check and whether multple blocks should be anticipated. The shortest path is the shortest pipeline distance on the NTS. With the exception of lines of latitude and longitude, great circle arcs do not follow a constant direction relative to true north and this means that as you travel along the arc your heading will vary. */ private static ArrayList shortestPath = new ArrayList(); /** * Finds the shortest path between two nodes (source and destination) in a graph. 0 the process of plotting an e ciently traversable path between points, called nodes. java from §4. Contributions are very welcome! If you see an problem that you’d like to see fixed, the best way to make it happen is to help out by submitting a pull request implementing it. , the obstacles) How can the robot find a collision-free path to the goal? × Search tree 17 × × × × Now, the robot can searchits model for a collision-free path to the goal 18 1000×1000 grid Æ1,000. A geodesic is the shortest path between two points on a curved surface, analogous to a straight line on a plane surface. Update: One day after writing this article I realized that glReadPixels was causing a gigantic bottlebeck. I am trying to write a recursive function (in C++) to count the number of north-east paths from one point to another in a rectangular grid. However, the line can touch any number of solid cubes. Consider the following graph. Use Dijkstra's algorithm to calculate the shortest distance from each node to the end. The API returns information based on the recommended route between start and end points, as calculated by the Google Maps API, and consists of rows containing duration and distance values for each pair. Any measured distances will use the great circle distance, which is the shortest path along the surface of the earth's ellipsoid between the two points. The contribution done by Zhihong, Xiangdan and Jizhou [4], focuses on inherent parallelism and scalability that make the algorithm suitable for grid computing task scheduling. In blue and green the two different, but equivalent, ways for reaching Node 9. (Try this with a string on a globe. There are two ways to reach the bottom-right corner: 1. In general, dist(v,w) 6= dist( w,v). The visibility graph of a polygon is a graph whose nodes corresponds to the vertices of the polygon, and whose edges correspond to the edges in the polygon formed by joining the vertices that can "see. There are situations were you might want your app to measure the distance and duration between two points on a Map, this can give an insight of how long the distance is and the time it will take to accomplish the journey. From this represen-tation, any query distance from a vertex in f can be obtained in O(logn) time. Path extraction is done by repeatedly following the parent point- ers from the goal vertex to the start vertex. the shortest path) between that vertex and eve-ry other vertex. In red the edges that are shared between the two shortest paths. Shortest path in a grid. The shortest path from one point to another.
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Math Central Quandaries & Queries
Question from Janie, a student: I have to State the excluded values for this equation and then solve, but not sure how to do this. Here is the problem (x+6)/x+3=(3)/(x+3)+2
Hi Janie,
I think you missed a pair of parentheses and the expression should be (x+6)/(x+3) = (3)/(x+3)+2.
I'm going to look at a very similar expression
Excluded values are values of x that result in the expression producing something which is not a real number. In my expression a value of x = 2 makes the denominators equal to zero and hence the expression evaluates to something that is not a real number. For any other value of x the two sides of the equation yield real numbers. Thus the only excluded value is x = 2.
To solve the equation I would multiply both values by x - 2 to get
x + 1 = 3 + 5(x - 2)
which on simplification becomes
x = 2.
But for x = 2 is excluded so the equation has no solution. Sometimes we say that x = 2 is an extraneous solution.
This is why it's important to verify your answer when you solve an algebraic equation. If you missed the fact that x = 2 is an excluded value and solved the equation as I did you might think that x = 2 is a solution. An attempt at verification by substitution of x = 2 into the original equation shows that x = 2 is not a solution.
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Image Details
Choose export citation format:
Removing the Impact of Correlated PSF Uncertainties in Weak Lensing
• Authors: Tianhuan Lu, Jun Zhang, Fuyu Dong, Yingke Li, Dezi Liu, Liping Fu, Guoliang Li, and Zuhui Fan
2018 The Astrophysical Journal 858 122.
• Provider: AAS Journals
Caption: Figure 6.
Residual additive bias correlations of three groups of galaxies: (1) σ = 2.1 (upper), (2) σ = 3.5 (middle), and (3) a mixture of σ = 2.1 and σ = 3.5 (lower). The results are measured based on CFHTLenS w2, including 16 pointings and 7 exposures on each pointing. The data points are offset by a small distance to improve visibility, and a dashed error bar means the value is negative.
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# Gravitational Attraction
Redbelly98
Staff Emeritus
Homework Helper
You're not far off. Something went awry when you plugged in the numbers, but you are close.
man big numbers frustrate me, i didnt even round so that id get a better answer but okay so here are my numbers:
mE: 5.98x10^24 kg
mM: 7.35x10^22 kg
R: 3.84x10^8 m
= 2(5.98x10^24)(3.84x10^8) +/- { 4(7.35x10^22)(5.98x10^24)(3.84x10^8)} /
2(5.98x10^24) - 2(7.35x10^22)
= 4.6x10^33 +/- {6.8x10^56} / 1.2x10^25
= 4.6x10^33 / 1.2x10^25
= 3.8x10^8
**i rounded the numbers i wrote here just to write less, but i plugged in complete numbers into the calculator
okay scrap that lol, i tried it a completely different way:
dE = {mE}(R) / {mM}+{mE}
= 3.457x10^8 m
** which sounds more reasonable as an answer
okay scrap that lol, i tried it a completely different way:
dE = {mE}(R) / {mM}+{mE}
= 3.457x10^8 m
** which sounds more reasonable as an answer
According to your equation dE = 3.79*10^8 wich is much too close to the moon.
$$\frac {2 m_e R \pm \sqrt {4 m_e m_m R^2}} { 2m_e - 2m_m}$$
this is really correct. You can still cancel a 2 and get R^2 out from under the square root sign. I hope you do not do the algebra in ascii, but write it out on paper.
A calculator that does variables is very useful. It's much harder to make mistakes if
you can enter:
>>> me = 5.98e24
>>> mm = 7.35e22
>>> R = 3.84e8
>>> me*R/(mm+me)
379337573.30469972
(this is in python)
Redbelly98
Staff Emeritus
Homework Helper
= 2(5.98x10^24)(3.84x10^8) +/- { 4(7.35x10^22)(5.98x10^24)(3.84x10^8)} /
2(5.98x10^24) - 2(7.35x10^22)
= 4.6x10^33 +/- {6.8x10^56} / 1.2x10^25
Here you forgot to square R inside the square root { } expression. It should be
(3.84x10^8)^2
Anyway, you got it to work out (perhaps by reading the other thread with this same problem).
p.s. Note to kammerling: he is using { } to signify square rooots.
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# How Do You Compute a Limit Using a Table?
In each of the next two examples, the value of the limit is the same as the value of the function at the point it approaches. This is typically the case for any polynomial.
For functions that are not polynomials, a table i often in order to evaluate the limit. In each of the following two examples, the output of the function grow more positive or more negative. This means the different limits do not exist.
However, a very similar looking fraction may also lead to a limit that does exist. In the next two examples, the limits do exist even though the functions are undefined at the point the x value is approaching.
