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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | I am doing a planet report and I am wondering how scientists know the weights of the planets. Do they calculate it? | | Question Date: 2002-04-24 | | | | Answer 1: You can measure the mass of something that orbits a larger body by how far away they are, and how long they take to make one complete orbit. Johannes Kepler, in the 1500's, predicted that the orbital sizes and periods of the planets should have a very definite proportion - that is: If we call T = time to make one orbit around the sun, and R = radius of the orbit, then (R3)/(T2) for all the planets = a constant, which includes the mass of the sun. So, to measure the mass of a planet, you can measure the periods and radii of the orbits of all the moons that orbit around it , graph R3/T2 for all the moons, and calculate the mass of the planet. This is one way to do it. Another way is by measuring the shape and the gravitational field, and then you can calculate the mass. | | | | Answer 2: This is a good question. What astronomers actually calculate is the mass of all the planets, not their weight. The mass tells you how much matter there is in the planet but the weight tells you how heavy something is on the Earth. The difference is that, on the moon, you are much lighter than on the Earth, but you still have the same mass (because you have the same amount of matter in you!). Anyway, the way to calculate the masses of the planets is using Newton's law of gravity The law of gravity is an equation that tells you how strong objects pull on each other due to gravity, and it depends on the masses of the two objects and on their distance apart. Using this law of gravity, you can calculate exactly where the planet should be and how fast it should be going at any time, and all of that information turns out to depend on the mass of the sun and the mass of the planet. By studying the exact orbit of the planets and sun in the solar system, you can calculate all of the masses of the planets. | | | | Answer 3: Yes. The masses of the planets are calculated most accurately from Newton's law of gravity, a = (GM)/(r2), which can be used to calculate how much gravitational acceleration (a) a planet of mass M will produce on objects at distance r away. One can solve for M once the other numbers are known. G is a universal number which determined here on Earth by measuring the attraction of objects of known mass at a known spacing r, then solving for G. (The law of gravity applies to all objects, no matter how small- we just don't notice the gravity of everyday objects, because it is so weak). The measurement is done with a Cavendish balance, which is surprisingly sensitive- I used one once, but got too close to the device and the experiment was ruined because of the gravitational pull of my head. A planet's mass can in principle be calculated by observing how its gravity causes small wobbles in the orbits of other planets. In fact, college student John Couch Adams realized that small wobbles in Uranus's orbit could be nicely explained by the presence of an undiscovered planet. Astronomers looked where Adams told them to, and found Neptune (which other astronomers, including Galileo, had seen but dismissed as just another faint star). Since a planet's gravity is stronger near the planet, a more accurate determination of its mass involves observing the orbits of its moons. All planets except Mercury and Venus have moons, but those two planets have been visited by space probes, along with all other planets except Pluto. Probes provide the most accurate measurement of the mass because the probe's motion can be measured to high accuracy by recording the times at which its radio pulses arrive on Earth. The probe's distance from Earth determines how long it takes the signal to reach us, and with high-accuracy clocks we can measure, within meters, how much the probe was gravitationally deflected by the planet. | | | | Answer 4: The short answer to your question is that yes, scientists calculate the mass of planets. Before describing it to you in detail, I should point out that in physics, the terms "mass" and "weight" have significantly different meanings. Mass is a measure of how much matter there is in an object. For any object, this number is the same anywhere in the universe. Weight, on the other hand, has to do with the attractive force between different objects. Every object with mass attracts every other body with mass according to Isaac Newton's equation F = G M m / (r2) rwhere M is the mass of one object, m is the mass of a different object, d is the distance between them, and G is the "gravitational constant." First off, we can see that as the mass of the objects increases, their product will increase and give bigger forces. However, G is a very, very small number (Side note: although G is extremely small, it is possible to measure it in a laboratory, as was originally done by Cavendish just before 1800). The other key part is that the attractive force (gravity) decreases as the inverse square of the distance between the objects. That means, if the distance between the two objects doubles, the force drops by 4; if it triples, the force drops by 9; if it's quadrupled, the force drops by 16, and so on. So how do we use this to measure the mass of the planets? For this, there are basically two options - you either 1. Need to be on the planet, or 2. Get some information about an object orbiting the planet. Let's talk about each one. 1. Calculating the mass of a planet while on that planet. It turns out that in the above equation, the group of terms GM / (r2) is what scientists think of as the acceleration due to gravity (taking M as the mass of the planet). This is something that can be experimentally measured. For example, on Earth, the value is about 10 m / s2. That means if I drop an object on earth (any object such that there is negligible air resistance), after 1 second, it will be falling at about 10 m / s; after 2 seconds it'll be about 20 m /s ; after 3 seconds it'll be about 30 m /s, and so on. So, if you drop an object on another planet, you could first measure how fast it accelerates towards the planet. Since you know what G is, and r would be the radius of the planet, you could then calculate M, or the mass of the planet! 2. Calculating the mass of a planet from earth Since we can't fly to the different planets to use method 1, we are forced to come up with the mass of planets a different way. To do this, you need to have some information about an object that is orbiting it. Without going into details, it turns out that starting from the first equation above, you can describe an object (say, a moon) orbiting a another object (say, a planet) by the equation M = 4 pi2 r3 / T( 2G) Here, M is the mass of the planet, pi is 3.14..., r is the distance between the center of the moon and the center of the planet, T is the period (the time it takes the moon to make one full orbit around the planet), and G is again the gravitational constant. The interesting part about this equation is that you do not need to know the mass of the orbiting moon to get the mass of the planet! If we see an object orbiting a planet, then all we need to do is figure out how far apart they are, and how long one orbit takes, and we can calculate the mass of that planet! | | | | Answer 5: The weight of the planet can be computed through several methods: 1) For our planet, the volume of the planet has been measured for several hundred years assuming it is a perfect sphere. Also the average density is known form measurements at the surface and in the depth of the earth. Once you have the 2 it is easy to compute the approximate weight of the earth. 2) For other planets it is more complex. Lately, radar waves are used to measure the density of the planet. The shape and dimensions of the planet are also measured using radar waves and astronomical observations. Once the shape, dimensions and density are known, it is trivial to get the weight. 3) Another technique is to use Newton's law which predicts the trajectory in space of a planet gravitating around another one. The trajectory can be measured precisely using astronomical techniques. The weight of one of the planet (eg. The earth) is known and using Newton's gravity law, and the known trajectory of the other planet (eg the moon) , its weight is deduced. I am sure there are many other techniques which I do not know. | | | | Answer 6: Well, the best way to do it is to measure the orbit of something around the planet, like a satellite. If the satellite is small enough compared to that of the planet, the path it travels around the planet will depend on the mass of only the planet. Many of the planets have moons orbiting around them. All you have to do is watch the moon travel around the planet and figure out how far it is from the planet as it orbits and how long it takes to make one circuit around the planet. You also, of course, need to know the distance to the planet. That can be measured by observing the planet's orbit around the Sun and by using parallax methods (observing the planet "shifting" in the sky compared to the background stars when observing from different locations on Earth). In reality, it's a bit more complicated than I've said, but this is the basic idea. | | | | Answer 7: Like much of astronomy, the masses of the planets are known by a set of linked assumptions about the universe. We know the mass of the earth since we can measure the gravitation constant (using a known mass). Since we also know the force of gravity on the surface of the earth, we can find its mass. Then, since we know the period of the revolution of the earth about the sun (1 year), we can find the mass of the sun as well. Finally, we know the periods of the planet orbits by observations from the earth -- so we can use the known solar mass to find the planet's masses. (The assumption is that newton's law of gravity applies everywhere...) If you think this is a long chain of reasoning, you should see how one finds the distance to far away galaxies and galaxy clusters... | | | | Answer 8: You probably know that massive objects, such as planets, attract each other. This effect is called gravity. Newton first understood the mathematical equations that describe this attraction. Kepler later extended Newton's theories and derived an equation know as Kepler's law. This equation relates the distance between a planet and its moon to the period of the moon?s orbit, and the planet?s mass. Therefore, if you know the distance to the moon, and how long it takes the moon to travel around the earth, you can use Kepler's law to 'weigh' the earth. | | | | Answer 9: The weight of planets is not easy to calculate, but we can get an idea with some simple geometry. If you know how big they are (i.e. treating them as a sphere, with a radius) then you can work out their volume (I think the rule is something like Volume = 4piradius2) Once you have the volume, you have to make some assumptions about what sort of rock or gas the planet is composed of, and how dense this material is. For instance, the radius of the Earth is about 6370 km. This is a volume of about 5,000,000,000 cubic kilometers. Most of the earth below 40 km deep is composed of an olivine rock called peridotite which has a density of about 3300 to 3500 kg per cubic meter. Therefore if we multiply this figure by 1000 to put it in cubic kilometers and multiply it by the volume of the Earth, we get a awesomely huge figure! This is only a low estimate though, as rocks become more dense as you go deeper into a planet (and the Earth is composed of different rock types with different densities, especially towards the core). We don't know much about what sort of rocks are on Mars or Venus, hopefully this will become clearer over the next decade as we send more probe missions there. Click Here to return to the search form. | | | --- | | | | | | | | | Copyright © 2020 The Regents of the University of California, All Rights Reserved. UCSB Terms of Use | | | | | |
16101 | https://math.stackexchange.com/questions/3630051/find-all-polynomials-px-such-that-2-p-left2-x2-1-right-px2-2 | proof explanation - Find all polynomials $P(x)$ such that $2 P\left(2 x^{2}-1\right)=P(x)^{2}-2$ - Mathematics Stack Exchange
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Find all polynomials P(x)P(x) such that 2 P(2 x 2−1)=P(x)2−2 2 P(2 x 2−1)=P(x)2−2
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(Romania, 1990)Find all polynomials P(x)P(x) such that 2 P(2 x 2−1)=P(x)2−2
2 P(2 x 2−1)=P(x)2−2
for all x∈R x∈R.
Solution
Putting x=1,x=1, we get a quadratic equation in P(1)P(1) and hence P(1)=1±√3.P(1)=1±3–√. If P(1)=1+√3,P(1)=1+3–√, we may write P(x)=(x−1)Q(x)+1+√3.P(x)=(x−1)Q(x)+1+3–√. This gives a relation for Q(x)Q(x). 4(x+1)Q(2 x 2−1)=(x−1)Q(x)2+2(1+√3)Q(x)
4(x+1)Q(2 x 2−1)=(x−1)Q(x)2+2(1+3–√)Q(x)
Taking x=1,x=1, we see that Q(1)=0.Q(1)=0. Thus (x−1)(x−1) divides Q(x).Q(x).
We may use the induction to prove that (x−1)n(x−1)n divides Q(x)Q(x) for all positive integers n.n. Thus Q(x)≡0 Q(x)≡0 and P(x)=1+√3 P(x)=1+3–√ for all x.x. Similarly, P(1)=1−√3 P(1)=1−3–√ gives P(x)=1−√3 P(x)=1−3–√ for all x x
now i did not understand how to prove that (x−1)n(x−1)n divides Q(x)Q(x) for all positive integers n.n. using induction....
Any help will be appreciated
thankyou
polynomials
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contest-math
functional-equations
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edited Jan 17, 2021 at 4:06
Eric Wofsey
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asked Apr 17, 2020 at 14:18
IshanIshan
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You have the relation P(x)=(x−1)Q(x)+(1+√3)P(x)=(x−1)Q(x)+(1+3–√). And you have also proved that Q(x)=(x−1)T(x)Q(x)=(x−1)T(x) with deg T(x)=deg Q(x)−1 deg T(x)=deg Q(x)−1, thus you plug in this value of Q(x)Q(x) in the relation to get P(x)=(x−1)2 T(x)+(1+√3)P(x)=(x−1)2 T(x)+(1+3–√) and you can apply the same trick as before (plugging in x=1 x=1) to get x−1∣T(x)x−1∣T(x) as well and thus (x−1)2|Q(x)(x−1)2|Q(x). Thus you can continue this process deg P(x)−1 deg P(x)−1 (degree of Q(x)Q(x)) times and degree of P P can be arbitrary so you have the conclusion.
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answered Apr 17, 2020 at 14:39
user732848 user732848
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Firstly, when using induction, you need the initial/base case. We conclude that (x−1)=(x−1)1(x−1)=(x−1)1 divides the polynomial Q(x)Q(x).
Note that I will be using the following notation; Q(x)=(x−1)⋅Q(1)(x)Q(x)=(x−1)⋅Q(1)(x), and in general Q(n)(x)=(x−1)⋅Q(n+1)(x)
Q(n)(x)=(x−1)⋅Q(n+1)(x)
or Q(x)=(x−1)n⋅Q(n)(x).
Q(x)=(x−1)n⋅Q(n)(x).
Now for induction-step we will assume that (x−1)n(x−1)n divides Q(x)Q(x). We get the following: P(x)=(x−1)n Q(n)(x)+1±√3
P(x)=(x−1)n Q(n)(x)+1±3–√
To answer one must prove "the inductive step". Prove that if we assume the statement holds for n n, it also holds for n+1 n+1.
So using the above, assumed identity, and plugging it into the original equation, we get: 2(((2 x 2−1)−1)n Q(n)(2 x 2−1)+1±√3)=((x−1)n Q(n)(x)+1±√3)2−2
You should work out the brackets to yield: 4(x−1)n(x+1)n Q(2 x 2−1)+2(1±√3)=(x−1)2 n Q(n)(x)2+(x−1)n Q(n)(x)+1±2√3+3−2 The constant cancels, we divide by (x−1)n and we get: 4(x+1)n Q(2 x 2−1)=(x−1)n Q(n)(x)2+Q(n)(x) Filling in x=1 gives: 2 n+2 Q(n)(1)=Q(n)(1)(2 n+2−1)Q(n)(1)=0 We conclude Q(n)(1)=0, thus (x−1) is a divisor of Q(n)(x).
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edited Apr 17, 2020 at 15:34
answered Apr 17, 2020 at 15:28
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Арифмогриф(или, иначе,криптограмма) - задача на отгадывание слов или текстов, где буквы закодированы цифрами, числами и др.
Лесная птица с сильным клювом - 3, 15, 4, 9, 17.
Детеныш свиньи - 12, 11, 14, 11, 16, 9, 7, 11, 6.
Небольшой грызун с острой мордочкой - 2, 10, 1, 5.
Сушеный виноград - 8, 13, 18, 2.
Заменив цифры в следующем числовом наборе на соответствующие буквы, получим полезную пословицу.
7 9 16 2 9 1 8 13 10 6 11 2,
4 1 14 11 12 8 16 5 3 9 17 11 2
(Ответы: дятел, поросенок, мышь, изюм. - Не спеши языком, торопись делом!)
Криптарифм (cryptarithm) - это математический ребус, в котором зашифрован пример на выполнение одного из арифметических действий. При этом одинаковые цифры шифруются одной и той же буквой, а разным цифрам соответствуют различные буквы. Считается, что никакое число не должно начинаться с нуля.
Криптарифм можно считать хорошим, если в результате шифрования получилась какая-то осмысленная фраза. Например, классическим криптарифмом является пример на сложение, придуманный Генри Э. Дьюдени ((Henry Ernest Dudeney). В 1924 году в июньском номере журнала "Strand Magazine" (в конце XIX векасэр Артур Конан ДойльписалШерлока Холмсаспециально для этого издания) Дьюдени публикует ребус: SEND + MORE = MONEY. В переводе с английского языка эта фраза означает "шлите больше денег" - лаконичный текст телеграммы, предположительно, посланной студентом колледжа к родителям)Кроме того, еще одно требование к правильному криптарифму: он должен иметь единственную возможную расшифровку. Например, единственным решением криптарифма Дьюдени является 9567+1085=10652.
Немного истории.
Наряду с классическими (вербальными) ребусами существуют математические ребусы. Они представляют собой примеры обычных арифметических действий (сложения, вычитания, деления и умножения), в которых часть или даже все цифры заменены на точки, звездочки, буквы или другие символы.В буквенных ребусах каждой буквой зашифрована одна определенная цифра. При этом одинаковые цифры шифруются одной и той же буквой, а разным цифрам соответствуют различные буквы.
В ребусах зашифрованных иными значками, например звездочками, каждый символ может обозначать любую цифру от 0 до 9. В таких ребусах некоторые цифры могут повторяться несколько раз, а другие не использоваться вовсе.
В целом, математические ребусы представляют собой логико-математические задачи, в которых путем логических рассуждений и математических вычислений требуется расшифровать значение каждого символа и восстановить числовую запись. Решить ребус - означает восстановить первоначальный вид математического равенства.
Математический буквенный ребус можно считать хорошим, если в результате шифрования получилась какая-то осмысленная фраза. Еще одно требование к хорошему ребусу: он должен иметь единственную возможную расшифровку.Математический ребус - довольно старая головоломка, изобретатель её не известен. Долгое время авторство ошибочно приписывали американскому шахматистуСэмюэлю Лойду(Samuel Loyd). Однако опубликованный (?) в декабрьском номере журнала "The American Agriculturist" за 1864 год математический ребус поставил под сомнение распространённое мнение.
Математический ребус в русском языке зачастую называют арифметическим, числовым или цифровым. В английском языке также используется несколько названий для обозначения данного вида головоломок: Verbal arithmetic, Alphametics, Cryptarithmetic (Crypt-arithmetic), Cryptarithm или Word addition.Термин "crypt-arithmetic" был придуман бельгийским композитором и составителем головоломок Симоном Ватриквантом (Simon Vatriquant), который под псевдонимом "Minos" с 1931 года и до началаВторой мировой войныпубликовал математические ребусы в бельгийском математическом журнале "Sphinx":
В 1945 году Алан Уэйн (Alan Wayne) представил особый вид математического ребуса именуемого "doubly-true". Такой ребус состоит из слов, обозначающих цифры или числа, которые также являются математическим равенством. Например: THREE + FOUR = SEVEN (три + четыре = семь), зашифровано 28566 + 7495 = 36061 либо 27566 + 8495 = 36061.
Позднее, в 1955 году,Джеймс Хантер(James Alston Hope Hunter) в октябрьском номере газеты "Toronto Globe and Mail" для обозначения буквенных математических ребусов в которых буквы образуют осмысленные слова или фразы использует слово "alphametic". Не смотря на то, что со временем, всё же, больше прижилось слово "криптарифм", Джеймса Хантера без преувеличения можно назвать "отцом" современного математического ребуса, создателем самых блестящих математических головоломок.
Первое упоминание об отечественных математических ребусах, автору проекта "Ребус № 1" удалось отыскать в книге выдающегося российского, советского учёного, популяризатора физики, математики и астрономии, одного из основоположников жанра научно-популярной литературыЯкова Исидоровича Перельмана"Занимательная арифметика. Загадки и диковинки в мире чисел", выпущенной Ленинградским издательством "Время" в 1926 году. В одной из глав автор пишет: "То, что я предлагаю назвать арифметическими ребусами - занимательная игра американских школьников, у нас пока еще совершенно неизвестная.
Английское название игры "div-al-et" - сокращение от "division by letters", т. е. деление c помощью букв. Она состоит в отгадывании задуманного словапосредством решения задачи (на деление)...". Позднее, в 1939 году также в Ленинграде выходит небольшая брошюра "Арифметические ребусы" (серия "Дом занимательной науки") целиком посвященная этой математической головоломке.
Источник:rebus1.com
Примеры КРИПТАРИФМОВ
1. КНИГА + КНИГА + КНИГА = НАУКА Ответ:28375+28375+28375=85125
ОДИН+ОДИН=МНОГО Ответ:6823+6823=13646
3.Задача
В этой задаче цифры заменены буквами. Одинаковыми буквами заменены одинаковые цифры. Восстановите зашифрованные цифры:
ПРИМЕР
РИМЕР
ИМЕР
МЕР
ЕР
Р
ЗАДАЧА Ответ ПРИМЕР = 851745, ЗАДАЧА = 906030
4.Вагоны
Замените буквы на цифры, чтобы добиться равенства:
ВАГОН + ВАГОН = СОСТАВ Ответ 85679 + 85679 = 171358
5.Нитки
НИТКА+НИТКА=ТКАНЬ Ответ
15306+15306=30612
6.ABCDE
Замените буквы цифрами так, чтобы пример на умножение был верен. Каждой букве соответствует только одна цифра. Разным буквам не могут соответствовать одинаковые цифры.
ABCDE
4
EDCBA
Ответ
21978
4
87912
Рассуждения такие: A<=2 (иначе перенос в след. разряд); а - четное (т.к. рез-т умножения на 4). => A=2 => E=8 => B<=2 (иначе перенос в старший разряд и E не будет равно 8). B - нечетное (т.к. перенос от 48=3) сл-но B=1. Отсюда D либо 7 либо 2 (последняя цифра 4D+3=1). D=2 быть не может, т.к. это говорит о том, что был перенос в старший разряд, сл-но D=7. Hу и C=9 (т.к. перенос во второй разряд = 3 и перенос из 4-ого разряда = 3, то С=7 или 8 или 9. Проверям - подходит только 9)
7.
ДЕРЕВО-ОПИЛКИ=ПАЛКИ=>569614-487307=82307
МУХА:ХА=УХА=>3125:25=125
КРОТЯ=ТРОЯК=>49738=39784
Допускается использование русских и латинских букв, круглых скобок, знаков сложения (+), вычитания (-), умножения (), деления (/), возведения в степень (^) и факториала (!). Также, вместо любой цифры в математическом выражении можно использовать символ?.
УМ^А=МЕШОК=>12 4=20736
(М+О+С+К+В+А)^4=МОСКВА=>(3+9+0+6+2+5)4=390625
А^РК^А=АРКА=>2 59 2=2592
Я!=АТЛЕТ=>8!=40320
НАГОЙ:ЙОГАН=?=>87912:21978=4
Пользователь может указать известные значения для некоторых букв. Это позволит существенно уменьшить время поиска.
Можно также указать, что гласные буквы в задании соответствуют четным цифрам, а согласные буквы - нечетным цифрам (и наоборот). Например, если гласные буквы соответствуют нечетным цифрам, а согласные буквы- четным, то следующая головоломка имеет единственное решение:
МУХА+МУХА=СЛОН=>2309+2309=4618
7.ШАХ+ШАХ+ШАХ+ШАХ+ШАХ+ШАХ=МАТ(6ШАХ=МАТ);
7ШАХ=МАТ;
7СЛОВО=ФРАЗА;
27СЛОВ=ФРАЗА;
4ОДИН=МНОГО;
6ОДИН=МНОГО;
КОМПЬЮ=ТЕРРА.
Вот несколько неправильных криптарифмов. Постарайтесь найти все ответы.
Так, ребус "РАЙОН+РАЙОН=ГОРОД" имеет два решения,
"ДЕДКА+БАБКА+РЕПКА=СКАЗКА"- четыре,
а "ЛАДЬЯ+ЛАДЬЯ+ФЕРЗЬ"- целых шестнадцать!
Тем же недостатком страдает и фраза-криптарифм "Merry xmas from Maxey"- у нее 24 различных решения.
КРИПТАРИФМЫ можно посмотреть ЗДЕСЬИли ЗДЕСЬИли ЗДЕСЬ!
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6 комментариев:
Unknown11 октября 2015 г. в 01:30
Как решить криптарифм:
FORTY+TEN+TEN=SIXTY. Может ли буква E принимать значение 5?
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Unknown11 октября 2015 г. в 01:30
Как решить криптарифм:
FORTY+TEN+TEN=SIXTY. Может ли буква E принимать значение 5?
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Ответы
1. Image 5 Рукалицо22 июня 2021 г. в 01:56
29786+850+850=31486
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Unknown28 ноября 2017 г. в 22:49
Помогите расшифровать математический термин 1201561516117191231
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Анонимный7 января 2018 г. в 10:11
песок+вода=оазис
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Анонимный10 января 2020 г. в 04:53
Нappy-New=Year
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Balanced coloring for $\mathbb{Z}^n$ with $m$ colors
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Call a coloring using $m$ colors on a finite number of points in $\mathbb{Z}^n$ balanced if for any line parallel to one of $n$ axes, the difference between the number of points for any 2 colors is at most 1. Does such a coloring exist for all $m, n \ge 2$?
Note that this problem is a generalization for IMO 1986 problem 6 where $m = n = 2$. I solved the original problem using induction on the number of points by removing one that lies on a line with an odd number of points and tackle the case when all lines have an even number of points by considering the simple graph with vertices as the horizontal and vertical lines that contain at least one point, which are finitely many, edges as the mentioned integer points connecting a horizontal and vertical line sharing one and use the fact that this graph is Eulerian to color the edges so that two adjacent edges in the cycle are colored differently. I have tried to create a similar reasoning when $m > 2$ or $n > 2$ but failed.
I would appreciate any improvement to my method or a solution to the above problem even if its just the case $m = 2$ or $n = 2$.
combinatorics
integer-lattices
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asked May 29, 2024 at 14:31
DÅ©ng Nguyá»
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When $n \ge 3$, such a coloring does not necessarily exist. First, here is a counterexample in $\mathbb Z^3$ when $m=2$:
We should think of this counterexample as consisting of three $L$ shapes strung together. A $L$ shape is a configuration
A | | X---B
which forces points $A$ and $B$ to have the same color. The diagram above has one of these in the plane $x=0$, one in the plane $z=0$, and one in the plane $x=1$, resulting in $\operatorname{color}(0,0,1) = \operatorname{color}(0,1,0) = \operatorname{color} (1,2,0) = \operatorname{color}(1,0,1)$. But the final two points are collinear and so no valid coloring can give them the same color.
For $m\ge 3$ and $n \ge 3$, we can find a counterexample by generalizing the one above. Instead of the L-shaped gadget, use the following (illustrated for $m=5$):
A | | | | | | | | | | | | --------B
Suppose point $A$ gets color $c$. Then the $m-1$ points directly below it cannot reuse color $c$, but they are each on a horizontal line of $m$ points that must use color $c$ at least once. Therefore the $(m-1)\times(m-1)$ grid of points to their right must contain $m-1$ points with color $c$: one in each row, and hence one in each column. This means that every vertical line in that grid already includes color $c$, so their bottom point cannot have color $c$. Finally, the horizontal line containing $B$ must have a point of color $c$, and the only point that can have this color is point $B$ itself.
Here is an illustration (again, for $m=5$) of how this gadget can be used to get a counterexample:
On the other hand, when $n=2$, such a coloring exists for all $n$. (This is essentially my strategy here, but generalized to $m$ colors and somewhat simplified.)
Let $P$ be our set of points; assume without loss of generality that $P \subseteq {1,2,\dots,N}^2$ for some integer $N$. We first modify the problem in two ways:
We replace $P$ by a larger point set $Q$ containing $m^2$ copies of $P$: the sets $(aN,bN) + P$, where $0 \le a,b \le m-1$.
After this operation, a horizontal or vertical line containing $k$ points in $P$ contains $km$ points in $Q$. We split every such line into $k$ segments, each containing $m$ consecutive points of that line. (This includes horizontal and vertical lines containing $1$ point in $P$; they contain $m$ points in $Q$, which we end up turning into just one segment. As a result, every point in $Q$ is on some horizontal and some vertical segment.)
Our goal is now to color so that all $m$ points on a segment get different colors.
If we solve this new coloring problem, then the copy of $P$ inside $Q$ is also correctly colored. A $k$-point horizontal or vertical line inside $P$ ends up being divided into $\lfloor km\rfloor$ complete segments and maybe a partial segment: so in the coloring of $Q$, all colors must get used $\lfloor km\rfloor$ times on the complete segments, and some may be used an additional time on the partial segment.
On to solving the problem for $Q$. We consider the bipartite graph $G$ with the horizontal segments in $Q$ on one side and the vertical segments in $Q$ on the other side; for each point in $Q$, we draw an edge between the horizontal and vertical segment containing it. The graph $G$ is $m$-regular, and by a theorem of KÅnig, we can partition it into $m$ perfect matchings. (Briefly: we can apply Hall's theorem to get a perfect matching in any regular graph, then we can remove the edges of that perfect matching and repeat.)
This partition into $m$ perfect matchings is exactly the coloring we want: for each $i=1,\dots,m$, take the $i^{\text{th}}$ perfect matching in $G$ and use the $i^{\text{th}}$ color on all points in $Q$ corresponding to its edges. Every segment corresponds to a vertex in $G$, which is incident to exactly one edge of the $i^{\text{th}}$ matching; therefore the segment contains exactly one point of the $i^{\text{th}}$ color.
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edited May 29, 2024 at 19:27
answered May 29, 2024 at 19:00
Misha LavrovMisha Lavrov
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This depression inventory can be self-scored. The scoring scale is at the end of the questionnaire.
0 I do not feel sad. 1 I feel sad 2 I am sad all the time and I can't snap out of it. 3 I am so sad and unhappy that I can't stand it.
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0 I don't feel I am being punished. 1 I feel I may be punished. 2 I expect to be punished. 3 I feel I am being punished.
0 I don't feel disappointed in myself. 1 I am disappointed in myself. 2 I am disgusted with myself. 3 I hate myself.
0 I don't feel I am any worse than anybody else. 1 I am critical of myself for my weaknesses or mistakes. 2 I blame myself all the time for my faults. 3 I blame myself for everything bad that happens.
0 I don't have any thoughts of killing myself. 1 I have thoughts of killing myself, but I would not carry them out. 2 I would like to kill myself. 3 I would kill myself if I had the chance.
0 I don't cry any more than usual. 1 I cry more now than I used to. 2 I cry all the time now. 3 I used to be able to cry, but now I can't cry even though I want to. 11. 0 I am no more irritated by things than I ever was. 1 I am slightly more irritated now than usual. 2 I am quite annoyed or irritated a good deal of the time. 3 I feel irritated all the time.
0 I have not lost interest in other people. 1 I am less interested in other people than I used to be. 2 I have lost most of my interest in other people. 3 I have lost all of my interest in other people.
0 I make decisions about as well as I ever could. 1 I put off making decisions more than I used to. 2 I have greater difficulty in making decisions more than I used to. 3 I can't make decisions at all anymore.
0 I don't feel that I look any worse than I used to. 1 I am worried that I am looking old or unattractive. 2 I feel there are permanent changes in my appearance that make me look
unattractive 3 I believe that I look ugly.
0 I can work about as well as before. 1 It takes an extra effort to get started at doing something. 2 I have to push myself very hard to do anything. 3 I can't do any work at all.
0 I can sleep as well as usual. 1 I don't sleep as well as I used to. 2 I wake up 1-2 hours earlier than usual and find it hard to get back to sleep. 3 I wake up several hours earlier than I used to and cannot get back to sleep.
0 I don't get more tired than usual. 1 I get tired more easily than I used to. 2 I get tired from doing almost anything. 3 I am too tired to do anything.
0 My appetite is no worse than usual. 1 My appetite is not as good as it used to be. 2 My appetite is much worse now. 3 I have no appetite at all anymore.
0 I haven't lost much weight, if any, lately. 1 I have lost more than five pounds. 2 I have lost more than ten pounds. 3 I have lost more than fifteen pounds. 20. 0 I am no more worried about my health than usual. 1 I am worried about physical problems like aches, pains, upset stomach, or constipation. 2 I am very worried about physical problems and it's hard to think of much else. 3 I am so worried about my physical problems that I cannot think of anything else.
0 I have not noticed any recent change in my interest in sex. 1 I am less interested in sex than I used to be. 2 I have almost no interest in sex. 3 I have lost interest in sex completely.
INTERPRETING THE BECK DEPRESSION INVENTORY
Now that you have completed the questionnaire, add up the score for each of the twenty-one questions by counting the number to the right of each question you marked. The highest possible total for the whole test would be sixty-three. This would mean you circled number three on all twenty-one questions. Since the lowest possible score for each question is zero, the lowest possible score for the test would be zero. This would mean you circles zero on each question. You can evaluate your depression according to the Table below.
Total Score________Levels of Depression
1-10____These ups and downs are considered normal 11-16__ Mild mood disturbance 17-20____Borderline clinical depression 21-30___Moderate depression 31-40__Severe depression over 40___Extreme depression |
16106 | https://mathworld.wolfram.com/PoissonDistribution.html | Poisson Distribution -- from Wolfram MathWorld
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Poisson Distribution
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Given a Poisson process, the probability of obtaining exactly successes in trials is given by the limit of a binomial distribution
(1)
Viewing the distribution as a function of the expected number of successes
(2)
instead of the sample size for fixed , equation (2) then becomes
(3)
Letting the sample size become large, the distribution then approaches
(4)
(5)
(6)
(7)
(8)
which is known as the Poisson distribution (Papoulis 1984, pp.101 and 554; Pfeiffer and Schum 1973, p.200). Note that the sample size has completely dropped out of the probability function, which has the same functional form for all values of .
The Poisson distribution is implemented in the Wolfram Language as PoissonDistribution[mu].
As expected, the Poisson distribution is normalized so that the sum of probabilities equals 1, since
(9)
The ratio of probabilities is given by
(10)
The Poisson distribution reaches a maximum when
(11)
where is the Euler-Mascheroni constant and is a harmonic number, leading to the transcendental equation
(12)
which cannot be solved exactly for .
The moment-generating function of the Poisson distribution is given by
(13)
(14)
(15)
(16)
(17)
(18)
so
(19)
(20)
(Papoulis 1984, p.554).
The raw moments can also be computed directly by summation, which yields an unexpected connection with the Bell polynomial and Stirling numbers of the second kind,
(21)
known as Dobiński's formula. Therefore,
(22)
(23)
(24)
The central moments can then be computed as
(25)
(26)
(27)
so the mean, variance, skewness, and kurtosis excess are
(28)
(29)
(30)
(31)
(32)
The characteristic function for the Poisson distribution is
(33)
(Papoulis 1984, pp.154 and 554), and the cumulant-generating function is
(34)
so
(35)
The mean deviation of the Poisson distribution is given by
(36)
The Poisson distribution can also be expressed in terms of
(37)
the rate of changes, so that
(38)
The moment-generating function of a Poisson distribution in two variables is given by
(39)
If the independent variables , , ..., have Poisson distributions with parameters , , ..., , then
(40)
has a Poisson distribution with parameter
(41)
This can be seen since the cumulant-generating function is
(42)
(43)
A generalization of the Poisson distribution has been used by Saslaw (1989) to model the observed clustering of galaxies in the universe. The form of this distribution is given by
(44)
where is the number of galaxies in a volume , , is the average density of galaxies, and , with is the ratio of gravitational energy to the kinetic energy of peculiar motions, Letting gives
(45)
which is indeed a Poisson distribution with . Similarly, letting gives .
See also
Binomial Distribution, Erlang Distribution, Poisson Process, Poisson TheoremExplore this topic in the MathWorld classroom
Explore with Wolfram|Alpha
More things to try:
poisson distribution
Poisson distribution with mean 10
Poisson distribution mean = 5
References
Beyer, W.H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p.532, 1987.Grimmett, G. and Stirzaker, D. Probability and Random Processes, 2nd ed. Oxford, England: Oxford University Press, 1992.Papoulis, A. "Poisson Process and Shot Noise." Ch.16 in Probability, Random Variables, and Stochastic Processes, 2nd ed. New York: McGraw-Hill, pp.554-576, 1984.Pfeiffer, P.E. and Schum, D.A. Introduction to Applied Probability. New York: Academic Press, 1973.Press, W.H.; Flannery, B.P.; Teukolsky, S.A.; and Vetterling, W.T. "Incomplete Gamma Function, Error Function, Chi-Square Probability Function, Cumulative Poisson Function." §6.2 in Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Cambridge, England: Cambridge University Press, pp.209-214, 1992.Saslaw, W.C. "Some Properties of a Statistical Distribution Function for Galaxy Clustering." Astrophys. J.341, 588-598, 1989.Spiegel, M.R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, pp.111-112, 1992.
Referenced on Wolfram|Alpha
Poisson Distribution
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Weisstein, Eric W. "Poisson Distribution." From MathWorld--A Wolfram Resource.
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16107 | https://www.math.unl.edu/~gledder1/Notes/842/Dominant%20Balance.pdf | Dominant Balance Example Problem • Find the asymptotic behavior as x →∞, including all non-vanishing terms, for the graph of the equation x2 + xy −y3 = 0.
The idea of a dominant balance argument is that no one term in an equation can dominate all of the others. While it is possible that more than two could be comparable, it is much more likely that there is a dominant balance between two terms and the rest are small. We therefore have to consider three cases: 1. Suppose y3 ≪x2, xy as x →∞. Then xy ∼−x2, so y ∼−x. Then y3 = O(x3), which means y3 ≫x2 as x →∞, which contradicts the initial assumption. Thus, we cannot have a dominant balance as x →∞that excludes y3.
2. Suppose x2 ≪xy, y3 as x →∞. Then y3 ∼xy, so y ∼±√x. Then xy = O(x3/2), which means x2 ≫xy, contradicting the initial assumption and forcing the conclusion that x2 cannot be omitted from the dominant balance.
3. Suppose xy ≪x2, y3 as x →∞. Then y3 ∼x2, so y ∼x2/3. So xy = O(x5/3), which is consistent with the initial assumption. There is no contradiction, so the behavior y ∼x2/3 is correct as x →∞.
Once we have one or more correct dominant balances, we can peel offthe asymptotic behavior to get additional terms. In these additional calculations, we can use what we already know about the orders of terms to restrict the dominant balance alternatives. Assume1 y = x2/3 + y1(x), y1 ≪x2/3.
(1) Then y3 = x2 + 3x4/3y1 + 3x2/3y2 1 + y3 1.
Grouping the two dominant terms yields h x2 + 3x4/3y1 + 3x2/3y2 1 + y3 1 i −x2 = x h x2/3 + y1 i , or 3x4/3y1 + 3x2/3y2 1 + y3 1 = x5/3 + xy1.
(2) Equation (2) is now a problem to be solved for the leading order behavior of y1. Because of the way we did the bookkeeping, we know in advance that the terms on each side of the equation are sorted into the proper asymptotic order, so we do not need to do a dominant balance argument and can immediately get 3x4/3y1 ∼x5/3, or y1 ∼1 3x1/3.
1The equation is merely a definition of y1, so there is no assumption needed for it. The assumption is on the ordering of y1, but that assumption is known to be consistent by our earlier argument that y ∼x2/3.
It appears likely that the next term in the asymptotic behavior is a constant, so we need to look for it in order to be sure of having all non-vanishing terms. To do that, we need to assume y1 = 1 3x1/3 + y2(x), y2 ≪x1/3.
(3) Rather than writing down every term, we can truncate each expansion so that we take at most two terms from the first term on each side of (2), one term from the second terms, and none from any subsequent terms. Thus, 3x4/3 1 3x1/3 + y2 + o(y2) + 3x2/3 1 9x2/3 + O x1/3y2 + [O(x)] = x5/3 + x 1 3x1/3 + O (y2) .
At this point, we expect the first term on each side to cancel because that follows from (3), but we actually get an unexpected second cancelation as well, leaving us with 3x4/3y2 + o(x4/3y2) + O(xy2) + O(x) = O(xy2).
Clearly we needed to take an additional term from at least some of the expansions. Nevertheless, we can still do a dominant balance. The terms of order xy2 are clearly smaller than the term 3x4/3y2, so we may conclude 3x4/3y2 = O(x).
We could track down the coefficient of the order x term, but it is enough to note that we have shown y2 = O x−1/3 →0.
Hence, the nonvanishing portion of the asymptotic behavior is y ∼x2/3 + 1 3x1/3.
(4) |
16108 | https://courses.lumenlearning.com/uvu-combinedalgebra/chapter/4-1-quadratic-functions-and-their-graphs/ | Chapter 4: Quadratic Functions
4.1: Quadratic Functions and Their Graphs
Learning Outcomes
Graph the quadratic function [latex]f(x)=x^2[/latex]
Identify the vertex, line of symmetry, [latex]x[/latex]-intercept(s), [latex]y[/latex]-intercept, and vertex of a parabola
Determine if a parabola has a minimum or maximum and find that value
Identify the symmetric point given a point on the parabola and the line of symmetry
Find the [latex]x[/latex]-coordinate of the vertex using two symmetric points
Identify the domain and range of a quadratic function
Graphing a Quadratic Function
One way to graph a function is to make a table of values and transfer the resulting coordinate pairs onto the coordinate plane. Then we can connect the plotted points to sketch the graph of the function.
The function [latex]f(x)=x^2[/latex] is a polynomial function of degree 2. Such polynomial functions are also called quadratic functions.To graph the quadratic function [latex]f(x)=x^2[/latex], we can generate the table of values in table 1. We can choose any [latex]x[/latex]-values we want, but it is best to cover both positive and negative values of [latex]x[/latex] so that the graph may be comprehensive rather than only showing part of the graph.
Table of values for the function [latex]f(x)=x^2[/latex]
| [latex]x[/latex] | [latex]f(x)=x^2[/latex] |
| -3 | 9 |
| -2 | 4 |
| -1 | 1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
| 3 | 9 |
| Table 1. Table of values for [latex]f(x)=x^2[/latex] |
Now, we take each latex[/latex] pair on the table as a coordinate pair latex[/latex] and plot them on the coordinate plane. We cannot connect these points using straight lines because the function is not linear (i.e., the exponent of the independent variable [latex]x[/latex] is not 1). That is, the slopes between any two points are not a constant. For example, we use a curve instead of a straight line to connect the two points (1,1) and (2,4) because you can find another point (1.5, 2.25) on the graph between the two points, and the slope between point (1,1) and (1.5, 2.25) is 2.5 but the slope between (1.5, 2.25) and (2, 4) is 3.5. Therefore, we cannot connect the two points (1,1) and (2,4) with a straight line because the slopes between the two points are different (i.e., [latex]2.5 \neq 3.5[/latex]). Since it is a polynomial, we will use a curve to connect these points on the coordinate plane (figure 1). The shape of this curve is called a parabola.
Figure 1. The graph of the function [latex]f(x)=x^2[/latex].
The function [latex]f(x)=x^2[/latex] is the parent function of all quadratic functions, which can be written in the form [latex]f(x)=ax^2+bx+c[/latex], where [latex]a, b, c[/latex] are real numbers with [latex]a\ne0[/latex]. The parabola will open upwards if [latex]a>0[/latex] and will open downwards if [latex]a<0[/latex] (figure 2).
Figure 2. Parabola opens up when [latex]a>0[/latex] and down when [latex]a<0[/latex] on graph of [latex]f(x)=ax^2+bx+c[/latex]
Features of the Graph of a Quadratic Function
One important feature of the graph of a quadratic function (a parabola) is that it has a turning point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric about a vertical line that passes through the vertex, called the line (or axis) of symmetry (figure 3).
Figure 3. The vertex is always on the axis of symmetry.
The intersection points between the graph and the [latex]x[/latex]-axis are the [latex]x[/latex]-intercepts. There can be 0, 1, or 2 [latex]x[/latex]-intercepts, depending where the graph lies on the coordinate plane. The intersection point between the graph and the [latex]y[/latex]-axis is the [latex]y[/latex]-intercept (figure 4).
| | | |
---
| Parabola opening up with a minimum function value and no [latex]x[/latex]-intercepts. | Parabola opening up with a minimum function value and one [latex]x[/latex]-intercepts. | Parabola opening down with a maximum function value and two [latex]x[/latex]-intercepts |
| Figure 4. Parabolas with 0, 1, and 2 [latex]x[/latex]-intercepts |
The parabola in figure 5 has [latex]x[/latex]-intercepts of (–1, 0) and (3, 0). The [latex]y[/latex]-intercept is (0, –3) and the vertex is (1, –4). The equation for the line of symmetry is [latex]x=1[/latex] because it is a vertical line that passes through the vertex (1, –4). The minimum value of the function occurs at the vertex and has a value of –4.
Figure 5. Features of a parabola
Example 1
Determine the vertex, line of symmetry, [latex]x[/latex]-intercepts, [latex]y[/latex]-intercept and minimum value of the parabola.
Solution
The graph crosses the [latex]y[/latex]-axis at 7, so the [latex]y[/latex]-intercept is (0, 7).
The graph never crosses the [latex]x[/latex]-axis, so there are no [latex]x[/latex]-intercepts.
The graph turns at (3, 1) so the vertex is (3, 1).
The graph has a minimum value at the vertex of 1.
The axis of symmetry is the vertical line through the vertex (3, 1), so is the line [latex]x=3[/latex].
Try It 1
Determine the vertex, line of symmetry, [latex]x[/latex]-intercepts, [latex]x[/latex]-intercept and maximum value of the parabola.
Show Answer
vertex = latex[/latex]
line of symmetry: [latex]x=1[/latex]
[latex]x[/latex]-intercepts = latex[/latex] and latex[/latex]
[latex]y[/latex]-intercept = latex[/latex]
maximum value = 4
Identifying Symmetric Points
Since the graph of a quadratic function is symmetric across the line of symmetry, every point on the graph has a reflection called its symmetric point. Given any point [latex]A[/latex] on a parabola, the symmetric point [latex]A^{\prime}[/latex] of the given point [latex]A[/latex] will be on the other side of the line of symmetry, the same distance from the line of symmetry as the distance of point [latex]A[/latex] from the line of symmetry. Since the line of symmetry is vertical, the point [latex]A[/latex] and [latex]A^{\prime}[/latex] lie on the same horizontal line that is perpendicular to the line of symmetry. Therefore, the point [latex]A^{\prime}[/latex] has the same [latex]y[/latex]-coordinate as the given point [latex]A[/latex] (figure 6).
Figure 6. Move point P to see its symmetric point.
Example 2
Find the symmetric point of a given point (2, 10) on a parabola with vertex (5, 1).
Solution
Draw a quick sketch of a parabola with its vertex at (5, 1) and that passes through the point (2, 10).
On the graph of the quadratic function, we have the line of symmetry [latex]x=5[/latex] passing through the vertex.
The symmetric point of (2, 10) will be on the other side of the line of symmetry where the [latex]y[/latex]-coordinate of the symmetric point will share the same [latex]y[/latex]-value of 10.
What is left now is to find the [latex]x[/latex]-coordinate of the symmetric point.
The distance between the given point (2, 10) and the line of symmetry [latex]x=5[/latex] is 3 (i.e., 5 – 2 = 3).
So starting at [latex]x=5[/latex] and moving to the right 3 units gives us the [latex]x[/latex]-coordinate of the symmetric point, 8 (i.e., 5 + 3 = 8).
Therefore, the symmetric point of the given point (2, 10) is (8, 10).
Try It 2
Find the symmetric point of a given point (0, –4) on a parabola with vertex (–3, 5).
Show Answer
Symmetric point of (0, –4) is (–6, –4).
Finding the [latex]x[/latex]-coordinate of the Vertex Using Two Symmetric Points
Any two symmetric points on a parabola are the same distance away from the line of symmetry. Consequently, any two symmetric points are the same distance away from the [latex]x[/latex]-coordinate of the vertex. Therefore, the [latex]x[/latex]-coordinate of the vertex is the midpoint between any two symmetric points on a parabola. This means that we can identify the [latex]x[/latex]-coordinate of the vertex if we know two symmetric points on a parabola. To find the halfway point between two given values, [latex]x_1[/latex] and [latex]x_2[/latex], we find the mean (or average) of the values: [latex]\dfrac{x_1+x_2}{2}[/latex]. To find the corresponding [latex]y[/latex]-coordinate of the vertex, we need the function equation so we can find the function value at the vertex.
For example, given two symmetric points (2, 10) and (8, 10) on the graph of the function [latex]f(x)=x^2-10x+26[/latex], the [latex]x[/latex]-coordinate of the vertex will be halfway between the two symmetric points (figure 7).
Figure 7. Axis of symmetry is halfway between two symmetric points.
The halfway point between [latex]x=2[/latex] and [latex]x=8[/latex] is:
[latex]x=\dfrac{2+8}{2} = \dfrac{10}{2} = 5[/latex]
Knowing that the [latex]x[/latex]-coordinate of the vertex is 5, the [latex]y[/latex]-coordinate of the vertex will be [latex]f(5)[/latex]:
[latex]\begin{aligned}f(5) &= (5)^2-10(5)+26 \&= 25 - 50 + 26 \&= 1\end{aligned}[/latex].
Therefore, the vertex of the function is at the point (5, 1).
Note. If we did not know the equation of the function, or were unable to find the equation, we would not have been able to determine the [latex]y[/latex]-coordinate of the vertex. This is because there are an infinite number of parabolas that pass through the two symmetric points (2, 10) and (8, 10) (figure 8).
Figure 8. Example of parabolas that pass through (2, 10) and (8, 10)
Example 3
Find the vertex of the parabola representing the function [latex]f(x)=x^2-8x+13[/latex], which passes through the symmetric points (2, 1) and (6, 1).
Solution
The axis of symmetry is halfway between [latex]x=2[/latex] and [latex]x=6[/latex]:
[latex]x=\dfrac{2+6}{2}=4[/latex]
Therefore, the [latex]x[/latex]-coordinate of the vertex is 4.
To find the [latex]y[/latex]-coordinate of the vertex, we need to find [latex]f(4)[/latex]:
[latex]\begin{aligned}f(4)&=(4)^2-8(4)+13\&=16-32+13\&=-3\end{aligned}[/latex]
The vertex is the point (4, –3).
Try It 3
Find the vertex of the parabola representing the function [latex]f(x)=-x^2-6x-3[/latex], which passes through the symmetric points (–7, –10) and (1, –10).
Show Answer
The vertex is (–3, 6).
Finding the Domain and Range of a Quadratic Function
Any number can be the input value of a quadratic function. Therefore the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum at the vertex, the range is restricted. Since the [latex]y[/latex]-coordinate of the vertex of a parabola will be either a maximum or a minimum, the range will be determined by the [latex]y[/latex]-coordinate of the vertex and whether or not the parabola opens upwards or downwards. If the parabola opens upwards, the range will consist of all [latex]y[/latex]-values greater than or equal to the [latex]y[/latex]-coordinate of the vertex. If the parabola opens downwards, the range will consist of all [latex]y[/latex]-values less than or equal to the [latex]y[/latex]-coordinate of the vertex.
Domain and Range of a quadratic function
For any quadratic function whose graph has a vertex at the point latex[/latex], the domain is all real numbers and the range depends on the whether the parabola opens up or down.
| | |
--- |
| Domain = latex[/latex] Range = latex[/latex] |
Example 4
State the domain and the range of a quadratic function whose graph is a parabola that opens downwards with a vertex at the point (–2, 5).
Solution
First we should recognize that there are an infinite number of parabolas that opens downwards with a vertex at the point (–2, 5)! However, they all have the same domain and range.
The domain is all real numbers: [latex]x\in(-\infty,+\infty)[/latex]
Because the parabola opens downwards it has a maximum value at the vertex. So the range is all real numbers less than or equal to 5: [latex]y\in(-\infty,5][/latex]
Try It 4
State the domain and the range of a quadratic function whose graph is a parabola that opens upwards with a vertex at the point (4, –3).
Show Answer
Domain: [latex]x\in(-\infty,+\infty)[/latex]
Range: [latex]y\in[-3,+\infty)[/latex]
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Graph the Quadratic Function f(x)=x^2. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Identify a Symmetric Point. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
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Graph the Quadratic Function f(x)=x^2. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Identify a Symmetric Point. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Find the x-coordinate of the Vertex Using Two Symmetric Points. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
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All graphs created using desmos graphing calculator. Authored by: Hazel McKenna and Leo Chang. Provided by: Utah Valley University. License: CC BY: Attribution
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16110 | https://ocw.mit.edu/courses/18-02-multivariable-calculus-fall-2007/7ac1ea75a9b0cd82d84297abe0958407_prac1a.pdf | MIT OpenCourseWare 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: ¢ ¢ ? µ ~ 18.02 Practice Exam 1 A Problem 1. (15 points) A unit cube lies in the first octant, with a vertex at the origin (see figure). −! −! a) Express the vectors OQ (a diagonal of the cube) and OR (joining O to the center of a face) in terms of ˆ ı, ˆ k.
´, ˆ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ O Q qR z y x b) Find the cosine of the angle between OQ and OR. Problem 2. (10 points) The motion of a point P is given by the position vector R ~ = 3 cos t ˆ ´ + t ˆ ı + 3 sin t ˆ k. Compute the velocity and the speed of P . Problem 3. (15 points: 10, 5) ⎤ 3 ⎤ 3 a) Let A = ⎥ 4 1 2 1 3 0 1 2 −1 0 7 ⎣; then det(A) = 2 and A−1 = 1 2 ⎥ 4 1 −1 2 a −2 2 b 5 −6 7 ⎣; find a and b. ⎤ 3 ⎤ 3 x 1 ⎥ 7 ⎥ 7 b) Solve the system A X = B, where X = 4 y ⎣ and B = 4 −2 ⎣ . z 1 c) In the matrix A, replace the entry 2 in the upper-right corner by c. Find a value of c for which the resulting matrix M is not invertible. For this value of c the system M X = 0 has other solutions than the obvious one X = 0: find such a solution by using vector operations. (Hint: call U, V and W the three rows of M, and observe that M X = 0 if and only if X is orthogonal to the vectors U, V and W .) Problem 4. (15 points) The top extremity of a ladder of length L rests against a vertical wall, while q the bottom is being pulled away. Find parametric equations for the midpoint P of the ladder, using as parameter the angle µ between the ladder and the q P ¢ horizontal ground. ¢ ¢ q ¾ Problem 5. (25 points: 10, 5, 10) a) Find the area of the space triangle with vertices P0 : (2, 1, 0), P1 : (1, 0, 1), P2 : (2, −1, 1). b) Find the equation of the plane containing the three points P0, P1, P2. c) Find the intersection of this plane with the line parallel to the vector V ~ = h1, 1, 1i and passing through the point S : (−1, 0, 0). Problem 6. (20 points: 5, 5, 10) a) Let R ~ = x(t)ˆ ´ + z(t)ˆ ı+ y(t)ˆ k be the position vector of a path. Give a simple intrinsic formula for d (R ~ · R) in vector notation (not using coordinates).
dt b) Show that if R ~ has constant length, then R ~ and V ~ are perpendicular. c) let A ~ be the acceleration: still assuming that R ~ has constant length, and using vector differ entiation, express the quantity R ~ · A ~ in terms of the velocity vector only. |
16111 | https://es.wikipedia.org/wiki/Plano_inclinado | Published Time: 2003-10-16T19:39:21Z
Plano inclinado - Wikipedia, la enciclopedia libre
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1 Visión general
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4 TerminologíaAlternar subsección Terminología
4.1 Pendiente
4.2 Ventaja mecánica
5 Plano inclinado sin fricción
6 Plano inclinado con rozamientoAlternar subsección Plano inclinado con rozamiento
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De Wikipedia, la enciclopedia libre
Rampa para silla de ruedas, Hotel Montescot, Chartres, Francia
El plano inclinado (también conocido como rampa o pendiente) es una máquina simple que consiste en una superficie plana que forma un ángulo agudo con el suelo y se utiliza para elevar cuerpos a cierta altura. Tiene la ventaja de que se requiere una fuerza menor a la empleada para levantar dicho cuerpo verticalmente (gracias a la descomposición de fuerzas), aunque se deba aumentar la distancia recorrida y vencer la fuerza de rozamiento.
Visión general
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Plano inclinado usado para la educación, Museo Galileo, Florencia
Plano inclinado y las fuerzas que actúan sobre el sólido
Un plano inclinado es una superficie de apoyo plana inclinada en ángulo, con un extremo más alto que el otro, que se utiliza como ayuda para subir o bajar una carga. El plano inclinado es una de las seis máquinas simples clásicas definidas por los científicos del Renacimiento. Se utilizan mucho para mover cargas pesadas sobre obstáculos verticales; los ejemplos varían desde una rampa utilizada para cargar mercancías en un camión, hasta una persona que camina por una rampa para peatones, hasta un automóvil o un tren que sube una pendiente.
Mover un objeto por un plano inclinado requiere menos fuerza que levantarlo hacia arriba, a costa de un aumento en la distancia recorrida. Su ventaja mecánica, el factor por el que se reduce la fuerza, es igual a la relación entre la longitud de la superficie inclinada y la altura que salva. Debido al principio de conservación de la energía, se requiere la misma cantidad de energía mecánica (trabajo) para levantar un objeto dado una distancia vertical dada, sin tener en cuenta las pérdidas por fricción, pero el plano inclinado permite realizar el mismo trabajo con una fuerza menor ejercida sobre una distancia mayor.
El ángulo de fricción,, también llamado a veces ángulo de reposo, es el ángulo máximo en el que una carga puede permanecer inmóvil en un plano inclinado debido a la fricción, sin deslizarse hacia abajo. Este ángulo es igual al arcotengente del coeficiente de fricciónμ s entre las superficies.
A menudo se considera que otras dos máquinas simples se derivan del plano inclinado. La cuña se puede considerar un plano inclinado en movimiento o dos planos inclinados conectados en la base. El tornillo consiste en un plano inclinado angosto envuelto alrededor de un cilindro.
El término también puede referirse a una aplicación específica, como en el caso de una rampa recta excavada en una ladera empinada para permitir el transporte de mercancías hacia arriba y hacia abajo. Puede incluir vagones sobre raíles o tirados por un sistema de cables; un funicular o ferrocarril por cable, como el Plano Inclinado de Johnstown.
Las leyes que rigen el comportamiento de los cuerpos en un plano inclinado fueron enunciadas por primera vez por el matemático Simon Stevin, en la segunda mitad del siglo XVI.
Para analizar las fuerzas existentes sobre un cuerpo situado sobre un plano inclinado, hay que tener en cuenta la existencia de varios orígenes en ellas.
En primer lugar se debe considerar la existencia de una fuerza de gravedad, también conocida como peso, que es consecuencia de la masa (M) que posee el cuerpo apoyado en el plano inclinado y tiene una magnitud de M.g con una dirección vertical y representada en la figura por la letra G.
Existe además una fuerza normal (N), también conocida como la fuerza de reacción ejercida sobre el cuerpo por el plano como consecuencia de la tercera ley de Newton, se encuentra en una dirección perpendicular al plano y tiene una magnitud igual a la fuerza ejercida por el plano sobre el cuerpo. En la figura aparece representada por N y tiene la misma magnitud que F 2= M.g.cos α y sentido opuesto a la misma.
Existe finalmente una fuerza de rozamiento, también conocida como fuerza de fricción (F R), que siempre se opone al sentido del movimiento del cuerpo respecto a la superficie, y cuya magnitud depende tanto del peso como de las características superficiales del plano inclinado y la superficie en contacto del cuerpo que proporcionan un coeficiente de rozamiento. Esta fuerza debe tener un valor igual a F 1=M.g.sen α, para que el cuerpo se mantenga en equilibrio. En el caso en que F 1 fuese mayor que la fuerza de rozamiento el cuerpo se deslizaría hacia abajo por el plano inclinado. Por tanto para subir el cuerpo se debe realizar una fuerza con una magnitud que iguale o supere la suma de F 1 + F R.
Historia
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Demostración de Stevin
En 1586, el ingeniero flamenco Simon Stevin (Stevinus) dedujo la ventaja mecánica del plano inclinado mediante un argumento que utilizaba un collar de cuentas. Imaginó dos planos inclinados de igual altura pero diferentes pendientes, colocados espalda con espalda (arriba) como en un prisma. Un collar con cuentas situadas a intervalos iguales se coloca sobre los planos inclinados, con una parte colgando por debajo. Las cuentas que descansan sobre los planos actúan como cargas sobre los planos, sostenidas por la fuerza de tensión que experimenta la cuerda en el punto "T". El argumento de Stevin es el siguiente:
La cadena debe estar quieta, en equilibrio mecánico. Si fuera más pesada de un lado que del otro, y comenzara a deslizarse hacia la derecha o hacia la izquierda por su propio peso, cuando cada cuenta se hubiera movido a la posición de la cuenta anterior, la cuerda sería indistinguible de su posición inicial y, por lo tanto, continuaría estando desequilibrada y deslizante. Este argumento podría repetirse indefinidamente, dando como resultado una situación de móvil perpetuo circular, lo que es absurdo. Por lo tanto, permanece estacionaria, con las fuerzas en los dos lados del punto T (arriba) iguales.
La parte de la cadena que cuelga debajo de los planos inclinados es simétrica, con igual número de cuentas a cada lado. Ejerce una fuerza igual en cada lado de la cuerda. Por lo tanto, esta porción de la cuerda se puede cortar en los bordes de los planos (puntos S y V), dejando solo las cuentas descansando en los planos inclinados, y esta porción restante aún estará en equilibrio estático.
Dado que las cuentas están situadas a intervalos iguales en el collar, el número total de cuentas soportadas por cada plano, la carga total, es proporcional a la longitud del plano. Dado que la fuerza de soporte de entrada, la tensión en el hilo del collar, es la misma a ambos lados, la ventaja mecánica de cada plano es proporcional a su longitud inclinada.
Como señaló Dijksterhuis, el argumento de Stevin no es completamente estricto. Las fuerzas ejercidas por la parte colgante de la cadena no necesitan ser simétricas porque la parte colgante “no necesita conservar su forma” cuando se suelta. Incluso si la cadena se suelta con un momento angular cero, el movimiento, incluidas las oscilaciones, es posible a menos que la cadena esté inicialmente en su configuración de equilibrio, una suposición que haría que el argumento fuera circular.
Se han utilizado planos inclinados desde tiempos prehistóricos para mover objetos pesados. Los caminos inclinados y los pedraplenes construidos por civilizaciones antiguas como los romanos son ejemplos de los primeros planos inclinados que han sobrevivido y muestran que entendieron el valor de este dispositivo para mover cargas cuesta arriba. Se cree que las piedras pesadas utilizadas en estructuras de piedra antiguas como Stonehenge se movieron y colocaron en su lugar usando planos inclinados hechos de tierra, aunque es difícil encontrar evidencia de tales rampas de construcción temporales. Las pirámides de Egipto se construyeron utilizando planos inclinados, y rampas de asedio permitieron a los ejércitos antiguos superar las murallas de las fortalezas. Los antiguos griegos construyeron una rampa pavimentada de 6 km (3,7 millas) de largo, el Diolkos, para arrastrar barcos por tierra a través del istmo de Corinto.
Sin embargo, el plano inclinado fue la última de las seis máquinas simples clásicas en ser reconocida como tal. Esto probablemente se deba a que es un dispositivo pasivo e inmóvil (la carga es la parte móvil), y también a que se encuentra en la naturaleza en forma de pendientes y colinas. Aunque entendieron su uso para levantar objetos pesados, los filósofos de la antigua Grecia que definieron las otras cinco máquinas simples, no incluyeron el plano inclinado. Este punto de vista persistió entre algunos científicos posteriores; y en una fecha tan tardía como 1826 Karl von Langsdorf escribió que un plano inclinado "... no es más una máquina que la pendiente de una montaña". El problema de calcular la fuerza requerida para empujar un peso hacia arriba en un plano inclinado (su ventaja mecánica) fue analizado por los filósofos griegos Herón de Alejandría (c. 10 - 60 EC) y Papo de Alejandría (c. 290 - 350 EC), pero ambos se equivocaron en su resolución.
No fue hasta el Renacimiento cuando el plano inclinado se resolvió matemáticamente y se clasificó con las otras máquinas simples. El primer análisis correcto del plano inclinado apareció en la obra del enigmático autor del siglo XIIIJordanus Nemorarius, aunque su solución del problema aparentemente no fue comunicada a otros filósofos de la época. Gerolamo Cardano (1570) propuso la solución incorrecta de que la fuerza a aplicar es proporcional al ángulo del plano. Posteriormente, a finales del siglo XVI, Michael Varro (1584), Simon Stevin (1586) y Galileo Galilei (1592) publicaron tres soluciones correctas a lo largo de diez años. Aunque no fue la primera, la deducción del ingeniero flamenco Simon Stevin es la más conocida, por su originalidad y el uso de un collar de cuentas (véase el recuadro). En 1600, el científico italiano Galileo incluyó el plano inclinado en su análisis de máquinas simples en Le Meccaniche ("Sobre la mecánica"), mostrando su similitud subyacente con las otras máquinas como un amplificador de fuerza.
Las primeras reglas elementales para deslizar objetos con fricción sobre un plano inclinado fueron descubiertas por Leonardo da Vinci (1452-1519), pero quedaron inéditas en sus cuadernos. Fueron redescubiertas por Guillaume Amontons (1699) y Charles-Augustin de Coulomb (1785) las desarrolló aún más. Leonhard Euler (1750) demostró que la tangente del ángulo de rozamiento interno en un plano inclinado es proporcional a la fricción.
Aplicaciones
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Los planos inclinados se utilizan mucho en forma de "rampas" para cargar y descargar mercancías en camiones, barcos y aviones. Las rampas para sillas de ruedas se utilizan para permitir que las personas con movilidad reducida superen obstáculos verticales sin exceder su fuerza. Las escalera mecánicas y las cintas transportadoras inclinadas también son formas de plano inclinado. En un funicular o ferrocarril por cable, un vagón de ferrocarril se sube por un plano inclinado con cables. Los planos inclinados también permiten que tanto personas como objetos pesados y frágiles salven de forma segura un desnivel vertical utilizando la fuerza normal del plano para distribuir el efecto de la gravedad. Las rampas de evacuación de las aeronaves permiten que las personas alcancen el suelo de forma rápida y segura desde la altura del compartimento de pasajeros de un avión comercial.
Uso de rampas para cargar un automóvil en un camión
Cargando un camión en un barco usando una rampa
Rampa de evacuación de emergencia de una aeronave
Rampa para silla de ruedas de un autobús japonés
Rampa de carga en un camión
Otros planos inclinados se construyen en estructuras permanentes. Las carreteras para vehículos y los ferrocarriles tienen planos inclinados en forma de pendientes graduales, rampas y pedraplenes para permitir que los vehículos superen obstáculos verticales como colinas sin perder tracción en la superficie de la carretera. De manera similar, los senderos para peatones y las aceras disponen de rampas suaves para limitar su pendiente, con el fin de garantizar que los peatones puedan mantener las condiciones de tracción necesarias. Los planos inclinados también se utilizan como entretenimiento para que las personas se deslicen hacia abajo de forma controlada, en toboganes, rampas acuáticas, esquí alpino y pistas de patinaje.
Rampa de tierra (derecha) construida por los romanos en el 72 d.C. para invadir Masada, Israel
Rampa peatonal, Palacio do Planalto, Brasilia
Plano Inclinado de Johnstown, un ferrocarril funicular
Carretera de Burma, Assam, India, desde Burma hacia China (1945)
Planos inclinados en una pista de "skate"
Terminología
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Pendiente
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El efecto de utilizar un plano inclinado depende de su pendiente, es decir, de su gradiente o inclinación. Cuanto menor sea la pendiente, mayor será la ventaja mecánica y menor la fuerza necesaria para levantar un peso dado. La pendiente s de un plano es igual a la diferencia de altura entre sus dos extremos, o "desnivel", dividida por su longitud horizontal, o "carrera". También se puede expresar por el ángulo que forma el plano con la horizontal, θ.
La geometría del plano inclinado se basa en un triángulo rectángulo. La longitud horizontal (H) a veces se llama carrera, y el desplazamiento en altura medido en vertical (V) se llama desnivel
θ=arctan(Desnivel Carrera)=arctan(V H){\displaystyle \theta =\arctan {\bigg (}{\frac {\text{Desnivel}}{\text{Carrera}}}{\bigg )}=\arctan {\bigg (}{\frac {\text{V}}{\text{H}}}{\bigg )}\,}
Ventaja mecánica
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La ventaja mecánicaVM de una máquina simple se define como la relación entre la fuerza resultante sobre la carga y la fuerza aplicada. Para el plano inclinado, la fuerza a vencer es solo la fuerza gravitatoria sobre el objeto cargado en el plano, su peso F p. La fuerza aplicada es F i ejercida sobre el objeto, paralelamente al plano, para moverlo hacia arriba remontando la rampa. Las ventajas mecánicas son...
V M=F p F ap.{\displaystyle \mathrm {VM} ={\frac {F_{p}}{F_{\text{ap}}}}.\,}
La ventaja mecánica VM de un plano inclinado ideal sin fricción se denomina a veces "ventaja mecánica ideal" (VMI), mientras que cuando se incluye la fricción se denomina "ventaja mecánica real" (VMR).
Plano inclinado sin fricción
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Plano inclinado instrumentado utilizado para la enseñanza de la física, alrededor de 1900. El peso de la izquierda proporciona la fuerza a vencer F p. El peso de la derecha proporciona la fuerza aplicada F ap tirando del rodillo hacia arriba del plano
Si no hay fricción entre el objeto que se mueve y el plano, el dispositivo se denomina "plano inclinado ideal". Se puede abordar esta condición si el objeto está rodando, como un barril, o apoyado sobre ruedas. Debido al principio de conservación de la energía, para un plano inclinado sin fricción, el trabajo realizado sobre la carga que se levanta, W res, es igual a la fuerza aplicada por la distancia recorrida, W ap
W r e s=W a p{\displaystyle W_{\rm {res}}=W_{\rm {ap}}\,}
El trabajo se define como la fuerza multiplicada por el desplazamiento que mueve un objeto. El trabajo realizado sobre la carga es igual a su peso multiplicado por el desplazamiento vertical que sube, que es la "elevación" del plano inclinado.
W r e s=F p⋅V{\displaystyle W_{\rm {res}}=F_{p}\cdot {\text{V}}\,}
El trabajo aplicado es igual a la fuerza F ap ejercida sobre el objeto multiplicada por la longitud de la diagonal del plano inclinado.
W a p=F a p⋅Long{\displaystyle W_{\rm {ap}}=F_{\rm {ap}}\cdot {\text{Long}}\,}
Sustituyendo estos valores en la ecuación de conservación de energía anterior y reorganizando los sumandos
VM=F p F a p=Long V{\displaystyle {\text{VM}}={\frac {F_{p}}{F_{\rm {ap}}}}={\frac {\text{Long}}{\text{V}}}\,}
Para expresar la ventaja mecánica en función del ángulo θ del plano, se puede ver en el diagrama de arriba que
sinθ=V Long{\displaystyle \sin \theta ={\frac {\text{V}}{\text{Long}}}\,}
Entonces
VM=F p F a p=1 sinθ{\displaystyle {\text{VM}}={\frac {F_{p}}{F_{\rm {ap}}}}={\frac {1}{\sin \theta }}\,}
En consecuencia, la ventaja mecánica de un plano inclinado sin fricción es igual al recíproco del seno del ángulo de inclinación. La fuerza aplicada F ap de esta ecuación es la fuerza necesaria para mantener la carga inmóvil en el plano inclinado o empujarla hacia arriba a una velocidad constante. Si la fuerza de entrada es mayor que el valor anterior, la carga acelerará hacia arriba en el plano; si la fuerza es menor, acelerará hacia abajo.
Plano inclinado con rozamiento
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Donde hay fricción entre el plano y la carga, como por ejemplo con una caja pesada que se desliza por una rampa, parte del trabajo aplicado por la fuerza de entrada se disipa como calor por fricción, W fric, por lo que se realiza menos trabajo en la carga. Debido a conservación de la energía, la suma del trabajo de salida y las pérdidas de energía por fricción es igual al trabajo de entrada
W ap=W fric+W res{\displaystyle W_{\text{ap}}=W_{\text{fric}}+W_{\text{res}}\,}
Por lo tanto, se requiere más fuerza de entrada y la ventaja mecánica es menor que si no hubiera fricción. Con la fricción, la carga solo se moverá si la fuerza neta paralela a la superficie es mayor que la fuerza de fricción F f que se opone a ella. La fuerza de fricción máxima viene dada por
F f=μ F n{\displaystyle F_{f}=\mu F_{n}\,}
donde F n es la fuerza normal entre la carga y el plano, en dirección normal a la superficie, y μ es la fricción entre las dos superficies, que varía con el material. Cuando no se aplica ninguna fuerza de entrada, si el ángulo de inclinación θ del plano es menor que un valor máximo φ, la componente de la fuerza gravitatoria paralela al plano será demasiado pequeña para superar la fricción y la carga permanecerá inmóvil. Este ángulo se denomina ángulo de rozamiento interno y depende de la composición de las superficies, pero es independiente del peso de la carga. A continuación se muestra que la tangente del ángulo de reposo φ es igual a μ
ϕ=arctanμ{\displaystyle \phi =\arctan \mu \,}
Con fricción, siempre hay algún rango de fuerza aplicada F ap para el cual la carga es estacionaria, (es decir, ni se desliza hacia arriba ni hacia abajo), mientras que con un plano inclinado sin fricción solo hay un valor particular de fuerza de entrada para la cual la carga está estacionaria.
Análisis
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Clave: F n = N = Fuerza normal que es perpendicular al plano, F ap = f = fuerza aplicada, F p = mg = peso del objeto, donde m = masa, g = gravedad
Una carga que descansa sobre un plano inclinado, cuando se considera como un cuerpo libre tiene tres fuerzas actuando sobre ella:
La fuerza aplicada, F ap ejercida sobre la carga para moverla, que actúa paralela al plano inclinado.
El peso de la carga, F p, que actúa verticalmente hacia abajo
La fuerza del plano sobre la carga, que se puede descomponer en dos componentes:
La fuerza normal F n del plano inclinado sobre la carga que la soporta. Esta se dirige perpendicular (normal) a la superficie.
La fuerza de fricción, F f del plano sobre la carga actúa paralela a la superficie, y siempre tiene una dirección opuesta al movimiento del objeto. Es igual a la fuerza normal multiplicada por el coeficiente de fricción μ entre las dos superficies.
Usando las leyes de Newton, la carga estará estacionaria o en movimiento constante si la suma de las fuerzas sobre ella es cero. Dado que la dirección de la fuerza de fricción es opuesta para el caso de movimiento cuesta arriba y cuesta abajo, estos dos casos deben considerarse por separado:
Movimiento cuesta arriba: La fuerza total sobre la carga es hacia el lado cuesta arriba, por lo que la fuerza de fricción se dirige hacia abajo en el plano, oponiéndose a la fuerza aplicada.
Demostración de la ventaja mecánica para el movimiento cuesta arriba
Las ecuaciones de equilibrio para fuerzas paralelas y perpendiculares al plano son
∑F‖=F a p−F f−F p sinθ=0{\displaystyle \sum F_{\|}=F_{\rm {ap}}-F_{f}-F_{p}\sin \theta =0\,}∑F⊥=F n−F p cosθ=0{\displaystyle \sum F_{\perp }=F_{n}-F_{p}\cos \theta =0\,}Sustituyendo F f=μ F n{\displaystyle F_{f}=\mu F_{n}\,} en la primera ecuación F a p−μ F n−F p sinθ=0{\displaystyle F_{\rm {ap}}-\mu F_{n}-F_{p}\sin \theta =0\,}Resolviendo la segunda ecuación para obtener F n=F p cosθ{\displaystyle F_{n}=F_{p}\cos \theta \,} y sustituyendo en la ecuación anterior F i−μ F p cosθ−F p sinθ=0{\displaystyle F_{i}-\mu F_{p}\cos \theta -F_{p}\sin \theta =0\,}F p F a p=1 sinθ+μ cosθ{\displaystyle {\frac {F_{p}}{F_{\rm {ap}}}}={\frac {1}{\sin \theta +\mu \cos \theta }}\,}Definiendo μ=tanϕ{\displaystyle \mu =\tan \phi \,}F p F a p=1 sinθ+tanϕ cosθ=1 sinθ+sinϕ cosϕ cosθ=cosϕ sinθ cosϕ+cosθ sinϕ{\displaystyle {\frac {F_{p}}{F_{\rm {ap}}}}={\frac {1}{\sin \theta +\tan \phi \cos \theta }}={\dfrac {1}{\sin \theta +{\dfrac {\sin \phi }{\cos \phi }}\cos \theta }}={\frac {\cos \phi }{\sin \theta \cos \phi +\cos \theta \sin \phi }}\,}Usando las identidades trigonométricas de la suma de ángulos en el denominador,
La ventaja mecánica es V M=F p F a p=cosϕ sin(θ+ϕ){\displaystyle \mathrm {VM} ={\frac {F_{p}}{F_{\rm {ap}}}}={\frac {\cos \phi }{\sin(\theta +\phi )}}\,}donde ϕ=arctanμ{\displaystyle \phi =\arctan \mu \,}. Esta es la condición para el "movimiento inminente" hacia arriba del plano inclinado. Si la fuerza aplicada F ap es mayor que la dada por esta ecuación, la carga se moverá hacia arriba en el plano.
Movimiento cuesta abajo: La fuerza total sobre la carga es hacia el lado cuesta abajo, por lo que la fuerza de fricción se dirige hacia arriba del plano.
Demostración de la ventaja mecánica para el movimiento cuesta abajo
Las ecuaciones de equilibrio son
∑F‖=F a p+F f−F p sinθ=0{\displaystyle \sum F_{\|}=F_{\rm {ap}}+F_{f}-F_{p}\sin \theta =0\,}∑F⊥=F n−F p cosθ=0{\displaystyle \sum F_{\perp }=F_{n}-F_{p}\cos \theta =0\,}Sustituyendo F f=μ F n{\displaystyle F_{f}=\mu F_{n}\,} en la primera ecuación F a p+μ F n−F p sinθ=0{\displaystyle F_{\rm {ap}}+\mu F_{n}-F_{p}\sin \theta =0\,}Resolviendo la segunda ecuación para obtener F n=F p cosθ{\displaystyle F_{n}=F_{p}\cos \theta \,} y sustituyendo en la ecuación anterior F a p+μ F p cosθ−F p sinθ=0{\displaystyle F_{\rm {ap}}+\mu F_{p}\cos \theta -F_{p}\sin \theta =0\,}F p F a p=1 sinθ−μ cosθ{\displaystyle {\frac {F_{p}}{F_{\rm {ap}}}}={\frac {1}{\sin \theta -\mu \cos \theta }}\,}Sustituyendo en μ=tanϕ{\displaystyle \mu =\tan \phi \,} y simplificando como arriba F p F a p=cosϕ sinθ cosϕ−cosθ sinϕ{\displaystyle {\frac {F_{p}}{F_{\rm {ap}}}}={\frac {\cos \phi }{\sin \theta \cos \phi -\cos \theta \sin \phi }}\,}Usando otro lista de identidades trigonométricas en el denominador,
La ventaja mecánica es V M=F p F a p=cosϕ sin(θ−ϕ){\displaystyle \mathrm {VM} ={\frac {F_{p}}{F_{\rm {ap}}}}={\frac {\cos \phi }{\sin(\theta -\phi )}}\,}Esta es la condición para el movimiento inminente hacia abajo del plano; si la fuerza aplicada F i es menor que la dada en esta ecuación, la carga se deslizará por el plano. Hay tres casos:
1. θ<ϕ{\displaystyle \theta <\phi \,}: La ventaja mecánica es negativa. En ausencia de fuerza aplicada, la carga permanecerá inmóvil y requiere alguna fuerza aplicada negativa (cuesta abajo) para deslizarse hacia abajo.
2. θ=ϕ{\displaystyle \theta =\phi \,}: El águlo de reposo. La ventaja mecánica es infinita. Sin fuerza aplicada, la carga no se deslizará, pero la más mínima fuerza negativa (cuesta abajo) hará que se deslice.
3. θ>ϕ{\displaystyle \theta >\phi \,}: La ventaja mecánica es positiva. En ausencia de fuerza aplicada, la carga se deslizará hacia abajo del plano y requiere una fuerza positiva (cuesta arriba) para mantenerla inmóvil.
Ventaja mecánica utilizando potencia
[editar]
Clave: N = Fuerza normal que es perpendicular al plano, p=mg, donde m=masa, g=gravedad, y θ(theta)=Ángulo de inclinación del plano
La ventaja mecánica de un plano inclinado es la relación entre el peso de la carga sobre la rampa y la fuerza requerida para hacerla subir. Si la energía no se disipa ni se almacena en el movimiento de la carga, esta ventaja mecánica puede calcularse a partir de las dimensiones de la rampa.
Para demostrar esto, sea r la posición de un vagón en una rampa con un ángulo θ por encima de la horizontal, dada por
r=R(cosθ,sinθ),{\displaystyle \mathbf {r} =R(\cos \theta ,\sin \theta ),}
donde R es la distancia en la rampa. La velocidad del vagón en la rampa es ahora
v=V(cosθ,sinθ).{\displaystyle \mathbf {v} =V(\cos \theta ,\sin \theta ).}
Debido a que no hay pérdidas, la potencia utilizada por la fuerza "F" para subir la carga por la rampa es igual a la potencia resultante, que es la elevación vertical del peso "p" de la carga.
La potencia de entrada que hace subir el vagón por la rampa está dada por
P a p=F V,{\displaystyle P_{\mathrm {ap} }=FV,!}
y la potencia resultante es
P r e s=p⋅v=(0,p)⋅V(cosθ,sinθ)=p V sinθ.{\displaystyle P_{\mathrm {res} }=\mathbf {p} \cdot \mathbf {v} =(0,p)\cdot V(\cos \theta ,\sin \theta )=pV\sin \theta .}
Igualando la potencia aplicada y la potencia resultante, se obtiene la ventaja mecánica como
V M=p F=1 sinθ.{\displaystyle \mathrm {VM} ={\frac {p}{F}}={\frac {1}{\sin \theta }}.}
La ventaja mecánica de una rampa inclinada también se puede calcular a partir de la relación entre la longitud de la rampa L y su altura vertical V, porque el seno del ángulo de la rampa está dado por
sinθ=V L,{\displaystyle \sin \theta ={\frac {V}{L}},}
por lo tanto,
V M=p F=L V.{\displaystyle \mathrm {VM} ={\frac {p}{F}}={\frac {L}{V}}.}
Disposición del sistema de accionamiento por cable del plano inclinado de Liverpool Minard
Ejemplo: si la altura de una rampa es V = 1 metro y su longitud es L = 5 metros, entonces la ventaja mecánica es
V M=p F=5,{\displaystyle \mathrm {VM} ={\frac {p}{F}}=5,}
lo que significa que una fuerza de 20 kg levantará una carga de 100 kg.
El plano inclinado Liverpool Minard tiene unas dimensiones de 1804 metros por 37,50 metros de desnivel, lo que proporciona una ventaja mecánica de
V M=p F=1804/37.50=48.1,{\displaystyle \mathrm {VM} ={\frac {p}{F}}=1804/37.50=48.1,}
por lo que una fuerza de tensión de 100 kp en el cable levantará una carga de 4810 kp. El grado de esta pendiente es del 2%, lo que significa que el ángulo θ es lo suficientemente pequeño como para que sen θ=tan θ.
Referencias
[editar]
↑Tipler, Paul Allen (1991). Física preuniversitaria. Reverte. ISBN9788429143751. Consultado el 20 de febrero de 2018.
↑ abCole, Matthew (2005). Explore science, 2nd Ed.. Pearson Education. p.178. ISBN978-981-06-2002-8.
↑Merriam-Webster's collegiate dictionary, 11th Ed.. Merriam-Webster. 2003. pp.629. ISBN978-0-87779-809-5. «inclined plane definition dictionary.»
↑ abcd«The Inclined Plane». Math and science activity center. Edinformatics. 1999. Consultado el 11 de marzo de 2012.
↑ abcSilverman, Buffy (2009). Simple Machines: Forces in Action, 4th Ed.. USA: Heinemann-Raintree Classroom. p.7. ISBN978-1-4329-2317-4.
↑ abcOrtleb, Edward P.; Richard Cadice (1993). Machines and Work. Lorenz Educational Press. pp.iv. ISBN978-1-55863-060-4.
↑ abReilly, Travis (24 de noviembre de 2011). «Lesson 04:Slide Right on By Using an Inclined Plane». Teach Engineering. College of Engineering, Univ. of Colorado at Boulder. Archivado desde el original el 8 de mayo de 2012. Consultado el 8 de septiembre de 2012.
↑Scott, John S. (1993). Dictionary of Civil Engineering. Chapman & Hill. p.14. ISBN978-0-412-98421-1. «angle of friction [mech.] in the study of bodies sliding on plane surfaces, the angle between the perpendicular to the surface and the resultant force (between the body and the surface) when the body begins to slide. angle of repose [s.m.] for any given granular material the steepest angle to the horizontal at which a heaped surface will stand in stated conditions.»
↑ abcdAmbekar, A. G. (2007). Mechanism and Machine Theory. PHI Learning. p.446. ISBN978-81-203-3134-1. «Angle of repose is the limiting angle of inclination of a plane when a body, placed on the inclined plane, just starts sliding down the plane.»
↑Rosen, Joe; Lisa Quinn Gothard (2009). Encyclopedia of Physical Science, Volume 1. Infobase Publishing. p.375. ISBN978-0-8160-7011-4.
↑Ignacio, Ramírez Vargas; Manuel, Palacios Pineda, Luis; E, Rodríguez C. , Mario. Estática para ingeniería. Grupo Editorial Patria. ISBN9786077442691. Consultado el 20 de febrero de 2018.
↑ abFisica Volumen i. Pearson Educación. 2006. ISBN9789702607762. Consultado el 20 de febrero de 2018.
↑Tipler, Paul Allen; Mosca, Gene (2005). Física para la ciencia y la tecnología. Reverte. ISBN9788429144116. Consultado el 20 de febrero de 2018.
↑ abcKoetsier, Teun (2010). «Simon Stevin and the rise of Archimedean mechanics in the Renaissance». The Genius of Archimedes – 23 Centuries of Influence on Mathematics, Science and Engineering: Proceedings of an International Conference Held at Syracuse, Italy, June 8–10, 2010. Springer. pp.94-99. ISBN978-90-481-9090-4.
↑Devreese, Jozef T.; Guido Vanden Berghe (2008). 'Magic is no magic': The wonderful world of Simon Stevin. WIT Press. pp.136-139. ISBN978-1-84564-391-1.
↑ abFeynman, Richard P.; Robert B. Leighton; Matthew Sands (1963). The Feynman Lectures on Physics, Vol. I. USA: California Inst. of Technology. pp.4.4-4.5. ISBN978-0-465-02493-3.
↑E.J.Dijksterhuis: Simon Stevin 1943
↑Therese McGuire, Light on Sacred Stones, in Conn, Marie A.; Therese Benedict McGuire (2007). Not etched in stone: essays on ritual memory, soul, and society. University Press of America. p.23. ISBN978-0-7618-3702-2.
↑Dutch, Steven (1999). «Pre-Greek Accomplishments». Legacy of the Ancient World. Prof. Steve Dutch's page, Univ. of Wisconsin at Green Bay. Archivado desde el original el 21 de agosto de 2016. Consultado el 13 de marzo de 2012.
↑Moffett, Marian; Michael W. Fazio; Lawrence Wodehouse (2003). A world history of architecture. Laurence King Publishing. p.9. ISBN978-1-85669-371-4.
↑Peet, T. Eric (2006). Rough Stone Monuments and Their Builders. Echo Library. pp.11-12. ISBN978-1-4068-2203-8.
↑Thomas, Burke (2005). «Transport and the Inclined Plane». Construction of the Giza Pyramids. world-mysteries.com. Archivado desde el original el 13 de marzo de 2012. Consultado el 10 de marzo de 2012.
↑Isler, Martin (2001). Sticks, stones, and shadows: building the Egyptian pyramids. USA: University of Oklahoma Press. pp.211–216. ISBN978-0-8061-3342-3.
↑Sprague de Camp, L. (1990). The Ancient Engineers. USA: Barnes & Noble. p.43. ISBN978-0-88029-456-0.
↑ abKarl von Langsdorf (1826) Machinenkunde, quoted in Reuleaux, Franz (1876). The kinematics of machinery: Outlines of a theory of machines. MacMillan. pp.604.
↑for example, the lists of simple machines left by Roman architect Vitruvius (c. 80 – 15 BCE) and Greek philosopher Herón de Alejandría (c. 10 – 70 CE) consist of the five classical simple machines, excluding the inclined plane. – Smith, William (1848). Dictionary of Greek and Roman antiquities. London: Walton and Maberly; John Murray. p.722., Usher, Abbott Payson (1988). A History of Mechanical Inventions. USA: Courier Dover Publications. pp.98, 120. ISBN978-0-486-25593-4.
↑Heath, Thomas Little (1921). A History of Greek Mathematics, Vol. 2. UK: The Clarendon Press. pp.349, 433–434.
↑ abcEgidio Festa and Sophie Roux, The enigma of the inclined plane in Laird, Walter Roy; Sophie Roux (2008). Mechanics and natural philosophy before the scientific revolution. USA: Springer. pp.195-221. ISBN978-1-4020-5966-7.
↑ abMeli, Domenico Bertoloni (2006). Thinking With Objects: The Transformation of Mechanics in the Seventeenth Century. JHU Press. pp.35-39. ISBN978-0-8018-8426-9.
↑ abBoyer, Carl B.; Uta C. Merzbach (2010). A History of Mathematics, 3rd Ed.. John Wiley and Sons. ISBN978-0-470-63056-3.
↑Usher, Abbott Payson (1988). A History of Mechanical Inventions. Courier Dover Publications. p.106. ISBN978-0-486-25593-4.
↑Machamer, Peter K. (1998). The Cambridge Companion to Galileo. London: Cambridge University Press. pp.47-48. ISBN978-0-521-58841-6.
↑ abArmstrong-Hélouvry, Brian (1991). Control of machines with friction. USA: Springer. p.10. ISBN978-0-7923-9133-3.
↑Meyer, Ernst (2002). Nanoscience: friction and rheology on the nanometer scale. World Scientific. p.7. ISBN978-981-238-062-3.
↑ abHandley, Brett; David M. Marshall; Craig Coon (2011). Principles of Engineering. Cengage Learning. pp.71-73. ISBN978-1-4354-2836-2.
↑Dennis, Johnnie T. (2003). The Complete Idiot's Guide to Physics. Penguin. pp.116-117. ISBN978-1-59257-081-2.
↑Nave, Carl R. (2010). «The Incline». Hyperphysics. Dept. of Physics and Astronomy, Georgia State Univ. Consultado el 8 de septiembre de 2012.
↑ abMartin, Lori (2010). «Lab Mech14:The Inclined Plane - A Simple Machine». Science in Motion. Westminster College. Consultado el 8 de septiembre de 2012.
↑Pearson (2009). Physics class 10 - The IIT Foundation Series. New Delhi: Pearson Education India. p.69. ISBN978-81-317-2843-7.
↑ abBansal, R.K (2005). Engineering Mechanics and Strength of Materials. Laxmi Publications. pp.165-167. ISBN978-81-7008-094-7.
↑ abThis derives slightly more general equations which cover force applied at any angle: Gujral, I.S. (2008). Engineering Mechanics. Firewall Media. pp.275-277. ISBN978-81-318-0295-3.
Enlaces externos
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Wikimedia Commons alberga una categoría multimedia sobre planos inclinados.
| Control de autoridades | Proyectos Wikimedia Datos:Q161462 Multimedia:Inclined planes / Q161462 Identificadores BNF:133360841(data) GND:4399805-7 LCCN:sh85064751 NLI:987007541075505171 NARA:10664117 Diccionarios y enciclopedias Britannica:url Treccani:url |
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16112 | https://www.studocu.com/en-us/messages/question/5558244/what-is-the-absorbance-a-of-a-solution-whose-transmittance-t-is-20-a-070-b-020-c | [Solved] What is the absorbance A of a solution whose transmittance T is - Intro To Chem Analysis I -Sl (CHEM 207) - Studocu
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What is the absorbance (A) of a solution whose % transmittance (%T) is 20%?A) -0.70B) 0.20C) 0.70D) 2.0E) none of the aboveIntro To Chem Analysis I -Sl (CHEM 207)
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Intro To Chem Analysis I -Sl
Question
What is the absorbance A of a solution whose transmittance T is
University of Louisville
Intro To Chem Analysis I -Sl
Question
### Charlie
1 year ago
What is the absorbance (A) of a solution whose % transmittance (%T) is 20%?
A) -0.70
B) 0.20
C) 0.70
D) 2.0
E) none of the above
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Answer Created with AI
1 year ago
The relationship between absorbance (A) and percent transmittance (%T) is given by the formula:
js
A = -log10(%T/100)
In this formula, %T is the percent transmittance, and A is the absorbance. The negative logarithm to the base 10 is used to convert the transmittance, which is a percentage, into absorbance, which is a unitless quantity.
Given that the %T is 20%, we can substitute this into the formula to find the absorbance:
js
A = -log10(20/100)
This simplifies to:
js
A = -log10(0.2)
Using a calculator, we find that the absorbance A is approximately 0.70. Therefore, the correct answer is C) 0.70.
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How to Easily Calculate a 20 Percent Discount
Learning to calculate a 20 percent discount is like having a superpower in your wallet. Whether you’re shopping for moving boxes in Chicago or browsing furniture stores in Melrose Park, this skill helps you spot real deals and avoid those sneaky “fake sales” that inflate original prices.
The best part? The math is surprisingly simple once you know the trick.
The Simple Formula and Steps
Here’s the secret formula that works every single time:
Discount Amount = Original Price × 0.20
Final Price = Original Price – Discount Amount
Let me walk you through this with a real example. Say you’re looking at a beautiful dining set in Des Plaines for $295. Here’s how to calculate your 20 percent discount:
Step 1: Convert 20% to a decimal by moving the decimal point two places left. So 20% becomes 0.20.
Step 2: Multiply the original price by 0.20 to find your savings. $295 × 0.20 = $59. That’s how much you’ll save!
Step 3: Subtract the discount from the original price. $295 – $59 = $236. That’s your final price.
Let’s try another example that hits closer to home. Imagine you’re planning a move from Park Ridge and find a service originally priced at $279 with a 20 percent discount. Your calculation would be: $279 × 0.20 = $55.80 in savings, making your final cost $223.20.
Pretty straightforward, right? Once you practice this a few times, you’ll be doing the math in your head while walking through stores. For more money-saving strategies during your move, check out our helpful Moving Tips.
Applying Your Discount: Online vs. In-Store
Now that you can calculate the savings, let’s talk about actually getting that 20 percent discount applied to your purchase. The process varies depending on whether you’re shopping online from your couch in Elmhurst or wandering through stores in downtown Chicago.
Online shopping makes discounts pretty painless. You’ll usually encounter promo codes – those little boxes during checkout where you type something like “SAVE20” or “DISCOUNT20.” The website does the math for you, but it’s still smart to double-check that your 20 percent discount actually shows up before you hit “buy.”
Some online stores offer automatic discounts that kick in when you meet certain requirements, like spending over $100 or buying from a specific category. These are nice because there’s no code to remember or type in.
In-store shopping in the Chicago area brings its own discount opportunities. You might have printed coupons from the Sunday paper or store flyers, or you could stumble upon a store-wide sale where the 20 percent discount is already marked on the tags.
Don’t forget about Chicago shopping events and local deals in communities like Franklin Park and Wood Dale. These neighborhood sales often feature deeper discounts than you’d find at big chain stores, and local businesses are usually happy to explain exactly how their discounts work.
Whatever method you use, always verify the final price matches your calculations before you pay. A quick glance at your receipt can save you from unpleasant surprises later. If you’re shopping for moving-related items in the Chicago area, our Local Movers Chicago page has information about local resources that might help.
Understanding a 20 Percent Discount vs. a Fixed-Amount Discount
Here’s where things get interesting. Not all discounts work the same way, and understanding the difference can save you serious money. A 20 percent discount gives you proportional savings – the more expensive the item, the more dollars you save. A fixed-amount discount (like “$20 off”) gives you the same dollar savings regardless of the original price.
Let me show you what I mean with a simple comparison:
| Original Price | 20% Discount | $20 Fixed Discount | Better Deal |
--- --- |
| $50 | Save $10, pay $40 | Save $20, pay $30 | $20 off wins |
| $100 | Save $20, pay $80 | Save $20, pay $80 | It’s a tie! |
| $200 | Save $40, pay $160 | Save $20, pay $180 | 20% discount wins |
See the pattern? For expensive items, a 20 percent discount usually beats fixed-amount discounts. For cheaper items, that “$20 off” coupon might be your best friend.
This knowledge becomes especially valuable when you’re making big purchases, like appliances for your new home or professional services. Speaking of which, if you want to dive deeper into discount calculations, Learn more about calculating discounts offers additional insights.
Common Mistakes to Avoid When Calculating
Even with the formula down pat, people still make mistakes that cost them money. Let me help you avoid the most common traps.
Forgetting about sales tax is probably the biggest mistake. Your 20 percent discount applies to the item price, but then Illinois adds sales tax on top of the discounted price. So if you buy that $295 dining set for $236 after your discount, you’ll still owe sales tax on the $236. In Chicago, that could add another $24 or so to your final bill.
Getting confused by multiple discounts is another common pitfall. If you see “20% off everything, plus an extra 10% off with this coupon,” don’t assume you’re getting 30% off. The math works like this: take your 20 percent discount first, then apply the 10% to the already-reduced price. You’ll end up with about 28% off total – still great, but not quite 30%.
Falling for fake “original prices” is the sneakiest trap of all. Some stores inflate their “regular prices” to make discounts look more impressive. If a store claims their $200 item is “normally $300,” do a quick price check elsewhere. That 20 percent discount might not be as sweet as it appears.
The key is staying alert and doing your homework. When you’re planning a major expense like a move, accurate calculations become even more important. Our Moving Estimate tool can help you plan your budget with confidence, so you know exactly what you’re paying and what you’re saving.
The Benefits of Smart Savings
There’s something deeply satisfying about mastering the art of the 20 percent discount. It’s not just about the math – though that’s important – it’s about becoming a smarter consumer who can stretch every dollar further. When you understand how discounts really work, you’re equipped to make better financial decisions that can impact everything from your weekly grocery budget to major life changes like relocating your family.
Why a 20 Percent Discount is So Appealing
A 20 percent discount hits that sweet spot where savings feel genuinely meaningful without seeming too good to be true. Unlike smaller discounts that barely move the needle or massive markdowns that make you wonder what’s wrong with the product, 20% off feels just right.
For you as a consumer, this discount level transforms your purchasing power in real ways. That furniture set you’ve been eyeing for your living room in Elmhurst suddenly becomes affordable. The moving supplies you need for your upcoming relocation from Chicago might finally fit within your budget. When you’re dealing with larger purchases, that 20% can translate into hundreds of dollars back in your pocket.
There’s also something psychologically rewarding about securing a good deal. It validates your shopping skills and makes you feel like you’ve beaten the system just a bit. This feeling encourages you to be more thoughtful about your purchases rather than buying impulsively at full price.
From a business perspective, offering a 20 percent discount is equally strategic. Retailers know this discount level attracts hesitant buyers without drastically cutting into their profit margins. It’s particularly effective for moving inventory – whether that’s seasonal items in a Des Plaines store or services that need to be scheduled during specific timeframes.
The discount also builds customer loyalty. When people consistently find good value at a business, they return. They tell their friends. They become advocates rather than just one-time customers. For service-based businesses like moving companies, this word-of-mouth marketing is invaluable in tight-knit communities throughout Illinois.
At Bernard Movers, we’ve seen how the right discount can help families make the leap to their dream move. Sometimes that 20 percent discount is exactly what transforms a “someday” relocation into a “let’s do this” decision. That’s why we’re committed to offering genuine value through our Affordable Moving Companies services.
Putting Your Savings Toward Your Next Big Move
Now that you’ve mastered calculating a 20 percent discount and spotted the common pitfalls, you have a valuable skill that extends far beyond everyday shopping. These smart savings habits can make a real difference when you’re planning something major – like relocating your life to a new state.
Think about it: every discount you secure on furniture, appliances, packing supplies, or household items adds up. That money you save can go directly toward making your move smoother and less stressful. Instead of stretching your budget thin across all the expenses that come with relocating, you can breathe a little easier knowing you’ve been strategic about your spending.
The principles you’ve learned about genuine versus misleading discounts become especially important during a move. You’ll encounter countless offers for moving-related services and supplies. Being able to quickly calculate real savings and spot inflated “original prices” protects you from overspending when you’re already dealing with a major life change.
At Bernard Movers, we believe in transparent pricing and genuine value. We understand that moving is already a significant investment, which is why we’re currently offering a 20 percent additional discount on moves to Georgia and Florida booked for delivery by March 15, 2025. This isn’t a gimmick or an inflated discount – it’s our way of helping families in Chicago, Melrose Park, and throughout Illinois make their relocation dreams more affordable.
Whether you’re moving across town in Park Ridge or starting fresh in sunny Florida, the money-saving mindset you’ve developed will serve you well. Every smart financial decision you make, from calculating discounts to choosing the right moving company, contributes to a smoother transition to your new home.
Ready to put your savings toward a move that’s both affordable and reliable? Find how we can help make your relocation seamless and budget-friendly with our affordable local and long-distance moving services. Your future self will thank you for making such a smart choice.
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World Bank Group/ V.Tremeau Community representatives come to visit a family in the outskirts of Beni to raise awareness about Ebola.
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Ebola virus disease
Overview
Ebola virus disease (EVD), formerly known as Ebola haemorrhagic fever, is a severe, often fatal illness affecting humans and other primates.
The virus is transmitted to people from wild animals (such as fruit bats, porcupines and non-human primates) and then spreads in the human population through direct contact with the blood, secretions, organs or other bodily fluids of infected people, and with surfaces and materials (e.g. bedding, clothing) contaminated with these fluids.
The average EVD case fatality rate is around 50%. Case fatality rates have varied from 25% to 90% in past outbreaks.
The first EVD outbreaks occurred in remote villages in Central Africa, near tropical rainforests. The 2014–2016 outbreak in West Africa was the largest and most complex Ebola outbreak since the virus was first discovered in 1976. There were more cases and deaths in this outbreak than all others combined. It also spread between countries, starting in Guinea then moving across land borders to Sierra Leone and Liberia.
It is thought that fruit bats of the Pteropodidae family are natural Ebola virus hosts.
Symptoms
The incubation period, that is, the time interval from infection with the virus to onset of symptoms, is from 2 to 21 days. A person infected with Ebola cannot spread the disease until they develop symptoms.
Symptoms of EVD can be sudden and include: fever, fatigue, muscle, pain, headache, and sore throat. This is followed by vomiting, diarrhea, rash, symptoms of impaired kidney and liver function, and in some cases internal and external bleeding (e.g. oozing from the gums, blood in the stools). Laboratory findings include low white blood cell and platelet counts and elevated liver enzymes.
It can be difficult to clinically distinguish EVD from other infectious diseases such as malaria, typhoid fever and meningitis. A range of diagnostic tests have been developed to confirm the presence of the virus.
Treatment and prevention
Supportive care - rehydration with oral or intravenous fluids - and treatment of specific symptoms improves survival. A range of potential treatments including blood products, immune therapies and drug therapies are currently being evaluated.
In the 2018-2020 Ebola outbreak in DRC, the first-ever multi-drug randomized control trial was conducted to evaluate the effectiveness and safety of drugs used in the treatment of Ebola patients under an ethical framework developed in consultation with experts in the field and the DRC.
Two monoclonal antibodies (Inmazeb and Ebanga) were approved for the treatment of Zaire ebolavirus (Ebolavirus) infection in adults and children by the US Food and Drug Administration in late 2020.
The Ervebo vaccine has been shown to be effective in protecting people from the species Zaire ebolavirus, and is recommended by the Strategic Advisory Group of Experts on Immunization as part of a broader set of Ebola outbreak response tools. In December 2020, the vaccine was approved by the US Food and Drug Administration and prequalified by WHO for use in individuals 18 years of age and older (except for pregnant and breastfeeding women) for protection against Ebola virus disease caused by Zaïre Ebola virus.
In May 2020, the European Medicines Agency recommended granting marketing authorization for a 2-component vaccine called Zabdeno-and-Mvabea for individuals 1 year and older.
The vaccine is delivered in 2 doses: Zabdeno is administered first and Mvabea is given approximately 8 weeks later as a second dose. This prophylactic 2-dose regimen is therefore not suitable for an outbreak response where immediate protection is necessary.
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16115 | https://principlesofcryptography.com/number-theory-primer-the-diophantine-equation-ax-by-c/ | Skip to the content
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Number Theory Primer : The Diophantine Equation ax + by = c
Post author By premmi
Post date July 15, 2024
authored by Premmi and Beguène
Previous Topic: The Euclidean Algorithm
Introduction
A Diophantine equation is any equation in one or more unknowns that is to be solved in the integers.
The simplest type of Diophantine equation that we shall consider is the linear Diophantine equation in two unknowns:
ax + by = c
where a, b, c are given integers and a, b are not both zero.
A solution of this equation is a pair of integers x_0, y_0 that, when substituted into the equation, satisfy it; that is, we ask that ax_0 + by_0 = c.
Diophantine equations frequently arise when solving certain types of problems, for example, while solving linear congruences, which we will encounter later during our study of modular arithmetic.
Motivation
Let us motivate a discussion on Diophantine equations through a problem posed by Euler in 1770.
“Divide 100 into two summands such that one is divisible by 7 and the other by 11.”
Framing this problem in terms of an equation leads us to the following Diophantine equation,
7x + 11y = 100
where 7x is divisible by 7 and 11y by 11 and consequently, x \text{ and } y are positive integers.
In order to find a solution to this problem we need to know whether this equation is solvable. Also, if the equation has a solution, we would like to how many solutions the equation has and what they are.
We will next discuss how to arrive at answers to these questions.
Intuition
The equation 7x + 11y = 100 is a specific instance of the Diophantine equation of the form ax + by = c, wherea = 7, b = 11 \text{ and } c = 100.
As usual, we will find a solution to this problem by relating it to something we already know. We have already encountered an equation of this form during our study of the greatest common divisor of two integers. We have seen that given two integers a \text{ and } b not both of which are zero, there exist integers x^\prime \text{ and } y^\prime such that ax^\prime + by^\prime = d, where d = \text{gcd}(a, b).
Suppose c is some multiple of d i.e., c = dr, where r is an integer.
If we multiply the equation ax^\prime + by^\prime = d with r, we get
arx^\prime + bry^\prime = dr
Since c = dr,
a(rx^\prime) + b(ry^\prime) = c
We can see that x = rx^\prime \text{ and } y = ry^\prime is a solution of the equation ax + by = c.
Therefore, we can see that a solution to the equation exists whenever d \,|\, c.
We will formalize this observation as a theorem and also derive all possible solutions of this equation, where such solutions exist.
Theorem
The linear Diophantine equation ax + by = c has a solution if and only if d \,|\, c, where d = \text{gcd}(a, b).
First we will prove the following conditional statement.
“If the linear Diophantine equation ax + by = c has a solution, then d \,|\, c, where d = \text{gcd}(a, b).”
Since d = \text{gcd}(a, b), by definition, d \,|\, a \text{ and } d \,|\, b. Therefore, from the definition of the divisibility of integers we know that there exist integers r \text{ and } s for which a = dr\text{ and } b = ds.
If a solution of ax + by = c exists, so that ax_0 + by_0 = c for suitable x_0 \text{ and } y_0, then
c = ax_0 + by_0 = drx_0 + dsy_0 = d(rx_0 + sy_0)
which implies that d \,|\, c.
Now lets prove the converse stated below.
“If d \,|\, c, where d = \text{gcd}(a, b), then there exists a solution to the linear Diophantine equation ax + by = c.”
Since d \,|\, c, from the definition of the divisibility of integers c = dt, for some integer t.
Also, since d = \text{gcd}(a, b), from this theorem we know that there exist integers x_0 \text{ and } y_0 such that d = ax_0 + by_0.
When we multiply this relation by t, we get
dt = (ax_0 + by_0)t = a(tx_0) + b(ty_0)
Since c = dt,
a(tx_0) + b(ty_0) = c
Hence, the Diophantine equation ax + by = c has x = tx_0 \text{ and } y = ty_0 as a particular solution.
How do we find all the other solutions to the Diophantine equation ax + by = c ?
Let us suppose that a solution x_0, y_0 of the given equation is known.
If x^\prime, y^\prime is any other solution, then
ax_0 + by_0 = c = ax^\prime + by^\prime
which is equivalent to,
a(x_0 - x^\prime) = b(y^\prime - y_0)
Substituting a = dr \text{ and } b = ds in the above equation, we get
dr(x_0 - x^\prime) = ds(y^\prime - y_0)
Cancelling the common factor d, we get
r(x_0 - x^\prime) = s(y^\prime - y_0)
From the definition of the divisibility of integers, the above equation implies that r \,|\, s(y^\prime - y_0).
Also, the equations a = dr \text{ and } b = ds can be rewritten as r = \dfrac{a}{d} \text{ and } s = \dfrac{b}{d}.
From corollary 1 of a theorem we have already proved, we know that \text{gcd}\bigg(\dfrac{a}{d}, \dfrac{b}{d}\bigg) = 1. Therefore, \text{gcd}(r, s) = 1.
From Euclid’s Lemma we know that r \,|\, (y^\prime - y_0) since r \,|\, s(y^\prime - y_0) and \text{gcd}(r, s) = 1.
From the definition of the divisibility of integers, this implies that y^\prime - y_0 = rt for some integer t.
Substituting the value of y^\prime - y_0 in the above equation, we get
r(x_0 - x^\prime) = srt
which is equivalent to
x_0 - x^\prime = st
This leads us to the formulas
\begin{equation} \begin{split} x^\prime = x_0 - st &= x_0 - \bigg(\frac{b}{d}\bigg)t \\ y^\prime = y_0 + rt &= y_0 + \bigg(\frac{a}{d}\bigg)t \\ \end{split} \end{equation}
It is easy to see that these values satisfy the Diophantine equation, regardless of the choice of the integer t; for
\begin{equation} \begin{split} ax^\prime + by^\prime &= a \Bigg[x_0 - \bigg(\frac{b} {d}\bigg)t\Bigg] + b\Bigg[y_0 + \bigg(\frac{a}{d}\bigg)t\Bigg]\\ &= (ax_0 + by_0) + \bigg(\frac{ab}{d} - \frac{ab}{d}\bigg)t \\ &= c + 0 \cdot t \\ &= c \\ \end{split} \end{equation}
Thus, there are an infinite number of solutions of the given equation, one for each value of t.
Summary
The linear Diophantine equation ax + by = c has a solution if and only if d \,|\, c, where d = \text{gcd}(a, b). If x_0, y_0 is any particular solution of this equation, then all other solutions are given by
x = x_0 - \bigg(\frac{b}{d}\bigg)t \quad \quad\quad y = y_0 + \bigg(\frac{a}{d}\bigg)t
where t is an arbitrary integer.
Example of an application of the theorem
We are now poised to solve the question posed by Euler in 1770, namely, dividing 100 into two summands such that one is divisible by 7 and the other by 11.
The solution to this problem can be framed in terms of a Diophantine equation as given below.
7x + 11y = 100
where x \text{ and } y are some positive integers.
First let us check whether this equation is solvable. From the theorem we have just proved, we know that this equation has a solution if \text{gcd}(7, 11) divides 100.
In this case, since 7 \text{ and } 11 are prime numbers, finding their greatest common divisor is quite straightforward; we know that their only common divisor is 1 and hence, \text{gcd}(7, 11) = 1.
However, we will still use Euclidean Algorithm to find \text{gcd}(7, 11) since we will need the steps in this algorithm to express \text{gcd}(7, 11) as a linear combination of 7 \text{ and } 11, which in turn will help us find the values of x \text{ and } y.
Since \text{gcd}(7, 11) = \text{gcd}(11, 7) and the Euclidean Algorithm requires the first operand to be greater than or equal to the second we will evaluate, \text{gcd}(11, 7).
Applying the Euclidean Algorithm to the evaluation of \text{gcd}(11, 7), we find that
\begin{equation} \begin{split} 11 &= 2 \times 7 - 3\\ 7 &= 2 \times 3 + 1\\ 3 &= 3 \times 1 + 0\\ \end{split} \end{equation}
and therefore, \text{gcd}(11, 7) = 1. Since 1 \,|\, 100, a solution to this equation exists.
To obtain 1 as a linear combination of 11 \text{ and } 7, we work backward through the previous calculations, as follows:
\begin{equation} \begin{split} 1 &= 7 - (2 \times 3)\\ &= 7 - [2 \times (2 \times 7 - 11)]\\ &= 7 - (2 \times 2 \times 7 ) + 2 \times 11 \\ &= 7 (1 - 4) + 2 \times 11 \\ &= 7 (-3) + 11 \times 2\\ \end{split} \end{equation}
Therefore,
7 (-3) + 11 \times 2 = 1
Multiplying this equation by 100, we get
\begin{equation} \begin{split} 100[7 (-3) + 11 \times 2] &= 100\\ \end{split} \end{equation}
which is equivalent to,
7(-300) + 11 \times 200 = 100
Hence, x = -300 \text{ and } y = 200 is one solution to the Diophantine equation in question.
All other solutions can be expressed by
\begin{equation} \begin{split} x &= -300 - \bigg(\frac{11}{1}\bigg)t = -300 - 11t \\ y &= 200 + \bigg(\frac{7}{1}\bigg)t = 200 + 7t\\ \end{split} \end{equation}
for some integer t.
Since we want to express 100 as a sum of two positive integers, we want the solution in positive integers.
From the above relations we can see that for x \text{ and } y to be positive, t must be chosen to satisfy simultaneously the following inequalities:
-300 - 11t > 0 \quad \quad \quad 200 + 7t > 0
Multiplying the inequality -300 - 11t > 0 with -1, we get
300 + 11t < 0
Adding -300 to both sides of the inequality results in
11t < -300
Dividing both sides of the inequality by 11, we get
t < \frac{-300}{11}
which is equivalent to,
t < -27\frac{3}{11}
Similarly, adding -200 to both sides of the inequality 200 + 7t > 0, we get
7t > -200
Dividing both sides of the inequality by 7, we get
t > \frac{-200}{7}
which is equivalent to,
t > -28\frac{4}{7}
Therefore,
-27\frac{3}{11} > t > -28\frac{4}{7}
Since t must be an integer, we are forced to conclude that t = -28.
Substituting this value of t in the following equations, we get
\begin{equation} \begin{split} x &= -300 - 11t =-300 - (11 \times -28) = -300 + 308 = 8 \\ y &= 200 + \bigg(\frac{7}{1}\bigg)t = 200 + 7t = 200 + (7 \times -28) = 200 - 196 = 4\\ \end{split} \end{equation}
Thus, our Diophantine equation has a unique positive solution x = 8, y = 4 corresponding to the value t = -28.
Therefore, 7x = 7 \times 8 = 56 \text{ and } 11y = 11 \times 4 = 44 are the two numbers that sum to 100 such that 56 is divisible by 7 and 44 by 11.
Diophantine equation ax + by = c, where a and b are relatively prime
As we have seen in the previous example, it might be helpful to record the form that the theorem we have just discussed takes when the coefficients are relatively prime.
Corollary
If \text{gcd}(a, b) = 1 and if x_0, y_0 is a particular solution of the linear Diophantine equation ax + by = c, then all solutions are given by
x = x_0 - bt \quad \quad \quad y = y_0 + at
for integral values of t.
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16116 | https://pubmed.ncbi.nlm.nih.gov/12219340/ | Is "shy bladder syndrome" a subtype of social anxiety disorder? A survey of people with paruresis - PubMed
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. 2002;16(2):84-7.
doi: 10.1002/da.10061.
Is "shy bladder syndrome" a subtype of social anxiety disorder? A survey of people with paruresis
Bavanisha Vythilingum1,Dan J Stein,Steven Soifer
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1 MRC Unit on Anxiety Disorders, University of Stellenbosch, Tygerberg, Cape Town, South Africa. bv@gerga.sun.ac.za
PMID: 12219340
DOI: 10.1002/da.10061
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Is "shy bladder syndrome" a subtype of social anxiety disorder? A survey of people with paruresis
Bavanisha Vythilingum et al. Depress Anxiety.2002.
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. 2002;16(2):84-7.
doi: 10.1002/da.10061.
Authors
Bavanisha Vythilingum1,Dan J Stein,Steven Soifer
Affiliation
1 MRC Unit on Anxiety Disorders, University of Stellenbosch, Tygerberg, Cape Town, South Africa. bv@gerga.sun.ac.za
PMID: 12219340
DOI: 10.1002/da.10061
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Paruresis is characterized by the fear of not being able to urinate in public bathrooms and has been classified by some to be a sub-type of social anxiety disorder (social phobia). Despite the existence of a consumer advocacy organization, the "Intentional Paruresis Association (www.paruresis.org)," there is sparse literature on this condition. A survey of people affiliated with the "International Paruresis Association" was undertaken using a self-report questionnaire with items that addressed demographic variables, the phenomenology of paruresis, comorbid disorders, and the impact of symptoms on quality of life. Sixty-three patients (59 M, 4 F) completed the questionnaire. The mean age of the subjects was 38.1+/-12 years, with the mean duration of symptoms being 24.5+/-13 years. Paruresis impacts significantly on sufferers' lives, with approximately one third limiting or avoiding parties, sports events, or dating and just over half of the sample limiting the job they choose to do. Social anxiety disorder (SAD) and depression are the most common comorbid disorders and the most common disorders in family members. Analysis of Liebowitz Social Anxiety Scale (LSAS) scores showed higher performance than social interaction subscale scores across the whole sample (whether suffering from SAD or not.) However, compared to subjects without co-morbid SAD, those with comorbidity had higher total, performance, and social interaction scores. Thus, paruresis can be a chronic and disabling symptom, and there seems to be an association between paruresis and other performance anxieties. Further research to characterize paruresis and to determine effective treatments is needed.
Copyright 2002 Wiley-Liss, Inc.
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How to minimize (x−a)2+(y−b)2(x−a)2+(y−b)2 subject to √a+√b=√2 a−−√+b√=2–√?
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Let x,y x,y be positive real numbers satisfying x y≤1 4 x y≤1 4.
I am trying to find min√a+√b=√2(x−a)2+(y−b)2
min a√+b√=2√(x−a)2+(y−b)2
as a function of x x and y y.
Is there any way to do that analytically?
Edit:
Is there a way to use a computer for this? I tried running Mathematica, but it never finishes the computation...
Lagrange’s multipliers method gives
(a−x)=λ√a(b−y)=λ√b.
(a−x)=λ a−−√(b−y)=λ b√.
In particular, (a−x)=√b a⋅(y−b).
I tried various ways to proceed but got stuck.
real-analysis
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edited May 1, 2020 at 10:31
Asaf ShacharAsaf Shachar
asked Apr 28, 2020 at 6:09
Asaf ShacharAsaf Shachar
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@PeterSzilas: the unknowns are a,b and x,y are given; they break the symmetry.user65203 –user65203 2020-04-28 06:59:57 +00:00 Commented Apr 28, 2020 at 6:59
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With a:=2(1 2+t)2,b:=2(1 2−t)2 we minimize
(x−2(1 2+t)2)2+(y−2(1 2−t)2)2.
By differentiating on t we get (after simplification)
8 t 3+2(3−x−y)t+y−x=0.
This is a general (depressed) cubic equation, and there is no hope for a nice solution.
Note that √a+√b=1 describes an arc of parabola (the axis is the first bissector), and you are looking for the shortest distance between the point (x,y) and this arc.
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edited Apr 28, 2020 at 7:39
answered Apr 28, 2020 at 7:27
user65203 user65203
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Thanks. I wonder if anything more can still be said about this cubic equation, e.g. how many real roots does it have? Wolframe alpha suggests that there is only one real root, but I am not sure.Asaf Shachar –Asaf Shachar 2020-04-28 14:27:34 +00:00 Commented Apr 28, 2020 at 14:27
@AsafShachar: when x=y, a solution is t=0 and there can be two other real ones of arbitrary value.user65203 –user65203 2020-04-28 14:30:38 +00:00 Commented Apr 28, 2020 at 14:30
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So we want to minimize (1 a+1 b)λ 2. If a b≤1 4, we can of course take λ=0, i.e., x=a, y=b and achieve the minimum value 0. So assume a b>1 4. From 1 4≥x y=(a−λ√a)(b−λ√b)=a b−(√a b+√b a)λ+λ 2√a b, we read off that we shold take the smaller of the two (necessarily positive) roots of the quadratic X 2−(a+b)X+(a b−1 4)√a b, i.e., λ=a+b−√a 2+b 2+2 a b−4 a b√a b+√a b 2=a+b−√(a−b)2+(2√a b−4√a b)2 2
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answered Apr 28, 2020 at 6:40
Hagen von EitzenHagen von Eitzen
384k 33 33 gold badges 379 379 silver badges 686 686 bronze badges
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1 I think that there is a problem with your answer. x,y are given in advance, you can't choose them to be x=a,y=b unless they satisfy √x+√y=√2. Also, since a,b satisfy √a+√b=√2, we know that a b≤1 4, by the AM-GM inequality, so the second option that you mentioned a b>1 4 cannot occur.Asaf Shachar –Asaf Shachar 2020-04-28 13:45:20 +00:00 Commented Apr 28, 2020 at 13:45
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16118 | https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(Ellis_and_Burzynski)/07%3A_Graphing_Linear_Equations_and_Inequalities_in_One_and_Two_Variables/7.05%3A_The_Slope-Intercept_Form_of_a_Line | Skip to main content
7.5: The Slope-Intercept Form of a Line
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7.4: Graphing Linear Equations in Two Variables
7.6: Graphing Equations in Slope-Intercept Form
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( \newcommand{\kernel}{\mathrm{null}\,})
The General Form of a Line
We have seen that the general form of a linear equation in two variables is ax+by=c (Section 7.4). When this equation is solved for y, the resulting form is called the slope-intercept form. Let's generate this new form.
This equation is of the form if we replace with and constant with . (Note The fact that we let is unfortunate and occurs because of the letters we have chosen to use in the general form. The letter occurs on both sides of the equal sign and may not represent the same value at all. This problem is one of the historical convention and, fortunately, does not occur very often.)
The following examples illustrate this procedure.
Example
Solve for .
The equation is of the form . In this case, and .
Example
Solve for .
This equation is of the form . In this case, and .
Example
Solve for .
This equation is of the form . In this case, and . Notice that we can write as
The Slope-Intercept Form of a Line
The Slope-Intercept Form of a Line
A linear equation in two variables written in the form is said to be in slope-intercept form.
Sample Set A
The following equations are in slope-intercept form:
Example
. In this case and
Example
. In this case and
Example
. In this case and
Example
. In this case and since we can write as .
The following equations are not in slope-intercept form.
Example
. The coefficient of is . To be in slope-intercept form, the coefficient of must be .
Example
. The equation is not solved for . THe and appear on the same side of the equal sign.
Example
. The equation is not solved for
Practice Set A
The following equation are in slope-intercept form. In each case, specify the slope and -intercept.
Practice Problem
Answer
Practice Problem
Answer
Practice Problem
Answer
Practice Problem
Answer
Practice Problem
Answer
Practice Problem
Answer
Slope and Intercept
When the equation of a line is written in slope-intercept form, two important properties of the line can be seen: the slope and the intercept. Let's look at these two properties by graphing several lines and observing them carefully.
Sample Set B
Example
Graph the line .
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| | | |
| | | |
Looking carefully at this line, answer the following two questions.
At what number does this line cross the y-axis? Do you see this number in the equation?
The line crosses the y-axis at −3.
Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?
After moving horizontally one unit to the right, we must move exactly one vertical unit up. This number is the coefficient of .
Example
Graph the line
| | | |
---
| | | |
| | | |
| | | |
| | | |
Looking carefully at this line, answer the following two questions.
At what number does this line cross the -axis? Do you see this number in the equation?
The line crosses the -axis at .
Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?
After moving horizontally one unit to the right, we must move exactly unit upward. This number is the coefficient of .
Practice Set B
Practice Problem
Graph the line
| | | |
---
| | | |
| - |
| - |
| - |
Looking carefully at this line, answer the following two questions.
At what number does the line cross the -axis? Do you see this number in the equation?
Answer
: The line crosses the -axis at . After moving horizontally unit to the right, we must move exactly units downward.
Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?
Answer
In the graphs constructed in Sample Set B and Practice Set B, each equation had the form . We can answer the same questions by using this form of the equation (shown in the diagram).
-Intercept
At what number does the line cross the -axis? Do you see this number in the equation?
Answer
: In each case, the line crosses the -axis at the constant . The number is the number at which the line crosses the -axis, and it is called the -intercept. The ordered pair corresponding to the -intercept is .
Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?
Answer
: To get back on the line, we must move our pencil exactly vertical units.
Slope
The number is the coefficient of the variable . The number is called the slope of the line and it is the number of units that changes when is increased by unit. Thus, if changes by unit, changes by units.
Since the equation contains both the slope of the line and the -intercept, we call the form the slope-intercept form.
The Slope-Intercept Form of the Equation of a Line
The slope-intercept form of a straight line is
The slope of the line is , and the -intercept is the point .
The Slope is a Measure of the Steepness of a Line
The word slope is really quite appropriate. It gives us a measure of the steepness of the line. Consider two lines, one with slope and the other with slope . The line with slope is steeper than is the line with slope . Imagine your pencil being placed at any point on the lines. We make a -unit increase in the -value by moving the pencil one unit to the right. To get back to one line we need only move vertically unit, whereas to get back onto the other line we need to move vertically units.
Sample Set C
Find the slope and the y-intercept of the following lines.
Example
The line is in the slope-intercept form . The slope is , the coefficient of . Therefore, . The -intercept is the point . Since , the -intercept is .
Slope:
-intercept:
Example
The line is in the slope-intercept form . The slope is , the coefficient of . Therefore, . The -intercept is the point . Since , the -intercept is .
Slope:
-intercept:
Example
.
This equation is written in general form. We can put the equation in slope-intercept form by solving for .
Now the equation is in slope-intercept form.
Slope:
-intercept:
Practice Set C
Practice Problem
Find the slope and -intercept of the line .
Answer
: Solving for we get . Now and .
The Formula for the Slope of a Line
We have observed that the slope is a measure of the steepness of a line. We wish to develop a formula for measuring this steepness.
It seems reasonable to develop a slope formula that produces the following results:
Steepness of line steepness of line 2.
Consider a line on which we select any two points. We’ll denote these points with the ordered pairs and . The subscripts help us to identify the points.
is the first point. Subscript indicates the first point.
is the second point. Subscript indicates the second point.
The difference in values gives us the horizontal change, and the difference in values gives us the vertical change. If the line is very steep, then when going from the first point to the second point, we would expect a large vertical change compared to the horizontal change. If the line is not very steep, then when going from the first point to the second point, we would expect a small vertical change compared to the horizontal change.
We are comparing changes. We see that we are comparing:
This is a comparison and is therefore a ratio. Ratios can be expressed as fractions. Thus, a measure of the steepness of a line can be expressed as a ratio.
The slope of a line is defined as the ratio
Mathematically, we can write these changes as
Finding the Slope of a Line
The slope of a nonvertical line passing through the points and is found by the formula:
Sample Set D
For the two given points, find the slope of the line that passes through them.
Example
and .
Looking left to right on the line we can choose to be , and to be . Then,
This line has slope . It appears fairly steep. When the slope is written in fraction form, , we can see, by recalling the slope formula, that as changes unit to the right (because of the ) changes units upward (because of the .
.
Notice that as we look left to right, the line rises.
Example
and .
Looking left to right on the line we can choose to be , and to be . Then,
This line has slope . Thus, as changes units to the right (because of the , changes unit upward (because of the ).
.
Notice that in examples 16 and 17, both lines have positive slopes, and , and both lines rise as we look left to right.
Example
and .
Looking left to right on the line we can choose to be , and to be . Then,
This line has slope .
When the slope is written in fraction form, , we can see that as changes unit to the right (because of the , changes unit downward (because of the ).
Notice also that this line has a negative slope and declines as we look left to right.
Example
and .
This line has slope. This means it has no rise and, therefore, is a horizontal line. This does not mean that the line has no slope, however.
Example
and .
Since division by is undefined, we say that vertical lines have undefined slope. Since there is no real number to represent the slope of this line, we sometimes say that vertical lines have undefined slope, or no slope.
Practice Set D
Practice Problem
Find the slope of the line passing through and . Graph this line on the graph of problem 2 below.
Answer
Practice Problem
Find the slope of the line passing through and . Graph this line.
Answer
: The line has slope
Practice Problem
Compare the lines of the following problems. Do the lines appear to cross? What is it called when lines do not meet (parallel or intersecting)? Compare their slopes. Make a statement about the condition of these lines and their slopes.
Answer
: The lines appear to be parallel. Parallel lines have the same slope, and lines that have the same slope are parallel
Before trying some problems, let’s summarize what we have observed.
The equation is called the slope-intercept form of the equation of a line. The number is the slope of the line and the point is the -intercept.
The slope, , of a line is defined as the steepness of the line, and it is the number of units that changes when changes unit.
The formula for finding the slope of a line through any two given points and is:
The fraction represents the \dfrac{\text{ change in } y}{\text{ change in } x}
As we look at a graph from left to right, lines with positive slope rise and lines with negative slope decline.
Parallel lines have the same slope.
Horizontal lines have 0 slope.
Vertical lines have undefined slope (or no slope).
Exercises
For the following problems, determine the slope and y-intercept of the lines.
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
Exercise
Exercise
Answer
: slope = ; -intercept=
For the following problems, find the slope of the line through the pairs of points.
Exercise
Exercise
Answer
Exercise
Exercise
Answer
Exercise
Exercise
Answer
Exercise
Exercise
Answer
Exercise
Exercise
Answer
Exercise
Exercise
Answer
Exercise
Exercise
Answer
Exercise
Exercise
Answer
Exercise
Exercise
Answer
: (horizontal line )
Exercise
Make a statement about the slopes of parallel lines.
Exercise
Exercise
Answer
: No slope (vertical line at )
Exercise
Exercise
Answer
: (horizontal line at )
Exercise
Do lines with a positive slope rise or decline as we look left to right?
Exercise
Do lines with a negative slope rise or decline as we look left to right?
Answer
For the following problems, determine the slope and y-intercept of the lines. Round to two decimal places.
Exercise
Answer
: slope=
−intercept=
Exercise
Exercise
Answer
: slope=
−intercept=
Exercise
For the following problems, find the slope of the line through the pairs of points. Round to two decimal places.
Exercise
Answer
Exercise
Exercise
Answer
: (horizontal line at )
Exercise
Exercise
Answer
: No slope (vertical line )
Exercise
Exercises for Review
Exercise
Simplify .
Answer
: if
Exercise
Solve the equation .
Exercise
When four times a number is divided by five, and that result is decreased by eight, the result is zero. What is the original number?
Answer
Exercise
Solve if .
Exercise
Graph the linear equation .
Answer
7.4: Graphing Linear Equations in Two Variables
7.6: Graphing Equations in Slope-Intercept Form |
16119 | http://physics.bu.edu/~duffy/semester2/c25_interference_doublesource.html | Interference from Two Sources
Two sources a distance d apart are sending out identical waves in phase. We observe an interference pattern with lines of constructive interference at particular angles and lines of destructive interference at other angles.
When we're at a point far from the sources (far relative to d) then to a good approximation the waves arriving from the two sources are parallel. In this case the path-length difference is d = d sin(θ). We can therefore predict where the lines of constructive interference occur, at angles satisfying:
Condition for constructive interference: d sin(θ) = mλ, where m is any integer.
Similarly the lines of destructive interference occur at angles satisfying:
Condition for destructive interference: d sin(θ) = (m + 1/2) λ
The lines of constructive and destructive interference get further apart if we increase the wavelength or if we move the sources closer together (i.e., reduce d). They get closer together if we do the opposite.
Let's say our two sources are emitting light, and we pick up the interference pattern on a screen a distance L from the slits. When the two sources are in phase there is always a bright spot at the center of the screen due to constructive interference. Other bright spots show up on the screen at distances ym from the center.
tan(θ) = ym/L
Combining this with the fact that the angles for the bright spots are given by the constructive interference equation:
sin(θ) = mλ/d
and that for small angles sin(θ) is approximately equal to tan(θ), the positions of the bright spots occur at:
ym = mλL/d
Similarly the dark spots, from destructive interference, occur at:
y = (m + 1/2) λL/d |
16120 | https://dcic-world.org/2022-08-28/Conditionals_and_Booleans.html | 6 Conditionals and Booleans
►A Data-Centric Introduction to Computing
IIntroduction
IIFoundations
IIIFoundations: Bonus Materials
IVAlgorithms
VAlgorithms: Bonus Materials
VIFrom Pyret to Python
VIIProgramming with State
VIIIAdvanced Topics
IXAppendices
▼IIFoundations
3Getting Started
4Naming Values
5From Repeated Expressions to Functions
6Conditionals and Booleans
7Introduction to Tabular Data
8Processing Tables
9From Tables to Lists
10Processing Lists
11Introduction to Structured Data
12Collections of Structured Data
13Recursive Data
14Trees
15Functions as Data
16Interactive Games as Reactive Systems
17Examples, Testing, and Program Checking
►6Conditionals and Booleans
6.1Motivating Example: Shipping Costs
6.2Conditionals: Computations with Decisions
6.3Booleans
6.4Asking Multiple Questions
6.5Evaluating by Reducing Expressions
6.6Composing Functions
6.7Nested Conditionals
6.8Recap: Booleans and Conditionals
On this page:
6.1Motivating Example: Shipping Costs
6.2Conditionals: Computations with Decisions
6.3Booleans
6.3.1Other Boolean Operations
6.3.2Combining Booleans
6.4Asking Multiple Questions
6.5Evaluating by Reducing Expressions
6.6Composing Functions
6.6.1How Function Compositions Evaluate
6.6.2Function Composition and the Directory
6.7Nested Conditionals
6.8Recap: Booleans and Conditionals
contents← prevupnext →
6Conditionals and Booleans
6.1 Motivating Example: Shipping Costs
6.2 Conditionals: Computations with Decisions
6.3 Booleans
6.3.1 Other Boolean Operations
6.3.2 Combining Booleans
6.4 Asking Multiple Questions
6.5 Evaluating by Reducing Expressions
6.6 Composing Functions
6.6.1 How Function Compositions Evaluate
6.6.2 Function Composition and the Directory
6.7 Nested Conditionals
6.8 Recap: Booleans and Conditionals
6.1Motivating Example: Shipping Costs
In Functions Practice: Cost of pens, we wrote a program (pen-cost) to compute the cost of ordering pens. Continuing the example, we now want to account for shipping costs. We’ll determine shipping charges based on the cost of the order.
Specifically, we will write a function add-shipping to compute the total cost of an order including shipping. Assume an order valued at $10 or less ships for $4, while an order valued above $10 ships for $8. As usual, we will start by writing examples of the add-shipping computation.
Do Now!
Use the is notation from where blocks to write several examples of add-shipping. How are you choosing which inputs to use in your examples? Are you picking random inputs? Being strategic in some way? If so, what’s your strategy?
Here is a proposed collection of examples for add-shipping.
add-shipping(10) is 10 + 4
add-shipping(3.95) is 3.95 + 4
add-shipping(20) is 20 + 8
add-shipping(10.01) is 10.01 + 8
Do Now!
What do you notice about our examples? What strategies do you observe across our choices?
Our proposed examples feature several strategic decisions:
Including 10, which is at the boundary of charges based on the text
Including 10.01, which is just over the boundary
Including both natural and real (decimal) numbers
Including examples that should result in each shipping charge mentioned in the problem (4 and 8)
So far, we have used a simple rule for creating a function body from examples: locate the parts that are changing, replace them with names, then make the names the parameters to the function.
Do Now!
What is changing across our add-shipping examples? Do you notice anything different about these changes compared to the examples for our previous functions?
Two things are new in this set of examples:
The values of 4 and 8 differ across the examples, but they each occur in multiple examples.
The values of 4 and 8 appear only in the computed answers—not as an input. Which one we use seems to depend on the input value.
These two observations suggest that something new is going on with add-shipping. In particular, we have clusters of examples that share a fixed value (the shipping charge), but different clusters (a) use different values and (b) have a pattern to their inputs (whether the input value is less than or equal to 10). This calls for being able to ask questions about inputs within our programs.
6.2Conditionals: Computations with Decisions
To ask a question about our inputs, we use a new kind of expression called an if expression. Here’s the full definition of add-shipping:
fun add-shipping(order-amt :: Number) -> Number:
doc: "add shipping costs to order total"
if order-amt <= 10:
order-amt + 4
else:
order-amt + 8
end
where:
add-shipping(10) is 10 + 4
add-shipping(3.95) is 3.95 + 4
add-shipping(20) is 20 + 8
add-shipping(10.01) is 10.01 + 8
end
In an if expression, we ask a question that can produce an answer that is true or false (here order-amt <= 10, which we’ll explain below in Booleans), provide one expression for when the answer to the question is true (order-amt + 4), and another for when the result is false (
order-amt +
8
). The else in the program marks the answer in the false case; we call this the else clause. We also need end to tell Pyret we’re done with the question and answers.
6.3Booleans
Every expression in Pyret evaluates in a value. So far, we have seen three types of values: Number, String, and Image. What type of value does a question like
order-amt
<= 10
produce? We can use the interactions prompt to experiment and find out.
Do Now!
Enter each of the following expressions at the interactions prompt. What type of value did you get? Do the values fit the types we have seen so far?
3.95 <= 10
20 <= 10
The values true and false belong to a new type in Pyret, called Boolean.Named for George Boole. While there are an infinitely many values of type Number, there are only two of type Boolean: true and false.
Exercise
What would happen if we entered order-amt <= 10 at the interactions prompt to explore booleans? Why does that happen?
6.3.1Other Boolean Operations
There are many other built-in operations that return Boolean values. Comparing values for equality is a common one: There is much more we can and should say about equality, which we will do later [Re-Examining Equality].
1 == 1
true
1 == 2
false
"cat" == "dog"
false
"cat" == "CAT"
false
In general, == checks whether two values are equal. Note this is different from the single = used to associate names with values in the directory.
The last example is the most interesting: it illustrates that strings are case-sensitive, meaning individual letters must match in their case for strings to be considered equal.This will become relevant when we get to tables later.
Sometimes, we also want to compare strings to determine their alphabetical order. Here are several examples:
"a" < "b"
true
"a" >= "c"
false
"that" < "this"
true
"alpha" < "beta"
true
which is the alphabetical order we’re used to; but others need some explaining:
"a" >= "C"
true
"a" >= "A"
true
These use a convention laid down a long time ago in a system called ASCII.Things get far more complicated with non-ASCII letters: e.g., Pyret thinks "Ł" is > than "Z", but in Polish, this should be false. Worse, the ordering depends on location (e.g., Denmark/Norway vs. Finland/Sweden).
Do Now!
Can you compare true and false? Try comparing them for equality (==), then for inequality (such as <).
In general, you can compare any two values for equality (well, almost, we’ll come back to this later); for instance:
"a" == 1
false
If you want to compare values of a specific kind, you can use more specific operators:
num-equal(1, 1)
true
num-equal(1, 2)
false
string-equal("a", "a")
true
string-equal("a", "b")
false
Why use these operators instead of the more generic ==?
Do Now!
Try
num-equal("a", 1)
string-equal("a", 1)
Therefore, it’s wise to use the type-specific operators where you’re expecting the two arguments to be of the same type. Then, Pyret will signal an error if you go wrong, instead of blindly returning an answer (false) which lets your program continue to compute a nonsensical value.
There are even more Boolean-producing operators, such as:
wm = "will.i.am"
string-contains(wm, "will")
true
Note the capital W.
string-contains(wm, "Will")
false
In fact, just about every kind of data will have some Boolean-valued operators to enable comparisons.
6.3.2Combining Booleans
Often, we want to base decisions on more than one Boolean value. For instance, you are allowed to vote if you’re a citizen of a country and you are above a certain age. You’re allowed to board a bus if you have a ticket or the bus is having a free-ride day. We can even combine conditions: you’re allowed to drive if you are above a certain age and have good eyesight and—either pass a test or have a temporary license. Also, you’re allowed to drive if you are not inebriated.
Corresponding to these forms of combinations, Pyret offers three main operations: and, or, and not. Here are some examples of their use:
(1 < 2) and (2 < 3)
true
(1 < 2) and (3 < 2)
false
(1 < 2) or (2 < 3)
true
(3 < 2) or (1 < 2)
true
not(1 < 2)
false
Exercise
Explain why numbers and strings are not good ways to express the answer to a true/false question.
6.4Asking Multiple Questions
Shipping costs are rising, so we want to modify the add-shipping program to include a third shipping level: orders between $10 and $30 ship for $8, but orders over $30 ship for $12. This calls for two modifications to our program:
We have to be able to ask another question to distinguish situations in which the shipping charge is 8 from those in which the shipping charge is 12.
The question for when the shipping charge is 8 will need to check whether the input is between two values.
We’ll handle these in order.
The current body of add-shipping asks one question: order-amt <= 10. We need to add another one for
order-amt
<= 30
, using a charge of 12 if that question fails. Where do we put that additional question?
An expanded version of the if-expression, using else if, allows you to ask multiple questions:
fun add-shipping(order-amt :: Number) -> Number:
doc: "add shipping costs to order total"
if order-amt <= 10:
order-amt + 4
else if order-amt <= 30:
order-amt + 8
else:
order-amt + 12
end
where:
...
end
At this point, you should also add where examples that use the 12 charge.
How does Pyret determine which answer to return? It evaluates each question expression in order, starting from the one that follows if. It continues through the questions, returning the value of the answer of the first question that returns true. Here’s a summary of the if-expression syntax and how it evaluates.
if QUESTION1:
<result in case first question true>
else if QUESTION2:
<result in case QUESTION1 false and QUESTION2 true>
else:
<result in case both QUESTIONs false>
end
A program can have multiple else if cases, thus accommodating an arbitrary number of questions within a program.
Do Now!
The problem description for add-shipping said that orders between 10 and 30 should incur an 8 charge. How does the above code capture “between”?
This is currently entirely implicit. It depends on us understanding the way an if evaluates. The first question is order-amt <= 10, so if we continue to the second question, it means order-amt > 10. In this context, the second question asks whether order-amt <= 30. That’s how we’re capturing “between”-ness.
Do Now!
How might you modify the above code to build the “between 10 and 30” requirement explicitly into the question for the 8 case?
Remember the and operator on booleans? We can use that to capture “between” relationships, as follows:
(order-amt > 10) and (order-amt <= 30)
Do Now!
Why are there parentheses around the two comparisons? If you replace order-amt with a concrete value (such as 20) and leave off the parenthesis, what happens when you evaluate this expression in the interactions pane?
Here is what add-shipping look like with the and included:
fun add-shipping(order-amt :: Number) -> Number:
doc: "add shipping costs to order total"
if order-amt <= 10:
order-amt + 4
else if (order-amt > 10) and (order-amt <= 30):
order-amt + 8
else:
order-amt + 12
end
where:
add-shipping(10) is 10 + 4
add-shipping(3.95) is 3.95 + 4
add-shipping(20) is 20 + 8
add-shipping(10.01) is 10.01 + 8
add-shipping(30) is 30 + 8
add-shipping(32) is 32 + 12
end
Both versions of add-shipping support the same examples. Are both correct? Yes. And while the first part of the second question (order-amt > 10) is redundant, it can be helpful to include such conditions for three reasons:
They signal to future readers (including ourselves!) the condition covering a case.
They ensure that if we make a mistake in writing an earlier question, we won’t silently get surprising output.
They guard against future modifications, where someone might modify an earlier question without realizing the impact it’s having on a later one.
Exercise
An online-advertising firm needs to determine whether to show an ad for a skateboarding park to website users. Write a function show-ad that takes the age and haircolor of an individual user and returns true if the user is between the ages of 9 and 18 and has either pink or purple hair.
Try writing this two ways: once with if expressions and once using just boolean operations.
Responsible Computing: Harms from Reducing People to Simple Data
Assumptions about users get encoded in even the simplest functions. The advertising exercise shows an example in which a decision gets made on the basis of two pieces of information about a person: age and haircolor. While some people might stereotypically associate skateborders with being young and having colored hair, many skateborders do not fit these criteria and many people who fit these criteria don’t skateboard.
While real programs to match ads to users are more sophisticated than this simple function, even the most sophisticated advertising programs boil down to tracking features or information about individuals and comparing it to information about the content of ads. A real ad system would differ in tracking dozens (or more) of features and using more advanced programming ideas than simple conditionals to determine the suitability of an ad (we’ll discuss some of these later in the book). This example also extends to situations far more serious than ads: who gets hired, granted a bank loan, or sent to or released from jail are other examples of real systems that depend on comparing data about individuals with criteria maintained by a program.
From a social responsibility perspective, the questions here are what data about individuals should be used to represent them for processing by programs and what stereotypes might those data encode. In some cases, individuals can be represented by data without harm (a university housing office, for examples, stores student ID numbers and which room a student is living in). But in other cases, data about individuals get interpreted in order to predict something about them. Decisions based on those predictions can be inaccurate and hence harmful.
6.5Evaluating by Reducing Expressions
In How Functions Evaluate, we talked about how Pyret reduces expressions and function calls to values. Let’s revisit this process, this time expanding to consider if-expressions. Suppose we want to compute the wages of a worker. The worker is paid $10 for every hour up to the first 40 hours, and is paid $15 for every extra hour. Let’s say hours contains the number of hours they work, and suppose it’s 45:
hours = 45
Suppose the formula for computing the wage is
if hours <= 40:
hours 10
else if hours > 40:
(40 10) + ((hours - 40) 15)
end
Let’s now see how this results in an answer, using a step-by-step process that should match what you’ve seen in algebra classes (the steps are described in the margin notes to the right): The first step is to substitute the hours with 45.
if 45 <= 40:
45 10
else if 45 > 40:
(40 10) + ((45 - 40) 15)
end
Next, the conditional part of the if expression is evaluated, which in this case is false.
=> if false:
45 10
else if 45 > 40:
(40 10) + ((45 - 40) 15)
end
Since the condition is false, the next branch is tried.
=> if 45 > 40:
(40 10) + ((45 - 40) 15)
end
Pyret evaluates the question in the conditional, which in this case produces true.
=> if true:
(40 10) + ((45 - 40) 15)
end
Since the condition is true, the expression reduces to the body of that branch. After that, it’s just arithmetic.
=> (40 10) + ((45 - 40) 15)
=> 400 + (5 15)
=> 475
This style of reduction is the best way to think about the evaluation of Pyret expressions. The whole expression takes steps that simplify it, proceeding by simple rules. You can use this style yourself if you want to try and work through the evaluation of a Pyret program by hand (or in your head).
6.6Composing Functions
We started this chapter wanting to account for shipping costs on an order of pens. So far, we have written two functions:
pen-cost for computing the cost of the pens
add-shipping for adding shipping costs to a total amount
What if we now wanted to compute the price of an order of pens including shipping? We would have to use both of these functions together, sending the output of pen-cost to the input of add-shipping.
Do Now!
Write an expression that computes the total cost, with shipping, of an order of 10 pens that say "bravo".
There are two ways to structure this computation. We could pass the result of pen-cost directly to add-shipping:
add-shipping(pen-cost(10, "bravo"))
Alternatively, you might have named the result of pen-cost as an intermediate step:
pens = pen-cost(10, "bravo")
add-shipping(pens)
Both methods would produce the same answer.
6.6.1How Function Compositions Evaluate
Let’s review how these programs evaluate in the context of substitution and the directory. We’ll start with the second version, in which we explicitly name the result of calling pen-cost.
Evaluating the second version: At a high level, Pyret goes through the following steps:
Substitute 10 for num-pens and "bravo" for message in the body of pen-cost, then evaluate the substituted body
Store pens in the directory, with a value of 3.5
As a first step in evaluating add-shipping(pens), look up the value of pens in the directory
Substitute 3.5 for order-amt in the body of add-shipping then evaluate the resulting expression, which results in 7.5
Evaluating the first version: As a reminder, the first version consisted of a single expression:
add-shipping(pen-cost(10, "bravo"))
Since arguments are evaluated before functions get called, start by evaluating pen-cost(10, "bravo") (again using substitution), which reduces to 3.5
Substitute 3.5 for order-amt in the body of add-shipping then evaluate the resulting expression, which results in 7.5
Do Now!
Contrast these two summaries. Where do they differ? What about the code led to those differences?
The difference lies in the use of the directory: the version that explicitly named pens uses the directory. The other version doesn’t use the directory at all. Yet both approaches lead to the same result, since the same value (the result of calling pen-cost) gets substituted into the body of add-shipping.
This analysis might suggest that the version that uses the directory is somehow wasteful: it seems to take more steps just to end up at the same result. Yet one might argue that the version that uses the directory is easier to read (different readers will have different opinions on this, and that’s fine). So which should we use?
Use whichever makes more sense to you on a given problem. There will be times when we prefer each of these styles. Furthermore, it will turn out (once we’ve learned more about nuances of how programs evaluate) that the two versions aren’t as different as they appear right now.
6.6.2Function Composition and the Directory
Let’s try one more variation on this problem. Perhaps seeing us name the intermediate result of pen-cost made you wish that we had used intermediate names to make the body of pen-cost more readable. For example, we could have written it as:
fun pen-cost(num-pens :: Number, message :: String)
-> Number:
doc:total cost for pens, each 25 cents
plus 2 cents per message charactermessage-cost = (string-length(message) 0.02)
num-pens (0.25 + message-cost)
where:
...
end
Do Now!
Write out the high level steps for how Pyret will evaluate the following program using this new version of pen-cost:
pens = pen-cost(10, "bravo")
add-shipping(pens)
Hopefully, you made two entries into the directory, one for message-cost inside the body of pen-cost and one for pens as we did earlier.
Do Now!
Consider the following program. What result do you think Pyret should produce?
pens = pen-cost(10, "bravo")
cheap-message = (message-cost > 0.5)
add-shipping(pens)
Using the directory you envisioned for the previous activity, what answer do you think you will get?
Something odd is happening here. The new program tries to use message-cost to define cheap-message. But the name message-cost doesn’t appear anywhere in the program, unless we peek inside the function bodies. But letting code peek inside function bodies doesn’t make sense: you might not be able to see inside the functions (if they are defined in libraries, for example), so this program should report an error that message-cost is undefined.
Okay, so that’s what should happen. But our discussion of the directory suggests that both pens and message-cost will be in the directory, meaning Pyret would be able to use message-cost. What’s going on?
This example prompts us to explain one more nuance about the directory. Precisely to avoid problems like the one illustrated here (which should produce an error), directory entries made within a function are local (private) to the function body. When you call a function, Pyret sets up a local directory that other functions can’t see. A function body can add or refer to names in either its local, private directory (as with message-cost) or the overall (global) directory (as with pens). But in no case can one function call peek inside the local directory for another function call. Once a function call completes, its local directory disappears (because nothing else would be able to use it anyway).
6.7Nested Conditionals
We showed that the results in if-expressions are themselves expressions (such as order-amt + 4 in the following function):
fun add-shipping(order-amt :: Number) -> Number:
doc: "add shipping costs to order total"
if order-amt <= 10:
order-amt + 4
else:
order-amt + 8
end
end
The result expressions can be more complicated. In fact, they could be entire if-expressions!. To see an example of this, let’s develop another function. This time, we want a function that will compute the cost of movie tickets. Let’s start with a simple version in which tickets are $10 apiece.
fun buy-tickets1(count :: Number) -> Number:
doc: "Compute the price of tickets at $10 each"
count 10
where:
buy-tickets1(0) is 0
buy-tickets1(2) is 2 10
buy-tickets1(6) is 6 10
end
Now, let’s augment the function with an extra parameter to indicate whether the purchaser is a senior citizen who is entitled to a discount. In such cases, we will reduce the overall price by 15%.
fun buy-tickets2(count :: Number, is-senior :: Boolean)
-> Number:
doc:Compute the price of tickets at $10 each with
senior discount of 15%if is-senior == true:
count 10 0.85
else:
count 10
end
where:
buy-tickets2(0, false) is 0
buy-tickets2(0, true) is 0
buy-tickets2(2, false) is 2 10
buy-tickets2(2, true) is 2 10 0.85
buy-tickets2(6, false) is 6 10
buy-tickets2(6, true) is 6 10 0.85
end
There are a couple of things to notice here:
The function now has an additional parameter of type Boolean to indicate whether the purchaser is a senior citizen.
We have added an if expression to check whether to apply the discount.
We have more examples, because we have to vary both the number of tickets and whether a discount applies.
Now, let’s extend the program once more, this time also offering the discount if the purchaser is not a senior but has bought more than 5 tickets. Where should we modify the code to do this? One option is to first check whether the senior discount applies. If not, we check whether the number of tickets qualifies for a discount:
fun buy-tickets3(count :: Number, is-senior :: Boolean)
-> Number:
doc:Compute the price of tickets at $10 each with
discount of 15% for more than 5 tickets
or being a seniorif is-senior == true:
count 10 0.85
else:
if count > 5:
count 10 0.85
else:
count 10
end
end
where:
buy-tickets3(0, false) is 0
buy-tickets3(0, true) is 0
buy-tickets3(2, false) is 2 10
buy-tickets3(2, true) is 2 10 0.85
buy-tickets3(6, false) is 6 10 0.85
buy-tickets3(6, true) is 6 10 0.85
end
Notice here that we have put a second if expression within the else case. This is valid code. (We could have also made an else if here, but we didn’t so that we could show that nested conditionals are also valid).
Exercise
Show the steps through which this function would evaluate in a situation where no discount applies, such as
buy-tickets3(2,
false)
.
Do Now!
Look at the current code: do you see a repeated computation that we might end up having to modify later?
Part of good code style is making sure that our programs would be easy to maintain later. If the theater changes its discount policy, for example, the current code would require us to change the discount (0.85) in two places. It would be much better to have that computation written only one time. We can achieve that by asking which conditions lead to the discount applying, and writing them as the check within just one if expression.
Do Now!
Under what conditions should the discount apply?
Here, we see that the discount applies if either the purchaser is a senior or more than 5 tickets have been bought. We can therefore simplify the code by using or as follows (we’ve left out the examples because they haven’t changed from the previous version):
fun buy-tickets4(count :: Number, is-senior :: Boolean)
-> Number:
doc:Compute the price of tickets at $10 each with
discount of 15% for more than 5 tickets
or being a seniorif (is-senior == true) or (count > 5):
count 10 0.85
else:
count 10
end
end
This code is much tighter, and all of the cases where the discount applies are described together in one place. There are still two small changes we want to make to really clean this up though.
Do Now!
Take a look at the expression is-senior == true. What will this evaluate to when the value of is-senior is true? What will it evaluate to when the value of is-senior is false?
Notice that the == true part is redundant. Since is-senior is already a boolean, we can check its value without using the == operator. Here’s the revised code:
fun buy-tickets5(count :: Number, is-senior :: Boolean)
-> Number:
doc:Compute the price of tickets at $10 each with
discount of 15% for more than 5 tickets
or being a seniorif is-senior or (count > 5):
count 10 0.85
else:
count 10
end
end
Notice the revised question in the if expression. As a general rule, your code should never include == true. You can always take that out and just use the expression you were comparing to true.
Do Now!
What do you write to eliminate == false? For example, what might you write instead of is-senior == false?
Finally, notice that we still have one repeated computation: the base cost of the tickets (count 10): if the ticket price changes, it would be better to have only one place to update that price. We can clean that up by first computing the base price, then applying the discount when appropriate:
fun buy-tickets6(count :: Number, is-senior :: Boolean)
-> Number:
doc:Compute the price of tickets at $10 each with
discount of 15% for more than 5 tickets
or being a seniorbase = count 10
if is-senior or (count > 5):
base 0.85
else:
base
end
end
6.8Recap: Booleans and Conditionals
With this chapter, our computations can produce different results in different situations. We ask questions using if-expressions, in which each question or check uses an operator that produces a boolean.
There are two Boolean values: true and false.
A simple kind of check (that produces a boolean) compares values for equality (==) or inequality(<>). Other operations that you know from math, like < and >=, also produce booleans.
We can build larger expressions that produce booleans from smaller ones using the operators and, or, not.
We can use if expressions to ask true/false questions within a computation, producing different results in each case.
We can nest conditionals inside one another if needed.
You never need to use == to compare a value to true or false: you can just write the value or expression on its own (perhaps with not to get the same computation).
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16121 | https://www.quora.com/What-is-the-formula-for-the-round-robin-format | Something went wrong. Wait a moment and try again.
Tournament Mode
Formula Name
Sports and Games
Draft Formats
Adult and Youth Round Rob...
Single Round Robin
5
What is the formula for the round-robin format?
2 Answers
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Raja Sekhar Telagathoty
Educator 117M + Content Views & 24.7M + Upvotes (2016–present)
·
Author has 16.7K answers and 95.4M answer views
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1y
26 October 2023
To determine the number of games for a single round robin tournament, as seen above, use the following formula,
Let us schedule a round robin format for 6 teams.
N x (N-1)/2.
With a tournament of 6 teams, the calculation would be :
6 x (6-1)/2 =
6 x 5/2 =
30/2 = 15 matches.
Total no of matches to be played : 15
Round Robin scheduling: Even number of teams.
Let N = number of teams in the tournament.
There will be N -1 rounds (each team will play N-1 games).
Since each team will play every other team once, no team will be idle during any of the rounds.
Raja Sekhar Telagathoty
26 October 2023
To determine the number of games for a single round robin tournament, as seen above, use the following formula,
Let us schedule a round robin format for 6 teams.
N x (N-1)/2.
With a tournament of 6 teams, the calculation would be :
6 x (6-1)/2 =
6 x 5/2 =
30/2 = 15 matches.
Total no of matches to be played : 15
Round Robin scheduling: Even number of teams.
Let N = number of teams in the tournament.
There will be N -1 rounds (each team will play N-1 games).
Since each team will play every other team once, no team will be idle during any of the rounds.
Raja Sekhar Telagathoty
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Related questions
How do you calculate the total number of outcomes in a double round-robin tournament with 4 teams (A,B, C,D) and no ties?
How many games are in a 4 team round robin?
What is the most effective 'round-robin' tournament format for all sports and what makes it the best?
What are the two types of round-robin tournaments?
How does a 5 team round robin work?
Assistant
Bot
·
1y
In a round-robin tournament format, each participant competes against every other participant. The formula for determining the total number of matches (or games) played in a round-robin tournament with n participants is:
Total Matches=n(n−1)2
Explanation:
Each participant plays against each of the other participants exactly once.
Since each match involves two participants, we divide the total number of pairings by 2.
If there are 4 teams in a tournament:
- The total number of matches would be:
Total Matches=4(4−1)2=4×32=6
Thus, there wou
In a round-robin tournament format, each participant competes against every other participant. The formula for determining the total number of matches (or games) played in a round-robin tournament with n participants is:
Total Matches=n(n−1)2
Explanation:
Each participant plays against each of the other participants exactly once.
Since each match involves two participants, we divide the total number of pairings by 2.
Example:
If there are 4 teams in a tournament:
- The total number of matches would be:
Total Matches=4(4−1)2=4×32=6
Thus, there would be 6 matches played in total.
Related questions
How do you calculate the total number of outcomes in a double round-robin tournament with 4 teams (A,B, C,D) and no ties?
How many games are in a 4 team round robin?
What is the most effective 'round-robin' tournament format for all sports and what makes it the best?
What are the two types of round-robin tournaments?
How does a 5 team round robin work?
How can I create a round robin tournament with 10 teams, each playing another team only once, only playing 5 rounds, and playing each of 5 games only once?
What is the other name of round robin tournament?
How is a round-robin tournament winner determined?
What is the origin of a round robin?
What is a round-robin tournament?
What is a 3-team round robin?
What are some of the best formats (e.g., round robin) you have come across in sports tournaments?
What are the disadvantages of a round-robin tournament?
How many games are played in a round-robin tournament?
What are the rules of a round-robin tournament?
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16122 | https://www.youtube.com/watch?v=NomGqYTnvyI | Chapter 1 Example Problem 1 | Weight and Volume | Engineering Fluid Mechanics
MajorMath
56 subscribers
4 likes
Description
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Posted: 10 Aug 2024
1.9) Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is 0.9 cm. The water density is 1.0 kg/L. The empty weight of the metal pipe is 2500 N/m. In kN, what is the total weight (pipe plus water)?
I will be solving this question from the textbook Engineering Fluid Mechanics by Donald F. Elger; Barbara A. LeBret; Clayton T. Crowe; John A. Roberson
Feel free to email majormathyt@gmail.com if you need help solving any fluid mechanic problems and I'll try my best to help!
Transcript:
what's up guys major math with another video on fluid mechanics if you have any questions drop it in the comments and if you're having trouble with other problems I've attached my email in the description so please feel free to reach out without further Ado let's get started question 1.9 in the textbook says water is flowing in a metal pipe the pipe outside diameter is 61 CM the pipe length is 120 M the pipe wealth thickness is 0.9 CM the water density is 1 kg per liter the empty weight of the metal pipe is 2,500 Newtons per meter in kons was the total weight Pipe Plus water um so with every single fluids question uh question and problem you're going to want to do write your goal write down your Givens and write down what you're assuming uh for the goal of this question you're trying to find the total weight so the weight of the pipe and the weight plus the water about this little formula so that's our goal our given we're given the outside diameter the length of the pipe the thickness of the pipe the density of the water inside the pipe and the pipe weight um with a distance per distance uh we're going to be assuming that the water is incompressible um all that means is that the density is just consistent throughout so wherever I take if I were to take a sample of water from anywhere in the pipe it's going to be that density of 1 kilogram per liter that's all that means um and then the other one is that this is gonna be important later to find the volume of the water inside the pipe we're we got to assume that the water fills the pipe completely there's no air pockets okay so the next step after we do this is figure out what equations we need so we're obviously going to need the weight equation because we have to find the weight of the water and the weight of the pipe so for this uh weight equals masstimes an accelerant in this case the only accelerant will be gravity so that weight equals m uh time mass time gravity we're also going to need to know the density formula um density we know the density formula is density equals mass over a volume I'm going to go ahead and rearrange this to mass equals volume time density because we're given the density the unknown here is mass right cuz we got to solve for this we want to solve for the weight of the water um we're going to have to solve for this Mass this is what's unknown here so how do we solve for the mass we got to use this density equation um so we got to solve for Mass so we know we know density it's already given and then we can solve for volume given using these variables outside diameter length of the pipe and thickness so this is what it look like um yeah so weight water equals mg right um we can also rearrange this because we know that mass is also equivalent from this density formula to volume time density so we're going to rewrite this as volume time density time gravity okay um so now we we know the density we know the gravity but we need to know what the volume of the water is so to find the volume of the water we're going to want to find the area of the the cross section of the pipe and then multiply by length of the pipe so if you think about it here's the pipe if you cut if you did a cross-section this is what the pipe would look like right because you got a thickness these your t's for thickness and then you got the outside diameter that goes across the whole entire pipe and then you got this inner diameter right for the smaller C right there um so if we want to find this is where the area the the water is going to be right this smaller pipe this smaller Circle um so if we want to find the volume of the water we can actually take the area of this circle right here where the water will fill it completely take get the area that and then multiply by the length of the pipe to get the whole volume of the water so the area of this Inner Circle would be pi over 4times the inner diameter squared um if it was uh if we were using the radius it would just be P pi r squ right or pi times the inner uh diameter or the inner radius squared but since we're using diameter we're going to we have to do pi over 4 and then we know that length I've written it up top is 120 M so to find the volume of the water inside the pipe we will do pi over 4 the inner diameter squared times the length but we don't have the inner di squar but we can solve that pretty easily here um so inner diameter will equal the outer diameter minus thickness twice right so you have this outer diameter that runs across and then if you minus the thickness on both sides you'll get this inner diameter um so that just be 61 CM - 2 0.9 C um that's going to give us 59.2 CM which if you convert that to meters be oh shoot 0.592 M right so now we have this inter diameter we can go ahead and solve this and this is going to equal 3303 M Cub right all right so we have the volume so now when we have the volume of the water we can solve for the weight of the water which is what we're going to do this whole time um let's go ahead and plug those numbers in so we got volume of water equals 33 I'm going to give myself more space right right then the density is one kilogram per liter and the gravity is 9.81 m/s squ okay but we got to convert we got to make these uh these numbers match up so if you look at the textbook and uh table F2 we know that 1 Newton equals kilogram time m/ second squar we also know that 1,000 l = 1 M cubed so using this we can uh help get our units right so we get the right number here um so we're going to multiply this by I'm multiplying this equation is what I'm doing to convert the units we'll do uh so the liters on bottom so we'll put liters on top here cancel those out we'll get one M Cub on the bottom [Music] times um we got to cancel out so we got put this on the bottom so that will cancel out this and cancel this right here then we got one newton um this meters Cub will uh cancel out and that will leave us with 324 about if I if you round Newtons um and then it says we the problem says it want kons so you just divide that by a th000 324 K okay so we found the weight of the water now we find the weight of the pipe so this one's pretty simple um just because of what the problem gives us uh it gives us that this 2500 Newtons per meter so for every meter of pipe it will weigh 2500 Newtons and we know that we have a length of 120 M so just doing 2500 Newtons per meter times times that 120 M will actually give us the weight of the pipe which is 300,000 Newtons but then if you divide it by a th000 will give us kiltons um and then from our goal from the equation we made we know that to find a weight total we have to add these together right so this is the weight of the pipe this is the weight of the water so 324 K plus 300 kons equals 624 K which equals the weight total and that is our answer thanks for watching soldiers don't forget to leave a like And subscribe for more content like this if you have any questions leave it in the comments and if you need help with any other problems feel free to contact me with the email shown on the screen |
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» Formulas
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#1 2006-04-22 18:32:15
Jai Ganesh
: Administrator
: Registered: 2005-06-28
: Posts: 51,869
Set Theory
Set:- A well defined collection of objects is called a set.
There is no repetition of elements in a set, each element appears only once.
Elements of a set:- The objects in a set are called its members or elements.
If an element a is prsent in setA, it is denoted by
If the element a is not present in set A, it is denoted by
Description of a set
I. Roster mehod:- In this method, a list of all the objects of the set is made and they are put within braces.
Example 1:- If A is the set of the first eight prime numbers,A={2, 3, 5, 7, 11, 13, 17, 19}
Example 2:- If B is the set of all the counting numbers between 20 and 30, B={21, 22, 23, 24, 25, 26, 27, 28, 29}
II. Set Builder form:- In this method, the properties common to all the elemnts of the set are listed.It is written in the form{x:x has properties P}
Example 1:- If B={3,5,7,9,11}, it is written asB={x:x=2n+1 where n∈N, n<6} orB={x|x=2n+1 where n∈N, n<6}
Singleton set:- A set consisting of a single elements is called a singleton set.Example :- A={2} or A={x:x is an even prime number}
Empty set:- A set which has no elements is called an empty set. It is represented by {} or
Finite and Infinite sets:- A finite set is one in which the number of elements is finite.Example:- A={x:x is a multiple of 5, x<1,000,000,000}
An infinite set is one in which the number of elements is not finite.Examples:- Set of all points on the arc of a circle, Set of all concentric circles with a given centre,{x|x∈Q, 0Equal sets:- Two sets A and B are equal if every element of set A is in set B and everyelement of set B is in set A. It is denoted byA=B.
Cardinal Number of a set:-The number of (distinct)elements in a set is called the Cardinal number of the set.If A is the set, n(A) is the caridanl number of set A.
Example:- If A={0,1,2,3,4,5,6,7,8,9}, n(A)=10
Equivalent sets:-Two sets are said to be equivalent if n(A)=n(B), that is, if their cardinal numbers are equal.
Remark:- Equal sets are always equivalent but equivalent sets are not always equal.Example:-if A={1,2,3,4,5}, B={a,b,c,d,e}A and B are equivalent sets because n(A)=n(B)=5, but A and B are not equal sets.
Subset:- Let A and B be two sets given in such a way that every element of A is in B, then it is said that A is a subset of B, written as
Superset:-If A is a subset of B, then B is a superset of A, denoted by
Proper subset:- If A is a subset of B and set A is not equal to set B, then A is called a proper subset of B, denoted by
Example, if A={1,2,3}, B={1,2,3,4,5,6,7}, then
Comparable sets:- Two sets A and B are comparable either ifA is a subset of B or B is a subset of A.
Properties of Subsets:-
{}, the empty set, is a subset of all sets.
Every set is a subset of itself.
The number of all subsets of a set containing n elements is
The number of all proper subsets of a set containing n elements is
The set of all subsets of a given set A is called the Power set of set A, denoted by P(A).If A has n elements, P(A) has
elements.
Operation on sets
Union of sets:- The union of two sets A and B is the set of all eleements in A or in B or in both A and B. It is denoted by
Example:- if A={0,2,4,6,8,10} and B={1,2,3,4,5},
Intersection of sets:- The intersection of two sets A and B is the set of elements common in A and B.
It is denoted by
If A={0,2,4,6,8,10} and B={1,2,3,4,5},
Difference of sets:-
The difference of two sets A and B is defined as the elements present in set A not present in set B.
It is denoted by A-B.
If A={0,2,4,6,8,10} and B={1,2,3,4,5},
A-B={0,6,8,10} and B-A={1,3,5}
It should be remembered that
Symmetric Difference of sets:- The symmetric difference of two sets A and B is defined as
Symmetric difference is commutative.
If A={0,2,4,6,8,10) and B={1,2,3,4,5},A-B={0,6,8,10}, B-A={1,3,5}
Universal Set:- A set which is a superset of all the given sets, denoted by U, is known as the Superset.
Complement of a Set:-If set A is the subset of Universal set U, then the complement of A, denoted by
is the set of all the elements in the Universal set not in A.
Example:-if U={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z} andA={a,e,i,o,u}, then
Important results on Complements
Laws of Operations
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#2 2006-04-22 19:24:45
Jai Ganesh
: Administrator
: Registered: 2005-06-28
: Posts: 51,869
Re: Set Theory
Commutative Laws:-
Associative Laws:-
Distributive Laws:-
De Morgan's Laws:-
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#3 2006-04-22 23:54:17
Jai Ganesh
: Administrator
: Registered: 2005-06-28
: Posts: 51,869
Re: Set Theory
Number of elements in a set
Example
Let
As per the formula,
Therefore, LHS=RHS.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#4 2006-04-23 00:53:41
Jai Ganesh
: Administrator
: Registered: 2005-06-28
: Posts: 51,869
Re: Set Theory
Some Important Results
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#5 2006-04-23 00:58:09
Jai Ganesh
: Administrator
: Registered: 2005-06-28
: Posts: 51,869
Re: Set Theory
Cartesian Product
The Cartesian Product of two sets A and B is defined to be the set of all ordered pairs with the first element in A and the second element in B. The Cartesian Product is denoted by
Example:- A={1,2,3}, B={a,b}
Imortant Results of Cartesian products
If A or B or both are infinite sets, A X B is also an infinite set.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#6 2006-11-24 02:28:50
knvsp123
: Member
: Registered: 2006-11-24
: Posts: 1
Re: Set Theory
how can we say minimum no of elements of AUB IF N(A)=8 AND N(B)=5
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#7 2007-02-21 12:20:49
Sekky
: Member
: Registered: 2007-01-12
: Posts: 181
Re: Set Theory
you said: "
denotes an element is not in the set"
this is wrong, the correct notation is
Last edited by Sekky (2007-02-21 12:28:13)
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#8 2007-02-21 16:45:33
Jai Ganesh
: Administrator
: Registered: 2005-06-28
: Posts: 51,869
Re: Set Theory
Hi Sekky,Welcome to the forum!Thanks for the post!I do agree, the symbol you have given is used to denote 'does not belong to'. But the one I had given is also used.Thanks again.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#9 2007-02-21 20:46:08
Sekky
: Member
: Registered: 2007-01-12
: Posts: 181
Re: Set Theory
ganesh wrote:
Hi Sekky,Welcome to the forum!Thanks for the post!I do agree, the symbol you have given is used to denote 'does not belong to'. But the one I had given is also used.Thanks again.
No, it isn't, it's used to denote containment, not lack thereof. It's simply back to front, equally as every single other binary relation can be written back to front, as in subset containment or have you ever heard of < and >? The symbol means the same, but the relation is reflexed.
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#10 2007-02-22 05:16:55
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
Sekky, this is a big world with many, many, mathematicians. Is it really that much of a surprise that two different mathematicians don't use exactly the same symbols?
Symbols are arbitrary. They can mean whatever you want them to. There are standards, but different people may conform to different standards.
Actually this list should really be in the algebra formulas thread, since a set is just a degenerate algebra.
Can you please define "degenerate algebra"? Wikipedia, MathWorld, and myself have no idea what you mean when you say that. Also, if you're going to go with that notion, then every thread in the formula section should be put into Sets, as sets are pretty much the basis of all math.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#11 2007-02-22 05:29:33
Sekky
: Member
: Registered: 2007-01-12
: Posts: 181
Re: Set Theory
Ricky wrote:
Sekky, this is a big world with many, many, mathematicians. Is it really that much of a surprise that two different mathematicians don't use exactly the same symbols?
Symbols are arbitrary. They can mean whatever you want them to. There are standards, but different people may conform to different standards.
Actually this list should really be in the algebra formulas thread, since a set is just a degenerate algebra.
Can you please define "degenerate algebra"? Wikipedia, MathWorld, and myself have no idea what you mean when you say that. Also, if you're going to go with that notion, then every thread in the formula section should be put into Sets, as sets are pretty much the basis of all math.
He's using the same symbol to denote both the relation and the opposite relation, it makes no sense. The subset relation means the same back to front providing the relatives are switched, as does the less than symbol, as do any mathematical relations.
Last edited by Sekky (2024-06-24 09:13:05)
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#12 2007-02-22 16:45:29
Jai Ganesh
: Administrator
: Registered: 2005-06-28
: Posts: 51,869
Re: Set Theory
Sekky,The formulas section is intended to help the members and all the students in general. The beginners identify set theory as a separate branch, when these threads werte created, the idea was to put all formulas useful to students. Whether a thread was a subset of one or superset was not a point of concern.Thanks for your ideas, and your contributions!Ricky and me were posting the formulas, and we wanted to create a databse. I think we have done fairly well in our endeavor!
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#13 2007-02-22 17:05:23
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
Morphisms, Operators, Relations ect, are not constructed from set algebras
A morphism is a mapping. A mapping is simply a special subset of AxB, where the mapping would be from A to B.
An operator is a mapping from AxA->A, which is in turn a subset of (AxA)xA.
A relation is again a subset of AxB.
He's using the same symbol to denote both the relation and the opposite relation, it makes no sense.
That's like saying:
and
Are the same symbol. I mean, they are just flipped around, right?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#14 2007-02-23 00:24:00
MathsIsFun
: Administrator
: Registered: 2005-01-21
: Posts: 7,713
Re: Set Theory
In my experience there are alternative notations (which often lead to confusion, but that is how the world is).
Perhaps we could include a note to that effect?
And then, sorry to say, we will need to clean up the discussion according to
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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#15 2007-02-23 03:29:56
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
Those are binary operators, which are commutative and therefore denoted such that it makes no different if you flip them horizontally because the operands won't need to change. we're talking about non-symmetric relations here.
Whats a binary operator? Intersection and union or inclusion? Cause neither are. A binary operator is a mapping from AxA to A.
And then, sorry to say, we will need to clean up the discussion according to
I will be doing so once this is resolved.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#16 2007-02-23 12:18:58
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
Given any set, both intersection and union map P|A| x P|A| -> P|A|
Union an intersection each take two different sets. Not the same set, though they can be.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#17 2007-02-23 12:23:28
Sekky
: Member
: Registered: 2007-01-12
: Posts: 181
Re: Set Theory
Ricky wrote:
Given any set, both intersection and union map P|A| x P|A| -> P|A|
Union an intersection each take two different sets. Not the same set, though they can be.
So you're saying any binary operator can only ever take the same element for it's operands? 2 + 3 = 5 last time I heard.
The mapping maps all elements of the set cart set to another element of the set, and they will forever be included in the power set of any given set, thusly.
Let A = {1,2,3}
P|A| = {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
Now take the union and intersection of all possible cartesian tuples of the power set, and the yield will ALWAYS be an element of the power set, hence P|A| x P|A| -> P|A|
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#18 2007-02-23 12:56:27
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
So you're saying any binary operator can only ever take the same element for it's operands? 2 + 3 = 5 last time I heard.
No, but the elements must belong to the same set. In other words, the integers.
Let A = {1,2,3}
P|A| = {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
Now take the union and intersection of all possible cartesian tuples of the power set, and the yield will ALWAYS be an element of the power set, hence P|A| x P|A| -> P|A|
Yes, but this is restricting the definition of union because I can take:
B = {4, 5, 6}
And do A U B.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#19 2007-02-23 13:06:49
Sekky
: Member
: Registered: 2007-01-12
: Posts: 181
Re: Set Theory
hence A and B are both elements of P|{1,2,3,4,5,6}| and no generality is lost, Z x Z -> Z necessarily for some Z = P|{1,2,3,4,5,6}|
Operators are associated with algebras to form more composite algebras, if I take a set, I can find an operator to form an algebra. Likewise if I find an operator and element I wish to perform, I can find a set containing them such that the operator will hold an algebra.
Last edited by Sekky (2007-02-23 13:10:43)
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#20 2007-02-23 13:15:40
Sekky
: Member
: Registered: 2007-01-12
: Posts: 181
Re: Set Theory
or better, why not just consider the power set of the universal set under both operators, and I believe that forms a boolean lattice.
Well, any set under both operators will be a boolean lattice, I believe, unless you can think of an axiom that fails.
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#21 2007-02-23 13:18:27
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
hence A and B are both elements of P|{1,2,3,4,5,6}|
Ah, I see. So now we're defining union by taking the union of two sets? And you don't see the problem in that?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#22 2007-02-23 13:32:17
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
And there in lies the problem. You are operating under what is called (I swear, this is the actual term, I'm not trying to be insulting), "naive set theory". It is the concept that any collection can be a set. Naive set theory is called so because it is, well, naive. It leads to contradictions and paradoxes.
On the other hand, I am going off of ZF set theory, which is the just about the standard when it comes to set theory in mathematics (although there are others). In ZF set theory, you can't just claim something is a set. The set must be constructed through the use of existing sets. The only set that is guaranteed to exist is the null set (which I will be denoting θ simply because of laziness). But through precise axioms, we can take the null set and form sets such as {θ} and {{θ}} or {θ, {{θ}}}. In ZF set theory, the Axiom of Union is used to define what we typically think of when we see union (though said in a completely different form).
So you need union to state that your set {1, 2, 3, 4, 5, 6} exists before you can ever even talk about it being a set. And this is the reason why we can never have union be an operator. Because the sets need to exist before we can call it an operator, but we need union to exist before we can say the sets exist.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#23 2007-02-23 16:45:14
MathsIsFun
: Administrator
: Registered: 2005-01-21
: Posts: 7,713
Re: Set Theory
Guys, this is such a great discussion, but it is in the Formulas section, so due for deletion at some point!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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#24 2007-02-23 18:32:13
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
Sekky, we're starting to beat around the bush here. Please come up with a binary operator definition for the union of two (possibly different) sets.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#25 2007-02-26 04:05:58
Ricky
: Moderator
: Registered: 2005-12-04
: Posts: 3,791
Re: Set Theory
I'm sorry, but that is not a binary operation. A binary operation is a mapping from AxA to A.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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16124 | https://medium.com/@vitoriodachef/validating-dates-with-symfony-8556468243f8 | Validating Dates With Symfony. Symfony has a quite powerful Validator… | by Victor Todoran | Medium
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Validating Dates With Symfony
Victor Todoran
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Feb 19, 2023
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Symfony has a quite powerful Validator Component which comes in handy in a myriad of situations. Lately I’ve been working a lot with Dates and Periods which need to be validated.
Symfony has some Date Constraints (things it can validate out of the box) but most of them are useful to validate if something is a valid date/date format or not.
I’ve noticed that people, which have not worked with Symfony in the past, quickly peruse the Date Constraints section of the documentation and then they start implementing their own Validators.
That is because theydon’t need to check whether something is a valid Date/Date format or not, they needto check if an endDate is greater than a startDate, if a Date is in the future or in the past and so on and so forth.
What people seem to overlook is that some of the Comparison Constraints also work with Dates.
Using these Comparison Constraints with Dates is something that we are going to explore today.
Let us imagine we have a very simple InvoicePeriod class
class InvoicePeriod
{
public readonly DateTimeImmutable $startDate;
public readonly DateTimeImmutable $endDate;
public function __construct(
DateTimeImmutable $startDate,
DateTimeImmutable $endDate
) {
$this->startDate = $startDate;
$this->endDate = $endDate;
}
}
One of the things we might want to validate is that both startDate and endDate are not future Dates. That is because in our fictional application you can not invoice a future period.
Let’s quickly run a failing test before we configure any Validation.
namespace App\Tests\Integration\DTO;
use App\DTO\InvoicePeriod;
use Symfony\Bundle\FrameworkBundle\Test\KernelTestCase;
use Symfony\Component\Validator\Validator\ValidatorInterface;
class InvoicePeriodTest extends KernelTestCase
{
public function testFutureInvoicePeriodIsInvalid(): void
{
$validator = self::getContainer()->get(ValidatorInterface::class);
$futureInvoicePeriod = new InvoicePeriod(
new \DateTimeImmutable('2030-01-01'),
new \DateTimeImmutable('2030-01-01')
);
$constraintViolationList = $validator->validate($futureInvoicePeriod);
$this->assertGreaterThan(
0,
$constraintViolationList->count()
);
}
}
Now we will configure a LessThanOrEqualConstraint to make that test pass. Supposing we use PHP attributes to configure this Constraint, our InvoicePeriod will look something like this:
use Symfony\Component\Validator\Constraints as Assert;
class InvoicePeriod
{
#[Assert\LessThanOrEqual('today')]
public readonly DateTimeImmutable $startDate;
#[Assert\LessThanOrEqual('today')]
public readonly DateTimeImmutable $endDate;
... unchanged ...
}
If you run the test again, it should pass. If it does not, try clearing the cache, validation configurations are cached.
The business logic of our fictional application also requires that the Invoice.startDate can not be greater than the Invoice.endDate.
For this we will use the GreaterThanOrEqualConstraint . Because we want to validate a property in relation to another we will also leverage the configuration called property path.
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But first let’s write a failing test:
class InvoicePeriodTest extends KernelTestCase
{
public function testInvoiceWithStartDateGreaterThanEndDateIsInvalid(): void
{
$validator = self::getContainer()->get(ValidatorInterface::class);
$invoicePeriod = new InvoicePeriod(
new \DateTimeImmutable('2023-01-31'),
new \DateTimeImmutable('2023-01-01')
);
$constraintViolationList = $validator->validate($invoicePeriod);
$this->assertGreaterThan(
0,
$constraintViolationList->count()
);
}
}
Now let us configure the new Constraint on the InvoicePeriod:
class InvoicePeriod
{
... unchanged ...
#[Assert\GreaterThanOrEqual(propertyPath: 'startDate')]
public readonly DateTimeImmutable $endDate;
... unchanged ...
}
If you run the test again, it should pass.
Before we wrap this up let’s consider another case.
At the moment I’m writing this it is 2023–02–19 14:45.
Consider the following Invoice Period and our configured constraints, should it fail or not?
$invoicePeriod = new InvoicePeriod(
new \DateTimeImmutable('2023-01-01 00:00:00'),
new \DateTimeImmutable('2023-02-19 20:45:00')
);
The endDate is the same as today, but the hour is six hours into the future. So, endDate is into the future, and the validation should and will fail.
Try plugging this data into InvoicePeriodTest::testFutureInvoicePeriodIsInvalid and the test will pass.
But business logic is messy. The dates might be coming from outside the app and we are not allowed to change them.
So Product comes and tells you that as long as the endDate is not into the future, the time is irrelevant, the InvoicePeriod must be considered valid.
First let us write a failing test for that.
class InvoicePeriodTest extends KernelTestCase
{
public function testTimeIsIgnored(): void
{
$validator = self::getContainer()->get(ValidatorInterface::class);
$startDate = (new \DateTimeImmutable())->setTime(0, 0);
$endDate = (new \DateTimeImmutable())->setTime(23, 59, 59);
$invoicePeriod = new InvoicePeriod($startDate, $endDate);
$constraintViolationList = $validator->validate($invoicePeriod);
$this->assertCount(0, $constraintViolationList);
}
}
Let us revisit the InvoicePeriod class
class InvoicePeriod
{
#[Assert\LessThanOrEqual('today')]
public readonly DateTimeImmutable $startDate;
#[Assert\LessThanOrEqual('today')]
#[Assert\GreaterThanOrEqual(propertyPath: 'startDate')]
public readonly DateTimeImmutable $endDate;
... constructor ...
}
When configuring the Constraint, we can pass any date string accepted by the DateTime constructor (e.g. today). Which means we can also use relative or time formats.
Let’s configure the LessThanOrEqual Constraint of the endDate as follows:
... unchanged ...
[Assert\LessThanOrEqual('today 23:59:59')]
[Assert\GreaterThanOrEqual(propertyPath: 'startDate')]
public readonly DateTimeImmutable $endDate;
... unchanged ...
If you run InvoicePeriodTest::testTimeIsIgnored it should now pass.
This only scratches the surface of what is possible with Dates and Comparison Constraints .
For example you could validate that a Date is in a given month by leveraging the Range Constraint and relative formats accepted by the DateTime constructor.
class PropertyTaxPayment
{
#[Assert\Range(
min: 'first day of January',
max: 'last day of January',
)]
private $paidAt;
}
Hopefully, if you are using Symfony, I’ve convinced you to take a hard look at the Validator Component before implementing some DateTime validation logic inside your App.
That is it! Thanks for reading!
Disclaimer: Consider this to be a living document, which means it’s subject to undocumented changes and it might even die in the future.
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Aurélien Tournayre
Mar 5, 2023
I didn't think to use validation for dates this way, it's a great approach, I would use it.
On the other hand, in this particular case (using constructor), I would have use assertions to validate the InvoicePeriod but it's not the topic!
--
1 reply
Reply
Calin Bolea
Aug 7, 2023
Great use of the GreaterThan{OrEqual} constraint! Would've loved to use it, but I've hit a case where the dates are string properties and timezones are different properties of the same object. The actual DateTime objects were instantiated in the…more
--
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16125 | https://en.front-sci.com/index.php/rerr/article/view/2980/3202 | 194
Region - Educational Research and Reviews , 20 24 , Volume6, Issue9
DOI: 10.32629/rerr.v 6i9. 2980
ISSN Online: 2661-4634
ISSN Print: 2661-4626
A pragmatic function study of the discourse
marker "I mean" in Friends
Xinyue YI
Applied Linguistics, Xi'an International Studies University, Xi 'an 710128, China
Abstract: This paper examines the discourse marker "I mean " in the first season of Friends , exploring its pragmatic
functions within dialogues. Analysis reveals that "I mean " facilitates politeness, moderates tone, and enhances
identification, while also serving functions such as clarification, correction, emphasis, and supplementation. These
functions support coherence and interaction, underscoring the marker 's role in promoting dialogue fluency. The findings
highlight how "I mean " contributes to character interactions and provide insights for learners on the pragmatic use of
discourse markers in communication.
Key words: Friends ; discourse markers; "I mean "
1 Introduction
Discourse markers can not only organize the dialogue and make it more coherent, but also convey the speaker 's
attitude, emotion or position, so that the listener can quickly understand the speaker's underlying meanings. As a classic
American sitcom, Friends is popular for its realistic and natural dialogues, and has attracted countless scholars to study its
language usage. In the show, the characters not only organize their conversations with authentic and coherent sentences or
idioms, but also frequently use a variety of discourse markers to facilitate communication, among which "I mean" is one of
the most common markers. In this paper, we will analyze the use of "I mean" in the first season of "Friends" to explore its
pragmatic function and its role in different contexts. By analyzing specific examples, the paper will reveal how "I mean"
helps the characters to make self-corrections, take positions or maintain interactions in the dialogues.
2 Discourse markers
In daily communication and usage, people often define discourse markers as words or phrases that play a connecting
and regulating role between or within sentences. They are not attached to any syntactic and semantic structure, but play an
important role in discourse coherence, attitude expression and interpersonal interaction, and help to form a coherent and
organized discourse . Since the 1970s, people began to devote themselves to the pragmatic study of discourse markers
(e.g. van Dijk 1979), and since then, discourse markers have gradually become one of the hot issues studied by domestic
and foreign academics. Levinson, a pragmatist, once said that "English, and undoubtedly most languages, have a lot of
words and phrases that mark a certain relationship between a certain discourse and the previous one, such as but, therefore,
in conclusion, however, anyway, well, actually, fortunately, so, you know, I mean, that is to say, in other words, you see
and so on ." Domestic scholar Ran Yongping reviewed the pragmatics of discourse markers in 2000, outlining the
Copyright ©20 24 by author(s) and Frontier Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
195
research status and trends at that time, contributing to the subsequent development and transformation of discourse
markers.
3 Pragmatic functions of the discourse marker "I mean" in Friends
As a discourse marker, "I mean" has the main functions of clarifying, correcting, emphasizing, supplementing,
organizing and expressing emotions . In the following, the basic pragmatic functions of "I mean" are analyzed in detail
with the specific example of Friends (Season 1).
3.1 Clarification
As a discourse marker, "I mean " is often used to clarify or explain the speaker 's point of view or position to help the
listener better understand what has been said before.
Example 1:
Phoebe: No, I don 't think this was your shot.
Phoebe: I mean, I don 't even think you just get one shot.
In Episode 6, when Joey feels he missed a valuable audition chance, Phoebe reassures him by saying "I mean, I don 't
even think you get just one shot. " Using "I mean ", she clarifies her optimism, emphasizing that Joey will have more
opportunities ahead. This reflects a positive outlook on career growth, underscoring the need to stay receptive to future
possibilities in competitive environments.
Example 2:
Ross: Rach, do you uh, are you gonna separate those?
Rachel: Oh God. Oh, am I being like a total laundry spaz? I mean, am I supposed to use like one machine for shirts
and another machine for pants?
In Episode 5, Rachel jokes about being a "laundry spaz " due to her inexperience. She uses "I mean " as a discourse
marker to clarify this, followed by questions that further explain her confusion about basic laundry principles. This usage
refines her statement, making her self-description more coherent and plausible.
3.2 Correction
"I mean " is used to correct previous statements, allowing the speaker to rephrase their message for accuracy.
Example 3:
Phoebe: Um, Monica, you 're scaring me.
Phoebe: I mean, you 're like, you 're like all chaotic and twirly.
In Episode 2, after seeing Monica 's anxious reaction to her parents ' arrival, Phoebe initially says "you 're scaring me ".
Realizing this is too direct, she correct s it with "I mean " to "you 're like all chaotic and twirly ". This euphemistic expression
provides a more accurate depiction of Monica 's behavior, shifting from emotion to description.
Example 4:
Ross: Helen Geller? I don 't think so.
Carol: It 's not gonna be Helen Geller.
Ross: Thank you!
Carol: No, I mean it 's not Geller.
In Episode 2, Ross accompanies his ex-wife Carol to a maternity checkup, where they discuss the baby 's names.
When Carol says the child won 't be named "Helen Geller ", Ross mistakenly thinks he has won the argument. Carol
clarifies with "No, I mean it 's not Geller ", correcting his misinterpretation about the last name. This exchange highlights a
semantic correction and reveals a deeper issue of naming rights, illustrating a father 's loss of naming authority after divorce. 196
3.3 Emphasis
"I mean " can also be used to emphasize a certain point or feeling, to make the speaker 's attitude clearer, and to
highlight the speaker 's opinion.
Example 5:
Chandler: Thank God you didn 't try to fan out the magazines.
Chandler: I mean, she 'll scratch your eyes right out.
In Episode 6, Rachel moves a green soft chair without Monica 's permission, raising concerns about Monica 's reaction.
When Chandler sees that Rachel hasn 't moved the magazines, he warns her, saying "I mean, she 'll scratch your eyes right
out. " This use of "I mean " directs Rachel 's attention and emphasizes the hyperbolic seriousness of the warning. Fortunately,
the situation doesn 't escalate. This discourse marker not only underscores the potential consequences but also enhances the
emotional intensity of the scene, making the dialogue more vivid and effective.
Example 6:
Rachel: Mrs. Bing, I have to tell you, I've read everything you 've ever written. No, I mean it! I mean, when I read
Euphoria at midnight, all I wanted to do was become a writer.
In Episode 11, Rachel expresses her love of Mrs. Bing 's books, making it clear that she has read all of Mrs. Bing 's
works. In this exchange, Rachel emphasizes her sincerity and determination by using the phrase "I mean it ". Subsequently,
Rachel uses "I mean " again to further elaborate her feelings, especially in reference to her experience of reading Euphoria
late at night. This expression not only deepens the credibility of the aforementioned praise, but also clearly demonstrates
the profound influence of the work on Rachel 's personal pursuits.
3.4 Supplementary
"I mean " is sometimes used to add additional information or ideas to help the speaker provide more details or
background on the original topic, making the conversation richer and the message more complete.
Example 7:
Monica: If you had to do it all over again, I mean, if she was here right now. Would you tell her?
Monica 's mother: Tell her what?
Monica: How she drove you crazy, picking on every little detail, like you hair... for example.
In Episode 8, Monica 's mother reminisces about some of the behaviors that her grandmother did when she was alive
that drove her crazy, and Monica uses this as an opportunity to ask a hypothetical question: if she could relive the past and
her grandmother was still alive, would her mother be honest with her grandmother about how she really felt? In effect,
through this hypothesis, Monica attempts to engage in an open and honest dialogue with her mother in order to express that
she is currently experiencing some similar distress. In this dialogue, "I mean " serves as an additional illustrative discourse
marker, leading to the hypothetical condition "if she was here right now ", which further refines the previous expression "all
over again ". This not only provides richer contextual information about Monica 's intentions and conveys her feelings and
thoughts more effectively, but also enhances the logical coherence and semantic integrity of the sentence.
Example 8:
Ross: It was the most elaborate filth you have ever heard.
Ross: I mean, there were characters, plot lines, themes, a motif.
In Episode 15, when Joey inquires about the progress of Ross 's relationship with Celia, Ross states that it is going
surprisingly well. He further explains that the reason it is going so well is because he is good at telling "dirty stories " and
believes that his stories are very clever and complex. The subsequent use of "I mean " to elaborate and supplement "the 197
most elaborate filth " by listing the elements of character, plot, theme, and motif in detail allows the listener to more clearly
understand what he means by "elaborate ".
3.5 Organizing discourse
In real communication, discourse is often not organized in advance. As the context changes or the message is different,
the speaker may not be able to organize what comes next, which may lead to hesitation or interruption. In this case, the
discourse marker "I mean " can be used as an organizing tool to help the speaker to introduce the next topic, so as to
express the intention more smoothly and avoid long pauses that affect normal communication.
Example 9:
Monica: There 's something that you should know.
Monica: I mean, there 's really no easy way to say this.
In Episode 3, Monica is having trouble telling her friends about her decision to break up with Alan. After saying
"There 's something that you should know " and realizing that she needs to find an appropriate way to convey the news of
the breakup to her friends . Monica uses "I mean " as a discourse organization device to play its discourse articulation role,
reflecting her internal hesitation and struggle, and at the same time obtaining a temporary pause for further thinking and
organization of the next words, thus making the speech more coherent.
Example 10:
Ross: OK, where am I talking to, here?
Ross: I mean, uh, well, there is one way that seems to offer a certain acoustical advantage, but ......
In Episode 9, Ross tries to have a conversation with his unborn child, but being new to fatherhood, he doesn 't know
how to communicate with the fetus through his ex-wife 's abdomen, so he asks "OK, where am I talking to, here? "
Subsequently, Ross uses "I mean " to guide his further explanation of the previous query. However, Ross makes frequent
use of discourse markers such as "I mean ", "uh " and "well " in this explanatory discourse, which provides him with extra
time to think, adjust his thoughts, and organize his language, avoid ing sudden interruptions and maintain ing the coherence
and logic of the dialogue.
6 Emotional expression
As a phrase expressed from the first-person perspective, "I mean " has a distinctly personal flavor, so it can convey or
enhance the speaker 's emotional color in different communicative contexts, and thus facilitate the expression of emotion.
Example 11:
Monica: Look at that face.
Monica: I mean, even sleeping, he looks smart.
In Episode 11, Monica and Phoebe are visiting a strange man who has been injured in a car accident. Monica looks at
that face and says "Look at that face. I mean, even sleeping, he looks smart ". In this context, "I mean " serves as a tool for
emotional expression, which not only effectively deepens Monica 's praise for the man 's appearance, but also deeply
conveys her positive feelings towards him, reflecting Monica 's sincerity and honesty, and increasing the credibility and
infectiousness of the dialog.
Example 12:
Rachel: Chandler, I gotta tell you, I love your mom 's books! I love her books! I can 't get on a plane without one. I
mean, this is so cool.
In Episode 11, Rachel emphasizes her love of Chandler 's mom 's books by repeating "I love her books ". The hyperbole
of "I can 't get on a plane without one " effectively conveys Rachel 's recognition and appreciation of the books, and then she 198
further deepens her emotional attitude towards Chandler 's mother 's books by saying "I mean, this is so cool ". This
expression not only complements the previous praise, but also strengthens the expression of emotion, making her love for
the books more real and significant.
4 The social function of the discourse marker "I mean " in Friends
The discourse marker "I mean " also has important social functions, especially in maintaining politeness, moderating
the tone and guiding identification. It provides a tactful way for speakers to avoid direct criticism when discussing
sensitive topics, maintains a friendly atmosphere, and reduces the other party 's defenses. In addition, the use of "I mean "
can lead the listener to agree with the point of view and enhance his/her acceptance, helping to establish more harmonious
communication and understanding.
4.1 Maintaining politeness
"I mean " helps speakers in social interactions to maintain politeness and ensure s that the conversation flows smoothly
to achieve the desired social effect.
Example 13:
Phoebe: I don 't know what you just said, so let 's get started.
Paolo: Uh, I am, uh, being naked?
Phoebe: Um, that 's really your decision, I mean, some people prefer, you know, to take off...oh whoops! You 're being
naked!
In Episode 12, when confronted by Paolo with the question of whether or not a massage requires full nudity, Phoebe
initially replies that the decision is up to Paolo 's personal choice. However, she then realizes that her answer may be too
direct and reorganizes the expression using "I mean " to make the subsequent explanation seem more polite and smooth,
showing respect for personal preference, avoiding direct offense, mitigating potential awkwardness in the exchange, and
enhancing the politeness of the conversation.
4.2 Moderating the tone
When expressing a personal opinion or making a criticism, a direct statement may seem too blunt or forceful. The use
of "I mean " allows for a more tactful tone, which reduces the likelihood of offending or starting an argument.
Example 14:
Ross: Yeah, who wants fair?
Ross: I mean, I just want things back. Y' know, the way they were.
In Episode 3, when Monica tells her friends that she 's decided to break up with Alan and they all show great
disapproval . Monica jokes that she can keep pretending to go out with Alan in order to maintain everyone 's relationship.
The friends agree, but Monica feels that it wouldn 't be fair to everyone, and Ross emotionally interject s, "Who wants fair ?"
Realizing that his tone was too intense, Ross soften s it with "I mean " and explain s that he just want s things to go back to
the way they used to be and everyone to get along like they used to.
4.3 Guiding identification
"I mean " can also guide the listener 's attention, increase the listener 's identification with a particular point of view, and
lead to the establishment of a mutual understanding.
Example 15:
Joey: Well, if she 's my friend, hopefully she 'll understand. I mean, wouldn 't you guys?
In Episode 17, when Joey plans to celebrate Ursula 's birthday with her, Ross poses a question: if Joey chooses to
celebrate with Ursula, what should be done about Phoebe 's birthday? Because both birthdays happen to fall on the same 199
night. In the end, Joey decides to stay with Ursula to avoid disappointing her, and says that if Phoebe is really his good
friend, she will understand his choice. Joey then attempts to get Ross and Candler to agree and empathize with his point of
view by saying "I mean ".
5 Conclusion
This paper analyzes the discourse function of the discourse marker "I mean " in the American comedy Friends , and
emphasizes the important role of "I mean " in clarifying, correcting, emphasizing, supplementing information and
organizing discourse. In addition, the marker has a clear potential for politeness, moderating the tone, and guiding
identification in social interactions. Together, these functions contribute to the harmony and effectiveness of conversations,
enabling participants to achieve better communication in complex social environments. Through an in-depth discussion of
discourse markers, this paper provides concrete practical guidance for learners to understand and utilize discourse markers,
highlighting their importance in language learning and communication.
Conflicts of interest
The author declares no conflicts of interest regarding the publication of this paper .
References
Levinson S. 1983 . Pragmatics . Cambridge: Cambridge University Press, 85-94.
Ran Y-P. 2003 . A pragmatic account of the discourse marker well. Journal of Foreign Languages , 3: 58-64.
Fraser B. 2009 . An account of discourse markers. International Review of Pragmatics , 1(2):1-28. |
16126 | https://www.teacherspayteachers.com/Product/Exterior-Angles-and-Triangle-Sum-Theorem-Self-Checking-Circuit-Angles-Worksheet-7623682 | Exterior Angles and Triangle Sum Theorem Self Checking Circuit Angles Worksheet
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Key Geometry Skills Covered:
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Solving for unknown angles in triangles
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16127 | https://www.kristakingmath.com/blog/rolles-theorem | Understanding Rolle's Theorem
What does Rolle’s Theorem say?
Rolle’s Theorem is a specific instance of the Mean Value Theorem, in which the endpoints of the function at the edges of the interval are equal to one another.
In the Mean Value Theorem lesson, we looked at a function in an interval, and the line that connected the endpoints was slanted.
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But to apply Rolle’s Theorem, the values of the function at the endpoints must be equal, which means the line that connects them will be perfectly horizontal.
Just like the Mean Value Theorem, Rolle’s Theorem tells us that, as long as the function is continuous (unbroken) and differentiable (smooth) inside the interval, then there must be a tangent line that’s parallel to the horizontal line that connects the endpoints.
Here’s where that tangent line exists for this particular function we’ve been using:
Notice how it’s parallel to the line that connects the endpoints. This is what Rolle’s Theorem guarantees: a tangent line, somewhere inside the interval, that’s parallel to the line that connects the endpoints.
Because the line that connects the endpoints is horizontal, the tangent line will also be horizontal. Remember that horizontal tangent lines exist wherever the derivative is equal to ???0???, so Rolle’s Theorem can prove all of the following:
The existence of a horizontal tangent line in the interval
A point at which the derivative is ???0??? in the interval
The existence of a critical point in the interval
A point at which the function changes direction in the interval, either from increasing to decreasing, or from decreasing to increasing
The point at which that tangent line intersects the curve (the point of tangency) will exist at ???x=c???. We call the interval ???[a,b]???, so ???c??? will fall between ???a??? and ???b???, ???a<c<b???.
We can prove Rolle’s Theorem if we start from the Mean Value Theorem. The Mean Value Theorem says
???f'(c)=\frac{f(b)-f(a)}{b-a}???
But if the endpoints are equal, then ???f(a)=f(b)???, so we could make a substitution, and get
???f'(c)=\frac{f(b)-f(b)}{b-a}???
???f'(c)=\frac{0}{b-a}???
???f'(c)=0???
This result tells us that, by assuming the endpoints were equal, we found a value of ???0??? for the derivative at ???x=c??? in the interval, which is exactly what Rolle’s Theorem states.
How to use Rolle’s Theorem to prove the existence of roots in the interval
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Finding roots with Rolle’s Theorem
Use Rolle’s Theorem to show that the function has a critical point in the interval ???[0,2]???.
???f(x)=-\frac12x^4+2x^2???
A polynomial function like this one will be continuous and differentiable everywhere in its domain. Then all we need to do is check to make sure that the function’s value is equal at the endpoints of the interval, ???x=0??? and ???x=2???.
???f(0)=-\frac12(0)^4+2(0)^2???
???f(0)=0???
and
???f(2)=-\frac12(2)^4+2(2)^2???
???f(2)=-\frac12(16)+2(4)???
???f(2)=-8+8???
???f(2)=0???
A polynomial function like this one will be continuous and differentiable everywhere in its domain. Then all we need to do is check to make sure that the function’s value is equal at the endpoints of the interval.
The value of the function is ???0??? at both endpoints. It’s not necessary, but we could verify this visually if we wanted to double-check ourselves.
Because the endpoints of the interval are equal to one another, Rolle’s Theorem tells us that there must be a critical point somewhere in the interval.
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16128 | https://math.stackexchange.com/questions/1313195/the-equation-of-a-plane-perpendicular-to-another-plane | Stack Exchange Network
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The equation of a plane perpendicular to another plane
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Suppose that I want to find the equation of the plane that is perpendicular to the plane $8x-2y+6z=1$ and passes through the points $P_1(-1,2,5)$ and $P_2(2,1,4)$.
I can think of two methods to find the normal vector to the plane.
Method 1: Since the plane is orthogonal to $8x-2y+6z=1$, then the normal vector of the plane should be orthogonal to $(8,-2,6)$. So one normal vector would be $(1,1,-1)$.
Method 2: Find the cross product of $(8,-2,6)$ and $\mathbf{P}_1\mathbf{P_2}$.
I think there is something wrong with method 1 but I can not spot it.
Could you please clarify this for me?