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# 6.34. sectionStiffness command¶
sectionStiffness(eleTag, secNum)
Returns the section stiffness matrix for a beam-column element. A list of values in the row order will be returned.
eleTag (int) element tag. secNum (int) section number, i.e. the Gauss integration number
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# Relation between $H^i_I(-)$ and $H^i_J(-)$ when $I\subset J$
What is the relation between $H^i_I(-)$ and $H^i_J(-)$ (cohomological functors) when $I\subset J$ are ideals of a (local) noetherian ring?
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What kinds of relations are you looking for, maps between them, containment or etc? – Youngsu May 20 '13 at 22:02
Any non-trivial relation! – QED May 21 '13 at 13:29
Well, I don't think you can get a good relation unless you specify something specific. For instance, let $(R,m)$ be a local ring and $I \subseteq m$ an ideal. Then $H^0_m(R) \subseteq H^0_I (R)$. However if you let $I = aR$ and $\dim R = 2$. Then $H^2_m(R) \neq 0$ whereas $H^2_I(R) = 0$. – Youngsu May 21 '13 at 16:37
An obvious special case is when $I$ and $J$ have the same radical. In that case the local cohomology groups are isomorphic. – Mahdi Majidi-Zolbanin May 22 '13 at 0:48
There's a map between them, and these do fit into a long exact sequence together. This is explained in the book on local cohomology by Hartshorne, see Lemma 1.8: You can even download this book if your institution has access...
I will sketch it briefly. Let $V(I) = Y \supseteq Z = V(J)$ and set $W = V(I) \setminus V(J) = Y \setminus Z$, this $W$ is only locally closed in $\text{Spec} R$.
We define $H^i_W(\bullet)$ to be the right derived functors of $\Gamma_W(\bullet)$. Here $\Gamma_W(M)$ for any $R$-module $M$ is defined to be the set of elements of $M$ which are zero in $M_{p}$ for all $p \in (\text{Spec }R) \setminus W$.
With this notation, for any $R$-module $M$ we have a long exact sequence $$... \to H^i_Z(M) \to H^i_Y(M) \to H^i_W(M) \to H^{i+1}_Z(M) \to ...$$ Note $H^i_Z(M) = H^i_J(M)$ and $H^i_Y(M) = H^i_I(M)$. I doubt this gives you much information unless $Y$ and $Z$ have some special relationship.
There are related things you can do too. If for example $Y = Z \cup X$ where $X$ is some other closed subset of $\text{Spec }R$, there's another long exact sequence but this is a special case. Can you tell me what $Y$ and $Z$ are in your case?
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Thanks for typing such a detailed answer. I was just curious if (and how) the inclusion of zeroth cohomologies affects the upper ones. – QED May 22 '13 at 13:53
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# University of Florida/Eml4507/s13.team4ever.R5
### Problem R5.1
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.
#### Description
We are to solve, by hand, the generalized eigenvalue for the mass-spring-damper system on p. 53-13 using the following data for the masses and stiffness coefficients:
${\displaystyle m_{1}=3}$
${\displaystyle m_{2}=2}$
${\displaystyle k_{1}=10}$
${\displaystyle k_{2}=20}$
${\displaystyle k_{3}=15}$
We are then to compare this result with those obtained using CALFEM. Finally, we are to verify the mass-orthogonality of the eigenvectors.
#### Given:
We are given that for a generalized eigenvalue problem:
${\displaystyle K\phi =\lambda M\phi }$
We also know that:
${\displaystyle M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}}$ and ${\displaystyle K={\begin{bmatrix}k_{1}+k_{2}&-k_{2}\\-k_{2}&k_{2}+k_{3}\\\end{bmatrix}}}$
Therefore, we have:
${\displaystyle M={\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}}$ and ${\displaystyle K={\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}}$
#### Solution:
The first step would be to find an eigenvalue, ${\displaystyle \lambda }$ that satifies:
${\displaystyle K\phi =\lambda M\phi }$
To do this, we must find an eigenvalue that satifies:
${\displaystyle det(K-\lambda M)=0}$
We then have:
${\displaystyle det({\begin{bmatrix}30-3\lambda &-20\\-20&35-2\lambda \\\end{bmatrix}}=0}$)
This gives us
${\displaystyle (30-3\lambda )(35-2\lambda )+400=0}$
${\displaystyle 6\lambda ^{2}-165\lambda +650=0}$
Solving this equation using the quadratic equation gives us eigenvalues of:
${\displaystyle \lambda _{1,2}=13.75\pm 8.985}$
${\displaystyle \lambda _{1}=4.765}$
${\displaystyle \lambda _{2}=22.735}$
We will first take the lowest eigenvalue. We'll now try to find the corresponding eigenvector. Plugging this eigenvalue into the equaiton
${\displaystyle K\phi =\lambda M\phi }$
${\displaystyle K\phi -\lambda M\phi =0}$
${\displaystyle (K-\lambda M)\phi =0}$
${\displaystyle (K-4.765M)\phi =0}$
We need to find a value of ${\displaystyle \phi }$ that satisfies the above equation.
${\displaystyle {\begin{bmatrix}30-(3)(4.765)&-20\\-20&35-(2)(4.765)\\\end{bmatrix}}{\begin{bmatrix}\phi _{1}\\\phi _{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}}$
${\displaystyle {\begin{bmatrix}15.705&-20\\-20&25.47\\\end{bmatrix}}{\begin{bmatrix}\phi _{1}\\\phi _{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}}$
Augmenting the right hand matrix onto the left hand matrix and putting this in row reduced echelon form, we obtain,
${\displaystyle {\begin{bmatrix}1&-1.273&0\\0&0&0\\\end{bmatrix}}}$
This means that the corresponding eigenvector is:
${\displaystyle \phi _{1}={\begin{bmatrix}1.273\\1\\\end{bmatrix}}}$
We will now take the higher eigenvalue. We'll now try to find the corresponding eigenvector. Plugging this eigenvalue into the equaiton
${\displaystyle K\phi =\lambda M\phi }$
${\displaystyle K\phi -\lambda M\phi =0}$
${\displaystyle (K-\lambda M)\phi =0}$
${\displaystyle (K-22.735M)\phi =0}$
We need to find a value of ${\displaystyle \phi }$ that satisfies the above equation.
${\displaystyle {\begin{bmatrix}30-(3)(22.735)&-20\\-20&35-(2)(22.735)\\\end{bmatrix}}{\begin{bmatrix}\phi _{1}\\\phi _{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}}$
${\displaystyle {\begin{bmatrix}-38.205&-20\\-20&-10.47\\\end{bmatrix}}{\begin{bmatrix}\phi _{1}\\\phi _{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}}$
Augmenting the right hand matrix onto the left hand matrix and putting this in row reduced echelon form, we obtain,
${\displaystyle {\begin{bmatrix}1&0.5234&0\\0&0&0\\\end{bmatrix}}}$
This means that the corresponding eigenvector is:
${\displaystyle \phi _{2}={\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}}$
We will now verify the mass orthogonality of the eigenvectors. To do this, we need to show that
${\displaystyle \phi _{i}^{T}M\phi _{j}=0}$ for i not equal to j. For our case, we have:
${\displaystyle \phi _{1}^{T}={\begin{bmatrix}1.273&1\\\end{bmatrix}}}$
${\displaystyle M={\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}}$ and
${\displaystyle \phi _{2}={\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}}$
We have then that
${\displaystyle \phi _{i}^{T}M\phi _{j}={\begin{bmatrix}1.273&1\\\end{bmatrix}}{\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}{\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}=0}$
Thus verifying the mass-orthogonality of the eigenvectors.