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asked Jun 5, 2015 at 11:29
user77791user77791
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$\begingroup$ In method 1, you find the normal of the vector and then you need to find $d$.You need only one point to do that (i.e $P1$ or $P2$), so if they gave you two points, then there is probably something wrong there. (Usually, there is no useless or unused data in math questions...) $\endgroup$
Eminem
– Eminem
2015-06-05 11:32:07 +00:00
Commented Jun 5, 2015 at 11:32
$\begingroup$ @Eminem there are an infinite number of planes through one point which are perpendicular to a given plane (rotate about the point with axis normal to the plane). $\endgroup$
Paul
– Paul
2015-06-05 11:42:13 +00:00
Commented Jun 5, 2015 at 11:42
$\begingroup$ Method 1 gives you the direction of a line in your original plane. There is no reason why this direction should be orthogonal to the plane you require. Method 2 will work though as you now have 2 directions in the plane you have to find. $\endgroup$
Paul
– Paul
2015-06-05 11:49:17 +00:00
Commented Jun 5, 2015 at 11:49
$\begingroup$ @Paul But if you have the normal of the plane, you have to find $d$ to complete the task. So here, if the normal is $(1,1,-1)$ than the equation is $x+y-z+d=0$. Then you put $P1$ or $P2$ there to find $d$, and that is it, right? $\endgroup$
Eminem
– Eminem
2015-06-05 14:19:20 +00:00
Commented Jun 5, 2015 at 14:19
$\begingroup$ @Eminem (1, 1, -1) is just one direction in the original plane. It is a normal to some plane alright, but there is no reason to suppose that it is normal to the plane through $P_1$ and $P_2$. Infact, it can't be, as those two points give different values for d. $\endgroup$
Paul
– Paul
2015-06-05 18:51:00 +00:00
Commented Jun 5, 2015 at 18:51
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The method (2) is correct since the normal vector to the searched plane have to be orthogonal to the vector $P_1-P_2= (-3,1,1)$ and to the normal vector to the given plane $ (8,-2,6)$.
For the method (1) note that an orthogonal vector to $ (8,-2,6)$ has components such that: $(8,-2,6)(x,y,z)^T=0$ so you can fix only one component from this equation and the other two can be fixed imposing that the plane pass through the two given points.
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answered Jun 5, 2015 at 11:57
Emilio NovatiEmilio Novati
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16129 | https://www.youtube.com/watch?v=nb7bY_1yACI | Binomial Theorem & Expansion Formula & Questions | JEE Main & Class 11 Maths Chapter 8 | Byju's
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A JEE Maths numerical from Binomial Theorem & expansion of class 11 was provided along with 4 incorrect solutions. The errors in the given solutions as well as the correct solution will be discussed in this video. Stay tuned to sharpen your own problem-solving skills and also clear your Misconception with this session.
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झाल हेलो हेलो एवरीवन क्या हाल है बच्चा लोग सोम आज हम लोग क्या कर रहे हैं आज हम लोग डिटेक्ट एंड क्लियर मिड टाउन उल्टे बुनेंगे और मतलब को सब्सक्राइब कीजिए और सबस्क्राइब नाउ यू मस्ट नॉट अवॉइड थिस चैप्टर [संगीत] सब्सक्राइब टो बिजली कंबाइंड विद पीएनसी और ब्रिटिश सब गलत डालोगे तो मिक्सचर इतने पैसे जाता है ठीक है अच्छा भाई चलते हैं हेलो रेहान हेलो गुड मॉर्निंग हेलो माइक हाइड सुक्लू सब्सक्राइब करें क्वेश्चन के साथ में मेज़ की बात करें तो काफी कुछ लिखा हुआ यहां पर भी this Video इन बोथ इंग्लिश एंड हिंदी में व्यस्त हैं ओके तो लैंग्वेज का कोई नहीं नॉट मेक एनी सिंगल प्वाइंट एंड क्लिक ओं सब्सक्राइब कर सकते हैं ठीक है तो चलिए शुरू करते हैं है प्रॉफिट लकीर विष्णु लैब्स क्वेश्चन लाइट ओन स्क्रीन क्वेश्चन सीधे रिड्यूस्ड टो ए पब्लिक - 06 2012 1.5 इंच सबस्क्राइब नाउ टू चेक एयरटेल डाटा एंट्री हो और सुनाओ गांव बिचपड़ी एक्सपेंशन विशेष दायित्व और चेक आउट फॉर एनी रिग्रेट्स टर्म इन पाइथन फॉर रिलेटिव वनप्लस 3T पावर एंड विजय को थिस एक्सप्रेशन एंड माइनस प्लस वन अपॉन और लहसुन और इक्वल टू वन अपऑन व्हाट आर द बेस्ट क्वालिटी क्वालिटी प्लस वन ओके वेरी गुड क्वालिटी क्वालिटी मी टू माइनस प्लस सीट पर क्विज फ्री मोड ऑफ दिस प्रॉब्लम पाचन क्रिया फिटिंग आउटपुट पावर टीम राइट तुम जब नोटिफिकेशन बाहर निकाला सुदेश विकास वन एंड डिस्कवर विथ मी राइट व्हेन कंपेयर्ड विद 150 तो यह तो मेरा मन हो गया मेरा टीम जो टाइम हो गया तो यह बन गया - 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प्राइवेट पॉलिटेक्निक मिट्टी की तरह से डिलीट करने लगते हैं ध्यान रखिए जवाबी हमला अप्लाई करने होते हैं आप यौवन प्रसिद्ध पॉवर होता है तो यह बहुत कम 108 बार ऑब्जेक्ट इन थे टाउन इस बार लेफ्ट डू इट मास में यह तो पागल है यार जैसा-जैसा नोटिफिकेशन कर सकते हो यह बॉयल्ड फ्रीडम यहां से फीड 2nd का इंडिमनिफिकेशन एक हाइब्रिड यू एवरीथिंग यू हैव ए स्ट्रांग और डिसाइड नो रिकॉर्ड्स फॉर इन थिस क्वालिटी टाइम जॉब 2019 जो विचार इक्वल टू 4 इसका मतलब यह हुआ कि मेरे पास आ रिक्वेस्ट फॉर गैलस्टोंस करने पर मुझे क्वालिटी मिल जाएगी सरकार विक्रम जी को 124 ए बटन इन क्वालिटी के अंदर मूवमेंट बैटरी क्वालिटी अलगिरदास पॉइंट कोट 40 ए लिमिटेड टाइम ऑफर को जब आर इक्वल टू फॉर हुआ यह मेरी क्वालिटी आई तो जब एक मिनट यहां पर आ रही है तो ओरिजिनल ऑप्शन था मोड ऑफ विहार फेज-1 ढेरों रिकॉर्ड मोटियार देर ऑलिव ऑयल भी कटिंग माइक क्वालिटी मोड ऑन टैप लक्ष्मण इक्वल टू व्हाट ईयर आईएफ ए रिक्वेस्ट फॉर दिस क्रीम इज द मोनोफॉर्मिक वॉल्यूम और 25dec 20.51 मिडस्ट आफ 120 विल नॉट बे ए गिवर जिस कैलकुलेशन पिछली एक पॉइंट को यह मिला कि भाई यहां पर 350 को बाहर निकालो ना वह सबसे बड़ी जरूरत होती है तो मैं एक और दूसरी बात यह है कि जब भी गारंटी जरा आएगा तुरंत में क्वालिटी दे रही होगी और क्वालिटी के हिसाब से दिए करना पड़ेगा नाइट ओके तो मिट्टी पॉइंटेड मिल गया अब मैं अपना एक बार उठा लेता हूं मैं को मिला किया था फ्री - 15th फ्लैश लाइट्स अप द फोन उठाइए और टिप्पणियों चाहिए तो टीवीपुर के लिए यू रिर्पोटिंग आईएस कॉल्ड फ्रीडम फाइटर प्लेन रिकॉर्डिंग रिकॉर्डिंग चाकू में डूबे सिक्योरिटी फोर्स डे रूटीन को y53 आयोडीन अधिक क्यों क्योंकि अब सबसे बड़ा इंपॉर्टेंट पॉइंट्स क्वेश्चन आता है कि आपने क्वेश्चन पड़ा था तो मेरे जान बूझकर इस बात को किया अब मुझे क्या चाहिए मुझे चाहिए था 2015 को यूज करते हुए से मिल गया पर मुझे चेक करना है कि मेघनाद डू सब्सक्राइब करें कि इस बार दीदी डब्लू सर्टिफिकेट मार राइट लेफ्ट साइड आदि कुल लुटा डू रिकॉर्डर - 4a पिछली बार डिफरेंट बट नियुक्त करने के लिए तो आपको क्वेश्चन में क्या पूछा था द क्वेश्चन वास क्लीयरली आज की हॉट गर्ल्स डिग्री कोल भाइयों नॉट बीन इंप्लिकेटेड 12345 आप इस रिपोर्ट नहीं कर सकते हो भाई आंसर दोगे कि आने वाला है और जब आप पिटी फॉर को चेक करोगे तो टीमों की वजह को क्या मिल रहा है 15232 - वह पावर थ्रू विच क्लीयरली स्टेटेड और आप जब ₹50 करोगे तो वहां पर मिलेगा आपको 11-12-13 प्वाइंट - अनमैरिड बारकोड स्पेशियली पॉजिटिव विच क्लीयरली मींस कि यह दोनों भले ही निकली इक्वल हूं मैंने Tubelight दोनों क्रीम वॉल्यूम को बट अल्टीमेटली ग्रेटेस्ट देवोटी 535 1035 एवरी स्टेज राइट यह एक सबसे पहला पॉइंट जो कि बहुत कॉमन कि लोग अक्सर जल्दी में मिक्स करते हैं सब्सक्राइब हेलो हेलो अपने फैसले में मुझे मिले ठीक है कि मुझे मेरा वॉइस मेल के लिए तो यह गलती करते करते है जहां पर आप - लेते हैं जबकि आपको बनाना होता है ताकि आप subscribe कर ले जाती है और मुझे दवा मिलती है है और सबसे बड़ी बात यह कि जब से एग्जाम लीडर पूछा गया है तो आप t45 दोनों को क्लिक करके आगे नहीं बढ़ सकते हैं आप बिजली चैक करते हैं जिस दिन रिमाइंड साइन इन बिटवीन थम राइट फ्री - फीडबैक तो इनमें से एक पॉजिटिव 1 लीटर होगा एलजी पिछली ग्रेट एस वोट देश क्लीयरली टीम फाइव डेज आफ बट t45 जरिए आदित्य राज अब मैं इसे लगा दो दोस्त वॉइस बन जाएगा ना तो आप चली अगर विकसित और एक्सटेंड करूं छुट्टी 500 पिछली लाद देते एंटी फ्लोर आईएस नथिंग बट अश्लील जेटली लिस्ट मेरे अगले फ्लाइट चुके रहना एग्जामिनेशन में सहायक है 2025 दोनों सकते हैं मल्टीपल का टाइम दिया हो सकता है तो ट्रैक पर सकते हो तो मैं 10 डोंट फॉर ट्रांसपोर्ट और गिफ्ट ओके सहायक राइट इन इंग्लिश वर्ष की छोटी सी ओनली बहुत हमारी इक्वल बट एवरी गर्ल ई डिपॉजिट 25.1 अधीर वीकली लार्जेस्ट एंड सिंसेरिटी आफ और इस नेगेटिव डिलीट डिलीट इस Twitter - एपिसोड विल बिकम अ यह पिछली लिस्ट हाइट एंड विड्थ के सभी बडे उद्योगपति प्वाइंट व्हेयर यू माइट फाइंड अट्रैक्टिव एंड मेक हेर फ्रॉम नावार्ड तुम्हारे भीतर अपनी परसों ठीक है मैं समय पर है लुटेरे बाइक यह विटामिन क्वेश्चन पूछे कि मुझे काफ़ी कुछ सिखाना था तुमको सारी ट्यूब स्टेप्स हो मासिक मासिक जाट आरक्षण 2011 - वर्ल्ड कप आंवला और भी है वन CR अपऑन थे प्लेसमेंट आफ वैरीयस मैथर्ड टो सॉल्व थिस क्वेश्चन आईएस ए इंटर्नशिप इन थे क्वेश्चन विच इंवॉल्वेस सीमा रोटेशन और सुबह बीम कि शो द फार्मूला प्लेलिस्ट से दूसरा व्यक्ति बटन क्लिक हो तो एनसीआर इज इक्वल टू ए बिहार एंड - 151 - वन ओके और लाइट ओन यूज पेरू है कि नोएडा प्लस वन प्लस वन प्लस वन प्लस वन इनटू ए बी सी डी लुट मेरा अ दाउद इब्राहिम इंच राइट द बेस्ट कोट 20 मिनट पावर यूनिट में 1138 बेसिकली प्लस टू प्लस टू गो एक तु 811 सी आर ए गुड प्लेस फॉर एनसीआर और प्लस वन प्वाइंट यहां मैंने यह यहां भी और 11-12-12 नॉट सब्सक्राइब टो और सुनाओ मौसम में अमरनाथ बॉलीवुड वापस लेगी उत्तम फिक्स्ड डिपॉजिट सोवु यह वीडियो विड्रॉल प्लस वन कैंडिडेट वास द प्रेसिडेंट ऑफ द नोटिफिकेशन सब्सक्राइब 2.1 raj2002 - 1000 सीआर इन 12th पार्ट प्लस वन 1000 subscribe Video plz subscribe करो यहां पर सबस्क्राइब नाउ टू प्लस टू प्लस वन प्लस वन है कि इतने चिपकाने वाला हूं तो मैसेज करो फिर देखो आप दिखना शुरू हुआ 8 प्लस पॉइंट यह प्लस टू प्लस पनियरा प्लस टू किस का वेट कर रहे हो प्लस टू वाले भाई साहब का नाम क्या था इलेवेंथ अचूक क्या होगा थर्टीन क्या नहीं है थर्टीन क्या करोगे मल्टीप्लाइंग डिवाइस बता 30th कि असली दूध स्पेसिफिकली मल्टीप्लाई 3010 डिवाइड बैटिंग अ कि अ व्हाट इज क्या हुआ यह वाले भाई का पूरा क्लब हो गए कि यहां पर इस मुद्दे का यह पूरा हो गया रिपोर्ट थर्टी प्लस टू इन 12th पार्ट प्लस 13237 13238 10 - 1the विड ऊ है अब गैस कांड 15 दिन को याद करो स्पेशल एडिशन क्या बोलता था मगर जनरल डोमिनिक स्ट्रॉस-कान पर सेक्स पॉवर है सुबह प्रसिद्ध पावर एंड इन टर्म्स आफ जैनिज्म एंड ईस्ट वेस्ट नॉर्थ सिक्किम और इक्वल टू जीरो दो एंड विटामिन सी आर इनटू एक्सल पावर प्राइवेट सिक्यूरिटी वे कैन चेंज हिस माइंड पावर और सब्सक्राइब आधे तो और लक्ष्मी पीठ गुड्डू यादव और पार्थो वॉच पवन साइंटिफिक ही वॉल्यूम ही यह इन टर्म्स आफ एक्सप्रेशन आफ मैंने मोडिफाइड किया है वह फ्राई करते हैं भाई क्या मिला मुझे साइकिल दुकान सज इंटेक्स लिए डूइंग ग्रेट वर्क 12231 सिगमा आलड्रिच 120 211 रिकॉर्ड इसे नथिंग बट - वर्ल्ड पावर प्लस टू 30cr प्लस टू इस तरह से कैसे लिख सकते हो आप वहां तो - बंटी पावर और था आपने - मंडप आप वसूली दिया क्या फर्क है भाई दोनों में - माइंड पावर आर्ट को अगर आप माइनस पॉइंट पॉवर टू से मल्टिप्लाई करोगे यह इंवाइट भगवान ओनली धुंध लो ना 15 2012 - 400 500 600 अब यह जो लोग करते हैं और वह यहां पर बता दूं क्या हुआ मेरे पास में श्योर कहानी दिस इज सबस्क्राइब करते हैं ठीक है इसके अलावा अगर आप MS Word प्लस माइनस प्लस नहीं बनाओगे संयुक्त संचालक मुझे पर थोड़ी सी गलती करता हूं ठीक है तो बीकाऐ बजे डॉक्टर के टॉर्च लाइट चॉप क्लीयरली सैंड आर्ट स्टार्टिंग फ्रॉम इक्वल टू जीरो इन थिस सेंचरी थॉट ई सी नो चेक फॉर द वैल्यू ऑफ आर्ट अमेज़ बिगिनिंग डेकोरेट योर सपोर्ट फॉर ए राइड शुरू यहां पर इंजीनियरिंग और डिसाइड 3293 हार्ड सो ओनली क्वालिटी प्लस माइनस साइनिफिकेंस आफ प्लेटफॉर्म पाठ अमरनाथ बॉलीवुड आसमान में हेलो हेलो हेलो हेलो हेलो हेलो अब तो क्या मेरे पास आप की डेट फ्रॉम 30 टो 40 आफ middle-class मिसिंग ऊ है इधर में मदरसा मोबाइल ऑल द फाइनल ऑडिशन फॉर ए सर्टेन इंडेक्सेस वोट स्टूडेंट राइट विशेष भट्ट सिगमा आलड्रिच 1202 एन एन सी आर स्मार्ट इनफ टू द पावर इन तस्वीरों तो अगर यह उसके अपने हिसाब से मैं चलूं से फास्ट फूड एंड रिकॉर्डिंग और 3230 300 विल गिव टो डिपार्ट अट 3am इसे यहां पर यह सब ओके और यह आपस में डिसाइड कर पाकिस्तान के साथ चलोगे तो मैं दूसरा सबसे लिखता हूं यहां पर ठीक है अब राइटिंग एनीथिंग जस्ट देखिए और द वॉल्यूम यहां बैठोगे प्रॉफिट्स आफ प्रॉब्लम्स विच वेंट फॉर इंडिया फॉर mid-day के दिल्ली - 110 टावर 2004 3153 हीरो - 80851 प्लस 32.1 - प्रति हे0 दो फ्लैश लाइट ओन नहीं यह ले करके आफ देखिए और शुरू किया तो मेरे पास था वह यहां से तो मैंने यह मुझे 36 लुटी आर्ट डिस्टर्व 3001 ओके खिलाफ यौन अपराध एवं सब्सक्राइब बटन राइट्स बेस्ड प्लस वक्तव्य के लिए 3233 - 100% सब्सक्राइब 3032 तो अब क्या हुआ यह पूरा एक्सपेंसिव तो तुम जानते हो ना यार यही तो लिखा मेरे प्रिय गौतम 1000 पर फिर से जल्दी इसको की कोशिश की कि किया या एक एडिशन यू विल रिग्रेट ठाट ओल्ड इज गोल्ड चैनल को सबस्क्राइब 238 - 30 - सब्सक्राइब टो ओम शांति डिफरेंट थिंग्स राइट आंसर आंसर इसमें यह की अनमोल ग्रेट सुधार 3130 कांग्रेस विन थे डे आफ थे फेस्टिवल आफ फॉलिंग इन थे ओके और 800 लाइट लिक्विड वॉटर फॉर एवरी बिट फ्रॉम 0 टो थे राइट थिंग बीइंग अप्लाइड फॉर योर लिमिटेशंस आफ लाइक दिस सिगमा आर0 सी0 बी0 सी0 कि गाइड शेयर गैस चालू करेंगे चलते हैं स्पेसिफिक डाट प्यार सिंह जी बहुत-बहुत धन्यवाद बहुत सारा प्यार देने के लिए चलिए बढ़ते हैं फाइनली Bigg Boss बंपर सुपर सुपर सुपर सुपर मॉडल क्वेश्चन पेपर subscribe to the Page if you liked The Video then subscribe to कि फाइंड द वैल्यू ऑफ एक्स पॉर्न विद माइनर 1592 - रिएक्शन पावर फॉर वेल्थ ओके नाम थिस को शेयर गेम्स ऑल अबाउट दिस फ्रेशनर पावरफुल आपको अच्छी तलब कल सुबह आफ पर्सनल पार्ट हुं केवल प्रथम नवरात्र स्पेशल अ कोई रिकॉर्ड यह 214 पूर्व सिंह तोमर नंबर 159 वनप्लस एक्स पावर इनवैलिडेटेड टीचर और अश्लील विडियो और यहां पर डेट व्यक्ति दबाएं Star Plus डूब स्किन व टॉर्च लाइट बिल ही दिन यूअर सपोज्ड टो ओपन अप डिजाइनों में एक्सपेंशन वे नीड टो इंक्लूड और प्रेजेंटेशन इस अपडेटर वेबसाइट पर दिस क्वेश्चन पर भी विजिट वैरीयस प्वाइंट्स 1515 महत्व बहुत सुपर डुपर ट्रिक देता हूं तो याद करने के इरादे दूं देखना तो आप जब इसका स्पांसर लिखने की कोशिश करते हैं वह राइट्स विल टाइटनिंग अ क्वेश्चन ऑब्जेक्टिव वाले सेक्शन पावर माइनस प्वाइंट मैं जॉब लिखोगे अगर आपको याद होगा और आपने पढ़ा होगा ठीक से रिवाइज किया होगा तो यह पाचन नली ऐसा दिखता है वनप्लस एक्स प्लस एक्स स्क्वायर प्लस एक्टिव डॉक्टर infinity है भी आयोडीन अरे बाप रे अनफोल्ड एंड पॉइंट्स पेड़ रिस्ट्रिक्शन फॉर इत्सेल्फ टो एग्जिट The Video then subscribe to the Page if you liked The Video then subscribe to the Page if you liked The Video then subscribe to the Page if you liked The Video then subscribe to do मजे में हूं कोट जाएगा आपको तुम शुरू करोगे पर अपने अश्लील पोटेंशियल बनी पहुंच पाओगे और वह तो मैं हमेशा खुलेगा Tubelight फ्री मोड 1328 इंपोर्टेंट सीरियल है मैं नागपुर इंतजार मत करो कि बाद में कर लेंगे तू इतना वक्त इजहार टाइम थिस राइट टाइम लाइफ मौजूद एसिड और करोगे दिस इज द टाइम सबस्क्राइब नाउ टू रिसीव न्यू अपडेट्स विडियो पॉसिबल ओनली व्हेन थे यहां यहां 2 - 4a कि यह भी मानते हैं कि आप लड्डू आप रेस्टोरेंट के सामने आ चुका है फ्लाइट मोड निकली लखनऊ कांड हुआ है विडमेट फंक्शन इन ऑर्डर टो वन प्रशिक्षित पावर एंड क्लासिफिकेशन दिया था मेरा तो वह चीज दो बार रिपीट करो टू द पावर फॉर वाईफाई को बाहर निकालो अंदर क्या बचेगा वांट माइन प्रोटेक्टेड बाय टू द पावर फॉर वाइफ इन दोनों देश - क्रिएट बाय टू इज द थिंग विच मस्ट बी यफ क्षण विच वास लेटर - 121 हेरिटेज दैट पॉइंट होता धुंधला डेलिगेशन सुदेश पट्टी कलेक्टर - फिर एक बार टू राइट मस्त लाइफ - 151 - 232 373 माइल्स 238 माइल्स 283 फर्स्ट फ्लाइट मोड ड्यूल दिन रुके डेट ओनली विजन एंड कंडीशन प्रेसिडेंट क्वेश्चन 56001 वांट टो लाइव विद नॉट मेड टो गिव फ्रीडम पैक एब्स है यह 1995 में आपको क्या करना है क्यों मैंने क्या किया मेरे यहां पर मैसेज पर लगाया उस मैंने क्वेश्चन को लिखा आप इस बारे में लिखने के बाद में लिमिटेड राइट जो लैपटॉप चैनल राधे-राधे ज्ञान क्वाइट इसे 0154 तो पॉवर हंड्रेड - 80851 6 पावर आ ओके ऑल राइट और सुनाओ मुझे चाहिए क्या क्वांटम फ्री ऑफ मेडिकल साइंस थोड़े प्रश्न पर क्लिक कर देना है और गिरावट पिछले स्विच ऑन डिस्ट्रिक्ट की जो भी पावर जा रहे सोडियम Vodafone नंबर ओनली राइट फॉर जानकारी सारे पॉइंट्स चाहिए ओके प्रेसिडेंट हमारा हैं इसको स्लाइडऐशो ट्वेंटी ईयर्स बैक ई हैव थे टाइम टो बे फ्री फ्रॉम रेडी कर राइट फॉर डेलायड ड्यू अल्सेजी अनुसार 2240 टॉर्च लाइट आज वाला है 1.6 वॉट इस द पावर ऑफ 8041 और 226 001 मैं दावे इक्विटी करना चाहिए उन्होंने कहा कि पूर्वी की जरूरत है मतलब आर को फोन का भी मल्टीपल होना है आर को कि जब वह पर होना है ओके लाइक इंवाइट गेवर जी बहुत कैसे करेंगे इस विडियो चैनल पर थोड़ी देर रात आवसीय रिक्वेस्ट करने पर लेता हूं ठीक है कि जहां पर क्या हो रहा है और बाइक वक्रीय होना है और बाकी 450 जरूर है तो आवाज़ फॉरवर्ड करो ना यार को पूरा लाइबिलिटी बनाना पड़ेगा सिर्फ बाह भाई पूरा एक्टिविटीज आफ विक्रम शर्मा टेंपल आफ और विनर बाइक हां 20180 बिस्मिल तथा ऑलमोस्ट भी मल्टीप्ल आफ सिक्स स्पीड आ बे कंफर्टेड फॉर इलेक्शंस वेयर फॉरेंसिबली 12-12-12 क्लियर फ्रॉम 100 को यह एक बड़ा हो जाता है अगर करते हैं कि 80 होता है मैं जॉब या परसों लेते हो यह साला रिकॉर्ड जीरो फ्रॉम से ज़ 20वां 600 लुकिंग फॉर इन थे वर्ल्ड विल नॉट वेल rights of your 10 अब यह टाइम जाता है इट राह पर चलो फतेह में बहुत अच्छे बहुत अच्छे स्टूडेंट प्लीज सब्सक्राइब करें और किसी को बताओगे तो नहीं करना चाहिए तो यह जितने अच्छे से अपनी तरफ अपने सिस्टम को और इस वीडियो की लिंक सब्सक्राइब बटन होगा वहीं परिचय मेरे चैनल को सबस्क्राइब कर लें को प्ले करो फिर फाइबर वीडियो वह टेंपर्ड ग्लास ढूंढ पड़ी टाइम्स हैव थे सेम क्वेश्चन ट्यूमर सी एवरीवन अराउंड यू नाइट श्यामलन क्योंकि जब इससे एक सिस्टम में आते हो जब तुम मुसलमान फ्री में आते हैं वह चटनी के साथ आते हो तो एवरीवन इस लर्निंग इन एवरी वन विल प्राय इनटू पुष्प डैंड्रफ बट फॉर द स्पीड ऑफ लाइट दो बहुत अच्छी क्वांटिटी स्टूडेंट बहुत अच्छे दोस्त बहुत अच्छा करते हैं सिर्फ जब दोनों एक दूसरे को इंप्रोवाइज्ड करने की कोशिश करते हैं हाईट आपके सुनाओ मेरे पास आ रहा कि वाकई शून्य से लेकर 1962 मेरे पास कितना गया रिकॉर्ड 90 तो वही तो पापा हो गया भाई साहब गजब हो गया तो क्या पापा हो गया 9 टर्म संगठन आगे भी लग रहा है अरे क्राइम कर दिया भाई क्या प्रैंक कर दिया ही सुनाई दे रहा है तुमको यह देख रहा है व्हेनेवर वे टॉक अबाउट थिस कैन हैव मेड योर प्वाइंट फॉर द गॉड तो यह डाट 9483 को पॉवर टू डू इट मींस ऑफ़ टॉम टॉम इन द प्राइम नंबर 9999 Please subscribe like और सब्सक्राइब करें तो मैंने लिया है है तो इमेजेस विद ब्रॉड ने पावर हाउस प्लस टू के पावर आफ पावर एंड थिस एक्सपेंशन इन अधिक 851 टूरिस्ट प्लेसिस इन टू टू टू टू टू टू टू है तो इसको थोड़ा और प्रॉपरली अधीक्षक रखोगे निशान डेढ सीआईए थर्ड पावर फी - आर राइट टू राइट इनटू थे पावर टो ऊ मैं अभी क्वेश्चन ज्यादा इसलिए होता है क्योंकि यू वांट टो बे फ्री फ्रॉम मेडिकल साइंस 10 क्लास नाइंथ थे वॉलिंटियर्स व्हो एंटर्स आर्बिट मनोवैज्ञानिक और 2001 ली राइटर डीआर बटु हरषहिं कीबोर्ड इंटीरियर तितली बाइक सो आदरणीय विपिन तीजा आर्ट फेस्टिवल कमेटी ने Bigg Boss कि वे वांट टो लुक फॉर 20th डेथ आर हॉट मूवी फ्रॉम 0 विच इंप्लाइज यूनियन फॉर ऑल 2000 की बात करोगे तो यह याद रखना चाहिए होता है तो मैं और तुम हो जाएंगे कृपया यह जाएंगे इफेक्टिव 1000 में दूसरी बात हमेशा अंधेरे में कि यह विड्रॉल है ताकि हर उस बच्चे को जिसको मतलब अगर आपने सब्सक्राइब नहीं किया तो बहुत और थिक चैनल है राइट यू एवरी इनफॉरमेशन रिलेटेड इनफॉरमेशन फ्रॉम मार्स और आप आसानी से उसके साथ यौन है तू लोग यह प्लेटफॉर्म फॉर एनी लाइक व समथिंग कंट्रोल चैनल अगर कुछ भी नोटिफिकेशन आएगा वी आर्य रवी अवॉर्ड अवॉर्ड फॉर एवरीथिंग ओके तो आप बस भरोसा रखिए अपने चैनल पर इस साल त्यागी वेरिफिकेशन घंटे तक ना तो कि हम निकल जाते थे इंटीग्रल पार्ट आफ वनप्लस 5t ट्वेल्व व एंड इमोशनल नंबर वो पूछ रहा है कि जो नंबर है वह यह कि और होगा ओके इंटीग्रल पार्ट आफ टोबैको नहीं तो ठीक है पर इसका टीचर पार्ट लोग इंडिकेट इधर आएगा यार इंटरेस्ट हुई व्यू आफ थे गिवर और ओके परफेक्ट नाव रोहित धवन जूते के अंदर लाइक यू आई लव यू टू यू टू मेक पदार्थों के 900 इस वर्ष के लिए बहुत जल्दी कि यहां पर ना देखो क्या हुआ और एवरी नंबर कैंडी क्रश ले ली है आधे जीरो टू वन आईडेंटिटी जो कितना डायबिटीज ऐप को ओपन करूंगी अब इसे मिडिंग मिडिंग की ली ना से होगा जिसमें यू टू आई लव यू टू सोना ही बेहतर हो तो प्लीज लाइक और सबस्क्राइब लाइक थिस कि आप इसको चैनल उपयोग करते हैं हम उसका कौन सी गाड़ी है तो कॉपीराइट क्या हो गया सेवन माइनस पाइंट टूटू तो पावर दोएस नॉट नंबर ताकि दर्शन अपडेट्स और अगर यह ब्लैक हो गया तो प्रयोग है तो आपको 307 तक लाइक करेगा ओके इसके अलावा चलें दिया मैंने दादी एयरप्लेन मोड 910 तो यह आइटम है इसमें इनको फारुक दिख रहा है ना तो प्रिंटर में फाइट दिख रहा है मैं इनको फ्राय करना चाहता हूं इसके प्लस आने लगा है यह भी है - साइंस क्या करोगे ऑलिव ऑयल वे इसलिए डुग डुग that I प्लज़ प्लज़ डोंट बिलीव थिस विल गिव सम इवेंट भेजो ओक तुझे क्या मिला आईपी एड्रेस पर भी नोटिफिकेशन गिवर नंबर बट फॉर डार्क जजमेंट क्या बनाओगे अधिक उत्सर्जन में साइड में बना रहा थोड़ा हेल्प यह याद रखना कि बहुत इंपोर्टेंट पॉइंट है तो मेरे पास क्या है मेरे पास आया है थिस लाइन फ्रॉम 021 टैब्स लाइन फ्रॉम 021 आपने दोनों को ऐड किया है प्लस डू लाइन फ्रॉम युटुब और हो क्या रहा है आई है तुम्हारा इंटर थे इंटीरियर में आप यह समझ कर रहे हो यदि आपको चैट कर रहे हैं और बाद में फ्रेंड्स मेरा मतलब हुआ कि अब वह क्या होगा इंटीरियर इसका मतलब ऐप अपडेट को हाथ में क्या होना होगा इंटीजर और फ्रॉम यू टू द ओन्ली बॉलीवुड कि डोंट वन विच मीन साइड आप व्वे चैंपियन डेविड वार्नर और राइट यह मिलेगा विधायक हुई 151 - 1994 गिफ्ट समय आंसर को अवोट और नो नो वगैरह भी बहुत जरूरी बात को समझना कहा बताओ गलती कहां हुई कि गलती कहां हुई कि वह बच्चे गलती हुई बोलो फटाफट कहां गए जी ठीक है और निर्मलता डोंट वरी अबाउट इट्स नोड वहीं फास्ट इज द टाइम राइटिंग एंड फास्ट बेबी बेबी डॉल हुए हो द वीडियो इस ग्रेट गोइंग टो इंटिरिम 9th YouTube हिंदी YouTube पर रहेगा आप आराम से इसको नेटवर्क एडिशन को मुझे कोई अश्लील कौन से पहन है गलती गलती हो गई होगी तो निर्मला बताइए मुझे कि कहां पर क्या गलती हुई है वह एग्जाम क्लियर राष्ट्रीय उद्यान हेल्प क्या ऐड हुआ है अमृतसर गोलू क्रिएशन बताइए क्या गलती हुई है क्या हुई है गैर कहां हो गई क्या गड़बड़ हुआ बोलो कहां गढ़वा गढ़वा यहां पर यह बात हो गया ए प्लेस इन स्टॉकपोर्ट 75230 के 5252 सुपरहिट विडियो प्रॉक्सिमिटी 1.4 कि दिसंबर निर्देशित द स्पाइडर विल गिव वोट तो सेवन - अनलॉक 1075 डिसाइड 72825 2107 प्रॉब्लम नेगेटिव वह निगेटिव हो यार हमने विधायकों के लिए 307 टाइम हो गया हम प्रेशर नहीं सके डिपार्चर नहीं सकते भाई तो क्या करोगे कौन सिगरेट को चूर स्माइली हरवंस उसे डिवाइस भी फ्लाइट क्या मिलने वाला है आपको यहां पर थी मैं मैसेजेस ऊपर आता हूं दोबारा हम सब चीजों को लिख रहा हूं आप में कैसे लिख रहा हूं इसको हम एडिलेड टेस्ट ₹5 लाख है इंटेंशनली अब एडिटर साइड इफेक्ट व्हाय डिड साय नो दैट इज क्वालिटी मिलेगा 529 सिर्फ ₹10 फैशन फॉर एनी पॉसिबल सेक्शन 2 मिनट के आसपास लिखा रुके कि नाव इसे पोजीशन क्या हुआ पाउडर - 7 विटामिंस जो मिट्टी अपडेट्स होंगे 1 मिनट ऑफ व्हाट्सएप खोलिए प्लीज लाइक 2021 एफिडेविट फील्ड ओं थे लाइन फ्रॉम टुमारो ओनली टिस है यह मेरा सबसे बड़ा हो गया तो यहां पर डाल दिस इज पॉज़िटिव मोस्ट मेंबर कि यह यकीन हो जा यार अच्छा ठीक है अब मैं इसको बहुत ही इजी और बहुत सारी लेकर आता हूं ठीक है सोना ध्यान से तो ना हो क्लियर लिए यहां पर फाइल रुकावट वन प्लस वन और नंबर तो रूट टू रह जाएगा हरा मुझे रूट को डायवर्ट करना है तो हमें लड्डू है subscribe this Video not support सिध्दि सिस्टम विटामिन हॉटस्पॉट ऑफ रूट टू धोवे नोटिफिकेशन फॉर एडमिट कार्ड 2019 हेयर ऑइल पावर 2012 डेट ऑफ बर्थ सर्टिफिकेट आफ 2002 में एक ऐसी रेमिडी प्रॉब्लम रहने हैं चाहिए फॉर थर्ड टाइम नहीं तक 15 दिन के बाद 1786 बटन आईएस प्रेस्ड बाय गेट अकैडमी डिफिकल्ट वो टाइम समथिंग वियर टाइम हो सके लेकिन अब टू टाइम्स व्हेन टियर्स विल गिव वॉटर एंड जब मैं अभी ऑफिस से क्या हो रहा है मुझे क्या मिला आई प्लस माइनस अवधेश इस गर्मी नथिंग बट ई कोल्ड नॉट कन्वे ए टीजर ओके तो इंटीरियर आयुर्वेद तथा अब मैं चीन के जो कलर चेंज कर रहा हूं टाइट फॉर दिस पॉइंट दिल्ली में भी शायद आंखों की रोशनी आप चाहे लाख रहेगा सुनिए ध्यान से आइए कि टीचर है और यहां यह लाइफ हैक्स रिजल्ट वहीं डीटेल्स आफ इंटीरियर में F - 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16130 | https://www.youtube.com/watch?v=Mvsau67PxFM | Can you prove this inequality? | Moscow Mathematical Olympiad Problem
Dr. Wang
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Posted: 16 Sep 2022
Given 𝒂, 𝒃 ∈ℝ^+, 𝒂 +𝒃 =𝟏, prove (𝒂+𝟏/𝒂)^𝟐 +(𝒃+𝟏/𝒃)^𝟐 ≥ 𝟐𝟓/𝟐. How to prove this inequality? We apply the GM-AM inequality to solve this problem.
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Transcript:
hello everyone I'm Dr Wang here is the question.
if a and b are positive real numbers a plus b equals one, prove that a plus one over
a squared plus b plus 1 over b squared is bigger than or equal to 25 over 2. let's derive a simple result first. if
x and y are positive real numbers, the square root of x minus the square root of y then squared is
bigger than or equal to zero. this implies x minus 2 square root of x times y plus y
is bigger than or equal to zero therefore the square root of x times y is
less than or equal to x plus y over 2. this result is also called the geometric
mean is less than or equal to arithmetic mean. if we replace x by x squared y by y squared in (1) then we have x times y is less
than or equal to x squared plus y squared over 2. both sides time two we have 2xy is less
than or equal to x squared plus y squared. on both sides adding x squared plus
y squared we have this inequality the left hand side is a complete
square which is x plus y squared therefore we have the second result. this is the most important
result to solve our problem. just replacing x by a plus one over a
y by b plus 1 over b in the second result we have this inequality simplify the inside part of the right hand side a plus b from this given condition which is one consider the common denominator for the last two
terms we have the common denominator a times b the numerator is a plus b, a plus b is 1 again. now we need to find the upper bound for
a times b. we use this formula again x equals a, y plus b. both sides take the square then we have a times b is less than or equal
to a plus b over 2 squared, a plus b plus one, one-half squared equals one-fourth. this means
a times b is less than or equal to 1/4. we replace it by one over four in this denominator,
the inequality is bigger than or equal to 1 over 1/4 which is four, one plus four is five,
five squared is 25, 25 over 4. this is the result. is 25 over 2 the best lower bound for the
left hand side? yes. the equal sign is true when a equals b equals one half. it's easy to check
it. just replace a by one half, one over a by one over one half which is two, two plus
one half is five halves, five halves squared, squared, you will get 25 over 2. that's
all thanks for watching and see you next time |
16131 | https://mechanicsmap.psu.edu/websites/16_one_dof_vibrations/16-2_viscous_damped_free/16-2_viscous_damped_free.html | Mechanics Map - Viscous Damped Free Vibrations
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Viscous Damped Free Vibrations
Viscous damping is damping that is proportional to the velocity of the system. That is, the faster the mass is moving, the more damping force is resisting that motion. Fluids like air or water generate viscous drag forces.
A diagram showing the basic mechanism in a viscous damper. As the system (mass) attached to the loop at the top vibrates up and down, the damper will resist motion in both directions due to the piston passing through the fluid. (CC-BY SA Egmason)
We will only consider linear viscous dampers, that is where the damping force is linearly proportional to velocity. The equation for the force or moment produced by the damper, in either x or θ, is:
→F c=c→˙x F c→=c x˙→
→M c=c→˙θ M c→=c θ˙→
Where c is the damping constant, which is a physical property of the damper (based on type of fluid, size of piston, etc.). Note that the units of c change depending on whether it is damps linear (N-s/m) or rotational motion (N-m s/rad).
Free-body diagram of the system in equilibrium position. The spring is at its equilibrium position, but it is stretched and does produce a force.
When the system is at rest in the equilibrium position, the damper produced no force on the system (no velocity), while the spring can produce force on the system, such as in the hanging mass shown above. Recall that this is the equilibrium position, but the spring is NOT at its unstretched length, as the static mass produces an extention of the spring.
Free-body diagram of the system in equilibrium position. The spring is at its equilibrium position, but it is stretched and does produce a force.
If we perturb the system (applying an initial displacement, or an initial velocity, or both), the system will tend to move back to its equilibrium position. What that movement looks like will depend on the system parameters (m, c, and k).
Free-body diagram of the system in a perturbed position. Since the spring is at its equilibrium position (neither stretched or unstretched), it does not produce a force.
To determine the equation of motion of the system, we draw a free-body diagram of the system with perturbation and apply Newton's second law.
Free-body diagram of the system with perturbation.
The process for finding the equation of motion of the system is again:
Sketch the system with a small positive perturbation (x or θ).
Draw the free-body diagram of the perturbed system. Ensure that the spring force and the damper force have directions opposing the perturbation.
Find the one equation of motion for the system in the perturbed coordinate using Newton's Second Law. Keep the same positive direction for position, and assign positive acceleration in the same direction.
Move all terms of the equation to one side, and check that all terms are positive. If all terms are not positive, there is an error in the direction of displacement, acceleration, and/or spring or damper force.
For the example system above, with mass m, spring constant k and damping constant c, we derive the following:
∑F x=m a x=m¨x∑F x=m a x=m x¨
−F k−F c=m¨x−F k−F c=m x¨
−k x−c˙(x)=m¨x−k x−c(˙x)=m x¨
m¨x+c˙(x)+k x=0 m x¨+c(˙x)+k x=0
This gives us a differential equation that describes the motion of the system. We can rewrite it in normal form:
m¨x+c˙x+k x=0 m x¨+c x˙+k x=0
⇒¨x+c m˙x+k m x=0⇒x¨+c m x˙+k m x=0
⇒¨x+2 ζ ω n˙x+ω 2 n x=0⇒x¨+2 ζ ω n x˙+ω n 2 x=0
As before, the term ω n is called the angular natural frequency of the system, and has units of rads/s.
ω 2 n=k m ω n 2=k m
ω n=√k m ω n=k m
ζ (zeta) is called the damping ratio. It is a dimensionless term that indicates the level of damping, and therefore the type of motion of the damped system.
ζ=c c c ζ=c c c
ζ=actual damping critical damping ζ=actual damping critical damping
The expression for critical damping comes from the solution of the differential equation. The solution to the system differential equation is of the form:
x(t)=a e r t x(t)=a e r t
Where a is a constant, and the value(s) of r can be obtained by differentiating this general form of the solution and substituting into the equation of motion.
m r 2 e r t+c r e r t+k e r t=0 m r 2 e r t+c r e r t+k e r t=0
⇒(m r 2+c r+k)e r t=0⇒(m r 2+c r+k)e r t=0
Because the exponential term is never zero, we can divide both sides by that term and get:
m r 2+c r+k=0 m r 2+c r+k=0
Using the quadratic formula, we can find the roots of the equation:
r 1,2=(−c±√c 2−4 m k 2 m)r 1,2=(−c±c 2−4 m k 2 m)
Critical damping occurs when the term under the square root sign equals zero:
c 2 c=4 m k c c 2=4 m k
c c=2√m k=2 m ω n c c=2 m k=2 m ω n
Four Viscous Damping Cases
There are four basic cases for the damping ratio. For the solutions that follow in each case, we will assume that the initial perturbaion displacement of the system is x 0 and the initial perturbation velocity of the system is v 0.
ζ = 0: Undamped.
c=0 c=0
This is the case covered in the previous section. Undamped systems oscillate about the equilibrium position continuously, unless some other force is applied.
Response of an undamped system.
ζ > 1: Overdamped.
c 2>4 m k c 2>4 m k
Roots are both real and negative, but not equal to each other. Overdamped systems move slowly toward equilibrium without oscillating.
Response of an overdamped system.
The solution for an overdamped system is:
x(t)=a 1 e(−c+√c 2−4 m k 2 m)t+a 2 e(−c−√c 2−4 m k 2 m)t x(t)=a 1 e(−c+c 2−4 m k 2 m)t+a 2 e(−c−c 2−4 m k 2 m)t
Where:
a 1=−v 0+r 2 x 0 r 2−r 1 a 1=−v 0+r 2 x 0 r 2−r 1
a 2=v 0+r 1 x 0 r 2−r 1 a 2=v 0+r 1 x 0 r 2−r 1
ζ = 1: Critically-damped.
c 2=4 m k(=c 2 c)c 2=4 m k(=c c 2)
Roots are real and both equal to -ω n. Critically-damped systems will allow the fastest return to equilibrium without oscillation.
Response of an critically-damped system.
The solution for a critically-damped system is:
x(t)=(A+B t)e−ω n t x(t)=(A+B t)e−ω n t
Where:
A=x 0 A=x 0
B=v 0+x 0 ω n B=v 0+x 0 ω n
ζ < 1: Underdamped.
c 2<4 m k c 2<4 m k
The roots are complex numbers. Underdamped systems do oscillate around the equilibrium point. Unlike undamped systems, the amplitude of the oscillations diminishes until the system eventually stops moving at the equilibrium position.
Response of an underdamped system.
The solution for an underdamped system is:
x(t)=[C 1 sin(ω d t)+C 2 cos(ω d t)]e−ω n ζ t x(t)=[C 1 sin(ω d t)+C 2 cos(ω d t)]e−ω n ζ t
Where:
C 1=v 0+ω n ζ x 0 ω d C 1=v 0+ω n ζ x 0 ω d
C 2=x 0 C 2=x 0
ζ=c 2 m ω n ζ=c 2 m ω n
This can alternately be expressed as:
x(t)=A sin(ω d t+ϕ)e−ω n ζ t x(t)=A sin(ω d t+ϕ)e−ω n ζ t
Where:
A=√(v 0+ω n ζ x 0)2+(x 0 ω d)2 ω 2 d A=(v 0+ω n ζ x 0)2+(x 0 ω d)2 ω d 2
ϕ=tan−1(x 0 ω d v 0+ω n ζ x 0)ϕ=tan−1(x 0 ω d v 0+ω n ζ x 0)
ζ=c 2 m ω n ζ=c 2 m ω n
ω d is called the damped natural frequency of the system. It is always less than ω n:
ω d=√1−ζ 2 ω n ω d=1−ζ 2 ω n
The period of the underdamped response differs from the undamped response as well.
Undamped:τ n=2 π ω n Undamped:τ n=2 π ω n
Underdamped:τ d=2 π ω d Underdamped:τ d=2 π ω d
Comparison of Viscous Damping Cases
Responses for all four types of system (or values of damping ratio) in viscous damping. All four systems have the same mass and spring values, and have been given the same initial perturbations (initial position and initial velocity: this is apparent because they start at the same y-intercept and have the same slope at x=0).
In the figure above, we can see that the critically-damped response results in the system returning to equilibrium the fastest. Also, we can see that the underdamped system amplitude is quite attenuated compared to the undamped case.
Video Lecture
Useful Resources
Units and Conversion Table
Wolfram Alpha Equation Solver
Wolfram Alpha Derivative and Integral Calculator
Wolfram Alpha Vector Operation Calculator
2D Centroid and Area Moment of Inertia Table
3D Centroid and Mass Moment of Inertia Table
Worked Problems:
Question 1:
A 15 kg block on a frictionless surface is attached to a spring (k = 300 N/m). Find the damping constant, c, that will make the system critically damped.
Solution:
PDF Solution
Question 2:
A 20 kg block on a frictionless surface is attached to a spring (k = 700 N/m) and a damper (c = 35 N-s/m). If the initial perturbation, x 0, is 0.2 m (v 0 = 0), how many half-cycles will it take for the amplitude of the oscillation to peak at half the original displacement or less (i.e. |x peak|<=|x 0/2|)?
Solution:
PDF Solution
Question 3:
A bar of length 1.5m mass of 2kg is pinned to the ceiling. A spring, k=50N/m, is attached to the bottom of the bar and a damper, c=10Ns/m, is attached halfway down. Given a small angle displacement, find the damped frequency and the roots.
Solution:
PDF Solution
Practice Problems:
Practice Problem 1:
Practice Problem 2:
Authors: Jacob Moore, Majid Chatsaz, Agnes d'Entremont, Joan Kowalski, and Douglas Miller have licensed all content under a Creative Commons Attribution-ShareAlike 4.0 International License.This work was partially supported by the U.S. National Science Foundation Award # TUES-1044790 as well as the Affordable Course Transformation Program at PSU. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the sponsoring organizations |
16132 | https://askfilo.com/user-question-answers-smart-solutions/if-a-number-is-represented-as-1000b-100b-10c-d-find-the-3336383531343734 | If a number is represented as 1000b + 100b + 10c + d, find the number of ..
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If a number is represented as 1000b + 100b + 10c + d, find the
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If a number is represented as 1000 b+100 b+10 c+d, find the number of digits in it.
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Updated on: Aug 12, 2025
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Text solutionVerified
Concepts
Place value, number of digits in an integer, addition of terms, simplification of expressions
Explanation
The given expression represents a number formed by digits represented by variables b, c, and d. We can simplify the number first and then analyze its magnitude to find the number of digits.
Step-By-Step Solution
Step 1
Write the expression clearly:
1000 b+100 b+10 c+d
Step 2
Combine like terms (terms with b):
1000 b+100 b=1100 b
So the number becomes
1100 b+10 c+d
Step 3
Note the variables b,c,d represent digits, so each is between 0 and 9.
Step 4
To find the number of digits, analyze the possible magnitude of the number
Minimum value is when b=0,c=0,d=0 which is 0 (1 digit)
Maximum value is when b=9,c=9,d=9
Calculate the maximum:
1100×9+10×9+9=9900+90+9=9999
Step 5
Since the maximum value is 9999 (which is a 4-digit number), the number can have from 1 to 4 digits depending on the digits b,c,d.
Final Answer
The number represented by the expression 1000 b+100 b+10 c+d can have up to 4 digits.
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Question Text If a number is represented as 1000 b+100 b+10 c+d, find the number of digits in it.
Updated On Aug 12, 2025
Topic All topics
Subject Smart Solutions
Class Class 9
Answer Type Text solution:1
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Big O confusion: log2(N) vs log3(N)
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Asked 11 years, 9 months ago
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Why is O(log 2 N) = O(log 3 N) ?
I don't understand this. Does big O not mean upper bound of something?
Isn't log 2 N bigger than log 3 N ? When I graph them, log 2 N is above log 3 N .
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edited Sep 22, 2016 at 4:24
Faheel
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asked Dec 11, 2013 at 7:03
Daver MuzaffarDaver Muzaffar
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Hint: Try converting to base 3, and remember that constants don't matter.msgambel –msgambel 2013-12-11 07:06:24 +00:00 Commented Dec 11, 2013 at 7:06
You can also use limits to help find the answer. lim_{x->infty}(log_2n/log_3n) = ln3/ln2 = approx 1.58 J'e –J'e 2020-09-01 17:19:10 +00:00 Commented Sep 1, 2020 at 17:19
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Big O doesn't deal with constant factors, and the difference between Log x(n) and Log y(n) is a constant factor.
To put it a little differently, the base of the logarithm basically just modifies the slope of a line/curve on the graph. Big-O isn't concerned with the slope of the curve on the graph, only with the shape of the curve. If you can get one curve to match another by shifting its slope up or down, then as far as Big-O notation cares, they're the same function and the same curve.
To try to put this in perspective, perhaps a drawing of some of the more common curve shapes would be useful:
As noted above, only the shape of a line matters though, not its slope. In the following figure:
...all the lines are straight, so even though their slopes differ radically, they're still all identical as far as big-O cares--they're all just O(N), regardless of the slope. With logarithms, we get roughly the same effect--each line will be curved like the O(log N) line in the previous picture, but changing the base of the logarithm will rotate that curve around the origin so you'll (again) have he same shape of line, but at different slopes (so, again, as far as big-O cares, they're all identical). So, getting to the original question, if we change bases of logarithms, we get curves that look something like this:
Here it may be a little less obvious that all that's happening is a constant change in the slope, but that's exactly the difference here, just like with the straight lines above.
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edited Jan 28, 2016 at 0:45
answered Dec 11, 2013 at 7:08
Jerry CoffinJerry Coffin
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I would also note that it's not just shifting that doesn't matter, as we disregard multiplication by constants as well. f is O(n) whether f = n or f = 2n, even though 2n is not a shift upwards of n. In fact, I'd say that disregarding constant factors is a more important feature of big O.
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It is because changing base of logarithms is equal to multiplying it by a constant. And big O does not care about constants.
log_a(b) = log_c(b) / log_c(a)
So to get from log2(n) to log3(n) you need to multiply it by 1 / log(3) 2.
In other words log2(n) = log3(n) / log3(2).
log3(2) is a constant and O(cn) = O(n), thus O (log2(n)) = O (log3(n))
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edited Dec 11, 2013 at 7:13
answered Dec 11, 2013 at 7:05
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There are some good answer here already, so please read them too. To understand why Log2(n) is O(log3(n)) you need to understand two things.
1) What is mean by BigO notation. I suggest reading this: If you understnad this,you will know 2n and 16n+5 are both O(N)
2) how logarithms work. the difference between log 2 (N) and log 10(N) will be a simple ratio, easily calculated if you want it as per luk32's answer.
Since logs at different bases differ only a by a constant ratio, and Big O is indifferent to minor things like constant multiplying factors, you will often find O(logN) actually omits the base, because the choice of any constant base (eg 2,3,10,e) makes no difference in this context.
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answered Dec 11, 2013 at 9:21
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It depends on the context in which O notation is used. When you are using it in algorithmic complexity reasoning you are interested in the asymptotic behaviour of a function, ie how it grows/decreases when it tends to (plus or minus) infinity (or another point of accumulation).
Therefore whereas f(n) = 3n is always less than g(n) = 1000n they both appear in O(n) since they grow linearly (according to their expressions) asymptotically.
The same reasoning pattern can be taken for the logarithm case that you posted since different bases logarithms differ for a constant factor, but share the same asymptotical behaviour.
Changing context, if you were interested in computing the exact performance of an algorithm given your estimates being exact and not approximate, you would prefer the lower one of course. In general all computational complexity comparisons are approximation thus done via asymptotical reasoning.
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answered Dec 11, 2013 at 7:16
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Find a equidistant point between two intersecting lines
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I want to draw a straight dashed line segment given an intersection point between two lines segments. I am wondering, what is the best way to do this since I don't really have a second point to draw this line segment?
What I have done so far is calculate the intersection point of the two line segments.
So now I have one point, my question is how do I get the second point or what should I set the second point to be if I want the line to be straight from the intersection point?
So, I was thinking, I want to find an equidistant point between two intersecting lines such that that point is not the intersecting point of those two lines.
Image drawn is not exact or to scale but something like below:
How can I find the equidistant point? What's the formula for it? How do I derive the formula?
EDIT: I need a way to do this algebraically because I want to translate this into code.
geometry
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edited Mar 18, 2015 at 18:33
Kala JKala J
asked Mar 18, 2015 at 18:07
Kala JKala J
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Do you want ruler and compass constructions? Or do you want to do it the algebraic way?Dasherman –Dasherman 2015-03-18 18:09:34 +00:00 Commented Mar 18, 2015 at 18:09
Bisect the angle Henry –Henry 2015-03-18 18:12:15 +00:00 Commented Mar 18, 2015 at 18:12
Assuming the use of protractors or analytical answers, it sounds as though you are interested in a point along a line which halves the angle where the line segments intersect. This should have the property you seek, seen easily with the notion of similar triangles (granted the one triangle will be a mirror image of the other, not a rotation). For a ruler&compass construction, see the applet hereJMoravitz –JMoravitz 2015-03-18 18:12:20 +00:00 Commented Mar 18, 2015 at 18:12
Sorry, guys. I want the algebraic way! Or a way to do it so that I can translate it into code. Thanks Kala J –Kala J 2015-03-18 18:31:44 +00:00 Commented Mar 18, 2015 at 18:31
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Let x 0 x 0 be the intersection point of two lines g g and h h, given by g(t)=x 0+t v,h(t)=x 0+t w,
g(t)=x 0+t v,h(t)=x 0+t w,
where v v is a vector pointing along g g and w w is a vector pointing along h h. Then the points which have the same distance to g g and h h lie on the lines given by x 0+t(v|v|±w|w|).
x 0+t(v|v|±w|w|).
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answered Mar 18, 2015 at 18:16
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If you write your two lines as c 1 x+c 2 y=c 3 c 1 x+c 2 y=c 3 and d 1 x+d 2 y=d 3 d 1 x+d 2 y=d 3 then the equation for the line that passes directly through the middle is (c 1+c 2)x+(d 1+d 2)y=c 3+d 3(c 1+c 2)x+(d 1+d 2)y=c 3+d 3.
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answered Mar 18, 2015 at 18:13
user2566092user2566092
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So if I have the intersection point, the second point can be found if I solve for x or y in that last equation?Kala J –Kala J 2015-03-18 18:36:09 +00:00 Commented Mar 18, 2015 at 18:36
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Assuming you want a ruler+compass construction:
-Draw a circle with the point of intersection as the middle point (radius does not matter). The circle will intersect the lines at four points. Let's call them, from upper left, to upper right, to lower right, to lower left, A, B, C, D.
-Draw a circle with A as the centre and draw another circle with B as the centre. Both circles should have the same radius and this radius can be any radius, as long as those two circles will intersect. This point of intersection is then your equidistant point.
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answered Mar 18, 2015 at 18:14
DashermanDasherman
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16136 | https://www.youtube.com/watch?v=W6fpR3W52gE | Prime Numbers 1-20 || Prime Numbers from 1 to 20
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In this video, we will find the prime numbers between 1 and 20.
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hey guys welcome to minstrel in this video we will learn to find the prime numbers between 1 and 20. to know more about prime numbers you can visit mathstudent.com or check the link given in the description box now let's start let us first understand the definition of a prime number we will call a number a prime if the number is divisible only by 1 and the number itself so according to this definition the prime numbers between 1 and 20 are 2 3 five seven eleven thirteen seventeen and nineteen so the total prime numbers between one and twenty is eight thank you for watching please like and share this video if you find it useful leave a comment in the comment box do not forget to subscribe this channel please hit the bell icon to get notification of new videos earlier |
16137 | https://www.omnicalculator.com/math/weighted-average | Board
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Weighted Average Calculator
To understand how a weighted average calculator works, you must first understand what a weighted average is. Weighted average has nothing to do with weight conversion, but people sometimes confuse these two concepts. The typical average, or mean, is when all values are added and divided by the total number of values. We can compute this using our average calculator, by hand, or by using a hand-held calculator since all the values have equal weights.
But what happens when values have different weights, which means that they're not equally important? Below you will see how to calculate the weighted mean using the weighted average formula. Also, you'll find examples where the weighted average method may be used - like e.g. calculation of the GPA, average grade, or your final grade.
Prefer watching over reading? Learn all you need in 90 seconds with this video we made for you:
Let's start from the beginning: what is a weighted average?
Weighted average (weighted arithmetic mean) is a concept similar to standard arithmetic mean (called simply the average), but in the weighted average, not all elements contribute equally to the final result. We can say that some values are more important than others, so they are multiplied by a coefficient called the weight.
For example, during your studies, you may encounter a situation where the grade from an exam is two times more important than the grade from a quiz — and that's exactly what we call the weighted average method. To define it in a more mathematical way, we can write the weighted average formula as:
xˉ=w1+w2+…+wnw1x1+w2x2+…+wnxn
where x1, x2,…, xn are our numbers, and w1, w2,…, wn are our weights - the importance of the numbers in averaging.
So, having an A from an exam and a C from a quiz, you'd get a B as a standard average, but assuming that the exam is two times more important, you should get a B+.
🙋 If you're running a business, you may be interested in checking our WACC calculator, which concerns the Weighted Average Cost of Capital.
How to calculate a weighted average
One type of average which is typically weighted is a grade point average. As the calculation of GPA may sometimes be tricky, we've created two dedicated tools: the high school GPA and the college GPA calculator — have you checked them yet?
Let's find out how to calculate a weighted average - the easiest way is to look at the simple example:
Suppose a student has two four-credit classes, a three-credit class, and a two-credit class. Assume that the grades of the courses are as follows:
A for a four-credit class;
B for the other four credit class;
A for the three credit class; and
C+ for the two credit class.
Then, we need to translate the letter grades into numerical values. Most schools in the US use a so-called 4.0 GPA scale, which is a 4-point grading scale. The table below shows a typical letter grade/GPA conversion system:
| Letter Grade | Percentile | 4.0 scale | +4.0 scale |
--- --- |
| A+ | 97-100 | 4 | 4.3 |
| A | 93-96 | 4 | 4 |
| A- | 90-92 | 3.7 | 3.7 |
| B+ | 87-89 | 3.3 | 3.3 |
| B | 83-86 | 3 | 3 |
| B- | 80-82 | 2.7 | 2.7 |
| C+ | 77-79 | 2.3 | 2.3 |
| C | 73-76 | 2 | 2 |
| C- | 70-72 | 1.7 | 1.7 |
| D+ | 67-69 | 1.3 | 1.3 |
| D | 65-66 | 1 | 1 |
| F | Below 65 | 0 | 0 |
So from the table, we know that A = 4.0, B = 3.0, and C+ = 2.3. Now that we have all the information, we can have a look at how to calculate the GPA using a weighted average method:
Sum the number of credits. 4 + 4 + 3 + 2 = 13, that was a really easy step.
Take the value assigned to the grade and multiply it by the number of credits. In our case, it will be:
A ⋅ 4 credits = 4.0 ⋅ 4 = 16;
B ⋅ 4 credits = 3.0 ⋅4 = 12;
A ⋅ 3 credits = 4.0 ⋅ 3 = 12;
C ⋅ 2 credits = 2.3 ⋅ 2 = 4.6.
Add all the values. 16 + 12 + 12 + 4.6 = 44.6.
Divide the sum by the total number of credits. So, for our example, it's equal to 44.6/13 = 3.43
We may write the whole weighted average formula as:
(4 ⋅ 4 + 4 ⋅ 3 + 3 ⋅ 4 + 2.3 ⋅ 2) / (4 + 4 + 3 + 2) = 3.43
Let's compare this result to an average that is not weighted. Then we don't take the credits into account, and we divide the sum of grades by the total number.
(4 + 3 + 4 + 2.3) / 4 = 3.33
Notice how the weighted average changed. Sometimes it may be a really significant difference — like a grade difference or even whether you pass or fail your course.
Weighted average formula
Let's repeat what the weighted average formula looks like:
xˉ=w1+w2+…+wnw1x1+w2x2+…+wnxn
But what does it mean? To figure out how to calculate a weighted average, we need to know the weight of each value. Typically, we present the weights in the form of a percentage or (in statistics) a probability of occurrence.
For example, let's suppose that exams, quizzes, and homework assignments all contribute to a class's grade. Each of the three exams is worth 25 percent of the grade, the quizzes are worth 15 percent, and the homework assignments are worth 10 percent. To calculate the average, you multiply the percentage by the grades and add them together. If the test scores are 75, 90, and 88, the quiz average is 70, and the homework grade is 86, the weighted average is as follows:
(0.25 ⋅ 75 + 0.25 ⋅ 90 + 0.25 ⋅ 88 + 0.15 ⋅ 70 + 0.10 ⋅ 86) / 1 = 82.35
Compare this to a non-weighted average of (75 + 90 + 88 + 70 + 86) / 5 = 81.8
In statistics, you will often encounter a discrete probability distribution that has values for x and their associated probabilities. Since the probabilities for each value of x will likely not all be the same, we can apply the weighted average formula. Simply multiply each x value by its probability of occurring and sum the values.
🙋 In case you need to estimate the geometric mean, Omni's geometric mean calculator will come in handy.
Weighted vs. unweighted GPA for high school
We often use a weighted average to calculate the so-called weighted GPA. It's a term that rarely appears in the context of college GPA (although college GPA is computed using a weighted average method, with courses credits as weights) but is usually used for high school GPA. Let's have a closer look at this topic.
The first thing we need to emphasize: you need to be precise about what you want to take into account during weighting — credits, course difficulty, or maybe both these factors?
Course difficulty is taken into account in most weighted GPA calculations. It rewards you for taking classes of a higher level by adding extra points to your grade. There are a couple of types of more demanding courses that influence your weighted GPA score:
AP Courses (Advanced Placement Courses) usually give you an additional 1 point to your standard GPA score;
IB Courses (International Baccalaureate Courses) are also rewarded with 1 extra point;
College Prep classes can also add 1 point to your grade; and
Honors Courses most often give you an additional 0.5 points (although you can find examples of schools where it's awarded with 1 point).
So, what are the options for weighing in High school GPA calculations? Let's define:
Unweighted GPA, as the GPA where we DON'T care about course difficulty:
a) and we DON'T care about course credits:
High School GPA = Σ grade value / Σ courses
b) and we DO care about course credits:
High School GPA = Σ (grade value · credits) / Σ credits
2. Weighted GPA, as the GPA where we DO care about course difficulty:
a) and we DON'T care about course credits:
High School GPA = Σ (weighted grade value) / Σ courses
b) and we DO care about course credits:
High School GPA = Σ (weighted grade value · credits)/ Σ credits
It may look a bit overwhelming, but let's have a look at a hypothetical results sheet, and everything should be clear:
| Course | Grade |
--- |
| Maths | A |
| Physics | B+ |
| Physics lab | C+ |
| English | A- |
1 a) Unweighted GPA: we DON'T care about course difficulty and credits.
All the courses have the same grade scale and credits, no matter the course difficulty. So we may convert our grades into numbers:
| Course | Grade | Scale |
---
| Maths | A | 4.0 |
| Physics | B+ | 3.3 |
| Physics lab | C+ | 2.3 |
| English | A- | 3.7 |
Then, we can calculate the unweighted GPA as follows:
Unweighted High School GPA = Σ grade value / Σ courses
= (4.0 + 3.3 + 2.3 + 3.7) / 4 = 13.3 / 4 = 3.325 ≈ 3.33
Did you notice that it's a standard average? It's just summing all scores and dividing the result by the total number of observations (4 courses).
1 b) Unweighted GPA: we DON'T care about course difficulty, but we DO care about credits.
Things are getting more complicated when we consider the course credits. Some sources ignore the course's credits for unweighted GPA scores, but others keep them. So, if your classes have some credits/points, you can calculate the weighted average of grades and credits (but still, it's not the thing we usually name the weighted GPA):
| Course | Credits | Grade | Scale |
--- --- |
| Maths | 0.5 | A | 4.0 |
| Physics | 1 | B+ | 3.3 |
| Physics lab | 0.5 | C+ | 2.3 |
| English | 1 | A- | 3.7 |
Then, the GPA will be equal to:
High School GPA = Σ (grade value · credits) / Σ credits
= (4.0 ⋅ 0.5 + 3.3 ⋅ 1 + 2.3 ⋅ 0.5 + 3.7 ⋅ 1) / (0.5 + 1 + 0.5 + 1)
= 10.15 / 3 = 3.38333… ≈ 3.38
The courses with higher credits value have better marks in our example, so the overall GPA is also higher.
2 a) Weighted GPA: we DO care about course difficulty and DON'T care about course credits.
Depending on the course type, the letter grades are translated to different numerical values:
| Letter Grade | Percentile | Regular GPA | Honors GPA | AP / IB / College Prep GPA |
--- ---
| A+ | 97-100 | 4 | 4.5 | 5 |
| A | 93-96 | 4 | 4.5 | 5 |
| A- | 90-92 | 3.7 | 4.2 | 4.7 |
| B+ | 87-89 | 3.3 | 3.8 | 4.3 |
| B | 83-86 | 3 | 3.5 | 4 |
| B- | 80-82 | 2.7 | 3.2 | 3.7 |
| C+ | 77-79 | 2.3 | 2.8 | 3.3 |
| C | 73-76 | 2 | 2.5 | 3 |
| C- | 70-72 | 1.7 | 2.2 | 2.7 |
| D+ | 67-69 | 1.3 | 1.8 | 2.3 |
| D | 65-66 | 1 | 1.5 | 2 |
| F | Below 65 | 0 | 0 | 0 |
Continuing with our example, now our four classes have the course type assigned:
| Course | Credits | Grade | Course Type | GPA Scale |
--- ---
| Maths | 0.5 | A | Honors | 4.5 |
| Physics | 1 | B+ | Regular | 3.3 |
| Physics lab | 0.5 | C+ | Regular | 2.3 |
| English | 1 | A- | AP | 4.7 |
As two courses are not standard classes, they get extra points (A from Maths - 4.5 instead of 4.0, as it's an Honors course, A- from English - 4.7 instead of 3.7, as it's an AP course).
The formula for the calculation of weighted GPA is:
Weighted High School GPA = Σ (weighted grade value) / Σ courses
= (4.5 + 3.3 + 2.3 + 4.7) / 4 = 14.8 / 4 = 3.7 ,
where weighted grade value is a:
grade value + 0 for Regular courses;
grade value + 0.5 for Honors courses; and
grade value + 1 for AP/IB/College Prep courses.
So we omitted the courses' credits, but we've considered the course's difficulty. And finally, we have
2 b) Weighted GPA: we DO care about course difficulty and DO care about course credits.
So if you're taking into account both credits and course difficulty, then the result is:
Weighted High School GPA = Σ (weighted grade value · credits) / Σ credits
= (4.5 ⋅ 0.5 + 3.3 ⋅ 1 + 2.3 ⋅ 0.5 + 4.7 ⋅ 1) / (0.5 + 1 + 0.5 + 1) = 11.4 / 3 = 3.8
That wasn't so hard, was it?
Different averages: arithmetic, geometric, harmonic
Now that you understood what a weighted average is let's compare different averages. We've prepared for you a table that sums up all the important information about four different means:
| | Arithmetic Mean | Geometric Mean | Harmonic Mean | Weighted Mean (Weighted Arithmetic Mean) |
--- ---
| Definition | Sum of observations divided by the total number of observations. | The 'n'th root product of 'n' observations. | The reciprocal of the arithmetic mean of the reciprocals of the given set of observations. | Numbers multiplied by a weight (based on relative importance), summed, divided by the sum of weights. |
| Examples: 4, 9 | A=24+9=6.5 | G=4⋅9=36=6 | H=41+912≈5.54 | Additional info - weights. w1=3, w2=1: W=44⋅3+9⋅1=5.25 |
| Applications | Many different fields, e.g. economics, physics (e.g. mean free path), biology, history, everyday life, and health (e.g., mean arterial pressure). | "Business (investment, CAGR), math (rectangle area in terms of square side, analogically volume), signal processing (spectral flatness, choosing an aspect ratio)". | Many situations involving rates and ratios in physics (e.g., average speed), averaging multiples in finance (such as the price–earnings ratio)), geometry, chemistry, and computer science. | Education (GPA, final grades, average grades), finances (e.g., WACC - Weighted Average Cost of Capital). |
| Relationship | Arithmetic mean ≥ Geometric mean ≥ Harmonic mean (for non-negative data) | | | |
General formulas for means look as follows:
Arithmetic mean:
A=na1+a2+…+an=n1i=1∑nai
Geometric mean:
G=nx1⋅x2⋅…⋅xn=(i=1∏nxi)n1
Harmonic mean:
H=x11+x21+…+xn1n= i=1∑nxi1n=⎝⎛ni=1∑nxi−1⎠⎞−1
Weighted mean
xˉ=w1+w2+…+wnw1x1+w2x2+…+wnxn=i=1∑nwii=1∑nwixi
FAQs
How to calculate my weighted average if my course work is worth 40%?
Assuming that your test score is worth 60%, and the coursework and test scores are expressed as fractions of 100, follow these steps to calculate the weighted average:
Multiply the coursework score by 2 and the test score by 3.
Add the results together and divide by the total of the weights: 5.
How do I calculate weighted average?
To calculate the weighted average, follow these steps:
Get the weight of each number.
Multiply each number by its weight.
Add all of the results from Step 2 together.
Add all of the weights together.
Divide the answer from Step 3 by the answer in Step 4.
How do I calculate the weighted average of my purchases?
If you purchased three products of different quantities:
5 packs of acrylic paint at $19.99;
3 packs of paint brushes at $13.99; and
2 art canvases at $25.00.
Use the following steps to calculate the weighted average of your spending:
Multiplying the price by the quantity:
5 × 19.99 = $99.95
3 × 13.99 = $41.97
2 × 25.00 = $50
Find the total spent:
99.95 + 41.97 + 50 = $191.92
Find the number of products sold:
5 + 3 + 2 = 10
Find the weighted average:
191.92/10 = $19.19
What is the weighted averages of the cost of my stationary?
Assuming that you purchased:
3 packs of pencils at $5 each;
2 packs of paper at $10.00 each; and
5 packs of pens at $15.00.
Your weighted average is $11.
To calculate this, we find the total amount of money spent by following these steps:
Find the amount of money spent.
Find the total amount of items purchased.
Divide the answer in Step 1 by the answer in Step 2.
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16138 | https://testbook.com/maths/perimeter-of-isosceles-triangle | Perimeter of Isosceles Triangle: Concept, Formula, Solved Example
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Perimeter of Isosceles Triangle
Perimeter of Isosceles Triangle: Learn the Concept, Related Formula and Solved Examples.
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Overview
Test Series
Perimeter is the sum of all the lengths or boundaries of a given shape, it is generally used for land survey purposes or development.It generally outlines a two dimensional shape like atriangle, square, rectangle etc. In this article we are going to talk about the perimeter of an isosceles triangle, an isosceles triangle is a type of triangle which has two equal and identical sides and a different third side. Here, we will learn about the concept of perimeter of an isosceles triangle in brief, the related formulas and a few solved examples for a better understanding.
Perimeter of an Isosceles Triangle
Consider an Isosceles triangle AOB, such that Side AO = Side AB
The perimeter of an isosceles triangle is the sum of the lengths of all its sides,
Hence, perimeter of triangle AOB = Side AO + Side OB + Side AB.
The perimeter of an isosceles triangle can be given by the formula,
Perimeter (P) = 2a + b
Where a and b are the sides of the triangle.
Perimeter of Isosceles Triangle Derivation
Consider the figure above,
Triangle AOB is an isosceles triangle,
We know a property of an isosceles triangle that has two equal and identical sides,
Hence, Side AO = Side AB = a,
Side OB = base = b
Perimeter is defined as the sum of the lengths of all the sides of a given two dimensional polygon,
Hence perimeter of a triangle = side 1 + side 2 + side 3
Perimeter of triangle AOB = side AO + side AB + side OB
= a + a + b
= 2a + b.
Hence the formula for perimeter of an isosceles triangle is 2a + b.
Learn about Perimeter of a Right Triangle
Perimeter of Isosceles Triangle on a Graph
Consider the figure above,
Triangle ABC is an isosceles triangle,
Where sides BA and BC are identical.
The coordinates of vertex A are (-2,3).
The coordinates of vertex B are (-5,-4).
The coordinates of vertex C are (2,-1).
Side AB = −5−(−2)2+(−4−3)2−−−−−−−−−−−−−−−−−−−√−5−(−2)2+(−4−3)2
Side AB = 58−−√58 = 7.61 units
Side BC = (−5−2)2+(−4−(−1)2−−−−−−−−−−−−−−−−−−−−√(−5−2)2+(−4−(−1)2
Side BC = 58−−√58 = 7.61 units
Side AC = (−2−2)2+(3−(−1))2−−−−−−−−−−−−−−−−−−−√(−2−2)2+(3−(−1))2
Side AC = 32−−√32 = 5.65 units
Perimeter of triangle ABC = Side AB + Side AC + Side BC = 2 x 7.61 + 5.65
= 20.87 units
Hence perimeter of triangle ABC = 20.87 units.
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Perimeter of an Isosceles Triangle in Different Scenarios.
We will learn about the perimeter of an isosceles triangle with respect to different conditions like,
Perimeter of isosceles triangle with missing side.
Perimeter of isosceles triangle with base and height.
Perimeter of Isosceles Triangle with Missing Side
In an isosceles triangle if a side is missing where the length of one side and the length of base is already given.
In an isosceles triangle,
We know two of its lengths are identical,
Then length of side 1 = side 2
If side 1 = a, then side 2 = a
And side 3 = base = b
Hence the perimeter of the given triangle would be = a + a + b = 2a + b.
Perimeter of Isosceles Triangle with Base and Height
In triangle ABC,
Consider the base and height of the triangle are given,
Where base BC = b
Height AD = h
The two identical sides = a
Height (AD) = a 2−(b 4)2−−−−−−−−√a 2−(b 4)2
We can find the value of “a” as height”h” and base “b” are known and then use it in the formula
Perimeter (P) = 2a + b.
Learn about Perimeter of Scalene Triangle
Perimeter of an Isosceles Right Triangle
We have already studied that the perimeter of a triangle is the sum of the length of its all sides
We have known that the perimeter of an isosceles triangle is equal to the sum of the identical lengths and the third side.
In a right angled triangle, there are two identical, adjacent sides, represented by “a” and an hypotenuse “h”.
The perimeter of a right angled isosceles triangle = a + a + h.
Also, by using pythagoras theorem,
h 2=(a 2+a 2)−−−−−−−√h 2=(a 2+a 2)
h 2=2 a 2−−−√h 2=2 a 2
h=2–√a h=2 a.
Hence the perimeter of a right angled isosceles triangle = 2 a+2–√a 2 a+2 a
= 2–√a(2–√+1)2 a(2+1).
Also the two angles in a right angled isosceles triangle should measure 45 degrees each.
Learn about Perimeter of a Right Triangle
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Perimeter of Isosceles Triangle Solved Examples
Example 1. Find the perimeter of an isosceles triangle if the length of the two identical sides is 5 cm each and the length of its base is 7 cm.
Solution 1.
Given data,
In an isosceles triangle,
Lengths of the identical sides (a) = 5cm.
Length of the base (b) = 7cm.
Perimeter of an isosceles triangle = 2a + b
= 2 x (5) + 7
= 10 + 7 = 17 cm
The perimeter of the given isosceles triangle is 17cm.
Example 2.In a right angled isosceles triangle the length of the identical adjacent sides is 6cm each, find the perimeter and the length of its hypotenuse.
Solution 2,
Given data,
In the given isosceles triangle,
Length of the adjacent sides (a) = 6cm.
Length of the hypotenuse can be found by using the pythagoras theorem,
(6)2+(6)2−−−−−−−−−√(6)2+(6)2
h=72−−√h=72
h = 8.485cm
The length of the hypotenuse is 8.485 cm.
Hence,
Perimeter = 2a + h
= 2 x 6 + 8.485
= 12 + 8.485 = 20.485 cm.
The perimeter of a right angled isosceles triangle is 20.485 cm.
Example 3. In an isosceles triangle,
The length of the identical sides is 5cm each and its base is 4cm.
Find the height of the given isosceles triangle.
Solution 3
Given data,
Length of the identical sides (a) = 5cm
Length of the base (b) = 4cm.
The height of an isosceles triangle is given by the formula,
h=a 2−(b 4)2−−−−−−−−√h=a 2−(b 4)2
h=5 2−(4 4)2−−−−−−−−√h=5 2−(4 4)2
h=25−1−−−−−√h=25−1
h=24−−√h=24
Height (h) = 4.89cm
Height of the given isosceles triangle is equal to 4.89 cm.
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If you are checking Perimeter of isoceles traingle article, also check the related maths articles in the table below:
Uniform distributionPerimeter of cube
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Perimeter of Isosceles Triangle FAQs
How do you find the perimeter of an isosceles triangle?
The perimeter of an isosceles triangle is equal to the sum of the lengths of its two adjacent sides (a) and the third side (b) = 2 x a + b.
What is the perimeter of an isosceles right angled triangle?
The perimeter of an isosceles right angled triangle is given by 2 x l + hWhere l = length of the two identical adjacent sides and h = hypotenuse.
How do you find the perimeter of an isosceles triangle using the Pythagorean Theorem?
From the Pythagorean Theorem, we can find the length of the hypotenuse from the given two adjacent sides in a right angled isosceles triangle,Once we have the length of the hypotenuse and the lengths of two identical sides, we can find the perimeter of an isosceles triangle.
Are all isosceles triangles 45 45 and 90?
Not all isosceles triangles have the angle measurement of 45 45 and 90 degrees,an isosceles triangle can have two identical angles and a third different angle.The triangle that measures 45 45 and 90 degrees is generally a right angled triangle, which can be isosceles if it has two equal and identical sides and a different third side.
What is the formula for the area of an isosceles triangle?
the area of an isosceles triangle is given by the formula,Area (A) = ½ x base x height.
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16139 | https://www.erdosproblems.com/sources_bib/Er81d | Erdős Problem #104
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OPEN This is open, and cannot be resolved with a finite computation. - $100
Given n n points in R 2 R 2 the number of distinct unit circles containing at least three points is o(n 2)o(n 2).
#104: [Er75h,p.2][Er81d,p.144][Er83b][Er92e,p.46][Er95,p.182]
geometry
In [Er81d] Erdős proved that ≫n≫n many circles is possible, and that there cannot be more than O(n 2)O(n 2) many circles. The argument is very simple: every pair of points determines at most 2 2 unit circles, and the claimed bound follows from double counting. Erdős claims in a number of places this produces the upper bound n(n−1)n(n−1), but Harborth and Mengerson [HaMe86] note that in fact this delivers an upper bound of n(n−1)3 n(n−1)3.
Elekes [El84] has a simple construction of a set with ≫n 3/2≫n 3/2 such circles. This may be the correct order of magnitude.
In [Er75h] and [Er92e] Erdős also asks how many such unit circles there must be if the points are in general position.
In [Er92e] Erdős offered £100 for a proof or disproof that the answer is O(n 3/2)O(n 3/2).
The maximal number of unit circles achieved by n n points is A003829 in the OEIS.
See also and .
External data from the database - you can help update this
Formalised statement? No (Create a formalisation here)
Related OEIS sequences: A003829
Additional thanks to: Desmond Weisenberg
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #104, accessed 2025-09-29
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Genotype vs Phenotype: Examples and Definitions
Article
Last Updated: September 12, 2022
Written by
Molly Coddington
Molly Coddington
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Molly Coddington is a Senior Writer and Newsroom Team Lead at Technology Networks. She holds a first-class honors degree in neuroscience. In 2021 Molly was shortlisted for the Women in Journalism Georgina Henry Award.
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Any organism is a by-product of both its genetic makeup and the environment. To understand this in detail, we must first appreciate some basic genetic vocabulary and concepts. Here, we provide definitions for the terms genotype and phenotype, discuss their relationship and take a look at why and how we might choose to study them.
Genotype vs phenotype: what’s the difference?
An individual’s genotype is the combination of alleles that they possess for a specific gene. An individual’s phenotype is the combination of their observable characteristics or traits. While an organism’s genotype is directly inherited from its parents, phenotype is merely influenced by genotype. Environmental factors can also affect phenotype.
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What is the definition of a genotype?
In biology, a gene is a section of DNA that encodes a trait. The precise arrangement of nucleotides (each composed of a phosphate group, sugar and a base) in a gene can differ between copies of the same gene. Therefore, a gene can exist in different forms across organisms. These different forms are known as alleles. The exact fixed position on the chromosome that contains a particular gene is known as a locus.A diploid organism either inherits two copies of the same allele or one copy of two different alleles from their parents. If an individual inherits two identical alleles, their genotype is said to be homozygous at that locus.However, if they possess two different alleles, their genotype is classed as heterozygous for that locus. Alleles of the same gene are either autosomal dominant or recessive. An autosomal dominant allele will always be preferentially expressed over a recessive allele. The subsequent combination of alleles that an individual possesses for a specific gene is their genotype.
Genotype examples
Let’s look at a classic example – eye color.
A gene encodes eye color.
In this example, the allele is either brown, or blue, with one inherited from the mother, and the other inherited from the father.
The brown allele is dominant (B), and the blue allele is recessive (b). If the child inherits two different alleles (heterozygous) then they will have brown eyes. For the child to have blue eyes, they must be homozygous for the blue eye allele.
Figure 1: Inheritance chart detailing how an individual may inherit blue or brown eyes depending on the alleles carried by their parents, with the brown eye color allele being dominant and the blue eye color allele being recessive. Other examples of genotype include:
Hair color
Height
Shoe size
What is the definition of a phenotype?
The sum of an organism’s observable characteristics is their phenotype. A key difference between phenotype and genotype is that, whilst genotype is inherited from an organism’s parents, the phenotype is not.Whilst a phenotype is influenced the genotype, genotype does not equal phenotype. The phenotype is influenced by the genotype and factors including:
Epigenetic modifications
Environmental and lifestyle factors
Figure 2: Flamingos are naturally white in color, it is only the pigments in the organisms that they eat that cause them to turn vibrantly pink.
Phenotype examples
Environmental factors that may influence the phenotype include nutrition, temperature, humidity and stress. Flamingos are a classic example of how the environment influences the phenotype. Whilst renowned for being vibrantly pink, their natural color is white – the pink color is caused by pigments in the organisms in their diet. A second example is an individual's skin color. Our genes control the amount and type of melanin that we produce, however, exposure to UV light in sunny climates causes the darkening of existing melanin and encourages increased melanogenesis and thus darker skin.
Genotype vs phenotype: observing
Observing the phenotype is simple – we take a look at an organism’s outward features and characteristics, and form conclusions about them. Observing the genotype, however, is a little more complex. Genotyping is the process by which differences in the genotype of an individual are analyzed using biological assays. The data obtained can then be compared against either a second individual’s sequence, or a database of sequences. Previously, genotyping would enable only partial sequences to be obtained. Now, thanks to major technological advances in recent years, state-of-the-art whole genome sequencing.
Figure 3: A workflow depicting the various steps of whole genome sequencing (WGS).(WGS) allows entire sequences to be obtained. An efficient process that is increasingly affordable, WGS involves using high-throughput sequencing techniques such as single-molecule real-time (SMRT) sequencing to identify the raw sequence of nucleotides constituting an organism’s DNA. WGS is not the only way to analyze an organism’s genome - a variety of methods are available.
Genotyping Techniques:
• PCR • DNA Microarray• Allele-specific Oligonucleotide (ASO) Probes • DNA Hybridization
Why is it important to study genotype vs phenotype?
Understanding the relationship between a genotype and phenotype can be extremely useful in a variety of research areas. A particularly interesting area is pharmacogenomics. Genetic variations can occur in liver enzymes required for drug metabolism, such as CYP450. Therefore, an individual’s phenotype, i.e. their ability to metabolize a specific drug, may vary depending on which form of the enzyme-encoding gene they possess. For pharmaceutical companies and physicians, this knowledge is key for determining recommended drug dosages across populations. Making use of genotyping and phenotyping techniques in tandem appear to be better than using genotype tests alone. In a comparative clinical pharmacogenomics study, a multiplexing approach identified greater differences in drug metabolism capacity than was predicted by genotyping alone. This has important implications for personalized medicine and highlights the need to be cautious when exclusively relying on genotyping.
How can we study the relationship between genotype and phenotype?
Using animal models such as mice, scientists can genetically modify an organism so that it no longer expresses a specific gene – known as knockout mice. By comparing the phenotype of this animal to the wild type phenotype (i.e. the phenotype that exists when the gene has not been removed), we can study the role of certain genes in delivering certain phenotypes. The Mouse Genome Informatics (MGI) initiative has compiled a database of thousands of phenotypes that can be created and studied, and the genes that must be knocked out to produce each specific phenotype.
Genotype vs phenotype chart:
| | |
--- |
| Genotype | Phenotype |
| Definition | The set of genes in our DNA which are responsible for a particular trait | An organism’s observable characteristics and traits |
| Characterized by | Genotyping techniques such as WGS | Observing an organism’s outward characteristics |
| Depends on | The gene sequences an organism possesses | Genotype, PLUS epigenetics and environmental factors |
| Inherited? | Yes | No |
| Example | Genes encoding eye color | An individual with brown eyes |
What is the definition of a genotype? In biology, a gene is a section of DNA that encodes a trait. The precise arrangement of nucleotides (each composed of a phosphate group, sugar and a base) in a gene can differ between copies of the same gene. Therefore, a gene can exist in different forms across organisms. These different forms are known as alleles. The exact fixed position on the chromosome that contains a particular gene is known as a locus. A diploid organism either inherits two copies of the same allele or one copy of two different alleles from their parents. If an individual inherits two identical alleles, their genotype is said to be homozygous at that locus. However, if they possess two different alleles, their genotype is classed as heterozygous for that locus. Alleles of the same gene are either autosomal dominant or recessive. An autosomal dominant allele will always be preferentially expressed over a recessive allele. The subsequent combination of alleles that an individual possesses for a specific gene is their genotype.
What is the definition of a phenotype? The sum of an organism’s observable characteristics is their phenotype. A key difference between phenotype and genotype is that, whilst genotype is inherited from an organism’s parents, the phenotype is not. Whilst a phenotype is influenced the genotype, genotype does not equal phenotype. The phenotype is influenced by the genotype and factors including: Epigenetic modifications Environmental and lifestyle factors
Why is it important to study genotype vs phenotype? Understanding the relationship between a genotype and phenotype can be extremely useful in a variety of research areas. A particularly interesting area is pharmacogenomics. Genetic variations can occur in liver enzymes required for drug metabolism, such as CYP450. Therefore, an individual’s phenotype, i.e. their ability to metabolize a specific drug, may vary depending on which form of the enzyme-encoding gene they possess. For pharmaceutical companies and physicians, this knowledge is key for determining recommended drug dosages across populations. Making use of genotyping and phenotyping techniques in tandem appear to be better than using genotype tests alone. In a comparative clinical pharmacogenomics study, a multiplexing approach identified greater differences in drug metabolism capacity than was predicted by genotyping alone. This has important implications for personalized medicine and highlights the need to be cautious when exclusively relying on genotyping.
How can we study the relationship between genotype and phenotype? Using animal models such as mice, scientists can genetically modify an organism so that it no longer expresses a specific gene – known as knockout mice. By comparing the phenotype of this animal to the wild type phenotype (i.e. the phenotype that exists when the gene has not been removed), we can study the role of certain genes in delivering certain phenotypes. The Mouse Genome Informatics (MGI) initiative has compiled a database of thousands of phenotypes that can be created and studied, and the genes that must be knocked out to produce each specific phenotype.
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Molly Coddington
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Molly Coddington is a Senior Writer and Newsroom Team Lead at Technology Networks. She holds a first-class honors degree in neuroscience. In 2021 Molly was shortlisted for the Women in Journalism Georgina Henry Award.
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Definite Integrals
Engineering Context
Integrals and the definite definition of integrals is one of the basic principles of engineering as they are used to calculate varia areas, volumes, and masses across all disciplines of engineering.
MAE: Mechanical engineers may be tasked with designing and implementing pipes and channels for a system. The definite integral is very important in calculating the pressure, velocity, and volume flow rate of fluids in these pipes and channels. These kind of designs can be used in plans for pumps, turbines, and other fluid handling systems.
ECE: Electrical and Computer Engineers use definite integrals to calculate the energy, power, and properties of signals over time. These calculations are crucial to the designing of communication systems, control systems, and other systems that transmit and process signals.
BE: In the development of prosthetic devices, a biological engineer will use the definite integral to calculate the different forces acting on systems such as the joints and muscles while they are moving.
CEE: The definite integral is important to civil and environmental engineers in the building of structures such as beams and bridges that will withstand forces of human use (civil) as well as nature (environmental). The definite integral is used to calculate forces and stress of these structures under loads.
The Essentials
Given an equation f that is continuous on an interval from a to b, the definite integral is the area under the curve f within the interval [a, b]. Compared with the integral of a function (called an indefinite integral to contrast it with a definite integral), a definite integral looks like this:
(a) Indefinite Integral
(b) Definite Integral
Evaluating an indefinite integral will yield a function representing the entire area under the curve. A definite integral will yield a numerical value.
A Deeper Dive
If
is continuous on an interval [a, b], the definite integral of f from a to b is given by
provided that the limit exists. In practice, the definite integral is usually evaluated by first evaluating the indefinite integral:
The function F(x) is then evaluated at a and b, and the difference is calculated to find the value of the definite integral:
Consider the first example on this page. The value of the equation is f(x) = x2, and we will find the definite integral over the interval [3, 6]. First, we find the indefinite integral:
We then evaluate the value of the integral from a to b:
When evaluating an indefinite integral, we add a constant of integration C because part of the original function is lost when we calculate the derivative. There is not enough information to determine the original function from its derivative alone
However, definite integrals do not need to account for the constant of integration because it is canceled out. Since the constant of integration is constant [citation needed], the C value in F(b) is the same as the C in F(a). When you evaluate F(b) - F(a), the Cs cancel out.
Practice
Evaluate the definite integral
:
The kinematic equations model the relationships of time, displacement, velocity, and acceleration. The function that describes acceleration is the derivative of the function for velocity (). The function that describes velocity is the derivative of the function for displacement (). Using this information and your knowledge of definite integrals, derive the following kinematic equations (t = time, a = constant acceleration, vt = velocity at time t, st = displacement at time t):
More Resources
Math Is Fun: Definite Integrals
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MIT OpenCourseWare: Single Variable Calculus - Lecture 18: Definite Integrals
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16142 | https://www.pbcsf.tsinghua.edu.cn/__local/C/A0/68/F11E0BA4690D3EB1C54BC42043B_05FFEEE6_84D63.pdf | Measuring Operating Leverage Huafeng (Jason) Chen FISF, Fudan University Jason V. Chen University of Illinois at Chicago Feng Li Shanghai Advanced Institute of Finance, Shanghai Jiaotong University Pengfei Li PBC School of Finance, Tsinghua University We examine a simple measure of operating leverage: the ratio of fixed costs (measured by depreciation and amortization plus selling, general, and administrative expenses) to the market (or book) value of assets. We find that this measure of operating leverage positively predicts returns. This operating leverage measure is not explained by common factors and performs better than the traditional measures of operating leverage. Furthermore, an exploratory two-factor model with the operating leverage factor works at least as well as, but does not subsume, the Fama and French five-factor model. (JEL G11, G12, G30) Received September 28, 2021; editorial decision July 30, 2021 by Editor Jeffrey Pontiff.
Authors have furnished an Internet Appendix, which is available on the Oxford University Press Web site next to the link to the final published paper online.
Operating leverage refers to the fact that firms’ fixed production costs repre-sent a fixed stream of cash outflow and therefore behave much like financial leverage in magnifying risk and affecting expected returns. When demands are low and some firm assets are idle, firms may still have to incur fixed production costs. This mechanism accentuates variations in profits and is a major source of risk. This important idea has been put forward since at least We thank Francisco Barillas, Viet Nga Cao, Murray Carlson, Luis Garc ıa-Feij oo, Jeffrey Pontiff (the editor), Ghon Rhee, Zhongzhi Song, Hong Yan, Hong Zhang, Xiaoyan Zhang, and Hao Zhou, as well as seminar participants at Fudan University, Renming University, Tsinghua University, and Zhejiang University for com-ments. H. Chen acknowledges financial support from the Youth Innovative Team on Humanities and Social Sciences of Fudan University, Shanghai Institute of International Finance and Economics, China, and the National Natural Science Foundation of China [Grant Number: 71790591]. The operating leverage factor (OLFactor) in this paper is available at Send correspondence to Huafeng (Jason) Chen, chenh@fudan.edu.cn.
The Review of Asset Pricing Studies 12 (2022) 112–154 The Author(s) 2021. Published by Oxford University Press on behalf of The Society for Financial Studies.
All rights reserved. For permissions, please email: journals.permissions@oup.com doi:10.1093/rapstu/raab025 Advance Access publication 4 October 2021 Downloaded from by Tsinghua University user on 21 February 2023 Rubinstein (1973) and Lev (1974). More recently, Carlson, Fisher, and Giammarino (2004) argue that the book-to-market effect can be driven by the operating leverage effect. In their model, operating leverage is measured by the present value of fixed costs over the market value of assets. They assume that fixed costs are proportional to book assets, and therefore the book-to-market ratio should capture operating leverage and be positively correlated with expected returns. In the absence of a mechanism, such as operating leverage, growth stocks are typically viewed as having more growth options and should therefore have higher risk and returns. Cooper (2006), Sagi and Seasholes (2007), Obreja (2013), Ozdagli (2012), and Zhang (2005), among others, explore related mechanisms in the research.
Although this idea is appealing, measuring operating leverage is not trivial.
The most widely used measure of operating leverage is provided in Mandelker and Rhee (1984) and has been utilized more recently in Garc ıa-Feij oo and Jorgensen (2010), Chen, Kacperczyk, and Ortiz-Molina (2011), and Cao (2015), who measure operating leverage by estimating a time-series regression of earnings on revenues. Although conceptually appealing, this measure suffers from at least two issues. First, because it comes from a re-gression, the estimate is typically not precise. Second, to obtain reasonable estimates, one needs a long time series (most empirical exercises only use annual data), but production technologies may change over a long period of time.
To mitigate these issues, other accounting-based measures of operating leverage have been used. Ferri and Jones (1979) use the ratio of net fixed assets to total assets as a measure for operating leverage, which is also utilized in Garc ıa-Feij oo and Jorgensen (2010) as a secondary measure. In a seminal paper, Novy-Marx (2011) directly utilizes total operating costs (cost of goods sold [COGS], plus selling, general, and administrative expenses [SGA]) in some of his tests for the operating leverage hypothesis, which is subsequently used in Cao (2015) as well. In a recent paper, Chen, Harford, and Kamara (2019) use SGA/assets as the measure for operating leverage and examine its relation to profitability and capital structure. Our measure is similar to theirs, but we focus on examining the asset pricing implications of operating leverage.
Although the theoretical link between operating leverage and fixed costs has long been discussed, empirical tests on the relation between operating leverage and stock returns have not directly utilized fixed costs. Motivated by existing theoretical studies and our own analysis, we examine a simple mea-sure of operating leverage: fixed costs over the market value of assets. Typical production costs provided by Compustat include DA (depreciation and am-ortization), SGA, and COGS. We argue that DA and SGA costs are more likely fixed costs, while COGS is a variable cost. In theory, fixed costs are Measuring Operating Leverage 113 Downloaded from by Tsinghua University user on 21 February 2023 those that a firm must incur even when it does not produce or sell any products. Depreciation should fall within the fixed costs category, since ma-chinery depreciates even when a firm is not producing any products. While some may argue that investment in property, plant, and equipment (PPE) is a sunk cost, we view depreciation and amortization as a proxy for the differ-ence between investment cost and resale value; thus, investment in PPE is not typically entirely a sunk cost. While part of depreciation may be related to production, vintage is often the most important determinant of resale value.
In this paper, we provide evidence that firms’ resale value decreases when DA increases. SGA typically covers company overhead costs that are not product specific. For example, SGA includes accounting expenses, research and de-velopment costs, corporate expenses, labor and related expenses, and mar-keting expenses. A firm is likely to incur much of these costs even when it is not producing or selling any products. On the contrary, COGS, cost of goods sold, is recorded only when a product is produced and sold. In the extreme event of low demand for the products leading to no production or sales, COGS would also drop to zero.
Results from our analyses are consistent with the notion that DA and SGA are more likely to be fixed costs, while COGS is more of a variable cost. We calculate the ratio of aggregate DA over total assets and find it to be relatively stable over time. The same pattern is found for the ratio of aggregate SGA over total assets. On the other hand, COGS as a fraction of total assets varies widely over time. Further, in a time-series regression of aggregate DA, SGA, and COGS on revenues and total assets, the coefficient for revenues for COGS is about one, which is higher than the coefficients for DA and SGA. The results on the firm-level data are the same. Furthermore, we pro-vide evidence that firms encounter more difficulty cutting DA and SGA than in cutting COGS, especially when sales decrease.
We then test whether the empirical measure of operating leverage can explain the return over and above known factors. Our primary variable is market operating leverage: the sum of DA and SGA over the market value of assets. Our use of market value as the denominator is a direct implication of all of the theoretical models, as the market value of assets, rather than book assets, provides a better proxy for the present value of all future operating profits. We find that market operating leverage predicts higher returns. A Fama-MacBeth (1973) regression and a portfolio analysis show that market operating leverage has predictive power over and above common factors.
To further examine this issue, we examine book operating leverage (the ratio of fixed costs over book assets). Because market operating leverage consists of information about book operating leverage and book-to-market assets, examining book operating leverage amounts to examining the part of Review of Asset Pricing Studies / v 12 n 1 2022 114 Downloaded from by Tsinghua University user on 21 February 2023 market operating leverage that is not mechanically related to book-to-market.
We find that book operating leverage is negatively correlated with book-to-market, yet positively predicts returns, contrary to the predictions of the Fama-French three-factor model. Furthermore, in Fama-MacBeth regres-sions, we cannot reject the hypothesis that the coefficient for the log book operating leverage equals the coefficient for the log book-to-market ratio.
These results suggest that market operating leverage contains information over and above book-to-market and therefore may be the more fundamental variable. We also find that the return spread of a long-short portfolio sorted by operating leverage is persistently positive before and after formation, consis-tent with the risk explanation.
We then examine whether operating leverage helps explain the returns of portfolios sorted by the book-to-market ratio. We find that in return regres-sions, the coefficients for book operating leverage and the book-to-market ratio are similar in magnitude. Further, using value-weighted and equal-weighted portfolios sorted by the book-to-market ratio,the GRS test cannotrejecta two-factor model with the market return and the operating leverage factor. These results suggest that operating leverage helps explain the book-to-market effect.
Finally, we explore the performance of a two-factor model with the market return factor and the operating leverage factor. We find that the two-factor model performs at least as well as the Fama and French five-factor model in terms of squared Sharpe ratio tests and numbers of anomalies explained.
However, the two-factor model does not subsume the Fama and French five-factor model, as it explains the size, value, and investment factors, but not the profitability factors. When the profitability factor from Hou, Xue, and Zhang (2015) is added to our two-factor model, the three-factor model is at least as good as the q-factor model from Hou, Xue, and Zhang (2015).