Using our own code from Report 3, we can compare this with CALFEM:
d0=[-1,2];
v0=[0,0];
alpha=1/4; delta=1/2;
nsnap=100;
m1=3;
m2=2;
alpha=1/4; delta=1/2;
d0=[-1,2];
v0=[0,0];
k1=10;
k2=20;
k3=15;
c1=1/2;
c2=1/4;
c3=1/3;
M=[m1,0;0,m2];
C=[(c1+c2),-c2;-c2,(c2+c3)];
K=[(k1+k2),-k2;-k2,(k2+k3)];
[L]=eigen(K,M);
w1=sqrt(L(1));
T1=(2*pi)/w1;
dt=T1/10;
T=[0:dt:5*T1];
Dsnap=step2x(K,C,M,d0,v0,ip,f,pdisp);
EDU>> L
L =
4.7651
22.7349
EDU>> w1
w1 =
2.1829
EDU>> T1
T1 =
0.3473
EDU>> dt
dt =
0.0347
### R5.2
Length of members 2-3 and 4-5 (truss height):
${\displaystyle L_{23}=L_{45}=1}$
Length of members 1-2, 2-4, 4-6 (truss length):
${\displaystyle L_{12}=L_{24}=L_{46}=1}$
Area of cross section:
${\displaystyle A=1/2}$
Young's modulus:
${\displaystyle E=5}$
Mass Density:
${\displaystyle \rho =2}$
Part One:
Solve the generalized eigenvalue problem of the above truss. Display the results for the lowest 3 eigenpairs. Plot the mode shapes.
Part Two:
Consider the same truss, but with 2 missing braces:
Solve the generalized eigenvalue for this truss, and plot the mode shapes.
### Solution
On my honor, I have neither given nor received unauthorized aid in doing this problem.
function R5p2
% Part One
E = 5;
p = 2; %kg/m3a
A = 0.5; %m2
L = 1; %m
h = p*A*L;
v = h;
d = (sqrt((L^2)+(L^2)))*A*p;
m = [(h/2)+(d/2);(h/2)+(d/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(d/2)+(d/2)+(v/2);(h/2)+(d/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(d/2)+(d/2)+(v/2);(h/2)+(d/2)+(d/2)+(v/2);
(h/2)+(d/2);(h/2)+(d/2)];
M = diag(m);
Coord = [0 0;1 0;1 1;2 0;2 1;3 0];
Dof = [1 2;3 4;5 6;7 8;9 10;11 12];
Edof = [1 1 2 3 4;2 1 2 5 6;3 3 4 5 6;4 5 6 9 10;5 5 6 7 8;
6 3 4 9 10;7 3 4 7 8;8 7 8 9 10;9 9 10 11 12;10 7 8 11 12];
bc = [1;2;12];
[Ex,Ey] = coordxtr(Edof,Coord,Dof,2)
K = zeros(12);
F = zeros(12,1);
ep = [E A];
for i=1:10
Ke = bar2e(Ex(i,:),Ey(i,:),ep);
K = assem(Edof(i,:),K,Ke);
end
[L,X] = eigen(K,M,bc);
eigval = L;
eigvect = X;
j1eig = eigval(1)
j1eigvx = eigvect(:,1)
j1eigvy = eigvect(:,2)
j2eig = eigval(2)
j2eigvx = eigvect(:,3)
j2eigvy = eigvect(:,4)
j3eig = eigval(3)
j3eigvx = eigvect(:,5)
j3eigvy = eigvect(:,6)
plotpar = [1 4 0];
scale = .5;
eldraw2(Ex,Ey);
for j=1:3
clear plot
ed = extract(Edof,X(:,j));
P = eldisp2(Ex,Ey,ed,plotpar,scale);
W(j) = getframe;
drawnow
end
Three lowest eigenpairs
j1eig =
0.0890
j1eigvx =
0
0
-0.0938
0.2910
-0.1706
0.2804
-0.1679
0.2946
-0.1131
0.2756
-0.2330
0
j1eigvy =
0
0
-0.1713
-0.2076
-0.2639
-0.1197
-0.2637
-0.1637
-0.3443
-0.1391
-0.2757
0
j2eig =
0.2779
j2eigvx =
0
0
-0.1902
0.2970
0.1621
0.2185
-0.2188
-0.3370
0.1475
-0.2327
-0.0779
0
j2eigvy =
0
0
0.0382
0.1640
-0.2909
0.1947
0.2640
-0.0109
-0.1970
-0.2662
0.4295
0
j3eig =
0.5582
j3eigvx =
0
0
0.2733
0.2621
-0.0575
-0.1395
0.1216
-0.3929
-0.1489
0.2738
-0.1309
0
j3eigvy =
0
0
0.3282
-0.2744
0.1584
0.4169
0.0086
-0.0685
-0.1419
-0.0349
-0.2114
0
Mode Shape 1
Mode Shape 2
Mode Shape 3
% Part Two
Coord = [0 0;1 0;1 1;2 0;2 1;3 0];
Dof = [1 2;3 4;5 6;7 8;9 10;11 12];
Edof = [1 1 2 3 4;2 1 2 5 6;3 3 4 5 6;4 5 6 9 10;5 9 10 11 12;
6 7 8 11 12;7 3 4 7 8;8 7 8 9 10];
[Ex,Ey] = coordxtr(Edof,Coord,Dof,2);
K = zeros(12);
bc = [1;2;12];
ep = [E A];
for i=1:8
Ke = bar2e(Ex(i,:),Ey(i,:),ep);
K = assem(Edof(i,:),K,Ke);
end
[L,X] = eigen(K,M,bc);
eigval = L;
eigvect = X;
j1eig = eigval(1)
j1eigvx = eigvect(:,1)
j1eigvy = eigvect(:,2)
j2eig = eigval(2)
j2eigvx = eigvect(:,3)
j2eigvy = eigvect(:,4)
j3eig = eigval(3)
j3eigvx = eigvect(:,5)
j3eigvy = eigvect(:,6)
plotpar = [1 4 0];
scale = .5;
eldraw2(Ex,Ey);
for j=3
clear plot
ed = extract(Edof,X(:,j));
P = eldisp2(Ex,Ey,ed,plotpar,scale);
W(j) = getframe;
drawnow
end
Three lower eigen pairs
j1eig =
5.5511e-17
j1eigvx =
0
0
0.0000
0.2666
-0.2666
0.2666
0.0000
-0.2666
-0.2666
-0.2666
0.0000
0
j1eigvy =
0
0
0.0872
-0.2534
0.1188
-0.2333
0.1675
-0.3552
0.0680
-0.3270
0.2344
0
j2eig =
0.0902
j2eigvx =
0
0
-0.1544
-0.3189
-0.2063
-0.2326
-0.2669
-0.0106
-0.3004
-0.0077
-0.3073
0
j2eigvy =
0
0
-0.2636
0.1334
0.2701
0.0540
-0.3702
-0.2297
0.2087
-0.0929
-0.2564
0
j3eig =
0.3066
j3eigvx =
0
0
-0.1315
-0.2916
-0.3174
0.2561
-0.0160
-0.1685
0.3129
0.1479
0.1295
0
j3eigvy =
0
0
-0.3520
-0.0374
-0.0322
0.0346
-0.0265
0.3638
0.0364
-0.3364
0.3501
0
Mode Shape 1
Mode Shape 2
Mode Shape 3
### Problem R.5.3
#### Honor Pledge:
On my honor I have neither given nor received unauthorized aid in doing this problem.