Our paper builds on earlier work, including that of Garc ıa-Feij oo and Jorgensen (2010) and Novy-Marx (2011). Garc ıa-Feij oo and Jorgensen (2010) primarily use Mandelker and Rhee’s (1984) operating leverage mea-sure but also use net fixed assets, as in Ferri and Jones (1979), to measure operating leverage. We find that portfolios sorted by fixed costs do exhibit different sensitivities of earnings to sales (or Mandelker and Rhee (1984) and Garc ıa-Feij oo and Jorgensen (2010) regressions-based operating leverage).
However, regressions-based measures of operating leverage likely exhibit significant estimation error, and when estimated at the firm level do not predict stock returns. We also find that our measure is a more robust pre-dictor of returns than net fixed assets. Novy-Marx (2011) uses COGS plus SGA over book assets to proxy for operating leverage. His measure is closer to total costs rather than fixed costs. To be fair, Novy-Marx (2011, p. 107) is aware of this and writes that “our empirical proxy for operating leverage, operating costs over book assets, is a better proxy for gearing ðVi A=ViÞ, than Measuring Operating Leverage 115 Downloaded from by Tsinghua University user on 21 February 2023 it is for operating leverage ððVi A=ViÞðbi R bi CÞÞ, and thus implicitly assumes that the level of gearing and the degree of operational inflexibility are uncor-related across firms.” We present evidence that DA and SGA behave like fixed costs, while COGS is closer to variable costs than fixed costs.1 We also differ from Novy-Marx (2011) in that our primary variable is scaled by mar-ket value instead of book value. Because market value is a better proxy for the present value of all future profits, all theoretical models point to market value as the right scaling variable; we therefore scale by market value in our pri-mary variable but do study book value as a scaling variable in robustness checks. Our measure differs from that in Chen, Harford, and Kamara (2019) in that we include depreciation and amortization as a source of fixed costs, in addition to selling, general, and administrative costs. Machinery depreciates even when a firm is not producing any products. We view depreciation and amortization as a proxy for the difference between investment cost and resale value, and vintage is often the most important determinant of resale value.
Therefore, we do not view depreciation and amortization as entirely a sunk cost. 2 We note that our measure of operating leverage is motivated by theory that is consistent with the conditional capital asset pricing model (CAPM).
However, we believe the conditional CAPM is a sufficient, but not a neces-sary, condition for operating leverage to matter. An analogy can be made about financial leverage. Financial leverage certainly matters in magnifying risk and returns in a model of conditional CAPM, but because its intuition is clear, we expect financial leverage to matter even in a more general setting, for example, in a multifactor model. Operating leverage is similar. While oper-ating leverage certainly works in these models with conditional CAPM, it is likely to hold in a more general setting. Given the evidence in the literature (e.g., Lewellen and Nagel 2006; Clementi and Palazzo 2019), we do not be-lieve that the conditional CAPM is literally true. Nonetheless, our evidence suggests that empirical measures of operating leverage are important predic-tors of returns.
1 Our measure of fixed costs is related to the gross profitability measure in Novy-Marx (2013). We find that fixed costs and the part of gross profitability not made up of fixed costs exhibits different cross-sectional properties. In particular, the effect of fixed costs on stock returns is greater among small stocks, while the effect of the part of gross profitability not made up of fixed costs is greater among large stocks. The Internet Appendix presents the results.
2 In a seminal paper, Eisfeldt and Papanikolaou (2013) show that firms with higher organizational capital have higher returns. Their measure of organizational capital is derived from the cumulative deflated value of SGA, which is related to our measure of operating leverage. In their model, organizational capital is firm specific and should not be related to systematic risk. We believe that our results are not explained by organizational capital because of the following three findings. First, we find that another fixed cost component, DA, also positively predicts stock returns, with the predictive power on the same order of magnitude as that of SGA. Second, while we also find a lack of correlation between our measure and the earnings-GDP sensitivities at the firm level, once we mitigate estimation errors and estimate the sensitivities at the portfolio level, high book operating leverage is associated with higher systematic risk. Third, the return predictability of our measure of operating leverage remains significant in the cross-sectional Fama-MacBeth regressions controlling for organizational capital.
Review of Asset Pricing Studies / v 12 n 1 2022 116 Downloaded from by Tsinghua University user on 21 February 2023 1. Data We collect monthly stock returns from CRSP and both annual and quarterly accounting data from Compustat. We include only U.S. common stocks (CRSP share code of 10 or 11) with nonmissing SIC codes and exclude financial stocks (SICCD: 6000–6999) and utility stocks (SICCD: 4900– 4999). Our sample covers the period July 1963 to June 2016.
The key explanatory variable in this paper is the operating leverage mea-sure. Our main operating leverage measure is defined as fixed costs over the market value of assets, where we use the sum of DA (depreciation and am-ortization) and SGA (selling, general, and administrative expenses) as a mea-sure of fixed costs during the previous fiscal year, and the market value of assets at the previous fiscal year-end (book assets plus market equity, minus book equity). We refer to this measure as market operating leverage, or simply, operating leverage. To address the concern regarding a potential price effect in the denominator of our operating leverage measure, we also examine results by using book operating leverage, that is, the ratio of fixed costs over the book value of assets.
In our analysis of the operating leverage effect on stock returns, we also control for other well-known variables related to stock returns. Monthly returns, in percentages, from July of year t to June of year t þ 1 are matched with accounting variables for fiscal years that end in year t1. Size is the capitalization of the firm at the end of June. BM is the book-to-market ratio of equity. FL is financial leverage, which is defined as total liabilities over total book assets.3 gBA is the annual growth rate of total book assets. Accrual is the change in current assets, minus the change in current liability, minus depreciation, all scaled by lagged total book assets, as in Sloan (1996). IK is the investment-to-capital ratio. Momentum is the past 6-month cumulative returns (skipping a month). We also control for firm’s age, lnAGE, which is defined as the log of one plus the number of years since the firm’s first ap-pearance in CRSP.
Table 1 presents simple summary statistics for our sample. Specifically, we calculate the cross-sectional mean, median, standard deviation, 1st and 99th percentiles, and the number of observations for each month. We then average these cross-sectional measures over time. For example, the average monthly stock return is 1.24%, and the average cross-sectional standard deviation of stock returns is 15.21%. Our market operating leverage measure has a mean of 0.265 and a cross-sectional standard deviation of 0.212. Book operating leverage has a mean of 0.355 and a standard deviation of 0.242. The average number of observations is 2,686 for Market OL and Book OL and ranges 3 In a robustness check, we also include market beta as a control variable, which is the measured stock’s five-year rolling window market beta. We require a firm to have 24 valid monthly returns to compute beta. The results are qualitatively the same.
Measuring Operating Leverage 117 Downloaded from by Tsinghua University user on 21 February 2023 from 2,685 (ln(FC/TC2), ln(TC2/BA)) to 2,928 (FL, lnAGE, and ln(BA/ BE)) for other variables.4 2. Our Measures of Operating Leverage 2.1 Theoretical motivation Carlson, Fisher, and Giammarino (2004) derive a simple formula for firm beta under the assumption of the capacity constraints and no variable cost in their Proposition 2 as follows: Table 1 Summary statistics Avg. obs.
Mean Median SD p1 p99 Return (%) 2,921 1.242 0.120 15.212 67.005 229.419 Market OL 2,686 0.265 0.206 0.212 0.025 1.173 Book OL 2,686 0.355 0.296 0.242 0.047 1.294 NMOL 2,689 1.246 1.078 0.837 0.097 4.877 Size 2,901 1570.83 170.51 7,502.16 1.22 183,836.42 BM 2,882 0.886 0.692 0.754 0.061 4.423 FL 2,928 0.455 0.458 0.203 0.045 0.914 gBA 2,806 0.194 0.085 0.453 0.385 2.842 Accrual 2,800 0.019 0.029 0.115 0.345 0.456 IK 2,761 0.377 0.235 0.477 0.015 3.250 Momentum 2,916 0.068 0.027 0.336 0.573 1.390 lnAGE 2,928 2.445 2.444 0.854 0.699 4.146 ln(BA/BE) 2,928 0.693 0.591 0.492 0.042 2.684 ln(ME/MA) 2,882 0.567 0.459 0.534 2.413 0.463 DA/BA 2,920 0.043 0.037 0.029 0.003 0.167 SGA/BA 2,689 0.311 0.255 0.239 0.017 1.229 COGS/BA 2,925 0.923 0.761 0.732 0.025 4.162 ln(SGA/TC1) 2,688 1.469 1.401 0.713 3.512 0.155 ln(FC/TC2) 2,685 1.301 1.267 0.611 3.037 0.128 ln(TC1/BA) 2,689 0.012 0.063 0.713 2.387 1.569 ln(TC2/BA) 2,685 0.049 0.103 0.657 1.986 1.579 This table reports summary statistics. The summary statistics (average number of observations, mean, median, standard deviation, and 1st and 99th percentiles) are calculated for each month, and the cross-sectional measures are then averaged over time. The sample covers all U.S. common stocks with nonmissing SIC codes, excluding financial and utility stocks, from July 1963 to June 2016. Return is the stock’s monthly raw return in percentage.
Market OL is operating leverage, defined as fixed costs divided by the market value of assets (book assets plus market equity, minus book equity), where we use the sum of DA and SGA in the previous year as a measure of fixed costs (FC). Book OL is book operating leverage, which is the ratio of fixed costs to BA. NMOL is the operating leverage measure from Novy-Marx (2011), which is the ratio of SGA and COGS to book assets. Size is the capitalization of the firm at the end of June (in $M). BM is the book-to-market equity ratio. FL is financial leverage, which is defined as the ratio of total liabilities to total book assets. gBA is the annual growth rate of total book assets. Accrual is measured as in Sloan (1996). IK is the investment-to-capital ratio. Momentum is the past 6-month cumulative returns (skipping a month). lnAGE is the log of one plus the number of years since the firm’s first appearance in CRSP. DA is depreciation and amortization; SGA is selling, general, and administrative expenses; COGS is cost of goods sold; BA is total book assets; BE is book equity; ME is the same as the size measure; and MA is the market value of assets at the fiscal year-end. The first total costs (TC1) are defined as the sum of SGA and COGS, and the second total costs (TC2) are defined as the sum of DA, SGA, and COGS.
4 For details of more summary statistics and the correlation matrix of the main variables, see Section 3 of the Internet Appendix.
Review of Asset Pricing Studies / v 12 n 1 2022 118 Downloaded from by Tsinghua University user on 21 February 2023 bi t ¼ 1 þ VG i Vi v 1 ð Þ þ VF i Vi ; (1) where VG i Vi denotes the fraction of the value of a growth option in the firm value, v is a constant, i denotes the firm stage (0 for juvenile, 1 for adolescent, and 2 for mature), and VF i Vi denotes the ratio of the present value of fixed costs to the firm value. The firm’s revenue beta is normalized to one. This model is a dynamic version of Rhee (1986), who shows a similar formulation that includes financial leverage. These analyses suggest that we measure operating leverage as the ratio of the present value of fixed costs to the value of the firm.
In the Internet Appendix, we extend the model of Carlson, Fisher, and Giammarino (2004) and explicitly explore the role of variable costs. We consider two specifications of the variable costs. In the first specification, the variable cost is a constant fraction of revenue. In the second specification, the variable cost is constant per unit of quantity, and we relax the capacity constraint. We find that in both specifications, the present value of fixed costs directly enters the numerator in the equation for beta, while variable costs do not appear directly in the numerator.5 Novy-Marx (2011) also analyzes operating leverage and decomposes it into two components. We give Equation (2) of his paper here: bi A ¼ bi R þ Vi C Vi A bi R bi C ; (2) where bi A refers to asset beta, bi R refers to revenue beta, and bi C refers to total cost beta. He refers to the ratio of the present value of total cost, Vi C Vi A, as the level of gearing, and bi R bi C as related to operational inflexibility. We show that under the assumption of a constant proportional variable cost, total cost can be expressed as C ¼ F þ v R, and because the beta of fixed costs is zero and the beta of the variable cost is the same as the revenue beta, operational inflexibility can be expressed as bi R bi C ¼ Vi F Vi C bi R: (3) Thus, bi A ¼ bi R 1 þ Vi C Vi A Vi F Vi C ¼ bi R 1 þ Vi F Vi A . This analysis suggests that Novy-Marx’s (2011) notion of the level of gearing can be captured by Vi C Vi A, and operational inflexibility can be captured by Vi F Vi C. This analysis also suggests that we measure operating leverage as the ratio of the present value of fixed costs to the value of the firm. We stress that Novy-Marx (2011, p. 107) is 5 We choose to interpret beta as a covariance with the pricing kernel, not necessarily that with the market return.
We also acknowledge that variable cost may play a role through its effect on market value.
Measuring Operating Leverage 119 Downloaded from by Tsinghua University user on 21 February 2023 aware of this, and he acknowledges that “our empirical proxy for operating leverage, operating costs over book assets, is a better proxy for gearing ðVi A=ViÞ, than it is for operating leverage ððVi A=ViÞðbi R bi CÞÞ , and thus implicitly assumes that the level of gearing and the degree of operational inflexibility are uncorrelated across firms.” Noxy-Marx (2011) attempts to go deeper into the mechanisms of how production costs affect a firm’s risk profile. We impose simplifying assumptions here to obtain a working empir-ical measure.
We therefore proceed to measure operating leverage as the ratio of fixed costs to the market value of assets. We also explore a version of our model in which we scale fixed costs by the book value of assets. For parsimony, we do not attempt to compute the present value of future fixed costs. Because mar-ket value is a better proxy for the present value of all future profits, all the-oretical models point to market value as the right scaling variable; we therefore use market operating leverage as our primary variable.6 2.2 Fixed cost measures As explained in the introduction, production costs typically include DA, SGA, and COGS. We view DA as largely fixed costs because machinery depreciates even when a firm is not producing any products. DA is a proxy for the difference between investment cost and resale value, and vintage is often the most important determinant of resale value. Therefore, we do not view DA as a purely sunk cost. SGA typically covers company overhead costs that are not product specific. A firm must incur these costs even when it is not producing or selling any products. Conversely, COGS is recorded only when a product is actually produced and sold.7 Our premise is that depreci-ation and selling, general, and administrative expenses are more likely fixed costs, while the cost of goods sold is a variable cost. To examine whether this assumption is reasonable, we look at the ratio of aggregate costs (across firms) as a fraction of total assets over time, the sensitivities of costs to sales revenue, and the stickiness of DA, SGA, and COGS.
Figure 1 plots aggregate DA (depreciation and amortization), SGA (sell-ing, general, and administrative expenses), COGS (cost of goods sold), and REVT (total revenue) as fractions of total assets in each year between 1963 and 2015. As is clear from Figure 1, depreciation and amortization are almost a constant fraction of total assets. Selling, general, and administrative 6 In Section 5.1, we also find that in return regressions, the coefficients for the book-to-market ratio and book operating leverage are almost the same, and the difference is not statistically significant. Additionally, these results are consistent with the notion that market operating leverage drives the empirical performance of the book-to-market effect and the book operating leverage effect.
7 In financial statements, some companies do not list depreciation as a separate item in expenses but break depreciation into SGA or COGS (or both). These cases raise the concern of double counting when we add DA and SGA to measure fixed costs. However, Compustat makes an adjustment that eliminates the possibility of double counting in Compustat XSGA, COGS, and DA. For details, see Section 2.1 of the Internet Appendix.
Review of Asset Pricing Studies / v 12 n 1 2022 120 Downloaded from by Tsinghua University user on 21 February 2023 costs are also very stable over time as a fraction of total assets. Cost of goods sold and total revenue, on the other hand, vary significantly over time as fractions of total assets. This figure is consistent with the view that DA and SGA are more likely fixed costs, whereas COGS can be viewed as a variable cost.
Panel A of Table 2 reports time-series ordinary least squares (OLS) regres-sions of aggregate DA (depreciation and amortization), SGA (selling, gen-eral, and administrative expenses), and COGS (cost of goods sold) on aggregate REVT (total revenue) and on aggregate TA (total assets) as well as firm-level panel analysis. Simply adding these accounting variables across firms may be problematic since firms may enter and exit the market. To address this issue, we follow Chen (2017) and compute these quantities across time for a $100 initial investment. This procedure first computes the portfolio value-weighted return with and without dividends. This approach allows us to compute the price series over time; we then multiply the fundamental-to-price ratio by this price series to obtain fundamental values that are compa-rable over time. We also control for the time trend in the regressions. The results in panel A show that the coefficients for market-level regressions on ln(REVT) for DA, SGA, and COGS are 0.108, 0.468, and 1.191, respec-tively. Columns 4 to 6 of panel A produce the firm-level counterparts to columns 1 to 3 and show that the respective coefficients are 0.429, 0.438, Figure 1 Costs and revenues as fractions of total assets over time This figure plots aggregate DA (depreciation and amortization), SGA (selling, general, and administrative expenses), COGS (cost of goods sold), and REVT (total revenue) as fractions of TA (total assets) in each year between 1963 and 2015. The ratios are the sums of respective items over the sum of total assets across firms.
Measuring Operating Leverage 121 Downloaded from by Tsinghua University user on 21 February 2023 and 0.984. Both market-level analysis and firm-level analysis show that COGS is more positively correlated with REVT, with a larger slope coeffi-cient and a higher significance relative to DA and SGA. Moreover, we in-vestigate the sensitivities of various costs to decreases in sales revenue in columns 7 to 9 of panel A. The coefficient for the interactor of ln(REVT) and the dummy for decreasing revenue (b2) is significantly negative for DA (0.053, t-statistic ¼ 3.26) and SGA (0.047, t-statistic ¼ 5.27).
However, the estimated b2 is statistically indistinguishable from zero for COGS. These results indicate that DA and SGA are less sensitive to sales revenue than COGS, especially when sales decrease.
We then follow Chen, Harford, and Kamara (2019) to test the stickiness of DA, SGA, and COGS by using the logarithmic change of costs and sales in the regressions and report the results in panel B of Table 2. In this specifi-cation, the estimated coefficient for Dln(REVT), b3, is a measure of how much the costs respond to changes in sales. The closer the estimate of b3 is to zero, the stickier are the costs. When b3 is positive, the negative b4 (the estimated coefficient for the interactor term of Dln(REVT) and the dummy variable for decreasing revenue) indicates that the firm cannot decrease its costs when its sales decrease as much as the firm increases them when its sales increase. In univariate regressions of logarithmic change in costs on logarith-mic change in sales, as shown in columns 1 to 3, the estimated b3 is 0.6185 for DA, 0.5448 for SGA, and 0.9377 for COGS, indicating that DA and SGA are stickier than COGS. In columns 4 to 6, the estimated b4 is -0.441 for DA and -0.172 for SGA, much larger in magnitude than that for COGS (-0.0498).
These results suggest that it is more difficult for firms to cut DA and SGA rather than COGS when facing declining sales. Overall, these results indicate higher stickiness of DA and SGA relative to COGS and are consistent with the view that COGS is a variable cost and that DA and SGA are more likely to be fixed costs.
If parts of DA and SGA are variable costs, then our measure may be viewed as the true fixed costs plus some noise, which may lead to some attenuation bias.
2.3 Depreciation and amortization cost and resale value In this paper, we view depreciation and amortization (DA) as a proxy for the difference between investment cost and resale value, while some may argue that investment in PPE is a sunk cost. Meanwhile, the results in the previous section show that DA is more likely to be a fixed cost, meaning that opera-tional flexibility would be less for firms with higher DA.8 Thus, we conjecture that, ceteris paribus, an increase in DA is associated with a decline in resale 8 In Section 1.3 of the Internet Appendix, we explore asimple extension of the theoretical model with depreciation, and we find that DA is equivalent to other fixed costs, using the cash-flow-based approach for estimating a firm’s value and asset beta, under the condition of asset resaleability.
Review of Asset Pricing Studies / v 12 n 1 2022 122 Downloaded from by Tsinghua University user on 21 February 2023 value. In practice, we collect firms’ asset redeployability score from Kim and Kung (2017).
We first test the correlation of depreciation and amortization to the asset redeployability score in Table IA.2 in the Internet Appendix. We find that DA/BA is negatively correlated with the asset redeployability score with an Table 2 Regressions of costs on revenue A. Regressions of log costs on log revenue Market-level regressions Firm-level regressions ln(DA) ln(SGA) ln(COGS) ln(DA) ln(SGA) ln(COGS) ln(DA) ln(SGA) ln(COGS) 1 2 3 4 5 6 7 8 9 ln(REVT) (b1) 0.1081 0.4680 1.1911 0.4288 0.4377 0.9842 0.4462 0.4530 0.9886 (0.70) (3.66) (26.31) (30.45) (35.43) (104.89) (28.65) (34.79) (97.28) ln(REVT) Decrease(b2) 0.0533 0.0470 0.0135 (3.26) (5.27) (1.58) ln(TA) 0.5274 0.2977 0.1399 0.4996 0.3175 0.0065 0.4963 0.3145 0.0073 (2.50) (2.68) (3.23) (38.68) (38.00) (0.82) (38.19) (37.12) (0.91) Year 0.0147 0.0086 0.0017 (2.04) (2.43) (2.49) b1 þ b2 0.393 0.406 0.9751 p-value .0000 .0000 .0000 Obs.
53 53 53 147,275 147,275 147,275 147,275 147,275 147,275 Adj. R2 .975 .993 .999 .509 .610 .832 .510 .610 .832 B. Stickiness of DA, SGA, and COGS Dependent variables Dln(DA) Dln(SGA) Dln(COGS) Dln(DA) Dln(SGA) Dln(COGS) 1 2 3 4 5 6 Dln(REVT) (b3) 0.6185 0.5448 0.9377 0.7592 0.5997 0.9536 (29.86) (44.45) (148.69) (30.81) (38.45) (122.81) Dln(REVT)Decrease (b4) 0.4412 0.1723 0.0498 (11.86) (8.61) (3.48) b3 þ b4 0.3180 0.4274 0.9038 p-value .0000 .0000 .0000 Obs.
130,791 130,791 130,791 130,791 130,791 130,791 Adj. R2 .273 .411 .730 .292 .417 .730 Panel A report regressions of various costs on revenues. Columns 1 to 3 report the time-series OLS regressions of the log of DA (depreciation and amortization), SGA (selling, general, and administrative expenses), and COGS (cost of goods sold) of the market portfolio, on the log of REVT (total revenue) of the market portfolio, the log of TA (total assets) of the market portfolio, and a time trend. We follow Chen (2017) to calculate the annual DA, SGA, COGS, REVT, and TA of the value-weighted market portfolio with an initial investment of $100.
Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorre-lation. Columns 4 to 9 report the firm-level panel regressions of the detrended log of DA, SGA, and COGS, on the detrended log of REVT and TA. Decrease equals one for decreasing REVT and zero otherwise. In panel B, we follow Chen, Harford, and Kamara (2019) to test the stickiness of DA, SGA, and COGS. Firm-specific variables are winsorized at the 1% and 99% levels every year. Standard errors are adjusted by two-way clus-tering in the dimensions of firm and year. We also report the p-values for the tests of H0: b1 þ b2 ¼ 0 in panel A and H0: b3 þ b4 ¼ 0 in panel B. Costs, revenue, and assets are adjusted to real values by the PCE deflator. Time fixed effects are controlled for in columns 4–9 in panel A and all columns in panel B. The sample is from 1963 to 2015.
Measuring Operating Leverage 123 Downloaded from by Tsinghua University user on 21 February 2023 overall Pearson coefficient of -0.127 and an overall Spearman coefficient of -0.116. The cross-sectional correlation is stronger in magnitude with an aver-age cross-sectional Pearson coefficient of -0.153 and an average cross-sectional Spearman coefficient of -0.139.
We then follow Kim and Kung (2017) to examine the effect of the changes in aggregate uncertainty on corporate investment conditional on firms’ de-preciation and amortization-to-asset ratio. Our hypothesis is that when un-certainty increases, corporate investment would decrease more for firms with less resalable assets, meaning higher DA/BA in this circumstance, since firm managers would delay investment if liquidating capital is costly. In Table 3, we examine the effect of firms’ depreciation and amortization-to-asset ratio on the response of corporate investment to changes in aggregate uncertainty.
The First Gulf War and the 9/11 terrorist attacks are two exogenous shocks to uncertainty for analysis in columns 1 to 3 and columns 4 to 6, respectively.
The dependent variable is the corporate investment measured by capital ex-penditure scaled by lagged total assets. For each event, After is a dummy variable that equals one for quarters ending after the occurrence of the event, and zero otherwise. Tobin’s q, sales growth, and cash flow are controlled in all specifications. In columns 1 and 4, we find that the corporate investment ratio decreases by 0.211 and 0.337 percentage points after the shocks of the First Gulf War and 9/11 terrorist attacks, respectively. When we include the variable DA/BA in the regressions, as shown in columns 2, 3, 5, and 6, the coefficients for the interactor of After and DA/BA are negative and statisti-cally significant at the 10% level with or without control variables, meaning that corporate investment decreases more for firms with higher DA/BA. In addition, we use two continuous measures, the VIX index and the economic policy uncertainty (EPU) index developed by Baker, Bloom, and Davis (2016), as proxies for aggregate economic uncertainty. The results are reported in columns 7 to 10. The coefficient for the interactor of VIX and DA/BA is -21.599 with a t-statistic of -4.32 in column 8 and for the interactor of EPU and DA/BA is -4.272 with a t-statistic of -4.79 in column 10. These results indicate that corporate investment would decrease more for firms with higher DA/BA when uncertainty increases, consistent with the conjecture that firms’ resale value decreases when DA increases.
In cross-sectional analyses, we focus on within-industry variation in costs for the following two reasons. First, in equilibrium models (e.g., Novy-Marx 2011), within-industry and across-industry market structures may be differ-ent, and cost behavior within and across industries may have different pricing implications. Second, different industries may have different accounting practices regarding how to classify certain cost items into different categories.
We therefore follow Novy-Marx (2011) and Eisfeldt and Papanikolaou (2013) and focus on within-industry variations. Nonetheless, we report in the Internet Appendix that the main results are the qualitatively the same if we do not use industry adjustment.
Review of Asset Pricing Studies / v 12 n 1 2022 124 Downloaded from by Tsinghua University user on 21 February 2023 3. Can Operating Leverage Be Explained by Book-to-Market?
3.1 Univariate sorts by operating leverage To look at the relation between our operating leverage measure and stock returns, we first sort stocks each June according to their operating leverage measures relative to their industry peers. Specifically, we use the 17-industry classification in Fama and French (1997), as in Eisfeldt and Papanikolaou (2013). In Section 16 of the Internet Appendix, we explore the robustness of our results with raw measures and the 49-industry adjustment. Within each industry, we assign firms a rank (1 to 10) based on the firm’s decile of the operating leverage measure. The average stock returns are then computed for Table 3 Depreciation and amortization-to-asset ratio, changes in aggregate uncertainty, and corporate investment First Gulf War 9/11 terrorist attacks Economic uncertainty 1 2 3 4 5 6 7 8 9 10 After 0.2109 0.3373 (6.54) (5.39) AfterDA/BA 2.38212.4955 3.16733.7279 (1.68) (1.85) (1.91) (2.31) VIX 0.4256 (2.91) VIXDA/BA 21.5987 (4.32) EPU 0.2143 (7.42) EPUDA/BA 4.2716 (4.79) DA/BA 1.9910 2.1079 (1.29) (1.47) Tobin’s q 0.3023 0.1636 0.2053 0.1756 0.1978 0.1757 (5.21) (4.34) (7.56) (6.70) (7.48) (6.71) Sales growth 0.2723 0.0315 0.3637 0.2828 0.3470 0.2836 (2.72) (0.44) (6.41) (5.54) (6.35) (5.58) Cash flow 0.0153 0.8874 1.2688 1.3185 1.2240 1.3126 (0.02) (1.83) (3.05) (3.44) (3.06) (3.39) Firm FE Y Y Y Y Y Y Y Y Y Y Year-month FE N Y Y N Y Y N Y N Y Obs 10,047 10,047 10,047 15,493 15,493 15,493 191,926 191,926 191,926 191,926 R2 .653 .665 .668 .671 .683 .687 .512 .534 .513 .534 In this table, we follow Kim and Kung (2017) in examining the effect of changes in aggregate uncertainty on corporate investment conditional on firms’ depreciation and amortization-to-asset ratio (DA/BA). The depen-dent variable is corporate investment measured by capital expenditure scaled by lagged total assets. The sample consists of firm-quarter observations in seven quarters surrounding the event date of the First Gulf War (August 2, 1990) in columns 1 to 3 and the event date of the 9/11 terrorist attack (September 11, 2001) in columns 4 to 6.
The dummy variable After equals one for quarters ending after an increase in aggregate economic uncertainty, and zero otherwise. DA/BA is held fixed for a given event window and measured in the latest fiscal year ending before or at the beginning of each event period. In columns 7 to 10, we use the VIX index and the economic policy uncertainty (EPU) index developed by Baker, Bloom, and Davis (2016) as proxies for aggregate economic uncertainty. VIX and EPU are the average levels of the VIX and the EPU indices during a 3-month period ending a month before a given quarter’s ending date. The sample consists of firm-quarter observations and is from 1989 to 2016. Tobin’s q is the ratio of the market value of assets to the book value of assets. Sales growth is computed as the first difference of the natural log of sales. Cash flow is income before extraordinary items plus depreciations and amortizations divided by lagged book assets. All standard errors are adjusted for sample clustering at the industry level, and t-statistics are in parentheses.
Measuring Operating Leverage 125 Downloaded from by Tsinghua University user on 21 February 2023 each operating-leverage-sorted portfolio and for each month between July and the following June.
Panel A of Table 4 shows results from 10 market operating-leverage-sorted deciles. As we move from decile 1 to decile 10, market operating leverage increases from 0.060 to 0.686. These 10 deciles have different returns.
The value-weighted portfolio returns are almost monotonically increasing in operating leverage, and the return spread between portfolio 10 (H) and port-folio 1 (L) is 0.771% per month with a t-statistic of 3.83. For the equal-weighted portfolio returns, the return spread is even greater at 1.430% per month between high and low operating leverage deciles with a t-statistic of 8.55. This return spread is both statistically and economically significant and shows a positive relation between firms’ operating leverage and their stock returns.
The above analysis uses market operating leverage defined as fixed costs over the market value of assets. Readers might be legitimately concerned that our operating leverage effect on stock returns is simply driven by prices in the denominator of market operating leverage. To address this concern, we redo all of the above analysis using book operating leverage, in which prices play no direct role.
Panel B of Table 4 reports results for portfolios sorted by book operating leverage. As we move from decile 1 to decile 10, book operating leverage increases from about 0.103 to 0.814. Book operating leverages are typically higher than market operating leverages because the book value of assets is typically lower than the market value of assets. Sorting on book operating leverage also produces dispersion in returns, although the dispersion tends to be smaller. The return spread between the high and low book operating leverage deciles is 0.527% per month for value-weighted deciles and 0.603% per month for equal-weighted deciles, and both are statistically and economically significant, albeit smaller than the return spread of the market operating leverage deciles.
We also examine the return persistence of long-short portfolios sorted by our operating leverage measures surrounding portfolio formation.
Figure 2 plots the long-short decile portfolio return spread. In Figure 2, panel A, we find that the return spread of long-short portfolios sorted by market operating leverage is mostly positive from 6 months prior to formation to 18 months after formation. The relative high return spreads 6 months prior to and 7 months after formation are likely due to the January effect. We find a similar result that the long-short spread is persistently positive before and after formation for book operating-leverage-sorted portfolios in Figure 2, panel B. The persistence of the long-short portfolio return spread surrounding formation is consistent with the risk explanation, if we believe that the return spread is compen-sation for persistent risk instead of transitory mispricing.
Review of Asset Pricing Studies / v 12 n 1 2022 126 Downloaded from by Tsinghua University user on 21 February 2023 One noteworthy finding in Table 9 is that our operating leverage measures decrease with size and increase with the operating leverage measure from Novy-Marx (2011). We will compare our operating leverage measures with the operating leverage measure from Novy-Marx (2011) in Section 7.1.
3.2 Evidence from cross-sectional regressions In the previous section, we have shown that returns are higher for stocks with higher market and book operating leverages. However, our operating lever-age measures might simply be proxies for other existing predictors of stock Table 4.
Operating-leverage-sorted deciles A. Portfolios sorted by market operating leverage Decile L 2 3 4 5 6 7 8 9 H H-L Market OL 0.060 0.103 0.136 0.167 0.201 0.240 0.287 0.349 0.443 0.686 0.626 (28.16) (31.24) (33.23) (34.65) (36.08) (37.62) (39.05) (40.49) (42.42) (43.78) (41.81) VWRet(%) 0.713 0.767 0.917 0.933 1.151 1.065 1.278 1.124 1.288 1.484 0.771 (3.11) (4.00) (4.98) (5.16) (5.94) (5.98) (6.43) (5.60) (5.86) (5.96) (3.83) EWRet(%) 0.545 0.803 0.962 1.072 1.190 1.210 1.454 1.560 1.641 1.975 1.430 (1.96) (3.05) (3.79) (4.19) (4.52) (4.66) (5.44) (5.60) (5.60) (5.96) (8.55) Size 3,189 2,682 2,745 2,418 1,827 1,266 750 435 293 114 3,074 (7.27) (7.51) (6.70) (6.66) (7.06) (7.00) (7.34) (9.46) (8.69) (12.68) (7.10) lnSize 5.496 5.420 5.342 5.163 4.965 4.726 4.449 4.144 3.794 3.118 2.378 (37.18) (38.38) (37.64) (38.40) (39.64) (39.33) (40.08) (39.63) (40.50) (36.80) (28.71) NMOL 0.823 0.945 1.036 1.094 1.171 1.252 1.338 1.429 1.565 1.827 1.004 (36.72) (52.24) (55.35) (60.20) (57.21) (61.77) (66.63) (71.22) (76.56) (92.01) (59.66) B. Portfolios sorted by book operating leverage Decile L 2 3 4 5 6 7 8 9 H H-L Book OL 0.103 0.163 0.208 0.249 0.292 0.336 0.389 0.456 0.557 0.814 0.711 (76.04) (105.07) (107.03) (98.96) (94.73) (91.69) (83.01) (75.06) (65.80) (48.91) (40.76) VWRet(%) 0.706 0.833 0.827 0.907 0.934 0.956 1.114 1.086 1.016 1.233 0.527 (3.52) (3.91) (4.22) (4.77) (4.79) (5.06) (6.06) (5.45) (4.75) (5.51) (3.52) EWRet(%) 0.885 1.025 1.127 1.149 1.290 1.260 1.315 1.396 1.440 1.489 0.603 (3.34) (3.94) (4.36) (4.54) (5.03) (4.78) (4.89) (4.99) (4.95) (4.37) (3.52) Size 2,270 1,894 2,289 2,222 2,051 1,919 1,144 933 684 359 1,910 (6.42) (7.06) (6.50) (6.18) (7.77) (7.89) (8.71) (8.83) (9.31) (8.68) (6.02) lnSize 5.098 5.095 5.042 4.961 4.852 4.721 4.590 4.414 4.179 3.702 1.396 (37.12) (39.25) (37.61) (37.32) (38.34) (38.75) (39.41) (40.30) (42.70) (39.32) (18.19) NMOL 0.771 0.930 1.028 1.097 1.178 1.247 1.310 1.429 1.572 1.911 1.140 (45.10) (52.23) (59.02) (54.40) (56.61) (56.24) (60.80) (66.67) (77.81) (105.96) (95.10) We sort stocks into 10 deciles in June of year t according to their operating leverage measures within each of the Fama and French (1997) 17 industries. Monthly returns are from July of year t to June of year tþ1. Accounting variables are for fiscal years that end in year t-1. The measures are calculated for each operating-leverage-sorted decile and for each month. Finally, we average over time to get the average values for each operating-leverage-sorted decile. Panel A reports results for portfolios sorted by market operating leverage (market OL, the ratio of fixed costs to market assets). Panel B reports results for portfolios sorted by book operating leverage (book OL, the ratio of fixed costs to book assets). Fixed costs are defined as the sum of DA (depreciation and amortization) and SGA (selling, general, and administrative expenses). VWRet (EWRet) is the value-weighted (equal-weighted) average portfolio return in percentage. Size is the capitalization of the firm at the end of June (in $M). NMOL is the operating leverage measure from Novy-Marx (2011), which is the ratio of SGA and COGS to book assets. Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
Measuring Operating Leverage 127 Downloaded from by Tsinghua University user on 21 February 2023 returns. We now control for other variables that are known to predict returns, such as size, book-to-market, and momentum. To do so, we estimate cross-sectional regressions of returns on operating leverage, while controlling for other variables.
Table 5 reports the results of Fama-MacBeth cross-sectional regressions.
For each month (from July in year t to June in year tþ1), we estimate a regression of stock returns on operating leverages, which uses accounting information from year t-1. To capture the influence of intra-industry differ-ences in operating leverage on stock returns, we include the industry fixed -2 -1 0 1 2 3 4 5 6 7 8 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Long-Short Portfolio Return(%) Month Market OL-sorted long-short portfolio spread surrounding formation Equal-Weighted Return Value-Weighted Return -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Long-Short Portfolio Return(%) Month Book OL-sorted long-short portfolio spread surrounding formation Equal-Weighted Return Value-Weighted Return A B Figure 2 Persistence of long-short return spread surrounding formation This figure plots the return spread of long and short decile portfolios sorted by market operating leverage (Market OL) and book operating leverage (Book OL). Market OL is operating leverage, defined as fixed costs divided by the market value of assets (book assets plus market equity, minus book equity), where we use the sum of DA (depreciation and amortization) and SGA (selling, general, and administrative expenses) in the previous year as a measure of fixed costs. Book OL is book operating leverage, which is the ratio of fixed costs to total book assets. We sort stocks into 10 deciles in June of year t according to their operating leverage measures within each of the Fama and French (1997) 17 industries. We calculate the value-weighted and equal-weighted return spreads between the highest decile and the lowest decile from 6 months prior to the formation to 18 months after the formation. Then we plot the average return spread in the figures. The sample period is from July 1963 to June 2016.
Review of Asset Pricing Studies / v 12 n 1 2022 128 Downloaded from by Tsinghua University user on 21 February 2023 effect in all specifications. We then report average coefficients and t-statistics from such cross-sectional regressions. Unless stated otherwise, all t-statistics in this paper are adjusted using the Newey-West (1987) procedure and ac-count for heteroscedasticity and autocorrelations. Column 1 shows that, consistent with the portfolio sort result, stocks with higher market operating leverages have higher future returns. The univariate regression coefficient of market operating leverage is 1.615. This effect is economically and statisti-cally significant (t-statistic ¼ 6.56). In column 2, we simply estimate a regres-sion of returns on lnSize and lnBM. Consistent with well-known results, small Table 5 Fama-MacBeth regressions Model 1 2 3 4 5 6 7 8 9 Market OL 1.6152 0.9610 0.7025 (6.56) (5.03) (3.82) Book OL 0.6734 0.7972 0.5654 (3.43) (4.96) (3.72) FC/lagMA 0.4050 (2.33) FC/lagBA 0.5027 (4.59) lnSize 0.1123 0.0807 0.0908 0.0811 0.0933 0.1005 0.0944 (2.58) (1.94) (2.53) (1.97) (2.63) (2.73) (2.63) lnBM 0.3303 0.2583 0.2064 0.3914 0.3025 0.2423 0.3053 (5.13) (3.91) (3.57) (6.28) (5.48) (4.21) (5.35) FL 0.1772 0.1327 0.2100 0.1260 (0.90) (0.69) (1.04) (0.64) gBA 0.3149 0.3017 0.6095 0.6525 (4.12) (3.94) (5.32) (5.72) Accrual 0.7791 0.8166 0.7155 0.7110 (3.98) (4.18) (2.42) (2.46) IK 0.1481 0.1534 0.0312 0.0490 (2.84) (3.03) (0.41) (0.63) Momentum 0.2485 0.2577 0.1302 0.1290 (1.12) (1.15) (0.54) (0.54) lnAGE 0.0072 0.0126 0.0095 0.0167 (0.20) (0.35) (0.24) (0.42) Avg. obs.
2,509 2,509 2,509 2,509 2,509 2,509 2,509 2,388 2,388 Adj. R2 .0264 .0397 .0406 .0528 .0250 .0406 .0528 .0509 .0504 We first estimate cross-sectional regressions for each month (July 1963 to June 2016). The time series of the coefficients from the first-stage regressions are then used to calculate the average coefficients and the t-statistics.
The adjusted R2 and the number of observations are the averages of the cross-sectional regressions. Monthly returns in percentages from July of year t to June of year tþ1 are matched with accounting variables for fiscal years that end in year t-1. Market operating leverage (market OL) is defined as fixed costs divided by the market value of assets (MA). Fixed costs (FC) are defined as the sum of depreciation and amortization and selling, general, and administrative expenses. Book operating leverage (book OL) is defined as fixed costs divided by the book value of assets (BA). Size is the capitalization of the firm at the end of June (in $M). BM is the book-to-market equity ratio. FL is financial leverage, which is defined as the ratio of total liabilities to total book assets.
gBA is the annual growth rate of total book assets. Accrual is measured as in Sloan (1996). IK is the investment-to-capital ratio. Momentum is the past 6-month cumulative returns (skipping a month). lnAGE is the log of one plus the number of years since the firm’s first appearance in CRSP. Note that we take the logarithm for both size and book-to-market equity in the regressions. In column 8, lagMA is the market value of assets at the end of fiscal year t-2. In column 9, lagBA is the book value of assets at the end of fiscal year t-2. Explanatory variables are winsorized at the 1% and 99% levels every month. The industry fixed effect is controlled for in all speci-fications. We use the 17-industry classification in Fama and French (1997). Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
Measuring Operating Leverage 129 Downloaded from by Tsinghua University user on 21 February 2023 stocks and value stocks have higher returns. In column 3, we jointly estimate a regression of returns on market operating leverage, lnSize, and lnBM.
Comparing columns 1 and 3 gives us two more important results. First, the operating leverage measure is still significant even after controlling for size and the book-to-market ratio. Second, both the operating leverage and book-to-market coefficients become smaller in column 3. This result indicates that operating leverage is positively correlated with the book-to-market ratio.
At the same time, operating leverage still has additional explanatory power beyond that of the book-to-market ratio. We further control for financial leverage, asset growth, accrual, investment, momentum, and firm age in col-umn 4. The results show that the operating leverage effect is still highly sig-nificant, with a coefficient of 0.703 and a t-statistic of 3.82, after controlling for all the other variables. This magnitude means that a one-standard-deviation increase in market operating leverage is associated with a 1.79% increase in annualized returns.
Columns 5 through 7 present the Fama-MacBeth regression results using book operating leverage. Similar to the market operating leverage results in column 1, book operating leverage is associated with higher returns. The average coefficient is positive at 0.673, which suggests that if book operating leverage increases, stock returns tend to increase as well. The effect is statis-tically significant in the univariate regression with a t-statistic of 3.43.
Moreover, controlling for size and the book-to-market ratio makes the effect of book operating leverage larger: the coefficient increases from 0.673 (col-umn 5) to 0.797 (column 6), and the t-statistic increases from 3.43 to 4.96.
Furthermore, the coefficient for book operating leverage remains significant, 0.565 (t-statistic ¼ 3.72), after controlling for other predictors of returns (column 7). This magnitude means that a one-standard-deviation increase in book operating leverage is associated with a 1.51% increase in annualized returns.
Considering that depreciation and amortization costs are included in our fixed costs, we scale fixed costs by the market value of assets in the previous fiscal year as an alternative market operating leverage measure and by the book value of assets in the previous fiscal year as an alternative book oper-ating leverage measure. In columns 8 and 9, the coefficient decreases from 0.703 to 0.405 for the alternative market operating leverage and decreases from 0.565 to 0.503 for the alternative book operating leverage, and both remain statistically significant. The results verify the robustness of our oper-ating leverage measure, although the economic magnitude for alternative measures of operating leverage declines somewhat.
3.3 Evidence from time-series regressions We now use a different methodology to examine whether operating leverage is subsumed by the known factors, such as the Fama-French three factors.
Review of Asset Pricing Studies / v 12 n 1 2022 130 Downloaded from by Tsinghua University user on 21 February 2023 Specifically, we estimate factor models on our operating-leverage-sorted dec-iles and examine the alphas.
We first examine deciles sorted by market operating leverage within each industry. Panel A of Table 6 presents, for the value-weighted highest decile and lowest decile as well as the high-minus-low portfolio,9 the excess return, one-factor (MktRf) alpha, Fama-French (1993) three-factor (MktRf, SMB, HML) alpha, four-factor (three-factor plus UMD) alpha, five-factor (four-factor plus liquidity factor LIQT) alpha, Fama-French (2015) five-factor (three-factor plus RMW and CMA) alpha, q-factor (MKT, ME, IA, and ROE) alpha, and Misp-factor (MktRf, SMB, MGMT, PERF) alpha. The LIQT is the tradable liquidity factor from P astor and Stambaugh (2003). The q-factor model is from Hou, Xue, and Zhang (2015).10 The Misp-factor model is with mispricing factors from Stambaugh and Yuan (2017).
Column 1 shows that deciles sorted by market operating leverage exhibit increasing returns (shown previously in panel A of Table 4). The difference between high-minus-low market operating leverage is 0.771% per month and is statistically significant. Column 2 shows that CAPM does not explain the returns of market operating leverage portfolios. The difference between the alphas of the high and low market operating leverage portfolios increases slightly to 0.783% per month and remains statistically significant. When we use Fama-French three factors, the alpha of the high-minus-low (H-L) mar-ket operating leverage portfolio decreases to 0.427% per month but is still statistically significant. The differences in alphas for the four-factor, five-fac-tor, FF5-factor, q-factor, and Misp-factor models are somewhat smaller than that for the three-factor model, but they are all statistically significant. These results suggest that value-weighted portfolios sorted by market operating leverages are not explained by common factor models.
Panel B reports results for the equal-weighted portfolios sorted by market operating leverage. The alpha for the equal-weighted H-L portfolio is higher than that for the value-weighted portfolio. The alpha spread across all models exceeds 1% per month and is highly statistically significant (t-statistics of all models exceed six).
Panel C of Table 6 also shows that market operating leverage contains information that cannot be completely explained by the common factors.
Specifically, one can view market operating leverage as containing informa-tion about book operating leverage and book-to-market assets. The book-to-market assets are mechanically related to book-to-market equity, but book operating leverage is likely to include information about market operating leverage that is not captured by book-to-market. We therefore examine port-folios sorted by book operating leverage in panel C. Similar to previous 9 We report the excess returnsand time-series regression alphas for each decile sorted bymarket or book operating leverage in Table IA.3 in the Internet Appendix.
10 We thank the authors for kindly providing us with the data.
Measuring Operating Leverage 131 Downloaded from by Tsinghua University user on 21 February 2023 panels, we present for each book operating leverage decile (value-weighted) the excess return, one-factor (MktRf), three-factor (MktRf, SMB, HML), four-factor (three-factor plus UMD), five-factor (four-factor plus liquidity factor LIQT), FF5-factor (three-factor plus RMW and CMA), q-factor (MKT, ME, IA, and ROE), and Misp-factor (MktRf, SMB, MGMT, Table 6 Alphas of portfolios sorted by operating leverages DecilesExcess return 1-factor Alpha 3-factor alpha 4-factor alpha 5-factor alpha FF5-factor alpha q-factor alpha Misp-factor alpha A. Value-weighted deciles sorted by market OL L 0.3169 0.2615 0.0698 0.0182 0.0111 0.0353 0.0573 0.0720 (1.36) (2.68) (0.97) (0.26) (0.15) (0.45) (0.55) (0.71) H 1.0881 0.5214 0.3568 0.3438 0.3150 0.3948 0.4234 0.4171 (4.35) (3.37) (2.89) (2.65) (2.23) (3.29) (2.86) (3.33) HL 0.7712 0.7829 0.4266 0.3256 0.3039 0.3595 0.3661 0.3451 (3.83) (3.73) (2.94) (2.17) (1.89) (2.60) (2.08) (2.26) B. Equal-weighted deciles sorted by market OL L 0.1487 0.5080 0.5495 0.2967 0.3250 0.3902 0.2092 0.1243 (0.53) (3.82) (6.50) (3.40) (3.59) (3.24) (1.38) (1.01) H 1.5789 1.0179 0.7396 0.9070 0.9067 0.8337 1.1902 0.9462 (4.72) (4.57) (5.13) (5.52) (5.19) (5.54) (5.82) (5.21) H-L 1.4303 1.5260 1.2891 1.2037 1.2317 1.2240 1.3994 1.0705 (8.55) (8.68) (8.31) (7.50) (7.26) (7.60) (7.44) (6.47) C. Value-weighted deciles sorted by book OL L 0.3102 0.2101 0.2319 0.1444 0.1057 0.1554 0.0404 0.0136 (1.52) (2.42) (2.74) (1.82) (1.24) (1.74) (0.41) (0.17) H 0.8371 0.2953 0.3135 0.2841 0.2149 0.3083 0.2669 0.2590 (3.68) (2.58) (2.99) (2.78) (2.05) (2.58) (2.00) (2.26) H-L 0.5270 0.5055 0.5453 0.4285 0.3206 0.4636 0.3073 0.2727 (3.52) (3.26) (4.13) (3.49) (2.44) (3.04) (1.83) (2.01) D. Equal-weighted deciles sorted by book OL L 0.4895 0.1026 0.3488 0.1330 0.1573 0.3378 0.1177 0.1047 (1.83) (0.76) (4.15) (1.69) (1.88) (3.23) (0.82) (1.12) H 1.0929 0.4808 0.3367 0.5689 0.5478 0.5305 0.8825 0.7029 (3.18) (2.17) (2.34) (3.18) (2.97) (3.09) (3.27) (3.19) H-L 0.6034 0.5834 0.6854 0.7019 0.7051 0.8683 1.0002 0.8076 (3.52) (3.47) (4.56) (4.12) (4.00) (5.42) (3.76) (3.81) This table reports the excess returns (raw returns minus the risk-free rate) of operating leverage (market OL and book OL) sorted deciles as well as the time-series regression alphas from (1) one-factor (MktRf); (2) three-factor (MktRf SMB HML); (3) four-factor (MktRf SMB HML UMD); (4) five-factor (MktRf SMB HML UMD LIQT); (5) Fama-French five-factor (MktRf SMB HML RMW CMA) models; (6) q-factors from Hou, Xue, and Zhang (2015); and (7) mispricing factors from Stambaugh and Yuan (2017). We sort stocks into 10 deciles in June of year t according to their operating leverage measures within each of the Fama and French (1997) 17 industries. Monthly returns are in percentages from July of year t to June of year tþ1. Market operating leverage (market OL) is defined as fixed costs divided by the market value of assets. Fixed costs are defined as the sum of depreciation and amortization and selling, general, and administrative expenses. Book operating leverage (book OL) is defined as fixed costs divided by the book value of assets. LIQT refers to the tradable liquidity factor from P astor and Stambaugh (2003). The sample period is from July 1963 to June 2016. Panels A and B report results for portfolios sorted by market OL. Panels C and D report results for portfolios sorted by book OL. Panels A and C are for value-weighted portfolios, and panels B and D are for equal-weighted portfolios. Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
Review of Asset Pricing Studies / v 12 n 1 2022 132 Downloaded from by Tsinghua University user on 21 February 2023 PERF) alphas. Column 1 of panel C shows that the excess return of the high-minus-low (H-L) book operating leverage portfolio is positive at 0.527% per month and statistically significant. The alpha of the H-L portfolio from the CAPM is similar at 0.506% per month. Once the SMB and HML factors are included, the alphas become slightly larger. For the H-L portfolio, the three-factor alpha is 0.545% per month (t-statistic ¼ 4.13). The fact that the three-factor alpha is larger than the one-factor alpha is interesting (we explore that fact later in Table 7). The four-factor, five-factor, FF5-factor, q-factor, and Misp-factor alphas are 0.429%, 0.321%, 0.464%, 0.307%, and 0.273% per month, respectively.
Panel D shows that the results are even stronger for equal-weighted book operating leverage deciles: the H-L portfolio excess return is 0.603% and statistically significant (t-statistic ¼ 3.52). The H-L portfolio has a CAPM alpha of 0.583% per month and is statistically significant (t-statistic ¼ 3.47).
Just like the value-weighted portfolios, including SMB and HML actually increases the alpha. The three-factor alpha for the H-L book operating le-verage portfolio is 0.685% per month (t-statistic ¼ 4.56). The four-factor, five-factor, FF5-factor, and Misp-factor alphas are 0.702% (t-statistic ¼ 4.12), 0.705% (t-statistic ¼ 4.00), 0.868% (t-statistic ¼ 5.42), and 0.808% (t-statistic ¼ 3.81) per month, respectively. The q-factor alpha is even larger at 1.000% (t-statistic ¼ 3.76).
In Table 7, we present the Fama-French three-factor loadings for each operating leverage decile. Panel A reports the results for the value-weighted portfolio sorted by market operating leverage relative to industry peers. The portfolio with high market operating leverage has a higher SMB loading, consistent with the finding in Table 4 that operating leverage decreases with firms’ size. Decile 10 has an SMB loading of 0.61, and decile 1 has an SMB loading of -0.02. The difference is 0.63 and statistically significant (t-statistic ¼ 7.64). The portfolio with high market operating leverage also has a higher HML loading. Decile 10 has an HML loading of 0.22, and decile 1 has an HML loading of -0.44. The difference is 0.66 and statistically significant (t-statistic ¼ 8.00). These results suggest that stocks with higher market oper-ating leverage are likely to be small and value stocks.
Panel B shows the results for equal-weighted portfolios sorted by market operating leverage. Again, the portfolio with high market operating leverage has higher SMB and HML loadings. Decile 10 has an SMB loading of 1.23, and decile 1 has an SMB loading of 0.78. The difference is 0.45 and statisti-cally significant (t-statistic ¼ 5.19). Decile 10 has an HML loading of 0.32, and decile 1 has an HML loading of -0.11. The difference is 0.43 and statis-tically significant (t-statistic ¼ 5.93).
Panel C of Table 7 reports Fama-French three-factor loadings for the value-weighted portfolios sorted by book operating leverage relative to in-dustry peers. The results are different from those for market operating lever-age. The portfolio with high book operating leverage has higher SMB but Measuring Operating Leverage 133 Downloaded from by Tsinghua University user on 21 February 2023 Table 7 Three-factor loadings of portfolios sorted by operating leverages Deciles L 2 3 4 5 6 7 8 9 H H-L A. Value-weighted deciles sorted by market OL Alpha 0.0698 0.0760 0.0645 0.0317 0.2693 0.1126 0.3409 0.1117 0.2260 0.3568 0.4266 (0.97) (1.23) (1.05) (0.54) (2.85) (1.49) (4.41) (1.29) (2.35) (2.89) (2.94) MktRf 1.0850 0.9897 0.9872 0.9749 0.9501 0.9729 0.9783 1.0072 1.0829 1.0449 0.0401 (37.71) (46.00) (45.43) (52.91) (37.24) (54.05) (35.00) (37.21) (30.90) (35.53) (1.00) SMB 0.0164 0.0399 0.0490 0.0265 0.1416 0.1212 0.1759 0.2456 0.3820 0.6115 0.6279 (0.40) (1.81) (1.37) (0.91) (2.29) (4.48) (2.95) (4.58) (6.85) (10.10) (7.64) HML 0.4406 0.1147 0.0766 0.0342 0.0591 0.1244 0.0390 0.1723 0.1193 0.2200 0.6606 (7.51) (2.81) (1.99) (0.94) (0.82) (3.09) (0.73) (2.77) (2.18) (4.26) (8.00) Adj. R2 .9031 .8887 .8967 .9055 .8535 .8735 .8373 .8240 .8131 .7762 .3317 B. Equal-weighted deciles sorted by market OL Alpha 0.5495 0.3410 0.1916 0.1025 0.0231 0.0121 0.2382 0.3150 0.4031 0.7396 1.2891 (6.50) (4.77) (2.84) (1.49) (0.32) (0.16) (3.17) (3.59) (3.77) (5.13) (8.31) MktRf 1.1344 1.1056 1.0863 1.0665 1.0661 1.0271 1.0114 1.0144 0.9936 0.9242 0.2102 (31.00) (38.54) (41.78) (41.20) (40.19) (36.62) (36.83) (38.27) (34.11) (26.81) (5.99) SMB 0.7822 0.7817 0.7787 0.8343 0.8713 0.9110 0.9989 1.0462 1.0484 1.2313 0.4491 (8.20) (8.54) (9.03) (10.73) (10.65) (10.99) (14.67) (15.88) (16.77) (20.54) (5.19) HML 0.1110 0.0742 0.1316 0.1873 0.2763 0.2634 0.2831 0.3335 0.3402 0.3200 0.4310 (1.26) (1.08) (1.99) (3.30) (4.43) (4.41) (5.08) (5.69) (5.51) (4.59) (5.93) Adj. R2 .8891 .9094 .9064 .9197 .9010 .9028 .8964 .8767 .8504 .7920 .2731 C. Value-weighted deciles sorted by book OL Alpha 0.2319 0.1119 0.0452 0.0211 0.0889 0.1460 0.2853 0.2619 0.1815 0.3135 0.5453 (2.74) (1.71) (0.65) (0.30) (1.40) (2.35) (3.68) (3.78) (2.18) (2.99) (4.13) MktRf 1.0589 1.0717 0.9969 1.0008 1.0326 0.9571 0.9603 0.9935 1.0005 0.9694 0.0894 (35.08) (61.39) (43.31) (50.58) (44.47) (47.91) (40.13) (59.11) (39.72) (37.40) (2.23) SMB 0.0437 0.0415 0.0591 0.0182 0.0547 0.0167 0.0742 0.0398 0.1165 0.4279 0.4716 (1.41) (1.45) (1.49) (0.57) (1.85) (0.63) (2.04) (1.20) (2.24) (10.60) (11.10) HML 0.0620 0.0096 0.1008 0.0431 0.1613 0.1756 0.0912 0.2225 0.2514 0.1554 0.2174 (1.43) (0.26) (1.52) (1.13) (3.67) (3.73) (1.58) (5.49) (4.63) (2.79) (2.78) Adj. R2 .8662 .8966 .8599 .8760 .8944 .8806 .8304 .8703 .8247 .8346 .2311 D. Equal-weighted deciles sorted by book OL Alpha 0.3488 0.1862 0.0789 0.0340 0.0937 0.0809 0.1426 0.2119 0.2501 0.3367 0.6854 (4.15) (2.34) (1.03) (0.54) (1.47) (1.18) (1.80) (2.68) (2.65) (2.34) (4.56) MktRf 1.0849 1.0680 1.0643 1.0486 1.0567 1.0407 1.0224 1.0406 1.0234 0.9839 0.1010 (35.21) (38.86) (40.87) (40.41) (44.91) (38.55) (43.06) (33.15) (34.68) (24.54) (3.32) SMB 0.7829 0.8163 0.8272 0.8415 0.8887 0.9125 0.9841 0.9833 1.0484 1.1917 0.4088 (8.61) (9.61) (9.90) (10.24) (12.37) (12.89) (15.40) (13.16) (15.94) (16.47) (5.76) HML 0.3640 0.3007 0.2849 0.2336 0.2310 0.1886 0.1509 0.1598 0.1589 0.0189 0.3451 (5.25) (4.59) (4.44) (3.50) (3.96) (3.74) (2.43) (2.77) (2.55) (0.19) (3.36) Adj. R2 .8974 .9101 .9032 .9051 .9114 .9136 .8950 .8929 .8688 .7725 .2149 We estimate time-series regressions of portfolio excess returns (raw returns minus the risk-free rate) on Fama-French threefactors:themarketfactor(MktRf),small-minus-big(SMB),andhigh-minus-low(HML).Wesortstocksinto10 deciles in June of year t according to their operating leverage measures within each of the Fama and French (1997) 17 industries. Monthly returns are in percentages from July of year t to June of year tþ1. Market operating leverage (market OL) is defined as fixed costs divided by the market value of assets. Fixed costs are defined as the sum of depreciationandamortizationandselling,general,andadministrativeexpenses.Bookoperatingleverage(bookOL)is defined as fixed costs divided by the book value of assets. The sample covers the period from July 1963 to June 2016.
Panels A and B report results for portfolios sorted by market OL. Panels C and D report results for portfolios sorted by book OL. Panels A and C are for value-weighted portfolios, and panels B and D are for equal-weighted portfolios.
Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
Review of Asset Pricing Studies / v 12 n 1 2022 134 Downloaded from by Tsinghua University user on 21 February 2023 lower HML loadings. Decile 10 has an SMB loading of 0.43, and decile 1 has an SMB loading of -0.04. The difference is 0.47 and statistically significant (t-statistic ¼ 11.10). However, decile 10 has an HML loading of -0.16, and decile 1 has an HML loading of 0.06. The difference is -0.22 and statistically significant (t-statistic ¼ -2.78). Therefore, stocks with higher book operating leverage are likely to be low book-to-market stocks. The results for equal-weighted returns for portfolios sorted by book operating leverage in panel D also show that book operating leverage is likely to be negatively correlated with book-to-market. For equal-weighted portfolios, decile 10 has an HML loading of 0.02 and decile 1 has an HML loading of 0.36. The difference is -0.35 and statistically significant (t-statistic ¼ -3.36).
The negative association between book operating leverage and book-to-market allows us to examine whether book-to-market or operating leverage is more fundamental. If book-to-market were the fundamental variable, then the Fama-French three-factor model suggests that stocks with higher book operating leverage should earn lower returns. But our earlier results show that stocks with higher book operating leverage actually earn higher returns than stocks with low book operating leverage. We make two remarks regard-ing this finding. First, this finding explains why the Fama-French three-factor model does a worse job than the CAPM in explaining the returns of portfo-lios sorted by book operating leverage. Second, this finding suggests that operating leverage is likely to be a more fundamental variable than book-to-market because part of the variation in operating leverage that is nega-tively correlated with book-to-market (book operating leverage) still predicts higher returns.11 In sum, we find that both market and book operating leverages predict higher returns. The returns of both value-weighted and equal-weighted port-folios sorted by operating leverage cannot be explained by common factors, even after adjustment for nonsynchronous price movements.12 More impor-tant, the information contained in market operating leverage that is not me-chanically related to book-to-market—that is, book operating leverage— cannot be explained by book-to-market. Therefore, we conclude that the association between operating leverage and returns cannot be fully explained by common factors, including the book-to-market effect.
4. Can Operating Leverage Help Explain the Book-to-Market Effect?
We now examine whether operating leverage can help explain the book-to-market effect and other promising anomalies in the literature. We first 11 Previous papers suggest that book-to-market simply may be a proxy for financial leverage (D/ME). However, financial leverage does not pass this test. We can view financial leverage as the product of book leverage and book-to-market (D/ME¼D/BEBE/ME). Part of the variation in financial leverage that is not directly related to book-to-market (D/BE) actually predicts lower returns (see Fama and French 1992).
12 For details, see Section 4 of the Internet Appendix.
Measuring Operating Leverage 135 Downloaded from by Tsinghua University user on 21 February 2023 provide tests from the operating leverage decomposition and then examine whether an operating leverage factor can explain returns of portfolios sorted by book-to-market.
4.1 Evidence from decomposing operating leverage In the previous section, we have shown that book operating leverage predicts higher returns despite its negative association with book-to-market. If oper-ating leverage is indeed the fundamental variable, then there is a quantitative restriction of the relative magnitude of the book operating leverage effect and the book-to-market effect. To show this, we decompose our logarithm of market operating leverage into four components: ln FC MA ¼ ln FC BA þ ln BE ME þ ln BA BE þ ln ME MA : (4) The above equation says that log market operating leverage is related to log book operating leverage, book-to-market equity, and two terms related to financial leverage (book financial leverage and market financial leverage). If book-to-market equity affects stock returns only to the extent that it affects market operating leverage, then the above decomposition implies that the coefficients for ln(FC/BA) and ln(BM) should be the same. We make no predictions about the signs of the other two variables that are related to financial leverages because financial leverages may affect stock returns on their own. We now test this implication in the Fama-MacBeth regressions.
Table 8 presents the results. Column 1 estimates univariate regressions of returns on log market operating leverage. The average coefficient is 0.533 with a t-statistic of 8.43. Column 2 estimates univariate regressions of returns on log book operating leverage. The average coefficient is 0.302 with a t-statistic of 4.31. Column 3 shows that the univariate effect of log book-to-market equity is 0.447 and is also highly statistically significant (t-statistic ¼ 7.31).
In column 4, we estimate bivariate regressions of returns on log book operating leverage and log book-to-market equity. The coefficient for log book operating leverage is 0.406, and the coefficient for log book-to-market is 0.494. These two coefficients are similar in magnitude. The difference is -0.088 and is statistically indistinguishable from zero. In column 5, we use all four variables in the above decomposition. The coefficient for log book operating leverage is 0.404, and the coefficient for log book-to-market is 0.585. These coefficients are on the same order of magnitude, although the difference of -0.181 is statistically significant. In column 6, we further control for other variables, such as size. The coefficient for log book operating leverage is 0.270, and the coefficient for log book-to-market is 0.340. The difference is -0.069 and is statistically insignificant with a t-statistic of -0.98. Again, the coefficients for log of operating leverage measures are still significant in Review of Asset Pricing Studies / v 12 n 1 2022 136 Downloaded from by Tsinghua University user on 21 February 2023 columns 7 and 8 when using the market value and book value of assets in the previous fiscal year as scaling variables for alternative market and book op-erating measures, respectively.
By and large, we conclude that the effect of book operating leverage is similar in magnitude to the book-to-market effect. In other words, this test supports the view that book operating leverage and book-to-market matter Table 8 Fama-MacBeth return regressions: Decomposition of market OL Model 1 2 3 4 5 6 7 8 ln(FC/MA) 0.5330 (8.43) ln(FC/BA) 0.3017 0.4061 0.4036 0.2701 (4.31) (5.68) (5.58) (5.22) ln(FC/lagMA) 0.3461 (6.29) ln(FC/lagBA) 0.2743 (5.60) lnBM 0.4468 0.4943 0.5848 0.3396 0.3589 (7.31) (7.96) (7.25) (4.85) (4.60) ln(BA/BE) 0.2659 0.0244 0.0640 (2.12) (0.24) (0.52) ln(ME/MA) 0.2051 0.0572 0.1283 (1.98) (0.64) (0.99) lnSize 0.0867 0.0962 (2.48) (2.63) gBA 0.2911 0.6648 (3.80) (5.96) Accrual 0.8016 0.7544 (4.08) (2.73) IK 0.1576 0.1095 (3.05) (1.53) Momentum 0.2496 0.1028 (1.12) (0.42) lnAGE 0.0055 0.0152 (0.15) (0.38) Avg. obs.
2,509 2,509 2,509 2,509 2,509 2,509 2,388 2,388 Adj. R2 .0267 .0249 .0285 .0303 .0343 .0538 .0324 .0503 ln(FC/BA)-lnBM 0.0882 0.1811 0.0694 (1.28) (2.14) (0.98) This table reports Fama-MacBeth return regression results. Monthly returns in percentages from July of year t to June of year tþ1 are matched with accounting variables for fiscal years that end in year t-1. Here, we break the logarithm of market operating leverage flnOL¼ ln[(DAþSGA)/MA]g into log book operating leverage ln(FC/ BA)¼ln[(DAþSGA)/BA], log book financial leverage ln(BA/BE), log book-to-market ln(BE/ME), and log market financial leverage ln(ME/MA). DA is depreciation and amortization; SGA is selling, general, and administrative expenses; BA is total book assets; BE is book equity; ME is the same as the size measure; and MA is the market value of assets at the fiscal year-end. We then test whether the two components lnFC/BA and lnBE/ME have equal coefficients in the Fama-MacBeth regression. Size is the capitalization of the firm at the end of June (in $M). BM is the book-to-market equity ratio. FL is financial leverage, which is defined by the ratio of total liabilities to total book assets. gBA is the annual growth rate of total book assets. Accrual is measured as in Sloan (1996). IK is the investment-to-capital ratio. Momentum is the past 6-month cumulative returns (skip-ping a month). lnAGE is the log of one plus the number of years since the firm’s first appearance in CRSP. Note that we take logarithms for both the size and the book-to-market-equity in the regressions. In column 7, lagMA is the market value of assets at the end of fiscal year t-2. In column 8, lagBA is the book value of assets at the end of fiscal year t-2. The sample coversthe period fromJuly 1963 to June 2016. The industry fixed effect is controlled in all specifications. We use the 17-industry classification in Fama and French (1997). Explanatory variables are winsorized at the 1% and 99% levels every month. Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
Measuring Operating Leverage 137 Downloaded from by Tsinghua University user on 21 February 2023 to the extent that they affect operating leverage, and operating leverage is the fundamental variable.
4.2 Evidence from time-series regressions We now examine whether the return spread of the high-minus-low operating leverage portfolios can serve as a factor and explain the returns of portfolios that are sorted by the book-to-market ratio. We describe the construction of the operating leverage factor (OLFactor) later in Section 6. In this test, we use the book-to-market deciles as our testing portfolios.
We test whether the 10 value-weighted or equal-weighted returns of port-folios sorted by book-to-market can be explained by our operating leverage factor. We use the market factor (MktRf) and our operating leverage factor to explain the returns of portfolios. Section 5.1 of the Internet Appendix (Table IA.5) reports the results, and we summarize the key findings here.
The GRS test fails to reject the hypothesis that all 10 value-weighted or equal-weighted alphas are jointly zero. In other words, the market factor and our operating leverage factor help explain the returns of the book-to-market deciles.
Overall, the decomposition test in Section 5.1 shows that the coefficient for book operating leverage is similar in magnitude to that of book-to-market.
The GRS test in Section 5.2 cannot reject the hypothesis that alphas are jointly zero for both value-weighted and equal-weighted portfolios.
Therefore, we conclude that operating leverage helps explain the book-to-market effect.
5. Exploring Operating Leverage as a Factor In this section, we empirically examine whether measures of operating lever-age are important in explaining returns in a multifactor model. We view these tests as exploratory. To construct the operating leverage factor, we sort stocks into two groups, small and big, by the median NYSE size. We sort stocks into three groups (low 30%, middle 40%, and high 30%) indepen-dently by market OL within each of the industries in the 17-industry classi-fication in Fama and French (1997). We calculate the equal-weighted return for the portfolios from the intersection of two size and three market OL groups.13 The operating leverage factor (OLFactor) is the return spread be-tween the simple average of the returns on two high market OL groups and the simple average of the returns on two low market OL groups.
In choosing factors, researchers typically use the right-hand-side (RHS) approach, where they focus on comparing factors, or the left-hand-side 13 We report the results of the value-weighted OL factor in Section 5 of the Internet Appendix. Also, as shown in Section 5.4 of the Internet Appendix, to make better use of information from our operating leverage measure, we mainly report the equal-weighted operating leverage factor (OLFactor) in the main text. The factor is available at
Review of Asset Pricing Studies / v 12 n 1 2022 138 Downloaded from by Tsinghua University user on 21 February 2023 (LHS) approach, where they focus on the alpha from LHS portfolios (Fama and French 2018). We examine both approaches and start with the LHS approach.
We now employ the LHS approach and examine whether our factor mod-els can explain the anomalies in the literature. This approach brings more economic content to the factor model tests.
To this end, we follow the procedure in Hou, Xue, and Zhang (2015) and construct the 80 anomaly variables because they comprehensively cover six categories of anomalies: momentum, value-versus-growth, investment, prof-itability, intangibles, and trading frictions. Table IA.10 of the Internet Appendix describes all 80 anomaly variables. Furthermore, the market op-erating leverage and book operating leverage in this paper are also included in the anomaly set. Finally, we select 39 anomaly variables in which the average return of the high-minus-low decile is significant at the 5% level during the sample period from July 1963 to June 2016.
Table 9 reports the excess returns and the intercepts from time-series regressions of CAPM, FF3, FF5, q-factor, OL2F, Misp-factor, and OL3F models. We find that 35 anomalies survive after being adjusted by the CAPM, 32 anomalies survive after being adjusted by the Fama and French three-factor model, and 25 anomalies survive after being adjusted by the Fama and French five-factor model. Ten anomalies survive after being adjusted by the q-factor model, and nine anomalies survive after being ad-justed by the Misp-factor model.
Despite having only two factors in the OL2F model, 24 anomalies remain after being adjusted by the market factor and our market operating leverage factor. This means that the OL2F model’s power in explaining anomalies is better than the Fama and French three-factor model (32 anomalies remain) and is at least as powerful as the Fama and French five-factor model (25 anomalies remain). In addition, after we add the ROE profitability factor, only 5 anomalies survive the OL3F model: cumulative abnormal returns around earnings announcements (Abr-1 and Abr-6), net operating assets (NOA), changes in property, plant, and equipment plus changes in inventory scaled by assets (DPI/A), and inventory changes (IvC). These five anomalies also survive the Fama and French models and the q-factor model, and three of them survive the Misp-factor model.
The overallfindings in thissection suggestthata simple two-factor model (the market factor, MktRf, and our operating leverage factor, OLFactor) is at least as good as, but does not subsume, the Fama and French five-factor model in explaining anomalies. When the profitability factor (ROE) is added to the two-factor model, the three-factor model is at least as good as the q-factor model.
Furthermore, when implementing GRS tests on portfolios sorted by 55 on any two variables of Size, BM, OP, and INV, we find a similar result that the two-Measuring Operating Leverage 139 Downloaded from by Tsinghua University user on 21 February 2023 Table 9.
Significant anomalies in the broad cross-section 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 SUE-1 R6-1 R6-6 R6-12 R11-1 Abr-1 Abr-6 RE-1 I-Mom B/M Rev CF/P NO/P Dur ACI I/A NOA DPI/A IG NSI CEI ExRet 0.38 0.74 0.93 0.65 1.29 0.81 0.36 0.72 0.64 0.44 0.47 0.39 0.55 0.40 0.34 0.39 0.59 0.55 0.48 0.66 0.51 t-stat 3.46 2.72 4.03 3.62 4.82 6.29 3.84 2.72 3.33 2.25 2.06 2.01 2.32 2.02 2.74 2.51 4.66 4.11 4.17 4.23 3.03 CAPM Alpha 0.45 0.89 1.00 0.69 1.42 0.84 0.37 0.75 0.70 0.48 0.47 0.48 0.79 0.55 0.32 0.50 0.61 0.61 0.53 0.76 0.73 t-stat 4.26 3.59 4.71 3.92 5.85 6.73 4.13 3.09 3.71 2.39 2.09 2.43 3.27 2.80 2.45 3.07 4.81 4.47 4.50 4.94 4.90 FF3 Alpha 0.50 1.04 1.18 0.91 1.62 0.92 0.44 0.94 0.77 0.12 0.03 0.04 0.43 0.03 0.32 0.18 0.69 0.45 0.35 0.61 0.53 t-stat 4.44 4.29 5.64 5.73 6.95 6.65 4.53 4.02 3.96 1.07 0.17 0.35 2.42 0.24 2.67 1.45 5.13 3.38 3.27 4.68 4.43 FF5 Alpha 0.37 0.84 1.08 0.87 1.39 0.93 0.50 0.65 0.67 0.18 0.05 0.13 0.01 0.09 0.28 0.03 0.66 0.39 0.21 0.30 0.26 t-stat 3.13 2.47 3.81 4.50 4.09 6.48 4.83 2.60 2.41 1.61 0.30 1.06 0.05 0.67 2.15 0.29 4.27 3.28 2.05 2.38 2.50 q-factor Alpha 0.09 0.03 0.32 0.20 0.41 0.75 0.32 0.08 0.11 0.12 0.21 0.05 0.12 0.07 0.12 0.02 0.55 0.30 0.13 0.25 0.30 t-stat 0.78 0.07 1.01 0.93 1.05 5.00 2.80 0.32 0.37 0.72 1.25 0.23 0.71 0.34 0.81 0.18 3.10 2.22 1.17 1.86 2.16 OL2F Alpha 0.47 0.80 0.99 0.81 1.35 1.00 0.48 0.99 0.48 0.36 0.26 0.31 0.09 0.17 0.32 0.16 0.54 0.32 0.15 0.33 0.28 t-stat 3.33 2.80 4.19 4.85 4.32 7.14 4.64 3.39 2.08 2.07 1.22 1.71 0.51 0.99 1.93 1.19 3.83 2.14 1.33 2.55 2.08 Misp-factor Alpha 0.17 0.29 0.10 0.26 0.07 0.68 0.31 0.31 0.13 0.12 0.18 0.00 0.25 0.11 0.12 0.21 0.22 0.14 0.11 0.08 0.11 t-stat 1.26 1.09 0.47 1.45 0.30 4.50 2.77 1.06 0.69 0.81 0.92 0.02 1.59 0.66 0.86 1.73 1.93 0.99 0.97 0.68 0.99 OL3F Alpha 0.14 0.03 0.31 0.19 0.38 0.83 0.36 0.08 0.00 0.12 0.38 0.15 0.15 0.05 0.24 0.01 0.47 0.47 0.20 0.10 0.18 t-stat 1.05 0.08 0.97 0.86 0.98 5.29 2.86 0.36 0.01 0.61 1.67 0.65 0.59 0.20 1.25 0.08 2.88 3.10 1.58 0.55 1.07 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Number of Significant Alphas IvG IvC OA POA PTA ROE ROA GP/A F TES RS NEI OC/A Ad/M RD/M OL MOL BOL ExRet 0.36 0.58 0.30 0.40 0.37 0.69 0.62 0.37 0.56 0.42 0.30 0.33 0.61 0.69 0.63 0.40 0.77 0.53 t-stat 2.88 4.51 2.47 3.02 2.99 2.79 2.85 2.67 2.21 2.90 2.01 2.92 5.05 2.78 2.36 2.44 3.83 3.52 CAPM Alpha 0.44 0.64 0.34 0.47 0.48 0.86 0.80 0.37 0.79 0.28 0.29 0.33 0.67 0.73 0.43 0.33 0.78 0.51 35 t-stat 3.75 4.79 2.75 3.62 4.03 3.60 3.91 2.57 3.49 1.93 1.90 2.95 5.19 2.85 1.69 1.87 3.73 3.26 FF3 Alpha 0.27 0.51 0.35 0.29 0.30 1.03 0.97 0.53 0.68 0.46 0.55 0.52 0.61 0.14 0.23 0.37 0.43 0.55 32 t-stat 2.32 3.81 3.01 2.66 2.72 4.73 5.27 3.87 3.56 3.96 4.22 4.98 5.56 0.81 0.92 2.36 2.94 4.13 FF5 Alpha 0.16 0.54 0.53 0.19 0.15 0.50 0.57 0.23 0.33 0.53 0.46 0.34 0.37 0.14 0.50 0.04 0.36 0.46 25 t-stat 1.46 4.55 4.47 1.74 1.40 3.44 3.85 1.91 1.79 4.08 3.14 3.30 3.31 0.75 1.93 0.30 2.60 3.04 q-factor Alpha 0.03 0.45 0.54 0.16 0.23 0.08 0.06 0.13 0.18 0.28 0.17 0.04 0.18 0.05 0.72 0.02 0.33 0.36 10 t-stat 0.25 2.75 3.94 1.39 2.02 0.53 0.46 0.90 0.90 1.64 1.13 0.42 1.47 0.19 2.66 0.13 1.91 2.21 OL2F Alpha 0.10 0.33 0.16 0.01 0.09 0.93 0.89 0.36 0.44 0.42 0.62 0.41 0.30 0.37 0.07 0.29 0.03 0.40 24 t-stat 0.86 2.48 1.25 0.08 0.77 4.05 4.37 2.49 2.44 2.80 3.83 3.39 2.35 1.61 0.24 1.56 0.18 2.35 Review of Asset Pricing Studies / v 12 n 1 2022 140 Downloaded from by Tsinghua University user on 21 February 2023 Misp-factor Alpha 0.01 0.28 0.39 0.17 0.05 0.31 0.26 0.05 0.22 0.43 0.36 0.23 0.19 0.01 0.07 0.09 0.35 0.27 9 t-stat 0.11 2.11 2.90 1.49 0.52 1.70 1.53 0.37 1.16 2.80 2.61 2.00 1.55 0.02 0.32 0.73 2.26 2.01 OL3F Alpha 0.12 0.49 0.28 0.04 0.14 0.19 0.11 0.08 0.17 0.24 0.18 0.07 0.04 0.08 0.63 0.03 0.12 0.31 5 t-stat 0.92 2.82 1.74 0.23 0.90 1.15 0.74 0.56 0.81 1.24 1.15 0.63 0.27 0.27 1.81 0.15 0.57 1.52 We construct the 80 anomaly variables16 from Hou, Xue, and Zhang (2015) and test the explanatory power of the three prominent asset pricing models and two base models in this paper on 37 significant anomaly variables as well as the market operating leverage (MOL) and book operating leverage (BOL). “Significant” in this table means statistical significance at the 5% level.
The models include (1) CAPM; (2) FF3 (MktRf SMB HML); (3) FF5 (MktRf SMB HML RMW CMA); (4) q-factors from Hou, Xue, and Zhang (2015); (5) OL2F (MktRf OLFactor); (6) Misp-factor (mispricing factors from Stambaugh and Yuan 2017); and (7) OL3F (MktRf OLFactor ROE). The operating leverage factor (OLFactor) is constructed as follows. In June of year t, we first sort stocks into two groups, small and big, by the median NYSE size. Independently, we sort stocks into three groups (low 30%, middle 40%, and high 30%) by market OL within each of the industries in the 17-industry classification in Fama and French (1997). Monthly returns are in percentages from July of year t to June of year tþ1. We calculate the equal-weighted return for the portfolios from the intersection of two size and three market OL groups. The operating leverage factor OLFactor is the return spread between the simple average of the returns on two high market OL groups and the simple average of the returns on two low market OL groups. The models are presented from top to bottom in order of increasing squared Sharpe ratios, as reported in Table IA.9 of the Internet Appendix. ExRet is the average return of the high-minus-low decile, and Alpha is the intercept of asset pricing models of the high-minus-low decile. The q-factors cover the period from January 1967 to June 2016, and the other factors cover the period from July 1963 to June 2016. The t-statistics are adjusted for heteroscedasticity and autocorrelation following Newey and West (1987).
Measuring Operating Leverage 141 Downloaded from by Tsinghua University user on 21 February 2023 factor model (the market factor, MktRf, and our operating leverage factor, OLFactor) works at least as well as the Fama and French five-factor model.14 In addition to above LHS approach, we next provide a RHS approach to testing the explanatory power of our operating leverage factor by spanning regressions and pairwise tests of squared Sharpe ratios. To save space, we report the results of the RHS approach in Section 5.3 (Table IA.8) and 5.4 (Table IA.9) of the Internet Appendix and only discuss the main findings here. We first examine whether our operating leverage factor spans the prom-inent factors from Barillas and Shanken (2018). We find that the market factor and our operating leverage factor can explain the size factors, value factors, and investment factors reasonably well. The profitability factors and momentum factors also could be explained when the ROE factor is added to our model. Since Barillas and Shanken (2017) develop the insight that the maximum Sharpe ratio of tradable factors from an asset pricing model can be directly used to compare factor models, we then calculate the Sharpe ratios and follow Barillas et al. (2020) to perform pairwise tests of equality of the squared Sharpe ratios. The results suggest that the OLFactor and the market factor have already performed as well as the Fama and French five-factor model. Overall, the results indicate that the OL2F does not subsume the Fama and French five-factor model as profitability is not explained by the OL2F. When the profitability factor ROE is added to the model, the OL3F model performs better than the q-factor model.
6. Relation to Previous Studies 6.1 Total operating costs We know of two accounting-based measures of operating leverage that have been used in the literature on stock returns. Novy-Marx (2011) uses (SGAþCOGS)/BA in his tests for the operating leverage hypothesis. Ferri and Jones (1979) use the ratio of net fixed assets to total assets as a measure for operating leverage, which is also utilized in Garc ıa-Feij oo and Jorgensen (2010) as a secondary measure.
In this section, we first compare the return predictability of our measure of operating leverage with the operating leverage measure from Novy-Marx (2011) using cross-sectional Fama-MacBeth regressions. Table 10 reports the results. In column 1, NMOL positively and significantly predicts future stock returns from the univariate regression with a coefficient of 0.189 (t-statistic ¼ 5.48). The coefficient of NMOL declines (0.115) and is still statis-tically significant with other control variables in column 2. Columns 3 and 4 report the regressions of stock returns on market operating leverage, repeated 14 Returns of the portfolios sorted by 55 on any two variables of Size, BM, OP, and INV are downloaded from Kenneth French’s website: The results of GRS tests on bivariate-sorted portfolios, which are not reported for simplicity, are available on request.
Review of Asset Pricing Studies / v 12 n 1 2022 142 Downloaded from by Tsinghua University user on 21 February 2023 from Table 5. We then jointly estimate regressions of returns on NMOL and market operating leverage in columns 5 and 6. In column 5, the coefficient of market operating leverage is 1.626, which is almost the same as 1.615 in column 3. However, the coefficient of NMOL becomes -0.004 and statisti-cally insignificant (t-statistic ¼ -0.13). When all other variables are controlled in column 6, the coefficient of market OL is still positive and significant at 0.614 (t-statistic ¼ 3.00), but the coefficient of NMOL is 0.050 and becomes statistically insignificant (t-statistic ¼ 1.40). Columns 7 and 8 report the regressions of stock returns on book operating leverage, repeated from Table 5. Similarly, in joint regressions of returns on NMOL and book op-erating leverage in columns 9 and 10, the coefficient for book operating Table 10 Fama-MacBeth return regressions with operating leverage measure from Novy-Marx (2011) Model 1 2 3 4 5 6 7 8 9 10 NMOL 0.1891 0.1150 0.0044 0.0496 0.1177 0.0466 (5.48) (3.72) (0.13) (1.40) (2.54) (1.21) Market OL 1.6152 0.7025 1.6256 0.6136 (6.56) (3.82) (6.13) (3.00) Book OL 0.6734 0.5654 0.5042 0.5131 (3.43) (3.72) (2.17) (2.91) lnSize 0.1053 0.0908 0.0891 0.0933 0.0915 (2.76) (2.53) (2.45) (2.63) (2.54) lnBM 0.2615 0.2064 0.2126 0.3025 0.3004 (4.63) (3.57) (3.53) (5.48) (5.52) FL 0.2372 0.1772 0.2046 0.1327 0.1733 (1.22) (0.90) (1.04) (0.69) (0.90) gBA 0.3375 0.3149 0.3170 0.3017 0.3040 (4.29) (4.12) (4.19) (3.94) (3.99) Accrual 0.9461 0.7791 0.8285 0.8166 0.8691 (4.60) (3.98) (4.34) (4.18) (4.60) IK 0.1381 0.1481 0.1405 0.1534 0.1457 (2.64) (2.84) (2.70) (3.03) (2.89) Momentum 0.2616 0.2485 0.2370 0.2577 0.2440 (1.17) (1.12) (1.07) (1.15) (1.09) lnAGE 0.0067 0.0072 0.0061 0.0126 0.0096 (0.19) (0.20) (0.17) (0.35) (0.28) Avg. obs.
2,509 2,509 2,509 2,509 2,509 2,509 2,509 2,509 2,509 2,509 Adj. R2 .0241 .0527 .0264 .0528 .0270 .0533 .0250 .0528 .0261 .0535 We first estimate cross-sectional return regressions for each month (July 1963 to June 2016). Time series of the coefficients from the first-stage regressions are then used to calculate the average coefficients and the t-statistics.
The adjusted R2 and the number of observations are the averages of the cross-sectional regressions. Monthly returns in percentages from July of year t to June of year tþ1 are matched with accounting variables for fiscal years that end in year t-1. NMOL is the operating leverage measure from Novy-Marx (2011), which is the ratio of SGA plus COGS to book assets. Market operating leverage (market OL) is defined as fixed costs divided by the market value of assets. Fixed costs are defined as the sum of depreciation and amortization and selling, general, and administrative expenses. Book operating leverage (book OL) is defined as fixed costs divided by the book value of assets. Size is the capitalization of the firm at the end of June (in $M). BM is the book-to-market equity ratio. FL isfinancial leverage, which is defined as the ratio of total liabilities to total book assets. gBA is the annual growth rate of total book assets. Accrual is measured as in Sloan (1996). IK is the investment-to-capital ratio. Momentum is the past 6-month cumulative returns (skipping a month). lnAGE is the log of one plus the number of years since the firm’s first appearance in CRSP. Note that we take logarithms for both size and book-to-market equity in the regressions. The industry fixed effect is controlled in all specifications. We use the 17-industry classification in Fama and French (1997). Explanatory variables are winsorized at the 1% and 99% level every month. Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedas-ticity and autocorrelation.
Measuring Operating Leverage 143 Downloaded from by Tsinghua University user on 21 February 2023 leverage is also significant at 0.504 (t-statistic ¼ 2.17) without other control variables and 0.513 (t-statistic ¼ 2.91) with all other control variables.
Meanwhile, the coefficient of NMOL becomes smaller (0.118, t-statistic ¼ 2.54) when only book operating leverage is added and becomes indistinguish-able from zero (0.047, t-statistic ¼ 1.21) when other control variables are included. These results suggest that our market operating leverage and book operating leverage measures are more robust predictors of future stock returns than NMOL. The results are similar when we employ the log of operating leverage measures, as shown in Table IA.12 in the Internet Appendix. The overall results verify that market operating leverage and book operating leverage in this paper contain information over and above the operating leverage measure from Novy-Marx (2011).
We have previously argued that DA and SGA are more likely fixed costs and that COGS is more likely a variable cost. In Section 3, we also provide evidence that is consistent with this view. We now further examine this issue by estimating Fama-MacBeth regressions of returns on three cost compo-nents: ln(DA/BA), ln(SGA/BA), and ln(COGS/BA), which are similar to those conducted in Table 8. Table 11 reports the results. In columns 1 through 3, we estimate regressions of returns on ln(DA/BA), ln(SGA/BA), and ln(COGS/BA), separately, in univariate regressions. The coefficients for the three variables are 0.233, 0.219, and 0.121, respectively. The coefficients are statistically significant with t-statistics of 4.84, 3.88, and 2.71. In columns 4 and 5, we estimate regressions of returns on the first two cost components and all three cost components, respectively. The coefficients for ln(DA/BA) and ln(SGA/BA) are positive, statistically significant, and similar in magni-tude to each other. The coefficient for ln(COGS/BA) becomes smaller (0.064) and statistically insignificant (t-statistic ¼ 1.24). Columns 6 through 10 repeat the exercise from columns 1 through 5, with all control variables included. In column 10, the coefficients for ln(DA/BA) and ln(SGA/BA) are both posi-tive (0.103 and 0.211, respectively) and statistically significant (with t-statis-tics of 2.78 and 4.67, respectively). The coefficient for ln(COGS/BA) becomes negative (-0.029) and statistically insignificant (t-statistic ¼ -0.73).
The coefficients for ln(DA/BA), ln(SGA/BA), and ln(COGS/BA) are all statistically different from each other. However, the coefficients for ln(DA/ BA) and ln(SGA/BA) are both positive, statistically significant, and on the same order of magnitude, while that for ln(COGS/BA) is negative and sta-tistically insignificant.
If we believe that operating leverage should predict returns, then these results are by and large consistent with the view that DA and SGA measure fixed costs while COGS measures variable costs. When comparing the long-term return predictive power of DA, SGA, and COGS, we find that DA and SGA perform better than COGS in predicting long-term future returns, con-sistent with the fact that DA and SGA are stickier than COGS. See Section 7 of the Internet Appendix for details.
Review of Asset Pricing Studies / v 12 n 1 2022 144 Downloaded from by Tsinghua University user on 21 February 2023 Table 11 Fama-MacBeth regressions of returns on various costs Model 1 2 3 4 5 6 7 8 9 10 11 12 ln(DA/BA) 0.2333 0.2127 0.1988 0.1268 0.1045 0.1027 (4.84) (4.77) (4.45) (3.41) (2.89) (2.78) ln(SGA/BA) 0.2186 0.2046 0.1878 0.2102 0.2043 0.2107 (3.88) (3.76) (3.10) (5.04) (4.97) (4.67) ln(COGS/BA) 0.1210 0.0639 0.0239 0.0294 (2.71) (1.24) (0.66) (0.73) ln(SGA/TC1) 0.2022 (3.74) ln(TC1/BA) 0.2103 (4.76) ln(FC/TC2) 0.2682 (3.85) ln(TC2/BA) 0.2697 (5.37) lnBM 0.2566 0.3385 0.2479 0.3362 0.3381 0.3370 0.3400 (3.46) (4.84) (3.40) (4.81) (5.05) (4.95) (4.96) ln(BA/BE) 0.0489 0.0223 0.0655 0.0191 0.0198 0.0131 0.0148 (0.46) (0.22) (0.62) (0.19) (0.20) (0.13) (0.14) ln(ME/MA) 0.0230 0.0583 0.0175 0.0530 0.0544 0.0559 0.0568 (0.25) (0.65) (0.19) (0.59) (0.62) (0.63) (0.64) lnSize 0.1183 0.0861 0.1161 0.0883 0.0889 0.0860 0.0859 (3.23) (2.46) (3.11) (2.51) (2.52) (2.43) (2.44) gBA 0.3616 0.3013 0.3688 0.2941 0.3026 0.3042 0.2926 (4.67) (3.84) (4.67) (3.80) (3.97) (3.90) (3.81) Accrual 0.6764 0.9313 0.9516 0.7316 0.7324 0.9542 0.8459 (3.43) (4.60) (4.86) (3.72) (3.95) (4.99) (4.62) IK 0.1134 0.1553 0.1215 0.1384 0.1280 0.1460 0.1449 (1.99) (2.85) (2.20) (2.52) (2.39) (2.71) (2.69) Momentum 0.2907 0.2653 0.2853 0.2599 0.2493 0.2567 0.2578 (1.31) (1.19) (1.28) (1.17) (1.13) (1.16) (1.16) lnAGE 0.0161 0.0001 0.0096 0.0032 0.0057 0.0007 0.0047 (0.45) (0.00) (0.29) (0.09) (0.17) (0.02) (0.13) (continued) Measuring Operating Leverage 145 Downloaded from by Tsinghua University user on 21 February 2023 Table 11 Continued Model 1 2 3 4 5 6 7 8 9 10 11 12 Avg. obs.
2,490 2,490 2,490 2,490 2,490 2,490 2,490 2,490 2,490 2,490 2,490 2,490 Adj. R2 .0245 .0251 .0248 .0262 .0279 .0540 .0541 .0542 .0546 .0553 .0547 .0547 ln(DA/BA)ln(SGA/BA) 0.0081 0.0110 0.0998 0.1080 (0.14) (0.18) (2.25) (2.40) ln(DA/BA)-ln(COGS/BA) 0.1349 0.1321 (2.28) (2.60) ln(SGA/BA)-ln(COGS/BA) 0.1239 0.2400 (1.51) (4.27) ln(SGA/TC1)-ln(TC1/BA) 0.0081 (0.19) ln(FC/TC2)-ln(TC2/BA) 0.0015 (0.03) This table reports Fama-MacBeth return regression results. Monthly returns in percentages from July of year t to June of year tþ1 are matched with accounting variables for fiscal years that end in year t-1. DA is depreciation and amortization; SGA is selling, general, and administrative expenses; COGS is cost of goods sold; fixed costs (FC) are defined as the sum of DA and SGA; the first total costs (TC1) are defined as the sum of SGA and COGS; the second total costs (TC2) are defined as the sum of DA, SGA, and COGS; BA is total book assets; BE is book equity; ME is the same as the size measure; and MA is the market value of assets at the fiscal year-end. We then test whether the ratios of different costs to total book assets have equal coefficients in the Fama-MacBeth regression. Size is the capitalization of the firm at the end of June (in $M). BM is the book-to-market equity ratio. FL is financial leverage, which is defined asthe ratio of total liabilities to total book assets. gBA is the annual growth rate of total book assets. Accrual ismeasured as in Sloan (1996). IK isthe investment-to-capital ratio. Momentum is the past 6-month cumulative returns (skipping a month). lnAGE isthe log of one plus the number of years since the firm’sfirst appearance in CRSP. Note that we take logarithms for both size and book-to-market equity in the regressions. The sample covers the period from July 1963 to June 2016. The industry fixed effect is controlled in all specifications. We use the 17-industry classification in Fama and French (1997). Explanatory variables are winsorized at the 1% and 99% levels every month. Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
Review of Asset Pricing Studies / v 12 n 1 2022 146 Downloaded from by Tsinghua University user on 21 February 2023 Next, we explore a different way to examine the relation between our measure and Novy-Marx’s measure. Specifically, we decompose our log book operating leverage measure into two components: ln FC BA ¼ ln FC TC þ ln TC BA ; (5) where TC is the total costs. Novy-Marx (2011) decomposes operating lever-age into the level of gearing and operational inflexibility. His proxy for op-erating leverage, TC/BA, basically captures the level of gearing. Novy-Marx (2011) is aware of this and acknowledges that this measure does not explicitly account for operational inflexibility in his empirical analysis. Both analyses in Novy-Marx (2011) and in our Section 3 show that operational inflexibility can be captured by FC/TC. We explore two specifications of fixed costs and total costs. In the first specification, we follow Novy-Marx (2011) and only use SGA as fixed costs and SGAþCOGS as total costs (denoted by TC1 in Table 11). In the second specification, we use DAþSGA as fixed costs and DAþSGAþCOGS as total costs 2 (denoted by TC2 in Table 11). We then estimate Fama-MacBeth regressions of monthly returns on both operational inflexibility and level of gearing and examine whether the difference between our measure and Novy Marx’s measure, ln(FC/TC), positively predicts returns and whether the coefficients for ln(FC/TC) and ln(TC/BA) are the same. The results are reported in columns 11 and 12 of Table 11. Column 11 shows that the coefficients for ln(SGA/TC1) and ln(TC1/BA) are both pos-itive (coefficients ¼ 0.202 and 0.210) and statistically significant (t-statistics ¼ 3.74 and 4.76). The difference between the coefficients is -0.008, which is not statistically significant (t-statistic of -0.19). The results are similar when we include DA in fixed costs as well. Column 12 shows that the coefficients for ln(FC/TC2) and ln(TC2/BA) are both positive (coefficients ¼ 0.268 and 0.270). The associated t-statistics are 3.85 and 5.37, respectively. The differ-ence between the coefficients is -0.002, which is not statistically significant (t-statistic of -0.03). These results suggest that ln(FC/TC), the difference be-tween our measure and Novy-Marx’s measure, positively predicts returns.
Furthermore, empirical results are consistent with the view that the coeffi-cients for operational inflexibility, ln(FC/TC), and level of gearing, ln(TC/ BA), are the same, and therefore what really matters simply may be ln(FC/ BA), or fixed costs.
6.2 Net fixed assets Next, we examine the relation between our measure and the measure used in Ferri and Jones (1979) and Garc ıa-Feij oo and Jorgensen (2010): net fixed assets to total assets. To do so, we focus on DA, the part of fixed costs that is directly related to net PPE. We decompose ln(DA/BA) into ln(DA/PPE) and ln(PPE/BA), with the latter being the traditional measure of operating Measuring Operating Leverage 147 Downloaded from by Tsinghua University user on 21 February 2023 leverage in Ferri and Jones (1979). Using these variables, we again estimate Fama-MacBeth regressions. To save space, we report the results in Table IA.17 in the Internet Appendix and summarize the main findings here. In the Fama-MacBeth regressions, ln(PPE/BA) becomes statistically significant only when we add ln(DA/PPE). If we believe that operating leverage should predict returns, then our measure is an improvement over the traditional measure of operating leverage.
6.3 Organizational capital and systematic risk In a seminal paper, Eisfeldt and Papanikolaou (2013) show that firms with higher organizational capital have higher returns. Their measure of organi-zational capital is derived from the cumulative deflated value of SGA, which is closely related to our measure of operating leverage. They argue that their results on organizational capital are not explained by operating leverage be-cause they find that the earnings of firms with high organizational capital do not exhibit a higher covariance with gross domestic product (GDP).
We believe that our results are not simply driven by organizational capital.
First, their notion of organizational capital is measured by the cumulated value of SGA. However, in Table 11, we find that another fixed cost com-ponent, DA, also positively predicts stock returns, with predictive power on the same order of magnitude as that of SGA.
Second, we now explore the relation between our operating leverage mea-sure and systematic risk. We first attempt to redo the analysis in table X of Eisfeldt and Papanikolaou (2013) on firms sorted by our measure of book operating leverage. Table IA.18 in the Internet Appendix reports the results, which are qualitatively the same as those in Eisfeldt and Papanikolaou (2013). While firms with higher book operating leverage appear to have higher earnings-to-sales sensitivities, they do not seem to exhibit higher earnings-to-GDP sensitivities.
We conjecture that this result is driven by the fact that firm-level estima-tions of these sensitivities are noisy. To investigate this conjecture, we now compute the relevant quantities at the portfolio level and then reestimate the sensitivities. We follow Chen (2017) and compute the relevant accounting variables for an initial investment of $100. Because we view depreciation as a cost, we do not add back depreciation to compute operating cash flows; rather, we compute EBIT. Because EBIT (ib) can be negative at the portfolio level, we use the transformation for logs of negative earnings from Ljungqvist and Wilhelm (2005): ln(1þib) if ib is positive and –ln(1-ib) if ib is negative. We also examine dividends, which are never negative, for robustness. Table 12 reports the results. Panel A1 shows that at the portfolio level, earnings are more sensitive to sales as book operating leverage increases. Panel A2 shows that once we estimate at the portfolio level, earnings of higher operating leverage portfolios do have higher sensitivities to GDP. Panel A3 displays Review of Asset Pricing Studies / v 12 n 1 2022 148 Downloaded from by Tsinghua University user on 21 February 2023 estimates of the same regression using the difference specification. Panels B1 through B3 show that the results are robust when we change EBIT to divi-dends. In fact, the sensitivities often have higher statistical precision, proba-bly because negative earnings create a specification difficulty. The specification in panel B2 is similar to that in Bansal, Dittmar, and Kiku (2009), who show that portfolios sorted by size and book-to-market equity differ in systematic risk. We find that portfolios sorted by fixed costs also differ in systematic risk.15 Table 12 Systematic risk A. Cash flow is measured by EBIT B. Cash flow is measured by Dividend Quintile 1 2 3 4 5 1 2 3 4 5 A1. ln(EBIT) on ln(Sale) B1. ln(Dividend) on ln(Sale) ln(Sale) 0.2117 0.4048 0.4482 0.6978 2.3745 0.1163 0.3703 0.4370 0.2332 0.7475 (0.33) (2.20) (3.73) (7.98) (2.46) (1.25) (1.54) (2.86) (1.60) (4.51) Obs.
53 53 53 53 53 53 53 53 53 53 Adj. R2 .026 .804 .917 .967 .224 .845 .937 .967 .982 .967 A2. ln(EBIT) on ln(GDP) B2. ln(Dividend) on ln(GDP) ln(GDP) 0.8310 0.0550 0.0716 0.2888 2.2678 0.3574 0.2656 0.5164 0.2641 1.4164 (0.79) (0.21) (0.30) (1.22) (1.64) (1.60) (1.00) (2.23) (1.64) (10.98) Obs.
53 53 53 53 53 53 53 53 53 53 Adj. R2 .020 .793 .904 .951 .138 .854 .932 .964 .981 .986 A3. Dln(EBIT) on Dln(GDP) B3. Dln(Dividend) on Dln(GDP) Dln(GDP) 1.1876 1.2327 2.6688 2.0584 7.7119 0.4981 0.3386 0.8677 0.4073 0.9550 (0.17) (1.00) (3.42) (3.19) (0.99) (1.58) (1.18) (2.29) (1.40) (2.33) Obs.
52 52 52 52 52 52 52 52 52 52 Adj. R2 .020 .007 .060 .077 .004 .009 .008 .074 .007 .059 In this table, we report regression results for annually rebalanced portfolios sorted by our measure of book operating leverage, the ratio of fixed costs to the book value of assets (FC/BA), relative to their industry peers at each June. Portfolios are held from July of year t to June of year tþ1. Fixed costs are defined as the sum of depreciation and amortization (DA) and selling, general, and administrative expenses (SGA). We use the 17-industry classification in Fama and French (1997). Within each industry, we assign firms a rank (1 to 5) based on the firm’s quintile of FC/BA. In the annually rebalanced value-weighted portfolios, we calculate the annual EBIT (panel A), dividend (panel B), and sale for an initial investment of $100, following Chen (2017). We estimate the sensitivity of the portfolio’s cash flows on outputs by the slope coefficient of regression of the log cash flow on the log sale, or on the log aggregate output, which is measured by GDP, with time trend controlled.
The cash flow is measured by EBIT or dividend. For negative earnings, we use the transformation for logs of negative earnings from Ljungqvist and Wilhelm (2005): ln(1þib) if ib is positive and –ln(1-ib) if ib is negative.
GDP is gross domestic product. We also report the slope coefficients of regressions of change in log cash flow on change in log aggregate output to capture the effects of aggregate shocks on the growth rate of cash flows. We keep the firms with fiscal year ending in December. The sample covers the period from 1963 to 2015. Variables are winsorized at the 5% and 95% levels for each portfolio. Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
15 To understand the difference between Table 7 and Table 12, we believe that it is important to distinguish between the regular market beta and cash flow beta. The market betas in Table 7 are regular market betas, and the earnings GDP sensitivities are essentially cash flow betas. Campbell and Vuolteenaho (2004) break the beta of stocks into cash flow beta and discount rate beta, and their intertemporal asset pricing theory suggests that cash flow beta should theoretically, and so empirically, have a higher price of risk. We believe their results are helpful to understand the contrast between Table 7 and Table 12.
Measuring Operating Leverage 149 Downloaded from by Tsinghua University user on 21 February 2023 When we compare the return predictability of organizational capital with our measure of operating leverage, as shown in Table IA.19 in the Internet Appendix, we find that our market operating leverage and book operating leverage measures are better at predicting stocks’ future returns than organi-zational capital.
Overall, we conclude that once we estimate the sensitivities at the portfolio level, high book operating leverage is associated with higher systematic risk.
But organizational capital is firm specific and should not be related to sys-tematic risk in Eisfeldt and Papanikolaou’s (2013) model. Additionally, DA also predicts higher stock returns, with predictive power on the same order of magnitude as that of SGA. Further, our measures of operating leverage perform well in predicting future stock returns when controlling for organi-zational capital. For these reasons, we believe that our results are not explained by organizational capital.
6.4 Operating leverage measures from regressions One classic measure of operating leverage is the sensitivity of firms’ earnings to firms’ output, as in Mandelker and Rhee (1984). Garc ıa-Feij oo and Jorgensen (2010) implement this measure by further controlling for the time trend when estimating operating leverage in the regression. In Table 12, our results show that portfolios sorted by fixed costs exhibit dif-ferent sensitivities of earnings to sales.
Although the portfolio-level regressions show a positive relation between fixed costs and regression coefficients, we conjecture that at the firm level, these kinds of regressions are likely to be ridden with estimation errors. To further examine this question, we estimate regression-based operating lever-age at the firm level. We then estimate firm-level Fama-MacBeth regressions of returns on the regression-based measure as well as our measure of oper-ating leverage. Similar to Garc ıa-Feij oo and Jorgensen (2010), we estimate regressions of the detrended natural logarithm of quarterly EBIT on the detrended natural logarithm of quarterly sales in the previous 20 quarters for each firm at the end of each fiscal year. The estimations require non-missing EBIT and sales data over the previous 20 quarters. Table 13 reports the results of Fama-MacBeth regressions of monthly returns on operating leverage. Because of the availability of quarterly data, the sample period for this table is from July 1972 to June 2016. The coefficients for the regression-based measure of operating leverage are statistically insignificant in both univariate and multivariate regressions.
After controlling for the regression-based operating leverage measure, the average coefficient for our measure of market operating leverage (FC/MA) is still positive and sig-nificant at 0.812 (t-statistic ¼ 4.02) with size and book-to-market ratio con-trolled and 0.582 (t-statistic ¼ 3.11) with all other control variables. After controlling for the regression-based operating leverage measure, the average Review of Asset Pricing Studies / v 12 n 1 2022 150 Downloaded from by Tsinghua University user on 21 February 2023 coefficient for our measure of book operating leverage (FC/BA) is also pos-itive and significant at 0.650 (t-statistic ¼ 3.50) with size and book-to-market ratio controlled and 0.460 (t-statistic ¼ 2.74) with all other control variables.
Overall, the results suggest that fixed costs are positively associated with the sensitivity of earnings to output. However, directly estimating operating leverage from regressions is likely to be ridden with estimation error prob-lems, and this problem is more severe at the firm level.
Table 13.
Fama-MacBeth regressions with regression-based operating leverage measure Model 1 2 3 4 5 6 7 RBOL 0.0079 0.0073 0.0012 0.0035 0.0029 0.0028 0.0036 (0.65) (0.84) (0.16) (0.42) (0.39) (0.34) (0.48) Market OL 0.8107 0.5822 (4.02) (3.11) Book OL 0.6497 0.4598 (3.50) (2.74) lnSize 0.0882 0.0888 0.0590 0.0675 0.0605 0.0700 (2.09) (2.43) (1.48) (1.94) (1.56) (2.06) lnBM 0.3275 0.2591 0.2733 0.2242 0.3806 0.2995 (4.60) (3.87) (3.74) (3.28) (5.55) (4.54) FL 0.0743 0.1036 0.0585 (0.32) (0.44) (0.25) gBA 0.4946 0.4465 0.4462 (4.63) (4.28) (4.35) Accrual 1.3339 1.2055 1.2277 (5.31) (5.04) (5.20) IK 0.0681 0.0769 0.0806 (0.95) (1.09) (1.17) Momentum 0.0327 0.0012 0.0062 (0.13) (0.00) (0.02) lnAGE 0.0187 0.0223 0.0173 (0.40) (0.47) (0.37) Avg. obs.
2,018 2,018 2,018 2,018 2,018 2,018 2,018 Adj. R2 .0259 .0407 .0525 .0417 .0532 .0417 .0532 We first estimate cross-sectional regressions for each month (July 1972 to June 2016). Time series of the coefficients from the first-stage regressions are then used to calculate the average coefficients and the t-statistics.
The adjusted R2 and the number of observations are the averages of the cross-sectional regressions. Monthly returns in percentages from July of year t to June of year tþ1 are matched with accounting variables for fiscal years that end in year t-1. RBOL is the regression-based measure of operating leverage. Similar to Garc ıa-Feij oo and Jorgensen (2010), we estimate the coefficients in regressions of the detrended natural logarithm of quarterly EBIT on the detrended natural logarithm of quarterly sales in the previous 20 quarters for each firm at the end of each fiscal year. The estimations require nonmissing EBIT and sales data over the previous 20 quarters. Market operating leverage (market OL) is defined as fixed costs divided by the market value of asset. Fixed costs are defined as the sum of depreciation and amortization and selling, general, and administrative expenses. The book operating leverage (book OL) is defined as fixed costs divided by the book value of assets. Size is the capital-ization of the firm at the end of June (in $M). BM is the book-to-market equity ratio. FL is financial leverage, which is defined as the ratio of total liabilities to total book assets. gBA is the annual growth rate of total book assets. Accrual is measured as in Sloan (1996). IK is the investment-to-capital ratio. Momentum is the past 6-month cumulative returns (skipping a month). lnAGE is the log of one plus the number of years since the firm’s first appearance in CRSP. Note that we take logarithms for both size and book-to-market equity in the regressions. The industry fixed effect is controlled in all specifications. We use the 17-industry classification in Fama and French (1997). Explanatory variables are winsorized at the 1% and 99% levels every month.
Reported in parentheses are Newey and West (1987) t-statistics adjusted for heteroscedasticity and autocorrelation.
Measuring Operating Leverage 151 Downloaded from by Tsinghua University user on 21 February 2023 6.5 Further robustness checks Our measure of fixed costs also might be related to the following literature: the gross profitability measure in Novy-Marx (2013), the operating profit-ability measure in Fama and French (2015), the cash-based operating prof-itability measure in Ball et al. (2016), the intangible assets effect (Chan, Lakonishok, and Sougiannis 2001), the retained earnings-to-market ratio in Ball et al. (2020), and the labor leverage (Favilukis and Lin 2016; Donangelo et al. 2019). For robustness checks, we compare the predictive ability of our operating leverage measure with these variables. We report the results of these robustness tests in Sections 10 to 13 (Tables IA.20 to IA.23) of the Internet Appendix. We find that book operating leverage and the remain-der of gross profitability have different cross-sectional properties, that the effect of book operating leverage is greater in small stocks, whereas the effect of the rest of gross profitability is greater in large stocks. The operating profitability measure, the cash-based operating profitability measure, and the retained earnings-to-market ratio do not subsume the information from book operating leverage. In addition, the predictive ability of our op-erating leverage measure does not come from intangible intensity or labor costs.
Finally, we examine whether our results are sensitive to industry adjust-ments. We perform robustness tests for the raw operating leverage measure (not industry adjusted) and the industry-adjusted operating leverage measure that uses the 49-industry classification in Fama and French (1997), as in Novy-Marx (2011). We report the results in Tables IA.27 to IA.32 of the Internet Appendix. The key regression coefficients of interest in the main text are qualitatively the same with no industry adjustment or with an adjustment from the 49-industry classification in Fama and French (1997).
7. Conclusion Although the idea of operating leverage has long been associated with fixed costs, recent empirical examinations of the relation between operating lever-age and stock returns have not directly utilized fixed costs. We examine a simple measure of operating leverage: the ratio of fixed costs to the market value of assets. We measure fixed costs as depreciation and amortization plus selling, general, and administrative expenses. We find that this measure of operating leverage positively predicts returns and that its predictive power is stronger than previous measures of operating leverage. Operating leverage is not explained by common factors. We also construct an operating leverage factor and find that it helps explain the returns of book-to-market portfolios.
Furthermore, an exploratory two-factor model with the operating leverage factor works at least as well as, but does not subsume, the Fama and French five-factor model in explaining anomalies.
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16143 | https://artofproblemsolving.com/wiki/index.php/2018_USAJMO_Problems/Problem_1?srsltid=AfmBOooxROixAc4VSKz6da8DwyzKQnJmE8IcwelDYJcob73dQxFThTMD | Art of Problem Solving
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2018 USAJMO Problems/Problem 1
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Video Solution
8 See also
Problem
For each positive integer , find the number of -digit positive integers that satisfy both of the following conditions:
no two consecutive digits are equal, and
the last digit is a prime.
Solution 1
The answer is .
Suppose denotes the number of -digit numbers that satisfy the condition. We claim , with . It is trivial to show that . Now, we can do casework on whether or not the tens digit of the -digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in ways and choose the units digit in ways, since it must be prime and not equal to the tens digit. Therefore, there are ways in this case.
If the tens digit is not prime, we can use complementary counting. First, we consider the number of -digit integers that do not have consecutive digits. There are ways to choose the first digit and ways to choose the remaining digits. Thus, there are integers that satisfy this. Therefore, the number of those -digit integers whose units digit is not prime is . It is easy to see that there are ways to choose the units digit, so there are numbers in this case. It follows that and our claim has been proven.
Then, we can use induction to show that . It is easy to see that our base case is true, as . Then, which is equal to as desired.
Solution by TheUltimate123.
Solution 2
The answer is .
As in the first solution, let denote the number of -digit numbers that satisfy the condition. Clearly and . We claim that for we have the recurrence .
To prove this, we split the -digit numbers satisfying the conditions into cases depending on whether or not the second digit is . If the second digit is nonzero, our number is formed from one of the numbers with one fewer digit satisfying the conditions, times possible choices of adding a digit to the left. If the second digit is zero, our number is formed from one of the numbers with two fewer digits satisfying the conditions, times possible choices of adding and then any nonzero digit to the left. This proves our claim.
This gives us a linear three-term recurrence. It is well-known that its solution is of the form , where the are constants to be determined from the initial conditions and , and and are the roots of the corresponding quadratic equation . We solve and get , so our roots are and . Now, we use our conditions and to derive the system of linear equations
Solving this system yields and , and we are done.
Solution by putnam-lowell.
Solution 3
The answer is .
As in the first solution, let denote the number of -digit numbers that satisfy the condition. Clearly and .
From here, we proceed by complementary counting. We first count the total amount of numbers that satisfy both bullets but that may have zero as its first digit. The units digit can be one of four primes, and each digit to the left will have 9 choices (any digit but the one that was just used). Then the total for this group of numbers is just .
Now we must subtract all numbers in the above group that have 0 as its first digit. This is just because for each number in the first group beginning with a zero, we could take away the zero, leaving us with a number that works for the case (this is true because the next number would not be zero, or the consecutive digit requirement would be violated). Then we have the recursive formula .
To simplify this, we take a look at the first few terms. We see that
We see a pattern where and we can prove that it holds for all because subtracting from is the same thing as reversing all previous signs of the preceding powers of 9. This constitutes an alternating pattern, which we can calculate as a geometric series. The first term is and the common ratio is , so
We are done.
Solution by aopsal
Solution 4
Let denote the number of -digit positive integers satisfying the conditions listed in the problem. Claim 1: To prove this, let be the leftmost digit of the -digit positive integer. When ranges from to the allowable second-to-leftmost digits is the set with excluded. Note that since are all repeated times and using our definition of (since -digit positive integers must begin with a positive digit), we have proved that part. Since is repeated times, and since digits that are allowed to be adjacent to the (the third-to-leftmost digit of our positive integer) are integers ranging from to we have proved the part similarly. We can see that and by simple computation. These will be our base cases. Now, let Claim 2: We can prove this using induction. Let and By plugging and into our formula we obtain Now, by plugging and into the same formula, we obtain Since with and we’re back where we started (this is since just like ), and using our base cases of and our induction is complete. Claim 3: We can prove that by simply using our base case of and repeatedly using the distributive property to obtain further values of For the case where is even, we can manipulate the summation into which equals by geometric series. For the case where is odd, we can manipulate the summation into which equals Thus, we have proved Claim 3 for both odd and even Since our answer is
-fidgetboss_4000
Solution 5
Let be the number of -digit numbers that satisfy the condition that most-left digit different than . Let be the number of -digit numbers that satisfy the condition that most-left digit is . We can write that and so we find that is the characteristic polynomial of the homogeneous recurrence relation . Then is a root. Also, by the non-homogeneous part of , we understand that is another root of the recurrence relation with the complementary solution idea. Therefore, sequence in the form We can calculate that with using . Thus we get:
Lokman GÖKÇE
Video Solution
-Osman Nal -inserted by srisainandan6
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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2018 USAJMO (Problems • Resources)
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All USAJMO Problems and Solutions
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16144 | https://www.youtube.com/watch?v=ptAw20kem90 | Naming ions and ionic compounds | Chemistry | Khan Academy
Khan Academy
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Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now:
Want to explore more? Check out the full Chemical bonding playlist here:
Ionic compounds are neutral compounds made up of positively charged ions called cations and negatively charged ions called anions. For binary ionic compounds (ionic compounds that contain only two types of elements), the compounds are named by writing the name of the cation first followed by the name of the anion. For example, KCl, an ionic compound that contains K⁺ and Cl⁻ ions, is named potassium chloride.
Sections:
00:00 - How ions form and compounds are named
00:12 - Group 1 alkali metals: Potassium
00:44 - Naming potassium ion
01:05 - Halogens and electron gain
01:36 - Naming chloride ion
02:12 - Forming ionic compound potassium chloride
02:34 - Naming conventions for ionic compounds
03:07 - Why potassium name remains unchanged
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45 comments
Transcript:
How ions form and compounds are named Let's get some practice now thinking about how ions typically form, how they might form compounds, and how we name those compounds. So, let's start with something in group one in this first Group 1 alkali metals: Potassium column. This first column is often known as alkali metals. And so, let's start with potassium. K is the symbol for potassium. Now, things in group one here, one way to think about it is their outermost shell has one electron in it. So they wouldn't mind losing that electron. So when they ionize, they tend to lose an electron and become a cation, a positive ion. And so let's look at a situation where I have some potassium Naming potassium ion that has been ionized. And I could I could write it just like this. We've seen that in previous videos. And we can refer to this just as a potassium ion. We could refer to this as potassium 1+. We could refer this refer to this as a potassium cation. Now let's go on to the other side of the periodic table. Things Halogens and electron gain that would really love to grab an electron. So things in a group in the hallides, which is this column right over here. So these are the hallides. They have seven electrons in their outermost shell. They would love to have eight. So they tend to be really good at grabbing electrons. And so let's say we're dealing with chlorine. And chlorine is able to ionize. So it's able to grab an electron. When chlorine grabs an Naming chloride ion electron, it will be a negatively charged ion. So you could write it as chlorine 1 minus. But the way that we generally refer to an annion, a negatively charged ion, instead of saying this is instead of just calling this the chlorine annion, we would call this chloride. So this we would refer to as chlor chloride. Now, as you can imagine, with potassium having a positive one charge or one plus charge and this having a negative charge, they're going to be Forming ionic compound potassium chloride attracted to each other and they can actually form an ionic compound. And the the ionic compound they would form, we would write as you'd write your positive ion first and then you would write your negative ion. And this right over here would be described as potassium chloride. Let me Naming conventions for ionic compounds write that down. Potassium potassium chloride. Now you might be saying well I just re let me rewrite the write the whole thing. So you know the chloride part you say okay this is going to be an annion because instead of writing chlorine which is the name of this element I wrote this ide. So I know that this is this is the chlorine annion. This is chloride. Why didn't I do something similar for potassium? Well, the way the convention Why potassium name remains unchanged works is if someone says potassium chloride, you know, you're dealing with an ionic compound. And if the chlorine has a negative one charge, an ionic compound, the whole thing is going to be neutral. So, if this one over here is one minus, then you know this over here, since they're one for one, this is going to be one plus. So you know that you're dealing with a potassium cation and a you could say a chlor and a chloride ion or a chlorine annion. You could refer to it other ways various ways but this is potassium chloride. You have a positively charged potassium and you have a negatively charged chlorine which we would call a chloride. In the next few videos I'll do many many more examples of this and ones that'll be a lot a little bit more complicated. |
16145 | https://www.youtube.com/watch?v=UJQkqV2zGv0 | Intervals and interval notation | Functions | Algebra I | Khan Academy
Khan Academy
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23 Jun 2015
Intervals describe specific sets of numbers and are very useful when discussing domain and range. In this video, Sal introduces the different types of intervals and their notations.
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193 comments
[Voiceover] What I hope
to do in this video is get familiar with the notion of an interval, and also think about ways
that we can show an interval, or interval notation. Right over here I have a number line. Let's say I wanted to talk
about the interval on the number line that goes from
negative three to two. So I care about this-- Let
me use a different color. Let's say I care about this
interval right over here. I care about all the numbers
from negative three to two. So in order to be more
precise, I have to be clear. Am I including negative three and two, or am I not including
negative three and two, or maybe I'm just including one of them. So if I'm including
negative three and two, then I would fill them in. So this right over here, I'm
filling negative three and two in, which means that
negative three and two are part of this interval. And when you include the endpoints, this is called a closed interval. Closed interval. And I just showed you how I can depict it on a number line, by actually
filling in the endpoints and there's multiple ways to talk about this interval mathematically. I could say that this is all of the... Let's say this number line is showing different values for x. I could say these are all of the x's that are between negative three and two. And notice, I have negative
three is less than or equal to x so that's telling us
that x could be equal to, that x could be equal to negative three. And then we have x is less
than or equal to positive two, so that means that x could
be equal to positive two, so it is a closed interval. Another way that we could
depict this closed interval is we could say, okay, we're
talking about the interval between, and we can use brackets because it's a closed interval, negative three and two, and once again I'm using brackets here,
these brackets tell us that we include, this
bracket on the left says that we include negative three,
and this bracket on the right says that we include
positive two in our interval. Sometimes you might see things written a little bit more math-y. You might see x is a member of
the real numbers such that... And I could put these curly
brackets around like this. These curly brackets say that we're talking about a set of values, and we're saying that the set of all x's that are a member of the real number, so this is just fancy math notation, it's a member of the real numbers. I'm using the Greek letter
epsilon right over here. It's a member of the
real numbers such that. This vertical line here means "such that," negative three is less x is less than-- negative three is less than or equal to x, is less than or equal to two. I could also write it this way. I could write x is a
member of the real numbers such that x is a member,
such that x is a member of this closed set, I'm
including the endpoints here. So these are all
different ways of denoting or depicting the same interval. Let's do some more examples here. So let's-- Let me draw
a number line again. So, a number line. And now let me do-- Let me
just do an open interval. An open interval just so that we clearly can see the difference. Let's say that I want to talk about the values between negative one and four. Let me use a different color. So the values between
negative one and four, but I don't want to include
negative one and four. So this is going to be an open interval. So I'm not going to include four, and I'm not going to include negative one. Notice I have open circles here. Over here had closed circles,
the closed circles told me that I included negative three and two. Now I have open circles here,
so that says that I'm not, it's all the values in
between negative one and four. Negative .999999 is going to be included, but negative one is not
going to be included. And 3.9999999 is going to be included, but four is not going to be included. So how would we-- What would be the notation for this? Well, here we could say
x is going to be a member of the real numbers
such that negative one-- I'm not going to say less than or equal to because x can't be equal to negative one, so negative one is strictly less than x, is strictly less than four. Notice not less than or
equal, because I can't be equal to four, four is not included. So that's one way to say it. Another way I could write it like this. x is a member of the
real numbers such that x is a member of... Now the interval is from
negative one to four but I'm not gonna use these brackets. These brackets say, "Hey,
let me include the endpoint," but I'm not going to include them, so I'm going to put the
parentheses right over here. Parentheses. So this tells us that we're
dealing with an open interval. This right over here,
let me make it clear, this is an open interval. Now you're probably
wondering, okay, in this case both endpoints were included,
it's a closed interval. In this case both endpoints were excluded, it's an open interval. Can you have things that
have one endpoint included and one point excluded, and
the answer is absolutely. Let's see an example of that. I'll get another number line here. Another number line. And let's say that we want to-- Actually, let me do it
the other way around. Let me write it first,
and then I'll graph it. So let's say we're thinking
about all of the x's that are a member of the
real numbers such that let's say negative four is
not included, is less than x, is less than or equal to negative one. So now negative one is included. So we're not going to
include negative four. Negative four is strictly less than, not less than or equal to, so x can't be equal to negative
four, open circle there. But x could be equal to negative one. It has to be less than
or equal to negative one. It could be equal to negative one so I'm going to fill
that in right over there. And it's everything in between. If I want to write it with
this notation I could write x is a member of the
real numbers such that x is a member of the interval, so it's going to go between
negative four and negative one, but we're not including negative four. We have an open circle here so I'm gonna put a
parentheses on that side, but we are including negative one. We are including negative one. So we put a bracket on that side. That right over there
would be the notation. Now there's other things that you could do with interval notation. You could say, well hey,
everything except for some values. Let me give another example. Let's get another example here. Let's say that we wanna talk
about all the real numbers except for one. We want to include all
of the real numbers. All of the real numbers except for one. Except for one, so we're gonna
exclude one right over here, open circle, but it can
be any other real number. So how would we denote this? Well, we could write x is a
member of the real numbers such that x does not equal one. So here I'm saying x can be
a member of the real numbers but x cannot be equal to one. It can be anything else, but
it cannot be equal to one. And there's other ways of
denoting this exact same interval. You could say x is a
member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that. Or you could do something interesting. This is the one that I would
use, this is the shortest and it makes it very clear. You say hey, everything except for one. But you could even do something
fancy, like you could say x is a member of the real
numbers such that x is a member of the set going from
negative infinity to one, not including one, or x is a
member of the set going from-- or a member of the
interval going from one, not including one, all
the way to positive, all the way to positive infinity. And when we're talking
about negative infinity or positive infinity, you
always put a parentheses. And the view there is you
could never include everything all the way up to infinity. It needs to be at least
open at that endpoint because infinity just
keeps going on and on. So you always want to put
a parentheses if you're talking about infinity
or negative infinity. It's not really an endpoint, it keeps going on and on forever. So you use the notation for open interval, at least at that end, and
notice we're not including, we're not including one
either, so if x is a member of this interval or that interval, it essentially could be
anything other than one. But this would have been
the simplest notation to describe that. |
16146 | https://www.youtube.com/watch?v=oYI-wyYoWxc | ❖The Closed Interval Method to Find The Absolute Maximum and Minimum ❖
Patrick J
1400000 subscribers
3211 likes
Description
509948 views
Posted: 19 Oct 2010
Title: Finding Absolute Maximums and Minimums with the Closed Interval Method
Description:
In this calculus tutorial, we explore the Closed Interval Method for finding absolute maximums and minimums of continuous functions on closed intervals. I’ll walk you through the full procedure while providing a short geometric insight into why this method works so effectively.
🔍 What You’ll Learn:
Step-by-step procedure for applying the Closed Interval Method
A graphical explanation to help you understand the concept
A specific example demonstrating the complete solution
Tips for identifying absolute extrema on closed intervals
This video is perfect for calculus students looking to grasp this essential technique. Don’t forget to like, subscribe, and turn on notifications for more math tutorials!
ClosedIntervalMethod #Calculus #AbsoluteMaximum #AbsoluteMinimum #MathTutorial #GraphicalUnderstanding #ContinuousFunctions #CalculusForBeginners #MathHelp #FindingExtrema #EducationalVideo
190 comments
Transcript:
all right in this video I want to talk about the closed interval method which is simply a way of finding absolute maximums and minimums of continuous functions on a closed interval and what we do is we simply find the critical numbers that fall within our interval and then uh we take those numbers along with the end points and we plug them back into the original function the biggest is the absolute Max the smallest is the absolute men and before I do a specific example um you know geometrically this makes I think sense if you think about it um you know suppose our function uh suppose our function does something like that um and then maybe it keeps going who knows well if we only look at you know maybe uh sort of one specific uh interval um where where do I want to chop it up at it doesn't really matter here um suppose I chop it up suppose this is my a and suppose we make B over here okay so really when we're looking at the closed interval we're only looking at sort of this section that I have in red okay well what happens you know if you kind of just restrict yourself you know to that little chunk of the graph well certainly um you know if we're trying to find the smallest and the largest values it certainly makes sense to look at sort of local minimums and local maximums and to find local minimums and local maximums that's uh at critical points either where the derivative zero or undefined maybe I even take you know kind of a longer section suppose we we stick B over here so now suppose I'm I'm looking at this whole section you know if you wanted to find the absolute maximum it certainly would occur you know at this place where the derivative doesn't exist but in this case the minimum would actually occur at an end point and you know in terms of the original graph that's certainly not going to be a critical number so that's why you have to look at you know uh the critical points which is where the derivative is zero are undefined but you also have to look at the end points so just kind of a little geometric argument why we're doing what we're doing nothing more complicated than that so the function that I want to do here is I'm going to find the absolute maximum and the absolute minimum of this function uh x^2 - 4 over x^2 + 4 on the interval -4 to four so the first thing I'm going to do is just find the critical points so I'm going to take the derivative and to take the derivative here we're just going to have to use the good old quotient rule so it says we take the denominator times the derivative of the numerator well the derivative of x^2 is going to be 2x the derivative of our constant is just zero so I'm going to leave that off minus uh then we'll just take the numerator and multiply that by the derivative of the denominator again we're just get uh 2x + 0 so I'm just going to Simply write 2X and then we simply take the denominator and we Square it so I'm going to simplify the numerator um this 2X is going to get distributed to both terms so 2x x^2 is going to be 2x cubed uh 2x pos4 is going to give us 8X um the 2x is going to get distributed um along with the negative you can think about it as really being - 2x if you want to think about it as being over there so -2X X will be - 2x cubed and -2X -4 will be POS 8x then we have our denominator still x^2 + 4^ 2 so one last thing here it looks like if we simply uh clean up the numerator 2x Cub - 2x Cub we've got 8 x + 8 x which is 16x and then we have our denominator x^2 + 4^ 2ar okay so now I've got to find the critical numbers and to find the critical numbers that's just where again the Der equals 0 or where the derivative is undefined but for the derivative to equal zero the numerator of the fraction is what has to equal zero for the derivative to be undefined the denominator of the fraction would have to equal zero well and then we'll just simply solve both equations for X so the first one's pretty easy we can just divide by 16 and that'll give us x equals 0 well the second one you know if you think about it um I think it's pretty clear there's no solution because what number can you square and then add four and then Square it and get zero at the end but likewise if you take a square root of both sides you know maybe if you didn't notice that you get x^2 + 4al 0 and again that makes sense it says the inside has to be zero then if you subtract -4 and try to take square roots well uh you can't take square roots of negative numbers without you using imaginary and we're just working with real numbers so the only critical number we find is xal 0 now I have to go back and make sure that that falls in the specified interval and certainly 0er does fall between -4 and 4 so we will use it um suppose we found two critical numbers suppose we found xal 0 and xal 30 well in this particular problem had we found another critical number if it's outside of this interval you know we don't we simply don't use it um and that just simply means you know xal 30 that's just simply a part of the graph we're not concerned with okay so my only critical number here was xal 0 our end points were at x = -4 and at X = pos4 so now what I have to do is I simply have to go back and plug all of these numbers uh back into the very original function we got to plug them back into this original function the biggest is going to be the maximum the smallest is going to be the minimum so I'm going to plug it back into FX = x^2 - 4 so let's see if we plug in 0o F of 0 would be 0^ 2 - 4 over 0^ 2 uh + 4 well it looks like we get -4 over pos4 which is going to give us -1 if we plug in F of -4 okay we'll get -4 2ar + 4 oops I was looking in the wrong place let's try that one more time F of -4 we'll get -4 squared but then in the numerator hey we've got aus4 and then in the denominator that's where we get our -42 + 4 so -4 2ar is going to be POS 16 - 4 16 - 4 is going to give us 12 in the numerator -42 that's 16 + 4 that's going to give us 20 okay so what's that I guess uh uh if we divide by two I guess we can divide by better than two we can divide by four if we divide the top by four we'll get three we divide the Bottom by four we'll get five notice if you plug in um positive four well since you're squaring them you're going to end up getting the same thing you're going to get 16 minus 4 on top and 16 + 4 on the bottom we'll get the exact same value which is 3 fths okay now it simply says uh all we have to do to pick out the absolute minimum is we just pick out the smallest value that occurs so in this case we would say there's an absolute minimum of -1 that's the smallest value and we would say there's an absolute maximum of well 3 fths it's okay that it occurs in two different places at the uh in this case it occurs at both the end points nothing wrong with that at all um but again there's only one maximum value it's 3 fifths again it just occurs in different places you know obviously if you had had you know two fifths instead uh well Nega -1 would still be the absolute minimum and three fifths would still be the absolute maximum two- fifths would just be uh you could probably go back well actually it's an endpoint so not much is going on there so again all you have to do is pick out the largest and the smallest again in this case uh the maximum occurred at two places but that's perfectly legal no problem there so nothing too crazy again you know it can be a little a little tedious especially when the derivative gets worse and you're finding all these critical points but again you're just taking the critical points that fall in the interval along with the end points of the interval you're plugging those back into the original function to get the maximum and the minimum |
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1.14
Subsets of Real Numbers
Written by:Neha Khandelwal
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
All numbers belong to the set of numbers known as the real number system. The real number system consists of every number you have ever dealt with since you were old enough to count.
There are several types of subsets of real numbers—numbers that can be expressed as a decimal. You are probably familiar with fractions, decimals, and counting numbers (1, 2, 3, . . .) from your daily life. All of these types of numbers are real numbers.
The sets of rational and irrational numbers constitute the set of real numbers. The set of real numbers is denoted by@$\begin{align}\mathbb{R}. \end{align}@$ Similar to integers, the real numbers can be divided into three subsets: negative real numbers, zero and positive real numbers. Each subset includes fractions, decimals and irrational numbers according to their algebraic sign (+ or -). Zero is considered neither positive nor negative.
The important subsets of the real numbers are described below.
What are Natural Numbers?
Natural numbers are the set of numbers we use for counting or enumerating items. The set of natural numbers is denoted by @$\begin{align}\mathbb{N}. \end{align}@$ Thus @$\begin{align}\mathbb{N} = { 1, 2, 3, 4, \dots }.\end{align}@$ The set of natural numbers is infinite.
What are Whole Numbers?
The set of whole numbers is the set of natural numbers plus zero. The set of whole numbers is denoted by @$\begin{align}\mathbb{W}. \end{align}@$ Thus @$\begin{align}\mathbb{W} = { 0, 1, 2, 3, 4, \dots }.\end{align}@$
What are Integers?
The set of integers adds the opposites of the natural numbers to the set of whole numbers: @$\begin{align}{ \dots -3, -2, -1, 0, 1, 2, 3, \dots }.\end{align}@$ Integers are made of three distinct subsets: negative integers, zero and positive integers. In this sense, the positive integers are just natural numbers. The set of integers is denoted by @$\begin{align}\mathbb{I}\end{align}@$ or @$\begin{align}\mathbb{Z}. \end{align}@$
What are Rational Numbers?
A rational number is any number that can be expressed in the form @$\begin{align}\frac{a}{b}\end{align}@$ where @$\begin{align}a\end{align}@$ and @$\begin{align}b\end{align}@$ are integers and @$\begin{align}b \neq 0. \end{align}@$ The set of rational numbers is denoted by @$\begin{align}\mathbb{Q}. \end{align}@$ Notice that rational numbers are fractions containing integers in both the numerator and the denominator, and the denominator is never 0. Every natural number, whole number, and integer is a rational number as they can be written in @$\begin{align}\frac{a}{b}\end{align}@$ form with@$\begin{align} b = 1 \end{align}@$ and @$\begin{align}a\end{align}@$ being the number itself. When a rational number is expressed as a decimal, then the decimals will terminate (end) or it will have a repeating pattern of digits. For example, @$\begin{align}\frac{1}{2}, \sqrt{81}, -7.456545654\end{align}@$ are all rational numbers.
What are Irrational Numbers?
The set of irrational numbers consists of all numbers that are not rational. This set of irrational numbers includes those numbers that cannot be written as the ratio of two integers, decimal numbers that are non-terminating and decimals that do not have a repeating pattern of digits. For example, @$\begin{align}\pi , \sqrt{2}, -2.345876921 \dots \end{align}@$ are irrational numbers.
Real Number System - Venn Diagram
The following diagram may help you to better understand the real number system.
Intervals as Subsets of @$\begin{align}\mathbb{R}\end{align}@$
Intervals are special subsets of the real numbers. Let @$\begin{align}a\end{align}@$ and @$\begin{align}b\end{align}@$ be two distinct real numbers where @$\begin{align}a<b.\end{align}@$ Intervals can be classified into two main types based on whether they are bounded or unbounded.
Finite Intervals
Finite intervals are those that are bounded by two finite numbers. They can be of different forms depending on whether the endpoints are included or excluded.
Open Interval @$\begin{align}\bf{(a,b):}\end{align}@$
The set of all real numbers lying between @$\begin{align}a\end{align}@$ and @$\begin{align}b,\end{align}@$ excluding the numbers @$\begin{align}a\end{align}@$ and @$\begin{align}b,\end{align}@$ is said to form an open interval.
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}(a, b) = { x:x ∈ R , a < x < b }\end{align}@$$The number @$\begin{align}a\end{align}@$ is called the left end point of the interval and the number @$\begin{align}b\end{align}@$ is called the right end point.
Note: The open interval @$\begin{align}(a, b)\end{align}@$ does not contain the left and right end points @$\begin{align}a\end{align}@$ and @$\begin{align}b.\end{align}@$
Geometrical Representation:
Let the points @$\begin{align}A\end{align}@$ and @$\begin{align}B\end{align}@$ on the real line represent the real numbers @$\begin{align}a\end{align}@$ and @$\begin{align}b\end{align}@$ respectively, then the open interval @$\begin{align}(a, b)\end{align}@$ is the set of all points to the right of @$\begin{align}A\end{align}@$ and to the left of @$\begin{align}B\end{align}@$.
It is represented on the real line by a line segment between @$\begin{align}A\end{align}@$ and @$\begin{align}B\end{align}@$ with hollow circles or parentheses at @$\begin{align}A(a)\end{align}@$ and @$\begin{align}B(b),\end{align}@$ indicating that the endpoints are not included.
Closed Interval @$\begin{align}\bf{[a, b]:}\end{align}@$
The closed interval @$\begin{align}[a, b]\end{align}@$ is the set of all points to the right of @$\begin{align}A\end{align}@$ (representing @$\begin{align}a\end{align}@$) and to the left of @$\begin{align}B\end{align}@$ (representing @$\begin{align}b\end{align}@$), including the endpoints @$\begin{align}a\end{align}@$ and @$\begin{align}b.\end{align}@$
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}[a, b] = { x:x ∈ R , a \le x \le b }\end{align}@$$
Geometrical Representation:
On the real line, this is represented by a solid line segment between @$\begin{align}A\end{align}@$ and @$\begin{align}B\end{align}@$ with solid circles or square brackets at @$\begin{align}A(a)\end{align}@$ and @$\begin{align}B(b),\end{align}@$ indicating the endpoints are included.
Half-open or Open-closed Interval @$\begin{align}\bf{(a,b]:}\end{align}@$
The half-open interval @$\begin{align}(a, b]\end{align}@$ is the set of all points to the right of @$\begin{align}A\end{align}@$ (representing @$\begin{align}a\end{align}@$) and to the left of @$\begin{align}B\end{align}@$ (representing @$\begin{align}b\end{align}@$), excluding @$\begin{align}a\end{align}@$ but including @$\begin{align}b.\end{align}@$
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}(a, b] = { x:x ∈ R , a < x \le b }\end{align}@$$
Geometrical Representation:
On the real line, this is represented by a line segment between @$\begin{align}A\end{align}@$ and @$\begin{align}B\end{align}@$ with a hollow circle or parenthesis at @$\begin{align}A(a)\end{align}@$ (excluded) and a solid circle or a square bracket at @$\begin{align}B(b)\end{align}@$ (included).
Half-closed or Closed-open Interval @$\begin{align}\bf{[a,b):}\end{align}@$
The half-closed interval @$\begin{align}[a, b)\end{align}@$ is the set of all points to the right of @$\begin{align}A\end{align}@$ (representing @$\begin{align}a\end{align}@$) and to the left of @$\begin{align}B\end{align}@$ (representing @$\begin{align}b\end{align}@$), including @$\begin{align}a\end{align}@$ but excluding @$\begin{align}b.\end{align}@$
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}[a, b) = {x: x ∈ R , a \le x < b }\end{align}@$$
Geometrical Representation:
On the real line, this is represented by a line segment between @$\begin{align}A\end{align}@$ and @$\begin{align}B\end{align}@$ with a solid circle or square bracket at @$\begin{align}A(a)\end{align}@$ (included) and a hollow circle or a parenthesis at @$\begin{align}B(b)\end{align}@$ (excluded).
Note: The length of any of the finite intervals @$\begin{align}(a,b), [a,b], [a,b), \text{ or } (a,b]\end{align}@$ is given by @$\begin{align}b-a.\end{align}@$
Infinite Intervals
Infinite intervals extend indefinitely in one or both directions. They are classified as follows:
Open Infinite Interval @$\begin{align}\bf{(a,\infty ):}\end{align}@$
The infinite interval @$\begin{align}(a, ∞)\end{align}@$ is the set of all points to the right of @$\begin{align}A\end{align}@$ (representing @$\begin{align}a\end{align}@$ ) and extending infinitely, excluding @$\begin{align}a\end{align}@$ .
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}(a, \infty) = { x:x ∈ R , x >a }\end{align}@$$
Geometrical Representation:
On the real line, this is represented by a ray starting from @$\begin{align}A(a)\end{align}@$ with a hollow circle or a paranthesis at @$\begin{align}A(a),\end{align}@$ extending to the right infinitely.
Closed Infinite Interval @$\begin{align}\bf{[a,\infty ):}\end{align}@$
The infinite interval @$\begin{align}[a, ∞)\end{align}@$ is the set of all points to the right of @$\begin{align}A\end{align}@$ (representing @$\begin{align}a\end{align}@$) and extending infinitely, including @$\begin{align}a\end{align}@$ .
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}[a, \infty) = { x:x ∈ R , x \ge a }\end{align}@$$
Geometrical Representation:
On the real line, this is represented by a ray starting from @$\begin{align}A(a)\end{align}@$ with a solid circle or a square bracket at @$\begin{align}A(a),\end{align}@$ extending to the right infinitely.
Open Infinite Interval @$\begin{align}\bf{(-\infty ,a):}\end{align}@$
The infinite interval @$\begin{align}(-\infty ,a)\end{align}@$ is the set of all points to the left of @$\begin{align}A\end{align}@$ (representing @$\begin{align}a\end{align}@$) and extending infinitely in the negative direction, excluding @$\begin{align}a\end{align}@$.
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}(-\infty ,a) = { x:x ∈ R , x < a }\end{align}@$$
Geometrical Representation:
On the real line, this is represented by a ray starting from @$\begin{align}A(a)\end{align}@$ with a hollow circle or a paranthesis at @$\begin{align}A(a),\end{align}@$ extending infinitely in the negative direction, excluding @$\begin{align}a.\end{align}@$
Closed Infinite Interval @$\begin{align}\bf{(-\infty,a ]:}\end{align}@$
The infinite interval @$\begin{align}[-∞,a)\end{align}@$ is the set of all points to the left of @$\begin{align}A\end{align}@$ (representing @$\begin{align}a\end{align}@$) and extending infinitely in the negative direction, including @$\begin{align}a\end{align}@$.
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}(-\infty,a] = { x:x ∈ R , x \le a }\end{align}@$$
Geometrical Representation:
On the real line, this is represented by a ray starting from @$\begin{align}A(a)\end{align}@$ with a solid circle or a square bracket at @$\begin{align}A(a),\end{align}@$ extending infinitely in the negative direction, including @$\begin{align}a.\end{align}@$
Infinite Interval @$\begin{align}(-∞, ∞)\end{align}@$ :
The infinite interval@$\begin{align}(-∞, ∞)\end{align}@$ represents the set of all real numbers. It extends infinitely in both directions.
Mathematical Description:
Mathematically, it is denoted by@$$\begin{align}(-\infty ,\infty ) = { x:x ∈ R }\end{align}@$$
Geometrical Representation:
On the real line, this interval is represented by the entire line, covering all points.
Examples of Subsets of Real Numbers
Example 1
Determine if @$\begin{align}x=-5\end{align}@$ belongs to interval @$\begin{align}(-\infty ,-3).\end{align}@$
The interval @$\begin{align}(-\infty,-3) \end{align}@$ includes all numbers less than @$\begin{align}-3,\end{align}@$ but not @$\begin{align}-3\end{align}@$ itself.
Since @$\begin{align}x=-5\end{align}@$ satisfies @$\begin{align}x<-3,\end{align}@$ it belongs to the interval @$\begin{align}(-\infty ,-3).\end{align}@$
Hence, @$\begin{align}x=-5\end{align}@$ belongs to @$\begin{align}(-\infty, -3). \end{align}@$
Example 2
Find the intersection of @$\begin{align}(-\infty, 3) \text{ and } (-\infty, 5].\end{align}@$
@$\begin{align}(-\infty,3)\end{align}@$ includes all numbers less than 3, excluding 3.
@$\begin{align}(-\infty, 5]\end{align}@$ includes all numbers less than or equal to 5.
The intersection is the set of elements common to both intervals:@$\begin{align}(-\infty, 3).\end{align}@$
Example 3
Verify if @$\begin{align}(-5,0]\end{align}@$ is a subset of @$\begin{align}(-\infty, 1].\end{align}@$
The interval @$\begin{align}(-\infty,1]\end{align}@$ includes all numbers less than or equal to 1.
The interval @$\begin{align}(-5,0]\end{align}@$ includes all numbers greater than @$\begin{align}-5\end{align}@$ and less than or equal to @$\begin{align}0.\end{align}@$
Since every element in @$\begin{align}(-5,0]\end{align}@$ is also in @$\begin{align}(-\infty,1], (-5,0]\end{align}@$ is a subset of @$\begin{align}(-\infty,1].\end{align}@$
Hence, @$\begin{align}(-5,0]\end{align}@$ is a subset of @$\begin{align}(-\infty,1].\end{align}@$
| |
| Summary of Subsets of Real Numbers |
| The real number system includes all numbers used in daily life, including fractions, decimals, and counting numbers. It is denoted by @$\begin{align}ℝ.\end{align}@$ The subsets of real numbers include rational and irrational numbers. Real numbers can also be divided into negative real numbers, zero, and positive real numbers. Finite intervals in @$\begin{align}ℝ\end{align}@$ include: + Open intervals: @$\begin{align}(a, b) = {x ∈ ℝ : a < x < b},\end{align}@$ excluding endpoints. + Closed intervals: @$\begin{align}[a, b] = {x ∈ ℝ : a ≤ x ≤ b},\end{align}@$ including endpoints. + Half-open intervals: @$\begin{align}(a, b] \text{ or }[a, b),\end{align}@$ including one endpoint. Infinite intervals in @$\begin{align}ℝ\end{align}@$ include: + Open infinite intervals: @$\begin{align}(a, ∞)\end{align}@$ and @$\begin{align}(-∞, a).\end{align}@$ + Closed infinite intervals: @$\begin{align}[a, ∞)\end{align}@$ and @$\begin{align}(-∞, a].\end{align}@$ + Infinite interval @$\begin{align}(-∞, ∞)\end{align}@$ includes all real numbers. Geometrically, intervals are represented on a real line: + Open intervals have parentheses or hollow circles at endpoints. + Closed intervals have square brackets or solid circles at endpoints. + Infinite intervals extend indefinitely in one or both directions. |
Review Question of Subsets of Real Numbers
Determine if @$\begin{align}x=0\end{align}@$ belongs to the interval @$\begin{align}(-\infty, 1).\end{align}@$
Find the intersection of @$\begin{align}(-\infty, 0]\end{align}@$ and @$\begin{align}[0, \infty).\end{align}@$
Identify if @$\begin{align}[1,\infty)\end{align}@$ is a subset of @$\begin{align}(-2,\infty).\end{align}@$
Solve @$\begin{align}{x}^{2}<9\end{align}@$ and express the solution in interval notation.
Determine the set of numbers that satisfy @$\begin{align}x>-4 \text{ and } x\leq2,\end{align}@$ and express it in interval notation.
If @$\begin{align}x^2>4,\end{align}@$ express the solution set in interval notation.
Express the set of all @$\begin{align}x≥−2\end{align}@$ in interval notation and geometrically describe it.
Solve @$\begin{align}2x-3<5\end{align}@$ and express the solution in interval notation.
Find the complement of @$\begin{align}[0,3)\end{align}@$ in @$\begin{align}(-\infty,\infty).\end{align}@$
Determine if the point @$\begin{align}x=-10\end{align}@$ lies in the union of @$\begin{align}(-\infty,-8) \text{ and } [0,\infty).\end{align}@$
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16148 | https://math.hawaii.edu/~dale/140/28o.pdf | Math 140 Lecture 28 is a vertical parabola with a vertical axis y ax2 bxc of symmetry. bOther parabolas have horizontal or slanted axes. axis focus directrix parabola vertex DEFINITION. A parabola consists of all points equidistant between a given focus point and a given directrix line. The axis is the line through the focus and perpendicular to the directrix. The vertex is the intersection of the parabola and the axis. It lies halfway between the focus and the directrix. ---------- VERTICAL PARABOLA THEOREM. For k = 0, is a vertical parabola with: x2 ky Axis = the y-axis. Vertex = (0, 0). Focus = (0, p). Directrix: where . y p p k 4 (0,p) = focus y=-p (x,y) directrix (0,0) = vertex axis bExchanging x and y gives ¥ HORIZONTAL PARABOLA THEOREM. For k = 0, is a horizontal parabola with: y2 kx Axis = the x-axis; Vertex = (0, 0). Focus = (p, 0). Directrix: where x p p k 4 ----------(p,0) = focus x=-p directrix axis (0,0) vertex focal-width point x2-parabolas like y = x2 are vertical; y 2-parabolas are horizontal. The focal-width points determine the parabola’s width.
Graph y2 4x Horizontal parabola with: Axis = the x-axis; k 4, p k 4 4 4 1 Vertex = (0, 0). Focus = (p, 0) or (0, p) = (-1,0). Directrix: is x p or y p x 1 Graph: (-1,0) = focus x=1 directrix axis (0,0) vertex focal-width point Find the focus, directrix and graph: .
y x2/8 . y2 4x The graph is a: (A) vertical parabola (B) horizontal parabola Find p. p k/4 Find the vertex: Find the focus: (0,p) or ( p,0). Find the directrix: . x p or y p Find the correct shape.
(A) (C) (B) (D) (E) # THEOREM. In any equation, down b units „ y by y+b up b units ‚ y by y- b left a units Î x by x+ a right a units ƒ x by x- a shifts the graph Replacing each Shifting a parabola also shifts its focus, directrix, vertex, and axis.
To graph a parabola: vGet occurrences of the squared variable on the left, everything else on the right.
v Complete the square if needed. vWrite the equation in one of these two forms: vertical parabola form: x a2 ky b horizontal parabola form: y b2 kx a Find the focus, directrix and graph: .
x2 2x98y 0 ` . y2 2y 4x5 The graph is a: (A) vertical parabola (B) horizontal parabola vAfter completing the square and factoring, the equation is: vThis is shifted y2 4x Find p. p k/4 Find the vertex. (0, 0) shifts to: Find the focus. (0,p) or ( p,0) shifts to: Find the directrix. shifts to: x p or y p Graph. b b |
16149 | https://openstax.org/books/anatomy-and-physiology-2e/pages/6-chapter-review | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Anatomy and Physiology 2e
Chapter Review
Anatomy and Physiology 2eChapter Review
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## 6.1 The Functions of the Skeletal System
The major functions of the bones are body support, facilitation of movement, protection of internal organs, storage of minerals and fat, and hematopoiesis. Together, the muscular system and skeletal system are known as the musculoskeletal system.
## 6.2 Bone Classification
Bones can be classified according to their shapes. Long bones, such as the femur, are longer than they are wide. Short bones, such as the carpals, are approximately equal in length, width, and thickness. Flat bones are thin, but are often curved, such as the ribs. Irregular bones such as those of the face have no characteristic shape. Sesamoid bones, such as the patellae, are small and round, and are located in tendons.
## 6.3 Bone Structure
A hollow medullary cavity filled with yellow marrow runs the length of the diaphysis of a long bone. The walls of the diaphysis are compact bone. The epiphyses, which are wider sections at each end of a long bone, are filled with spongy bone and red marrow. The epiphyseal plate, a layer of hyaline cartilage, is replaced by osseous tissue as the organ grows in length. The medullary cavity has a delicate membranous lining called the endosteum. The outer surface of bone, except in regions covered with articular cartilage, is covered with a fibrous membrane called the periosteum. Flat bones consist of two layers of compact bone surrounding a layer of spongy bone. Bone markings depend on the function and location of bones. Articulations are places where two bones meet. Projections stick out from the surface of the bone and provide attachment points for tendons and ligaments. Holes are openings or depressions in the bones.
Bone matrix consists of collagen fibers and organic ground substance, primarily hydroxyapatite formed from calcium salts. Osteogenic cells develop into osteoblasts. Osteoblasts are cells that make new bone. They become osteocytes, the cells of mature bone, when they get trapped in the matrix. Osteoclasts engage in bone resorption. Compact bone is dense and composed of osteons, while spongy bone is less dense and made up of trabeculae. Blood vessels and nerves enter the bone through the nutrient foramina to nourish and innervate bones.
## 6.4 Bone Formation and Development
All bone formation is a replacement process. Embryos develop a cartilaginous skeleton and various membranes. During development, these are replaced by bone during the ossification process. In intramembranous ossification, bone develops directly from sheets of mesenchymal connective tissue. In endochondral ossification, bone develops by replacing hyaline cartilage. Activity in the epiphyseal plate enables bones to grow in length. Modeling allows bones to grow in diameter. Remodeling occurs as bone is resorbed and replaced by new bone. Osteogenesis imperfecta is a genetic disease in which collagen production is altered, resulting in fragile, brittle bones.
## 6.5 Fractures: Bone Repair
Fractured bones may be repaired by closed reduction or open reduction. Fractures are classified by their complexity, location, and other features. Common types of fractures are transverse, oblique, spiral, comminuted, impacted, greenstick, open (or compound), and closed (or simple). Healing of fractures begins with the formation of a hematoma, followed by internal and external calli. Osteoclasts resorb dead bone, while osteoblasts create new bone that replaces the cartilage in the calli. The calli eventually unite, remodeling occurs, and healing is complete.
## 6.6 Exercise, Nutrition, Hormones, and Bone Tissue
Mechanical stress stimulates the deposition of mineral salts and collagen fibers within bones. Calcium, the predominant mineral in bone, cannot be absorbed from the small intestine if vitamin D is lacking. Vitamin K supports bone mineralization and may have a synergistic role with vitamin D. Magnesium and fluoride, as structural elements, play a supporting role in bone health. Omega-3 fatty acids reduce inflammation and may promote production of new osseous tissue. Growth hormone increases the length of long bones, enhances mineralization, and improves bone density. Thyroxine stimulates bone growth and promotes the synthesis of bone matrix. The sex hormones (estrogen and testosterone) promote osteoblastic activity and the production of bone matrix, are responsible for the adolescent growth spurt, and promote closure of the epiphyseal plates. Osteoporosis is a disease characterized by decreased bone mass that is common in aging adults. Calcitriol stimulates the digestive tract to absorb calcium and phosphate. Parathyroid hormone (PTH) stimulates osteoclast proliferation and resorption of bone by osteoclasts. Vitamin D plays a synergistic role with PTH in stimulating the osteoclasts. Additional functions of PTH include promoting reabsorption of calcium by kidney tubules and indirectly increasing calcium absorption from the small intestine. Calcitonin inhibits osteoclast activity and stimulates calcium uptake by bones.
## 6.7 Calcium Homeostasis: Interactions of the Skeletal System and Other Organ Systems
Calcium homeostasis, i.e., maintaining a blood calcium level of about 10 mg/dL, is critical for normal body functions. Hypocalcemia can result in problems with blood coagulation, muscle contraction, nerve functioning, and bone strength. Hypercalcemia can result in lethargy, sluggish reflexes, constipation and loss of appetite, confusion, and coma. Calcium homeostasis is controlled by PTH, vitamin D, and calcitonin and the interactions of the skeletal, endocrine, digestive, and urinary systems.
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Book title: Anatomy and Physiology 2e
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Absolute value
The absolute value of a real number, denoted , is the unsigned portion of . Geometrically, is the distance between and zero on the real number line.
The absolute value function exists among other contexts as well, including complex numbers.
Contents
1 Real numbers
2 Complex numbers
3 Examples
4 Problems
5 See Also
Real numbers
When is real, is defined as For all real numbers and , we have the following properties:
(Alternative definition)
(Non-negativity)
(Positive-definiteness)
(Multiplicativeness)
(Triangle Inequality)
(Symmetry)
Note that
and
Complex numbers
For complex numbers, the absolute value is defined as , where and are the real and imaginary parts of , respectively. It is equivalent to the distance between and the origin, and is usually called the complex modulus.
Note that , where is the complex conjugate of .
Examples
If , for some real number , then or .
If , for some real numbers , , then or , and therefore or .
Problems
Find all real values of if .
Find all real values of if .
(AMC 12 2000) If , where , then find .
See Also
Magnitude
Norm
Valuation
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16152 | https://www.quora.com/What-is-a-geometric-sequence-Is-it-always-increasing-or-decreasing-If-yes-then-why | Something went wrong. Wait a moment and try again.
Special Sequences
Geometric Patterns
Algebraic Concepts
Series and Sequences
Mathematical Ideas
Function Sequences
Mathematical Sciences
5
What is a geometric sequence? Is it always increasing or decreasing? If yes, then why?
Vesal Ea
A geometric sequence is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. (the Wikipedia definition)
The sequence can be increasing, constant, decreasing, or neither (I’ll call those irregulars). Here are some examples:
(r is the common ratio of the sequence)
1, 2, 4, 8, … (r = 2)
-8, -4, -2, -1/2, … (r = 1/2)
9, 3, 1, 1/3, … (r = 1/3)
-1/3, -1, -3, -9, … (r = 3)
5, 5, 5, 5, … (r = 1)
2, -4, 8, -16, … (r = -2)
As you can see, the ratio and the first term to
A geometric sequence is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. (the Wikipedia definition)
The sequence can be increasing, constant, decreasing, or neither (I’ll call those irregulars). Here are some examples:
(r is the common ratio of the sequence)
Increasing:
1, 2, 4, 8, … (r = 2)
-8, -4, -2, -1/2, … (r = 1/2)
Decreasing:
9, 3, 1, 1/3, … (r = 1/3)
-1/3, -1, -3, -9, … (r = 3)
Constant:
5, 5, 5, 5, … (r = 1)
Irregular:
2, -4, 8, -16, … (r = -2)
As you can see, the ratio and the first term together determine which category the sequence falls into.
I’ll write the more structured way to determine whether a sequence is increasing or decreasing in the end, but really, making some examples for yourself and trying to figure out how a sequence changes is the best way to learn how this works.
The more structured way:
If the first term is positive:
0<r<1 : decreasing
r>1: increasing
If the first term is negative:
0<r<1 : increasing
r>1 : decreasing
In both cases:
r=1 : constant
r<0 : the sequence increases and decreases repeatedly (the irregular category mentioned before)
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Related questions
What is the sum of the first six terms of a geometric sequence whose first term is 5 and whose common ratio is 2?
Is the first term in a geometric sequence interchangeable?
How do I fill in a geometric sequence?
Why is a geometric sequence useful?
Which of the following statements is true to all geometric sequences? If the common ratio is negative, is it increasing or decreasing?
Ajay Sreenivas
Former Aerospace Engineer/ Staff Consultant at Ball Aerospace (1980–2010)
·
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2y
A geometric series is one in which every term after its first term is equal to the product of the previous term and a constant multiplier f. The terms in the series always increase if the multiplier is greater than unity and decrease if the multiplier is less than unity.
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Is the sequence decreasing or increasing?
From n=1. The expression =1/(1+1)=1/2 or 0.5
n=100 the expression= 100/(100+1)=100/101 or 0.99..
You can see that the solution approaches 1
If n is between 1 and 0
Try n=0.9 Now n/(n+1)=0.9/(1.9)=0.47
Try n=0.8 It = 0.8/1.8=0.44
0.7 Gives 0.7/1.7= 0.41
0.6 Gives 0.375
0.5 Gives 0.33…
0.4 Gives 0.285
0.3 Gives 0. 23.
0.2 Gives 0.166.
0.1 Gives 0.09.
0.01 Gives 0.0099..
So the solution approaches 0 as n moves from 1 to 0
If n=—1 the solution is undefined —1/(—1+1)=1/0
If n =— 0.1 Solution is —0.11
n=—0.5 Solution is —1
n=—0.9 Solution is —9
n=—0.99. Solution is —99 So as n approaches—1 the solution approaches nega
From n=1. The expression =1/(1+1)=1/2 or 0.5
n=100 the expression= 100/(100+1)=100/101 or 0.99..
You can see that the solution approaches 1
If n is between 1 and 0
Try n=0.9 Now n/(n+1)=0.9/(1.9)=0.47
Try n=0.8 It = 0.8/1.8=0.44
0.7 Gives 0.7/1.7= 0.41
0.6 Gives 0.375
0.5 Gives 0.33…
0.4 Gives 0.285
0.3 Gives 0. 23.
0.2 Gives 0.166.
0.1 Gives 0.09.
0.01 Gives 0.0099..
So the solution approaches 0 as n moves from 1 to 0
If n=—1 the solution is undefined —1/(—1+1)=1/0
If n =— 0.1 Solution is —0.11
n=—0.5 Solution is —1
n=—0.9 Solution is —9
n=—0.99. Solution is —99 So as n approaches—1 the solution approaches negative infinity
n=—1.01 Solution is 101
n= —2 Solution is 2
n= —10 Solution is 1.11
n=—100 Solution is 1.01 So as n becomes more negative the solution approaches 1
A graph will show asymptotes x=—1 and y= 1
So you can see that it does both
Scooby
I’ve done math before
·
3y
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When is a sequence geometric?
A sequence is geometric when you multiply by a number (denoted by r) to get to the next number.
Take the sequence 1, 2, 4, 8, 16.
This sequence is geometric because you have to multiply by 2 to get to the next.
So, if we start at one we can generalize this into the form you might have seen.
Take one, which is the first number of the sequence (represented by the letter a)
1 = 1 (1 2^0 = 1)
1 2 = 2
1 2 2 = 4
1 2^3 = 8
1 2^4 = 16
And this is the original sequence.
So, let’s have {math} x_{n} {/math} represent the nth term for the sequence - if n = 1, x = 1, if n = 2, x = 2, if n = 3, x = 4 (Sinc
A sequence is geometric when you multiply by a number (denoted by r) to get to the next number.
Take the sequence 1, 2, 4, 8, 16.
This sequence is geometric because you have to multiply by 2 to get to the next.
So, if we start at one we can generalize this into the form you might have seen.
Take one, which is the first number of the sequence (represented by the letter a)
1 = 1 (1 2^0 = 1)
1 2 = 2
1 2 2 = 4
1 2^3 = 8
1 2^4 = 16
And this is the original sequence.
So, let’s have {math} x_{n} {/math} represent the nth term for the sequence - if n = 1, x = 1, if n = 2, x = 2, if n = 3, x = 4 (Since we’re doubling to get to the next number). Recall that the first term is represented by “a” and the factor is represented by “r.”
{math} x_{n} = ar^{n-1} {/math}
This is the general form of a geometric sequence.
Plugging in for this specific sequence:
{math} x_{n} = 2^{n-1} {/math}
Pretty boring, but still geometric. Keep in mind that the “a” can be any number, and so can the “r.” This includes fractions.
More sequences:
5, 20, 80, 320, 1280, … (a = 5, r = 4)
10, 5/2, 5/8, 5/32, 5/128 … (a = 10, r = 1/4)
See if you can write the general form for these sequences.
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Related questions
What is geometric sequence, and how is it applied?
What is the definition of an increasing or decreasing sequence? Can you check if a sequence is increasing or decreasing without actually calculating it?
How do you find the eighth term of a geometric sequence?
What is the sum of the geometric sequence in #2?
What are the common differences in a geometric sequence and is there a relationship between them and their terms?
Daniel Melchor
IB Math Higher Level student, 2nd year
·
5y
Related
What is a geometric sequence in math?
A geometric sequence is a sequence of numbers where the common difference between each of them is a multiplication or division.
Let’s say for example that we have this sequence:
1,2,4,8,16,32,64…
To find out what the common factor is we can simply divide two consecutive numbers (the second by the first one) so, for example, we could find out that the common factor in this series is 2 by dividing 64/32.
We could also have a sequence that always gets smaller instead of bigger. Let’s say, for example, that we have a sequence where the common factor is less than one. So 0<r<1:
1,1/2,1/4,1/8,1/16…
To fin
A geometric sequence is a sequence of numbers where the common difference between each of them is a multiplication or division.
Let’s say for example that we have this sequence:
1,2,4,8,16,32,64…
To find out what the common factor is we can simply divide two consecutive numbers (the second by the first one) so, for example, we could find out that the common factor in this series is 2 by dividing 64/32.
We could also have a sequence that always gets smaller instead of bigger. Let’s say, for example, that we have a sequence where the common factor is less than one. So 0<r<1:
1,1/2,1/4,1/8,1/16…
To find out the common factor we can do the same thing and we get 1/2.
Some formulas we can use are the following:
To find any term (n) in the sequence we can use [math]a_n=a_1r^{n-1}[/math]. So for example, if we wanted to find the 3rd term in the first example, we could multiply the first term times the common factor to the power of 3–1. So we would have [math]a_n=12^{3–1} \rightarrow a_n=2^2 \rightarrow a_n=4[/math]
If a sequence has a common factor where [math]-1<r<1[/math], we can use the formula [math]\frac{a_1}{1-r}[/math] to find out what the sum of the infinite terms in the sequence is. For example, in the second example where [math]-1<\frac{1}{2}<1[/math], the infinite sum would be [math]\frac {1}{1–0.5}=2[/math]
Finally, if we want to sum only X terms, we can use the formula [math]\frac{a_1(r^{n}-1)}{r-1}[/math]. Where the sum of the first 10 numbers in the first example would be: [math]\frac {1(2^{10}–1)}{1}=1023[/math]
Edit: equations
Steve Smith
Orchestrator and Audio Engineer (1979–present)
·
Author has 1.3K answers and 635.1K answer views
·
6y
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Why are geometric sequences exponential functions?
In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 3, 9, 27, 81 ... is a geometric progression with common ratio 3.
Put this sequence into Cartesian coordinates to make a function (a series of corrdinates expressed as (x,y) pairs
(1,3) (2,9) (3,27)(4,81)
Joining these points shows a curve which is common to exponential functions.
And then we can see - upon analysis - that the function is y=x^3
the p
In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 3, 9, 27, 81 ... is a geometric progression with common ratio 3.
Put this sequence into Cartesian coordinates to make a function (a series of corrdinates expressed as (x,y) pairs
(1,3) (2,9) (3,27)(4,81)
Joining these points shows a curve which is common to exponential functions.
And then we can see - upon analysis - that the function is y=x^3
the presence of the 3 as a power of x denotes an exponential function
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Ram Kushwah
Up and coming Most viewed writer
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6y
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What is the meaning of geometric (geometry) in geometric progression or how is geometry related to geometric progression?
let a=1/2 ,r=1/2
then sum of infinite GP series
S∞=1/2 +1/4 +1/8 +1/16+,…………………………….∞ ————————-(1)
S∞=a(r^n-1)/r-1
as r =1/2 <1 so If n=∞ then r^n=0
S∞=a(0-1)/r-1
S∞=a/1-r ——————————(2)
S∞=a/1-r
putting a=1/2
S∞=1/2/1–1/2
=1/2 / 1/2
S∞= 1—————————(3)
In above figure
Let the side of big square =1 unit
area of square=1^2 =1 square unit
Sum of area of all small squares:=1/2+1/4+1/8+1/+……………………+∞———-(4)
and from eq
let a=1/2 ,r=1/2
then sum of infinite GP series
S∞=1/2 +1/4 +1/8 +1/16+,…………………………….∞ ————————-(1)
S∞=a(r^n-1)/r-1
as r =1/2 <1 so If n=∞ then r^n=0
S∞=a(0-1)/r-1
S∞=a/1-r ——————————(2)
S∞=a/1-r
putting a=1/2
S∞=1/2/1–1/2
=1/2 / 1/2
S∞= 1—————————(3)
In above figure
Let the side of big square =1 unit
area of square=1^2 =1 square unit
Sum of area of all small squares:=1/2+1/4+1/8+1/+……………………+∞———-(4)
and from equation (1)
S∞=1/2 +1/4 +1/8 +1/16+,…………………………….∞——————-(5)
From equation(4) and (5) we conclude
Area of big square =Sum of area of all small ...
Rajeev Raj
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Is the sequence decreasing or increasing?
Hint : → for increasing f(n+1) ≥ f(n)
→ for decreasing f(n+1) ≤ f(n)
………….
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What is a geometric sequence? What happens if a geometric sequence has negative terms?
A geometric sequence is a sequence of terms/numbers where the next term is taken from the previous term mutliplied by something called a common ratio.
The general term would be [math]ar^{n-1}[/math], where
a is the first term
r is the common ratio
n is the term number/position
So, it goes like [math]a,\ ar,\ ar^2[/math] and so on.
A geometric sequence can have negative terms if the first term is negative. If the common ratio is positive, then every term afterwards would be negative too.
If the common ratio is negative, then the sign of the terms would alternate between positive and negative.
It would be the same as a normal ge
A geometric sequence is a sequence of terms/numbers where the next term is taken from the previous term mutliplied by something called a common ratio.
The general term would be [math]ar^{n-1}[/math], where
a is the first term
r is the common ratio
n is the term number/position
So, it goes like [math]a,\ ar,\ ar^2[/math] and so on.
A geometric sequence can have negative terms if the first term is negative. If the common ratio is positive, then every term afterwards would be negative too.
If the common ratio is negative, then the sign of the terms would alternate between positive and negative.
It would be the same as a normal geometric sequence, nothing much happens if they are negative.
Aaron Dunbrack
Just about ran UMass out of math courses (undergrad & grad)
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Could someone explain to me why a geometric series is called geometric?
Historically, I don’t know, but one motivation is
this image
:
You can see that splitting a square repeatedly in half and adding the parts gives you back the original square. Each part is a half of the previous part, thus giving you a geometric conception of the sum you just made. (This particular usage is specific to 1/2, but the idea generalizes with a few additional technicalities, like an overall normalization factor.)
Historically, I don’t know, but one motivation is
this image
:
You can see that splitting a square repeatedly in half and adding the parts gives you back the original square. Each part is a half of the previous part, thus giving you a geometric conception of the sum you just made. (This particular usage is specific to 1/2, but the idea generalizes with a few additional technicalities, like an overall normalization factor.)
Timothy Norfolk
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Is the sequence decreasing or increasing?
Q: Is the sequence decreasing or increasing?
Writing [math]\frac{n}{n+1} = 1 - \frac{1}{n+1}[/math], we see that the quantity is clearly increasing in [math]n[/math].
Ron Davis
I earn my living with mathematics.
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Is there a geometric sequence that contains each of the numbers 1, 2 and 3 (geometric series, math)?
Say that a geometric series exists that contains the numbers 1, 2, and 3. Say the common ratio between successive terms is [math]a.[/math]
By the terms of the question, the zeroth term in the sequence is
[math]2[/math] is the [math]n^{\text{th}}[/math] term, for some integer [math]n,[/math] so
for that integer, and [math]3[/math] is the [math]m^{\text{th}}[/math] term, so
From Equations (2) and (3),
[math]3=2^{\dfrac m n},~~~~~~(4)[/math]
and so
However, [math]3^n[/math] is a multiple of [math]3,[/math] and an odd number, but [math]2^m[/math] is an even number and no multiple of [math]3.[/math] Thus, there is no pair of integers [math]n[/math] and [math]m[/math] that will satisfy Equation (5).
Say that a geometric series exists that contains the numbers 1, 2, and 3. Say the common ratio between successive terms is [math]a.[/math]
By the terms of the question, the zeroth term in the sequence is
[math]a^0=1,~~~~~~(1)[/math]
[math]2[/math] is the [math]n^{\text{th}}[/math] term, for some integer [math]n,[/math] so
[math]a^n=2~~~~~~(2)[/math]
for that integer, and [math]3[/math] is the [math]m^{\text{th}}[/math] term, so
[math]a^m=3.~~~~~~(3)[/math]
From Equations (2) and (3),
[math]3=2^{\dfrac m n},~~~~~~(4)[/math]
and so
[math]3^n=2^m.~~~~~~(5)[/math]
However, [math]3^n[/math] is a multiple of [math]3,[/math] and an odd number, but [math]2^m[/math] is an even number and no multiple of [math]3.[/math] Thus, there is no pair of integers [math]n[/math] and [math]m[/math] that will satisfy Equation (5).
Consequently, the answer is no, there is no geometric sequence that contains the numbers 1, 2, and 3.
Dmitry Volovich
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Is it true that every infinite geometric series converges to a positive value? If not, can you provide an example? What is the reason behind this?
No, it is not true. A geometric series converges if and only if the ratio is strictly between [math]-1[/math] and [math]1[/math]. In other words, not all geometric series converge at all: for example,
clearly diverges (at least, in the real numbers). Furthermore, a geometric series can easily converge to a negative value: notice how
converges to a positive value [math]2[/math], but just like math^n[/math] is a geometric progression, so is [math]-(1/2)^n[/math], and we have
No, it is not true. A geometric series converges if and only if the ratio is strictly between [math]-1[/math] and [math]1[/math]. In other words, not all geometric series converge at all: for example,
[math]\sum\limits_{n=0}^{+\infty}2^n=1+2+4+8+16+\cdots[/math]
clearly diverges (at least, in the real numbers). Furthermore, a geometric series can easily converge to a negative value: notice how
[math]\sum\limits_{n=0}^{+\infty}\left(\frac12\right)^n=1+\frac12+\frac14+\frac18+\frac1{16}+\cdots=2[/math]
converges to a positive value [math]2[/math], but just like math^n[/math] is a geometric progression, so is [math]-(1/2)^n[/math], and we have
In other words, not all geometric series converge, and those that converge don't necessarily do so towards a positive value (unless you provide further restrictions, like all of the terms in the series being positive).
Related questions
What is the sum of the first six terms of a geometric sequence whose first term is 5 and whose common ratio is 2?
Is the first term in a geometric sequence interchangeable?
How do I fill in a geometric sequence?
Why is a geometric sequence useful?
Which of the following statements is true to all geometric sequences? If the common ratio is negative, is it increasing or decreasing?
What is geometric sequence, and how is it applied?
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16153 | https://www.sciencedirect.com/topics/computer-science/chessboard-pattern | Chessboard Pattern - an overview | ScienceDirect Topics
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Chessboard Pattern
In subject area:Computer Science
A 'Chessboard Pattern' refers to a regular arrangement of squares on a chessboard, which is used for camera calibration in computer vision applications.
AI generated definition based on:Emerging Trends in Image Processing, Computer Vision and Pattern Recognition, 2015
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You might find these chapters and articles relevant to this topic.
Chapter
A topological approach for detection of chessboard patterns for camera calibration
2015, Emerging Trends in Image Processing, Computer Vision and Pattern RecognitionGustavo Teodoro Laureano, ... Clarimar José Coelho
Abstract
This work aims to automatically identify chessboard patterns for camera calibration. The method uses a fast x-shaped corner detector and a geometric mesh to represent the relative association between features. The mesh allows considering the regularity of the chessboard pattern and a topological filter is presented. The matching between real world points and their image projections is done using neighboring properties in a filtered mesh. The point location is locally updated to the subpixel precision with a specific x-corner detector. The calibration points are determined even when the pattern is partially occluded. The experiments show that the proposed algorithm provides a robust detection of the chessboard patterns and take great advantage of image frames. The method is applicable for both online and off-line detection of chessboard patterns.
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Book2015, Emerging Trends in Image Processing, Computer Vision and Pattern RecognitionGustavo Teodoro Laureano, ... Clarimar José Coelho
Chapter
ACQUISITION
2007, Point-Based GraphicsRusinkiewicz Szymon, ... Hanspeter Pfister
Calibration Patterns
OpenCV's detection of the calibration pattern inside the images is done in two steps. First, approximative guesses of the chessboard corners are computed. Then, the corners are detected at subpixel accuracy from the guesses. The second step works well, but the first step fails under difficult light or viewing conditions.
To ease the pattern detection, identical color marks are attached onto the four corners of the chessboard pattern and an additional one for defining the origin. It has been chosen to use colors instead of spatial marks because of the invariance of color with respect to projective transforms. A detailed description of the detection procedure and the subsequent calibration steps is given in Sadlo et al. [SWPG05].
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Book2007, Point-Based GraphicsRusinkiewicz Szymon, ... Hanspeter Pfister
Chapter
Evoked Potentials
2005, Bioelectrical Signal Processing in Cardiac and Neurological ApplicationsLeif Sörnmo, Pablo Laguna
Recording setup.
The recording of a VEP is often based on a pattern reversal stimulus, generated by a chessboard pattern being displayed on a video screen. During the investigation, the patient is required to focus on a point in the center of the screen while the black-and-white squares are reversed at a fixed repetition rate so that the white squares become black, and vice versa. Typically, a rate of two reversals per second is used. The size of the chessboard squares, the luminance and contrast of the squares, and the repetition rate exemplify factors which influence the waveform amplitude and latency. These factors must therefore be taken into account when interpreting the VEP; the factors can obviously be manipulated in order to infer additional information from the VEP.
The use of flash stimulus is considered when the patient is unable to either focus or maintain the level of fixation required for pattern reversal stimulation. For example, flash stimulation can be helpful when suspected vision disorders are investigated in neonates. Flash stimulus is delivered at a rate of five to seven flashes per second. Although the eyes are closed during this procedure, a sufficient amount of light will pass through the eyelids to activate the retina.
The recording electrodes are positioned at locations close to the visual cortex, and the reference electrode is placed at the vertex.
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Book2005, Bioelectrical Signal Processing in Cardiac and Neurological ApplicationsLeif Sörnmo, Pablo Laguna
Chapter
Precision distortion correction technique based on FOV model for wide-angle cameras in automotive sector
2015, Emerging Trends in Image Processing, Computer Vision and Pattern RecognitionHaijung Choi, ... Eui Sun Kang
4 Experiment and evaluation
In order to verify the result of the distortion correction algorithm, the following distortion correction experiment device was developed. The experiment environment was configured using notebook computers, USB cameras, wide-angle lenses, and chessboard patterns as the hardware configuration. The target image was a 2D black and white chessboard pattern with 6 horizontal rows and 11 vertical columns. One square of the chessboard was 163 pixel×163 pixel while the actual size was 45 mm 2.
As shown in Figure 5, the experimental environment was prepared to perform real-time processing for the proposed algorithm, in which a 2D chessboard was positioned at the center and a real-time camera image taken with a wide-angle lens was sent to the personal computer for distortion correction.
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Figure 5. Experiment environment for distortion correction using wide-angle lens camera.
Once the accurate distortion center (C) is set in the FOV model and the distortion is corrected, all straight lines can be restored almost completely as shown in Figure 6. This result is obtained because the distortion center was set repeatedly in the distortion correction process.
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Figure 6. Distortion correction using the FOV model.
However, additional distortion was generated as shown in Figure 7 when distortion correction was performed again assuming that the distortion center was the center point of the image where there was an error from the lens distortion center. The analysis result of the distortion center showed that when a fine error of −30 pixels in the X direction and +30 pixels in the Y direction was applied to the same image, a phenomenon representing a fine curve with a specific directivity was found. This means that more severe distortion was found in proportion to the distortion center. The reason for this fine center error while projecting it onto the 2D plane is due to a mismatch of the center point with regard to the optical axis occurred during the camera manufacturing process in the case of cameras with a number of layered lenses. Therefore, while applying the FOV model, which does not estimate the center of distortion separately, there is a problem of deterioration of distortion correction accuracy as the error of the distortion center of the lens and the center point of the image becomes larger.
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Figure 7. Result of the distortion center error (X axis: −30 pixels, Y axis: +30 pixels).
The experiment to determine the error of the measurement angle using the Zhang algorithm , which is regarded as a representative method of distortion correction, produced four measurement results as shown in Table 2. In addition, it was verified that deflection of the lens center was discovered in the X direction, to the right of the Y axis, and in upper portion of the image.
Table 2. Estimation Result of Distortion Center Using the Zhang Algorithm
| Empty Cell | Cod x | Cod y |
--- |
| First measurement | 12.1 | −1.13 |
| Second measurement | 12.03 | −1.01 |
| Third measurement | 6.96 | −1.87 |
| Fourth measurement | 10.43 | −0.78 |
Accordingly, this chapter solved the problems that straight lines were expressed as curves due to no estimation of the distortion center in the FOV model and deflected representation by means of precise correction of the distortion center. Figures 8 and 9 show the correction result after the distortion center in the FOV model was estimated in the horizontal and vertical directions.
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Figure 8. Estimation of the distortion center in the (a) vertical and (b) horizontal directions using the FOV model.
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Figure 9. Correction result using the distortion center estimation method.
In the case of incorrect distortion correction due to the displaced distortion center, the image was corrected only in either the vertical or horizontal direction as shown in Figure 8(a) and (b), respectively. This phenomenon occurred because of the incorrect designation of the distortion center axis for the distortion correction. To minimize this error, the precise center axis can be found using the distortion center estimation method by means of 2D patterns such as a chessboard pattern. This resulted in minimizing the distortion correction error that might occur owing to instrument error while configuring the experiment or camera mounting (Table 3).
Table 3. Result of the Proposed Distortion Estimation Method
| Empty Cell | Cod x | Cod y |
--- |
| First measurement | 10 | −11 |
Figure 10 shows the distortion correction results as image data were received in real time from the 190° wide-angle camera. As shown in Figure 11(a), distorted correction can be found in the distorted monitor image in the horizontal and vertical directions. Figure 11(b) shows the distorted result image when the original image was corrected while (c) and (d) show the full unwrapped images of (b). If images collected from a wide-angle lens are corrected in a distorted manner, they are transformed into images of infinite size so that cropping at an appropriate scale should be used to produce images for monitoring and composition.
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Figure 10. Result of the distortion center correction using 2D planar patterns.
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Figure 11. Experiment result using actual image. (a) Distorted image, (b) distorted correction result, (c) distorted correction result, (d) distorted correction result.
As shown in Figure 12, there was a significant difference in the image interface matching before and after the distortion center estimation application. The distortion center estimation can be more influential on the composition of a number of camera inputs than on a single camera input.
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Figure 12. Comparison between image compositions before and after distortion center estimation. (a) Image composition prior to application of the distortion center estimation value, (b) image composition after application of the distortion center estimation value.
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Book2015, Emerging Trends in Image Processing, Computer Vision and Pattern RecognitionHaijung Choi, ... Eui Sun Kang
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ACQUISITION
2007, Point-Based GraphicsRusinkiewicz Szymon, ... Hanspeter Pfister
Turntable Axis Calibration
Calibration of turntables is approached in different ways, such as using markers permanently attached to the turntable [SPMS04], or by fitting an axis to rotated reconstructions. A color-marked chessboard pattern is put horizontally on the turntable and a full rotation is done in a given number of steps, typically 12. For each step, the position and orientation of the camera relative to the pattern is computed by extrinsic calibration. Then a circle is fitted to the resulting ring of virtual camera positions. The rotational axis of the circle represents the axis of the turntable.
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Book2007, Point-Based GraphicsRusinkiewicz Szymon, ... Hanspeter Pfister
Chapter
Precision distortion correction technique based on FOV model for wide-angle cameras in automotive sector
2015, Emerging Trends in Image Processing, Computer Vision and Pattern RecognitionHaijung Choi, ... Eui Sun Kang
3.4 Distortion Center Estimation Method Using 2D Patterns
The distortion center estimation method using 2D patterns is proposed to overcome the limitation of the distortion correction method using the distortion coefficient ω of the FOV model. In this chapter, a chessboard pattern was shot to estimate the center of distortion followed by projecting it onto an actual straight line in the image data of corrected distortion, thereby estimating the center of distortion using the value of the distance difference between the straight lines.
In order to estimate the center of distortion, first the distortion in the 2D patterns was corrected using the distortion coefficient of the FOV model. Then, a certain range surrounding the center of distortion, which was determined while applying the FOV model, was set to the detection window of the center of distortion.
Next, straight lines are generated by following the detected points using the method described in Ref. in the detection window of the center of distortion. Figure 3 shows the straight lines used to compute the distortion error and detection window in which the distortion center is expected. In the chessboard pattern, where a number of crossing points of straight lines can be found; crossing points of M×N representing every corner point are found, and then vertical M straight lines and horizontal N straight lines using the main outer points are generated based on the outermost points, thereby determining whether crossing points are present in the center within the straight lines. The smaller the vertical and horizontal distortions are, the closer the point is to the center of distortion, which also means it is closer to a straight line. Thus, a center of distortion can be estimated by computing the error between the straight line and the distortion. To estimate the center of distortion, distances in the row and column directions are added, and an equation for the straight line that is closest to the center of distortion and C x,C y is solved. Therefore, the distances to the distortion points p ij,c x,c y, where there are error points with the straight line L i,c x,c y are computed.
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Figure 3. Example of the detection window of the distortion center and straight-line components.
Figure 4 shows an example of errors between the distortion and a straight line, which represents distances d n between the blue-colored straight line and the distortion in the 2D chessboard pattern. The points detected in the figure are used to determine the degree of distortion. That is, the distances between the crossing points and the straight line are computed in the vertical and horizontal directions, thereby summing the distances of the inner points (d 1+⋯+d n) to calculate the distortion error.
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Figure 4. Distortion error.
(8)E ij c x,c y=||L i,c x,c y−p ij,c x,c y||2 E ji c x,c y=||L j,c x,c y−p ji,c x,c y||2
(9)arg min c x c y∑i=1 n∑j=1 m E ij c x,c y+∑j=1 m∑i=1 n E ji c x,c y
In other words, it finds the minimum sum-of-squares difference between the actual straight-line component and the projected line component. By using this function for distance computation, the minimum distance to the straight-line component of row (m) and column (n) is calculated to estimate the distortion center (c x, c y).
The distortion center estimation method using 2D patterns performs precise distortion correction by finding the minimum distortion distance. Precise distortion correction can be achieved by applying the least distortion distance estimation method to the FOV distortion correction model using 2D patterns. An LUT is produced with regard to the distortion locations on the 2D plane using the estimated distortion center, thereby being applied to the actual image.
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Book2015, Emerging Trends in Image Processing, Computer Vision and Pattern RecognitionHaijung Choi, ... Eui Sun Kang
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ACQUISITION
2007, Point-Based GraphicsRusinkiewicz Szymon, ... Hanspeter Pfister
3.2.3 CALIBRATION
Although often underestimated, precise calibration of the structured-light system is one of the main prerequisites for a successful and accurate reconstruction. This is especially true for multiview reconstructions, where a given surface point is reconstructed several times and the corresponding reconstructions have to match. In order to increase accuracy and to ease calibration, separate intrinsic calibration of camera and projector is performed and then an extrinsic calibration step is applied.
For intrinsic and extrinsic calibration of the camera and projector, the Zhang calibration method [Zha00] is used as implemented by OpenCV's cvCalibrateCamera. This calibration model consists of focal length with respect to pixel widths and heights, the principal point, and a radial and a tangential lens distortion modeled by two parameters each. As proposed by OpenCV, a chessboard pattern is used for calibration. However, the pattern detection algorithm of OpenCV fails under difficult lighting or oblique viewing conditions. This conflicts with the experience that strong perspective views at oblique angles increase the accuracy of intrinsic and especially extrinsic calibration. It also conflicts with the need to detect a printed pattern and a (distorted) chessboard projection in the same view (for intrinsic projector calibration). Therefore, an alternative detection procedure based on projective geometry is applied.
Calibration Patterns
OpenCV's detection of the calibration pattern inside the images is done in two steps. First, approximative guesses of the chessboard corners are computed. Then, the corners are detected at subpixel accuracy from the guesses. The second step works well, but the first step fails under difficult light or viewing conditions.
To ease the pattern detection, identical color marks are attached onto the four corners of the chessboard pattern and an additional one for defining the origin. It has been chosen to use colors instead of spatial marks because of the invariance of color with respect to projective transforms. A detailed description of the detection procedure and the subsequent calibration steps is given in Sadlo et al. [SWPG05].
Intrinsic Camera Calibration
The camera is calibrated using different views of a printed chessboard pattern as described above. The pattern is illuminated by the projector in order to create similar lighting conditions as in the case of projector calibration. This allows the use of identical camera settings and HSV segmentation ranges for detection in all calibration steps.
Intrinsic Projector Calibration
The projector is calibrated as an inverse camera. This means that instead of taking pictures of a chessboard with known geometry and detecting the corners inside the images, a chessboard pattern with known geometry is projected to different orientations and positions of a plane and the projections are measured with the calibrated camera.
Extrinsic Calibration
The orientation of projector and camera relative to each other is determined in a similar way as in the projector calibration, but with a single camera image. This time both projector and camera are calibrated extrinsically. The reference coordinate frame is also determined in this step.
Turntable Axis Calibration
Calibration of turntables is approached in different ways, such as using markers permanently attached to the turntable [SPMS04], or by fitting an axis to rotated reconstructions. A color-marked chessboard pattern is put horizontally on the turntable and a full rotation is done in a given number of steps, typically 12. For each step, the position and orientation of the camera relative to the pattern is computed by extrinsic calibration. Then a circle is fitted to the resulting ring of virtual camera positions. The rotational axis of the circle represents the axis of the turntable.
Luminous Projector Calibration
Because the object is illuminated by the projector for texture acquisition, the irradiance from the projector has to be known at a given point in space. For each new projector, the luminous intensityI of the projector pixels is to be initially calibrated. This could be done using a calibrated reflection target, for example, Spectralon.
As our application does not require absolute physical quantities, a gray cardboard is used instead and I is scaled to map color intensities into a useful range. Two calibration modes have been implemented. I is either assumed to be identical for all pixels, or I is determined on a per-pixel basis, capturing spatial intensity variations at the cost of more noise. The irradiance E at a given surface point is then computed as follows: E = I/d 2, with d its distance to the projector's center of projection.
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Precision distortion correction technique based on FOV model for wide-angle cameras in automotive sector
2015, Emerging Trends in Image Processing, Computer Vision and Pattern RecognitionHaijung Choi, ... Eui Sun Kang
3.1 Distortion Correction Method Considering Distortion Center Estimation
The process of distortion correction considering the distortion center is shown in Figure 1. In this chapter, a 2D planar image of a chessboard pattern was used first to correct distortion of an image from a fish-eye lens quickly. Using the distorted chessboard pattern, a distortion coefficient was estimated by applying the FOV model.
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Figure 1. Distortion correction algorithm of 2D planar pattern.
The center of distortion was found using the estimated distortion coefficient and distorted curve component. Based on the center of distortion (C x, C y), a lookup table (LUT) is produced, which represents the relationship between distorted location and 2D planar location using the FOV model. Distortion can be corrected by applying the LUT produced offline to real images.
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Computer Simulations
2011, Statistical Mechanics (Third Edition)R.K. Pathria, Paul D. Beale
16.2.A Metropolis Monte Carlo algorithm
The Metropolis method can be implemented in a computer program by using a pseudorandom number generator rand() that returns pseudorandom numbers that are uniformly distributed on the open unit interval (0.0,1.0); see Appendix I for a discussion of how pseudorandom numbers are generated. First, initialize the system by choosing a starting state q 0 from the set of all microstates of the model. It is helpful if q 0 is not atypical of the states in the equilibrium ensemble. This reduces the number of steps needed for the system to equilibrate. For example, a disordered liquid-like state would not be the best starting point for a simulation of a crystalline solid.
The Metropolis algorithm is defined by the following steps:
1.
Generate a random trial state q trial that is “nearby” the current state q j of the system. “Nearby” here means that the trial state should be almost identical to the current state except for a small random change made, usually, to a single particle or spin. For example, one can create a trial state of a particle simulation by randomly moving one particle to a nearby location
(8)x i trail=x i+Δ x(rand()-0.5),
with two more calls to rand() to generate y i trail and z i trail. The trial state of a spin system usually involves a spin flip or a random rotation of a single spin.3 2.
Determine the change in the energy of the trial state compared to the previous state, namely Δ E = E(q trial) - E(q j). If Δ E ≤ 0, accept the trial state, that is, set q j+1 = q trial. If Δ E> 0, then accept the trial state with probability exp(–βΔ E). This is accomplished by using an additional call to the pseudorandom number generator. If rand() < exp(–βΔ E), then accept the trial state. If the interactions are short-ranged, the calculation of the energy change will only involve interactions with a few nearby particles or spins. If the trial state is illegal in some way, that is, it is not an allowed state in the set of all configurations, then the state should be rejected. This is equivalent to setting the energy change at + ∞. If the trial state is rejected for either reason, then set the new state of the system equal to the previous state q j+1 = q j, that is, leave the state at the old value q j, throw away the trial state, and move on.
3.
Perform steps 1 and 2 once for each particle or spin in the system. This is often done randomly to ensure detailed balance.4 Steps 1 through 3 define one Monte Carlo sweep.
4.
Repeat steps 1 through 3 for M eq Monte Carlo sweeps to let the system equilibrate. The proper choice of M eq is not obvious a priori. At the very least, all the measures A(q) studied in the simulation should no longer have any obvious monotonic drift by the end of equilibration. This does not guarantee that the system has reached equilibrium since the system could well be trapped in the vicinity of a long-lived metastable state.
5.
Repeat steps 1 through 3 for M Monte Carlo sweeps while keeping track of all the thermodynamic variables one wants to measure, namely {A(q j)}f=1 M. Use equations (16.1.1) and (16.1.2) to determine the equilibrium averages and uncertainties.
To determine averages at a different set of parameters (temperature, density, etc.), change the parameters by a small amount and repeat steps 1 through 5, including the equilibration step 4.5 Using the last configuration of the previous run as the first configuration of the next run can often reduce the equilibration time. Figure 16.1 shows a Monte Carlo calculation of the specific heat of the two-dimensional Ising model on a 128 × 128 square lattice, as compared to the exact solution presented in Section 13.4.A.
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FIGURE 16.1. Monte Carlo specific heat (x′s) of the two-dimensional Ising model on a 128 × 128 lattice, as compared to the exact solution (solid line) from Section 13.4.A; see Kaufman (1948), Ferdinand and Fisher (1967), and Beale (1996). The MC error bars are smaller than the symbols used, except near the bulk critical temperature T c(∞). Each data point represents an average using 10 5 Monte Carlo sweeps, except at the bulk critical point where 10 6 Monte Carlo sweeps were used to mitigate critical slowing down.
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ACQUISITION
2007, Point-Based GraphicsRusinkiewicz Szymon, ... Hanspeter Pfister
3.2.1 OVERVIEW
The previous section gave an overview of state-of-the-art acquisition techniques. This section describes in detail an implementation of a temporal structured-light system based on binary Gray codes. It allows robust and accurate acquisition of objects with arbitrary geometry and a wide range of materials. A benefit of this method is that geometry and texture can be acquired with the same camera, resulting in texture that is consistent with the geometry. As stated in Section 3.1.2, another benefit is the low cost compared to other systems, since video projectors are often available and because only a single camera is needed. During the whole acquisition stage, the data are represented and processed in point-based format, resulting in an unorganized point-cloud model.
The object is rotated in small known steps by a turntable and for each position the view is reconstructed using structured light. Unlike many approaches, the demanding and error-prone task of mutually registering the single reconstructions is avoided. Instead, precise calibration of the projector, the camera, and the axis of the turntable is assured. This allows them to produce consistent multiview reconstructions (rings of overlapping views). Then some methods for the removal of artifacts are applied. After that, an efficient method for merging the overlapping reconstructions into a single-layered surfel representation is applied. For the reconstruction of texture, photometric calibration is added to the already computed geometric calibration of the projector. This way, a calibrated light source is obtained that is used for the per-surfel reconstruction of either Lambertian texture or texture according to the Phong reflectance model. At the end of this section a method for analyzing the geometric accuracy of the system is described.
3.2.2 SYSTEM OVERVIEW
The acquisition system (Figure 3.9) consists of custom hardware, namely a video projector that projects the structured-light patterns, a turntable that rotates the object, a camera that takes images of the projections, and a computer that controls the hardware and does the reconstruction.
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Figure 3.9. System overview: turntable, camera, projector, and control computer (from left to right).
In our setup, the system contains an analog-input 1, 024 × 768 DLP video projector. It is beneficial if the projector allows for disabling features such as automatic synchronization and image size adaption. This way the calibration is not lost when the projector is turned off between scans. Another aspect is the projector's minimal image diagonal, or in other words, the minimal focal distance. This becomes important if small objects have to be acquired. Currently, an IEEE-1394 video camera with a resolution of uncompressed 1, 024 × 768 pixels is used to acquire the images. It is preferred to use monochrome cameras and to illuminate the object (or calibration pattern) with red, green, and blue projector light in order to acquire colors. The reason is that cameras using a Bayer pattern for color acquisition (each pixel has a red, green, or blue filter attached) exhibit increased blur and artifacts due to undersampling and interpolation. This complicates calibration, smoothes the reconstruction, and can produce artifacts comparable to those in Figure 3.14 shown later. For similar reasons, it is also avoided to rectify the images and instead the lens distortion is modeled.
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Figure 3.14. Removal of shadows (seam artifacts). (a) No removal. (b) Local range = 1 and minimal distance = 2 mm (18,452 surfels removed).
Often several rings of reconstructions are needed in order to get a complete reconstruction. This is accomplished by either tilting the object, or by moving projector and camera to a new position between the scans. The optimal solution would be to perform these transformations again around a calibrated axis.
But since this tends to be mechanically demanding, the object is tilted by hand and each time a ring of reconstructions is acquired by rotation of the turntable. Therefore, the resulting rings have to be mutually registered, treating each ring as a separate rigid model. Some methods for mutual registration of reconstructions are geometry based, such as iterated closest points (ICP) described in Section 3.1.4, and others are image based, such as that by Bernardini et al. [BMR01]. The current system uses ICP for registration. Some challenges of registration, such as error accumulation, play a minor role in this case because the rings are already consistent at high accuracy, leading to a simpler registration problem. After the rings have been merged to a single-layered surfel representation, reflectance samples for each surfel are collected into lumitexels [LGK+01] and used for texture reconstruction according to the Phong reflectance model, which is then rendered and edited using Pointshop3D [ZPKG02] as described in Section 5.2.
3.2.3 CALIBRATION
Although often underestimated, precise calibration of the structured-light system is one of the main prerequisites for a successful and accurate reconstruction. This is especially true for multiview reconstructions, where a given surface point is reconstructed several times and the corresponding reconstructions have to match. In order to increase accuracy and to ease calibration, separate intrinsic calibration of camera and projector is performed and then an extrinsic calibration step is applied.
For intrinsic and extrinsic calibration of the camera and projector, the Zhang calibration method [Zha00] is used as implemented by OpenCV's cvCalibrateCamera. This calibration model consists of focal length with respect to pixel widths and heights, the principal point, and a radial and a tangential lens distortion modeled by two parameters each. As proposed by OpenCV, a chessboard pattern is used for calibration. However, the pattern detection algorithm of OpenCV fails under difficult lighting or oblique viewing conditions. This conflicts with the experience that strong perspective views at oblique angles increase the accuracy of intrinsic and especially extrinsic calibration. It also conflicts with the need to detect a printed pattern and a (distorted) chessboard projection in the same view (for intrinsic projector calibration). Therefore, an alternative detection procedure based on projective geometry is applied.
Calibration Patterns
OpenCV's detection of the calibration pattern inside the images is done in two steps. First, approximative guesses of the chessboard corners are computed. Then, the corners are detected at subpixel accuracy from the guesses. The second step works well, but the first step fails under difficult light or viewing conditions.
To ease the pattern detection, identical color marks are attached onto the four corners of the chessboard pattern and an additional one for defining the origin. It has been chosen to use colors instead of spatial marks because of the invariance of color with respect to projective transforms. A detailed description of the detection procedure and the subsequent calibration steps is given in Sadlo et al. [SWPG05].
Intrinsic Camera Calibration
The camera is calibrated using different views of a printed chessboard pattern as described above. The pattern is illuminated by the projector in order to create similar lighting conditions as in the case of projector calibration. This allows the use of identical camera settings and HSV segmentation ranges for detection in all calibration steps.
Intrinsic Projector Calibration
The projector is calibrated as an inverse camera. This means that instead of taking pictures of a chessboard with known geometry and detecting the corners inside the images, a chessboard pattern with known geometry is projected to different orientations and positions of a plane and the projections are measured with the calibrated camera.
Extrinsic Calibration
The orientation of projector and camera relative to each other is determined in a similar way as in the projector calibration, but with a single camera image. This time both projector and camera are calibrated extrinsically. The reference coordinate frame is also determined in this step.
Turntable Axis Calibration
Calibration of turntables is approached in different ways, such as using markers permanently attached to the turntable [SPMS04], or by fitting an axis to rotated reconstructions. A color-marked chessboard pattern is put horizontally on the turntable and a full rotation is done in a given number of steps, typically 12. For each step, the position and orientation of the camera relative to the pattern is computed by extrinsic calibration. Then a circle is fitted to the resulting ring of virtual camera positions. The rotational axis of the circle represents the axis of the turntable.
Luminous Projector Calibration
Because the object is illuminated by the projector for texture acquisition, the irradiance from the projector has to be known at a given point in space. For each new projector, the luminous intensityI of the projector pixels is to be initially calibrated. This could be done using a calibrated reflection target, for example, Spectralon.
As our application does not require absolute physical quantities, a gray cardboard is used instead and I is scaled to map color intensities into a useful range. Two calibration modes have been implemented. I is either assumed to be identical for all pixels, or I is determined on a per-pixel basis, capturing spatial intensity variations at the cost of more noise. The irradiance E at a given surface point is then computed as follows: E = I/d 2, with d its distance to the projector's center of projection.
3.2.4 GEOMETRY RECONSTRUCTION
Geometry is reconstructed using structured light according to the Gray code and phase-shifting method described below. Normals are computed from the weighted positions of neighboring samples by plane fitting. They are computed separately for each view before merging the reconstructions. This avoids influence of registration errors.
Gray Code and Phase Shifting
Structured-light methods make use of a projection device to determine the z-depth of every illuminated camera pixel. This is done by optical triangulation of the camera ray with the corresponding projector ray that illuminated the surface element. There are many possibilities to structure light in order to allow the identification of a projector pixel by its light. It has been chosen to do time-multiplexing of gray level codes. This allows a wide range of materials but limits the system to static objects.