#### Problem Description:
We are tasked with finding the eigenvector ${\displaystyle x_{1}}$ corresponding to the eigenvalue ${\displaystyle \lambda _{1}}$ for the same system as described in R4.1. We must also plot the mode shapes and compare with those from R4.1. Lastly we are tasked with creating an animation for each mode shape.
#### Given:
${\displaystyle K={\begin{bmatrix}3&-2\\-2&5\\\end{bmatrix}}}$
${\displaystyle \gamma _{1}=4-{\sqrt {5}}}$
${\displaystyle \gamma _{2}=4+{\sqrt {5}}}$
and, from R4.1
${\displaystyle X_{2}={\begin{bmatrix}-0.618\\1\\\end{bmatrix}}}$
#### Solution:
${\displaystyle {\begin{bmatrix}K-\gamma _{1}*I\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}}$
So,
${\displaystyle {\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}}$
Combining and reducing to row echelon form.
${\displaystyle {\begin{bmatrix}1&-0.618&0\\0&0&0\\\end{bmatrix}}}$
we may set ${\displaystyle x_{1}=1}$ and obtain:
${\displaystyle X_{1}={\begin{bmatrix}1\\1.618\\\end{bmatrix}}}$
Proving that they are orthogonal by ensuring the dot product between the two is zero:
${\displaystyle X_{1}}$ ${\displaystyle {\dot {}}}$ ${\displaystyle X_{2}=(1)(-0.618)+(1)(1.618)=0}$
The eigenvalue corresponding to this eigenvector is the smallest possible and therefore this must be a fundamental mode unlike the one from R.4.1.
Mode shape plotted in picture below as well as Matlab graph showing the mode 1 in blue and mode 2 in green. The oscillation moves from zero out to the displacement for each respective mass. It can be seen that the oscillation for mode 2 is indeed fundamental as both masses move in the same direction.
## Problem 5.4 Eigenvalues and Eigenvectors for Mode Shape
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.
### Problem Statement
Assume the mass density to be 5,000 ${\displaystyle kg/m^{3}}$. Contruct the diagonal mass matrix and find the eigenpairs. With this, design a matlab and calfem program to determine the deformation shape.
### Matlab Solution
%Constants
p = 5000;
E = 100000000000;
A = 0.0001;
L = 0.3;
%Degree of Freedom
dof = zeros(25,5);
for i = 1:6
j = 2*i-1;
dof(i,:) = [i,j,j+1,j+2,j+3];
end
for i =7:12
j = i*2+1;
dof(i,:) = [i,j,j+1,j+2,j+3];
end
for i = 13:19
j = (i-12)*2-1;
dof(i,:) = [i,j,j+1,j+14,j+15];
end
for i = 20:25
j = (i-19)*2-1;
dof(i,:) = [i,j,j+1,j+16,j+17];
end
%coordinates
pos = zeros(2,14);
for i=1:7
pos(:,i) = [(i-1)*L;0];
pos(:,i+7) = [(i-1)*L;L];
end
%Connections
conn = zeros(2,25);
for i = 1:6
conn(1,i)= i;
conn(2,i)= i+1;
end
for i = 7:12
conn(1,i) = i+1;
conn(2,i) = i+2;
end
for i = 13:19
conn(1,i) = i-12;
conn(2,i) = i-5;
end
for i = 20:25;
conn(1,i) = i-19;
conn(2,i) = i-11;
end
%seperates into x and y coordinates
x = zeros(14,2);
y = zeros(14,2);
for i = 1:25
x(i,:) = [pos(1,conn(1,i)),pos(1,conn(2,i))];
y(i,:) = [pos(2,conn(1,i)),pos(2,conn(2,i))];
end
%Set up K and M matrix
K = zeros(28);
M = zeros(28);
for i = 1:25
xm = x(i,2)-x(i,1);
ym = y(i,2)-y(i,1);
L = sqrt(xm^2+ym^2);
l = xm/L;
m = ym/L;
k = E*A/L*[l^2 l*m -l^2 -l*m;l*m m^2 -l*m -m^2;-l^2 -l*m l^2 l*m;-l*m -m^2 l*m m^2;];
m = L*A*p;
m = [m/2 0 0 0;0 m/2 0 0;0 0 m/2 0;0 0 0 m/2];
Dof = dof(i,2:5);
K(Dof,Dof) = K(Dof,Dof)+k;
M(Dof,Dof) = M(Dof,Dof)+m;
end
%Makes an alternate K and M matrix to calncel rows and columns for boundary
%conditions.
Kr = K;
Mr = M;
Kr([1 15 16],:) = [];
Kr(:,[1 15 16]) = [];
Mr([1 15 16],:) = [];
Mr(:,[1 15 16]) = [];
%Find Eigenvalue and Eigenvector
[Vec Val] = eig(Kr,Mr);
Vec1 = zeros(25,1);
Vec2 = zeros(25,1);
Vec3 = zeros(25,1);
for i = 1:25
Vec1(i) = Vec(i,1);
Vec2(i) = Vec(i,2);
Vec3(i) = Vec(i,3);
end
%Insert back in the values for initial conditions.
Vec1 = [0;Vec1(1:13);0;0;Vec1(14:25)];
Vec2 = [0;Vec2(1:13);0;0;Vec2(14:25)];
Vec3 = [0;Vec3(1:13);0;0;Vec3(14:25)];
%Add the Eigenvectors to the nodes
X1 = zeros(14,1);
Y1 = zeros(14,1);
X2 = zeros(14,1);
X3 = zeros(14,1);
Y2 = zeros(14,1);
Y3 = zeros(14,1);
X = zeros(14,1);
Y = zeros(14,1);
for i = 1:14
X1(i) = Vec1((i*2)-1,1);
Y1(i) = Vec1((i*2),1);
X(i) = pos(1,i);
Y(i) = pos(2,i);
X2(i) = Vec2((i*2)-1,1);
Y2(i) = Vec2((i*2),1);
X3(i) = Vec3((i*2)-1,1);
Y3(i) = Vec3((i*2),1);
end
%plot origional shape
for i = 1:25
xc = [X(conn(1,i)),X(conn(2,i))];
yc = [Y(conn(1,i)),Y(conn(2,i))];
axis([0 2 -1 1])
plot(xc,yc)
hold on
end
X1 = X+-0.1*X1;
Y1 = Y+-0.1*Y1;
%plot mode 1
for i = 1:25
xc = [X1(conn(1,i)),X1(conn(2,i))];
yc = [Y1(conn(1,i)),Y1(conn(2,i))];
axis([0 2 -1 1])
plot(xc,yc,'r')
hold on
end
X2 = X+X2;
Y2 = Y+Y2;
X3 = X-0.9*X3;
Y3 = Y-0.9*Y3;
%plot mode 2
for i = 1:25
xc = [X2(conn(1,i)),X2(conn(2,i))];
yc = [Y2(conn(1,i)),Y2(conn(2,i))];
axis([0 2 -1 1])
plot(xc,yc,'r')
hold on
end
%plot mode 3
for i = 1:25
xc = [X3(conn(1,i)),X3(conn(2,i))];
yc = [Y3(conn(1,i)),Y3(conn(2,i))];
axis([0 2 -1 1])
plot(xc,yc,'r')
hold on
end
References for Matlab code: Team 3 Report 4 Team 7 Report 4
### CALFEM Solution
function R4p4
p = 5000; %kg/m3
A = 0.01; %m2
L = 0.3; %m
h = p*A*L;
v = h;
d = (sqrt((L^2)+(L^2)))*A*p;
m = [(h/2)+(d/2)+(v/2);(h/2)+(d/2)+(v/2);
(h/2)+(v/2);(h/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(h/2)+(d/2)+(v/2);(h/2)+(h/2)+(d/2)+(v/2);
(h/2)+(v/2);(h/2)+(v/2);
(h/2)+(d/2)+(v/2);(h/2)+(d/2)+(v/2)];
M = diag(m);
Coord = [0 0;0 0.3;0.3 0;0.3 0.3;0.6 0;
0.6 0.3;0.9 0;0.9 0.3;1.2 0;1.2 0.3;
1.5 0;1.5 0.3;1.8 0;1.8 0.3];
Dof = [1 2;3 4;5 6;7 8;9 10;1 12;13 14;15 16;
17 18;19 20;21 22;23 24;25 26;27 28];
Edof = [1 1 2 5 6;2 5 6 9 10;3 9 10 13 14;4 13 14 17 18;5 17 18 21 22;
6 21 22 25 26;7 3 4 7 8;8 7 8 11 12;9 11 12 15 16;10 15 16 19 20;
11 19 20 23 24;12 23 24 27 28;13 1 2 3 4;14 5 6 7 8;15 9 10 11 12;
16 13 14 15 16;17 17 18 19 20;18 21 22 23 24;19 25 26 27 28;20 1 2 7 8;
21 5 6 11 12;22 9 10 15 16;23 13 14 19 20;24 17 18 23 24;25 21 22 27 28];
[Ex,Ey] = coordxtr(Edof,Coord,Dof,2);
K = zeros(28);
ep = [100000000000 0.01];
for i=1:25
Ke = bar2e(Ex(i,:),Ey(i,:),ep);
K = assem(Edof(i,:),K,Ke);
end
[L,X] = eigen(K,M);
eigval = L;
eigvect = X;
Three lowest eigenvalues and eigenvectors
j1eig = eigval(1)
j1eigvx = eigvect(:,1)
j1eigvy = eigvect(:,2)
j2eig = eigval(2)
j2eigvx = eigvect(:,3)
j2eigvy = eigvect(:,4)
j3eig = eigval(3)
j3eigvx = eigvect(:,5)
j3eigvy = eigvect(:,6)
j1eig =
-4.1672e-08
j1eigvx =
-0.0018
0.0853
0.0136
0.0853
-0.0018
0.0699
0.0136
0.0699
-0.0018
0.0545
0.0136
0.0545
-0.0018
0.0391
0.0136
0.0391
-0.0018
0.0238
0.0136
0.0238
-0.0018
0.0084
0.0136
0.0084
-0.0018
-0.0070
0.0136
-0.0070
j1eigvy =
-0.0485
0.0068
-0.0488
0.0068
-0.0485
0.0071
-0.0488
0.0071
-0.0485
0.0074
-0.0488
0.0074
-0.0485
0.0077
-0.0488
0.0077
-0.0485
0.0080
-0.0488
0.0080
-0.0485
0.0083
-0.0488
0.0083
-0.0485
0.0086
-0.0488
0.0086
j2eig =
-1.0740e-08
j2eigvx =
0.0155
-0.0334
-0.0053
-0.0334
0.0155
-0.0127
-0.0053
-0.0127
0.0155
0.0081
-0.0053
0.0081
0.0155
0.0289
-0.0053
0.0289
0.0155
0.0496
-0.0053
0.0496
0.0155
0.0704
-0.0053
0.0704
0.0155
0.0912
-0.0053
0.0912
j2eigvy =
-0.0242
0.0747
0.0200
0.0761
-0.0197
0.0156
0.0197
0.0202
-0.0092
-0.0374
0.0148
-0.0335
0.0040
-0.0572
0.0040
-0.0572
0.0148
-0.0335
-0.0092
-0.0374
0.0197
0.0202
-0.0197
0.0156
0.0200
0.0761
-0.0242
0.0747
j3eig =
1.2133e-07
j3eigvx =
0.0206
-0.0518
-0.0370
-0.0552
0.0097
0.0333
-0.0347
0.0284
-0.0031
0.0552
-0.0190
0.0633
0.0002
-0.0083
-0.0002
0.0083
0.0190
-0.0633
0.0031
-0.0552
0.0347
-0.0284
-0.0097
-0.0333
0.0370
0.0552
-0.0206
0.0518
j3eigvy =
0.0616
-0.0146
0.0727
-0.0164
0.0453
0.0010
0.0647
-0.0047
0.0125
0.0180
0.0453
0.0165
-0.0204
-0.0017
0.0204
0.0017
-0.0453
-0.0165
-0.0125
-0.0180
-0.0647
0.0047
-0.0453
-0.0010
-0.0727
0.0164
-0.0616
0.0146
Plotting the three lowest mode shapes
for j=1:3
figure
plot(eigvect(:,j));
title(['Mode Shape ',num2str(j)])
end
## Problem R5.5a: Eigen vector plot for zero evals of the 2 bar truss system p.21-2 (fead.f08.mtgs.[21])
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
### Given: Two member zero evals
Consider a 2 member system. Plot the eigen vectors corresponding to zero evals and interpret the results.
Use standard node and element naming conventions.
Figure 1: Two member system (fig.JPG)
### Solution: Plot the eigen values and interpret
This is the general stiffness matrix for a two bar system as given.