In the standard Gray code algorithm (see Section 3.1.2), the ray defined by a camera pixel is intersected with the plane defined by the corresponding projector column. However, this assumes no lens distortion inside the projector because otherwise the plane would be distorted. To make the reconstruction process more robust against lens distortion and decoding errors, both projector columns and rows are encoded. The plane-ray intersection problem becomes an overdetermined ray-ray intersection problem that also allows for the removal of artifacts by the ray skew criterion as described below.
Since the projected Gray codes are binary, the achievable precision is limited to integer projector coordinates. Therefore, the detected pixel relations are subsequently improved using a variant of the phase-shifting technique (Section 3.1.2) based on a grayscale sine pattern as shown in Figure 3.10. This also reduces decoding errors in the least significant bits of the Gray code. In addition, phase-shift reconstruction even allows to determine the projector coordinates at subpixel accuracy. It has also been experimented with line-shifting [GBBF00], which achieves subpixel accuracy in camera coordinates rather than in the projector domain. However, in our setup it generally produced inferior results.
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Figure 3.10. (a) Gray-coded structured light. The temporal signal (blue box) at a given object point represents the column of the corresponding projector pixel. The procedure is repeated with horizontal stripes to also encode the projector rows. (b) Phase shifting. Shifted sinusoidal stripe patterns are projected to the object, producing a temporal sine signal on a given object point. The phase of this signal represents the projector column. The procedure is repeated with horizontal stripes to also encode the projector rows.
Due to errors in calibration and decoding, the two corresponding rays usually do not intersect. The method presented by Guehring [Gue01] addresses the problem by nonlinear least squares and analysis of the residual, while Hartley and Sturm [HS94] give an overview and introduce a polynomial method. In our system the point of intersection is computed as the point on the camera ray that is closest to the projector ray. The solution is constrained to the camera ray because its calibration can be assumed more accurate than that of the projector rays. This way the projector only contributes depth information, which meets the original intention. The distance between the intersecting rays (ray skew) is used for the removal of artifacts as described below. The ray skew is also visualized by color coding the reconstructions for visual verification of the quality of calibration. Figure 3.11 shows an example.
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Figure 3.11. Single-ring reconstruction colorized by the ray skew for verification of calibration quality. Red means small skew; surfels with ray skew larger than 0.6 mm (outliers) have already been removed.
Artifact Removal
Here some of the implemented methods for the elimination of geometric artifacts are described. They are all applied to the single-view reconstructions before they are merged.
Signal Strength
A black reference image with all projector pixels set to black and a white reference image with all projector pixels set to white are taken for each view. They are used for elimination of camera pixels that receive no or too weak signals and they are also used for normalization of the structured-light signal. A camera pixel is eliminated if the white reference differs from the black reference by less than a user-defined threshold. The threshold is chosen in order to reject mainly background and part of the shadows.
Ray Skew
This method detects artifacts caused by decoding errors as well as artifacts that are produced by reflected or scattered codes. Assuming accurate calibration, it is unlikely that the projector ray corresponding to the falsely decoded code intersects with the camera ray as well as the correct projector ray would do. In other words, the ray skew tends to increase on decoding errors. A threshold for the minimal distance between intersecting rays is used and the reconstruction is rejected if the threshold is exceeded (Figure 3.12).
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Figure 3.12. Removal of artifacts based on ray skew and component size. (a) Ray skew and component size are unlimited (2,527,362 surfels). (b) Ray skew is limited to 0.6 mm and connected component size to six surfels (2,053,543 surfels). Only the last few outliers were eliminated by component size.
Subpixel Variance
This method addresses artifacts that originate during phase shifting from object regions with varying reflectance and curved (or discontinuous) surface as described by Curless and Levoy [CL95]. The phase-shift signal is spatially integrated over the area of each camera pixel during acquisition. Assuming that the left half of the camera pixel looks, say, at black material, while the right half observes a bright material, the signal integrated over the pixel will contain only codes from the right half, leading to wrong projector rays usually visible as depth errors. As overlapping views have been acquired, it is possible to address this problem by eliminating surfels that lie on edges in image space. The pixel corresponding to the surfel is resampled at subpixel resolution using Lanczos interpolation, and its variance is computed. It is rejected if its variance exceeds a threshold. The method also removes sharp shadow boundaries.
Outliers are also removed based on other geometric criteria, such as small connected component size of surfels, and photometric criteria, such as saturated pixels.
View Merging
After the rings of reconstructions have been registered, we want to merge the overlapping reconstructions into a single-layered surfel representation in order to reduce storage and visualization cost. Doing so, care has to be taken that texture quality and geometric accuracy remain as high as possible.
Blending of the overlapping textures and averaging the overlapping surfel positions would require that the reconstructions fit together at subpixel accuracy, otherwise the resulting reconstruction would get blurred or doubled. Therefore, combining original patches of the overlapping reconstructions, as described later in this section, is tried to preserve detail. This method relates to Turk and Levoy [TL94] and it also has a thinning effect because the patches have original resolution.
Bounded Projective Nearest Neighbors
The system does a nearest-neighbor search to get the overlapping surfels of a given surface element of the object. Instead of performing a standard nearest-neighbor search, it is more efficient to perform the search in the M camera images. Because every point is reconstructed by a corresponding camera pixel, one can make use of the calibrated setup. To find for a given point p and a given search radius r the nearest (or all) point reconstructions, p is projected to all camera images, yielding a p′i for each view i. Then all point reconstructions that have been reconstructed by camera pixels within a search radius i′i around p′i in image space are tested for whether they lie within distance r to p in world space. The search radius i′i is computed from r by projection.
The complexity of the search is O(M·r'2), where M is the number of camera images (r'2 is limited by the number of pixels per image). In the current application, the search radius is small and constant. Therefore r' has a small upper bound, resulting in an algorithmic complexity of O(M).
Best Surfels
A simple greedy approach is used to select patches from overlapping reconstructions in order to get a single layer of points and for each object region the best patch regarding the quality of geometry and texture. The current approach is purely surfel based, hence it does not maintain a volumetric representation such as by Pulli et al. [PDH+97].
The algorithm defines and selects the patches implicitly in two steps. First, the homologous surfels representing the same point on the surface are determined using the nearest-neighbor search just presented. Then the best of these candidates is chosen for the resulting reconstruction.
The method is based on a simple idea: for each object point, take the surfel that has been reconstructed most orthogonally by its camera (regarding the surface normal computed from its neighbors). This addresses the fact that texture and geometry usually have the highest resolution when acquired by perpendicular view. At the same time, this simple criterion generates the patches. Figure 3.13 shows an example result. There are holes due to the greedy nature of the algorithm. However, the holes are small enough to disappear when the surfels are rendered.
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Figure 3.13. Best surfels. (a) Before merging (2,053,543 surfels). (b) After merging (226,664 surfels). Search radius in world space = 1.2 mm. Search radius in image space = 1 pixel in order to remove outliers.
A modification is applied to the described algorithm in order to remove outliers (additionally to the methods described above). This is achieved by modification of the nearest-neighbor search. Instead of computing the search range in image space r′i using projection, it is set by the user to a smaller range. This way, the region where best surfel candidates are sampled is not spherical any more (as would result from a true 3D distance test); instead, the search range is restricted in directions perpendicular to the views. This eliminates outliers because it allows us to choose a larger search radius that captures the outliers without thinning out too many neighboring surfels. Figure 3.13 shows an example result.
3.2.5 TEXTURE RECONSTRUCTION
Most materials exhibit a certain amount of specular reflection. In the case of strong specular reflection, the effects are confined to a small region on the object for a given view and illumination by the projector. Since many overlapping reconstructions are acquired, one could remove the specularly reflecting parts in all of them before view merging. Theoretically this would lead to consistent diffuse reconstructions of specular objects. However, the specular information would be lost. Additionally, many materials have moderate specular reflection that leads to specular effects that cover large parts of the object. There it would not be possible to remove them without producing holes. Consequently, a specular reflectance model has to be fitted in these cases.
The diffuse reflectance model can however be used to generate high-resolution reconstructions of objects that do not possess too strong of a specular reflection. The surfel reconstruction preserves the subpixel information discretized by the camera if the surfels are located on the viewing rays of the camera pixels and if each surfel gets its uniform color only from the corresponding pixel. This can be achieved for the diffuse reflectance model, because it can be robustly fitted to a single reflectance sample.
Shadow Removal
The acquisition system can generally not acquire shadowed object regions, because the structured light does not reach these parts. However, the reconstruction still may yield points in shadows if they are indirectly illuminated by interreflection, subsurface scattering, and other effects. It is important to detect and remove shadows when computing texture. Furthermore, structured-light reconstruction of indirectly illuminated regions and at shadow boundaries leads to geometric artifacts (see Figure 3.14a). These artifacts include texture embossing (see Section 3.1.2) and additional depth error from indirect illumination. The texture embossing phenomena can be reduced by avoiding Bayer tiling in the camera and by the above subpixel variance method.
The shadows are detected using a depth test relative to the projector. For each position of the turntable a reconstruction with corresponding viewpoint and projector position is obtained. After view merging, the remaining surfels are projected into each virtual view of the projector. Projector pixels that get more than one surfel projection are tested for shadow. If the surfel producing the projection belongs to the same depth map as the virtual view and if it is not nearest to the camera, it is removed. In order to preserve surfels at silhouettes, the surfel is only removed if it is farther away from the nearest surfel than a user-defined small threshold.
The generated holes are filled with other surfels by repeating the view-merging procedure.
Although some shadow surfels are eliminated by the described method, there are usually shadow surfels remaining due to calibration errors, reconstruction errors, and the fact that the sampled surface elements have an extent. Therefore, instead of projecting each surfel to the projector and storing it only in the corresponding projector pixel, it is stored to a user-defined range around that projector pixel. This way the sensitivity for depth discontinuities is increased, and at the same time, surfels that have been reconstructed at grazing angles are removed.
Reflectance Sampling
In order to reconstruct the texture, it is necessary to collect for each surface point all available and reliable reflectance samples from the acquired texture views. Reflectance is computed from the image intensities using the calibrated illumination model described in Section 3.2.3. According to Lensch et al. [LGK+01], a surface point together with the corresponding samples is called a lumitexel. Not all views contribute a sample for a given surface point. There might be effects like occlusion and insufficient illumination or shadows that invalidate a sample in a given view. Insufficient illumination is detected using a user-defined threshold that rejects samples illuminated at grazing angles, in addition to the signal strength threshold described above.
The shadow and occlusion tests are based on the surfel representation. Only surfels that belong to the final merged reconstruction represent surface points and hence lumitexels. Therefore, the points that are removed by the merging procedure are only marked as such, since they are needed for sample selection.
Occlusion/Shadow Test
The structured-light method already performs a kind of implicit occlusion test in the sense that it can only acquire surface points that are visible by the camera. The same holds for the light source, so theoretically no shadows can be acquired. Therefore, it can be decided if a point p gets occluded in a given view v i by projecting p to that view (onto the camera pixel p′i) and by examining the surfel s p'i that has been reconstructed by the pixel p′i of view v i. Possible cases are:
•
There has been no s p'i reconstructed. This means that reconstruction failed for that pixel or that it has been removed by one of the presented methods. Another possibility is that it has been removed by the shadow removal process. In any case, the corresponding sample is rejected.
•
s p'i has been reconstructed (and either selected as best surfel or not). This means either that v i has an unobstructed view to the point p or that s p'i represents a surface point different to p. To test if s p'i represents a different point, the distance between p and pvi (pvi is the camera position of v i) is measured as well as the distance between s p'i. and pvi. If the difference |||p−p v i||−||s p′i−p v i||| exceeds a user-defined threshold, the points are assumed to be distinct and the sample is rejected due to occlusion.
Lambertian Texture
Diffuse texture is computed from a single reflectance sample according to the Lambertian reflectance model:
(3.13)ρ d=L n T l E,
with, for a given object point, ρ d the diffuse albedo, L the radiance observed by the camera, E the irradiance from the projector at that point, n its normalized surface normal, and l the normalized vector toward the light source. The surface normal is actually computed from the neighbors of the reconstructed point. The direction l is computed from the geometric projector calibration and E from the luminous projector calibration. Figures 3.15 and 3.16a show respective results.
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Figure 3.15. Single view reconstruction, Lambertian texture.
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Figure 3.16. Reconstructed Lambertian texture (a), Phong texture (b), blended overlapping reconstructions (c), region of original camera view (d), diffuse component of Phong texture (e), and specular component (f).
Phong Texture
Alternatively, the diffuse term is used together with a specular Phong lobe as the reflectance model:
(3.14)L E=ρ d n T l+ρ s(l T(2 nn T−I)v)n,
with ρ s the maximum specular albedo, v the normalized vector toward the camera, and n the specular exponent. Nonlinear least squares fitting of Equation (3.14) to the reflectance samples is done simultaneously for red, green, and blue ρ d and ρ s, but with a single exponent n using the Levenberg-Marquardt method as suggested by Lafortune et al. [LFTG97]. An initial estimate is computed by linear fitting. This is achieved by fixing n for optimization and by solving the resulting linear least squares problem. This is repeated for exponentially increasing n and each time the residual is computed. The fit leading to the smallest error is chosen as a result of the linearized fitting. Linear fitting can also be used as fall back to nonlinear fitting.
Figure 3.16 shows a result of the Phong fit. The reconstruction still shows some artifacts in regions of sparse sampling and at points with erroneous normals due to outliers. However, the diffuse part of the specular fit provides a significantly improved texture compared to the Lambertian fit, as the Lambertian fit overestimates brightness in specular regions.
Further improving the Phong fit is difficult due to the limited number of samples per lumitexel. One way to come around this would be to apply material clustering as presented by Lensch et al. [LGK+01]. Another possibility is to improve the accuracy of the normals by including them into the texture-fitting process, but this also requires a well-distributed reflectance sampling.
3.2.6 RESULTS
Figure 3.17 shows a single-ring reconstruction of a clay pot acquired from only 15 views. Figure 3.18 shows a telephone also reconstructed from 15 rotated views using the same setup. It can be seen that the clay pot produces better quality regarding texture and geometry because its material is mostly diffuse, in contrast to plastic material of the telephone.
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Figure 3.17. Clay pot reconstructed from 15 rotary views (384,492 surfels). Lambertian texture (a) and synthetic Phong texture (b) for showing geometric detail. The Lambertian texture exhibits stripe artifacts due to specular reflection of the clay surface.
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Figure 3.18. Telephone reconstructed from 15 rotary views (317,681 surfels). Lambertian texture (a) and no texture (b).
Because the chosen objects contain both diffuse and specular materials and relatively much occlusion, the achieved results still contain artifacts such as outliers, false normals, and holes. These are addressed by the postprocessing stage as described by Weyrich et al. [WPK+04], although in our case, most of the holes originate from suboptimal setup and could be addressed by interactive control of the turntable and more appropriate tilting of the object in order to generate optimal views.
Accuracy
The accuracy of the system is evaluated using a steel sphere of 150 mm in diameter that had been manufactured with a tolerance of 0.1 mm and painted in white color. A single ring of 10 partial reconstructions at a uniform angular step was taken and merged according to Section 3.2.4. For the merged reconstruction, a center was computed by fitting a least squares sphere to the surfels. The mean distance to the center was 74.947 mm and the maximal errors were less than 0.5 mm, which is roughly the surfel spacing (red histogram in Figure 3.19a). For further analysis of these errors, a sphere of the same radius was fitted to each partial reconstruction. The resulting centers (Figure 3.19b) reveal a small systematic error in the turntable calibration, plus a mechanical effect happening between the first two scans. The blue histograms show the higher accuracy of each partial reconstruction. The green histogram is obtained by translating the partial reconstructions to a common center. This step cannot be done for general objects, it just illustrates that there is some potential left in the merging process. However, registration of surfel objects based on geometry or texture to extreme subsurfel precision is difficult.
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Figure 3.19. (a) Histograms of surfel distances to sphere center (red: full reconstruction; blue: partial reconstruction; green: partial reconstructions merged by optimal translations). (b) Centers of least squares fits for partial reconstructions. All units given in millimeters.
Performance
The angel figurine of 180 mm height was reconstructed from 3 × 30 views. Each view contributed in average 28,082 surfels, leading to a raw reconstruction consisting of 2,527,362 surfels. From these, 473,819 erroneous surfels were removed by the ray skew criterion or because they formed too-small connected components (outliers). Another 1,826,879 surfels were discarded during merging. Finally, 18,452 surfels got discarded by the shadow removal process. Removing the shadows is the computationally most expensive step—it took 38 minutes. Texture reconstruction took 16 minutes and computation of the normals another 6 minutes. The best surfels merging procedure took again 6 minutes and the cost of the remaining operations is negligible.
3.2.7 CONCLUSION
This section described a completely point-based acquisition system for geometry and texture. The system consists of a projector as a calibrated light source, a single grayscale camera, and a turntable. Methods for robust reconstruction and removal of artifacts have been presented as well as methods for merging multiple point-based reconstructions into a single-layered representation. Additionally, to the Lambertian reflectance model, the Phong model was fitted to the reflectance samples, utilizing the underlying structured-light approach.
The described system is oriented for low cost, simple operation, and arbitrary object geometry, and can handle cavities due to the structured-light approach. The next section presents a high-end system based on the visual hull method, aiming at the acquisition of objects with difficult surface properties, such as fuzzy and highly specular objects.
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Book2007, Point-Based GraphicsRusinkiewicz Szymon, ... Hanspeter Pfister
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Horizontal Direction
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Salt Per Water Brine Calculator
This brine table calculator will help you work out the ratios of salt and water.
Use the table to select your preferred units of measurement.
Sodium Chloride (Salt) Brine Tables for Brine at 60 F (15° C) in US Gallons At 100 degrees brine is fully saturated and contains 26.395 % of salt.
1 US gallon of water weighs 8.33 lbs, 1 US gallon = 3.8 liters = 3.8 kilograms, 1 pound =.453kg, 1kg = 2.2 lbs.
Salometer
Degrees Pounds of Salt per Gallon of
Water Percent of Sodium Chloride (Salt) by
Weight Pounds of Salt per Gallon of
Brine Pounds of Water per Gallon of
Brine
0 0.000 0.000 0.000 8.328
1 0.022 0.264 0.022 8.323
2 0.044 0.526 0.044 8.317
3 0.066 0.792 0.066 8.307
4 0.089 1.056 0.089 8.298
5 0.111 1.320 0.111 8.292
6 0.134 1.584 0.133 8.286
7 0.157 1.848 0.156 8.280
8 0.180 2.112 0.178 8.274
9 0.203 2.376 0.201 8.268
10 0.226 2.640 0.224 8.262
11
Washed Rinds 0.249 2.903 0.247 8.256
12 0.272 3.167 0.270 8.250
13 0.296 3.431 0.293 8.239
14 0.320 3.695 0.316 8.229
15 0.343 3.959 0.339 8.222
16 0.367 4.223 0.362 8.216
17 0.391 4.487 0.386 8.209
18 0.415 4.751 0.409 8.202
19 0.440 5.015 0.433 8.195
20 0.464 5.279 0.456 8.188
21 0.489 5.543 0.480 8.181
22 0.513 5.807 0.504 8.174
23 0.538 6.071 0.528 8.167
24 0.563 6.335 0.552 8.159
25 0.588 6.599 0.576 8.152
26 0.614 6.863 0.600 8.144
27 0.639 7.127 0.624 8.137
28 0.665 7.391 0.649 8.129
29 0.690 7.655 0.673 8.121
30 0.716 7.919 0.698 8.113
31 0.742 8.162 0.722 8.105
32 0.768 8.446 0.747 8.097
33 0.795 8.710 0.772 8.089
34 0.821 8.974 0.797 8.081
35 0.848 9.238 0.822 8.073
36 0.874 9.502 0.847 8.064
37 0.901 9.766 0.872 8.056
38 0.928 10.030 0.897 8.047
39 0.956 10.294 0.922 8.038
40 0.983 10.558 0.948 8.030
41 1.011 10.822 0.973 8.021
42 1.038 11.086 0.999 8.012
43 1.066 11.350 1.025 8.003
44 1.094 11.614 1.050 7.994
45 1.123 11.878 1.076 7.985
46 1.151 12.142 1.102 7.975
47 1.179 12.406 1.128 7.966
48 1.208 12.670 1.154 7.957
49 1.237 12.934 1.181 7.947
50 1.266 13.198 1.207 7.937
51 1.295 13.461 1.233 7.928
52 1.325 13.725 1.260 7.918
53 1.355 13.989 1.286 7.908
54 1.384 14.253 1.313 7.898
55 1.414 14.517 1.340 7.888
56 1.444 14.781 1.368 7.878
57 1.475 15.045 1.393 7.867
58 1.505 15.309 1.420 7.857
Salometer
Degrees Pounds of Salt per Gallon of
Water Percent of Sodium Chloride (Salt) by
Weight Pounds of Salt per Gallon of
Brine Pounds of Water per Gallon of
Brine
59 1.536 15.573 1.447 7.847
60 1.567 15.837 1.475 7.836
61 1.598 16.101 1.502 7.826
62 1.630 16.365 1.529 7.815
63 1.661 16.629 1.557 7.804
64 1.693 16.893 1.584 7.793
65 1.725 17.157 1.612 7.782
66 1.757 17.421 1.639 7.771
67 1.789 17.685 1.668 7.764
68 1.822 17.949 1.697 7.756
69 1.854 18.213 1.725 7.744
70 1.887 18.477 1.753 7.733
71 1.921 18.740 1.781 7.721
72 1.954 19.004 1.809 7.710
73 1.988 19.268 1.837 7.698
74 2.021 19.532 1.866 7.686
75 2.056 19.796 1.895 7.678
76 2.090 20.060 1.925 7.669
77 2.124 20.324 1.953 7.657
78 2.159 20.588 1.982 7.645
79 2.194 20.852 2.011 7.633
80 2.229 21.116 2.040 7.621
81 2.265 21.380 2.069 7.608
82 2.300 21.644 2.098 7.596
83 2.336 21.908 2.128 7.586
84 2.372 22.172 2.159 7.577
85 2.409 22.436 2.188 7.584
86 2.446 22.700 2.217 7.551
87 2.482 22.964 2.248 7.542
88 2.520 23.228 2.279 7.532
89 2.557 23.492 2.309 7.519
90 2.595 23.756 2.338 7.505
91 2.633 24.019 2.368 7.492
92 2.671 24.283 2.398 7.479
93 2.709 24.547 2.430 7.468
94 2.748 24.811 2.461 7.458
95 Saturated Brine 2.787 25.075 2.491 7.444
96 2.826 25.339 2.522 7.430
97 2.866 25.603 2.552 7.416
98 2.908 25.867 2.570 7.409
99 2.948 26.131 2.616 7.394
99.6 2.970 26.289 2.634 7.385
100 2.986 26.395 2.647 7.380
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16155 | http://mathcentral.uregina.ca/qq/database/qq.09.03/keisha1.html | What is a Fact Family?
Quandaries and Queries
Name: Keisha
Who is asking: Parent
Level: Elementary
Question:
I am helping my child with his homework. The worksheet we are trying to do ask the child to choose a number from inside of a circle and then one from inside of a triangle and use the number to build a fact family. The title of the worksheet is "Building a Fact Family House". I noticed that Silver Burdett Ginn,Inc. was the publisher of the book the worksheet came from so I was on the website trying to find an answer to my question. My question is "What is a Fact Family?"
Dear Keisha,
Thank you for your question. Let me begin by giving an example. The numbers 5,7, and 12 form a fact family. Notice that:
5+7=12
7+5=12
12-5=7
12-7=5
There are several ideas that we are trying to build when we introduce students to fact families. Here are some of them:
Notice that 5+7=12 and 7+5=12, thus the first two equations
illustrate that if we add two numbers in any order, we get the same answer.
2. Notice that 12-7=5 and 7+5=12, so subtraction and addition are
related. Similarly 12-5=7 and 5+7=12.
3. Notice that 12-7=5 and 12-5=7, so if you subtract one of the
members of the fact family from 12, you get the other.
You and your child can look for more relationships together. Fact families are really about building a conceptual understanding of how addition and subtraction work.
So, informally, you can think of a fact family as consisting of three numbers, two of which add up to the third.
If you are given two numbers, there are two possible fact families you can build. For example, if we are given 5 and 7, we can build the family 5,7, and 12 as above, or the family 2,5, and 7 since
2+5=7
5+2=7
7-2=5
7-5=2
Hopefully there will be something in the directions of your worksheet that will tell you if you should find the third member of the family using addition (ie add the two numbers (5+7=12)) or using subtraction (ie subtract the larger from the smaller (7-5=2)).
I hope this helps you with the worksheet your child was given, as well as gives you an idea of what sorts of relationships you and your child can observe as you look at fact families together. They're great for illustrating "patterns" in mathematics.
Take care,
Judi
Go to Math Central |
16156 | https://scetcivil.weebly.com/uploads/5/3/9/5/5395830/fluids_chap04.pdf | ã2005 Pearson Education South Asia Pte Ltd Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and Bernoulli’s Equation 7. General Energy Equation 8. Reynolds Number, Laminar Flow, Turbulent Flow and Energy Losses Due to Friction ã2005 Pearson Education South Asia Pte Ltd Applied Fluid Mechanics 9. Velocity Profiles for Circular Sections and Flow in Noncircular Sections 10.Minor Losses 11.Series Pipeline Systems 12.Parallel Pipeline Systems 13.Pump Selection and Application 14.Open-Channel Flow 15.Flow Measurement 16.Forces Due to Fluids in Motion ã2005 Pearson Education South Asia Pte Ltd Applied Fluid Mechanics 17.Drag and Lift 18.Fans, Blowers, Compressors and the Flow of Gases 19.Flow of Air in Ducts 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Chapter Objectives 1. Compute the force exerted on a plane area by a pressurized gas.
2. Compute the force exerted by any static fluid acting on a horizontal plane area.
3. Compute the resultant force exerted on a rectangular wall by a static liquid.
4. Define the term center of pressure.
5. Compute the resultant force exerted on any submerged plane area by a static liquid.
6. Show the vector representing the resultant force on any submerged plane area in its proper location and direction.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Chapter Objectives 7. Visualize the distribution of force on a submerged curved surface.
8. Compute the total resultant force on the curved surface.
9. Compute the direction in which the resultant force acts and show its line of action on a sketch of the surface.
10.Include the effect of a pressure head over the liquid on the force on a plane or curved surface.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Chapter Outline 1.
Introductory Concepts 2.
Gases Under Pressure 3.
Horizontal Flat Surfaces Under Liquids 4.
Rectangular Walls 5.
Submerged Plane Areas-General 6.
Development of the General Procedure for Forces on Submerged Plane Areas 7.
Piezometric Head 8.
Distribution of Force on a Submerged Curved Surface 9.
Effect of a Pressure Above the Fluid Surface 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Chapter Outline 10. Forces on a Curved Surface with Fluid Below It 11. Forces on Curved Surfaces with Fluid Above and Below 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.1 Introductory Concepts • Fig 4.1 shows the examples of cases where forces on submerged areas must be computed.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.1 Introductory Concepts • In each case, the fluid exerts a force on the surface of interest that acts perpendicular to the surface, considering the basic definition of pressure, p = F/A and the corresponding form, F = pA.
• We apply these equations directly only when the pressure is uniform over the entire area of interest. • An example is when the fluid is a gas for which we consider the pressure to be equal throughout the gas because of its low specific weight.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.2 Gases Under Pressure • Fig 4.2 shows the fluid power cylinder.
• The air pressure acts on the piston face, producing a force that causes the linear movement of the rod. • The pressure also acts on the end of the cylinder, tending to pull it apart.
• Therefore, we can calculate the force on the piston and the cylinder ends directly from F = pA.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.2 Gases Under Pressure 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.1 If the cylinder in Fig. 4.2 has an internal diameter of 51 mm and operates at a pressure of 2070 kPa (gage), calculate the force on the ends of the cylinder.
( ) kN m m m N F m m D A pA F 41 .
4 002 .
0 / 10 2070 002 .
0 4 051 .
0 4 2 2 2 3 2 2 2 = ´ ´ = = = = = p p 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.1 Notice that gage pressure was used in the calculation of force instead of absolute pressure. The additional force due to atmospheric pressure acts on both sides of the area and is thus balanced. If the pressure on the outside surface is not atmospheric, then all external forces must be considered to determine a net force on the area.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.3 Horizontal Flat Surfaces Under Liquids • Fig 4.3 shows the cylindrical drum.
• The pressure in the water at the bottom of the drum is uniform across the entire area because it is a horizontal plane in a fluid at rest. • We can simply use F = pA to calculate the force on the bottom 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.2 If the drum in Fig. 4.3 is open to the atmosphere at the top, calculate the force on the bottom.
To use we must first calculate the pressure at the bottom of the drum and the area of the bottom: 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.2 Would there be any difference between the force on the bottom of the drum in Fig. 4.3 and that on the bottom of the cone-shaped container in Fig. 4.4?
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.2 The force would be the same because the pressure at the bottom is dependent only on the depth and specific weight of the fluid in the container. The total weight of fluid is not the controlling factor. Comment: The force computed in these two example problems is the force exerted by the fluid on the inside bottom of the container. Of course, when designing the support structure for the container, the total weight of the container and the fluids must be considered. For the structural design, the cone-shaped container will be lighter than the cylindrical drum.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.4 Rectangular Walls • Fig 4.5 shows the rectangular walls.
• They are walls which exposed to a pressure varying from zero on the surface of the fluid to a maximum at the bottom of the wall. • The force due to the fluid pressure tends to overturn the wall or break it at the place where it is fixed to the bottom.
• The actual force is distributed over the entire wall, but for the purpose of analysis it is desirable to determine the resultant force and the place where it acts, called the center of pressure.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.4 Rectangular Walls 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.4 Rectangular Walls • Fig 4.6 shows the vertical rectangular wall.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.4 Rectangular Walls • The total resultant force can be calculated from the equation where pavg is the average pressure and A is the total area of the wall.
• But the average pressure is that at the middle of the wall and can be calculated from the equation where h is the total depth of the fluid.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.4 Rectangular Walls • Therefore, we have • The center of pressure is at the centroid of the pressure distribution triangle, one third of the distance from the bottom of the wall. • The resultant force acts perpendicular to the wall at this point.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.4 Rectangular Walls • Below are the procedures for computing the force on a rectangular wall: 1. Calculate the magnitude of the resultant force from 2. Locate the center of pressure at a vertical distance of from the bottom of the wall.
3. Show the resultant force acting at the center of pressure perpendicular to the wall.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.4 In Fig. 4.6, the fluid is gasoline and the total depth is 3.7 m. The wall is 12.2 m long. Calculate the magnitude of the resultant force on the wall and the location of the center of pressure.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.4 Step 1 ( ) ( )( ) ( )( ) kN m m m m kN F m m m A m kN m kN A h F R R 0 .
557 14 .
45 2 7 .
3 / 67 .
6 14 .
45 2 .
12 7 .
3 / 67 .
6 / 81 .
9 68 .
0 2 / 2 3 2 2 3 3 = ´ ´ = = = = = = g g 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.4 Step 2 Step 3 The force acts perpendicular to the wall at the center of pressure as shown in Fig. 4.6.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.5 Figure 4.7 shows a dam 30.5 m long that retains 8 m of fresh water and is inclined at an angle of 60°. Calculate the magnitude of the resultant force on the dam and the location of the center of pressure.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.5 Step 1 To calculate the area of the dam we need the length of its face, called L in Fig. 4.7: Then, the area of the dam is 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.5 Step 1 Now we can calculate the resultant force: 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.5 Step 2 The center of pressure is at a vertical distance of from the bottom of the dam, or, measured from the bottom of the dam along the face of the dam, the center of pressure is at 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.5 Step 2 Measured along the face of the dam we define We show FR acting at the center of pressure perpendicular to the wall.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General • Fig 4.85 shows the force on a submerged plane area.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General • The standard dimensions and symbols used in the procedure described later are shown in the figure and defined as follows: 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General • Fig 4.9 shows the properties of a rectangle.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General • Below are the procedure for computing the force on a submerged plane area: 1. Identify the point where the angle of inclination of the area of interest intersects the level of the free surface of the fluid. This may require the extension of the angled surface or the fluid surface line. Call this point S.
2. Locate the centroid of the area from its geometry.
3. Determine hc as the vertical distance from the level of the free surface down to the centroid of the area.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General 4. Determine Lc as the inclined distance from the level of the free surface down to the centroid of the area. This is the distance from S to the centroid. Note that hc and Lc are related by 5. Calculate the total area A on which the force is to be determined.
6. Calculate the resultant force from 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General where γ is the specific weight of the fluid. This equation states that the resultant force is the product of the pressure at the centroid of the area and the total area.
7. Calculate Ic the moment of inertia of the area about its centroidal axis.
8. Calculate the location of the center of pressure from 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General Notice that the center of pressure is always below the centroid of an area that is inclined with the horizontal. In some cases it may be of interest to calculate only the difference between Lp and Lc from 9. Sketch the resultant force FR acting at the center of pressure, perpendicular to the area.
10. Show the dimension Lp on the sketch in a manner similar to that used in Fig. 4.8.
11. Draw the dimension lines for Lp and Lc from a reference line drawn through point S and perpendicular to the angle of inclination of the area.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.5 Submerged Plane Areas - General 12. If it is desired to compute the vertical depth to the center of pressure, hp either of two methods can be used. If the distance Lp has already been computed, use Alternatively, Step 8 could be avoided and hp can be computed directly from We will now use the programmed instruction approach to illustrate the application of this procedure.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.6 The tank shown in Fig. 4.8 contains a lubricating oil with a specific gravity of 0.91. A rectangular gate with the dimensions B=1.2 m and H=0.6 m is placed in the inclined wall of the tank (θ=60°).The centroid of the gate is at a depth of 1.5 m from the surface of the oil. Calculate (a) the magnitude of the resultant force FRon the gate and (b) the location of the center of pressure.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.6 The terms Lc and hc are related in this case by Therefore, we have Because the area of the rectangle is BH, We need the specific weight of the oil: 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Then we have We find that for a rectangle, Because Ic=0.02 m4, Lc=1.73 m, and A=0.72 m2, Example 4.6 kN m m m kN A h F c R 63 .
9 72 .
0 5 .
1 92 .
8 2 3 0 = ´ ´ = = g ( )( ) m m m L m m m m A L I L L p c c c p 75 .
1 016 .
0 73 .
1 72 .
0 73 .
1 02 .
0 73 .
1 2 4 = + = + = + = 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd This means that the center of pressure is 0.016 m (or 16 mm) below the centroid of the gate.
Be sure you understand how the dimension Lp is drawn from the reference line.
Example 4.6 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6 Development of the General Procedure for Forces on Submerged Plane Areas • The resultant force is defined as the summation of the forces on small elements of interest.
• Fig 4.11 shows the development of the general procedure for forces on submerged plane areas.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6 Development of the General Procedure for Forces on Submerged Plane Areas • The resultant force is defined as the summation of the forces on small elements of interest.
• Fig 4.11 shows the development of the general procedure for forces on submerged plane areas.
• On any small area dA, there exists a force dF acting perpendicular to the area owing to the fluid pressure p. But the magnitude of the pressure at any depth h in a static liquid of specific weight is p = γh.
• Then, the force is 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6 Development of the General Procedure for Forces on Submerged Plane Areas 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6 Development of the General Procedure for Forces on Submerged Plane Areas • Because the area is inclined at an angle it is convenient to work in the plane of the area, using y to denote the position on the area at any depth h. Note that where y is measured from the level of the free surface of the fluid along the angle of inclination of the area. Then, 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6 Development of the General Procedure for Forces on Submerged Plane Areas • The summation of forces over the entire area is accomplished by the mathematical process of integration, • From mechanics we learn that is equal to the product of the total area times the distance to the centroid of the area from the reference axis. That is, 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6 Development of the General Procedure for Forces on Submerged Plane Areas • Then, the resultant force is • This is the same form as Eq. (4–4). Because each of the small forces dF acted perpendicular to the area, the resultant force also acts perpendicular to the area.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6.1 Center of Pressure • The center of pressure is that point on an area where the resultant force can be assumed to act so as to have the same effect as the distributed force over the entire area due to fluid pressure.
• The moment of each small force dF is • But 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6.1 Center of Pressure • Now, if we assume that the resultant force acts at the center of pressure, its moment with respect to the axis through S is FRLp. Then • Substituting FR for from Eq. (4–10) gives 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6.1 Center of Pressure • A more convenient expression can be developed by using the transfer theorem for moment of inertia from mechanics.
• Equation (4–12) then becomes 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6.1 Center of Pressure • Rearranging gives the same form as Eq. (4–6): • We now continue the development by creating an expression for the vertical depth to the center of pressure hp. Starting from Eq. (4–13), note the following relationships: 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.6.1 Center of Pressure • Then 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.7 Piezometric Head • In all the problems demonstrated so far, the free surface of the fluid was exposed to the ambient pressure where p = 0 (gage).
• Fig 4.12 shows the illustration of piezometric head.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.7 Piezometric Head • A convenient method uses the concept of piezometric head, in which the actual pressure above the fluid, is converted into an equivalent depth of the fluid, that would create the same pressure (Fig. 4.12): • This depth is added to any depth h below the free surface to obtain an equivalent depth, That is, • The equivalent depth to the centroid is 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.7 Repeat Example Problem 4.6, except consider that the tank shown in Fig. 4.8 is sealed at its top and that there is a pressure of 10.3 kPa(gage) above the oil.
Several calculations in the solution to Example Problem 4.6 used the depth to the centroid, given to be 1.5 m below the surface of the oil. With the pressure above the oil, we must add the piezometric head from Eq. (4–14). Using ?
=8.92 kN/m3, we get m m kN m kN p h a a 15 .
1 / 92 .
8 / 3 .
10 3 2 = = = g 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.7 Then, the equivalent depth to the centroid is The resultant force is then Compare this with the value of 9.63 kN found before for the open tank. The center of pressure also changes because the distance Lc changes Lce to as follows: ( )( ) mm m m m m A L I L L m m h L ce c ce pe ce cce 9 009 .
0 72 .
0 0 .
3 02 .
0 0 .
3 60 sin / 65 .
2 sin / 2 4 = = = = -= ° = = q 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.7 Then, the equivalent depth to the centroid is The resultant force is then Compare this with the value of 9.63 kN found before for the open tank. The center of pressure also changes because the distance Lc changes Lce to as follows: 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8 Distribution of Force on a Submerged Curved Surface • Fig 4.13 shows the tank with a curved surface containing a static fluid.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8 Distribution of Force on a Submerged Curved Surface • One way to visualize the total force system involved is to isolate the volume of fluid directly above the surface of interest as a free body and show all the forces acting on it, as shown in Fig. 4.14.
• Fig 4.14 shows the free-body diagram of a volume of fluid above the curved surface.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8 Distribution of Force on a Submerged Curved Surface 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.1 Horizontal Force • The vertical solid wall at the left exerts horizontal forces on the fluid in contact with it in reaction to the forces due to the fluid pressure.
• The magnitude of F2b and its location can be found using the procedures developed for plane surfaces. That is, where hc is the depth to the centroid of the projected area.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.1 Horizontal Force • Calling the height of the rectangle s, you can see that hc = h + s/2. Also, the area is sw, where w is the width of the curved surface. Then, • Again using the principles developed earlier, we get • For the rectangular projected area, however, 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.1 Horizontal Force • Then, 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.2 Vertical Force • The vertical component of the force exerted by the curved surface on the fluid can be found by summing forces in the vertical direction. • Only the weight of the fluid acts downward, and only the vertical component acts upward.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.3 Resultant Force • The total resultant force is • The resultant force acts at an angle relative to the horizontal found from 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.4 summary of the Procedure for Computing the Force on a Submerged Curved Surface 1. Isolate the volume of fluid above the surface.
2. Compute the weight of the isolated volume.
3. The magnitude of the vertical component of the resultant force is equal to the weight of the isolated volume. It acts in line with the centroid of the isolated volume.
4. Draw a projection of the curved surface onto a vertical plane and determine its height, called s.
5. Compute the depth to the centroid of the projected area from the below, where h is the depth to the top of the projected area.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.4 summary of the Procedure for Computing the Force on a Submerged Curved Surface 6. Compute the magnitude of the horizontal component of the resultant force from 7. Compute the depth to the line of action of the horizontal component from 8. Compute the resultant force from 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.8.4 summary of the Procedure for Computing the Force on a Submerged Curved Surface 9. Compute the angle of inclination of the resultant force relative to the horizontal from 10.Show the resultant force acting on the curved surface in such a direction that its line of action passes through the center of curvature of the surface.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 For the tank shown in Fig. 4.13, the following dimensions apply: Compute the horizontal and vertical components of the resultant force on the curved surface and the resultant force itself. Show these force vectors on a sketch.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 Using the steps outlined above: 1. The volume above the curved surface is shown in Fig. 4.15.
2. The weight of the isolated volume is the product of the specific weight of the water times the volume. The volume is the product of the area times the length w.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 3. Then Fv = 153.7N, acting upward through the centroid of the volume. The location of the centroid is found using the composite-area technique. Refer to Fig. 4.15 for the data. Each value should be obvious except the location of the centroid of the quadrant. From Appendix L, Then, the location of the centroid for the composite area is 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 4. The vertical projection of the curved surface is shown in Fig. 4.15. The height s equals 1.50 m.
5. The depth to the centroid of the projected area is 6. The magnitude of the horizontal force is 7. The depth to the line of action of the horizontal component is found from 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 8. The resultant force is computed from 9. The angle of inclination of the resultant force relative to the horizontal is computed from 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 10. The horizontal component, the vertical component, and the resultant force are shown in Fig. 4.16. Note that the line of action FR of is through the center of curvature of the surface. Also note that the vertical component is acting through the centroid of the volume of liquid above the surface. The horizontal component is acting through the center of pressure of the projected area at a depth hp from the level of the free surface of the fluid.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd Example 4.8 10. The horizontal component, the vertical component, and the resultant force are shown in Fig. 4.16. Note that the line of action FR of is through the center of curvature of the surface. Also note that the vertical component is acting through the centroid of the volume of liquid above the surface. The horizontal component is acting through the center of pressure of the projected area at a depth hp from the level of the free surface of the fluid.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.9 Effect of Pressure above the Fluid Surface • If an additional pressure exists above the fluid or if the fluid itself is pressurized, the effect is to add to the actual depth a depth of fluid equivalent to p/γ.
• This is the same procedure, called piezometric head.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.10 Forces on a Curved Surface with Fluid Below It • Fig 4.17 shows the curved surface restraining a liquid below it.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.10 Forces on a Curved Surface with Fluid Below It • The pressure in the fluid at any point is dependent on the depth of fluid to that point from the level of the free surface.
• This situation is equivalent to having the curved surface supporting a volume of liquid above it, except for the direction of the force vectors.
• Fig 4.18 shows the forces exerted by a curved surface on the fluid.
• As before, the horizontal component of the force exerted by the curved surface on the fluid is the force on the projection of the curved surface on a vertical plane.
• The vertical component is equal to the weight of the imaginary volume of fluid above the surface.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.10 Forces on a Curved Surface with Fluid Below It 4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.11 Forces on Curved Surfaces with Fluid above and below • Fig 4.19 shows the semicylindrical gate.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.11 Forces on Curved Surfaces with Fluid above and below • The force due to fluid pressure would have a horizontal component acting to the right on the gate. • This force acts on the projection of the surface on a vertical plane and is computed in the same manner as used in Section 4.8.
• However, there is also a force acting upward on the bottom surface of the gate equal to the total weight of the fluid, both real and imaginary, above that surface. • The net vertical force is the difference between the two forces, equal to the weight of the semicylindrical volume of fluid displaced by the gate itself.
4. Forces due to Static Fluid ã2005 Pearson Education South Asia Pte Ltd 4.11 Forces on Curved Surfaces with Fluid above and below • Fig 4.19 shows the volumes used to compute the net vertical force on the gate. |
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16158 | https://testbook.com/maths/properties-of-complex-numbers | Properties of Complex Numbers, Properties with Proof & Examples
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HomeMaths
Properties of Complex Numbers
Properties of Complex Numbers are explained with Proof.
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Overview
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Operations of Complex Numbers are mathematical operations that we perform on complex numbers. Operations of Complex Numbers are more or less like operations on normal numbers except for division. In this article, we are going to learn about the Operations of Complex Numbers, Addition & Subtraction of Complex Numbers, Properties of Addition and Subtraction of Complex Numbers, Multiplication Properties of Multiplication of Complex Number, Division, Conjugate, Modulus of Complex Number, and Solved Examples. There are 5properties of natural numbers: Closure Property, Commutative Property, Associative Property, Identity Property and Distributive Property.
Introduction to Complex Number
Complex number is an element of a number system that contains the real numbers and a specific element denoted i, called the imaginary unit, and satisfying the equation i 2=−1 i 2=−1. Moreover, every complex number can be expressed in the form a + bi, where a and b are real numbers.
If we solve any quadratic equationwe get its two roots. The two roots of the quadratic equation a x 2+b x+c=0 a x 2+b x+c=0 are the Sridharacharya formula.
Solving a quadratic equation using the Sridharacharya formula can be tricky in terms of calculations but it is comparatively faster.
Sridharacharya Formula is x=−b±b 2−4 a c√2 a x=−b±b 2−4 a c 2 a
When b 2−4 a c>0 b 2−4 a c>0 then these two roots are real and distinct; graphically they are where the curve y=a x 2+b x+c y=a x 2+b x+c cuts the x-axis. When b 2−4 a c=0 b 2−4 a c=0 then we have one real root and the curve just touches the x-axis. But when b 2−4 a c<0 b 2−4 a c<0 there are no real solutions to the equation as no real squares to give the negative b 2−4 a c b 2−4 a c.
In order to aid the calculation process further, mathematicians took the help of the imaginary number i where, i=−1−−−√i=−1
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Properties of Complex Numbers
Here are the basic properties of complex numbers with proof. The properties of a complex number are the same for the normal form and polar form of complex numbers.
When x, y are real numbers and x + iy = 0 then x = 0, y = 0
Proof:
According to the property,
x + iy = 0 = 0 + i ∙ 0,
Therefore, from the definition of equality of two complex numbers, we conclude that, x = 0 and y = 0.
When a, b, c and d are real numbers and x + iy = u + iv then x = u and y = v.
Proof:
According to the property,
x + iy = u + iv and u, v, x and y are real numbers.
Therefore, from the definition of equality of two complex numbers, we conclude that x = u and y = v.
For any three the set complex numbers u, v and z satisfy the commutative, associative and distributive laws.
u + v = v + u (Commutative law for addition).
u ∙ v = v ∙ u (Commutative law for multiplication).
(u + v) + z = u + (v + z) (Associative law for addition)
(u.v)z = x(y.z) (Associative law for multiplication)
x(v + z) = x.v + x.z (Distributive law).
Learn about De Moivre’s Theorem
The sum of two conjugate complex numbers is real.
Proof:
Let, z = x + iy (a, b are real numbers) be a complex number. Then, the conjugate of z is z¯z¯ = x – iy.
Now, z + z¯z¯ = x + iy + x – iy = 2x, which is real.
The product of two conjugate complex numbers is real.
Proof:
Let, z = x + iy (x, y are real number) be a complex number. Then, the conjugate of z is z¯=x–i y.z¯=x–i y.
z∙z¯=(x+i y)(x–i y)=x 2–i 2 y 2=x 2+y 2 z∙z¯=(x+i y)(x–i y)=x 2–i 2 y 2=x 2+y 2, (Since i 2=−1 i 2=−1), which is real.
you should also read about theDeterminants here.
Properties of Addition and Subtraction of Complex Numbers
Consider the complex numbers z 1=a 1+b 1 i,z 2=a 2+b 2 iz 1=a 1+b 1 i,z 2=a 2+b 2 i and latex] z_3=a_3+b_3i) . The following properties hold at all times:
Commutativity of Addition and Subtraction for Complex Numbers: z 1+z 2=z 2+z 1.z 1+z 2=z 2+z 1.
Associativity of Addition and Subtraction for Complex Numbers: z 1+(z 2+z 3)=(z 1+z 2)+z 3.z 1+(z 2+z 3)=(z 1+z 2)+z 3.
Existence of an Additive Identity for Complex Numbers: z 1+0=0+z 1=z 1.z 1+0=0+z 1=z 1.
Existence of an Additive Inverse for Each Complex Number: z 1+(−z 1)=(−z 1)+z 1=0.z 1+(−z 1)=(−z 1)+z 1=0.
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Properties of Multiplication of Complex Number
Consider the complex numbers z 1=a 1+b 1 i,z 2=a 2+b 2 i z 1=a 1+b 1 i,z 2=a 2+b 2 i and z 3=a 3+b 3 i z 3=a 3+b 3 i. The following properties hold at all times:
Commutativity of Multiplication for Complex Numbers: z 1⋅z 2=z 2⋅z 1.z 1⋅z 2=z 2⋅z 1.
Associativity of Multiplication for Complex Numbers: z 1⋅(z 2⋅z 3)=(z 1⋅z 2)⋅z 3.z 1⋅(z 2⋅z 3)=(z 1⋅z 2)⋅z 3.
Existence of a Multiplicative Identity for Complex Numbers: z 1⋅1=1⋅z 1=z 1.z 1⋅1=1⋅z 1=z 1.
Existence of a Multiplicative Inverse for Each Complex Number: If z≠0 then z⋅z−1=z−1⋅z=1.z⋅z−1=z−1⋅z=1.
Distributive Property: z 1⋅(z 2+z 3)=z 1⋅z 2+z 1⋅z 3.z 1⋅(z 2+z 3)=z 1⋅z 2+z 1⋅z 3.
Properties of Conjugate of Complex Number
(z¯)¯=z(z¯)¯=z
z+z¯z+z¯ if z is purely real.
z−z¯z−z¯ if z is purely imaginary.
R e(z)=z+z¯2 R e(z)=z+z¯2
I m(z)=z−z¯2 i I m(z)=z−z¯2 i
z 1+z 2¯=z 1¯+z 2¯z 1+z 2¯=z 1¯+z 2¯
z 1−z 2¯=z 1¯−z 2¯z 1−z 2¯=z 1¯−z 2¯
z 1.z 2¯=z 1¯.z 2¯z 1.z 2¯=z 1¯.z 2¯
z 1 z 2¯=z 1¯z 2¯,z 2≠0 z 1 z 2¯=z 1¯z 2¯,z 2≠0
Properties of Division of Complex Number
The Properties of Division of a Complex Number are as follows:
The division of two complex numbers is, by definition, a complex number.
Commutative and associative properties are not true for the division of complex numbers.
Also, learn aboutVector Algebra here.
Properties of Modulus of a Complex Number
|z|>0
|z|= 0, then z = 0 i.e., Re(z) = 0 = Im (z)
-|z|≤ Re(z) ≤ |z| and -|z| ≤ Im(z) ≤ |z|
|z| = |z¯||z¯| =|-z| = |−z¯||−z¯|
z−z¯z−z¯ = |z|2|z|2
|z 1 z 2|=|z 1||z 2|I n g e n e r a l|z 1 z 2 z 3 z 4…z n|=|z 1||z 2||z 3|…|z n||z 1 z 2|=|z 1||z 2|I n g e n e r a l|z 1 z 2 z 3 z 4…z n|=|z 1||z 2||z 3|…|z n|
|z 1||z 2|=|z 1 z 2||z 1||z 2|=|z 1 z 2| provided z 2≠0 z 2≠0
|z 1±z 2|≤|z 1|−|z 2||z 1±z 2|≤|z 1|−|z 2|
|z n|=|z|n|z n|=|z|n
||z 1|−|z 2||≤|z 1+z 2|≤|z 1|+|z 2|||z 1|−|z 2||≤|z 1+z 2|≤|z 1|+|z 2| greatest possible value of |z 1+z 2|i s|z 1|+|z 2||z 1+z 2|i s|z 1|+|z 2| and least possible value of |z 1+z 2||z 1+z 2| is ||z 1|−|z 2||||z 1|−|z 2||
|z 1+z 2|2=(z 1+z 2)(z 1¯+z 2¯)=|z 1|2+|z 2|2+z 1 z 2¯+z 2 z 1¯=|z 1|2+|z 2|2+2 R e(z 1 z 2¯)=|z 1|2+|z 2|2+2|z 1||z 2|c o s θ|z 1+z 2|2=(z 1+z 2)(z 1¯+z 2¯)=|z 1|2+|z 2|2+z 1 z 2¯+z 2 z 1¯=|z 1|2+|z 2|2+2 R e(z 1 z 2¯)=|z 1|2+|z 2|2+2|z 1||z 2|c o s θ
|z 1−z 2|2=(z 1−z 2)(z 1¯−z 2¯)=|z 1|2+|z 2|2−(z 1 z 2¯+z 2 z 1¯)=|z 1|2+|z 2|2−2 R e(z 1 z 2¯)=|z 1|2+|z 2|2−2|z 1||z 2|c o s θ|z 1−z 2|2=(z 1−z 2)(z 1¯−z 2¯)=|z 1|2+|z 2|2−(z 1 z 2¯+z 2 z 1¯)=|z 1|2+|z 2|2−2 R e(z 1 z 2¯)=|z 1|2+|z 2|2−2|z 1||z 2|c o s θ
z 1 z 2¯+z 2 z 1¯=2|z 1||z 2|c o s(θ 1−θ 2)z 1 z 2¯+z 2 z 1¯=2|z 1||z 2|c o s(θ 1−θ 2)|z 1+z 2|2+|z 1−z 2|2=2|z 1|2+|z 2|2|z 1+z 2|2+|z 1−z 2|2=2|z 1|2+|z 2|2
|z 1+z 2|2=|z 1|2+|z 2|2|z 1+z 2|2=|z 1|2+|z 2|2 if z 1 z 2 z 1 z 2 is purely imaginary
|a z 1−b z 2|2+|a z 1+b z 2|2=(a 2+b 2)(|z 1|2+|z 2|2)|a z 1−b z 2|2+|a z 1+b z 2|2=(a 2+b 2)(|z 1|2+|z 2|2) where a,b ϵ ϵ R
z is unimodulus, if |z|=1
Hope this article on Properties of Complex Numbers was informative. Get some practice of the same on our free Testbook App. Download Now!
If you are checking Properties of Complex Numbers article, also check the related maths articles in the table below:
Application of vectorModulus of a complex number
Circular permutationSum of cubes of first n-natural numbers
Mean and variance of binomial distributionPermutation with repetition
More Articles for Maths
Modulus of a Complex Number
De Moivres Theorem
Operations of Complex Numbers
Complex Numbers
Decimal Expansion of Rational Numbers
Number System
Number Line
Irrational Numbers
Operations of Integers
Argument of Complex Numbers
Properties of Complex Numbers FAQs
What are the properties of the Addition and Subtraction of complex numbers?
The properties of Addition and Subtraction for Complex Numbers are as follows:Commutativity of Addition and Subtraction for Complex Numbers: z 1+z 2=z 2+z 1 z 1+z 2=z 2+z 1.Associativity of Addition and Subtraction for Complex Numbers: z 1+(z 2+z 3)=(z 1+z 2)+z 3 z 1+(z 2+z 3)=(z 1+z 2)+z 3.Existence of an Additive Identity for Complex Numbers: z 1+0=0+z 1=z 1 z 1+0=0+z 1=z 1.Existence of an Additive Inverse for Each Complex Number: z 1+(−z 1)=(−z 1)+z 1=0 z 1+(−z 1)=(−z 1)+z 1=0.
Prove the following properties of complex numbers: Commutative Property?
z 1+z 2=z 2+z 1 z 1+z 2=z 2+z 1. Let z 1=x+i y z 1=x+i y and z 2=u+i v z 2=u+i v. z 1+z 2=x+i y+u+i v=(x+u)+i(v+y)=z 2+z 1 z 1+z 2=x+i y+u+i v=(x+u)+i(v+y)=z 2+z 1
How do you do Division of Complex Numbers?
To divide a complex number u + iv by x + iy, multiply the numerator and denominator of the fraction (u + iv) / (x + iy) by (c − id) and simplify. The conjugate of the complex z = (x + iy) is (x − iy).
How do you do Addition of Complex Numbers?
To add two complex numbers we just add the corresponding real and imaginary parts. For the addition of three or more complex numbers, we follow the same process of adding the corresponding real and imaginary parts.
Title
Complex multiplication is a more difficult operation to understand from either an algebraic or a geometric point of view. In the multiplication of Complex Numbers, the real part of the product is the product of the real parts minus the product of the imaginary parts and the imaginary part of the product, is the sum of the two products of one real part and the other imaginary part.
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16159 | https://www.youtube.com/watch?v=dMqAU-xtq9Q | IB MYP Math eAssessment Review - Unit 4 - Statistics and Probability
DSS Math eAssessment Prep
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Hello, this is the fourth (and last) video of our series of four about reviewing for the IB MYP Math eAssessment, both Standard and Extended! Hopefully this content review and included tips and tricks assist you a lot with your revision!
The videos of the main eAssessment video is now complete. Congratulations on making it this far, I hope our efforts make it so you guys can take your exams with much more confidence!!
Images in the video are from the textbooks: Mathematics 9 (MYP 4) (3rd Edition), Mathematics 10 (MYP 5 Extended) (3rd Edition), both by Haese Mathematics.
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Transcript:
so now we're into our final chapter of statistics and probability in this chapter we'll be learning about the applications of how to calculate different statistics and calculate the probability of different situations given by the question so I'll be start by defining a few terms but before we uh before I start explaining different things I have to let you know that this unit is comparably easier than the other units that you will be studying in this curriculum however as you reach towards DP uh this unit becomes very difficult as uh different kind kinds of series comes in and probability functions come in so I recommend you guys really um built up a sufficient pillar in order to study and guide your way through the whole unit throughout up to DP so all right let's get started first let's look at the definition so probability is the mathematical chance of something happening yes I mean sure we all know what probability is so let's look into the specific terms used in math which are sets and set operations so first of all set is a collection of objects or things example numbers so set of all factor is in of 12 is 1 2 3 4 6 12 notice we place the factor within curly bracket with commas between we often use a capital letter to represent a set so that we can refer to it easily for example we might let f equals set 1 2 3 4 6 12 we can say that f is the set of all factors of 12 So currently what this is trying to explain is that um set is literally just sets of numbers for I mean for instance in math we will call it sets of numbers so any number series can be a set for inance instance uh here we go this would also be a set and this would also be a set of numbers so this is not something really complicated we just call these things sets and each uh object in a set is called element or member so these things are called element or member all right that's pretty straightforward now then what is subsets uh let's look at the let's read The Prompt here suppose p and Q are two sets p is a subset of Q if every element of p is also an element of q that is the most complicated way to express subsets but don't worry I'll make it more easier um Let's ignore the notation for now and I'll start by explaining what subsets are so we just said that uh set is just sets of numbers let's say uh we have a set called f which contains the value 1 2 3 4 five and six and let's see uh let's call a another set of C with the numbers two 4 5 and C notice how each num is within the set of C is also a value within the number of set f as can be seen here now because all the elements of C is a member of f c is a subset to F so f is f uh F holds C within uh within its own set all right then that so that is basically what subsets are saying if for instance C had another value of seven C would not be a subset of f all right then here are a few notations that you guys have to understand when you're solving mathematical questions about statistics and probability first is this e looking guy which uh read is which means the element or member of it of the set and e with a knot normally when you have a line like in the original dation it means not so you know how this is a equal sign and this is a not equal sign so um this this notation reads that is not an element or is not a member of the set these notations are used to express that the uh set is empty so if I just said f equals just blank then F does not have a set of numbers and this reads that it's a subset of so in this case we if we bring back c as a subset of f we will call it f is a Subs I we call it C is a subset of f and then next we have NS which meets the number of element in set s all right I this is this can be easily explained as we look at the example here so four is an F so these these statements written here are all examples and write examples in expressing the number so for instance 4 F uh four is an element of f because because we have four right here so this statement is true seven is not here in The Element Section so 7 is not a member of f 2 46 are all values within the uh within the subset of f so we can say 2 46 is a subset of F and finally N means the number of elements within set s so one 2 3 four five six so the answer will be six not too difficult yet however it gets a little confusing as we are introduced to Union and instruction let's look at what these are if p and Q are two set but it gets quite difficult as we are introduced to Union and intersection so let's look it I call it n but it's actually called intersection so P intersection Q is literally the intersection of p and Q and consists of all element which are in both p and Q so the key word here is both let's give it another highlight it's important and then PQ is the union of p and Q consisting of all elements which are in P or Q all right this is very confusing if you put it into words but it's very easy when you uh look into the example all right um all right before we move on to the example I'll first tell you what mutually exclusive mean sometimes the questions are going to tell you that P and Q are mutually exclusive or set F and C from the previous question are mutually exclusive what mutually exclusive mean is that the two sets are disjoint uh if they have no elements in common so for instance if I have um yeah reading the red notes here so if I have a set of a which consists consists the number of 1 3 and four while B there uh there's a set of there's a set called B which has the number two five and six you can see how no numbers here overlap now that's why we can call that A and B are mutually exclusive so it means that no same values are between the sets so in the example my friend gave here because twos overlap these two sets are not mutually exclusive all right then what does Union and inter uh intersection mean first of all before I uh introduce you to Union and intersection I first want to point out how to memorize this because because these two symbols look very similar it gets quite confusing as we are introduced to this in a exam situation so I really want to uh I really want you guys guys to remember that p u q is always Union and p n q is intersection you can memorize it through the alphabets they use because Union obviously starts with a u so you guys can understand that it's a union intersection in the other hand has n on the second letter so I want you guys to remember in this fashion so that um it's a little more easier to understand then what are these Union and intersections well Union is when uh there are overlapping numbers between two sets this can be looked through the question here so for example if there's a set named p with the value 1 2 4 5 6 8 and a set called Q with the value 256 then p and q p intersection Q would be the the same numbers that exist within the same uh within the different sets So currently uh number 256 are both in the two sets so these two are the values of p and q p intersection Q as 256 are in both sets however PQ is like a universal PQ is are are just all the elements within the two sets so it would be all the numbers listing from here so all the numbers from P of one 2 4 5 68 and 0 2 3 5 6 8 set six seven so all the numbers combined to create a set right a set that consists all these numbers together would be P Union Q so as you can see number one number one 2 4 five 6 and eight is in here and zero two is already in there and three five is also already in there six is already in there and seven a set that consists all the element of the two sets are p uh are union of the two sets as these are the elements within which are in P or Q this can be a little confusing but if you do a few um practice question about three or four in the Hai mathematics textbook it will Pro it will very help uh it will you will understand in a short time uh the page numbers are written here it's the great n textbook of page 102 so please do like a five minute practice and I believe that's that's all the time it's going to take for you to understand the whole concept now moving on special number sets remember from unit one yes yes I have told you I have already introduced this topic uh in the unit one video so I'll skim through it real quickly but uh these are just a set of numbers that mathematicians came to agree that this letter rep normally represents this set obviously if the question tells you otherwise this it can change but normally if there's a question saying that uh if there's a question noted this way it means that X is just a number of uh natural is a is a natural number so it's telling you the region of the answer can be so um n normally uh notes natural number that's pretty easy z uh means integers so every uh every values that are not a decimal imaginary number and Etc and z+ uh because we have the plus here is positive integer Z minus in the other hand is negative integer while Q is set of all rational numbers which are real numbers and which can be written in the form of p over Q where p and Q are integers so just generally all rational numbers are noted in Q while Q not normally this is a coding term but if you have a comma here it means not so Q not is called All of irrational numbers that I also introduced the unit one video so if this confuses you this is a sign that you have to review unit one again so it's numbers like root2 and cube root 7 and Pi belong uh and pi as these numbers belong to Q showing that these are not real numbers which are numbers that can be placed on the number line like this way these units are dealt more heavily in unit one video so please check that out compliment of set all right so this this unit um in this section we're we're going to be learning about Universal set and complimentary sets first of all Universal sets set Universal set is very important Whenever there is a set represented by you it means the whole set of numbers so uh normally when we do these type of statistic questions I like to express it in a box of whole number series and this is going to be you and when we draw Vin diagrams these diagrams are going to represent the number um the set of numbers I will I will do I'll explain this more in depth uh as these are introduced later on this video but basically Universal set is uh all the all the set of numbers so if if z uh has the number one two 3 and F has the number 3 2 six the universal statement would be the union where it's one two three and six so it's just all the numbers that are given within the question the the maximum set of numbers and then complimentary set is all elements of U that are not in a so uh as I introduced this notation means not for so if they if the question told you if the question informed you that you the Universal set has numbers 1 2 3 4 5 6 7 8 and nine and a set a has elements of 1 3 5 7 9 then the complimentary set of a not would be every number in U that is not in a so if I try just erasing as what my friend did here uh 1357 9 the numbers 2 4 68 would be a complement of a which is the denoted as a uh Apostrophe a not so something useful to look over uh there is no intersection between a and a KN because a is a set and a not is set of numbers that are not a a so there would no there would there would never be an intersection between this equation so this is a way of representing n a not available so just understand that a intersection a not does not exist in the other hand a union a KN is the universal statement because a not includes every element in universal statement that is not a while a includes itself a so a union a not is the universal statement and the number of elements in a and the number of elements in a not is also same as number of elements in the universal statement I think this concept is pretty okay then we'll be moving on to an example questions to show you how to solve these type of questions uh let's try to cover the answers for right now we're looking here all right if you if there's a set of numbers called U uh consisting actually I can't cover the answers anymore you know just ignore the part of set cover answer if there's a set called U which U with the elements -3 - 2 -1 0 1 two and three and there's a set a of - 2 0 to 3 and set B of Min - 3 - 213 will be able to answer the questions listed here so we they want they first want us to draw I want us to list a notot so um all right first of all a not is all the elements that are not within a so I'm going to that are in the universal statement so I'm going to erase each of this uh sets together to easily notate and now all the numbers left over would be a KN so a KN would be Min - 3 -1 1 that is pretty straightforward and just got to refresh these numbers and we're going to look for B notot in this case uh we're going to erase all the numbers Within um this looks like a three so annoying okay so I'm going to erase all the numbers within uh B in the universal statement and the leftover number is going to be B not so - 3 - 2 1 and 3 so we only have minus1 0 and two in b b equal min-2 - two oh and that goes minus1 02 yeah all right so next question is going to be a intersection B so a and intersection B is going to be the same values that a and b share so in this case it's going to be minus 2 and 3 so a intersection b a n b would be - 2 and 3 there we go and then we have Au U which is a union b and a union b as I introduced earlier is all sets of numbers together Within A or B so first I am going to write all the numbers of set a first which is going to be min-2 0 two and three now I'm going to write all the numbers within set B so minus 3 one and three is already in there so I don't have to add another three or minus two you don't have to write the same set of numbers many multiple times when we're think we're talking about sets of numbers so this would be a union B okay then this is a special question a intersection B KN a inter intersection not B all right so all elements that are not B that we are got from the previous answer we can use this here so it's the numbers that uh intersect between union A and B not so it's going to be zero and two so a intersection B KN is going to be0 and two and and then for a knot and B not uh we can see so next the question wants to get the union of a knot and B knot because we already have the a kn and the vot set values from the previous question we just have to get the union value of this as I did before before I am going to attempt to write the values of a notot first which are minus three minus one and one and then now I'm going to add the bot values which is minus one's already there 0o and two so these values are going to be a not Union B KN now finally we have the question of n what is the number of elements for a KN and B not Union so in this case we just have to count which is 1 2 3 four five so that's going to be five elements there we go and one thing to uh advise is that this is going to be very easier when you understand this vend diagram however um this V diagram is going to be explained later in this video just stand tight and we will explain everything to you don't worry uh it may be a little confusing because of the wordings that they use but probability and statistics in my opinion is more about how you gain your experience and the experience is equivalent to knowledge so if you just practice again and again you'll eventually get better and better so don't worry all right then let's move on now we'll be looking into sample space um sample space is definition given by the textbook is the set of possible outcomes of an exper um experiment uh experiment in this case is the set of numbers or the probability we have um I'll be looking more into that as I explain about Ben diagrams and 2D grid tree diagrams and so on okay ways to visualize sample space and probability now the first way to do it is list and this is pretty much the most easy not most um this is the most common way of doing a probability question which is listing out the all possible outcomes so for instance as the question gives when two coins are Tau lost the possible outcomes are two heads heads and tail tail and head and two tails now um so the probability representing it would if you list it out would be in in this format so head head head tail tail head and tail tail and um obviously this method will not work for uh probability of counting like bazillion coins but uh it's a really easy way to solve simple problem probability questions now another way to solve um probability question is using the tree diagram and this is uh favor in a lot of different math questions especially probability questions it's useful for listing a sequence of events so if you if you look at the example my friend gave I'm choosing a short pants and a sock so oh shirt pants and sock oh my God okay um first there's the one over two chance of choosing a green shirt or a blue shirt and then out of the green um after you choose the green sh you can think about the chances of um getting a gray pants or a black pants and then inside that choice you also have three more choices of three different color socks so uh as is written here choosing a blue the probability of choosing a blue shirt gray pants and green socks would be one over two here one over two here and one over one over three here so multiplying that all together since they're dependent events uh since they're independent events it will give us a one over 12th chance another way um I can make a question here is what is the probability of choosing green shirt but not black pants cuz Okay so because we want to choose the probability of green shirt first we look into the one over two chance here so I'll be highlighting this way so we go this way and then here we don't want to choose uh the black pants so we don't want this guy then um we go over to this uh Road however um because the question states that uh doesn't talk anything about stocks we assume that any color type of color of stocks is fine so all of this is one probability so if you write that down it would become one over two from choosing green shirt over blue shirt and then multiplying one over two again it's because we want to choose gray pants over um instead of black pants and then you multiply one because this whole Cho Choice adds up to one that would give you four over one chance so it would be 25% um this tree diagram uh can come a little difficult if they start asking you about uh assuming something happens but I'll be deal dealing with that a little later so let's move on to vend diagrams vend diagrams is an alternative way to visualize combination two sets or subsets and I kind of implied um B diagrams important uh up in the unit here this while while explaining complement sets and um vend diagram is um first you have to before you learn the specific things about vend diagram here uh the definition of vend diagram is that vend diagram consists of universal set u represented by a rectangle and set within is that are generally represented by circles this is not like a universal law that um everybody does have a square Universal set and then you have circles in the middle you can have squares in the middle like this but it looks ugly so they choose to use circles so uh before we move on to learning what a Bend diagram how to use vend diagram first you have to know the different terms Union and disjoint of M exclusive terms and if you noticed um we kind of learned this uh words over here uh no not here um here we have Union and then we have mutually exclusive here so these are these terms should be familiar and if you if it's not I recommend you go back and check on the names but just to briefly go over um Au consists of all elements in a or b or both it is a shaded region which includes everywhere in either Circle this joint or mutually exclusive um have no common elements and they're represented by non overlapping circles before we look the um the writing in red let's look over the writing on red here first a combination of these four main areas can be used for any combination of two sets uh this may be a little confusing but I'll be look this will become more easier to understand as we solve the questions so just you wait and for all of these calculation combination try not to memorize them rather identify the set circles how the operation you and and Slack not and remember Brack bracketed terms uh to go together A and B means not A and B yep these things are straightforward and I kind of explained them before so I'm just going to move on to explaining intersections so intersection is again um I said it before it's a m b it consists of all elements common to both A and B and I said it's a because it's a and b they have to have an overlapping value and using the vend diagr we can understand that that value is right here the overlapping side and um yeah let's just try looking into that uh is that shaded area where the circle representing A and B overlap yes a is shaded A and B is shaded B not since um everything besides B has to be a p b not is shaded and a and b not is shaded I think these are straightforward uh on separate diagram shade it shade these region for two overlapping sets of a a a not and B A and B not yes uh you guys can try this by yourself um I don't think this is a major issue that's going to come out in your tests but let's see at this question because this question is a little fun yeah I'm going to try to cover the answer um given Nu = 40 n a = 24 and B = 27 and a n a n a n Bal 13 find n a u b n a n b not all right uh first uh since this notation gives us the universal set we know that um the total number of set we have is is 40 this um this guy will be n and na a here inside a circle we have 24 the whole thing is 24 and the MB is 27 however it says that a n b so a and a so n a and NB equals 13 which means that they have a overlapping side in the middle so here and here we have a and b and in the middle as it says we have 13 so what we know is that since a total has 24 this would be 11 since the whole circle since the whole since the whole circle has to add up to be 24 and we also know that um since B also have to add up to become 27 we know that uh 27 minus 13 is 14 so this part of the number would be 14 so now using this if you want to get n Au U that would give us uh Since U represent or we just have to add all the numbers here here so it' be 11 + 13 + 14 which gives us 38 and n a n b not would give us so um N means and and B not means we have to do every side besides the uh I'll use this color besides B so every part here that overlaps with a which means that it's going to be the area here so it be exactly 11 wait let me just erase all the highlights and then we have the final oh wa no that was the final question yes so this is the way you can get it and if my explanation is not uh is a little confusing you can look at the solutions and let's look at the tip written here don't get uh confused between the value of N and the actual value with these sets oh so what this means that um if they give you this number set like 4 One 2 3 four five um this doesn't mean that uh like you have one number in one set one two number in two sets it just means that n has like inside n there is one two three four and five so yeah don't he just means that don't get confused with having like five numbers in N um yeah so something like this and another red note here for myp any V diagram four up to three circles will come out four okay four up to three circles will come out write down all the information given organized into the vend diagram and you should be fine uh yes um if actually ven diagram in my opinion is one of the most difficult things that can come out if it comes out into three sets of circle um so make sure you guys practice this vend diagram thoroughly and we will be providing some um example questions especially regarding this um three Circle Vin diagram so check that out and then study tightly and 2D grid honestly in my opinion this is very very very this is very very opinionated but I don't think um 2D grid is really needed in solving probability because every everything can 2D GD can be like done with every other um things we have learned but because since NP includes this part of the curriculum and maybe the question will ask you to draw to the grid I recommend you study and know what this concept is so 2D grid is basically representing um one event on the x- axis and another event on the y axis so this is very restricted to having two independent U um events if it goes over two you have to write three dimension and that is starting that's going to nauseate you like crazy so each do represents a different outcome but can only be useful used for two events at once yeah as I said so for instance rolling two dice or rolling two coins um uh what What's another probability think that could happen uh drawing a ball like from two different boxes so the maximum amount of events we can have is two events so I'll just give you example of let's say rolling to six faced dice so in x-axis I'm going to do dice one and then the y- axis I'm going to do dice two and obviously because the six faes di six face dice the only outcome that we can have is one two three four five and six actually let's draw this in a straight line give me a second yeah so one two three 4 five six and then you can draw another six layer since the second dice would only have um six chances too since it's a six face dice as you can see this steps takes a lot of time and if you uh I just recommend rather using the uh here listing method if you're guys going to try to use the grid method so for instance if you want to get like um outcome of probability not probability uh let's do chance chance of drawing at least one three from the two dice so like we want um at least one dice rolling a three out of the two dice they can overlap so what we'll do is looking at the grid diagram we will look into here as we'll look into here uh uh because we because we want to get oh tou so we look into the 2D grid and we look into number three and we see one 2 3 four five six outcomes and looking because we have another dice we look into the three in here and that would give us one 2 3 four five six but since there's one overlapping if we count the number of out because that be 1 2 3 four five 6 7 8 9 10 11 so the chance would be 11 so this is a way you would use 2D grid but as I say stated again just know um what this is uh because uh the the test might ask you to draw a 2d grid but normally if you're just calculating uh probability use the listing method instead of the 2D grid [Music] method okay next we have range inter interqual range and standard standard deviation fun fact this guy is going to come out in dp2 and it's going to drive you nuts if you're a going to do AO yes any who that's a future future statement and knowing the middle point of the data set can be quite useful but for a more complete picture of the data set we also need to know its spread or variation so three commonly used statistics that indicate the spread of the data sets are range interqual range and standard deviation first of all let's start with range the range is the difference between maximum data value and minimum data value that is pretty straightforward later if you go on to functions there's another definition of range but until then in a normal set of numbers uh the range is highest value minus the lowest value so in this case uh my friend made a mistake but the since the highest value let me just where is that since the highest value is nine and the lowest value is two the range is seven yes maybe he was a little sleepy making this document don't be so harsh on it and then we have the inter quota range uh the median divides and Order and then we have the interqual range where the median divides an order set in halves by finding the median of each half we divide the original data and into quartiles so as my friend read in red must be ordered just like for just like for the median yes the interor range is the range of the median have 50% of the data so inter quartile range is upper quartile minus the lower quartile quartile quartile and I'll be explaining this through an example question but you can also check out this page in the text Haze textbook if it confuses you a lot let's see for the data set y y y wow that is a lot of set of numbers um find the median lower up and upper qu range inter range so first um if you how what did I say when we want to find the mediate the first step is always to list the numbers in increasing order so I see three which is the lowest number so three four uh four five five five five six six six six six H 7 7 7 8 and8 and nine Let's uh and always make sure to check if the number of sets and your increased order set have the same number so two three four 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 and I should also have 19 numbers 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 16 17 18 as you can see I have five found I have made a mistake in the number of sets so I have to look back and see what's the mistake I actually know the mistake because I actually intent intentionally made the mistake so that I can tell you how to check if you're missing a number or not I missed a six so I'm going to add back a six [Music] and yeah uh next what did I say we do we so what you do for the intercal range is that you find the um median but you do it with the lower half of the data besides um compared to the median and then you use it with the upper hand side of the median so we agreed that the median is six since we know that median is six um we're going to eliminate that six in the middle so I'll say I'll eliminate this guy because we have 1 2 3 4 5 6 7 8 N9 and then the other side we have 1 2 3 four five six seven eight nine yeah so we're going to eliminate this six now what we're going to do is we found that this number is the median what we're going to do is we're going to eliminate this number we're going to move it outside of the set and now what we have is we have two sets of value without the median now what we to find the lower quot range we're going to find another median value from this newly um created individual set and then to find the upper quartile we're going to find the median from this set so let's see um using the elimination method one two three four five six seven eight so we have five here and then let's see one 2 3 4 five six seven eight we have seven here so the lower cortile is five and the upper quartile is seven let's see so now that we have the lower and the upper quartile if you want to find the inter quartile range as the equation as as the formula States here it's the upper cortile minus the lower cortile um so this is not a by the way I just noticed how confusing this line can be um I'm just going to erase this line because you might think that it's a division line no it's not I drew that line so don't worry the book is never wrong I mean sometimes is but ha hbook tends not to be wrong if we use the formula to get the inter qu range by minusing the lower Cal from the upper Cal we get 7 minus 5 which is two so the inter qual range is two okay so understand the definition of these different type of methods and then you'll be okay but we're not done because we have standard deviation the most annoying one and even my friend wrote this usually doesn't comes out but it did once so maybe we should learn it um honestly uh this standard deviation is as I said pretty Out Of Reach like you don't need to specifically know and understand fully about this topic and then you don't have to be a master because you learn in DP anyways but as my friend said it has came out in the E assessment once so if that really worries you and if don't have other things to study I know how e assessments can be tough and then you have other subjects to study besides math so I won't be going over this topic but if you're really okay and you really want to get a good score in math I also recommend studying this part two by by yourself and the book pages are here so uh if there's anything confusing just email me uh we'll put the email on the um our YouTube descriptions and you can also ask your teacher anytime I know they're all experts about it so you just know that this topic exists and then if it really worries you ask your teachers and try to learn it by yourself because it is pretty difficult since it comes out in DPN you already see Sigma limitations that you don't need to learn and then you have um the different deviation numbers and then you look at these difficult calculations so let's move on uh but correlation is something that comes out and it has came out every not every year that's an exaggeration um correlation also oftenly comes out so you have to learn this guy correlation is the measure of strength of the relationship or association between two variables so the most um not most so the thing you have to know is look at the spread points to make a judgment about the strength of the correlation so linear relationship between the variables and non and let's see um as you can see we have for the values we have strong correlations there's uh okay as you can see since a lot of the data points are very near to the linear um linear correlation here we say we have a strong correlation and this guy seems moderate because it's scatter out a little and this is weak because there is no pattern now some people like question like to what extent is moderate and strong like how do we Define the two difference well the short answer is you can't like it's your standard if you feel like oh there's a really good correlation you just say strong maybe for you this type of things uh also shows a good strong correlations maybe the the data set is like 0.01 cm and that's not a big difference so actually this shows a pretty good correlation maybe that's the case um the judgment is yours I don't think the iob will ever um DeMark you for calling um these two type of graphs uh one strong and one moderate but you cannot call a weak correlation graph strong correlation that's for sure and I also recommend don't I actually just don't recommend using the word moderate because it's such a broad term so I rather just stick with strong or weak and then justify the reason and then you'll be fine and my one tip my friend wrote is that strong correlation doesn't necessarily mean the values are increasing or there's a positive or negative correlation it just means there are a strong relationship with the database and the correlation graph and yes not um some the graph doesn't necessarily have to be always always linear there can be nonlinear graphs like this guy and let's see you are not required to calculate the numerical value for the strength of the correlation yes don't look at this psychotic equation do you want to calculate this no even a DP student like me I don't want to use that ever in my life so yes just see the definition I recommend you don't really need to see this part just read this part out and then you're okay just understand there's a strong um there's strong positive correlation and strong negative correlation so in this case this is a strong positive correlation because the datas are line together and it's a positive graph while this guy is a strong negative graph because um it's pointing down and then there are strong correlation so something like that uh finally we have line of best fit oh yes I love this part um yes for the line of best fit you're never required to calculate you only are required to sketch by I so what is mean what this means that is very relative and unless you don't draw a super weird graph you would be okay so um there are pretty uh mathematically s um Advanced written steps but the real thing about line of best fit is just just see a good looking graph that's going to fit every data point so for instance if I just give a small example oh my God what is that maybe the data is like this obviously I don't know the specific uh best fit data that is mathematically proven but because in your level you just have to use the eyes any graph that looks like this would be okay so how the IB does it is like they set a like a boundary like if the graph gra is pointing towards inside of these range you get the point so you don't have to like forcefully think oh my gosh this is the best of best fit equation and uh best fit um graph and I have to draw it very very closely touching every point and freak out something like that just make it look good and like it make it look persuasive and it's fine okay let's see uh yes you won't be uh asked to calculate the exact line of best fit using linear regression method linear regression method is the ugly looking equation like this guy right here and then good to know how to do it on your calculator though uh yes that's a tip but not for your test not yet so this is an example of how you would draw a um line of best fit and one thing I also Rec wanted to recommend oh even though I my friend did say we don't have to find the exact line I in actually in my year we were asked to draw a line of best fit and then find the slope of it so this also has a range of answers and if you're fit in that range you're okay so um you don't have to freak out and then suddenly start thinking about how should I get the uh uh rad gradient of the slope of the graph you just have to Simply look at the graph you drew and then find the slope from there and then maybe you can modify your graph a little so that it's easier to get the um slope and that would make your life a little more easier to test since Sometimes some for some people um you don't have enough time for your test so this is the end of statistics and probability uh one thing to make sure is that this part of the unit has a lot of definitions and different terms so you have to all understand understand and I gave a lot of many tips throughout the whole uh questions and just understand that we're trying our best to give the best type of tips so that you can go through the E assessment so I recommend you look through the video one more time and then take notes thoroughly and for the probability part I really recommend um practicing again and again till you understand you get a sense of how to solve these type of questions because these questions um can be very confusing when you mute in a test all right so that's the tip and I hope you guys do great on your test thank you |
16160 | https://www.chegg.com/homework-help/questions-and-answers/part-1-20-marks-rotating-cylinder-viscometer-shown-figure--inner-cylinder-radius-r-height--q40765626 | Solved Part 1-a [20 Marks] A rotating cylinder viscometer is | Chegg.com
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Part 1-a [20 Marks] A rotating cylinder viscometer is shown in the figure below. The inner cylinder has radius, R, and height, H. An incompressible, viscous, Newtonian fluid of density, p, and viscosity, u, is contained between the cylinders. The narrow gap between the cylinders has width, Ar (<< R and H i.e the velocity distribution inside the gap Ar
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Question: Part 1-a [20 Marks] A rotating cylinder viscometer is shown in the figure below. The inner cylinder has radius, R, and height, H. An incompressible, viscous, Newtonian fluid of density, p, and viscosity, u, is contained between the cylinders. The narrow gap between the cylinders has width, Ar (<< R and H i.e the velocity distribution inside the gap Ar is
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Identify the relationship between torque T and shear stress τ using the equation τ A=T, where A is the surface area of the inner cylinder. View the full answer Previous questionNext question
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Part 1-a [20 Marks] A rotating cylinder viscometer is shown in the figure below. The inner cylinder has radius, R, and height, H. An incompressible, viscous, Newtonian fluid of density, p, and viscosity, u, is contained between the cylinders. The narrow gap between the cylinders has width, Ar (<< R and H i.e the velocity distribution inside the gap Ar is assumed linear). A torque, T, is required to rotate the inner cylinder at constant speed 22, the outer cylinder is stationery. Determine the fluid viscosity, u, in terms of the other system parameters. , T Newtonian fluid of density, p, and viscosity, u H.
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mmmmmmmmmmlli mmmmmmmmmmlli mmmmmmmmmmlli |
16161 | https://pubmed.ncbi.nlm.nih.gov/37417783/ | Angiotensin-converting enzyme inhibitor induced cough compared with placebo, and other antihypertensives: A systematic review, and network meta-analysis - PubMed
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. 2023 Aug;25(8):661-688.
doi: 10.1111/jch.14695. Epub 2023 Jul 7.
Angiotensin-converting enzyme inhibitor induced cough compared with placebo, and other antihypertensives: A systematic review, and network meta-analysis
Yiyun Hu1,Ling Liang23,Shuang Liu4,Janice Y Kung5,Hoan Linh Banh6
Affiliations Expand
Affiliations
1 Department of Pharmacy, Second Xiangya Hospital of Central South University, Changsha, China.
2 Department of Cardiology, The Third Clinical Medical College, Fujian Medical University, Fuzhou, China.
3 Department of Cardiology, The First Affiliated Hospital of Xiamen University, Xiamen, China.
4 Medical Affairs Management Department, Second Xiangya Hospital of Central South University, Changsha, China.
5 University of Alberta, John W. Scott Health Sciences Library, Edmonton, Canada.
6 Faculty of Medicine and Dentistry, Department of Family Medicine, University of Alberta, Edmonton, Canada.
PMID: 37417783
PMCID: PMC10423763
DOI: 10.1111/jch.14695
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Angiotensin-converting enzyme inhibitor induced cough compared with placebo, and other antihypertensives: A systematic review, and network meta-analysis
Yiyun Hu et al. J Clin Hypertens (Greenwich).2023 Aug.
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. 2023 Aug;25(8):661-688.
doi: 10.1111/jch.14695. Epub 2023 Jul 7.
Authors
Yiyun Hu1,Ling Liang23,Shuang Liu4,Janice Y Kung5,Hoan Linh Banh6
Affiliations
1 Department of Pharmacy, Second Xiangya Hospital of Central South University, Changsha, China.
2 Department of Cardiology, The Third Clinical Medical College, Fujian Medical University, Fuzhou, China.
3 Department of Cardiology, The First Affiliated Hospital of Xiamen University, Xiamen, China.
4 Medical Affairs Management Department, Second Xiangya Hospital of Central South University, Changsha, China.
5 University of Alberta, John W. Scott Health Sciences Library, Edmonton, Canada.
6 Faculty of Medicine and Dentistry, Department of Family Medicine, University of Alberta, Edmonton, Canada.
PMID: 37417783
PMCID: PMC10423763
DOI: 10.1111/jch.14695
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Abstract
Studies have shown that angiotensin converting enzyme inhibitors (ACEIs) are superior in primary and secondary prevention for cardiac mortality and morbidity to angiotensin receptor blocker (ARBs). One of the common side effects from ACEI is dry cough. The aims of this systematic review, and network meta-analysis are to rank the risk of cough induced by different ACEIs and between ACEI and placebo, ARB or calcium channel blockers (CCB). We performed a systematic review, and network meta-analysis of randomized controlled trials to rank the risk of cough induced by each ACEI and between ACEI and placebo, ARB or CCB. A total of 135 RCTs with 45,420 patients treated with eleven ACEIs were included in the analyses. The pooled estimated relative risk (RR) between ACEI and placebo was 2.21 (95% CI: 2.05-2.39). ACEI had more incidences of cough than ARB (RR 3.2; 95% CI: 2.91, 3.51), and pooled estimated of RR between ACEI and CCB was 5.30 (95% CI: 4.32-6.50) Moexipril ranked as number one for inducing cough (SUCRA 80.4%) and spirapril ranked the least (SUCRA 12.3%). The order for the rest of the ACEIs are as follows: ramipril (SUCRA 76.4%), fosinopril (SUCRA 72.5%), lisinopril (SUCRA 64.7%), benazepril (SUCRA 58.6%), quinapril (SUCRA 56.5%), perindopril (SUCRA 54.1%), enalapril (SUCRA 49.7%), trandolapril (SUCRA 44.6%) and, captopril (SUCRA 13.7%). All ACEI has the similar risk of developing a cough. ACEI should be avoided in patients who have risk of developing cough, and an ARB or CCB is an alternative based on the patient's comorbidity.
Keywords: ACE inhibitors; angiotensin receptor blocker; calcium channel blockers; network meta-analysis.
© 2023 The Authors. The Journal of Clinical Hypertension published by Wiley Periodicals LLC.
PubMed Disclaimer
Conflict of interest statement
The authors declare no conflicts of interest.
Figures
FIGURE 1
PRISMA diagram.
FIGURE 1
PRISMA diagram.
FIGURE 1
PRISMA diagram.
FIGURE 2
Bias risk assessment for the…
FIGURE 2
Bias risk assessment for the studies.
FIGURE 2
Bias risk assessment for the studies.
FIGURE 3
Network map. A: The network…
FIGURE 3
Network map. A: The network map of single ACEIs comparison for cough risk…
FIGURE 3
Network map. A: The network map of single ACEIs comparison for cough risk ratios. The width of the black line is positively proportional to the number of trials including every pair of treatments, whereas every circle size is positively proportional to the total number of participants for each treatment. A: Placebo; B: Benazepril; C: Captopril; D: Enalapril; E: Fosinopril; G: Lisinopril; H: Moexipril; I: perindopril; J: Quinapril; K: Ramipril; L: Spirapril; M: Trandolapril. Network map B: The network map of different groups comparisons for cough risk ratios. A: Placebo; B: ACEIs group; C: ARBs group; D: CCBs group.
FIGURE 4
Single ACEI interventions network meta‐analysis…
FIGURE 4
Single ACEI interventions network meta‐analysis for cough. League table showing results of the…
FIGURE 4
Single ACEI interventions network meta‐analysis for cough. League table showing results of the network meta‐analysis comparing cough of all treatments including RR and 95% credible intervals. RR>1 means the top‐left treatment is better. The league table represents the relative risk with 95% confidence interval of single ACEIs compared with placebo. The probabilities beside the ACEIs are the treatment ranking based on SUCRA from left to right. The treatment drugs divide the figure into upper (blue colored) and lower (green colored) sections. For the lower section, the efficacy estimate is the ratio of the column defining treatment to the row defining treatment. For the upper part, the efficacy estimate was the ratio of the row defining treatment to the column defining treatment. The lower and the upper portions’ results are mutually reciprocal. The relative risk ratio in each treatment is compared to the treatment to the right in the same row.
FIGURE 5
Five different types of anti‐hypertension…
FIGURE 5
Five different types of anti‐hypertension drugs network meta‐analysis for cough. League table showing…
FIGURE 5
Five different types of anti‐hypertension drugs network meta‐analysis for cough. League table showing results of the network meta‐analysis comparing cough of five types of drugs including RR and 95% credible intervals. RR>1 means the top‐left treatment is better. The league table represents the relative risk with 95% confidence interval of single ACEIs compared with placebo. The probabilities beside the ACEIs are the treatment ranking based on SUCRA from left to right. The treatment drugs divide the figure into upper (blue colored) and lower (green colored) sections. For the lower section, the efficacy estimate is the ratio of the column defining treatment to the row defining treatment. For the upper part, the efficacy estimate was the ratio of the row defining treatment to the column defining treatment. The lower and the upper portions’ results are mutually reciprocal. The relative risk ratio in each treatment is compared to the treatment to the right in the same row.
FIGURE 6
Forest plot comparing ACEI vs.…
FIGURE 6
Forest plot comparing ACEI vs. placebo.
FIGURE 6
Forest plot comparing ACEI vs. placebo.
FIGURE 7
Forest plot comparing ACEI vs.…
FIGURE 7
Forest plot comparing ACEI vs. ARB.
FIGURE 7
Forest plot comparing ACEI vs. ARB.
FIGURE 8
Forest plot comparing ACEI vs.…
FIGURE 8
Forest plot comparing ACEI vs. CCB.
FIGURE 8
Forest plot comparing ACEI vs. CCB.
FIGURE 9
Comparison‐adjusted funnel plots of cough…
FIGURE 9
Comparison‐adjusted funnel plots of cough development. A: single ACEI versus placebo network meta‐analysis.…
FIGURE 9
Comparison‐adjusted funnel plots of cough development. A: single ACEI versus placebo network meta‐analysis. A: Placebo; B: Benazepril; C: Captopril; D: Enalapril; E: Fosinopril; G: Lisinopril; H: Moexipril; I: Perindopril; J: Quinapril; K: Ramipril; L: Spirapril; M: Trandolapril. B: the five groups of anti‐hypertension drugs comparisons network meta‐analysis. A: placebo; B: ACEIs; C: ARBs; D: CCBs.
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References
Nasution SA. The use of ACE inhibitor in cardiovascular disease. Acta Med Indones. 2006;38(1):60‐64. - PubMed
Timmermans PB, Wong PC, Chiu AT, et al. Angiotensin II receptors and angiotensin II receptor antagonists. Pharmaco Rev. 1993;45(2):205‐251. - PubMed
Beckett NS, Peters R, Fletcher AE, et al. Treatment of hypertension in patients 80 years of age or older. N Engl J Med. 2008;358(18):1887‐1898. - PubMed
Staessen JA, Fagard R, Thijs L, et al. Randomised double‐blind comparison of placebo and active treatment for older patients with isolated systolic hypertension. Lancet. 1997;350(9080):757‐764. - PubMed
SHEP . Prevention of stroke by antihypertensive drug treatment in older persons with isolated systolic hypertension: final results of the systolic hypertension in the Elderly Program (SHEP). JAMA. 1991;265(24):3255‐3264 - PubMed
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16162 | https://metanumbers.com/490 | 490 (Number)
MetaNumbers
+-
?
| Max | 9223372036854775807 |
| 25835 | Positive integer |
| 0b10001100 | Binary number |
| 01771452 | Octal number |
| 0x129abf | Hex number |
| 5.012 10^7 | Scientific notation |
| 2^4 5 | Factorized form |
| | Random number |
More examples
Search
490 (number)
490 is an even three-digits composite number following 489 and preceding 491. In scientific notation, it is written as 4.9 × 10 2. The sum of its digits is 13. It has a total of 4 prime factors and 12 positive divisors. There are 168 positive integers (up to 490) that are relatively prime to 490.
Properties
Name
Notation
Prime Factorization
Divisors
Other Arithmetic Functions
Divisibility test
Base converter
Classification
Basic calculations
Geometrical shape
Hash Functions
Basic properties
Is Prime?no
Number parity even
Number length 3
Sum of Digits 13
Digital Root 4
Name
| Name | four hundred ninety |
Notation
| Scientific notation | 4.9 × 10 2 |
| Engineering notation | 490 × 10 0 |
Prime Factorization of 490
Prime Factorization2 × 5 × 7 2
Composite number
| ω | Distinct Factors | 3 | Total number of distinct prime factors |
| Ω | Total Factors | 4 | Total number of prime factors |
| rad | Radical | 70 | Product of the distinct prime numbers |
| λ | Liouville Lambda | 1 | Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) |
| μ | Mobius Mu | 0 | Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor |
| Λ | Mangoldt function | 0 | Returns log(p) if n is a power p k of any prime p (for any k>= 1), else returns 0 |
The prime factorization of 490 is 2 × 5 × 7 2. Since it has a total of 4 prime factors, 490 is a composite number.
Divisors of 490
1, 2, 5, 7, 10, 14, 35, 49, 70, 98, 245, 490
12 divisors
| Even divisors | 6 |
| Odd divisors | 6 |
| 4k+1 divisors | 4 |
| 4k+3 divisors | 2 |
| τ | Total Divisors | 12 | Total number of the positive divisors of n |
| σ | Sum of Divisors | 1026 | Sum of all the positive divisors of n |
| s | Aliquot Sum | 536 | Sum of the proper positive divisors of n |
| A | Arithmetic Mean | 85.5 | Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) |
| G | Geometric Mean | 22.135943621179 | Returns the nth root of the product of n divisors |
| H | Harmonic Mean | 5.7309941520468 | Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors |
The number 490 can be divided by 12 positive divisors (out of which 6 are even, and 6 are odd). The sum of these divisors (counting 490) is 1026, the average is 85.5.
Other Arithmetic Functions (n = 490)
1 φ(n)n
| φ | Euler Totient | 168 | Total number of positive integers not greater than n that are coprime to n |
| λ | Carmichael Lambda | 84 | Smallest positive number such that a λ(n) ≡ 1 (mod n) for all a coprime to n |
| π | Prime Pi | ≈ 95 | Total number of primes less than or equal to n |
| r 2 | Sum of 2 squares | 8 | The number of ways n can be represented as the sum of 2 squares |
There are 168 positive integers (less than 490) that are coprime with 490. And there are approximately 95 prime numbers less than or equal to 490.
Divisibility of 490
| m | n mod m |
--- |
| 2 | 0 |
| 3 | 1 |
| 4 | 2 |
| 5 | 0 |
| 6 | 4 |
| 7 | 0 |
| 8 | 2 |
| 9 | 4 |
The number 490 is divisible by 2, 5 and 7.
Classification of 490
By Arithmetic functions
Abundant
Expressible via specific sums
Polite
Base conversion 490
| Base | System | Value |
---
| 2 | Binary | 111101010 |
| 3 | Ternary | 200011 |
| 4 | Quaternary | 13222 |
| 5 | Quinary | 3430 |
| 6 | Senary | 2134 |
| 8 | Octal | 752 |
| 10 | Decimal | 490 |
| 12 | Duodecimal | 34a |
| 16 | Hexadecimal | 1ea |
| 20 | Vigesimal | 14a |
| 36 | Base36 | dm |
Basic calculations (n = 490)
Multiplication
n×y
| n×2 | 980 |
| n×3 | 1470 |
| n×4 | 1960 |
| n×5 | 2450 |
Division
n÷y
| n÷2 | 245.000 |
| n÷3 | 163.333 |
| n÷4 | 122.500 |
| n÷5 | 98.000 |
Exponentiation
n y
| n 2 | 240100 |
| n 3 | 117649000 |
| n 4 | 57648010000 |
| n 5 | 28247524900000 |
Nth Root
y√n
| 2√n | 22.135943621179 |
| 3√n | 7.8837351631052 |
| 4√n | 4.7048850805496 |
| 5√n | 3.4517490659801 |
490 as geometric shapes
Circle
Radius = n
| Diameter | 980 |
| Circumference | 3078.760800518 |
| Area | 754296.39612691 |
Sphere
Radius = n
| Volume | 492806978.80291 |
| Surface area | 3017185.5845076 |
| Circumference | 3078.760800518 |
Square
Length = n
| Perimeter | 1960 |
| Area | 240100 |
| Diagonal | 692.96464556282 |
Cube
Length = n
| Surface area | 1440600 |
| Volume | 117649000 |
| Space diagonal | 848.70489570875 |
Equilateral Triangle
Length = n
| Perimeter | 1470 |
| Area | 103966.34972432 |
| Altitude | 424.35244785437 |
Triangular Pyramid
Length = n
| Surface area | 415865.39889729 |
| Volume | 13865067.616636 |
| Height | 400.08332465459 |
Cryptographic Hash Functions
| md5 | c410003ef13d451727aeff9082c29a5c |
| sha1 | 1b0a69d74c5ab68f9e3505f103f40618a51e5987 |
| sha256 | cbd02d97b0731d88c78d30c20d90492b2d4c3f2f983931c38fef2dedc7ce48d4 |
| sha512 | e25c04bd6ee67e4f1550798ebc8e3a13df2e668d314a9ff9cbe7d4657d5ef35541535ef5b58458a90e19c04d673392fca7d11c46265320e3bcc463f47c864efe |
| ripemd-160 | 12f5005815a127699fbe130ef45d80d0a9a9b677 |
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16163 | https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/12%3A_Vectors_in_Space/12.02%3A_Vectors_in_Three_Dimensions | 12.2.1
12.2.1
12.2.2
12.2.2
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12.2: Vectors in Three Dimensions
Last updated
: Jan 17, 2025
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12.1E: Exercises for Section 12.1
12.2E: Exercises for Section 12.2
Page ID
: 2587
Gilbert Strang & Edwin “Jed” Herman
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Describe three-dimensional space mathematically.
Locate points in space using coordinates.
Write the distance formula in three dimensions.
Write the equations for simple planes and spheres.
Perform
vector
operations in R3.
Vectors are useful tools for solving two-dimensional problems. Life, however, happens in three dimensions. To expand the use of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space. For example, although a two-dimensional map is a useful tool for navigating from one place to another, in some cases the topography of the land is important. Does your planned route go through the mountains? Do you have to cross a river? To appreciate fully the impact of these geographic features, you must use three dimensions. This section presents a natural extension of the two-dimensional Cartesian
coordinate plane
into three dimensions.
Three-Dimensional Coordinate Systems
As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontal x-axis and the verticaly-axis. We can add a third dimension, the z-axis, which is perpendicular to both the x-axis and the y-axis. We call this system the
three-dimensional rectangular coordinate system
. It represents the three dimensions we encounter in real life.
Definition: Three-dimensional Rectangular Coordinate System
The
three-dimensional rectangular coordinate system
consists of three perpendicular axes: the x-axis, the y-axis, and the z-axis. Because each axis is a number line representing all real numbers in ℝ, the three-dimensional system is often denoted by ℝ3.
In Figure 12.2.1a, the positivez-axis is shown above the plane containing the x- and y-axes. The positive x-axis appears to the left and the positive y-axis is to the right. A natural question to ask is: How was this arrangement determined? The system displayed follows the right-hand rule. If we take our right hand and align the fingers with the positive x-axis, then
curl
the fingers so they point in the direction of the positivey-axis, our thumb points in the direction of the positive z-axis (Figure 12.2.1b). In this text, we always
work
with coordinate systems set up in accordance with the
right-hand rule
. Some systems do follow a left-hand rule, but the
right-hand rule
is considered the standard representation.