${\displaystyle {\vec {K}}={\begin{bmatrix}&1&2&3&4&5&6\\1&(l^{(1)})^{2}*K^{(1)}&l^{(1)}m^{(1)}*K^{(1)}&-(l^{(1)})^{2}*K^{(1)}&-l^{(1)}m^{(1)}*K^{(1)}&0&0\\2&l^{(1)}m^{(1)}*K^{(1)}&(m^{(1)})^{2}*K^{(1)}&-l^{(1)}m^{(1)}*K^{(1)}&-(m^{(1)})^{2}*K^{(1)}&0&0\\3&-(l^{(1)})^{2}*K^{(1)}&-l^{(1)}m^{(1)}*K^{(1)}&(l^{(1)})^{2}*K^{(1)}+(l^{(2)})^{2}*K^{(2)}&l^{(1)}m^{(1)}*K^{(1)}+l^{(2)}m^{(2)}*K^{(2)}&-(l^{(2)})^{2}*K^{(2)}&-l^{(2)}m^{(2)}*K^{(2)}\\4&-l^{(1)}m^{(1)}*K^{(1)}&-(m^{(1)})^{2}*K^{(1)}&l^{(1)}m^{(1)}*K^{(1)}+l^{(2)}m^{(2)}*K^{(2)}&(m^{(1)})^{2}*K^{(1)}+(m^{(2)})^{2}*K^{(2)}&-(l^{(2)})^{2}*K^{(2)}&-l^{(2)}m^{(2)}*K^{(2)}\\5&0&0&-(l^{(2)})^{2}*K^{(2)}&-l^{(2)}m^{(2)}*K^{(2)}&(l^{(2)})^{2}*K^{(2)}&l^{(2)}m^{(2)}*K^{(2)}\\6&0&0&-l^{(2)}m^{(2)}*K^{(2)}&-(m^{(2)})^{2}*K^{(2)}&l^{(2)}m^{(2)}*K^{(2)}&(m^{(2)})^{2}*K^{(2)}\\\end{bmatrix}}}$
The global stiffness matrix in numerical form match to this specific problem is as follows.
${\displaystyle {\vec {K}}={\begin{bmatrix}0.5625&0.3248&-0.5625&-0.3248&0&0\\0.3248&0.1875&-0.3248&-0.1875&0&0\\-0.5625&-0.3248&3.0625&-2.1752&-2.5000&2.5000\\-0.3248&-0.1875&-2.1752&2.6875&2.5000&-2.5000\\0&0&-2.5000&2.5000&2.5000&-2.5000\\0&0&2.5000&-2.5000&-2.5000&2.5000\\\end{bmatrix}}}$
The eigen vector matrix is as follows.
${\displaystyle {\vec {V}}={\begin{bmatrix}-0.1118&0.5043&-0.0000&-0.5931&0.6174&-0.0139\\-0.0814&-0.8634&0.0000&-0.3476&0.3565&-0.0080\\-0.4628&0.0089&0.0000&-0.4803&-0.5409&0.5123\\0.5266&-0.0053&-0.0000&-0.5429&-0.4330&-0.4904\\-0.4947&0.0071&-0.7071&0.0313&-0.0765&-0.4984\\0.4947&-0.0071&-0.7071&-0.0313&0.0765&0.4984\\\end{bmatrix}}}$
${\displaystyle {\vec {D}}={\begin{bmatrix}0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&1.4705&0\\0&0&0&0&0&10.0295\\\end{bmatrix}}}$
The eigen vectors correspond to the columns of the eigen vector matrix shown above. The eigen value mode shapes are a linear combination of the pure mode shapes (pure rigid body and pure mechanism).
Figure 1: The Vectors plotted from the eigen vector matrix ([1])
function R5_5e2
K= [9/16 (3*sqrt(3))/16 -9/16 -(3*sqrt(3))/16 0 0;
(3*sqrt(3))/16 3/16 -(3*sqrt(3))/16 -3/16 0 0;
-9/16 -(3*sqrt(3))/16 49/16 ((3*sqrt(3)-40)/16) -5/2 5/2;
-(3*sqrt(3))/16 -3/16 ((3*sqrt(3)-40)/16) 43/16 5/2 -5/2;
0 0 -5/2 5/2 5/2 -5/2;
0 0 5/2 -5/2 -5/2 5/2]
Eval=eig(K)
[V,D]=eig(K)
V
D=abs(D)
%Each column of V is a mode
for i=1:4
figure
plot(V(:,i));
title(['Mode ',num2str(i)])
end
end
As a thought experiment, we plotted the position plus the deformation derived from the eigen vector matrix. This will supply 4 shapes but they are not constrained to the boundary conditions. Applying the boundary conditions to the K matrix reduces the number of non zero eigen vector outputs to two. The results are shown below. The figure plots came from the eigen vectors in the fifth and sixth columns of the constrained eigen vector matrix. Using the other columns produced a figure that was not constrained to the fixed supports.
Figure 1: Mode shape constrained to the boundary conditions. The dotted line represents the original shape. ([2])
Figure 1: Mode shape constrained to the boundary conditions. The dotted line represents the original shape. ([3])
function R5_5e
K= [9/16 (3*sqrt(3))/16 -9/16 -(3*sqrt(3))/16 0 0;
(3*sqrt(3))/16 3/16 -(3*sqrt(3))/16 -3/16 0 0;
-9/16 -(3*sqrt(3))/16 49/16 ((3*sqrt(3)-40)/16) -5/2 5/2;
-(3*sqrt(3))/16 -3/16 ((3*sqrt(3)-40)/16) 43/16 5/2 -5/2;
0 0 -5/2 5/2 5/2 -5/2;
0 0 5/2 -5/2 -5/2 5/2]
%K_red=K([3 4],[3 4]);
K1=[0 0 0 0 0 0;
0 0 0 0 0 0;
0 0 K(3,3) K(3,4) 0 0;
0 0 K(4,3) K(4,4) 0 0;
0 0 0 0 0 0;
0 0 0 0 0 0]
[V D] = eig(K1)
N=[1 2;
2 3];
P=[0 cos(pi/6)*4 cos(pi/6)*4+cos(pi/4)*2;
0 sin(pi/6)*4 sin(pi/6)*4-sin(pi/4)*2];
for i=1:2:5
x_1((i+1)/2)=P(1,((i+1)/2))+V(i,1);
end
for i=2:2:6
y_1((i/2))=P(2,(i/2))+V(i,1);
end
%Plot deformed system with constraints
hold on
for i=1:2
l1=N(1,i);
l2=N(2,i);
x1=[x_1(l1),x_1(l2)];
y1=[y_1(l1),y_1(l2)];
axis([-1 5 -1 5])
plot(x1,y1,'b')
title(['Constrained Shape'])
hold on
end
%Plot undeformed system
for i=1:2:5
x_o((i+1)/2)=P(1,((i+1)/2));
end
for i=2:2:6
y_o((i/2))=P(2,(i/2));
end
for i=1:2
l1=N(1,i);
l2=N(2,i);
x1=[x_o(l1),x_o(l2)];
y1=[y_o(l1),y_o(l2)];
axis([-1 5 -1 5])
plot(x1,y1,'-.or')
hold on
end
end
## Problem R5.5b: Eigen vector plot for zero evals of the square 3 bar truss system p.21-3 (figure a) (fead.f08.mtgs.[21])
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
### Given: Square 3 bar truss system
Consider a 3 member system. Plot the eigen vectors corresponding to zero evals and interpret the results.
a=b=1 E=2 A=3
Figure 1: Two member system (b.JPG)
### Find:
#### 1.Plot the eigen values
Use standard node and element naming conventions.