In two dimensions, we describe a point in the plane with the coordinates (x,y). Each coordinate describes how the point aligns with the corresponding axis. In three dimensions, a new coordinate, z, is appended to indicate alignment with the z-axis: (x,y,z). A point in space is identified by all three coordinates (Figure 12.2.2). To plot the point (x,y,z), go x units along thex-axis, theny units in the direction of the y-axis, then z units in the direction of the z-axis.
Example 12.2.1: Locating Points in Space
Sketch the point (1,−2,3) in three-dimensional space.
Solution
To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in the positive x direction, 2 units in the negative y direction, and 3 units in the positive zdirection. Complete the prism to plot the point (Figure 12.2.3).
Exercise 12.2.1
Sketch the point (−2,3,−1) in three-dimensional space.
Hint
: Start by sketching the coordinate axes. e.g., Figure 12.2.3. Then sketch a rectangular prism to help find the point in space.
Answer
In two-dimensional space, the
coordinate plane
is defined by a pair of perpendicular axes. These axes allow us to name any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just as in two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms a
coordinate plane
: the xy-plane, thexz-plane, and the yz-plane (Figure 12.2.4). We define the xy-plane formally as the following set: {(x,y,0):x,y∈ℝ}. Similarly, thexz-plane and the yz-plane are defined as {(x,0,z):x,z∈ℝ} and {(0,y,z):y,z∈ℝ}, respectively.
To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished. (Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finished walls meet, facing out into the room, the floor is the xy-plane, the wall to your right is thexz-plane, and the wall to your left is theyz-plane.
In two dimensions, the coordinate axes
partition
the plane into four quadrants. Similarly, the coordinate planes divide space between them into eight regions about the origin, called octants
. The
octants
fill ℝ3 in the same way that quadrants fill ℝ2, as shown in Figure 12.2.5.
Most
work
in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. In this section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to three dimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space.
If two points lie in the same
coordinate plane
, then it is straightforward to calculate the distance between them. We know that the distance d between two points (x1,y1) and (x2,y2) in the xy-
coordinate plane
is given by the formula
d=√(x2−x1)2+(y2−y1)2.
The formula for the distance between two points in space is a natural extension of this formula.
The Distance between Two Points in Space
The distance d between points (x1,y1,z1) and (x2,y2,z2) is given by the formula
d=√(x2−x1)2+(y2−y1)2+(z2−z1)2.
The proof of this theorem is left as an exercise. (Hint: First find the distance d1 between the points (x1,y1,z1) and (x2,y2,z1) as shown in Figure 12.2.6.)
Example 12.2.2: Distance in Space
Find the distance between points P1=(3,−1,5) and P2=(2,1,−1).
Solution
Substitute values directly into the distance formula (Equation 12.2.1):
d(P1,P2)=√(x2−x1)2+(y2−y1)2+(z2−z1)2=√(2−3)2+(1−(−1))2+(−1−5)2=√(−1)2+22+(−6)2=√41.
Exercise 12.2.2
Find the distance between points P1=(1,−5,4) and P2=(4,−1,−1).
Hint
: d=√(x2−x1)2+(y2−y1)2+(z2−z1)2
Answer
: 5√2
Before moving on to the next section, let’s get a feel for how ℝ3 differs from ℝ2. For example, in ℝ2, lines that are not parallel must always intersect. This is not the case in ℝ3. For example, consider the lines shown in Figure 12.2.8. These two lines are not parallel, nor do they intersect.
Figure 12.2.8: These two lines are not parallel, but still do not intersect.
You can also have circles that are interconnected but have no points in common, as in Figure 12.2.9.
Figure 12.2.9:These circles are interconnected, but have no points in common.
We have a lot more flexibility working in three dimensions than we do if we stuck with only two dimensions.
Writing Equations in ℝ3
Now that we can represent points in space and find the distance between them, we can learn how to write equations of geometric objects such as lines, planes, and curved surfaces in ℝ3. First, we start with a simple equation. Compare the graphs of the equation x=0 in ℝ, ℝ2,and ℝ3 (Figure 12.2.10). From these graphs, we can see the same equation can describe a point, a line, or a plane.
In space, the equation x=0 describes all points (0,y,z). This equation defines the yz-plane. Similarly, the xy-plane contains all points of the form (x,y,0). The equation z=0 defines thexy-plane and the equation y=0 describes the xz-plane (Figure 12.2.11).
Understanding the equations of the coordinate planes allows us to write an equation for any plane that is parallel to one of the coordinate planes. When a plane is parallel to thexy-plane, for example, the z-coordinate of each point in the plane has the same constant value. Only the x- and y-coordinates of points in that plane vary from point to point.
Equations of Planes Parallel to Coordinate Planes
The plane in space that is parallel to the xy-plane and contains point (a,b,c) can be represented by the equation z=c.
The plane in space that is parallel to the xz-plane and contains point (a,b,c) can be represented by the equation y=b.
The plane in space that is parallel to the yz-plane and contains point (a,b,c) can be represented by the equation x=a.
Example 12.2.3: Writing Equations of Planes Parallel to Coordinate Planes
Write an equation of the plane passing through point (3,11,7) that is parallel to the yz-plane.
Find an equation of the plane passing through points (6,−2,9),(0,−2,4), and (1,−2,−3).
Solution
When a plane is parallel to the yz-plane, only the y- and z-coordinates may vary. The x-coordinate has the same constant value for all points in this plane, so this plane can be represented by the equation x=3.
Each of the points (6,−2,9),(0,−2,4), and (1,−2,−3) has the same y-coordinate. This plane can be represented by the equation y=−2.
Exercise 12.2.3
Write an equation of the plane passing through point (1,−6,−4) that is parallel to the xy-plane.
Hint
: If a plane is parallel to the xy-plane, the z-coordinates of the points in that plane do not vary.
Answer
: z=−4
As we have seen, in ℝ2 the equation x=5 describes the vertical line passing through point (5,0). This line is parallel to the y-axis. In a natural extension, the equation x=5 in ℝ3 describes the plane passing through point (5,0,0), which is parallel to theyz-plane. Another natural extension of a familiar equation is found in the equation of a
sphere
.
Definition: Sphere
A
sphere
is the set of all points in space equidistant from a fixed point, the center of the
sphere
(Figure 12.2.12), just as the set of all points in a plane that are equidistant from the center represents a circle. In a
sphere
, as in a circle, the distance from the center to a point on the
sphere
is called the radius.
The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a
sphere
is based on the three-dimensional formula for distance.
Standard Equation of a Sphere
The
sphere
with center (a,b,c) and radius r can be represented by the equation
(x−a)2+(y−b)2+(z−c)2=r2.
This equation is known as the standard equation of a sphere.
Example 12.2.4: Finding an Equation of a Sphere
Find the standard equation of the
sphere
with center (10,7,4) and point (−1,3,−2), as shown in Figure 12.2.13.
Figure 12.2.13:The
sphere
centered at (10,7,4) containing point (−1,3,−2).
Solution
Use the distance formula to find the radius r of the
sphere
:
r=√(−1−10)2+(3−7)2+(−2−4)2=√(−11)2+(−4)2+(−6)2=√173
The standard equation of the
sphere
is
(x−10)2+(y−7)2+(z−4)2=173.
Exercise 12.2.4
Find the standard equation of the
sphere
with center (−2,4,−5) containing point (4,4,−1).
Hint
: First use the distance formula to find the radius of the
sphere
.
Answer
: (x+2)2+(y−4)2+(z+5)2=52
Example 12.2.5: Finding the Equation of a Sphere
Let P=(−5,2,3) and Q=(3,4,−1), and suppose line segment ¯PQ forms the diameter of a
sphere
(Figure 12.2.14). Find the equation of the
sphere
.
Solution:
Since ¯PQ is a diameter of the
sphere
, we know the center of the
sphere
is the midpoint of ¯PQ.Then,
C=(−5+32,2+42,3+(−1)2)=(−1,3,1).
Furthermore, we know the radius of the
sphere
is half the length of the diameter. This gives
r=12√(−5−3)2+(2−4)2+(3−(−1))2=12√64+4+16=√21
Then, the equation of the
sphere
is (x+1)2+(y−3)2+(z−1)2=21.
Exercise 12.2.5
Find the equation of the
sphere
with diameter ¯PQ, where P=(2,−1,−3) and Q=(−2,5,−1).
Hint
: Find the midpoint of the diameter first.
Answer
: x2+(y−2)2+(z+2)2=14
Example 12.2.6: Graphing Other Equations in Three Dimensions
Describe the set of points that satisfies (x−4)(z−2)=0, and graph the set.
Solution
We must have either x−4=0 or z−2=0, so the set of points forms the two planes x=4 and z=2 (Figure 12.2.15).
Exercise 12.2.6
Describe the set of points that satisfies (y+2)(z−3)=0, and graph the set.
Hint
: One of the factors must be zero.
Answer
: The set of points forms the two planes y=−2 and z=3.
Example 12.2.7: Graphing Other Equations in Three Dimensions
Describe the set of points in three-dimensional space that satisfies (x−2)2+(y−1)2=4, and graph the set.
Solution
The x- andy-coordinates form a circle in thexy-plane of radius 2, centered at (2,1). Since there is no restriction on thez-coordinate, the three-dimensional result is a circular
cylinder
of radius 2 centered on the line with x=2 and y=1. The
cylinder
extends indefinitely in the z-direction (Figure 12.2.16).
Exercise 12.2.7
Describe the set of points in three dimensional space that satisfies x2+(z−2)2=16, and graph the
surface
.
Hint
: Think about what happens if you plot this equation in two dimensions in the xz-plane.
Answer
: A
cylinder
of radius 4 centered on the line with x=0 and z=2.
Working with Vectors in ℝ3
Just like two-dimensional vectors, three-dimensional vectors are quantities with both
magnitude
and direction, and they are represented by directed line segments (arrows). With a three-dimensional
vector
, we use a three-dimensional arrow.
Three-dimensional vectors can also be represented in
component
form. The notation ⇀v=⟨x,y,z⟩ is a natural extension of the two-dimensional case, representing a
vector
with the
initial point
at the origin, (0,0,0), and
terminal point
(x,y,z). The
zero vector
is ⇀0=⟨0,0,0⟩. So, for example, the three dimensional
vector
⇀v=⟨2,4,1⟩ is represented by a directed line segment from point (0,0,0) to point (2,4,1) (Figure 12.2.17).
Vector addition
and
scalar multiplication
are defined analogously to the two-dimensional case. If ⇀v=⟨x1,y1,z1⟩ and ⇀w=⟨x2,y2,z2⟩ are vectors, and k is a
scalar
, then
⇀v+⇀w=⟨x1+x2,y1+y2,z1+z2⟩
and
k⇀v=⟨kx1,ky1,kz1⟩.
If k=−1, then k⇀v=(−1)⇀v is written as −⇀v, and
vector
subtraction is defined by ⇀v−⇀w=⇀v+(−⇀w)=⇀v+(−1)⇀w.
The
standard unit vectors
extend easily into three dimensions as well, ˆi=⟨1,0,0⟩, ˆj=⟨0,1,0⟩, and ˆk=⟨0,0,1⟩, and we use them in the same way we used the
standard unit vectors
in two dimensions. Thus, we can represent a
vector
in ℝ3 in the following ways:
⇀v=⟨x,y,z⟩=xˆi+yˆj+zˆk
.
Example 12.2.8: Vector Representations
Let −−⇀aPQ be the
vector
with
initial point
P=(3,12,6) and
terminal point
Q=(−4,−3,2) as shown in Figure 12.2.18. Express −−⇀aPQ in both
component
form and using
standard unit vectors
.
Solution
In
component
form,
−−⇀aPQ=⟨x2−x1,y2−y1,z2−z1⟩=⟨−4−3,−3−12,2−6⟩=⟨−7,−15,−4⟩.
In standard unit form,
−−⇀aPQ=−7ˆi−15ˆj−4ˆk.
Exercise 12.2.8
Let S=(3,8,2) and T=(2,−1,3). Express →ST in
component
form and in standard unit form.
Hint
: Write −−⇀aST in
component
form first. T is the
terminal point
of −−⇀aST.
Answer
: −−⇀aST=⟨−1,−9,1⟩=−ˆi−9ˆj+ˆk
As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of
vector addition
, for example, is the same in both two- and three-dimensional space (Figure 12.2.19).
We have already seen how some of the algebraic properties of vectors, such as
vector addition
and
scalar multiplication
, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.
Properties of Vectors in Space
Let ⇀v=⟨x1,y1,z1⟩ and ⇀w=⟨x2,y2,z2⟩ be vectors, and let k be a
scalar
.
Scalar multiplication: k⇀v=⟨kx1,ky1,kz1⟩
Vector addition: ⇀v+⇀w=⟨x1,y1,z1⟩+⟨x2,y2,z2⟩=⟨x1+x2,y1+y2,z1+z2⟩
Vector
subtraction: ⇀v−⇀w=⟨x1,y1,z1⟩−⟨x2,y2,z2⟩=⟨x1−x2,y1−y2,z1−z2⟩
Vector
magnitude: ‖⇀v‖=√x21+y21+z21
Unit vector
in the direction of ⇀v: 1‖⇀v‖⇀v=1‖⇀v‖⟨x1,y1,z1⟩=⟨x1‖⇀v‖,y1‖⇀v‖,z1‖⇀v‖⟩,if⇀v≠⇀0
We have seen that
vector addition
in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of
vector
operations are valid for three-dimensional vectors as well.
Scalar multiplication
of vectors satisfies the distributive property, and the
zero vector
acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.
Example 12.2.9: Vector Operations in Three Dimensions
Let ⇀v=⟨−2,9,5⟩ and ⇀w=⟨1,−1,0⟩ (Figure 12.2.20). Find the following vectors.
3⇀v−2⇀w
5‖⇀w‖
‖5⇀w‖
A
unit vector
in the direction of ⇀v
Solution
a. First, use
scalar multiplication
of each
vector
, then subtract:
3⇀v−2⇀w=3⟨−2,9,5⟩−2⟨1,−1,0⟩=⟨−6,27,15⟩−⟨2,−2,0⟩=⟨−6−2,27−(−2),15−0⟩=⟨−8,29,15⟩.
b. Write the equation for the
magnitude
of the
vector
, then use
scalar multiplication
:
5‖⇀w‖=5√12+(−1)2+02=5√2.
c. First, use
scalar multiplication
, then find the
magnitude
of the new
vector
. Note that the result is the same as for part b.:
‖5⇀w‖=∥⟨5,−5,0⟩∥=√52+(−5)2+02=√50=5√2
d. Recall that to find a
unit vector
in two dimensions, we divide a
vector
by its
magnitude
. The procedure is the same in three dimensions:
⇀v‖⇀v‖=1‖⇀v‖⟨−2,9,5⟩=1√(−2)2+92+52⟨−2,9,5⟩=1√110⟨−2,9,5⟩=⟨−2√110,9√110,5√110⟩.
Exercise 12.2.9:
Let ⇀v=⟨−1,−1,1⟩ and ⇀w=⟨2,0,1⟩. Find a
unit vector
in the direction of 5⇀v+3⇀w.
Hint
: Start by writing 5⇀v+3⇀w in
component
form.
Answer
: ⟨13√10,−53√10,83√10⟩
Example 12.2.10: Throwing a Forward Pass
A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of 30° (see the following figure). Write the
initial velocity
vector
of the ball, ⇀v, in
component
form.
Solution
The first thing we want to do is find a
vector
in the same direction as the
velocity vector
of the ball. We then scale the
vector
appropriately so that it has the right
magnitude
. Consider the
vector
⇀w extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of 30° (see the following figure). This
vector
would have the same direction as ⇀v, but it may not have the right
magnitude
.
The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is
Dist from QB to receiver =√152+202=√225+400=√625=25 yd.
We have 25‖⇀w‖=cos30°. Then the
magnitude
of ⇀w is given by
‖⇀w‖=25cos30°=25⋅2√3=50√3 yd
and the vertical distance from the receiver to the
terminal point
of ⇀w is
Vert dist from receiver to
terminal point
of ⇀w=‖⇀w‖sin30°=50√3⋅12=25√3 yd.
Then ⇀w=⟨20,15,25√3⟩, and has the same direction as ⇀v.
Recall, though, that we calculated the
magnitude
of ⇀w to be ‖⇀w‖=50√3 yd, and ⇀v has
magnitude
60 mph. So, we need to multiply
vector
⇀w by an appropriate constant, k. We want to find a value of k so that ∥k⇀w∥=60 mph. We have
‖k⇀w‖=k‖⇀w‖=k50√3 yd,
so we want
k(50√3 yd)=60 mph
k=60√350 mph / yd
k=6√35 mph / yd.
Then
⇀v=k⇀w=k⟨20,15,25√3⟩=6√35⟨20,15,25√3⟩=⟨24√3,18√3,30⟩.
Let’s double-check that ‖⇀v‖=60 mph. We have
‖⇀v‖=√(24√3)2+(18√3)2+(30)2=√1728+972+900=√3600=60 mph.
So, we have found the correct components for ⇀v.
Note
Readers who have been watching the units of measurement may be wondering what exactly is going on at this point: haven't we just mixed yards and miles per hour? We haven't, but the reason is subtle. One way to understand it is to realize that there are really two parallel coordinate systems in this problem: one gives positions down the field, across the field, and up into the air in units of yards; the other gives speeds down the field, across the field, and up into the air in units of miles per hour. The
vector
⇀w is calculated in the position coordinate system;
vector
⇀v will be in the system. Because corresponding axes in each system are parallel, directions in the two systems are also parallel, so the claim that ⇀w and ⇀v point in the same direction is correct. The constant k that we're looking for is a conversion factor between the magnitudes of these two vectors, converting from the position system to the
speed
one in the process. And as seen above, our calculation of k produces the right units for such a conversion, namely miles per hour per yard.
Exercise 12.2.10
Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of 40 mph and an angle of 45°. Write the
initial velocity
vector
of the ball, ⇀v, in
component
form.
Hint
: Follow the process used in the previous example.
Answer
: v=⟨16√2,12√2,20√2⟩
Key Concepts
The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples (x,y,z) are used to describe the location of a point in space.
The distance d between points (x1,y1,z1) and (x2,y2,z2) is given by the formula d=√(x2−x1)2+(y2−y1)2+(z2−z1)2.
In three dimensions, the equations x=a,y=b, and z=c describe planes that are parallel to the coordinate planes.
The
standard equation of a sphere
with center (a,b,c) and radius r is (x−a)2+(y−b)2+(z−c)2=r2.
In three dimensions, as in two, vectors are commonly expressed in
component
form, ⇀v=⟨x,y,z⟩, or in terms of the
standard unit vectors
, ⇀v=xˆi+yˆj+zˆk.
Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let ⇀v=⟨x1,y1,z1⟩ and ⇀w=⟨x2,y2,z2⟩ be vectors, and let k be a
scalar
.
Scalar multiplication
:
k⇀v=⟨kx1,ky1,kz1⟩
Vector addition
:
⇀v+⇀w=⟨x1,y1,z1⟩+⟨x2,y2,z2⟩=⟨x1+x2,y1+y2,z1+z2⟩
Vector
subtraction:
⇀v−⇀w=⟨x1,y1,z1⟩−⟨x2,y2,z2⟩=⟨x1−x2,y1−y2,z1−z2⟩
Vector
magnitude
:
‖⇀v‖=√x21+y21+z21
Unit vector
in the direction of ⇀v:
⇀v‖⇀v‖=1‖⇀v‖⟨x1,y1,z1⟩=⟨x1‖⇀v‖,y1‖⇀v‖,z1‖⇀v‖⟩,⇀v≠⇀0
Key Equations
Distance between two points in space:
d=√(x2−x1)2+(y2−y1)2+(z2−z1)2
Sphere
with center (a,b,c) and radius r:
(x−a)2+(y−b)2+(z−c)2=r2
Glossary
coordinate plane
: a plane containing two of the three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the xy-plane, xz-plane, or the yz-plane
right-hand rule
: a common way to define the of the three-dimensional coordinate system; when the right hand is curved around the z-axis in such a way that the fingers
curl
from the positive x-axis to the positive y-axis, the thumb points in the direction of the positive z-axis
octants
: the eight regions of space created by the coordinate planes
sphere
: the set of all points equidistant from a given point known as the center
standard equation of a sphere
: (x−a)2+(y−b)2+(z−c)2=r2 describes a
sphere
with center (a,b,c) and radius r
three-dimensional rectangular coordinate system
: a coordinate system defined by three lines that intersect at right angles; every point in space is described by an ordered triple (x,y,z) that plots its location relative to the defining axes
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at
Example 12.2.10 has been modified by Doug Baldwin and Paul Seeburger to clarify the units of measurement that it uses and how it uses them.
Paul Seeburger also created dynamic versions of Figures 12.2.8,12.2.9 and 12.2.13 using CalcPlot3D.
12.1E: Exercises for Section 12.1
12.2E: Exercises for Section 12.2 |
16164 | https://www.astro.uni-bonn.de/hisurvey/euhou/HI_Spirals_HowTo.pdf | Peter M.W. Kalberla Argelander-Institut für Astronomie Bonn Astropeiler Stockert e.V. Milky Way rotation from 21cm line profiles HI 2 Why is HI important?
• Hydrogen is the most abundant element • Neutral hydrogen (HI) fills most of the space between the stars • Because of the Doppler effect HI lines supply kinematical information for the observed HI clouds (center velocities), even information about internal motions (profile width) 3 Element abundance, Solar & Cosmic 4 Hydrogen is by far the most abundant element • Very basic ideas: – Assuming that the hydrogen atoms are uniformly distributed throughout the galaxy, we might expect to find hydrogen for each line of sight – Stars are expected to be born out of gas condensations but some gas should have been left over (the interstellar medium, ISM) – Hydrogen is a gas and free to move. This gas however should have settled in the gravitational potential (caused by the stellar disk). In analogy to planetary atmospheres we even might expect to find a “Galactic atmosphere” around the disk. – Line emission would be best for a diagnosis • Each of the observed lines would be Doppler shifted. Hence, one could be able to see the rotation of the Milky.
– Radio lines would be important since they would not be obscured by dust 5 The quest for a hydrogen line (1944) • Oort (Professor in Leiden) had spent many years studying the rotation and structure of the Galaxy using optical means. He was frustrated by the extensive clouds of dust lying in the galactic plane, which block visible light. But radio waves will penetrate the dust and show us the galactic center and indeed the opposite side of the galaxy. One can then study the differential rotation of the galaxy and estimate distances to gas clouds, and thus map the distribution of matter in the galaxy. • Radio science has in 1944 advanced to a promising field after detections of radio emission from the Milky Way by Jansky and Reber.
• Oort assigned his student, H.C. Van de Hulst, the job of figuring out what radio spectral lines might exist and what their frequencies would be. Since hydrogen is the most abundant element in the universe, he started his studies with hydrogen. He found that a "hyperfine" transition in the ground state of neutral hydrogen would produce radiation in the radio range, at a frequency of 1420 MHz, or about 21 cm wavelength. 6 1944: Van der Hulst reports about the HI line… van de Hulst & Oort 7 The hyperfine splitting of the ground state 2S is the source of the 21 cm hydrogen line. Frequency 1420.40575177 MHz 8 Properties of the HI 21-cm line • Magnetic dipole emission is a “forbidden” transition. This means, the transition probability is extremely low. • The lifetime of an exited state can be derived from the Einstein coefficient A10 – τ = 1/ A10 ~ 3.5·1014 s ~ 11 million years – The frequency is extremely well defined • Observed line reflects motion of the gas (Doppler effect) – Center velocity - rotation curve – radial motion – Line width - temperature and turbulence in the ISM • Excitation by collisions • Line detected 1951 in US, Australia, and the Netherlands 9 “Doc” Ewen – receiver and feed (1951) 10 The potential of 21-cm line data for 3D studies of the Milky Way was realized first by Oort 1958 11 Today it is easy to browse HI data 12 Data access through AIfA web interface Access to profiles and to the rotation curve 13 How to get profiles Simply select a position in equatorial or Galactic coordinates Example: l = 90o, b = 0o in Galactic coordinates 14 You get the observed HI profile 15 Now you may explore the Milky Way in HI But: you need to take the geometry into account The Sun is offset from the Galactic center 16 Galactic coordinates in longitude l and latitude b 17 l = 90o, b = 0o, we look in direction Cepheus l = 90o Cepheus l = 0o Sag A 18 The thin disk - looking perpendicular to the disk Radial velocity Vr (km/s) T (K) 19 Exercise • Select several positions in the Galactic plane ( l = 0o ) – For each constant longitude vary the latitude b and compare profiles and calculated column densities – Plot the observed column densities as function of latitude |b| – Explain the result • What causes the emission around v = 0 km/s?
• What is the reason for the other lines?
• Can you find a function that fits best to the observed column densities?
20 Rotation - looking around in the disk 21 HI in differential rotation • For a rotating disk we expect different velocities for clouds along the line of sight.
• Assuming circular rotation, the distance of HI clouds from the Galactic center (or from the Sun) could then be calculated. • The rotation curve needs to be known.
• In the inner Galaxy the rotation at the tangent points is unambiguous (terminal velocity).
van Dishoek lecture 8 22 Calculate observed radial velocity for cloud at M Most important is the knowledge of the Galactic constants R0 and V0 (called Oort constants) A good approximation is V = V0 (flat rotation curve) From with one gets: 23 Rotation curve of the Milky Way Fich et al. 1989 24 Rotation curves of a few other galaxies: flat Flat rotation inconsistent with Kepler law Implies Dark Matter 25 MW rotation decomposition: Dark Matter needed Dark matter brings the rotation curve up 26 Search for spiral arms • Spiral arms cause density enhancements in the HI distribution • Such enhancements are observable as peaks in the HI line, hence peaks may indicate spiral structure – Caution: gas along a long distance can be "crowded" into a small velocity interval, mimicking also spiral structure 27 Exercise • Search for HI peaks in profiles at latitude b = 0o • Determine their velocities and positions according to the rotation curve – Use ASCII tables provided by the web interface • Plot positions found – Use the Galactic grid • Trace spiral arm features in longitude • Can you also trace spiral arms in latitude?
28 Step by step • Select profile (e.g. l = 90o, b = 0o) • Check velocity of a peak in the profile – Download HI profile as ASCII file (For Windows use wordpad) – Read off velocity • From rotation curve get corresponding position in x, y – Download rotation curve as ASCII file (For Windows use wordpad) – Read x, y • Plot position and think about a new position…. 29 How to - using ASCII data for l=90o, b=0o v_lsr T_B freq. wavel. -80.38 33.84 1420.787 21.10046 -79.35 34.78 1420.782 21.10053 -78.32 35.19 1420.777 21.10060 -77.29 35.37 1420.772 21.10067 -76.26 35.53 1420.767 21.10075 -75.23 35.59 1420.762 21.10082 -74.20 35.54 1420.757 21.10089 -73.17 35.01 1420.752 21.10097 -72.14 34.07 1420.747 21.10104 Dist v_rot R z azi x y 8.5 -65.0 12.0 0.0 135.0 8.5 8.5 9.0 -69.5 12.4 0.0 133.3 9.0 8.5 9.5 -73.9 12.8 0.0 131.8 9.5 8.5 10.0 -78.1 13.1 0.0 130.4 10.0 8.5 10.5 -82.2 13.5 0.0 129.0 10.5 8.5 11.0 -86.2 13.9 0.0 127.7 11.0 8.5 v = 75 km/s x = 9.5 kpc, y = 8.5 kpc 30 EU Hands-On Universe exercise Sonne Galaktisches Zentrum Perseus Arm mm paper is available at: 31 More accurate fitting with Gaussians 32 Fitted Parameters (gnuplot fit, 6 components) Final set of parameters Asymptotic Standard Error ======================= ========================== a1 = 35.1037 +/- 0.828 (2.359%) v1 = 6.10903 +/- 0.06736 (1.103%) s1 = 3.73012 +/- 0.0891 (2.389%) a2 = 49.9281 +/- 0.572 (1.146%) v2 = -2.81713 +/- 0.1982 (7.035%) s2 = 10.2931 +/- 0.1194 (1.16%) a3 = 37.9548 +/- 0.9601 (2.53%) v3 = -24.1013 +/- 0.0942 (0.3908%) s3 = 4.82079 +/- 0.1146 (2.377%) a4 = 59.9751 +/- 1.073 (1.789%) v4 = -43.6551 +/- 0.18 (0.4123%) s4 = 10.5102 +/- 0.2965 (2.821%) a5 = 16.7286 +/- 1.439 (8.6%) v5 = -76.383 +/- 0.2554 (0.3343%) s5 = 6.18121 +/- 0.3675 (5.945%) a6 = 19.8456 +/- 1.284 (6.471%) v6 = -80.443 +/- 1.815 (2.256%) s6 = 19.2644 +/- 0.7706 (4%) The v = -75 km/s estimate was not too bad!
33 The Milky Way spiral pattern, artist impressions Caveat: check orientation and position of Sun and Galactic center Radio Infrared Early data 1958 Spitzer Space Telescope 34 Care about astronomical conventions…..
35 Spitzer presentation • Missing Spiral Arms: Following the demotion of Pluto, it seems that rewriting the nature of the Universe has become part and parcel of an astronomer's job. Well, now scientists working with the Spitzer Space Telescope have removed two arms from the Milky Way Galaxy. Hello, I'm Daniel Brennan. ts/20080613transcript.pdf • Also putting the Galaxy upside down, great!
• For critical discussion see 36 This might be a better MW spiral model (in agreement with asymmetries found in HI) |
16165 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6565404/ | An overview of urethral injury - PMC
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Can Urol Assoc J
. 2019 Jun;13(6 Suppl4):S61–S66. doi: 10.5489/cuaj.5931
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An overview of urethral injury
R Christopher Doiron
R Christopher Doiron, MD
1 Division of Urology, University of Alberta, Edmonton, AB, Canada
Find articles by R Christopher Doiron
1, Keith F Rourke
Keith F Rourke, MD, FRCSC
1 Division of Urology, University of Alberta, Edmonton, AB, Canada
Find articles by Keith F Rourke
1,✉
Author information
Copyright and License information
1 Division of Urology, University of Alberta, Edmonton, AB, Canada
✉
Correspondence: Dr. Keith F. Rourke, Division of Urology, University of Alberta, Edmonton, AB, Canada; krourke@ualberta.ca
✉
Corresponding author.
Copyright: © 2019 Canadian Urological Association or its licensors
PMC Copyright notice
PMCID: PMC6565404 PMID: 31194929
Case.
A 37-year old male involved in a high-speed motor vehicle accident is brought to the emergency department for assessment. As the belted passenger of the vehicle, he sustained blunt trauma to the lower abdomen. On initial assessment by the trauma team, the patient is hemodynamically stable but unable to void. A plain-film assessment of the pelvis reveals right-sided pubic rami fractures (Fig. 1) with some associated soft tissue swelling. The trauma team seeks urological assessment.
On physical exam, the patient’s bladder is palpable and distended in the midline. There is blood at the urethral meatus; the prostate is impalpable on digital rectal exam, while a “butterfly” hematoma is observed over the scrotum and perineum.
Retrograde urethrogram (RUG) performed at the bedside reveals extravasation of contrast at the level of the bulbomembranous urethra — a complete urethral disruption with no contrast visualized in the bladder (Fig. 2).
Introduction
Urethral injuries can first be classified based on location as either anterior or posterior. Anterior urethral injuries are often the result of blunt or penetrating trauma1 with the bulbar urethra the most common location affected.2 The mechanisms of injury ranges from gunshot wounds to self-inflicted sexual misadventures. In contrast, posterior urethral injuries are most commonly the result of pelvic fracture or iatrogenic trauma during pelvic surgeries. Pelvic-fracture associated urethral injuries (PFUI) are present in 1.5% –10%3,4 of injuries resulting in fracture of the pelvis. High-speed motor vehicle accidents are the most common mechanism of injury here,4–6 though fall from height is another common mechanism resulting in pelvic fracture and PFUI. Given the relative complexity of PFUI and fact that it’s three times more common in the setting of trauma than anterior urethral injury,1 PFUI will be the focus of the current review.
Diagnosis of a suspected urethral injury
Clinical signs/symptoms
Assessment of a patient suspected of urethral trauma should follow a standard approach to any trauma patient. Given that patients suspected of PFUI are often complex in presentation with multiple associated injuries, additional referring services are often involved in care. The most responsible team at time of initial assessment is often the trauma service, therefore, thorough communication and collaboration is of utmost importance in providing the best urological care.
After hemodynamic stability is confirmed, a thorough genitourinary (GU) exam is required. Blood at the urethral meatus is a common finding7 in the presence of a urethral injury and PFUI, although its absence does not rule one out. Other findings on inspection may include ecchymosis of the scrotum and/or perineum, while in women, urethral injury should be suspected in the presence of labial edema or blood at the introitus. Given that urinary retention is another common presenting symptom of PFUI,7 the bladder may be distended and palpable on abdominal exam. Digital rectal exam (DRE) findings in the setting of urethral trauma may vary from impalpable, particularly in the acute phase of PFUI7,8 secondary to presence of a pelvic hematoma obscuring the prostate to palpation,9 or “high-riding” in the case of a partial or complete posterior urethral disruption. Furthermore, DRE should be performed to rule out a rectal injury.8
Imaging and flexible cystoscopy
Plain-film or scout fluoroscopy imaging of the pelvis can identify pelvic fracture or presence of foreign bodies in the GU tract10 but has no role in diagnosing urethral injury directly. Furthermore, although potentially useful in the acute trauma setting for ruling out concomitant injuries or in the scenario where a suprapubic catheter (SPC) is to be placed,1 there is no role for use of ultrasound, computed tomography (CT), or magnetic resonance imaging (MRI) in the diagnosis of PFUI in the acute setting.
Retrograde urethrogram (RUG) is the diagnostic imaging study of choice in the setting of suspected urethral injury. Partial disruption of the urethra is suggested with active extravasation of contrast with simultaneous bladder filling, while a complete disruption is suggested in the setting of contrast extravasation without visualization of contrast in the bladder.7 While the American Association for the Surgery of Trauma (AAST) staging of urethral injury11 remains the gold standard staging system, Chapple et al1 suggested a simpler, practical classification system based on RUG findings in their 2004 consensus statement on urethral injury. A RUG will help identify the location and extent of urethral injury and can be helpful in guiding subsequent management decisions.
Bedside flexible cystoscopy may be performed in the acute setting of suspected urethral injury as a diagnostic adjunct with potential for intervention; it can help identify a partial vs. complete urethral injury while primary realignment (PR) with a urethral catheter may be performed with placement of a wire across the injured area.12 Bladder drainage is paramount in the acute management of PFUI and PR has been associated with improved outcomes compared with SPC drainage only in terms of eventual stricture severity.13,14 Thus, an attempt at PR via bedside flexible cystoscopy is a reasonable and commonly used manoeuvre, recognizing that in the setting of an unstable patient, bladder drainage with SPC placement may be more appropriate (see section on “Immediate primary realignment vs. SPC placement” below).
Guideline review
In the described case, a complete PFUI is confirmed with RUG following an assessment with clinical signs and symptoms consistent with a urethral injury. The American Urological Association (AUA) urotrauma guideline15 is clear on their statement suggesting assessment with RUG when there is blood at the urethral meatus in the setting of pelvic trauma.
The European Association of Urology (EAU) urological trauma guideline similarly recommends evaluation of suspected urethral injury with RUG as the gold standard.16
From a Canadian perspective, the authors suggest evaluation of suspected urethral injury with RUG in the acute setting is appropriate; alternatively, the use of flexible cystoscopy is acceptable in appropriately experienced hands. These findings help guide subsequent management decisions.
Management of PFUI
When treating urethral injury, timing of the various interventions is often described as immediate: <48 hours; delayed: 2–14 days; deferred: >3 months.1,9 In the immediate setting, when treating a suspected urethral injury, there is no evidence to suggest that a partial urethral injury may be converted to complete disruption. Furthermore, no evidence exists to suggest that attempts at catheterization increase subsequent risk of infection or stricture.7,9 In practice, urethral catheterization has often been attempted by the initial assessment team. An attempt at passage of a urethral catheter by an experienced practitioner in the setting of suspected urethral injury is a reasonable first step.
Partial posterior urethral injury
Placement of a urethral catheter in the setting of a partial urethral injury may result in complete healing without need for further intervention. The injury should be evaluated with urethrography at two-week intervals until healing has occurred.17,18
Complete posterior urethral injury
Case cont’d: Acute treatment (complete posterior urethral injury).
Bedside cystoscopy is unsuccessful at placing an aligning catheter and the patient is brought to the operating room, where an open cystotomy is performed with combined retrograde and antegrade endoscopic placement of an aligning urethral catheter. Postoperatively, a SPC is left in place in addition to an aligning catheter.
Immediate urethroplasty
Immediate urethroplasty in the setting of PFUI results in an unacceptably high rate of stricture, erectile dysfunction (ED), and incontinence18 and should not be attempted. Furthermore, exploring PFUI in the acute setting risks disrupting a stable pelvic hematoma and has the potential to cause significant bleeding.9
Immediate primary realignment vs. suprapubic catheter placement
There is controversy among urologists involved in managing GU trauma with regard to the optimal acute management of PFUI. Immediate PR via an endoscopic approach by placing a wire into the bladder and Foley catheter placement over the wire establishes continuity of the lower urinary tract. This can be achieved using flexible and/or rigid cystoscopy, using retrograde or combined retrograde and antegrade cystoscopic approaches.19
Evidence guiding practice in this domain is largely from small case series and retrospective studies. Literature supporting PR suggests that achieving realignment in the immediate setting lowers the degree and severity of stricture formation, minimizes fibrosis and scarring at the distraction site, improves the success and ease of future urethroplasty efforts and, in some cases, may negate the need for urethroplasty altogether.19–25 Those critical of this approach, however, have pointed out that a selection bias may be introduced into these studies, as those in whom immediate PR is successful may represent a population of PFUI patients with less severe injuries, who may have experienced improved outcomes either way.26
On the contrary, proponents of SPC placement alone suggest this technique minimizes prolonged SPC drainage and minimizes length of time to definitive urethroplasty. It has also been shown to be associated with less ED and urinary incontinence.19,27 We await the results of a prospective, randomized cohort study designed to help answer this challenging clinical question.26
Delayed primary urethroplasty
Although there is minimal evidence in this space, some small, retrospective case series have shown reasonable results in terms of stricture recurrence, ED, and continence outcomes.28–31 Given the excellent results with a deferred approach, further evidence in larger patients populations, ideally in a prospective manner, would be required before making recommendations on this challenging clinical approach.
Guideline review
In the scenario described in our case — complete traumatic urethral disruption — the AUA guideline states that urinary drainage should be achieved promptly in patients with PFUI, and that an attempt at PR may be performed in hemodynamically stable patients. They warn, however, that prolonged attempts at PR should be avoided and that the primary goal should remain prompt drainage, if not by PR then by SPC.15
The EAU guideline has a more thorough discussion of treatment options in the setting of PFUI with complete disruption, as in the case presented. They do not, however, commit to recommending one approach over the other, and discuss either PR or SPC with deferred definitive treatment as acceptable options. They state that immediate urethroplasty cannot be recommended given the current state of the evidence and potential risk of suboptimal outcomes and complications.16
Case followup.
The patient is discharged from the hospital and brought to the office for urological evaluation six weeks following his injury. His urethral catheter is removed, his SPC is plugged, and the patient undergoes an attempted trial of void. The patient is unable to void and subsequently the SPC is left in situ for drainage.
There are no specific statements or recommendations from either the AUA or EAU guideline regarding length of catheterization following PR. The EAU guideline, however, does note that in most reported series of PR, catheterization lasted from 4–8 weeks prior to an observed trial of void. The authors would agree that a period of catheterization of at least four weeks following PR is warranted.
At 12 weeks post-injury, a RUG is performed. RUG combined with an attempted voiding cysto-urethrogram reveals complete urethral obliteration at the bulbomembranous junction with suggestion of a long urethral defect (Fig. 3). Subsequent combined RUG and antegrade cystoscopy through the suprapubic tract — termed an “up-and-downogram” — delineates a 2 cm stenosis at the bulbomembranous junction (Fig. 4). With the stricture length and location confirmed, the patient is booked for a posterior urethroplasty via a perineal approach, which he undergoes uneventfully at five months post-injury.
The authors acknowledge the ongoing debate regarding PR vs. SPC drainage alone and agree that one cannot be considered superior over the other at this point. From a Canadian perspective, in a resource-limited operative setting, the authors recognize that there may be barriers to PR should it require an operative setting, and support either strategy as an acceptable approach as we await results from a prospective trial29 looking to provide further guidance in this clinical scenario.
Deferred urethroplasty
An approach of deferred urethroplasty — considered as primary urethroplasty >3 months following initial PFUI — would be considered the favoured approach to complete urethral disruption and obliteration secondary to PFUI by most,32 with success rates reported as high as 85– 97%.33,34 A well-established perineal approach to posterior urethroplasty for PFUI35 describes four manoeuvres in order to bridge the distraction defect: extensive bulbar urethral mobilization, splitting of the corporal bodies, inferior pubectomy, and corporal rerouting, performed in a stepwise approach as necessary to achieve a tension-free anastomosis.
Guideline review
The AUA guidelines do not specifically make a statement regarding treatment for stricture following PR. However, they do make a vague suggestion that either urethroplasty or direct visual internal urethrotomy (DVIU) would be acceptable options. The approach selected would depend on the length, location, and density of the stricture, as well as patient preference.15
The EAU guidelines suggest that for those requiring deferred urethroplasty, standard treatment would be a single-stage perineal approach, after waiting for a minimum of three months. They further suggest that short, non-obliterative strictures following PR can be managed endoscopically with DVIU or urethroplasty, and that it should be recognized that repeated attempts at endoscopic management or dilation would be considered palliative and should be avoided when possible.16
In the current case, the patient has a completely obliterated urethral stenosis and in this specific instance, endoscopic procedures are very unlikely to offer a durable response. The AUA guideline on urethral stricture36 recommends that delayed urethroplasty should be performed instead of endoscopic procedures after urethral obliteration due to PFUI. In the case of complete urethral obliteration after PFUI, “cut to the light” procedures are rarely successful outside of the short-term and have a significant risk of transfusion and rectal injury. Repeated endoscopic maneuvers, including intermittent catheterization, should be avoided because they are not successful in the majority of PFUI, increase patient morbidity, and may delay the time to anastomotic reconstruction.
The authors suggest that treatment of stricture recurrence following PR should be a shared decision with an informed patient, recognizing the risks and benefits of formal urethroplasty vs. endoscopic management vs. prolonged catheterization. From a Canadian perspective, the authors recognize the potentially prolonged catheterization time many patients experience as they await referral to a urologist experienced in performing posterior urethroplasty. This may result in a disproportionate amount of attempts at endoscopic management but is the reality of our resource-limited system. Presuming an informed discussion with the patient and recognition of the palliative nature of repeated endoscopic attempts, this may be an appropriate approach. However, for those in whom formal urethroplasty is a viable option, and should the patient be in agreement, the authors prefer this approach, given the high success rate and definitive nature of the treatment.
Complications of PFUI
It is of utmost importance that PFUI patients receive adequate followup, as many will develop GU complications as a result of their injury; these complications may include stricture formation, urethrocutaneous fistula, ED, urinary incontinence, and pain. In their guidelines, the AUA states that patients should be surveyed for at least one year using some combination of uroflowmetry, RUG, and/or cystoscopy. The EAU guidelines make no specific reference to followup after PR for PFUI. The authors concur with the AUA recommendation that patients should receive close urological followup for at least one year following PFUI.
The debate regarding potential complications of PFUI is perhaps most present in the setting of the discussion around immediate SPC placement vs. primary realignment for treatment of PFUI with complete posterior urethral disruption (see section on “Immediate primary realignment vs. suprapubic catheter placement” for discussion and review of evidence).
Anterior urethral injury
Blunt injuries of the anterior urethra
Blunt injuries to the anterior urethra, often the result of a “straddle injury” to the bulbar urethra, result in significant contusion to the spongiosus with possible significant hematoma of the perineum. For these reasons, immediate repair is contraindicated.1 The AUA and EAU trauma guidelines15,16 are in agreement that SPC drainage or PR with urethral catheter are both options in the acute setting. Depending on the degree of injury, patients may re-canalize with PR alone, however, they should be followed closely given the risk of anterior urethral stricture formation.
Penetrating anterior urethral injury
Penetrating injuries of the anterior urethra, in the absence of other complicating factors, should be explored immediately with attempted primary repair. Injured tissue should be debrided and urethral ends spatulated with primary reanastomosis. 15,16 Some have suggested the defects of 2–3 cm of the bulbar urethra and up to 1.5 cm of the penile urethra are amenable to spatulation and primary anastomosis.9
Summary
Posterior urethral injuries in the setting of pelvic fracture can be challenging cases to manage in the acute setting, with challenging clinical decision-making in potentially unstable patients with significant comorbid injuries. These patients’ subsequent risk of complications, regardless of immediate treatment decisions, can be devastating and they require close attention and followup. Given that long-term outcomes with respect to urethral patency remain acceptable in the hands of those experienced in managing posterior urethral injury, these practitioners should be sought out for patient care assistance when they are not available in the acute setting.
Fig. 1.
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Pelvic x-ray obtained during the initial trauma assessment demonstrating pelvic rami fractures raising the suspicion for a potential pelvic fracture urethral injury (PFUI).
Fig. 2.
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Results from a retrograde urethrogram performed at initial assessment of the patient in the emergency department. Contrast extravasation can be seen at the level of the bulbomembranous urethra, while no contrast is seen beyond the site of injury, suggesting complete urethral disruption.
Fig. 3.
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Results of a combined retrograde urethrogram and attempted voiding cystourethrogram 12 weeks following pelvic fracture urethral injury and six weeks following removal of aligning urethral catheter. There is complete bulbomembranous urethral obliteration with the suggestion of a lengthy but poorly delineated urethral defect due to closure of the bladder neck.
Fig. 4.
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Results of combined antegrade cystoscopy (via the suprapubic tract) with simultaneous retrograde urethrogram, known as an “up-and-downogram.” This confirms complete bulbomembranous urethral obliteration with a 2 cm posterior urethral defect.
Footnotes
Competing interests: Dr. Rourke has participated in advisory board meetings for and is a shareholder of Boston Scientific; and has participated in clinical trials supported by Red Leaf Medical. Dr. Doiron reports no competing personal or financial conflicts.
This paper has been peer reviewed.
References
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16166 | https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Quantum_Mechanics_(Walet)/12%3A_Quantum_Mechanics_of_the_Hydrogen_Atom/12.05%3A_Smaller_Effects/12.5.02%3A_The_Lamb_Shift | Skip to main content
12.5.2: The Lamb Shift
Last updated
: Nov 3, 2024
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12.5.1: Hyperfine Structure
12.5.3: The Zeeman Effect
Page ID
: 5163
Niels Walet
University of Manchester
( \newcommand{\kernel}{\mathrm{null}\,})
According to Dirac and Schrödinger theory, states with the same and quantum numbers but different quantum numbers ought to be degenerate. However, a famous experiment by Lamb and Retherford in 1947 showed that the and states of the hydrogen atom were not degenerate, but that the state had slightly higher energy by an amount now known to be . The effect is explained by the theory of quantum electrodynamics, in which the electromagnetic interaction itself is quantized. Some of the effects of this theory which cause the Lamb shift are shown in the Feynman diagrams of figure 5. Table 3 shows how much each of these contribute to the splitting of and . The most important effect is illustrated by the center diagram, which is a result of the fact that the ground state of the electromagnetic field is not zero, but rather the field undergoes "vacuum fluctuations" that interact with the electron. Any discussion of the calculation is beyond the scope of this paper, so the answers will merely be given. For ,
where is a numerical factor which varies slightly with from 12.7 to 13.2. For ,
for , where is a small numerical factor which varies slightly with and . Notice that the Lamb shift is very small except for .
Table : Contribution of different effects to the energy splitting of and in hydrogen. Numbers are given in units of frequency .
| Effect | Energy contribution |
| Vacuum polarization | -27 MHz |
| Electron mass renormalization | +1017 MHz |
| Anomalous magnetic moment | +68 MHz |
| Total | +1058 MHz |
Contributors and Attributions
Randal Telfer (JWST Astronomical Optics Scientist, Space Telescope Science Institute)
12.5.1: Hyperfine Structure
12.5.3: The Zeeman Effect |
16167 | http://www.icas.org/icas_archive/ICAS2010/PAPERS/434.PDF | 27TH INTERNATIONAL CONGRESS OF THE AERONAUTICAL SCIENCES 1 Abstract Various propeller theories are treated in developing a model that analyses the aerodynamic performance of an aircraft propeller along with the construct and behaviour of the resultant slip-stream. Blade element momentum theory is used as a low-order aerodynamic model of the propeller and is coupled with a vortex wake representation of the slip-stream to relate the vorticity distributed throughout the slip-stream to the propeller forces. Nomenclature a axial induction factor ′ a tangential induction factor c chord length (m) CD drag coefficient CL lift coefficient CT thrust coefficient dr blade element and annulus width (m) dQ torque of element or annulus (Nm) dT thrust of element or annulus (Nm) F combined tip- and hub-loss coefficient I momentum of inertia N in-plane normal force (N) p pressure (P) r radial position (m) R propeller radius (m) Rhub hub radius (m) S propeller disk/cross-sectional slip-stream area (m2) t time (s) U resultant velocity at blade element (m/s) V flow velocity (m/s) x thrust axis location downstream from propeller (m) α local angle of attack (rad) β local element pitch angle (rad) γ strength of distributed annular vorticity (m/s) γ s strength of distributed swirl vorticity (m/s) Γ strength of bound vorticity on propeller (m2/s) blade φ angular coordinate around propeller disk (rad) ϕ local inflow angle (rad) ′ σ local solidity ρ air density (kg/m3) Ω propeller rotational speed (rad/s) Subscripts: ∞ free-stream x at station x p propeller/aircraft body s fully-developed slip-stream 0 at φ = π/2 F between thrust line and direction of resultant propeller force 1 Introduction An understanding of the propeller slipstream is important in the design phase of propeller-driven aircraft. Aerodynamic performance, aircraft stability, and noise and vibration can be significantly affected by it’s interactions with PROPELLER BLADE ELEMENT MOMENTUM THEORY WITH VORTEX WAKE DEFLECTION M. K. Rwigema School of Mechanical, Industrial and Aeronautical Engineering University of the Witwatersrand, Private Bag 3, Johannesburg, 2050, South Africa Keywords: Blade Element Momentum, Propeller, Vortex Wake, Wake Deflection M.K. Rwigema 2 structural and aerodynamic control surfaces. During the early stages of design, one of the inputs is the thrust curve of the power plant - available thrust versus flight velocity. The iterative optimisation procedure requires rapid analysis of prospective engines and propellers to compare aircraft performance and discover configuration characteristics. A propeller producing thrust by forcing air behind the aircraft produces a slip-stream. This may rudimentarily be considered as a cylindrical tube of spiraling air propagating rearward over the fuselage and wings – having detrimental and beneficial effects. Flow within the slip-stream is faster than free-stream flow resulting in increased drag over the parts exposed to it’s trajectory. The rotary motion of the slip-stream also causes the air to strike the tailplane at indirect angles, having an effect of the stability and control of the aircraft. This analysis shows compatibility between different approaches to propeller modeling using blade element theory to predict the propeller forces, momentum theory to relate the flow momentum at the propeller to that of the far wake, and a vortical wake model to describe the slipstream deflection. 2 Mathematical Models 2.1 Momentum Theory Classical momentum theory, introduced by Froude as a continuation of the work of Rankine , imposes five simplifying assumptions. The flow is assumed to be: 1) inviscid, 2) incompressible, and 3) irrotational; and both 4) the velocity and 5) the static pressure are uniform over each cross section of the disk and stream-tube. Fig. 1: Propeller stream-tube For the axial direction, the change in flow momentum along a stream-tube starting upstream, passing through the propeller, and then moving off into the slip-stream must equal the thrust produced by this propeller, T = ρπ Rs 2Vs(Vs −V∞) (1.1) Applying the conservation of mass to the stream-tube control volume yields, A ∞V∞= AV = A sVs (1.2) Far upstream the static pressure is p∞ and the velocity V∞. The static pressure falls to a value p- immediately upstream of the actuator disk and rises discontinuously through the disk, to a value p+ immediately downstream, although the velocity remains constant at V between these two planes, which are an infinitesimal distance apart. The thrust is therefore given by, T = ( p+ −p−)A (1.3) Since the flow is inviscid, the total pressure - in accordance with Bernoulli’s equation - is constant along any streamline, except for those that pass through the disk. The axial and tangential velocities are continuous as no fluid is created within the disk; the increase in total pressure is therefore experienced as an increase in static pressure. Applying Bernoulli’s equation upstream and downstream of the actuator disk, p+ −p−= 1 2 ρ(Vs 2 −V∞ 2) (1.4) From the equations above it can be shown that the velocity of the flow through the actuator disk is the average of the upstream and downstream velocities. V = V∞+Vs 2 (1.5) 3 PROPELLER BLADE ELEMENT MOMENTUM THEORY WITH VORTEX WAKE DELFECTION An axial induction factor, a, is customarily defined as, V = V∞(1+ a) (1.6) and from Equation (1.5), Vs = V∞(1+ 2a) (1.7) A real propeller, however, is never uniformly loaded as assumed by the Rankine-Froude actuator disk model. In order to analyse the radial load variation along the blades, the angular momentum being imparted to the wake by the propeller must be considered. 2.1.1 Effects of Wake Rotation The conservation of angular momentum necessitates rotation of the slip-stream if the propeller is to impart useful torque. Jonkman describes how the Rankine-Froude actuator disk model may be modified to account for this rotation. As such, assumptions 4) and 5) from the actuator disk model may be relaxed and three supplementary assumptions added: 1. The flow entering the control volume far upstream remains purely axial and uniform 2. The slip-stream can be split into a series of non-interacting, annular stream-tube control volumes 3. The angular velocity of the slip-stream flow far downstream of the propeller is low so the static pressure can be assumed to be the unobstructed ambient static pressure. Although wake rotation is now included in the analysis, the assumption that the flow is irrotational has not been lifted. Conserving angular momentum about an axis consistent with the slip-stream’s axis of symmetry can be applied to determine the torque. Q = dL dt = dIω dt = dm dt ωr 2 (1.8) 2.2 Blade Element Theory Unless some state of flow is assumed, momentum theory does not provide enough equations to solve for the differential propeller thrust and torque at a given span location. Additional equations , governing the state of the flow are dependent on the characteristics of the propeller blades, such as airfoil shape and twist distribution. Blade Element (BE) theory uses these geometrical properties to determine the forces exerted by a propeller on the flow-field. Fig. 2: Blade element aerodynamic forces As the word element in the title suggests, BE theory, again, uses several annular stream-tube control volumes. At the propeller plane, the boundaries of these control volumes effectively split the blade into a number of distinct elements, each of width dr. At each element, blade geometry and flow-field properties can be related to a differential propeller thrust, dT, and torque, dQ, if the following assumptions are made: 1. Just as the annular stream-tube control volumes used in the slip-stream rotation analysis were assumed to be non-interacting [assumption (1)], it is assumed that there is no interaction between the analyses of each blade element 2. The forces exerted on the blade elements by the flow stream are determined solely by the two-dimensional lift and drag characteristics of the blade element airfoil shape and orientation relative to the incoming flow. As we are required to obtain the local angle-of-attack to determine the aerodynamic forces on a blade element, we must first determine the inflow angle based on the two components of the local velocity vector. tanϕ = V∞(1+ a) Ωr(1−′ a ) (2.1) From Fig. 2 the following relation is apparent, M.K. Rwigema 4 U = V∞(1+ a) sinϕ (2.2) The induced velocity components in Equations (2.1) and (2.2) are a function of the forces on the blades and blade element momentum theory is used to calculate them. From Fig. 2, the thrust distributed around an annulus of width dr is equivalent to, dT = 1 2 BρU 2(CL cosϕ −CD sinϕ)cdr (2.3) and the torque introduced by each annular section is given by, dQ = 1 2 BpU 2(CL sinϕ + CD cosϕ)crdr (2.4) 2.3 Blade Element Momentum Theory Combining the results of the previous two analyses enables a model of the performance of a propeller whose airfoil properties, size, and twist distribution are known. This analysis is based on the differential propeller thrust, dT, and torque, dQ, derived from momentum theory and blade element theory, being equivalent. Equations (2.3) and (2.4) are made more useful by noting that ϕ and U can be expressed in terms of induction factors etc , . Substituting Equation (2.2) and carrying out some algebra, dT = ′ σ πρV∞ 2(1+ a)2 sin2ϕ CL cosϕ −CD sinϕ ( )rdr (3.1) dQ = ′ σ πρV∞ 2(1+ a)2 sin2ϕ CL sinϕ + CD cosϕ ( )r 2dr (3.2) 2.3.1. Prandtl Loss Correction The effect on induced velocity in the propeller plane is most pronounced near the tips of the blades. The original blade element momentum theory does not consider the influence of vortices shed from the blade tips into the slip-stream on the induced velocity field. These tip-vortices create multiple helical structures in the wake, (as seen in Fig. 3), and play a major role in the induced velocity distribution along the propeller. To compensate for this deficiency in BEM theory, a tip-loss (or correction) factor, F, originally developed by Prandtl is used. As a blade has a suction-surface and a pressure-surface, air tends to flow over the blade tip from the lower (pressure) surface to the upper (suction) surface, effectively reducing the resulting forces in the vicinity of the tip. This theory is summarized by a correction to the induced velocity field and can be expressed simply by the following: F = 2 π cos−1 e−f (3.3) where, ftip = B 2 R −r r sinϕ (3.4) Much like the tip-loss model, the hub-loss model serves to correct the induced velocity resulting from a vortex being shed near the hub of the propeller. The model uses a similar implementation to that of the tip-loss to describe the effect of this vortex, replacing Equation (3.4) with, fhub = B 2 r −Rhub r sinϕ (3.5) For a given element, the local aerodynamics may be affected by both the tip- and hub-loss, in which case the two correction factors are multiplied to create the total loss factor. Now, to relate the induced velocities in the propeller plane to the elemental forces of Equations (3.1) and (3.2), the conservation of momentum in the axial direction (between the far upstream and the far downstream section) must be considered. This states that the thrust introduced by each propeller annulus is equivalent to the change in axial momentum flow rate. The correction factor is used to modify the momentum segment of the BEM equations. From Equations (1.3), (1.4), and (1.7) , considering a radial differential, dT = 4πrρV∞ 2(1+ a)aFdr (3.6) Considering a radial differential of Equation (1.8) and substituting Equation (1.6), 5 PROPELLER BLADE ELEMENT MOMENTUM THEORY WITH VORTEX WAKE DELFECTION dQ = 4πr3ρV∞Ω(1+ a) ′ a Fdr (3.7) Equalising Equations (3.6) and (3.7) with Equations (2.3) and (2.4), respectively, it is possible to obtain, a = 4F sin2ϕ σ(CL cosϕ −CD sinϕ) −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ −1 (3.8) ′ a = 4F sinϕ cosϕ σ(CL sinϕ −CD cosϕ) +1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ −1 (3.9) Thus, when we include two-dimensional airfoil tables of lift and drag coefficient as a function of the angle-of-attack, α, we have a set of equations that can be iteratively solved for the induced velocities and the forces on each blade element. 2.4 Vortex Structure of the Wake Propeller theory generally presumes that, after an initial distortion, the vortex sheets shed from the trailing edges of the propeller blades form a set of interleaved helicoidal sheets which propagate uniformly downstream parallel to the slip-stream axis without further deformation as if they were rigid surfaces. In reality, these sheets will soon roll up into a set of helical vortex filaments and a central vortex filament of opposite sense on the axis. The helicoidal vortex sheets are floating freely in an irrotational field with equal velocity on either side of the sheet, hence equal pressure. There is also no discontinuity of normal velocity, only a discontinuity of tangential velocity with magnitude equivalent to the vortex strength of the sheet. Since there is no pressure discontinuity across the sheets, we deduce that sheets move axially backward without deformation. To approximate this we assume that the discrete bound vorticity is transformed in the slip-stream onto the surface of a vortex tube. There will be two components of this vorticity: a tube of elemental annular vortices that account for the thrust, and lines of elemental axial velocity along the tube that account for swirl . Fig. 3: Vortex tube Only the annular vorticity is considered initially. This has strength, γ, and has to meet certain compatibility conditions: 1. The total pressure inside the wake is taken to be higher than the free-stream by the loading on the propeller. However, the inner and outer flows must have the same static pressure, which yields the condition, 1 2 ρVs 2 = 1 2 ρV∞ 2 + T π R2 (4.1) 2. From a consideration of the continuity of mass, the radius of the fully-developed slip-stream, Rs, is related to that of the propeller, R, by, VR2 = VsRs 2 (4.2) 3. By considering vorticity, the velocity jump across the wake boundary, γ = Vs −V∞ (4.3) 2.4.1. The Propeller related to the Vortex Wake From the previous description of air flowing from the aerodynamic pressure side to aerodynamic suction, this is now described with the trailing vorticity which includes a reduction of the inflow angle towards the blade tip and, resultantly, a reduction in blade lift . The induced velocity may be considered to be the resultant velocity at a point due to the entire system of bound and free vorticity. For simplicity we assume that the propeller blades are narrow and are distributed along equally spaced radial lines. Considerations of symmetry further reveal that equally spaced radial vortex M.K. Rwigema 6 lines of equal strength induce no overall velocity on any one of the lines. These conditions therefore satisfy that the effect on each blade due to bound vorticity on the other blades can be ignored and only trailing vorticity contributes to the resultant velocity at a blade. The fundamental expressions for the forces developed by the propeller may be described most conveniently by applying the Kutta-Joukowsky theorem, dL = ρU × Γdr (4.4) The cross-product relationship can be used to separate the two components of the resultant force, the thrust and the torque. dT = ρΓΩr(1−′ a )dr (4.5) dQ = ρΓV∞(1+ a)rdr (4.6) 2.4.2. The Propeller Slip-stream The deflection of the wake is determined by relating the propeller forces to the characteristics of the fully-developed slipstream. The aerodynamic forces generated by a propeller under any deviation from uniform flight parallel to the thrust axis will lead to deflection of the slip-stream. We begin by considering the swirl component in the wake, represented by the vorticity γs. The axial component of the bound vorticity shed from the tip of the propeller blades per unit time is related to elemental axial vorticity lines along the stream-tube by, BΓV Δt = γ s2π RsV Δt (4.7) so that, 2π Rsγ s = BΓ (4.8) This satisfies the condition that the total vorticity shed at the tips, BΓ, must also be shed in the opposite sense at the roots as a discrete vortex down the axis. Behind the propeller the vortex tube translates downstream as it is embedded between the inner and outer flows with a velocity, Vw. The annular component of the bound vorticity shed per unit time is related to the elemental annular vorticity distributed over the surface of the tube by, γ 2πVw = BΓΩ (4.9) But from a simplification of Equation (4.5), integrated across the blade, and for a number of blades, B, BΓΩ = 2T ρR2 (4.10) such that, ργVw = T π R2 (4.11) This combined vortex model, ensures that the swirl is contained within the stream-tube and induces no flow outside the wake . 2.4.3. Propeller at Incidence We now consider a propeller at incidence to the slip-stream axis. In general, this will generate thrust and a tangential in-plane force N. Initially, thrust alone is considered; the inclusion of the in-plane force will follow as an incremental effect. Relative to the slip-stream, the down-going blade will have a forward velocity component, whereas the up-going side will be retreating. Fig. 4: Propeller slip-stream Considering a simple propeller blade at a rotation angle φ (where φ = 0 is at the top) with constant bound vorticity shed at the tip, the in-plane force acting normal to the blade will be ρΓVR (for αp = 0). Taking an angular position φ around the propeller disk, the vertical component of the tip velocity is ΩRssin(φ), and the forward component of this is ΩRsαssin(φ) for small αs. Equation (4.7) now becomes, BΓ[V + ΩRsαs sin(φ)]Δt = γ s2π RsV Δt (4.12) from which, 2π Rsγ s = BΓ 1+ ΩRsαs sin(φ) V ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (4.13) Compared with Equation (4.8), this shows an incremental sinusoidal variation of swirl velocity that will induce a downwash. Now a 7 PROPELLER BLADE ELEMENT MOMENTUM THEORY WITH VORTEX WAKE DELFECTION sinusoidal vorticity distribution of the form γ0sin(φ) induces a uniform downwash of strength γ0/2 inside the wake, which governs the deflection angle of the wake as it is counterbalanced by the up wash from the free-stream, V∝θp. From Equation (4.13), with φ = π/2, γ0 = BΓΩαs / (2πV), so that V∞θ p = γ 0 2 = BΓΩαs 4πV (4.14) From Equation (4.10) we deduce that, V∞θ p = Tαs 2ρπ R2V (4.15) From Fig. 4, αs = αp - θp. Considering this along with Equation (1.1), Tα p = ρπ R2V (Vs +V∞)θ p (4.16) Where Tαp is the tangential component of thrust and must be equivalent to the tangential momentum flux in the slip-stream, which is represented by ρπR2VVsθp and resolved through θp. The other term, ρπR2VV∝θp, can be re-written, from Equation (4.2), as ρπRs 2VsV∝θp, and represents the momentum flux of the part of the free-stream displaced by the cylindrical wake. From the above it can be shown that θ p α p = 1−V∞ V = CT 1+ 1−CT ( ) 2 (4.17) When αp ≠ 0, to account for the net normal force it is assumed that, in addition to any constant bound vorticity that may be present to generate thrust, there is a component of bound vorticity that varies sinusoidally around the propeller disk, Γ = Γ0sin(φ). The effect on the slip-stream by the tangential in-plane force may be determined by resolving the normal-to-blade force vertically, N(φ) = ρVRΓ0 sin2(φ) (4.18) Integration around the propeller disk, allowing for the number of blades, B, gives an average of N = 1 2 ρVBRΓ0 (4.19) Following the analysis above, this is dispersed over the trailing wake stream0tube with a form γ = γ0sin(φ), where, from Equation (4.8), 2π Rsγ 0 = BΓ0 (4.20) This will add to the downwash in the wake, balanced by the up wash from the free-stream, V∝θp as above. The incremental effect is therefore, ΔV∞θ p = N 2πρRRsV (4.21) Expressing the above equation in terms of the force coefficient and the propeller incidence, αp, and adding this to the right hand side of Equation (4.15), the analysis proceeds until Equation (4.17) becomes, θ p α p = 1−V∞ V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟+ 1 4 dCN dα p Vs 2 V 2 (4.22) But, from actuator theory, Vs 2 V 2 = 4Vs 2 Vs +V∞ ( ) 2 = 4 1+ V∞ Vs ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 (4.23) The incremental term due to the in-plane force now has a similar form to the contribution from thrust alone, and Equation (4.17) becomes , θ p α p = CT + dCN dα p 1+ 1−CT ( ) 2 (4.24) 2.4.4. Slip-stream Trajectory The trajectory of the slip-stream, and hence it’s lateral displacement at any streamwise station, is governed by it’s angle to the thrust line (or free-stream). The deflection from the datum can then be obtained by integration. The incidence case is considered and all angles are assumed small. The approach adopted is to consider momentum in a direction perpendicular to the resultant force F, see Fig. 4 . In this direction there is no force and the momentum must remain constant. M.K. Rwigema 8 For a segment of slip-stream at station x downstream of the propeller, the longitudinal moment is ρSxVxdx, and the tangential momentum due to the outer flow is (ρSxV∝dx)θx. Momentum is constant along the normal to F, such that ρSxdx[Vx(α x +θF ) +V∞θx] = ρSsdx[Vs(αs +θF ) +V∞θ p] (4.25) where the conditions on the right are those in the fully-developed slip-stream. Writing Sx = πRx 2, Ss = πRs 2 and S = πR2, SxVx = SsVs = SV (4.26) and ρS pVdx α x +θF ( ) + V∞ Vx θx ⎡ ⎣ ⎢ ⎤ ⎦ ⎥= ρS pVdx αs +θF ( ) + V∞ V θ p ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (4.27) Now, θx = αp - αs and θp = αp - αs, so that α x 1−V∞ Vx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟+ V∞ Vx α p = αs 1−V∞ Vs ⎛ ⎝ ⎜ ⎞ ⎠ ⎟+ V∞ Vs α p (4.28) From the equations of motion the streamwise distribution of velocity in the slip-stream is derived in as, Vs V∞ = 1+ s (4.29) in which s = a 1+ x x2 + d 2 4 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ (4.30) Equations (4.29) and (4.30) can be re-arranged to give Vx Vs = V∞ Vs + 1−V∞ Vs ⎛ ⎝ ⎜ ⎞ ⎠ ⎟f (x) (4.31) where f (x) = 1 2 1+ x D 1 4 + x D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ (4.32) Equation (4.28) can be re-written as α x(Vx −V∞)Vs = αs(Vs −V∞)Vx + (Vx −Vs)V∞α p (4.33) and from Equation (4.31) Vx −V∞= (Vs −V∞) f (x) (4.34) So finally, α x α p = αs α p 1 f (x) Vx Vs −V∞ Vs 1 f (x) −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= αs α p 1+ V∞ Vs 1 f (x) −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥−V∞ Vs 1 f (x) −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (4.35) The diameter of the slip-stream at any station x is given by Dx D = 1+ a 1+ s (4.36) In order to calculate the displacement, z, of the slip-stream it is only necessary to integrate the angular deflection. z = D α x dx 0 x ∫ (4.37) 3. References Froude RE. Trans Inst Naval Architects, 1889;30:390. Rankine WJM. On the mechanical principles of the action of propellers. Trans Inst Naval Architects (British), 1865;6(13). Glauert H. Airplane Propellers, vol iv, div. L. In: Durand WF, editor. Aerodynamic Theory. Berlin: Julius Springer, 1935 [Reprinted 1963 by Dover Publications, Inc., New York]. Jonkman JM. Modeling of the UAE Wind Turbine for Refinement of FAST_AD. NREL Technical Report [NREL/TP-500-34755], 2003. Drzewiecki S. Bulletin de L’Association Technique Maritime, Paris. A Second Paper in 1901, 1892. 9 PROPELLER BLADE ELEMENT MOMENTUM THEORY WITH VORTEX WAKE DELFECTION Betz A. with Appendix by L. Prandtl, 1919, Schraubenpropeller mit Geringstem Energieverlust, Göttinger Nachrichten, Göttingen, p. 193-217. Reprinted in Vier Abhandlungen uber Hydro- und Aerodynamik, L. Prandtl & A. betz, 1927 Wald QR. The distribution of circulation on propellers with finite hubs. ASME Paper 64-WA/UNT-4, Winter Annual Meeting, New York, 1964. Wald QR. The aerodynamics of propellers. Progress in Aerospace Sciences 42, pg 85-128, 2006. ESDU. Propeller slipstream modeling for incidence and sideslip. Item No. 06013, Engineering Sciences Data Unit, London, 2006. ESDU. Introduction to installation effects on thrust and drag for propeller-driven aircraft. Item No. 85015, Engineering Sciences Data Unit, London, 1985. Contact Author nasi@me.com; +27 83 324 2097 Copyright Statement The authors confirm that they, and/or their company or organization, hold copyright on all of the original material included in this paper. The authors also confirm that they have obtained permission, from the copyright holder of any third party material included in this paper, to publish it as part of their paper. The authors confirm that they give permission, or have obtained permission from the copyright holder of this paper, for the publication and distribution of this paper as part of the ICAS2010 proceedings or as individual off-prints from the proceedings. |
16168 | https://books.google.com/books/about/Respiratory_Physiology.html?id=eLRjk-VDF3cC | Respiratory Physiology: The Essentials - John Burnard West - Google Books
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Respiratory Physiology: The Essentials ====================================== John Burnard West Lippincott Williams & Wilkins, 2012 - Medical - 200 pages Widely considered the "gold standard" for the teaching and learning of respiratory physiology, this fully updated Ninth Edition includes key points for each chapter and multiple-choice review questions and answers with full explanations. Available online via thePoint, animations help to clarify particularly difficult concepts and provide a visual component for use during instruction or review. --NEW Presents rationales for all questions, as well as explanations for each answer choice --Provides 82 essential-to-know, multiple-choice review questions which appear at the end of each chapter --Features an Appendix of important equations --Supports learning through chapter-opening learning objectives and introductory material, as well as Key Concepts summaries at the end of each chapter --Includes online resources such as question bank, animations, and full text for students --Includes animations online--8 in total--via thePoint to illustrate particularly challenging concepts More » Preview this book »
Selected pages
Title Page
Table of Contents
Index
Contents
Structure and Function 1
Ventilation 12
Diffusion 24
Blood Flow and Metabolism 36
VentilationPerfusion Relationships 56
Gas Transport by the Blood 77
Mechanics of Breathing 95
Control of Ventilation 125
Respiratory System Under Stress 141
Tests of Pulmonary Function 159
Symbols Units and Equations 173
Answers 180
Figure Credits 193
Index195
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airway resistancealkalosisalveolar gasalveolar gas equationalveolar pressurealveolar ventilationalveolar wallalveolianatomic dead spaceapexarterial bloodarterial Pcoarterial pressurebicarbonateblood flowblood-gas barrierbreathingcapillarycapillary bloodcarbon monoxidecardiac outputcausedcentral chemoreceptorsChapterchest wallchoices are incorrectcm waterCO₂complianceconcentrationcorrectdecreasesdiaphragmdiffusing capacitydissociation curveelasticequationexerciseextra-alveolar vesselsFigureflowflfluidforced expirationgas exchangegaseshemoglobinhigh altitudehypoventilationhypoxemiahypoxiainhaledinspired gasintrapleural pressurelung diseaselung unitlung volumemeasuredmixed venous bloodml-¹normal subjectsO₂occursoxideoxygenpartial pressurepatientsPco₂perfusionperipheral chemoreceptorspulmonary arterypulmonary capillariespulmonary circulationpulmonary vascular resistancereceptorsreducedshuntsmooth musclesurface tensionterminal bronchiolestidal volumetiontissueuptakevenous pressureventilation and bloodventilation-perfusion inequalityventilation-perfusion ratiozone
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Title Respiratory Physiology: The Essentials
AuthorJohn Burnard West
Edition illustrated
Publisher Lippincott Williams & Wilkins, 2012
ISBN 1609136403, 9781609136406
Length 200 pages
SubjectsMedical
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Allied Health Services
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Respiratory Therapy
Medical / Allied Health Services / Respiratory Therapy
Medical / Pulmonary & Thoracic Medicine
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16169 | https://en.wiktionary.org/wiki/perpendicular | perpendicular - Wiktionary, the free dictionary
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1 EnglishToggle English subsection
1.1 Etymology
1.2 Pronunciation
1.3 Adjective
1.3.1 Derived terms
1.3.2 Translations
1.4 Noun
1.4.1 Translations
1.5 See also
2 CatalanToggle Catalan subsection
2.1 Etymology
2.2 Pronunciation
2.3 Adjective
2.3.1 Derived terms
2.4 Noun
2.5 Further reading
3 PortugueseToggle Portuguese subsection
3.1 Etymology
3.2 Pronunciation
3.3 Adjective
3.4 Noun
3.4.1 Derived terms
4 RomanianToggle Romanian subsection
4.1 Etymology
4.2 Noun
4.2.1 Declension
5 SpanishToggle Spanish subsection
5.1 Etymology
5.2 Pronunciation
5.3 Adjective
5.3.1 Derived terms
5.4 Further reading
perpendicular
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From Wiktionary, the free dictionary
See also:Perpendicular
English
[edit]
perpendicular on Wikipedia
Etymology
[edit]
Derived from Middle Frenchperpendiculaire, from Old Frenchperpendiculer, from Latinperpendiculum(“plumb line”).
Pronunciation
[edit]
(UK)IPA(key): /ˌpɜː.pənˈdɪk.jə.lə(ɹ)/enPR: pû"pəndĭ'kyələ(r),
(US)IPA(key): /pɝ.pɛnˈdɪk.ju.lɚ/, /pɝ.pənˈdɪk.jə.lɚ/
Audio (US):Duration: 2 seconds.0:02(file)
Rhymes: -ɪkjʊlə(ɹ)
Adjective
[edit]
perpendicular (comparativemore perpendicular, superlativemost perpendicular)
(geometry) At or forming a right angle (to something). synonyms▲quotations▼Synonyms:normal, orthogonalIn most houses, the walls are perpendicular to the floor.
2012 March, Henry Petroski, “Opening Doors”, in American Scientist, volume 100, number 2, pages 112–3:A doorknob of whatever roundish shape is effectively a continuum of levers, with the axis of the latching mechanism—known as the spindle—being the fulcrum about which the turning takes place. Applying a force tangential to the knob is essentially equivalent to applying one perpendicular to a radial line defining the lever.
Exactly upright; extending in a straight line toward the centre of the earth, etc.
Independent of or irrelevant to each other; orthogonal. quotations▼
2019 May 31, David M. Willis, “Wrangled”, in Dumbing of Age:Hey, I'm not unsabotaging anything! This is completely perpendicular sabotage!
Derived terms
[edit]
perpendicular axis theorem
perpendicular distance
perpendicularity
perpendicularly
perpendicularness
perpendicular pronoun
perpendicular recording
perpendicular universe
quasiperpendicular
show more ▼
Translations
[edit]
show ▼±at or forming a right angle to
[Select preferred languages] [Clear all]
Arabic: عَمُودِيّ(ar)(ʕamūdiyy), رَأْسِيّ(raʔsiyy)
Armenian: ուղղահայաց(hy)(uġġahayacʻ)
Belarusian: перпендыкуля́рны(pjerpjendykuljárny)
Bulgarian: перпендикуля́рен(bg)(perpendikuljáren)
Catalan: perpendicular(ca)
Chinese: Mandarin: 垂直(zh)(chuízhí)
Czech: kolmý(cs)
Danish: vinkelret(da)
Dutch: loodrecht(nl), haaks(nl), perpendiculair(nl)
Esperanto: perpendikulara
Estonian: risti(et)
Finnish: kohtisuora(fi), kohtisuorassa, pystysuorassa
French: perpendiculaire(fr)
Galician: perpendicular(gl)
Georgian: პერპენდიკულარული(ṗerṗendiḳularuli), მართობი(martobi)
German: rechtwinklig(de); lotrecht(de), senkrecht(de), perpendikular(de), perpendikulär(de)
Greek: κάθετος(el)m(káthetos)
Hebrew: מאונך \ מְאֻנָךְ(he)(meunákh), ניצב \ נִצָּב(he)(nitsáv)
Hungarian: merőleges(hu)
Indonesian: tegak lurus(id)
Italian: perpendicolare(it)
Japanese: 垂直の(ja)(すいちょくの, suichoku no), 縦の(ja)(たての, tate no), 直立の(ja)(ちょくりつの, chokuritsu no)
Korean: 수직의(ko)(sujigui)
Latin: perpendiculāris
Latvian: perpendikulārs(lv)
Malay: serenjang
Maori: hāngai
Norwegian: Bokmål: loddrettNynorsk: loddrett
Persian: عمودی(fa)('amudi)
Polish: prostopadły(pl)
Portuguese: perpendicular(pt)
Romanian: perpendicular(ro)
Russian: перпендикуля́рный(ru)(perpendikuljárnyj)
Sanskrit: लम्ब(sa)m(lamba)
Slovak: kolmý
Spanish: perpendicular(es)
Swedish: vinkelrät(sv)
Tagalog: tadlong
Thai: ตั้งฉาก(táng-chàak)
Turkish: dik(tr)
Ukrainian: перпендикуля́рний(perpendykuljárnyj), простопа́дний(prostopádnyj)
Welsh: perpendiciwlar
West Frisian: leadrjocht
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
Noun
[edit]
perpendicular (pluralperpendiculars)
(geometry) A line or plane that is perpendicular to another.
A device such as a plumb line that is used in making or marking a perpendicular line.
(obsolete,slang) A meal eaten at a tavernbar while standing up.
Translations
[edit]
show ▼±line or plane
[Select preferred languages] [Clear all]
Bulgarian: перпендикулярm(perpendikuljar)
Catalan: perpendicular(ca)f
Chinese: Mandarin: 垂线(zh)(chuíxiàn), 垂面(zh)
Czech: kolmice(cs)f(line)
Dutch: loodlijn(nl)f
Finnish: kohtisuora(fi)
French: perpendiculaire(fr)f
Georgian: პერპენდიკულარი(ṗerṗendiḳulari), მართობი(martobi)
German: Lotrechtef, Senkrechtef
Hebrew: מאונך \ מְאֻנָךְ(he)m(meunákh), ניצב \ נִצָּב(he)m(nitsáv)
Hindi: लंब(hi)(lamb), अभिलंब(abhilamb)
Hungarian: merőleges(hu)
Navajo: kʼéhézdongo ahidiiʼá
Polish: prostopadłaf
Portuguese: perpendicular(pt)f
Romanian: perpendiculară(ro)f
Russian: перпендикуля́р(ru)m(perpendikuljár)
Swedish: normal(sv)c
Thai: เส้นตั้งฉาก(séen-táng-chàak)
Turkish: dik(tr)
Ukrainian: простопа́дm(prostopád)
West Frisian: leadline
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
show ▼±device
[Select preferred languages] [Clear all]
Bulgarian: отвесm(otves)
Dutch: schietlood(nl)n
Finnish: luotilanka(fi), luoti(fi)
French: fil à plomb(fr)m
German: Bleilotn, Senkblei(de)n
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
See also
[edit]
⟂ (This symbol can be pronounced “perp” when used as a subscript of a letter representing a vector.)
Catalan
[edit]
Etymology
[edit]
Borrowed from Late Latinperpendiculāris, from perpendiculum.
Pronunciation
[edit]
IPA(key): (Central)[pər.pən.di.kuˈlar]
IPA(key): (Balearic)[pər.pən.di.kuˈla]
IPA(key): (Valencia)[peɾ.pen.di.kuˈlaɾ]
Audio (Barcelona):Duration: 2 seconds.0:02(file)
Adjective
[edit]
perpendicularm or f (masculine and feminine pluralperpendiculars)
perpendicular
Derived terms
[edit]
perpendicularment
Noun
[edit]
perpendicularf (pluralperpendiculars)
perpendicular
Further reading
[edit]
“perpendicular”, in Diccionari de la llengua catalana [Dictionary of the Catalan Language] (in Catalan), second edition, Institute of Catalan Studies [Catalan: Institut d'Estudis Catalans], April 2007
“perpendicular”, in Gran Diccionari de la Llengua Catalana, Grup Enciclopèdia Catalana, 2025
“perpendicular” in Diccionari normatiu valencià, Acadèmia Valenciana de la Llengua.
“perpendicular” in Diccionari català-valencià-balear, Antoni Maria Alcover and Francesc de Borja Moll, 1962.
Portuguese
[edit]
Etymology
[edit]
Borrowed from Late Latinperpendiculāris, from perpendiculum.
Pronunciation
[edit]
more ▼
(Brazil)IPA(key): /peʁ.pẽ.d͡ʒi.kuˈlaʁ/[peh.pẽ.d͡ʒi.kuˈlah]
(Brazil)IPA(key): /peʁ.pẽ.d͡ʒi.kuˈlaʁ/[peh.pẽ.d͡ʒi.kuˈlah]
(São Paulo)IPA(key): /peɾ.pẽ.d͡ʒi.kuˈlaɾ/
(Rio de Janeiro)IPA(key): /peʁ.pẽ.d͡ʒi.kuˈlaʁ/[peχ.pẽ.d͡ʒi.kuˈlaχ]
(Southern Brazil)IPA(key): /peɻ.pẽ.d͡ʒi.kuˈlaɻ/
more ▼
(Portugal)IPA(key): /pɨɾ.pẽ.di.kuˈlaɾ/
(Portugal)IPA(key): /pɨɾ.pẽ.di.kuˈlaɾ/
(Southern Portugal)IPA(key): /pɨɾ.pẽ.di.kuˈla.ɾi/
Hyphenation: per‧pen‧di‧cu‧lar
Adjective
[edit]
perpendicularm or f (pluralperpendiculares)
perpendicular
Noun
[edit]
perpendicularf (pluralperpendiculares)
perpendicular
Derived terms
[edit]
perpendicularidade
perpendicularmente
Romanian
[edit]
Etymology
[edit]
Borrowed from Frenchperpendiculaire.
Noun
[edit]
perpendicularf (pluralperpendiculare)
perpendicular
Declension
[edit]
show ▼Declension of perpendicular| | singular | | plural |
---
| | indefinite | definite | indefinite | definite |
| nominative-accusative | perpendicular | perpendiculara | perpendiculare | perpendicularele |
| genitive-dative | perpendiculare | perpendicularei | perpendiculare | perpendicularelor |
| vocative | — | — |
Spanish
[edit]
Etymology
[edit]
Borrowed from Late Latinperpendiculāris, from perpendiculum.
Pronunciation
[edit]
IPA(key): /peɾpendikuˈlaɾ/[peɾ.pẽn̪.d̪i.kuˈlaɾ]
Rhymes: -aɾ
Syllabification: per‧pen‧di‧cu‧lar
Adjective
[edit]
perpendicularm or f (masculine and feminine pluralperpendiculares)
perpendicular
Derived terms
[edit]
perpendicularmente
Further reading
[edit]
“perpendicular”, in Diccionario de la lengua española [Dictionary of the Spanish Language] (in Spanish), online version 23.8, Royal Spanish Academy [Spanish: Real Academia Española], 10 December 2024
Retrieved from "
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perpendicular
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16170 | https://www.wikiwand.com/simple/articles/Bacillus_(shape) | Bacillus (shape) - Wikiwand
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Bacillus (shape)
any rod-shaped bacterium or archaeon From Wikipedia, the free encyclopedia
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A bacillus (plural bacilli) is a rod-shaped cylindrical bacterium. Bacilli are found in many different taxonomic groups of bacteria.
When the name Bacillus is capitalized and italicized, it refers to a specific genus of bacteria.
If it is not capitalized or italicized the name refers to any rod-shaped bacterium.
A typical bacillus, Gram stained: click for enlarged view
There is no connection between the shape of a bacterium and its colors in the gram staining. In other words, some of them are gram negative and some are gram positive.
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16171 | https://courses.lumenlearning.com/wm-developmentalemporium/chapter/simplifying-expressions-using-the-order-of-operations/ | Simplifying Expressions Using the Order of Operations
Learning Outcomes
Use the order of operations to simplify simple mathematical expressions
Simplify complex mathematical expressions involving addition, subtraction, multiplication, division, and exponents
Correctly using the order of operations
What is 3+5×2 ? Is it 13 or 16 ? This may seem like a trick question, but there is actually only one correct answer.
Many years ago, mathematicians developed a standard order of operations that tells you which calculations to make first in an expression with more than one operation. In other words, order of operations simply refers to the specific order of steps you should follow when you solve a math expression. Without a standard procedure for making calculations, two people could get two different answers to the same problem, like the one above. So which is it, 13 or 16 ? By the end of this module you’ll know!
We’ve introduced most of the symbols and notation used in algebra, but now we need to clarify the order in which they will be carried out. Otherwise, expressions may have different meanings, and they may result in different values.
For example, consider the expression:
4+3⋅7
Some students say it simplifies to 49.Some students say it simplifies to 25.4+3⋅7Since 4+3 gives 7.7⋅7And 7⋅7 is 49.494+3⋅7Since 3⋅7 is 21.4+21And 21+4 makes 25.25
Imagine the confusion that could result if every problem had several different correct answers. The same expression should give the same result. So mathematicians established some guidelines called the order of operations, which outlines the order in which parts of an expression must be simplified.
Order of Operations
When simplifying mathematical expressions perform the operations in the following order:
1. Parentheses and other Grouping Symbols
Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
Exponents
Simplify all expressions with exponents.
Multiplication and Division
Perform all multiplication and division in order from left to right. These operations have equal priority.
Addition and Subtraction
Perform all addition and subtraction in order from left to right. These operations have equal priority.
Students often ask, “How will I remember the order?” Here is a way to help you remember: Take the first letter of each key word and substitute the silly phrase. Please Excuse My Dear Aunt Sally or PEMDAS.
| Order of Operations | |
--- |
| Please | Parentheses |
| Excuse | Exponents |
| My Dear | Multiplication and Division |
| Aunt Sally | Addition and Subtraction |
It’s good that ‘My Dear’ goes together, as this reminds us that multiplication and division have equal priority. We do not always do multiplication before division or always do division before multiplication. We do them in order from left to right.
Similarly, ‘Aunt Sally’ goes together and so reminds us that addition and subtraction also have equal priority and we do them in order from left to right.
example
Simplify the expressions:
4+3⋅7
(4+3)⋅7
Solution:
| | |
--- |
| 1. | |
| | 4+3⋅7 |
| Are there any parentheses? No. | |
| Are there any exponents? No. | |
| Is there any multiplication or division? Yes. | |
| Multiply first. | 4+3⋅7 |
| Add. | 4+21 |
| | 25 |
| | |
--- |
| 2. | |
| | (4+3)⋅7 |
| Are there any parentheses? Yes. | (4+3)⋅7 |
| Simplify inside the parentheses. | (7)7 |
| Are there any exponents? No. | |
| Is there any multiplication or division? Yes. | |
| Multiply. | 49 |
try it
example
Simplify:
18÷9⋅2
18⋅9÷2
Show Solution
Solution:
| | |
--- |
| 1. | |
| | 18÷9⋅2 |
| Are there any parentheses? No. | |
| Are there any exponents? No. | |
| Is there any multiplication or division? Yes. | |
| Multiply and divide from left to right. Divide. | 2⋅2 |
| Multiply. | 4 |
| | |
--- |
| 2. | |
| | 18⋅9÷2 |
| Are there any parentheses? No. | |
| Are there any exponents? No. | |
| Is there any multiplication or division? Yes. | |
| Multiply and divide from left to right. | |
| Multiply. | 162÷2 |
| Divide. | 81 |
try it
example
Simplify: 18÷6+4(5−2).
Show Solution
Solution:
| | |
--- |
| | 18÷6+4(5−2) |
| Parentheses? Yes, subtract first. | 18÷6+4(3) |
| Exponents? No. | |
| Multiplication or division? Yes. | |
| Divide first because we multiply and divide left to right. | 3+4(3) |
| Any other multiplication or division? Yes. | |
| Multiply. | 3+12 |
| Any other multiplication or division? No. | |
| Any addition or subtraction? Yes. | 15 |
try it
In the video below we show another example of how to use the order of operations to simplify a mathematical expression.
When there are multiple grouping symbols, we simplify the innermost parentheses first and work outward.
example
Simplify: 5+23+3[6−3(4−2)].
Show Solution
Solution:
| | |
--- |
| | 5+23+3[6−3(4−2)] |
| Are there any parentheses (or other grouping symbol)? Yes. | |
| Focus on the parentheses that are inside the brackets. | 5+23+3[6−3(4−2)] |
| Subtract. | 5+23+3[6−3(2)] |
| Continue inside the brackets and multiply. | 5+23+3[6−6] |
| Continue inside the brackets and subtract. | 5+23+3 |
| The expression inside the brackets requires no further simplification. | |
| Are there any exponents? Yes. | |
| Simplify exponents. | 5+23+3 |
| Is there any multiplication or division? Yes. | |
| Multiply. | 5+8+3 |
| Is there any addition or subtraction? Yes. | |
| Add. | 5+8+0 |
| Add. | 13+0 |
| | 13 |
try it
In the video below we show another example of how to use the order of operations to simplify an expression that contains exponents and grouping symbols.
example
Simplify: 23+34÷3−52.
Show Solution
Solution:
| | |
--- |
| | 23+34÷3−52 |
| If an expression has several exponents, they may be simplified in the same step. | |
| Simplify exponents. | 23+34÷3−52 |
| Divide. | 8+81÷3−25 |
| Add. | 8+27−25 |
| Subtract. | 35−25 |
| | 10 |
try it
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Ex: Evaluate an Expression Using the Order of Operations. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Example 3: Evaluate An Expression Using The Order of Operation. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Question ID: 144748, 144751, 144756, 144758, 144759, 144762. Authored by: Alyson Day. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
CC licensed content, Specific attribution
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CC licensed content, Shared previously
Ex: Evaluate an Expression Using the Order of Operations. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Example 3: Evaluate An Expression Using The Order of Operation. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Question ID: 144748, 144751, 144756, 144758, 144759, 144762. Authored by: Alyson Day. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
CC licensed content, Specific attribution
Prealgebra. Provided by: OpenStax. License: CC BY: Attribution. License Terms: Download for free at |
16172 | http://ncvm4.books.nba.co.za/chapter/unit-5-sketch-cubic-functions/ | Skip to content
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Functions and algebra: Investigate and use instantaneous rate of change of a variable when interpreting models both in mathematical and real life situations
Unit 5: Sketch cubic functions
Natashia Bearam-Edmunds
Unit outcomes
By the end of this unit you will be able to:
Determine the shape of a cubic function.
Determine the x and y intercepts.
Find the turning points of the graph.
Find the maximum and minimum values of the graph.
Find the point of inflection and discuss concavity using second derivatives.
What you should know
Before you start this unit, make sure you can:
Find the derivative by using the rules of differentiation.
Sketch and interpret information from graphs of functions. Revise level 3 subject outcome 2.1 for a refresher on functions.
Introduction
We have worked with cubic polynomials in this subject outcome. Cubic functions are of the form [latex]\scriptsize y=a{{x}^{3}}+b{{x}^{2}}+cx+d[/latex]. The highest power or degree of a cubic function is three. This also tells us that the graph will have at most three x-intercepts.
There are two general shapes for cubic functions depending on the value of [latex]\scriptsize a[/latex] the co-efficient of [latex]\scriptsize {{x}^{3}}[/latex].
When [latex]\scriptsize a \gt 0[/latex], the graph starts out concave down (sad [latex]\scriptsize \cap[/latex]) and ends up concave up (happy [latex]\scriptsize \cup[/latex]).
When [latex]\scriptsize a \lt 0[/latex] the graph starts out concave up (happy) and ends up concave down (sad).
Sketching a cubic function
To sketch a cubic function you need the:
shape
x- intercepts
y-intercept
turning points.
To find the x-intercepts:
let [latex]\scriptsize \displaystyle y=0[/latex]
factorise by using the factor theorem
solve.
To find the y-intercept:
let [latex]\scriptsize \displaystyle x=0[/latex] and solve for [latex]\scriptsize \displaystyle y[/latex].
Example 5.1
Find the x and y-intercepts of [latex]\scriptsize f(x)=-{{x}^{3}}+4{{x}^{2}}+x-4[/latex].
Solution
To find the y-intercept let [latex]\scriptsize x=0[/latex] and solve for [latex]\scriptsize y[/latex].
[latex]\scriptsize \begin{align}f(0)&=-{{(0)}^{3}}+4{{(0)}^{2}}+(0)-4\&=-4\end{align}[/latex]
The coordinates of the y-intercept are[latex]\scriptsize (0;-4)[/latex].
To find the x-intercepts you will need to use the factor theorem since we must factorise a cubic expression.
[latex]\scriptsize f(x)=-{{x}^{3}}+4{{x}^{2}}+x-4[/latex]
We use trial and error to find factors of [latex]\scriptsize f(x)[/latex]. Remember that a factor will leave no remainder when it divides into an expression.
[latex]\scriptsize \begin{align}f(1)&=-{{(1)}^{3}}+4{{(1)}^{2}}+(1)-4\&=0\end{align}[/latex]
[latex]\scriptsize x=1[/latex] leaves no remainder so one factor of the expression is [latex]\scriptsize (x-1)[/latex].
Factorise further by inspection (or other methods as shown in level 4 subject outcome 2.1).
[latex]\scriptsize \begin{align}f(x)&=-{{x}^{3}}+4{{x}^{2}}+x-4\&=-({{x}^{3}}-4{{x}^{2}}-x+4)\&=-(x-1)({{x}^{2}}-3x-4)\&=-(x-1)(x-4)(x+1)\end{align}[/latex]
To find the x-intercepts let [latex]\scriptsize f(x)=0[/latex] and solve for [latex]\scriptsize x[/latex].
[latex]\scriptsize \displaystyle \begin{align}0&=-(x-1)(x-4)(x+1)\x&=1,\text{ }x=4\text{ or }x=-1\end{align}[/latex]
The coordinates of the x-intercepts are [latex]\scriptsize \displaystyle (-1;0),\text{ }(1;0)\text{ and }(4;0)[/latex] .
Exercise 5.1
Determine the x and y-intercepts of the following functions:
[latex]\scriptsize f(x)=-{{x}^{3}}-5{{x}^{2}}+9x+45[/latex]
[latex]\scriptsize y={{x}^{3}}+3{{x}^{2}}-10x[/latex]
[latex]\scriptsize y=2{{x}^{3}}-32x[/latex]
The full solutions are at the end of the unit.
Stationary points
The turning points of a graph are called the stationary points. A cubic graph will have at most two turning points. At the stationary points a tangent drawn to graph will have gradient of zero, therefore the gradient of the graph is zero and the derivative also will be zero. You can think of a stationary point as a point where the function stops increasing or decreasing.
There are three types of stationary points; local maximum (maxima), local minimum (minima) and horizontal points of inflection.
Local maxima or minima are called relative minimum and maximum values, as there are other points on the graph with lower and higher function values.
The gradient of the function changes on either side of local maxima or minima values as shown in figure 3. The function changes from decreasing to increasing at the local minima and changes from increasing to decreasing at the local maxima.
The gradient on either side of a horizontal point of inflection stays constant as shown in figure 4. The graph is increasing on either side of the point of inflection.
To determine the coordinates of the stationary point(s) of [latex]\scriptsize \displaystyle f\left( x \right)[/latex]:
Find the derivative.
Let [latex]\scriptsize {f}'(x)=0[/latex] and solve for the x-coordinate(s) of the stationary point(s).
Substitute value(s) of [latex]\scriptsize \displaystyle x[/latex] into [latex]\scriptsize \displaystyle f\left( x \right)[/latex] to calculate the y-coordinate(s) of the stationary point(s).
Example 5.2
Calculate the stationary point(s) of the graph [latex]\scriptsize f(x)=-{{x}^{3}}+4{{x}^{2}}+3x-4[/latex].
Solution
Step 1: Determine the derivative of [latex]\scriptsize \displaystyle f\left( x \right)[/latex].
[latex]\scriptsize {f}'(x)=-3{{x}^{2}}+8x+3[/latex]
Step 2: Let [latex]\scriptsize \displaystyle {f}'\left( x \right)=0[/latex] and solve for the x-values of the turning point(s).
[latex]\scriptsize \displaystyle \begin{align}-3{{x}^{2}}+8x+3&=0\3{{x}^{2}}-8x-3&=0\(3x+1)(x-3)&=0\x&=-\displaystyle \frac{1}{3}\text{ and }x=3\end{align}[/latex]
Step 3: Substitute the x-values into [latex]\scriptsize \displaystyle f\left( x \right)[/latex] to calculate the corresponding y-coordinates of the stationary points.
[latex]\scriptsize \displaystyle \begin{align}f(-\displaystyle \frac{1}{3})&=-{{(-\displaystyle \frac{1}{3})}^{3}}+4{{(-\displaystyle \frac{1}{3})}^{2}}+3(-\displaystyle \frac{1}{3})-4\&=-\displaystyle \frac{{122}}{{27}}\\f(3)&=-{{(3)}^{3}}+4{{(3)}^{2}}+3(3)-4\&=14\end{align}[/latex]
Step 4: Write the final answer.
The stationary points are [latex]\scriptsize \displaystyle (-\displaystyle \frac{1}{3}\text{;-}\displaystyle \frac{{122}}{{27}}\text{) and (3;14)}[/latex].
Exercise 5.2
Find the turning points of the following functions:
[latex]\scriptsize f(x)=-{{x}^{3}}-3{{x}^{2}}+9x-10[/latex]
[latex]\scriptsize y={{x}^{3}}+3{{x}^{2}}-10x[/latex]
[latex]\scriptsize y=2{{x}^{3}}-54x+40[/latex]
The full solutions are at the end of the unit.
Now, we are ready to sketch a cubic function.
General method for sketching cubic graphs:
Use the sign of [latex]\scriptsize a[/latex] to determine the general shape of the graph.
Determine the y-intercept by letting [latex]\scriptsize x=0[/latex].
Determine the x-intercepts by letting [latex]\scriptsize y=0[/latex] and solving for [latex]\scriptsize x[/latex].
Find the x-coordinates of the turning points by letting [latex]\scriptsize {f}'(x)=0[/latex] and solving for [latex]\scriptsize x[/latex].
Determine the y-coordinates of the turning points by substituting their x-values into [latex]\scriptsize f(x)[/latex].
Plot the points and join as a smooth curve.
Example 5.3
Sketch the graph [latex]\scriptsize f(x)=-{{x}^{3}}-3{{x}^{2}}+9x+27[/latex].
Solution
Step 1: Determine the shape of the graph.
The coefficient of the [latex]\scriptsize {{x}^{3}}[/latex] term is less than zero, therefore the graph will have the following shape:
Step 2: Determine the intercepts.
To find the y-intercept let [latex]\scriptsize x=0[/latex] and solve for [latex]\scriptsize y[/latex].
[latex]\scriptsize \begin{align}f(0)&=-{{x}^{3}}-3{{x}^{2}}+9x+27\&=27\end{align}[/latex]
y-intercept: [latex]\scriptsize (0;27)[/latex]
Find the x-intercepts by letting [latex]\scriptsize f(x)=0[/latex] and solving for [latex]\scriptsize x[/latex]:
[latex]\scriptsize \begin{align}-{{x}^{3}}-3{{x}^{2}}+9x+27&=0\{{x}^{3}}+3{{x}^{2}}-9x-27&=0\(x+3)({{x}^{2}}-9)&=0\(x+3)(x+3)(x-3)&=0\x&=-3\text{ or }x=3\end{align}[/latex]
x-intercepts: [latex]\scriptsize (-3;0)[/latex]and [latex]\scriptsize (3;0)[/latex]
Step 3: Calculate the stationary/turning points.
[latex]\scriptsize \begin{align}{f}'(x)&=-3{{x}^{2}}-6x+9\-3{{x}^{2}}-6x+9&=0\{{x}^{2}}+2x-3&=0\(x+3)(x-1)&=0\x&=-3,\text{ }x=1\f(-3)&=-{{(-3)}^{3}}-3{{(-3)}^{2}}+9(-3)+27\&=0\f(1)&=-{{(1)}^{3}}-3{{(1)}^{2}}+9(1)+27\&=32\end{align}[/latex]
Turning points: [latex]\scriptsize (-3;0)\text{ }[/latex] and [latex]\scriptsize (1;32)[/latex]
Step 4: Draw a neat sketch (not necessarily drawn to scale).