### Solution: Plot the eigen values
${\displaystyle {\vec {V}}={\begin{bmatrix}-0.7071&0&0&-0.7071\\0&1.0000&0&0\\-0.7071&0&0&0.7071\\0&0&1.0000&0\\\end{bmatrix}}}$
${\displaystyle {\vec {D}}={\begin{bmatrix}0&0&0&0\\0&6&0&0\\0&0&6&0\\0&0&0&12\\\end{bmatrix}}}$
Figure 1: The only shape corresponding to zero evals. ([4])
function R5_5ec
E=2;
A=3;
L=1;
k=(E*A)/L;
k1=k*[0 0 0 0;
0 1 0 -1;
0 0 0 0;
0 -1 0 1]
k3=k*[0 0 0 0;
0 1 0 -1;
0 0 0 0;
0 -1 0 1]
k2=k*[1 0 -1 0;
0 0 0 0;
-1 0 1 0;
0 0 0 0]
K=zeros(8,8)
K(1:4,1:4)=k1
K(5:8,5:8)=k3
Kt=zeros(8,8)
Kt(3:6,3:6)=k2
Kg=K+Kt
%boundary conditions applied
K1=[0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0;
0 0 Kg(3,3) Kg(3,4) Kg(3,5) Kg(3,6) 0 0;
0 0 Kg(4,3) Kg(4,4) Kg(4,5) Kg(4,6) 0 0;
0 0 Kg(5,3) Kg(5,4) Kg(5,5) Kg(5,6) 0 0;
0 0 Kg(6,3) Kg(6,4) Kg(6,5) Kg(6,6) 0 0;
0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0]
Kred=[Kg(3,3) Kg(3,4) Kg(3,5) Kg(3,6);
Kg(4,3) Kg(4,4) Kg(4,5) Kg(4,6);
Kg(5,3) Kg(5,4) Kg(5,5) Kg(5,6);
Kg(6,3) Kg(6,4) Kg(6,5) Kg(6,6)]
[V D] = eig(Kred)
for i=1:6
figure
plot(V(:,i));
title(['Mode ',num2str(i)])
end
end
### Problem R5.6
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.
#### Description
We are to solve Pb-53.6 p.53-13b over again using the modal superposition method.
#### Given:
We are given from Pb-53.6 that:
${\displaystyle m_{1}=3}$, ${\displaystyle m_{2}=2}$
${\displaystyle k_{1}=10}$, ${\displaystyle k_{2}=20}$, ${\displaystyle k_{3}=15}$
${\displaystyle c_{1}=1/2}$, ${\displaystyle c_{2}=1/4}$, ${\displaystyle c_{3}=1/3}$
${\displaystyle F_{1}(t)=0}$, ${\displaystyle F_{2}(t)=0}$
We also know from Pb-53.9 that:
${\displaystyle \lambda _{1}=4.765}$
${\displaystyle \lambda _{2}=22.735}$
${\displaystyle \phi _{1}={\begin{bmatrix}1.273\\1\\\end{bmatrix}}}$
${\displaystyle \phi _{2}={\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}}$
${\displaystyle M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}}$ and ${\displaystyle K={\begin{bmatrix}k_{1}+k_{2}&-k_{2}\\-k_{2}&k_{2}+k_{3}\\\end{bmatrix}}}$ and ${\displaystyle C={\begin{bmatrix}c_{1}+c_{2}&-c_{2}\\-c_{2}&c_{2}+c_{3}\\\end{bmatrix}}}$
Therefore, we have:
${\displaystyle M={\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}}$ and ${\displaystyle K={\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}}$ and ${\displaystyle C={\begin{bmatrix}3/4&-1/4\\-1/4&7/12\\\end{bmatrix}}}$
#### Solution:
From Pb-53.9, we already know the eigenvectors. To solve the problem using superposition, we know only need to find the modal coordinates of d(t) corresponding to the mode shapes. That is to say, we need to find z so as to solve the equation:
${\displaystyle d(t)=z_{1}\phi _{1}+z_{2}\phi _{2}}$
We can get this from:
${\displaystyle z_{j}^{''}+{\bar {\phi }}_{i}^{T}C{\bar {\phi }}_{j}z_{j}^{'}+(w_{j})^{2}z_{j}={\bar {\phi }}_{j}^{T}F(t)}$
Since F(t) = 0, the above differential equation becomes homogeneous. However, we still need to find ${\displaystyle {\bar {\phi }}_{1}}$ and ${\displaystyle {\bar {\phi }}_{2}}$. These can be found from the equation:
${\displaystyle {\bar {\phi }}_{i}={\frac {\phi _{i}}{\sqrt {\phi _{i}^{T}M\phi _{i}}}}}$, therefore,
${\displaystyle {\bar {\phi }}_{1}={\frac {\begin{bmatrix}1.273\\1\\\end{bmatrix}}{\sqrt {{\begin{bmatrix}1.273&1\\\end{bmatrix}}{\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}{\begin{bmatrix}1.273\\1\\\end{bmatrix}}}}}={\frac {\begin{bmatrix}1.273\\1\\\end{bmatrix}}{\sqrt {6.86}}}={\frac {\begin{bmatrix}1.273\\1\\\end{bmatrix}}{2.619}}={\begin{bmatrix}0.485\\0.381\\\end{bmatrix}}}$
${\displaystyle {\bar {\phi }}_{2}={\frac {\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}{\sqrt {{\begin{bmatrix}-0.5234&1\\\end{bmatrix}}{\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}{\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}}}}={\frac {\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}{\sqrt {2.82}}}={\frac {\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}{1.68}}={\begin{bmatrix}-0.311\\0.595\\\end{bmatrix}}}$
Remembering that ${\displaystyle (w_{j})^{2}=\lambda _{j}}$, our first differential equation becomes:
${\displaystyle z_{1}^{''}+{\begin{bmatrix}0.485&0.381\\\end{bmatrix}}{\begin{bmatrix}3/4&-1/4\\-1/4&7/12\\\end{bmatrix}}{\begin{bmatrix}0.485\\0.381\\\end{bmatrix}}z_{1}^{'}+4.765z_{1}=0}$
${\displaystyle z_{1}^{''}+0.1687z_{1}^{'}+4.765z_{1}=0}$
Using the method of undetermined coefficients for solving differential equations, we arrive at:
${\displaystyle z_{1}=Ae^{-0.0843t}cos(2.18t)}$
We now need to find the modal coordinate initial condition using the following equaiton:
${\displaystyle z_{i}(0)={\bar {\phi }}_{i}^{T}Md(0)}$
${\displaystyle z_{1}(0)={\begin{bmatrix}0.485&0.381\\\end{bmatrix}}{\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}{\begin{bmatrix}-1\\2\\\end{bmatrix}}=0.069}$
We can easily find A from:
${\displaystyle z_{1}(0)=A(1)(1)=0.069}$
so
${\displaystyle z_{1}=0.069e^{-0.0843t}cos(2.18t)}$
Our second differential equation is:
${\displaystyle z_{2}^{''}+{\begin{bmatrix}-0.311&0.595\\\end{bmatrix}}{\begin{bmatrix}3/4&-1/4\\-1/4&7/12\\\end{bmatrix}}{\begin{bmatrix}-0.311\\0.595\\\end{bmatrix}}z_{2}^{'}+22.735z_{2}=0}$
${\displaystyle z_{2}^{''}+0.371z_{2}^{'}+22.735z_{2}=0}$
Using the method of undetermined coefficients for solving differential equations once again, we obtain:
${\displaystyle z_{2}=Be^{-0.185t}cos(4.76t)}$
We now need to find the modal coordinate initial condition using the following equaiton:
${\displaystyle z_{i}(0)={\bar {\phi }}_{i}^{T}Md(0)}$
${\displaystyle z_{2}(0)={\begin{bmatrix}-0.311&0.595\\\end{bmatrix}}{\begin{bmatrix}3&0\\0&2\\\end{bmatrix}}{\begin{bmatrix}-1\\2\\\end{bmatrix}}=3.313}$
We can easily find B from:
${\displaystyle z_{2}(0)=B(1)(1)=3.313}$
so
${\displaystyle z_{2}=3.313e^{-0.0843t}cos(2.18t)}$
The solution using modal superposition can then be displayed as:
${\displaystyle d(t)=0.069e^{-0.0843t}cos(2.18t){\begin{bmatrix}1.273\\1\\\end{bmatrix}}+3.313e^{-0.185t}cos(4.76t){\begin{bmatrix}-0.5234\\1\\\end{bmatrix}}}$
## Problem R5.7: Consider the free vibration of truss system p.53-22b (fead.f13.sec53b.[21])
On our honor, we did this assignment on our own.
### Given: Truss system under free vibration with applied force and zero initial velocity
Consider the truss system. The initial force is applied at node four and the force equals five.
Use standard node and element naming conventions.
Figure 1: Truss system (fead.f13.sec53b p.53-22b). ([5])
### Solution: Solve for the truss motion
K =
Columns 1 through 5
3.3839 0.8839 -2.5000 0 -0.8839
0.8839 0.8839 0 0 -0.8839
-2.5000 0 5.8839 0.8839 0
0 0 0.8839 3.3839 0
-0.8839 -0.8839 0 0 4.2678
-0.8839 -0.8839 0 -2.5000 0
0 0 -2.5000 0 -0.8839
0 0 0 0 0.8839
0 0 -0.8839 -0.8839 -2.5000
0 0 -0.8839 -0.8839 0
0 0 0 0 0
0 0 0 0 0
Columns 6 through 10
-0.8839 0 0 0 0
-0.8839 0 0 0 0
0 -2.5000 0 -0.8839 -0.8839
-2.5000 0 0 -0.8839 -0.8839
0 -0.8839 0.8839 -2.5000 0
4.2678 0.8839 -0.8839 0 0
0.8839 5.8839 -0.8839 0 0
-0.8839 -0.8839 3.3839 0 -2.5000
0 0 0 4.2678 0
0 0 -2.5000 0 4.2678
0 -2.5000 0 -0.8839 0.8839
0 0 0 0.8839 -0.8839
Columns 11 through 12
0 0
0 0
0 0
0 0
0 0
0 0
-2.5000 0
0 0
-0.8839 0.8839
0.8839 -0.8839
3.3839 -0.8839
-0.8839 0.8839
p = 2;
E = 5;
A = 0.5;
L = 1;
Ld = 1*sqrt(2);
%degree of freedom for each element
dof = zeros(10,5);
dof(1,:) = [1 1 2 3 4];
dof(2,:) = [2 1 2 5 6];
dof(3,:) = [3 3 4 5 6];
dof(4,:) = [4 5 6 9 10];
dof(5,:) = [5 5 6 7 8];
dof(6,:) = [6 3 4 9 10];
dof(7,:) = [7 3 4 7 8];
dof(8,:) = [8 7 8 9 10];
dof(9,:) = [9 9 10 11 12];
dof(10,:) = [10 7 8 11 12];
%position of each node
pos = zeros(2,6);
pos(:,1) = [0;0];
pos(:,2) = [L;0];
pos(:,3) = [L;L];
pos(:,4) = [2*L;0];
pos(:,5) = [2*L;L];
pos(:,6) = [3*L;0];
%Connections
conn = zeros(2,10);
conn(:,1) = [1;2];
conn(:,2) = [1;3];
conn(:,3) = [2;3];
conn(:,4) = [3;5];
conn(:,5) = [3;4];
conn(:,6) = [2;5];
conn(:,7) = [2;4];
conn(:,8) = [4;5];
conn(:,9) = [5;6];
conn(:,10) = [4;6];
%seperates into x and y coordinates
x = zeros(10,2);
y = zeros(10,2);
for i = 1:10
x(i,:) = [pos(1,conn(1,i)),pos(1,conn(2,i))];
y(i,:) = [pos(2,conn(1,i)),pos(2,conn(2,i))];
end
%Set up K and M matrix
K = zeros(12);
M = zeros(12);
for i = 1:10
xm = x(i,2)-x(i,1);
ym = y(i,2)-y(i,1);
L = sqrt(xm^2+ym^2);
l = xm/L;
m = ym/L;
k = E*A/L*[l^2 l*m -l^2 -l*m;l*m m^2 -l*m -m^2;-l^2 -l*m l^2 l*m;-l*m -m^2 l*m m^2;];
m = L*A*p;
m = [m/2 0 0 0;0 m/2 0 0;0 0 m/2 0;0 0 0 m/2];
Dof = dof(i,2:5);
K(Dof,Dof) = K(Dof,Dof)+k;
M(Dof,Dof) = M(Dof,Dof)+m;
end
## Table of Assignments R5
Problem Assignments R5
Problem # Solved&Typed by Reviewed by
1 David Bonner, Vernon Babich All
2 Chad Colocar, David Bonner All
3 Bryan Tobin, Tyler Wulterkens All
4 Tyler Wulterkens, Vernon Babich All
5 Vernon Babich, Tyler Wulterkens All
6 David Bonner, Vernon Babich All
7 Tyler Wulterkens, Chad Colocar, Vernon Babich All
|
{}
|
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