Show all key points on the graph.
Exercise 5.3
Given [latex]\scriptsize f(x)={{x}^{3}}+{{x}^{2}}-5x+3[/latex]:
Show that [latex]\scriptsize (x-1)[/latex] is a factor of [latex]\scriptsize f(x)[/latex] and hence find the x-intercepts.
Determine the y-intercept and turning points.
Sketch the graph.
Sketch the graph of [latex]\scriptsize f(x)=-{{x}^{3}}+4{{x}^{2}}+11x-30[/latex]. Show all the turning points and intercepts with the axes.
The full solutions are at the end of the unit.
Second derivative and concavity
A change in the sign of the second derivative shows that there is a change in the direction of the gradient of the original function. To find the second derivative of a function you take the derivative of the first derivative.
The second derivative tells us about the concavity of a graph. Concavity relates to the rate of change of a function’s derivative. Concavity indicates whether the gradient of a curve is increasing, decreasing or constant.
Take note!
If [latex]\scriptsize {{f}'}'(x) \lt 0[/latex], the graph is concave down and there is a local maximum. Tangent lines drawn to the graph lie above the graph. The gradient of the curve decreases as [latex]\scriptsize x[/latex] increases.
If [latex]\scriptsize {{f}'}'(x) \gt 0[/latex], the graph is concave up and there is a local minimum. Tangent lines lie below the graph. The gradient of the curve increases as [latex]\scriptsize x[/latex] increases.
If [latex]\scriptsize {{f}'}'(x)=0[/latex], the graph may have a point of inflection. If [latex]\scriptsize {{f}'}'(x)[/latex] changes sign from one side of the point of inflection to the other, the concavity changes. For a cubic function, the graph will have a point of inflection when [latex]\scriptsize {{f}'}'(x)=0[/latex].
Figure 5 shows how the concavity tells us about the shape of a graph.
Concave down: [latex]\scriptsize {{f}'}'(x) \lt 0[/latex].
Concave up: [latex]\scriptsize {{f}'}'(x) \gt 0[/latex].
Point of inflection: [latex]\scriptsize {{f}'}'(x)=0[/latex] and changes sign at this point.
Exercise 5.4
Determine if the following graphs have a local maximum, minimum or point of inflection at the given points, by finding the second derivative:
[latex]\scriptsize f(x)=-{{x}^{3}}-3{{x}^{2}}+9x-10[/latex]at [latex]\scriptsize x=1[/latex]
[latex]\scriptsize y={{x}^{3}}+3{{x}^{2}}-10x[/latex]at [latex]\scriptsize x=-1[/latex]
[latex]\scriptsize y=2{{x}^{3}}-54x+40[/latex]at [latex]\scriptsize x=3[/latex]
The full solutions are at the end of the unit.
Summary
In this unit you have learnt the following:
How to identify the shape of a cubic function.
How to find the x and y-intercepts of a cubic function.
How to find the stationary points.
How to find the point of inflection of a cubic function.
How to test concavity.
How to sketch a cubic function.
Unit 5: Assessment
Suggested time to complete: 45 minutes
Given [latex]\scriptsize g(x)={{x}^{3}}-2{{x}^{2}}-3x+6[/latex]:
Determine the intercepts with the axes.
Find the stationary points.
Calculate the point of inflection.
Sketch the graph.
Answer the following questions for [latex]\scriptsize f(x)={{(x-2)}^{3}}[/latex]:
Determine the concavity of the graph.
Find the inflection point.
Sketch the graph.
The full solutions are at the end of the unit.
Unit 5: Solutions
Exercise 5.1
. [latex]\scriptsize f(x)=-{{x}^{3}}-5{{x}^{2}}+9x+45[/latex] [latex]\scriptsize \begin{align}f(3)&=0\\therefore &(x-3)\text{ is a factor}\end{align}[/latex] [latex]\scriptsize \begin{align}f(x)&=-({{x}^{3}}+5{{x}^{2}}-9x-45)\&=-(x-3)({{x}^{2}}+8x+15)\&=-(x-3)(x+3)(x+5)\end{align}[/latex] Find the x-intercepts: [latex]\scriptsize \displaystyle \begin{align}&-(x-3)(x+3)(x+5)=0\&\therefore x=3,\text{ }-3\text{ or }-5\&\text{x-intercepts :}\&(-5;0),\text{ }(-3;0)\text{ and }(3;0)\text{ }\end{align}[/latex] y-intercept: [latex]\scriptsize (0;45)[/latex].
. [latex]\scriptsize \begin{align}y&={{x}^{3}}+3{{x}^{2}}-10x\&=x({{x}^{2}}+3x-10)\&=x(x+5)(x-2)\&\text{x-intercepts}:\&(0;0),(2;0)\text{ and }(-5;0)\end{align}[/latex] y-intercept: [latex]\scriptsize (0;0)[/latex].
. [latex]\scriptsize \displaystyle \begin{align}y&=2{{x}^{3}}-32x\&=2x({{x}^{2}}-16)\&=2x(x-4)(x+4)\&\text{x-intercepts:}\&(0;0),(-4;0)\text{ and }(4;0)\end{align}[/latex] y-intercept: [latex]\scriptsize (0;0)[/latex].
Back to Exercise 5.1
Exercise 5.2
. [latex]\scriptsize \begin{align}{f}'(x)&=-3{{x}^{2}}-6x+9\-3{{x}^{2}}-6x+9&=0\{{x}^{2}}+2x-3&=0\(x+3)(x-1)&=0\x&=-3,\text{ }x=1\f(-3)&=-{{(-3)}^{3}}-3{{(-3)}^{2}}+9(-3)-10\&=-37\f(1)&=-{{(1)}^{3}}-3{{(1)}^{2}}+9(1)-10\&=-5\\text{Turning points: }&(-3;-37)\text{ and }(1;-5)\end{align}[/latex]
. [latex]\scriptsize \displaystyle \begin{align}{y}' &=3{{x}^{2}}+6x-10\3{{x}^{2}}+6x-10&=0\x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\&=\displaystyle \frac{{-6\pm \sqrt{{{{{(6)}}^{2}}-4(3)(-10)}}}}{{2(3)}}\&=\displaystyle \frac{{-3\pm \sqrt{{39}}}}{3}\x&=\displaystyle \frac{{-3+\sqrt{{39}}}}{3}\text{ or }x=\displaystyle \frac{{-3-\sqrt{{39}}}}{3}\x&=0.180...\text{ or }x=-3.081...\end{align}[/latex] [latex]\scriptsize \begin{align}f(0.180...)&=-6.04\f(-3.081...)&=30.04\\text{Turning points: }&(0.18;-6.04)\text{ and }(-3.08;30.04)\end{align}[/latex]
. [latex]\scriptsize \begin{align}{y}' &=6{{x}^{2}}-54\{{x}^{2}}-9&=0\x&=3,\text{ }x=-3\f(3)&=2{{(3)}^{3}}-54(3)+40\&=-68\f(-3)&=2{{(-3)}^{3}}-54(-3)+40\&=148\\text{Turning points: }&(3;-68)\text{ and }(-3;148)\end{align}[/latex]
Back to Exercise 5.2
Exercise 5.3
[latex]\scriptsize f(x)={{x}^{3}}+{{x}^{2}}-5x+3[/latex]
. [latex]\scriptsize \begin{align}f(1)&={{(1)}^{3}}+{{(1)}^{2}}-5(1)+3\&=0\\therefore (x-1)&\text{ is a factor}\end{align}[/latex] [latex]\scriptsize \begin{align}{{x}^{3}}+{{x}^{2}}-5x+3&=(x-1)({{x}^{2}}+2x-3)\(x-1)({{x}^{2}}+2x-3)&=0\{{(x-1)}^{2}}(x+3)&=0\x&=1\text{ or }x=-3\\text{x-intercepts:}&\text{ }(1;0)\text{ and }(-3;0)\end{align}[/latex]
[latex]\scriptsize \text{y-intercept }(0;3)[/latex][latex]\scriptsize \begin{align}{f}'(x)&=3{{x}^{2}}+2x-5\(3x+5)(x-1)&=0\x&=-\displaystyle \frac{5}{3}\text{ and }x=1\f(-\displaystyle \frac{5}{3})&=9.48\f(1)&=0\\text{Turning points:}&\text{ }(-\displaystyle \frac{5}{3};9.48)\text{ and }(1;0)\end{align}[/latex]
.
[latex]\scriptsize f(x)=-{{x}^{3}}+4{{x}^{2}}+11x-30[/latex]
Back to Exercise 5.3
Exercise 5.4
[latex]\scriptsize {f}'(x)=-3{{x}^{2}}-6x+9[/latex][latex]\scriptsize \begin{align}{{f}'}'(x)&=-6x-6\{{f}'}'(1)&=-6(1)-6\&=-12\{{f}'}'(1)& \lt 0\\therefore &\text{ local maximum at }x=1\end{align}[/latex]
[latex]\scriptsize \displaystyle {y}'=3{{x}^{2}}+6x-10[/latex][latex]\scriptsize \displaystyle \begin{align}{{y}'}' &=6x+6\{{y}'}'(-1)&=0\\therefore &\text{ there is a point of inflection at }x=-1\\text{Note:}&\text{ we can make the above statement since y is a cubic function}\text{.}\end{align}[/latex]
[latex]\scriptsize {y}'=6{{x}^{2}}-54[/latex][latex]\scriptsize \begin{align}{{y}'}' &=12x\{{y}'}'(3)&=36\{{y}'}'(3)& \gt 0\\therefore &\text{ local minimum at }x=3\end{align}[/latex]
Back to Exercise 5.4
Unit 5: Assessment
.
. [latex]\scriptsize \begin{align}g(0)&=6\\text{y-intercept }(0;6)\g(2)&=0\&\therefore (x-2)\text{ is a factor}\{{x}^{3}}-2{{x}^{2}}-4x+6&=(x-2)({{x}^{2}}-3)\x&=2\text{ or }x=\pm \sqrt{3}\\text{x-intercepts:}&\text{ }(2;0),\text{ }(1.73;0)\text{ and }(-1.73;0)\end{align}[/latex]
. [latex]\scriptsize \begin{align}{g}'(x)&=3{{x}^{2}}-4x-3\3{{x}^{2}}-4x-3&=0 &&\text{ Use the quadratic formula}\x&=-0.54\text{ or }1.87\g(-0.54)&=6.88\g(1.87)&=-0.06\\text{Stationary points:}&\text{ }(-0.54;6.88)\text{ and }(1.87;-0.06)\end{align}[/latex]
. [latex]\scriptsize \begin{align}{g}'(x)&=3{{x}^{2}}-4x-4\{{g}'}'(x)&=6x-4\\text{Note: since }&g\text{ is a cubic function we can let }{g}''(x)=0\text{ to find the point of inflection}\text{.}\{{g}'}'(x)&=0\6x-4&=0\x&=\displaystyle \frac{2}{3}\\text{Point of inflection:}&\&(\displaystyle \frac{2}{3};3.41)\end{align}[/latex]
.
.
. [latex]\scriptsize \displaystyle \begin{align}{f}'(x)&=3{{(x-2)}^{2}}(1)\&=3({{x}^{2}}-4x+4)\&=3{{x}^{2}}-12x+12\{{f}'}'(x)&=6x-12\\text{For }x& \lt 2,\text{ the graph is concave down}\text{.}\\text{For }x& \gt 2,\text{ the graph is concave up}\text{.}\end{align}[/latex]
. Note that the function is a cubic: [latex]\scriptsize \displaystyle \begin{align}{{f}'}'(x)&=6x-12\&=0\\therefore x&=2\f(2)&=0\\text{ point of inflection: }&(2;0)\end{align}[/latex]
.
Back to Unit 5: Assessment
Media Attributions
Figure 1 positive cubic © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Figure 2 negative cubic © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Figure 3 maxima and minima © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Figure 4 point of inflection © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Figure 6 Example 5.3 © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Figure 5 concavity © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Exercise 5.3 Q1 © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Exercise 5.3 Q2 © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Assess Q1 Ans © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
Assess Q2 Ans © Geogebra is licensed under a CC BY-SA (Attribution ShareAlike) license
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National Curriculum (Vocational) Mathematics Level 4 by Natashia Bearam-Edmunds is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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16173 | https://www.jogc.com/article/S1701-2163(19)30452-9/fulltext | Guideline No. 390-Classification and Management of Endometrial Hyperplasia - Journal of Obstetrics and Gynaecology Canada
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Joint GOC-SOGC Clinical Practice GuidelineVolume 41, Issue 12p1789-1800 December 2019
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Guideline No. 390-Classification and Management of Endometrial Hyperplasia
Marie-Hélène Auclair, MDCM
Marie-Hélène Auclair, MDCM
Affiliations
Montréal QC
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∙ Paul J.Yong, MD
Paul J.Yong, MD
Affiliations
Vancouver, BC
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∙ Shannon Salvador, MD
Shannon Salvador, MD
Affiliations
Montreal, QC
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∙ Jackie Thurston, MD
Jackie Thurston, MD
Affiliations
Calgary, AB
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∙ Terence (Terry) J.Colgan, MD
Terence (Terry) J.Colgan, MD
Affiliations
Toronto ON
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∙ Alexandra Sebastianelli, MD
Alexandra Sebastianelli, MD
Correspondence
Corresponding author: Dr. Alexandra Sebastianelli
alexandra.sebastianelli@fmed.ulaval.ca
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Québec, QC
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alexandra.sebastianelli@fmed.ulaval.ca
Affiliations & Notes Article Info
1 Montréal QC
2 Vancouver, BC
3 Montreal, QC
4 Calgary, AB
5 Toronto ON
6 Québec, QC
Footnotes:
This document reflects clinical and scientific consensus on the date issued and is subject to change. The information should not be construed as dictating an exclusive course of treatment or procedure to be followed. Local institutions can dictate amendments to these opinions. They should be well-documented if modified at the local level. None of these contents may be reproduced in any form without prior written permission of the publisher.
All people have the right and responsibility to make informed decisions about their care in partnership with their health care providers. In order to facilitate informed choice, patients should be provided with information and support that is evidence-based, culturally appropriate, and tailored to their needs.
This guideline was written using language that places women at the centre of care. The SOGC is committed to respecting the rights of all people–including transgender, gender non-binary, and intersex people–for whom the guideline may apply. We encourage health care providers to engage in respectful conversation with patients regarding their gender identity and their preferred gender pronouns to be used as a critical part of providing safe and appropriate care. The values, beliefs, and individual needs of each patient and their family should be sought and the final decision about the care and treatment options chosen by the patient should be respected.
DOI: 10.1016/j.jogc.2019.03.025 External LinkAlso available on ScienceDirect External Link
Copyright: © 2019 The Society of Obstetricians and Gynaecologists of Canada/La Société des obstétriciens et gynécologues du Canada. Published by Elsevier Inc.
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Errata
Erratum to “Guideline No. 390-Classification and Management of Endometrial Hyperplasia” [Journal of Obstetrics and Gynaecology Canada 41/12 (2019) 1789–1800]
August 25, 2020
Abstract
Objective
The aim of this guideline is to aid primary care physicians and gynaecologists in the initial evaluation of women with suspected endometrial hyperplasia, to recommend the use of the 2014 World Health Organization classification for endometrial hyperplasia by all health care providers, and to guide the optimal treatment of women diagnosed with endometrial hyperplasia.
Intended Users
Physicians, including gynaecologists, obstetricians, family physicians, general surgeons, emergency medicine specialists; nurses, including registered nurses and nurse practitioners; medical trainees, including medical students, residents, and fellows; and all other health care providers.
Target Population
Adult women (18 years and older) presenting with suspected or confirmed endometrial hyperplasia.
Options
The discussion relates to the medical therapy as well as surgical treatment options for women with and without atypical endometrial hyperplasia.
Evidence
For this guideline, relevant studies were searched in PubMed, Cochrane Wiley, and the Cochrane Systematic Reviews using the following terms, either alone or in combination, with the search limited to English language materials, human subjects, and published since 2000: (endometrial hyperplasia, endometrial intraepithelial neoplasia, endometrial sampling, endometrial curettage, diagnosis) AND (treatment, progestin therapy, surgery, LNG-IUS, aromatase inhibitors, metformin ), AND (obesity). The search was performed in April 2018. Relevant evidence was selected for inclusion in the following order: meta-analyses, systematic reviews, guidelines, randomized controlled trials, prospective cohort studies, observational studies, non-systematic reviews, case series, and reports. Additional significant articles were identified through cross-referencing the identified reviews. The total number of studies identified was 2152, and 82 studies were included in this review.
Validation Methods
The content and recommendations were drafted and agreed upon by the authors. The Executive and Board of the Society of Gynecologic Oncology of Canada reviewed the content and submitted comments for consideration, and the Board of the Society of Obstetricians and Gynaecologists of Canada approved the final draft for publication. The quality of evidence was rated using the criteria described in the Grading of Recommendations Assessment, Development and Evaluation (GRADE) methodology framework. The interpretation of strong and weak recommendations was also included. The Summary of Findings is available upon request.
Benefits, Harms, and/or Costs
It is expected that this guideline will benefit women with endometrial hyperplasia. This should guide patient informed consent before both medical and surgical management of this condition.
Guideline Update
Evidence will be reviewed 5 years after publication to decide whether all or part of the guideline should be updated. However, if important new evidence is published prior to the 5-year cycle, the review process may be accelerated for a more rapid update of some recommendations.
Summary Statements
1
In addition to epidemiologic risk factors related to estrogen exposure, intermenstrual bleeding and postmenopausal bleeding are associated with increased risk of endometrial hyperplasia. Endometrial sampling should be carried out as per published algorithms with particular attention to women 40 years or older or with a body mass index of 30 kg/m 2 or greater (moderate).
2
Since the majority of cases of endometrial hyperplasia without atypia are successfully managed medically, hysterectomy is not considered first-line treatment and surgery is reserved for specific circumstances (moderate).
3
A minimally invasive approach to hysterectomy is preferred for endometrial hyperplasia as it decreases perioperative morbidity and mortality (high).
4
If hysterectomy is indicated for endometrial hyperplasia without atypia then postmenopausal women should also be offered bilateral salpingo-oophorectomy. This decision is individualized for premenopausal women due to increased mortality and morbidity associated with removal of the ovaries in young women with benign disease (moderate).
5
Hysterectomy and bilateral salpingo-oophorectomy are the recommended treatment for atypical endometrial hyperplasia due to the underlying risk of malignancy or progression to endometrial cancer. Retention of the ovaries in premenopausal women may be considered (low).
6
There is no evidence to support routine intraoperative frozen section analysis in cases of endometrial hyperplasia (low).
7
There is no evidence to support routine lymphadenectomy for atypical endometrial hyperplasia (moderate).
8
There is insufficient evidence to support endometrial ablation as first-line surgical treatment for endometrial hyperplasia without atypia (low).
9
Endometrial hyperplasia found in endometrial polyps should be treated according to its histologic classification (low).
Recommendations
1
Health care providers should use the 2014 World Health Organization histopathologic classification of endometrial hyperplasia (strong, low). If endometrial cancer is suspected, endometrial tissue sampling using a Pipelle device in an outpatient setting is the most appropriate first step for diagnosis (strong, high).
2
Those with recurrent symptoms of abnormal uterine bleeding after initial observation or medical treatment should be reassessed with an endometrial biopsy (strong, high).
3
Patients with endometrial hyperplasia should be assessed for reversible risk factors and receive education and support from their clinicians in order to treat and reverse those conditions (strong, high).
4
Patients with endometrial hyperplasia without atypia can be observed. They can be offered hormonal treatment if hyperplasia does not resolve with observation or experience abnormal uterine bleeding (weak, low).
5
The levonorgestrel intrauterine system should be used as the first-line treatment for endometrial hyperplasia without atypia due to its effectiveness and favourable side effect profile (strong, high) and due to the fact that it can be kept in place for 5 years in patients showing treatment response (strong, moderate).
6
Low-dose oral and injectable progestins remain an acceptable treatment option for women with endometrial hyperplasia with and without atypia desiring an alternative treatment modality (strong, high). For patients on oral progestins, we suggest starting on a low dose for a minimum of 6 months. We suggest that assessment of the endometrium be done mid-therapy as well as 3 weeks after completion of treatment to ensure proper interpretation (strong, very low).
7
Surgical treatment of endometrial hyperplasia without atypia should be reserved for patients who do not want to preserve their fertility and experience progression to atypical hyperplasia or carcinoma during follow-up, whose hyperplasia fails to regress after 12 months of medical treatment or relapses after completing treatment with progestins, who continue to experience abnormal uterine bleeding despite treatment, or who decline endometrial surveillance or medical treatment (strong, high).
8
If surgery is indicated for endometrial hyperplasia without atypia, the procedure should include total hysterectomy with opportunistic salpingectomy, with or without bilateral oophorectomy depending on menopausal status (strong, moderate).
9
Total hysterectomy with bilateral salpingo-oophorectomy is recommended for treatment of atypical hyperplasia in premenopausal and postmenopausal women (strong, moderate). In premenopausal women, ovarian preservation should be discussed (strong, moderate).
10
We recommend that subtotal (supracervical) hysterectomy and morcellation be avoided in all cases of endometrial hyperplasia (strong, low).
Key Words
Endometrial hyperplasia
endometrial cancer
endometrial biopsy
progestin
hysterectomy
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Tromino Puzzle: Deficient Squares
S. Golomb gave an inductive proof to the following fact: any 2n×2n board with one square removed can be tiled by right (or L-) trominoes - a piece formed by three adjacent squares in the shape of an L. Golomb's proof of the theorem became a model of elegance in elementary mathematics.
I-Ping Chu and Richard Johnsonbaugh extended Golomb's result to squares with sides not necessarily a power of 2. More accurately:
Theorem
If n ≠ 5, then a deficient n×n board can be tiled with L-trominoes if and only if n is not a multiple of 3.
(The board is deficient if one of the small unit squares is missing.)
Some 5×5 deficient boards can be tiled, others cannot. Although 5 appears as an exceptional case in the theorem, the 5×5 board with a removed corner can be tiled and plays an important role in the proof. The proof does exploits other special cases.
The applet below helps you test your understanding of the theorem by tiling the board manually. It takes three clicks to place a tromino piece on the board: click on three adjacent squares in sequence. (Do not drag the mouse.)
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| What if applet does not run? |
Note that for small (2×2 and 4×4) boards the solution is unique and follows Golomb's proof. For larger boards, viz. starting with 8×8, this is no longer true: there is a good deal of solutions. However, some thinking is still required to tile the board and, more often than not, careless tiling will produce 1 and 2 squares pockets.
Another applet provides additional insight into the tiling with L-trominoes.
Interstingly, we run into an entirely different situation if we try to cover the chessboard with straight trominos. Now, we'll have to consider very carefully which single square may or may not be removed!
References
I-Ping Chu, Richard Johnsonbaugh, Tiling Deficient Boards with Trominoes, Mathematics Magazine, Vol. 59, No. 1 (Feb., 1986), pp. 34-40
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| Related material Read more... | |
| | |
| | - Covering A Chessboard With Domino |
| | - Dominoes on a Chessboard |
| | - Tiling a Chessboard with Dominoes |
| | - Vertical and Horizontal Dominoes on a Chessboard |
| | - Straight Tromino on a Chessboard |
| | - Golomb's inductive proof of a tromino theorem |
| | - Tromino Puzzle: Interactive Illustration of Golomb's Theorem |
| | - Tromino as a Rep-tile |
| | - Tiling Rectangles with L-Trominoes |
| | - Squares and Straight Tetrominoes |
| | - Covering a Chessboard with a Hole with L-Trominoes |
| | - Tiling a Square with Tetrominoes Fault-Free |
| | - Tiling a Square with T-, L-, and a Square Tetrominoes |
| | - Tiling a Rectangle with L-tetrominoes |
| | - Tiling a 12x12 Square with Straight Trominoes |
| | - Bicubal Domino |
| | |
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Copyright © 1996-2018 Alexander Bogomolny
Theorem
If n ≠ 5, then a deficient n×n board can be tiled with L-trominoes if and only if n is not a multiple of 3.
The proof is a consequences of several partial results.
Proposition 1
A 5×5 board with one corner square removed can be tiled
The proof is by inspection:
Observe that the board with any of the squares next to a corner one removed could not be tiled with L-trominoes.
Proposition 2
A (2i)×(3j) board, i, j > 1, can be tiled.
Such a board could be tiled by 2×3 rectangles each composed of 2 trominoes:
More general rectangles can be tiled with L-trominoes, but the above is the only case used below.
Proposition 3
Every deficient 7×7 board can be tiled.
The proof is again by inspection. There are just a few cases to consider. Enumerate squares starting in the upper left corner of the board with two integers (row, column). Because of the symmetry, we only have to check the cases of missing squares for 1 ≤ row ≤ column ≤ 4.
The untiled portion of the board is the deficient 5×5 board from Proposition 1 and can be tiled. If we enumerate squares starting in the upper left corner of the board with two integers (row, column), the diagram tells us that the 7×7 board with the (3, 3) removed could be tiled. With a slight modification it also shows how to tile the board with squares (2, 2), (3, 2), (2, 3) removed.
This diagram serves to show tilings with one of the squares - (1, 1), (1, 2), (2, 2), (2, 1) - removed.
This diagram serves to show tilings with one of squares - (1, 3), (2, 3), (1, 4), (2, 4) - removed.
Two additional tilings (without (4, 4) and (3, 4) squares) complete the proof:
Proposition 4
Any deficient n×n board, with n is odd, n > 5, and not divisible by 3, can be tiled by L-trominoes.
First, we examine a deficient 11×11 board. As the diagram suggests, the board can be split into the a 7×7 piece (that is assumed to contain the missing square), two 4×6 pieces, and one 5×5 piece, with one corner square removed.
As it was already shown, all four can be tiled with L-trominoes.
From here we proceed by induction. Assume that n is odd, not divisible by 3, and greater than 11. For the inductive step, assume that the statement holds for all smaller odd n not divisible by 3.
Split the board as shown below. The (n - 6)×(n - 6) pieces is assumed to contain the missing square.
If n is odd and not divisible by 3, so is (n - 6), making the (n - 6)×(n - 6) tilable by the inductive assumption. On the other hand, for n odd, n - 7 is even, so that the (n - 7) & times; 6 pieces can be tiled by Proposition 2. The deficient 7×7 pieces can, in turn, be tiled due to Proposition 3.
Proposition 5
Any deficient n×n board, with n even, greater than 1 and not divisible by 3 can be tiled with L-trominoes.
We start the induction with the observation that the 2×2, 4×4, and 8×8 deficient boards can be tiled (either by inspection or from Golomb's theorem).
Assume n is even, greater than 8, and is not divisible by 3. Suppose also that all smaller boards with such n can be tiled. Split the board into 4 pieces as shown:
We assume that the missing square is located in the (n - 3)×(n - 3) piece. If n is even and not divisible by 3, then (n - 3) is odd and not divisible by 3. Given that n > 8, n - 3 > 5 so that Proposition 4 applies. The (n - 4)×3 pieces can be tiled by Proposition 2. So the whole board can be tiled and we are done.
A combination of the 5 propositions proves the theorem:
If n ≠ 5, then a deficient n×n board can be tiled with L-trominoes if and only if n is not a multiple of 3.
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Open AccessReview
Update on the Diagnosis of Behçet’s Disease
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Fatma Alibaz-Oner
Fatma Alibaz-Oner
SciProfiles Scilit Preprints.org Google Scholar
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Haner Direskeneli
Haner Direskeneli
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Division of Rheumatology, Department of Internal Medicine, School of Medicine, Marmara University, 34722 Istanbul, Turkey
Author to whom correspondence should be addressed.
Diagnostics 2023, 13(1), 41;
Submission received: 21 October 2022 / Revised: 16 December 2022 / Accepted: 16 December 2022 / Published: 23 December 2022
(This article belongs to the Special Issue Challenges in the Diagnosis and Management of Autoimmune Diseases)
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Abstract
Behçet’s disease (BD) is a systemic inflammatory disease with unknown etiology. It is characterized by recurrent mucocutaneous lesions and major organ disease such as ocular, neurologic, vascular, and gastrointestinal manifestations. The diagnosis of BD is mainly based on clinical manifestations after ruling out other potential causes. There are no specific laboratory, histopathologic, or genetic findings for the diagnosis of BD. The International Study Group (ISG) criteria set is still the most widely used set for the diagnosis. The main limitation of this criteria set is the lack of major organ manifestations such as vascular, neurologic, and gastrointestinal involvement. The ICBD 2014 criteria are more sensitive, especially in early disease. However, patients with such as spondyloarthritis can easily meet this criteria set, causing overdiagnosis. Diagnosing BD can be a big challenge in daily practice, especially in patients presenting with only major organ involvement such as posterior uveitis, neurologic, vascular, and gastrointestinal findings with or without oral ulcers. These patients do not meet ISG criteria and can be diagnosed with “expert opinion” in countries with high BD prevalence. The pathergy test is the only diagnostic test used as diagnostic or classification criteria for BD. Our recent studies showed that common femoral vein (CFV) thickness measurement can be a valuable, practical, and cheap diagnostic tool for BD with sensitivity and specificities higher than 80% for the cut-off value of 0.5 mm. However, the diagnostic accuracy of CFV measurement should be investigated in other disease groups in the differential diagnosis of BD and in also different ethnic populations.
Keywords:
Behçet’s disease; diagnosis; venous wall thickness; ultrasonography
1. Introduction
Behçet’s disease (BD) is a systemic inflammatory disease with unknown etiology. BD was defined as a variable-vessel vasculitis in the revised Chapel Hill Consensus Conference due to the involvement of both arterial and venous vessels of all sizes . BD is first described with “triple symptom complex” as oral ulcers, genital ulcers, and uveitis by Professor Hulusi Behçet in 1937 . It is characterized by recurrent mucocutaneous lesions and major organ involvement, such as ocular, neurologic, vascular, and gastrointestinal manifestations. BD has a distinct geographical distribution. It is more prevalent in countries around the Mediterranean basin and East Asia such as Japan, Korea and China (ancient silk road). The prevalence was reported to be 370/10,000 in Turkey, 11.9/10,000 in Israel, 13.5/10,000 in Japan, 0.64/10,000 in England, 5.2/10,000 in the United States, and 3.8–15.9/10,000 in Italy . BD frequently starts in second and third decades of the life and has a remitting–relapsing course with decreasing disease activity during older age. Complete remission was observed in at least 60% of patients during 20 years of follow-up. The disease course is more severe in male BD patients, especially those younger than 25 years old .
The diagnosis of BD is mainly based on clinical manifestations after ruling out other potential causes. There are no specific laboratory, histopathologic, or genetic findings for the diagnosis of BD. Furthermore, there is a large geographical variation both in the disease prevalence and the disease manifestations. Therefore, the diagnosis of BD may be difficult in patients presenting with only major organ involvement such as posterior uveitis, neurologic, vascular, and gastrointestinal manifestations. The emergence of other disease manifestations aiding the definite diagnosis of BD can take months and even years in this group of patients. The disease can also remain limited in some patients, which causes diagnostic difficulty. In a recent study of 189 suspected or probable BD patients followed up for 3.2 years, only 71 (37.6%) patients were classified as BD according to the International Study Group (ISG) criteria during follow-up . In fact, recent data highlight an increase in the frequency of incomplete BD in Far Eastern countries, such as Japan and Korea . In this group of patients, diagnosis is made according to the presence of specific clinical manifestations of BD by ‘expert opinion’. The specific clinical findings such as genital ulcers, ocular, vascular, and parenchymal neurological involvement were proposed to be defined as strong elements by Yazici et al. for the differential diagnosis of Behçet’s disease . The aim of this brief review is to discuss the new developments in the diagnosis of BD, mainly diagnostic criteria and tools, in light of the current data.
2. Clinical Manifestations of Behçet’s Disease
2.1. Mucocutaneous Manifestations
Recurrent aphthous (oral) ulcers are seen in 95–97% of patients and are usually the first disease manifestation of BD. It precedes the diagnosis by an average of 6–7 years (Table 1). In long-term, routine follow-up, we observed that oral ulcers were the main cause of ongoing clinical activity in BD . Genital ulcers are another major manifestation of BD. They are also the most specific (95%) mucocutaneous sign of BD and seen in 50–85% of BD patients. Genital ulcers are generally located on the scrotum in males and on the major and minor labiae in females. They leave scars in about half of the patients . Papulopustular lesions are seen at usual acne sites, such as the face, upper chest, and back, and additionally on the legs and arms. They are generally indistinguishable from ordinary acne . Erythema nodosum-like lesions are red, painful, erythematous non-ulcerating nodules that are frequently located on the legs. They are seen in approximately 50% of patients and more frequently present in females. Superficial thrombophlebitis is a frequent type of venous involvement. It presents as palpable, painful subcutaneous nodules that are string-like hardenings with reddening of the overlying skin . In some cases, the differentiation of superficial thrombophlebitis and erythema nodosum-like lesions can be very difficult in daily practice.
2.2. Musculoskeletal Involvement
Arthritis or arthralgia is seen in about 50% of patients with BD. Musculoskeletal involvement of BD is a non-deforming, non-erosive peripheral oligoarthritis. Most frequently involved joints are the knees, ankles, hands, and wrists. It usually resolves in days or weeks. Sacroiliitis may be a part of the musculoskeletal involvement of BD in some cases .
2.3. Major Organ Involvement
Ocular involvement is one of the main causes of morbidity in BD. It is observed in up to 50% of patients, and is generally bilateral. Ocular inflammation is commonly panuveitis and retinitis. However, isolated anterior uveitis can be seen in some cases. At the end of a 20-year follow-up, ocular involvement was bilateral in 87% of males and 71% of females .
Vascular involvement is the one of the main causes of mortality and morbidity, especially in young males . Vessels of all sizes can be involved, in both the arterial and venous systems, in the vascular involvement of BD . Major vessel involvement types are vein thromboses, arterial occlusions, and arterial aneurysms. Venous involvement is reported to be more common than arterial involvement (up to 80%). Vascular involvement is observed in up to 40% of the patients with BD. Deep venous thrombosis in lower extremity is the most frequent form of vascular involvement . Although venous thrombosis is seen primarily in lower extremities, it may also affect different sites, including the inferior and superior vena cava, pulmonary artery, suprahepatic vessels, and cardiac cavities. Up to 17% of the mortality in BD is reported to be associated with venous involvement such as pulmonary embolism or Budd–Chiari syndrome .
Neurologic involvement is seen in 5% of patients with BD. There are two main forms: vascular and parenchymal. Parenchymal neuro-BD leads to inflammatory lesions in the brain stem, diencephalon, basal ganglia and, less frequently, the spinal cord and cerebellum. It usually presents with bilateral pyramidal signs, unilateral hemiparesis, behavioral changes, sphincter disturbances, and headache. The vascular neuro-BD is dural sinus thrombosis, mainly characterized by headache and papilledema. It usually associates with venous thrombosis in lower extremities, and has better prognosis compared to parenchymal neuro-BD [17,18].
Gastrointestinal involvement is seen in one-third of BD patients in Japan and Korea, but it is rare in Mediterranean countries (<5%). Mucosal aphthous ulcers, primarily in the terminal ileum and the cecum, are characteristic signs of intestinal involvement of BD. The most frequent symptoms are vomiting, abdominal pain, and diarrhea . Gastrointestinal involvement of BD and Crohn’s disease can be indistinguishable in some cases.
2.4. Other Clinical Findings
Fever is not a typical sign in BD. It can be seen in major vascular, neurological involvement, and arthritis . Testicular pain and epididymitis can be observed in male patients . In a study from Turkey, it was reported that incidence of varicocele was increased in Behcet’s disease . Glomerulonephritis was rarely reported in BD. AA-type amyloidosis can occasionally be seen in longstanding disease .
3. Laboratory
There is no characteristic or pathognomonic laboratory finding in BD. Erythrocyte sedimentation rate and C-reactive protein levels are usually mildly elevated, mainly in cases with arthritis, erythema nodosum-like lesions, or vascular disease. Autoantibodies such as rheumatoid factor, antinuclear, anticardiolipin, and antineutrophil cytoplasmic antibodies are generally absent. However, it was reported that BD patients with gastrointestinal involvement had higher levels of anti-Saccharomyces cerevisiae antibodies compared to BD patients without gastrointestinal involvement . BD has also no specific histopathologic features. The strongest genetic association between HLA-B51 and BD was first reported by Ohno et al. . Then, it was shown in different ethnic populations to have positivity ranges around 40–60% in BD . However, HLA-B51 has low diagnostic value for daily practice usage due to its high frequency in the general population in the countries with high BD prevalence . The genome-wide association studies showed associations with interleukin (IL)-10 and IL23R-IL12RB2 loci, ERAP-1, CCR1-CCR3, KLRC4 and STAT4 [27,28].
4. Diagnostic and Classification Criteria Sets for Behçet’s Disease
The ISG criteria, which are the most widely used for diagnosis, were published in 1990 (Table 2) . This criteria set was generated by a group of experts following large number of BD patients in daily practice. The presence of oral ulcers is accepted as sine qua non. Additionally, two of the following—genital ulceration, eye lesions, skin lesions, and positive pathergy test—are needed for diagnosis of BD. These criteria had been shown to have 95% sensitivity and 98% specificity. The main limitation of this criteria set is the exclusion of major organ involvement such as vascular, neurologic, and gastrointestinal involvement. Low positivity of the pathergy (skin prick) test, possibly related to less traumatic punctures and changing microbial skin flora, is another limitation of the ISG criteria set . Fourteen years later, revised Japanese diagnostic criteria for BD were published in 2004. Clinical manifestations were divided into two groups of main and additional symptoms. Patients with four main symptoms were defined as complete BD; patients with three main symptoms, two main symptoms and two additional symptoms, typical ocular lesions with one main symptom or typical ocular lesions with two additional symptoms were defined as incomplete BD (Table 3) . Although not widely accepted, the definition of incomplete BD led to the early diagnosis of patients presenting with limited disease manifestations, and possibly prevented the delay in the treatment. In 2014, international criteria for BD (ICBD) were published and included vascular and neurological involvement (Table 4) . The ICBD criteria set is based on a scoring system attributing 2 points for oral ulcer, genital ulcer, and ocular lesions; 1 point for positive pathergy test, neurologic, and vascular involvement. Patients having ≥4 points are classified as BD. In a collaborative study of 27 countries, the ICBD criteria set demonstrated an unbiased estimate of sensitivity of 94.8%, which is considerably higher than that of the ISG criteria (85.0%). However, the specificity (90.5%) was lower compared to ISG (96%) . However, the ICBD criteria, which seem more sensitive, especially in early disease, may cause overdiagnosis and patients with spondyloarthropathic features can be mislabeled as BD .
There are separate efforts for the diagnosis of specific disease manifestations of BD. In 2014, expert consensus recommendations were published for the diagnosis and management of Neuro-Behçet’s Disease. In the light of the clinical, laboratory, and neuroimaging findings, disease was defined as “definite” or “probable” Neuro-Behçet’s Disease. The novelty of these criteria is the definition of “probable disease” by expert consensus. According to these criteria, probable Neuro-Behçet’s disease was defined as the presence of typical neurological findings suggesting Neuro-Behçet’s disease without fulfillment of ISG criteria or non-characteristic neurological findings for Neuro-Behçet’s disease with the fulfillment of ISG criteria . Although providing opportunity for the diagnosis of patients lacking typical features, this criteria set has not yet been validated. Recently, Tugal-Tutgun et al. reported an algorithm for the diagnosis of ocular involvement of Behçet disease in 2020. Superficial retinal infiltrate, signs of occlusive retinal vasculitis, and diffuse retinal capillary leakage, as well as the absence of granulomatous anterior uveitis or choroiditis in patients with vitritis were the items with the highest accuracy in classification and regression tree analysis . These organ-specific diagnostic approaches can be helpful, especially for the diagnosis of patients presenting with only major organ involvement in daily practice.
For pediatric BD patients, classification criteria were proposed in 2016 by an international expert consensus group (PEDBD). This criteria set contains six categories (oral, genital, skin, ocular, neurological, and vascular involvement). At least three findings (each from a different category) are needed to define pediatric BD . This new international PEDBD criteria set has higher sensitivity (91.7%) but lower specificity (42.9%) when compared to ISG in the pediatric population .
5. Pathergy Test
The skin pathergy reaction (SPR) is the only diagnostic test currently existing for BD. It is a nonspecific hyperinflammatory response to sterile needle-induced tissue damage, first described by Blobner and Jensen . Despite the lack of consensus on the methodology of pathergy testing, in total, 4–6 sharp or blunted needle pricks are performed. Twenty-gauge needles are inserted either perpendicularly or oblique through the glabrous skin of both forearms after cleaning with an antiseptic. Fresko et al. showed that surgical cleaning of the skin with disinfectants reduces the positivity rate of pathergy test in BD . When 20 G and 26 G needles were tested in BD, higher positivity rate with 20G needles was acquired . The positivity of SPR is defined as the presence of an erythematous papule ≥2 mm or a pustule after 24–48 h. The positivity is affected by many factors such as the usage of sharp or blunted needles, prick number, gender, and disease activity. There is also great geographic variation. The positivity rate is reported to range between 7.7 and 84%. While BD patients had a low positivity rate in Northern Europe, patients from Turkey, China, and Middle Eastern countries had higher positivity rates. SPR can also be positive in other diseases such as Sweet’s syndrome, Crohn’s disease, pyoderma gangrenosum, A20 haploinsufficiency, and a few others. Although SPR is quite specific for BD, the sensitivity decreased in the last few decades, possibly due to less traumatic punctures and changing microbial skin flora . Davatchi et al. assessed the affect of the positivity of pathergy test on the 16 available classification/diagnosis criteria sets for Behcet’s disease. They found that the sensitivity and the accuracy decreased, the specificity increased without positive pathergy test in 15 out of 16 criteria sets . Very recently, Deniz et al. reported that 23 valent polysaccharide pneumococcal vaccine application to the skin increased pathergy positivity both in active (sensitivity: 64.3%, specificity: 100%) and inactive BD patients (sensitivity: 80.3%, specificity: 100%) .
6. Femoral Vein Wall Thickness Measurement as a Diagnostic Tool for Behcet’s Disease
Vascular involvement is seen in about one-third of BD patients . Venous vessels are involved in 67–84% of all vascular manifestations [15,44]. Arterial vessels are involved in less than 15% of vascular BD patients . Deep vein thrombosis (DVT) in the lower extremity is the most frequent form of vascular BD (around 80%). Despite the dominant venous involvement, the data investigating veins directly are very limited in BD. Few data pointing out the vein wall inflammation also in skin, ocular, and neurologic involvement of BD have been previously published [46,47,48,49].
The first study directly assessing veins in BD was reported by Ambrose N et al. In this magnetic resonance imaging study, increased vein wall thickness (VWT) was reported in popliteal veins of BD patients . Boulon et al. later published a case presenting with acute calf pain (without thrombosis). Increased VWT in the right great saphenous vein was detected in this case by ultrasound (US) . We recently published the first controlled doppler US study showing increased VWT of lower extremity veins in male BD patients independent of vascular involvement [52,53]. Among all thickened lower extremity veins, the common femoral vein (CFV), which is the largest vein of the lower extremity, was chosen as the primary site in the US assessment. Bilateral CFV measurements had high area under the receiver operating characteristic curves (>0.8) with sensitivities of 81–82.8% and specificities of 78.4–81.1% for the cut-off values of 0.5 mm. Positive (PPV) and negative (NPV) predictive values for this cut-off were also acceptable (PPV: 85.7–87.5%, NPV: 72.5–75%) in our study [52,53] (Figure 1). Other studies from Turkey reported increased VWT in BD, and confirmed our observations [54,55,56]. Recently, we assessed the wall thickness of the pulmonary artery, which is most frequently involved artery in BD. Pulmonary artery had a similar structure to systemic veins in terms of width and thin-walled vessels with increased compliance. We found that PA wall thickness is increased in patients with only major organ involvement, which can be a sign of more severe disease in BD patients .
We later assessed the diagnostic performance of CFV thickness measurement in BD compared with multiple disease control groups such as ankylosing spondylitis, systemic vasculitides, antiphospholipid syndrome, venous insufficiency, and non-inflammatory DVT. Our findings indicated an increased VWT, a distinctive feature of BD, rarely present in other inflammatory or vascular diseases. The cut-off value of ≥0.5 mm, determined in our first study, performed quite well against all control groups with sensitivity and specificity (with the exception of antiphospholipid syndrome) higher than >80%. We also found that values especially higher than 0.75 mm seem to indicate a very high probability of BD . In our recent study, CFV measurement was also found to be a distinctive diagnostic tool for the differentiation of BD and Crohn’s disease, which is an important challenge in daily practice, in especially BD patients with GI involvement . Most recently, increased venous wall thickness was found in childhood BD with and without vascular involvement, and Atalay et al. suggested that increased VWT may be a new criterion for the diagnosis in both definite and incomplete pediatric BD patients . Taken together, assessing VWT seems to be a cheap, easy, and widely available tool aimed at the diagnosis of BD.
7. Conclusions
The diagnosis of BD is mainly based on clinical manifestations after excluding other potential mimickers. There is no specific laboratory or genetic test. Diagnosing BD can be a big challenge in daily practice, especially in patients presenting with only major organ involvement with or without oral ulcers. The early diagnosis of BD has a critical value, especially in patients presenting with only vascular thrombosis to prevent the delay in the treatment. The treatment of vascular thrombosis in BD is different from noninflammatory DVT, as it needs immunosuppressive treatment rather than anticoagulants. Until recently, the pathergy test was the only diagnostic test that was used as diagnostic or classification criteria for BD. Our recent studies showed that CFV thickness measurement can be a valuable, practical, and cheap diagnostic tool for BD with the sensitivity and the specificities higher than 80% for the cut-off value of 0.5 mm. However, the diagnostic accuracy of CFV measurement should be investigated in other disease groups in the differential diagnosis of BD and in different ethnic populations.
Author Contributions
Conceptualization, F.A.-O.; writing—original draft preparation, F.A.-O.; writing—review and editing, F.A.-O. and H.D.; supervision, H.D. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
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Figure 1. Measurement of common femoral vein thickness by ultrasonography.
Figure 1. Measurement of common femoral vein thickness by ultrasonography.
Table 1. Frequency of Behçet’s disease manifestations.
Table 1. Frequency of Behçet’s disease manifestations.
| Manifestation | Frequency (%) |
--- |
| Oral ulcers | 97–99 |
| Genital ulcers | 85 |
| Genital scar | 50 |
| Papulopustular lesions | 85 |
| Erythema nodosum | 50 |
| Pathergy reaction | 40–60 |
| Uveitis | 50 |
| Arthritis | 30–50 |
| Subcutaneous thrombophlebitis | 25 |
| Deep vein thrombosis | 15 |
| Arterial occlusion/aneurysm | 5–10 |
| Central nervous system involvement | 20 |
| Epididymitis | 5 |
| Gastrointestinal lesions | 1–58 |
Table 2. International Study Group criteria for the diagnosis of Behçet’s disease.
Table 2. International Study Group criteria for the diagnosis of Behçet’s disease.
| Manifestation | Definition |
--- |
| Recurrent oral ulceration | Observed by a physician or reported reliably by patient, recurring at least three times in one 12-month period |
| Plus any two of the following findings: | |
| Recurrent genital ulceration | Recurrent genital aphthous ulceration or scarring, observed by a physician or reported reliably by patient |
| Eye lesions | Anterior uveitis, posterior uveitis, or cells in vitreous on slit lamp examination; or retinal vasculitis observed by qualified physician (ophthalmologist) |
| Skin lesions | Erythema nodosum-like lesions, observed by a physician or reported reliably by patient; Pseudofolliculitis or papulopustular lesions; or acneiform nodules observed by a physician in post adolescent patients not receiving glucocorticoids |
| Positive pathergy test | Test interpreted as positive by a physician at 24–48 h, performed with oblique insertion of a 20-gauge needle or smaller under sterile conditions |
Table 3. Revised diagnostic criteria proposed by the Behcet’s Disease (BD) Research Committee of Japan.
Table 3. Revised diagnostic criteria proposed by the Behcet’s Disease (BD) Research Committee of Japan.
| |
| Main Symptoms |
| Recurrent aphthous ulcers on oral mucosaSkin lesions a. Skin lesion with erythema nodosum b. Subcutaneous thrombophlebitis c. Follicular papules, acneiform papules c. cf.) Skin hypersensitivity Ocular lesions a. Iridocyclitis b. Posterior uveitis (retinochoroiditis) c. If the patients have the following eye symptoms after (a) and (b), diagnose as BD lesions in accordance with (a) and (b) c. Posterior adhesion of iris, pigmentation on lens, retinochoroid atrophy, atrophy of optic nerve, complicated cataract, secondary c. glaucoma, leakage of bulbus oculi Genital ulcer |
| Additional symptoms |
| Arthritis without deformity or sclerosis Epididymitis Gastrointestinal lesion represented by ileocecal ulceration Vascular lesions Central nervous system lesions, moderate or severe |
| Criteria for diagnosis of disease types |
| Complete type: The four main symptoms appeared during the clinical course Incomplete types: Three of the main four symptoms, or two main symptoms and two additional symptoms, appeared during the clinical course Typical ocular lesion and another main symptom, or two additional symptoms appeared during the clinical course BD suspected: Although some main symptoms appear, the case does not meet the criteria for the incomplete type Typical additional symptom is recurrent or becomes more severe Special lesions: Gastrointestinal lesions—presence of abdominal pain and occult blood should be confirmed Vascular lesions—vasculitis of aorta, artery, large veins, or small veins should be differentially diagnosed Neuronal lesions—presence of headache, paresis, lesions of brain and spinal cord, mental symptoms, and other symptoms should be confirmed |
Table 4. International criteria for Behçet’s disease.
Table 4. International criteria for Behçet’s disease.
| BD Manifestations | Score Assigned |
--- |
| Ocular lesions | 2 |
| Oral aphthosis | 2 |
| Genital aphthosis | 2 |
| Skin lesions | 1 |
| Neurological manifestations | 1 |
| Vascular manifestations | 1 |
| Positive pathergy test | 1 |
Patients are classified as having BD with scores ≥4. Pathergy test is optional.
| |
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16176 | https://www.youtube.com/watch?v=fCai4-l8YVo | Can you Solve this using Parity? | Georgia TST 2005 Problem 7 | Cheenta
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Transcript:
Introduction hi everyone so today we're gonna learn about something called as parity which is something very fundamental to mathematics especially number Theory but um it's a very important problem solving tool when it comes to Olympiad mathematics so we're gonna see how we can actually employ a parity a very powerful chromosome technique in one of our problems so without wasting any time let's get started so this is a question from the team selection test for the country of Georgia and it is day three problem number seven in the year 2005. Outline and in this video we're going to be looking at how we can maybe exploit this parity and this perfect squares a couple of things over there then we're obviously going to look at induction and then we have some book systems so senior Math olympiads and at the end a similar will challenge problem this video is sponsored by chinta.com since 2010 Chinta has trained thousands of students from all around the world olympiads physics olympiads computer science and informatics olympiads isi CMI entrances and research projects for school and college students okay so we are very short but uh interesting question really and we need to determine all positive integers and positive integers or otherwise natural numbers and for which this quantity is a perfect square right so 2 raised for n minus 1 times n plus 1 is M Square for some natural number m right it's it's a perfect square so whenever I have this um a setup like this I can take this one to the other side and just kind of factorize this right the 2 is for n minus 1 times n is equal to M Square minus 1 which is M plus 1 times M minus 1. and I think this is uh very common to all questions in which we have perfect squares factorization parity you know reducing mod M these are a couple of ideas that you should keep in your mind because really it's uh very rarely that we see a question that goes beyond these ideas but anyways so if we actually check at n equals to 1 what do we get we'll have 2 raised power 1 minus 1 which is 0 times 1. is equal to M Square minus 1 right on other words M Square minus 1 is equal to 1 so we see that M square is equal to 2 which is obviously not ten integer it's not it's not natural number right so n is equal to 1 gives no solution and we know that N is a natural number so therefore we can safely conclude that n is 1. or or you know in other words N is greater than or equal to 2. okay perfect now then if you actually notice this left hand side 2 raised for n minus 1 times n is equal to M plus 1 times M minus 1. now left hand side this thing is always even for all n greater than 1 it is always and always even because we have this power of 2 which is always going to be even so this entire quantity also needs to be even now now what we notice is that M minus 1 and M plus 1 both of these quantities have the same parity right which means that either both of them are ordered both of them are even parity essentially ordinance or evenness of a particular number right if if two numbers are either both out of both even we say that both are the same parity if one of them is odd and the other one is even we say that they have different parity okay so M plus 1 and M minus will have the same parity and we want their product to be even so therefore M plus 1 and M minus 1 are both even or in other words that you can just imply that m is all [Music] now because m is odd I can write M as 2A plus 1 for some a belonging to some let's say natural number and when I substitute this into our original equation I'll get 2 raised for n minus 1 times n is equal to 2A plus 1 plus 1 times 2A plus 1 minus 1 M plus 1 M minus 1 you know now what I'm going to do is I can write this as I can take a 4 common over here and I can write this as a times a plus 1. and if I divide by 4 on both sides of this equation I'll get a times a plus 1 is equal to 2 raised power n minus 3 times n now again we're going to employ parity the parity is involved in a lot of places over here okay now we notice that A and A plus 1 these two quantities have opposite parity right if one of them is even the other one will obviously be odd if one of them is odd the other one will obviously be even but the right hand side is always even what does that mean it essentially means that the even term is gobbling up the entire evenness of the right hand side right so you can just distinctly split this up into two two cases right so case one will be let's say where a is even so that implies a gobbles up all the even Powers so 2 raised for n minus 3 times B and obviously a plus 1 is C such that [Music] BC is equal to n right this is the construction for our this particular case now now what I'm going to do is I'm going to make a claim so my claim is that 2 raised power n minus 3 is in fact greater than n for all n greater than equal to 6 and how are we going to prove this we're going to prove this via induction okay so let's first test the base case what is the base case it is n is equal to 6 so 2 is power 6 minus 3 greater than 6 we can see that this is very true 2 is power 3 is greater than 6 which is true 8 is greater than 6. now now what do we do next we just claim our induction hypothesis foreign hypothesis which we essentially intend to prove so we suppose that K greater than equal to 6 for all K greater than equal to 6 we have 2 raised for K minus 3 greater than K right this is our obviously an induction hypothesis and once we assume this for K we need to prove for K plus 1 so 2 raised power K plus 1 minus 3 is greater than K plus 1 or otherwise I need to prove that 2 is power K minus 2 is greater than K plus 1 this is what I need to prove with the assumption that this is true right so 2 raised per K minus C greater than K this is true where induction hypothesis I can multiply by 2 on both sides without any change in the inequality so I'll get 2 is power K minus 2 is greater than K plus k or in other words I can write 2 is for K minus 2 is greater than K plus 1 because K is strictly greater than 1. right and do you notice what is this this is what I actually had to prove over here and uh so yeah our induction hypothesis is true and it is really fascinating because all we really added towards multiply by 2 and simplify it and we see that our induction hypothesis is true therefore our claim is true so therefore um 2 raised power n minus 3 is indeed greater than n for all n greater than or equal to 6. but you might think that this is something very very uh vague right it's it's very abstract it's not related to the question you know what I basically I just wrote this case and I just went down to some abstract claim but you'll actually see that this is very important because we have two response n minus 3 over here as well and obviously BC is equal to here so there's probably some some similarity over here and let's see what that is okay so we Proof have this this claim that I just proved by induction now notice that a was 2 is power n minus 3 times B so effectively a is greater than n right cos a is essentially equal to 2 is power n minus c times P now this quantity is greater than n and a is obviously uh basically a quantity greater than n multiplied by another quantity so obviously a will always be greater than n right now A is greater than n now what next now we also know that a plus 1 was equal to C let me just remind that whether it go yes this a plus 1 was equal to C obviously we see that c is greater than AC is 1 More Than A and C is greater than a greater than n so we can essentially imply that c is greater than n so we have these two results a greater than n and C greater than n if you also remember b c was actually equal to n or another was accurate B is equal to n by C but since C is greater than n b cannot be an integer B is not an integer right because n divided a number divided by a larger number will never be an integer right it will always be a fraction or a rational number it can never be an integer if you think about it so therefore n greater than or equal to 6 will have no Solutions because essentially our claim this this entire this entire claim is only valid for greater than equal to 6. before that is not valid right so this entire construction this entire inequality that I'm forming are all are all only valid for n greater than equal to 6 that means if solution exists it will only be 2 3 4 5 I'm not including one cause is already seen n is greater than one n is equal to one does not work we've already seen that before so n is equal to 2 3 4 5 will give our only possible solution so if a solution X is it will be at this four particular values right and we can check them you can check them at the end so I'll just move on to case number two and uh you can check them or we'll just check them at the end I think they'll be more comfortable now now what I'm going to do is I'm going to assume that a is odd right basically the other case uh in case one I assume is even right so here A Plus One will gobble up all the even Parts because n minus times p and a will obviously gobble up everything else right and in this case BC is obviously equal to n now here we use a similar argument uh 2 raised for n minus 3 is greater than n plus 1 for all n greater than or equal to 6 and this is basically the same thing what I've done before and this I'm Gonna Leave You I'm going to leave this to you actually maybe try proving this by yourself same induction based on the same process and uh try doing this yourself it's very simple actually and you will prove it by induction and once you have proved this claim you will notice that essentially a plus 1 is 2 is for n minus 3 times p and because this quantity is greater than n plus 1 that implies that a plus 1 is greater than n plus 1 you can just subtract one for both sides so it essentially implies that a greater than n and because a is equal to C that implies C is greater than n or I can just write a is equal to 3 greater than n but BC was equal to n and by a similar assertion B is n by C and B can never be integer right this is the same argument C is greater than n you have a smaller number divided by bigger number obviously can never be an integer can never ever be an integer right so no Solutions again for all n greater than equal to 6 because again this assertion this assertion was only for n greater than or equal to 6 so therefore if a solution exists it will be at n two three four five what are we seeing from here we are seeing that both in case number two and up top in case number one only solutions are n two three four five and that's all that we have to check right so we just check these four values and see at which of them the quantity 2 is for n minus 1 times n plus 1 is indeed a perfect square and we see that n is equal to 2 is not a perfect square at n is equal to 3 it is not above x square and N is equal to 4 also it's not a perfect square but at n is equal to 5 we get um 2 raised power 5 minus 1 4 times 5 plus 1 16 times size 80 plus 1 81 which is 9 squared so n is equal to 5 is our Only Solution actually [Music] right so that's quite fascinating I think there was a very uh lucrative discussion of thing called parity and that concept of parity really helped us to kind of simplify this down to two distinct cases after which it was some really good induction and we powered through the solution so I hope you learned something from that and hopefully you can think about this parity in number three problems Books okay so moving on we are send books lessons of senior Math olympiads I'm a compendium Paul lome's Elementary number three secrets and inequalities and functional equations and how to solve them by Christopher G small Summary okay so at the end it was sort of a challenging problem and I wanted to find all positive integers M comma n such that this product 2 is for M minus 1 times 3 raised power n minus 1 is a perfect square now admittedly this is a challenging problem okay this is not the easiest um at least the methods I found were not easy if you found a easy method let me know and maybe try this out we try reducing mod M maybe try using some parity ideas whatever comes to your brain just give it a go and if you're able to do it let me know until then I'll see the next video thank you very much and bye-bye in the Outro programs are designed for students who are passionate about mathematics and they are personalized with one-on-one training individual evaluation and remedial sessions the reason Chinta students are successful over the last 10 years because they are taught by mathematicians and real olympiads from leading universities in India United States and Europe [Music] some of our students come back to teach at Chinta from Oxford Cambridge Harvard MIT UCLA isi CMI iits TFR and iisc for more information visit chinta.com |
16177 | https://physics.stackexchange.com/questions/588747/for-discrete-optimization-problems-why-does-continuous-approximation-solving | ising model - For discrete optimization problems, why does continuous approximation - solving - discretization work? - Physics Stack Exchange
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For discrete optimization problems, why does continuous approximation - solving - discretization work?
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I was wondering about the following question.
In maths/computer science, many optimization problems ask for integer optimizations. A prominent example would be spin-glass (Ising) systems for simulated annealing (which encompasses quadratic unconstrained binary optimisation problems, such as Max-Cut etc.). Here, we have an optimisation function
and the optimization values are binary, i.e. .
Commonly, to solve this problem (e.g. with simmulated annealing, or simulated quantum annealing ; or with gradient-based methods) we approximate the values as real numbers to achieve differentiability of .
My question now is why this actually works / when this gives the right solution. In problems like the one mentioned above, but also for e.g. training of binary neural networks, the continuous outcome parameters will be discretized in a final step, mostly according to some threshold function . Then, a common choice is to set the continuous value to +1 if it is above some treshold and to -1 otherwise.
My question concerns the following: I do not see how
discrete problem --> optimization --> discrete-valued solution
gives the same optimal output as
discrete problem --> continuous relaxation --> optimization --> discretization of continuous solution --> discrete-valued solution
for a couple of reasons. Depending on the discretization function we bias our real-valued answers towards some direction. For the above example of spin-glass, this seems to induce spin-flip errors. Yet, depending on the problem, nearest-neighbour flips can already lead to strongly sub-optimal solutions.
I would be happy if you could share your insight here. Specifically, (1) is such a continuous-variable relaxation generally "good", in particular for spin-glass sytems (and if yes, how good? and why?), (2) if not, when is continuous relaxation a good thing to do and (3) apart from computational convenience, what are the reasons to do the relaxations? Are there any maths papers / theorems that discuss this issue?
Best regards, Dan
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edited Oct 22, 2020 at 2:54
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asked Oct 22, 2020 at 2:39
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Find all intersections for m timespans of n parties
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I just can't figure out a nice way to solve the following problem:
For an event I have n parties (party_id) taking part. Each party has m availabilities for said event in the form of start_date and end_date.
What I would like to know is all possible combinations of overlapping availabilities containing exactly one availability for each party_id. I discovered Interval Trees ( which might be utilized, but as I said I can't quite figure it out.
So thanks for any thoughts on that!
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asked Sep 11, 2017 at 19:13
Daniel BeckerDaniel Becker
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Just to get it right, what's meant with all possible combinations etc etc.... Could you provide an example just for clarity?yacc –yacc 2017-09-11 19:20:43 +00:00 Commented Sep 11, 2017 at 19:20
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The straightforward way is sort all your events (start_date and end_date alike) by time. Then go through this list in chronological order, while keeping track of all "active" parties (i.e., those which have started, but not stopped).
The above method is a batch process, which should let you find all possible combinations of a given schedule easily. The interval tree you mention is useful in cases where you want to incrementally update your set of parties -- the datastructure can make individual queries about a particular time much cheaper than re-running the batch process for each update or query.
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Daniel Becker
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Events do not have times. They are available for any timeslot all parties are available at the same time. On second thought, that might be possible for all event availabilities.
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16179 | https://www.youtube.com/watch?v=ImwVg6OQXD8 | Complex Numbers : Equations - Equating real imag. parts : ExamSolutions
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hi now in this tutorial what I want to do is show you how you can solve problems involving complex numbers and the method we're going to use is where we equate the real parts and the imaginary parts and at the end of this tutorial got a couple of questions here you might like to try uh two is harder than one but uh they follow the same principles as I'm going to show you and I'll give you the work Solutions of these also so how do we solve an equation like this a + 2 bi i - b + a i all divid 7 - 2 I = 1 and we've got to find these values of A and B what we need to do is clean up the top for starters group together the real parts and the imaginary parts so on the top here we've got a minus B so we'll put therefore a minus B and we'll bracket that together as the real part and then for the imaginary part we've got an AI and 2 bi so we could put plus and in Brackets a + 2 B and then that that's the imaginary part and it's all divided then by 7 - 2 I and it equals 1 now what I'll do is get rid of this fraction here 7 - 2 I by multiplying both sides by 7 - 2 I so we have a minus B+ a + 2 B and that's the I part the imaginary part equals 1 7 - 2 I so it's just going to be 7 - 2 I and when you get equations like this when you've reduced it to this form just on one line what we can then start to do is compare the real parts and the imaginary Parts because this is the real part A minus B and it must be equal to this real part here seven and the imaginary part A + 2 B must be equal to the imaginary part over here minus 2 so let's just write that down that if we equate the real parts we just put R we've got a minus B = 7 then and if we equate the imaginary Parts I then we've got a + 2 b equal the minus 2 and what we have here is a couple of simultaneous equations we'll call them one and two and to solve simultaneous equations then we can either use substitution or elimination method I'm going to use elimination method here because we've got the two A's that are exactly the same so I'm going to do equation 1 minus equation 2 if we do that the A's get eliminated and we got Min - B Min - 2 B which is - 3 B = 7 - -2 which is 9 IDE by -3 and we then end up with B equaling -3 and then what we need to do is substitute that back into say equation one so Sub in one and what we end up with then is a - - 3 = 7 so a - - 3 = 7 so a + 3 = 7 subtract three from both sides and you end up with a = 4 so we end up then if we just summarize with a = 4 B = -3 so what we need to do then is reduce this to a single line like this compare the real parts and the imaginary parts and if necessary solve some simultaneous equations so hopefully that will give you an idea then of starting on uh example one here and then going on to two so I'll give you a few moments just to copy those down have a go at them and we'll run through them uh in a moment okay welcome back if you had a go let's have a look then at the first question so the first thing that I'd want to do here is expand out these two brackets so if we were to do that we get a 5 which is 5 a then you've got minus a i and + 5 b and then minus bi and that equals all of this over here I'm going to start to Clump this together we've got 10 a plus b + 1 that's the real part and then Min - 2 a + 1 which is the imaginary part and so we need to clean this up here group together again the real Parts we've got 5 a + 5 b that's the real part there and as for the imaginary part we've got minus and then a bracket A + B and that's the imaginary part so it's going to be equal to the 10 a plus b + 1 minus the 2 a + 1 I so what we can do now is equate the real Parts together and if we do that we've got 5 a + 5 b equal the 10 A + B + 1 now if we were to subtract 10 a from both sides and B from both sides we'd end up with 4 b - 5 a = 1 and we can't go much further with that so we'll just call that one and if we now equate the imaginary Parts together then we've got minus A + B minus braet a + b must equal minus then the 2 a + 1 well clearly that means that a + b must equal the 2 a + 1 so A + B must equal 2 a + 1 that's if we multiply both sides by minus one and clean this up we could take a from both sides and therefore we get B = A + 1 and if we call this two all we need to do now is just substitute 2 into one so if we just say sub 2 into 1 what do we get well we're going to have four lots of a + 1 so that's going to be 4 a + 4 if we expand that minus the 5 a = 1 and if we carry on with this we'll just come down here we could do 4 A minus 5 a which is min - A therefore got minus a and then if we take four from both sides we've got 1 - 4 which is min -3 leaving us with a = 3 sub this now back into say 2 and that gives us that b equals a + 1 3 + 1 which is going to be 4 if we summarize we've got a = 3 then and B = 4 so I hope you're able to get that one now I did say question two is a little harder so we'll have a quick look at uh that one now in two what I'd want to do again is to expand this bracket out so if we do that we've got a 2 which is 2 a and then a - 3 I which is - 3 a i bi I 2 that's going to be + 2 bi and then plus bi - 3 I well that's going to be + 3B remember i^ 2 is min -1 and then we've got AB + 6 + AI group together the real parts and the imaginary Parts we've got 2 A+ 3B so 2 a + 3 B I'll put that in Brackets and as for the imaginary Parts we've got 2 B and then Min - 3 a and that's the imaginary part and over here we've got ab+ 6 that's the real part there I'll put that in Brackets and then plus AI so if we now equate the real Parts what have we got it's going to be 2 a + 3 B equals AB + 6 and can't really get much mileage out of this so we'll call that equation one and if we now equate the imaginary parts then we've got 2 B- 3 a must equal the A and what we could do now is add 3A to both both sides and so therefore we've got 2 B = A + 3 a which is 4 a and that means that b equal 2 a if we divide both sides by two so we'll call that equation two and to solve this all I've got to do now is substitute b = 2 a into equation one so if we say sub b = 2 a into 1 then what are we going to get well we're going to have 2 a so we just put there 2 A plus three lots of B three lots of B is going to give us 6 a equal a B so that's a the 2 a so that's 2 a 2 + 6 so we've got a quadratic equation here this is going to be 8 a = 2 A2 + 6 if we rearrange this we've got 2 a 2 take away the 8 a and then + 6 = 0 we could divide both sides now by two and we could thin this out then to give us a^ 2 - 4 a + 3 = 0 which will factorize couple of brackets then and we've got an A and an A and we've got a minus 3 and and a minus1 so each of these factors could equal zero giving us a = 3 or a = 1 and then we just need to substitute these values back into two so if we just say sub a = 3 and a = 1 into 2 then what we have is that when a = 3 b = 2 2 3 which is 6 and when a = 1 B will equal 2 1 so that's going to be two so couple of solutions here you could check them out if you like there a little exercise to check that they work but you'll find that they do okay so I hope that's given you some idea then how you can go about solving equations like this just by comparing the real parts and the imaginary parts which can often lead into simultaneous equations right |
16180 | https://physics.stackexchange.com/questions/494344/constraint-forces-motion-of-a-bead-on-a-wire | Skip to main content
Constraint Forces - Motion of a bead on a wire
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I am trying to understand the transition from Newtonian Mechanics to Lagrangian Mechanics. I have been looking at various examples of physical problems, starting from Newtonian Mechanics, and trying to 'naturally' transition to Lagrangian Methods.
I'm currently looking at the motion of a bead on a wire with no applied forces or friction. So only the constraint force acts to keep the bead on the wire.
The bead is started off with some velocity.
For simplicity, the wire is fixed in space.
Thinking about conservation of energy, there is nowhere the kinetic energy can reasonably go to - no fields, no friction, etc, so the absolute scalar speed should be constant. Also velocity is tangential to the bead, so the tangential velocity component should be constant. This tallies with the idea that constraint forces are perpendicular to a constraint surface - those constraint forces cannot change the tangential velocity at any point.
So the bead should just move along the wire at constant speed. i.e. if the wire had length intervals marked onto it, the speed in those units should be constant.
I find this counterintuitive - surely around a sharp bend in a wire, the speed should slow down? Or speed up? It's hard to accept that it just moves along at constant speed, the structure of the wire being more or less irrelevant to the problem. i.e. the bead can experience unlimited amounts of force, but because it is perpendicular to the wire, its tangential speed never changes.
Edit:
The root cause of my confusion I think is this - it is often said that movement along a wire is caused by a constraint force which is always perpendicular to the wire. Is there any clear argument to support this?
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edited Jul 30, 2019 at 16:13
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Concerning your edit: a better expression would be "the reason the bead follows the wire rather than going straight is because of constraint forces ...". The reason for longitudinal motion at all is conservation of energy.
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The root cause of my confusion I think is this - it is often said that movement along a wire is caused by a constraint force which is always perpendicular to the wire. Is there any clear argument to support this?
You have already stated that the only force in this system is caused by interactions with the wire. Saying the force is always perpendicular to the wire equivalently means that the force never has a component parallel to the wire. It might be easier to see why this is the case. Any force component parallel to the wire on the bead caused by the wire would mean the wire is somehow pulling on the bead due to some sort of friction force, or pushing on the bead along the wire due to some mechanism (not sure what that would be exactly). There is no mechanism (ideally) where the wire would have this capability. Therefore, the force must always be perpendicular to the wire.
Addressing your faulty intuition, you might be thinking about how when you go around a sharper curve you have a larger magnitude of acceleration. And this intuition is true. However, as you have correctly reasoned this acceleration is always perpendicular to the velocity, so the speed along the wire cannot change.
the bead can experience unlimited amounts of force, but because it is perpendicular to the wire, its tangential speed never changes.
I wouldn't say "unlimited amounts of force". If your wire does have a kink in it to require an infinite acceleration then I think we are going to be in trouble in thinking about an infinite force. If instead you mean you could contrive in your mind a wire with an infinite number of bends, then that is fine. The motion of the bead will not remember how many turns it has encountered, and it does not know how many more turns it will encounter. You have set up a system where kinetic energy can never change, so it should not be surprising that you have found the only force present cannot do any work on the bead.
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answered Jul 30, 2019 at 18:23
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Walker HK, Hall WD, Hurst JW, editors. Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition. Boston: Butterworths; 1990.
Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition.
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Walker HK, Hall WD, Hurst JW, editors.
Boston: Butterworths; 1990.
Contents
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Chapter 73The Plantar Reflex
H. Kenneth Walker.
Definition
Stroking the lateral part of the sole of the foot with a fairly sharp object produces plantar flexion of the big toe; often there is also flexion and adduction of the other toes. This normal response is termed the flexor plantar reflex.
In some patients, stroking the sole produces extension (dorsiflexion) of the big toe, often with extension and abduction ("fanning") of the other toes. This abnormal response is termed the extensor plantar reflex, or Babinski reflex.
Technique
Place the patient in a supine position and tell him or her that you are going to scratch the foot, first gently and then more vigorously. Fixate the foot by grasping the ankle or medial surface with the examiner's hand that will be closest to the midline of the patient: examiner's left hand when the patient's left foot is being tested, and vice versa with the right foot. Begin with light stroking, using your finger; then use a blunt object such as the point of a key. Finally, if no abnormal response has been obtained, take a tongue blade and break it in half longitudinally and use the sharp point. The reason for the graded stimuli is twofold: (1) Light touch, as with a finger, frequently obtains the reflex without causing a withdrawal response that on occasion makes interpretation of the response difficult; (2) one cannot conclude the response is normal until a noxious stimulus is indeed used, since the reflex is a cutaneous nociceptive reflex.
The first line to be stroked begins a few centimeters distal to the heel and is situated at the junction of the dorsal and plantar surfaces of the foot (Figure 73.1). The line extends to a point just behind the toes and then turns medially across the transverse arch of the foot. Stroke slowly, taking 5 or 6 seconds to complete the motion. Do not dig into the sole, but stroke.
Figure 73.1
Testing the plantar reflex.
Successive lines are stroked, each about 1 cm medial to the preceding stroke, until the examiner is stroking the midline of the foot. The reason for beginning laterally is that in some cases the response is abnormal laterally and then becomes normal as the midline is approached. The occurrence of the extensor response on any of these lines is abnormal, even if the response is flexor on another line of stroking. This variation relates to variability in the receptive field of the reflex, undoubtedly due to the extent of corticospinal involvement as well as individual differences.
The reflex is normal if the abnormal response is not obtained from any of the stroke lines using all of the stimuli described.
A good habit is to describe whatever response is obtained in addition to noting whether in your opinion the response is normal or abnormal.
Basic Science
The neurophysiology of this reflex has not been completely elucidated. The account given here follows the suggestions made by Kugelberg, Eklund, and Grimby (1960) and is the result of their electromyographic studies. Each area of the skin of the body appears to have a specific reflex response to noxious stimuli. The purpose of the reflex is to cause the withdrawal of the area of the skin from the stimulus. This reflex is mediated by the spinal cord, but influenced by higher centers. The area of skin from which the reflex can be obtained is known as the receptive field of the reflex. To be more specific, a noxious stimulus to the sole of the foot, which is the receptive field, causes immediate flexion of the toes, ankle, knee, and hip joints with attendant withdrawal of the foot from the stimulus. Remember your own experiences with this reflex, an example being stepping on a sharp object while barefoot. There is an instant involuntary flexion of all joints with withdrawal. Another reflex in the normal individual is the great toe reflex: Stimulation of the ball of the toe, which is the receptive field, causes extension (dorsiflexion) of the toe with flexion at ankle, knee, and hip joints. The differences between these two reflexes are in the receptive fields and the fact that the great toe is flexed in one and extended in the other. The reason for the extension in the toe reflex is to remove the toe from the stimulus.
The abnormal plantar reflex, or Babinski reflex, is the elicitation of toe extension from the "wrong" receptive field, that is, the sole of the foot. Thus a noxious stimulus to the sole of the foot produces extension of the great toe instead of the normal flexion response. The essential phenomenon appears to be recruitment of the extensor hallucus longus, with consequent overpowering of the toe flexors (Landau, 1971). The movements of the other joints remain the same.
The corticospinal tract influences the segmental reflex in the spinal cord. When the corticospinal tract is not functioning properly, the result is that the receptive field of the normal toe extensor reflex enlarges at the expense of the receptive field for toe flexion. Toe extension is consequently elicited from what is normally the receptive field for toe flexion. The maintenance of territorial integrity of the receptive fields is apparently one way in which the cortex exerts its influence under normal conditions.
Clinical Significance
The plantar reflex is a nociceptive segmental spinal reflex that serves the purpose of protecting the sole of the foot. The clinical significance lies in the fact that the abnormal response reliably indicates metabolic or structural abnormality in the corticospinal system upstream from the segmental reflex. Thus the extensor reflex has been observed in structural lesions such as hemorrhage, brain and spinal cord tumors, and multiple sclerosis, and in abnormal metabolic states such as hypoglycemia, hypoxia, and anesthesia.
There is disagreement about whether the response is plantar flexion or dorsiflexion in the majority of newborns (Hogan and Milligan, 1971; Ross et al., 1976). In all cases, however, the response does become flexor by the sixth to twelfth month of life.
On rare occasions the extensor reflex has been reported in individuals who were otherwise normal; however, there is no long-term follow-up of these cases reported in the literature. In summary, there is widespread agreement among neurologists that an extensor plantar response about the sixth to twelfth month of life indicates structural or metabolic dysfunction of the corticospinal system.
The receptive field for the extensor plantar response can be quite extensive. On occasion the extensor reflex has been elicited by stimulating as high as the face. Even in the same individual there is often shrinkage in the receptive field as time passes after the occurrence of the lesion.
The reflex response is on occasion equivocal. For example, there may be flexion of the toes before extension. Landau has addressed very nicely this question of an initial flexor movement of the toe followed by extension:
But even when the abnormal response is maximally developed, as in our illustrated case of paraplegia, early flexion may occur, especially with threshold stimulation. What these observations amount to practically is that competent clinical judgment of this peculiar behavior, God's gift to the neurologist, has more validity than an arbitrary rule concerning the initial direction of hallux movement. (Landau and Clare, 1959)
On occasion, the reflex may be unequivocally flexor on one side and the toe remains neutral without movement on the other side. Under these circumstances the question is whether there is evidence of corticospinal tract dysfunction, and not whether the response is flexor or extensor.
This question can often be answered by looking for other evidence of corticospinal dysfunction, such as repeating deep tendon reflexes. Table 73.1 gives a number of other signs that indicate corticospinal dysfunction, just as the Babinski sign. Occasionally one of them may be positive when the Babinski is not present. Sometimes it is helpful to do two of them together, such as the Babinski and the Oppenheim, or the Babinski and the Gordon. The Babinski sign is the most reliable, and the one most likely to be present initially. Excellent discussions of these various signs may be found in DeJong (1979) and Van Gijn (1977).
Table 73.1
Variants of the Babinski Sign.
References
Babinski JF, in Wilkins RH, Brody IA, eds Babinski's sign. Arch Neurol. 1967;17:441–46. [PubMed: 4860271]
Brain R, Wilkinson M. Observations on the extensor plantar reflex and its relationship to the functions of the pyramidal tract. Brain. 1959;82:297–320. [PubMed: 13803783]
Brodal A. Neurological anatomy in relation to clinical medicine. 3rd ed. New York: Oxford University Press, 1981.
DeJong RN. The neurologic examination. 4th ed. New York: Harper & Row. 1979;451–63.
Dohrmann GJ, Nowack WJ. The upgoing great toe: optimal method of elicitation. Lancet. 1973;1:339–41. [PubMed: 4121935]
Estanol B. Temporal course of the threshold and size of the receptive field of the Babinski sign. J Neurol Neurosurg Psychiatry. 1983;46:1055–57. [PMC free article: PMC491746] [PubMed: 6606699]
Fulton JF, Keller AD. The sign of Babinski. Springfield: Charles C Thomas, 1932.
Hogan GR, Milligan JE. The plantar reflex of the newborn. N Engl J Med. 1971;285:502–3. [PubMed: 5558889]
Kugelberg E, Eklund K, Grimby L. An electromyographic study of the nociceptive reflexes of the lower limb: mechanism of the plantar responses. Brain. 1960;83:394–410. [PubMed: 13754935]
Landau W. Clinical definition of the extensor plantar reflex (letter). N Engl J Med. 1971;285:1149–50. [PubMed: 5095750]
Landau WM, Clare MH. The plantar reflex in man with special reference to some conditions where the extensor response is unexpectedly absent. Brain. 1959;82:321–55. [PubMed: 14413775]
Ross ED, Velez-Borras J, Rosman NP. The significance of the Babinski sign in the newborn: a reappraisal. Pediatrics. 1976;57:13–15. [PubMed: 1082122]
VanGijn J. The plantar reflex. Rotterdam: Krips Repro-Meppel, 1977.
Walshe F. The Babinski plantar response: its form and its physiological and pathological significance. Brain. 1956;79:529–56. [PubMed: 13396062]
Copyright © 1990, Butterworth Publishers, a division of Reed Publishing.
Bookshelf ID: NBK397PMID: 21250238
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Walker HK. The Plantar Reflex. In: Walker HK, Hall WD, Hurst JW, editors. Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition. Boston: Butterworths; 1990. Chapter 73.
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3: Symmetry and Conservation
Part II- Discarding Complexity Without Losing Information
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3.5: Drag Using Conversion of Energy
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3.4: Box Models and Conservation
3.6: Lift Using Conservation of Momentum
Page ID 24097
Sanjoy Mahajan
Olin College of Engineering via MIT OpenCourseWare
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. 3.5.1 Box model for drag
2. 3.5.2 Testing the analysis with a home experiment
3. 3.5.3 Cycling
4. 3.5.4 Fuel efficiency of automobiles
A box model will next help us estimate drag forces. Drag, one of the most difficult subjects in physics, is also one of the most important forces in everyday life. If it weren’t for drag, bicycling, flying, and driving would be a breeze. Because of drag, locomotion requires energy. Rigorously calculating a drag force requires solving the Navier–Stokes equations:
(3.5.1)(v⋅∇)v+δv δt=−1 ρ∇p+v∇2 v
They are coupled, nonlinear, partial-differential equations. You could read many volumes describing the mathematics to solve these equations. Even then, solutions are known only in a few circumstances—for example, a sphere moving slowly in a viscous fluid or moving at any speed in a nonviscous fluid. However, a nonviscous fluid—what Feynman [14, Section II-40-2], quoting John von Neumann, rightly disparages as “dry water”—is particularly irrelevant to real life because viscosity is the cause of drag, so a zero-viscosity solution predicts zero drag! Using a box model and conservation of energy is a simple and insightful alternative.
3.5.1 Box model for drag
We will first estimate the energy lost to drag as an object moves through a fluid, as in Section 3.2.1. From the energy, we will find the drag force. To quantify the the problem, imagine pushing an object of cross-sectional area A cs at speed v for a distance d. The object sweeps out a tube of fluid. (The tube length d is arbitrary, but it will cancel out of the force.)
How much energy is consumed by drag?
Energy is consumed because the object gives kinetic energy to the fluid (say, water or air); viscosity, as we will model in Section 6.4.4, then turns this energy into heat. The kinetic energy depends on the mass of the fluid and on the speed it is given. The mass of fluid in the tube is ρA csd, where ρ is the fluid density. The speed imparted to the fluid is roughly the speed of the object, which is v. Therefore, the kinetic energy given to the fluid is roughly ρA csv 2d:
(3.5.2)E kinetic∼ρA csd⏟mass×v 2=ρA csv 2d.
This calculation ignores the factor of one-half in the definition of kinetic energy. However, the other approximations, such as assuming that only the swept-out fluid is affected or that all the swept-out fluid gets speed v, are at least as inaccurate. For this rough calculation, there is little point in including the factor of one-half.
This kinetic energy is roughly the energy converted into heat. Therefore, the energy lost to drag is roughly ρA csv 2d. The drag force is then given by
(3.5.3)energy lost to drag⏟∼ρA csv 2d=drag force⏟F drag×distance⏟d.
Now we can solve for the drag force:
(3.5.4)F drag∼ρA csv 2.
As expected, the arbitrary distance d has canceled out.
3.5.2 Testing the analysis with a home experiment
To test this analysis, try the following home experiment. Photocopy or print this page at 200 percent enlargement (a factor of 2 larger in width and height), cut out the template, and tape the two straight edges together to make a cone:
We could use many other shapes. However, a cone is easy to construct, and also falls without swishing back and forth (as a sheet of paper would) or flipping over (as long as you drop it point down).
We’ll test the analysis by predicting the cone’s terminal speed: that is, its steady speed while falling. When the cone is falling at this constant speed, its acceleration is zero, so the net force on it is, by Newton’s second law, also zero. Thus, the drag force F drag equals the cone’s weight mg (where m is the cone’s mass and g is the gravitational acceleration):
(3.5.5)ρ airv 2A cs∼mg
The terminal speed thus reveals the drag force. (Even though the drag force equals the weight, the left side is only an approximation to the drag force, so we connect the left and right sides with a single approximation sign ~.) The terminal speed v term is then
(3.5.6)v term∼mg A csρ air.
The mass of the cone is
(3.5.7)m=A paper=areal density of paper⏟σ paper.
Here, A paper is the area of the cone template; and the areal density σ paper, named in analogy to the regular (volume) density, is the mass per area of paper. Although areal density seems like a strange quantity to define, it is used worldwide to describe the “weight” of different papers.
The quotient m/A cs contains the ratio A paper/A cs. Rather than estimating both areas and finding their ratio, let’s estimate the ratio directly.
How does the cross-sectional area A cs compare to the area of the paper?
How does the cross-sectional area A cs compare to the area of the paper?
Because the cone’s circumference is three-quarters of the circumference of the full circle, its cross-sectional radius is three-quarters of the radius r of the template circle. Therefore,
(3.5.8)A cs=π(3 4r)2
Because the template is three-quarters of a full circle,
(3.5.9)A paper=3 4πr 2.
The paper area has one factor of three-quarters, whereas the cross-sectional area has two factors of three-quarters, so A paper/A cs=4/3. Now v term simplifies as follows:
(3.5.10)v term∼(A paperσ paper⏞m×g A csρ air)1/2=(4 3σ paperg ρ air)1/2.
The only unfamiliar number is the areal density σ paper, the mass per area of paper. Fortunately, areal density is used commercially, so most reams of printer paper state their areal density: typically, 80 grams per square meter.
Is this σ paper consistent with the estimates for a dollar bill in Section 1.1?
There we estimated that the thickness t of a dollar bill, or of paper in general, is approximately 0.01 centimeters. The regular (volume) density ρ would then be 0.8 grams per cubic centimeter:
(3.5.11)ρ paper=σ paper t≈80gm−2 10−2cm×1m 2 10 4cm 2=0.8g cm 3.
This density, slightly below the density of water, is a good guess for the density of paper, which originates as wood (which barely floats on water). Therefore, our estimate in Section 1.1 is consistent with the proposed areal density of 80 grams per square meter.
After putting in the constants, the cone’s terminal speed is predicted to be roughly 0.9 meters per second:
(3.5.12)v term∼(4 3×8×10−2kg m−2⏞σ paper×10m s−2⏞g 1.2kg m−3⏟ρ air)1/2∼0.9m s−1.
To test the prediction and, with it, the analysis justifying it, I held the cone slightly above my head, from about 2 meters high. After I let the cone go, it fell for almost exactly 2 seconds before it hit the ground—for a speed of roughly 1 meter per second, very close to the prediction. Box models and conservation triumph again!
3.5.3 Cycling
In introducing the analysis of drag, I said that drag is one of the most important physical effects in everyday life. Our analysis of drag will now help us understand the physics of a fantastically efficient form of locomotion—cycling (for its efficiency, see Problem 3.34).
What is the world-record cycling speed?
The first task is to define the kind of world record. Let’s analyze cycling on level ground using a regular bicycle, even though faster speeds are possible riding downhill or on special bicycles. In bicycling, energy goes into rolling resistance, friction in the chain and gears, and air drag. The importance of drag rises rapidly with speed, due to the factor of v 2 in the drag force, so at high-enough speeds drag is the dominant consumer of energy.
Therefore, let’s simplify the analysis by assuming that drag is the only consumer of energy. At the maximum cycling speed, the power consumed by drag equals the maximum power that the rider can supply. The problem therefore divides into two estimates: the power consumed by drag (P Drag) and the power that an athlete can supply (P athlete).
Power is force times velocity:
(3.5.13)power=energy time=force×distance time=force×velocity
Therefore,
(3.5.14)P drag=F dragv m a xsimρv 3A cs.
Setting P drag=P athlete allows us to solve for the maximum speed:
(3.5.15)v m a x∼(P athlete ρ airA cs)1/3,
where A cs is the cyclist’s cross-sectional area. In Section 1.7.2, we estimated P athlete as 300 watts. To estimate the cross-sectional area, divide it into a width and a height. The width is a body width—say, 0.4 meters. A racing cyclist crouches, so the height is roughly 1 meter rather than a full 2 meters. Then A cs is roughly 0.4 square meters. Plugging in the numbers gives
(3.5.16)v m a x∼(300W 1kgm−3×0.4m 2)1/3.
That formula, with its mix of watts, meters, and seconds, looks suspicious. Are the units correct?
Let’s translate a watt stepwise into meters, kilograms, and seconds, using the definitions of a watt, joule, and newton:
(3.5.17)W≡J s,J≡NM,N≡kgm s 2.
The three definitions are represented in the next divide-and-conquer tree, one definition at each nonleaf node. Propagating the leaves toward the root gives us the following expression forthe watt in terms of meters, kilograms, and seconds (the fundamental units in the SI system):
(3.5.18)W≡kgm 2 s 3.
The units in v max become
(3.5.19)(kg m 2s−3⏞W kgm−3×m 2)1/3=(s−3 m−3)1/3.
The kilograms cancel, as do the square meters. The cube root then contains only meters cubed over seconds cubed; therefore, the units for v max are meters per second.
Let’s estimate how many meters per second. Don’t let the cube root frighten you into using a calculator. We can do the arithmetic mentally, if we massage (adjust) the numbers slightly. If only the power were 400 watts (or instead the area were 0.3 square meters)! Instead of wishing, make it so—and don’t worry about the loss of accuracy: Because we have neglected the drag coefficient, our speed will be approximate anyway. Then the cube root becomes easy an calculation:
(3.5.20)v m a x∼(300400W 1kgm−3×0.4m 2)1/3=(1000)1/3ms−1=10ms−1.
In more familiar units, the record speed is 22 miles per hour or 36 kilometers per hour. As a comparison, the world 1-hour record—cycling as far as possible in 1 hour—is 49.7 kilometers or 30.9 miles, set in 2005 by Ondřej Sosenka. Our prediction, based on the conservation analysis of drag, is roughly 70 percent of the actual value.
How can such an estimate be considered useful?
High accuracy often requires analyzing and tracking many physical effects. The calculations and bookkeeping can easily obscure the most important effect and its core idea, costing us insight and understanding. Therefore, almost everywhere in this book, the goal is an estimate within a factor of 2 or 3. That level of agreement is usually enough to convince us that our model contains the situation’s essential features.
Here, our predicted speed is only 30 percent lower than the actual value, so our model of the energy cost of cycling must be broadly correct. Its main error arises from the factor of one-half that we ignored when estimating the drag force—as you can check by doing Problem 3.33.
3.5.4 Fuel efficiency of automobiles
Bicycles, in many places, are overshadowed by cars. From the analysis of drag, we can estimate the fuel consumption of a car (at highway speeds). Most of the world measures fuel consumption in liters of fuel per 100 kilometers of driving. The United States uses the reciprocal quantity, fuel efficiency—distance per volume of fuel—measured in miles per US gallon. To develop unit flexibility, we’ll do the calculation using both systems.
For a bicycle, we compared powers: the power consumed by drag with the power supplied by an athlete. For a car, we are interested in the fuel consumption, which is related to the energy contained in the fuel. Therefore, we need to compare energies. For cars traveling at highway speeds, most of the energy is consumed fighting drag. Therefore, the energy consumed by drag equals the energy supplied by the fuel.
Driving a distance d, which will be 100 kilometers, consumes an energy
(3.5.21)E drag∼ρ airv 2A csd.
The fuel provides an energy
(3.5.22)E fuel∼energy density⏟ε fuel×fuel mass⏟ρ fuelV fuel=ε fuelρ fuelV fuel.
Because E fuelsimE drag, the volume of fuel required is given by
(3.5.23)V fuel∼E drag ρ fuelε fuel∼ρ air ρ fuelv 2A cs ε fuel⏟A consumptiond.
Because the left-hand side, V fuel, is a volume, the complicated factor in front of the travel distance d must be an area. Let’s make an abstraction by naming this area. Because it is proportional to fuel consumption, a self-documenting name is A consumption. Now let’s estimate the quantities in it.
1.Density ratio ρ air/rho fuel. The density of gasoline is similar to the density of water, so the density ratio is roughly 10−3.
Speed v. A highway speed is roughly 100 kilometers per hour (60 miles per hour) or 30 meters per second. (A useful approximation for Americans is that 1 meter per second is roughly 2 miles per hour.)
Energy density ε fuel. We estimated this quantity Section 2.1 as roughly 10 kilocalories per gram or 40 megajoules per kilogram.
Cross-sectional area A cs. A car’s cross section is about 2 meters across by 1.5 meters high, so A cs ∼ 3 square meters
With these values,
(3.5.24)A consumption∼10−3×10 3m 2s−2⏞v 2×3m 2⏞A cs 4×10 7J kg−1⏟ε fuel≈8×10−8m 2.
To find the fuel consumption, which is the volume of fuel per 100 kilometers of driving, simply multiply A consumption by d = 100 kilometers or 10 5 meters, and then convert to liters to get 8 liters per 100 kilometers:
(3.5.25)V fuel≈8×10−8m 2⏟A consumption×10 5m⏟d×10 3l 1m 3=8l.
For the fuel efficiency, we use A consumption in the form d=V fuel/A consumption to find the distance traveled on 1 gallon of fuel, converting the gallon to cubic meters:
[d \sim \frac{\overbrace{1 \cancel{\textrm{ gallon}}}^{V_{fuel}}}{\underbrace{8 \times 10^{-8} \cancel{m^{2}}{A{consumption}}} \times \frac{4 \cancel{l}}{1 \cancel{\textrm{ gallon}}} \times \frac{10^{-3} m^{\cancel{3}}}{1 \cancel{l}} = 5 \times 10^{4} m.]
The struck-through exponent of 3 in the m 3 indicates that the cubic meters became linear meters, as a result of cancellation with the m 2 in the A consumption. The resulting distance is 50 kilometers or 30 miles. The predicted fuel efficiency is thus roughly 30 miles per gallon.
This prediction is very close to the official values. For example, for new midsize American cars (in 2013), fuel efficiencies of nonelectric vehicles range from 16 to 43 miles per gallon, with a mean and median of 30 miles per gallon (7.8 liters per 100 kilometers)
The fuel-efficiency and fuel-consumption predictions are far more accurate than we deserve, given the many approximations! For example, we ignored all energy losses except for drag. We also used a very rough drag force ρ airv 2A cs, derived from a reasonable but crude conservation argument. Yet, like Pippi Longstocking, we came out right anyway.
What went right?
The analysis neglects two important factors, so such accuracy is possible only if these factors cancel. The first factor is the dimensionless constant hidden in the single approximation sign of the drag force:
(3.5.26)F drag∼ρ airA csv 2.
Including the dimensionless prefactor (shown in gray), the drag force is
(3.5.27)F drag=1 2c dρ airA csv 2,
where c d is the drag coefficient (introduced in Section 3.2.1). The factor of one-half comes from the one-half in the definition of kinetic energy. The drag coefficient is the remaining adjustment, and its origin is the subject of Section 5.3.2. For now, we need to know only that, for a typical car, c d ≈ 1/2. Therefore, the dimensionless prefactor hidden in the single approximation sign is approximately 1/4.
Based on this more accurate drag force, will cars use more or less than 8 liters of fuel per 100 kilometers?
Including the c d/2 reduces the drag force and the fuel consumption by a factor of 4. Therefore, cars would travel 120 miles on 1 gallon of fuel or would consume only 2 liters per 100 kilometers. This more careful prediction is far too optimistic—and far worse than the original, simpler estimate.
What other effect did we neglect?
The engine efficiency—a typical combustion engine, whether gasoline or human, is only about 25 percent efficient: An engine extracts only one-quarter of the combustion energy in the fuel; the remaining three-quarters turns into heat without doing mechanical work. Including this factor increases our estimate of the fuel consumption by a factor of 4.
The engine efficiency and the more accurate drag force together give the following estimate of the fuel consumption, with the new effect in gray:
(3.5.28)V fuel≈1 2c d 0.25×ρ airv 2A cs ρ fuelε fueld.
The 0.25 in the denominator, from the engine efficiency, cancels the 1 2c d in the numerator. That is why our carefree estimate, which neglected both factors, was so accurate. The moral, which I intend only half jokingly: Neglect many factors, so that the errors can cancel one another out.
Exercise 3.5.1: Adjusting the cycling record
Our estimate of the world 1-hour record as roughly 35 kilometers (Section 3.5.3) ignored the drag coefficient. For a bicyclist, c d≈1. Will including the drag coefficient improve or worsen the prediction in comparison with the actual world record (roughly 50 kilometers)? Answer that question before making the new prediction! What is the revised prediction?
Exercise 3.5.2: Bicyclist fuel efficiency
What is the fuel consumption and efficiency of a bicyclist powered by peanut butter? Express your estimate as an efficiency (miles per gallon of peanut butter) and a consumption (liters of peanut butter per 100 kilometers). How does a bicycle compare with a car?
This page titled 3.5: Drag Using Conversion of Energy is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Sanjoy Mahajan (MIT OpenCourseWare) .
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16185 | https://study.com/academy/lesson/phase-diagrams-critical-point-triple-point-and-phase-equilibrium-boundaries.html | Critical Point & Triple Point Phase Diagrams | What is a Phase Diagram? - Lesson | Study.com
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Science Courses / Chemistry 101: General Chemistry Course
Critical Point & Triple Point Phase Diagrams | What is a Phase Diagram?
Lesson
Transcript
Carolyn LaRoche, Amy Meyers
Author Carolyn LaRoche
Carolyn LaRoche has been a high school science teacher for twenty three years. She is also an adjunct professor for the local community college. Her specialties include biology, chemistry, forensic science and anatomy and physiology. She also has laboratory research experience. Carolyn holds a BA in Biological Sciences/Premed and a MS in Forensic Chemistry.
View bio
Instructor Amy Meyers
Amy holds a Master of Science. She has taught science at the high school and college levels.
View bio
Learn what a phase diagram is. Discover the importance of the triple point and critical point in a phase diagram, and study examples of phase diagrams. Updated: 11/21/2023
Table of Contents
What Is a Phase Diagram?
Critical Point on a Phase Diagram
What Is a Triple Point?
Critical and Triple Point Phase Diagram Examples
Lesson Summary
Show
Frequently Asked Questions
What is the difference between triple point and critical point?
The triple point and critical point are both found on a phase diagram. The triple point represents a temperature and pressure combination where all three states of matter exist in equilibrium. Critical point is the temperature and pressure combination where the gas form of a substance can no longer be condensed back to a liquid, which becomes a supercritical fluid.
What is the critical point in chemistry?
The critical point in chemistry is a temperature and pressure combination in which a given substance no longer can condense back into a gas. This point can be an important application in thermodynamics. Going beyond the critical point creates a supercritical fluid.
Can there be multiple critical points on a phase diagram?
There is only one critical point on a phase diagram. It can be found at the end of the equilibrium line between liquid and gas. This is the point that once passed, the substance becomes a supercritical fluid.
What is triple point and why is it important?
Triple point is found on a phase diagram where the three lines of equilibrium between states of matter converge. The triple point is a temperature and pressure combination. At this point, all three states of solid, liquid, and gas will exist simultaneously. Knowing this temperature and pressure combination helps with identifying the compounds in a substance.
What is phase diagram and how can one interpret it?
A phase diagram is a representation of all the temperature and pressure combinations that create the different phases in a substance. Generally, the solid is left and top, liquid is in the middle and gas fills in the rest, until critical point is reached. The phase diagram has lines between the states indicating equilibrium points between two states of matter. The point where the lines converge is known as the triple point, which is a state of equilibrium where solid, liquid and gas exist together.
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Table of Contents
What Is a Phase Diagram?
Critical Point on a Phase Diagram
What Is a Triple Point?
Critical and Triple Point Phase Diagram Examples
Lesson Summary
Show
What Is a Phase Diagram? ------------------------
In thermodynamics a phase diagram is a very useful tool to scientists. Used in many scientific fields, a phase diagram is a graph that uses temperature and pressure combinations to illustrate when a substance will occur as solid, liquid or gas. At several points on the graph, the temperature and pressure combinations will create a state of phase equilibrium between two phases. This is illustrated by a line along those temperature-pressure combinations called, "phase equilibrium lines." At every point along that line, there is an equilibrium of both phases. For example, solid and liquid phase of a substance will exist together.
At one place on the graph, the lines will converge, offering an equilibrium state where all three phases of solid, liquid and gas exist simultaneously. This temperature and pressure combination is known as the triple point. There is also a point in which the vapor and the liquid of the substance becomes indistinguishable from each other. This is called the critical point.
Every substance has temperature and pressure combinations where the solid form transitions into the liquid form. This occurs when enough energy is absorbed by the solid, under the corresponding pressure conditions that the solid melts. The series of phase diagram melting points can be found on the equilibrium line between solid and liquid.
Beyond the critical point, in the area above and to the right of the critical point are the temperature and pressure combinations that create a supercritical fluid. Supercritical fluids have unique properties that make them very useful in industrial applications.
A phase diagram shows temperature and pressure combinations of a substance that create each of the three phases. It also shows equilibrium lines, the triple point, and critical point of the substance.
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Critical Point on a Phase Diagram ---------------------------------
The critical point on a phase diagram represents the temperature and pressure combination in which the liquid and vapor form of the substance in question both become indistinguishable from each other. The critical point is generally labeled at the end of the equilibrium line between the liquid and gas phase on a phase diagram.
At the critical point of a substance, scientists have to decide if they want to go on and create a supercritical fluid. Pushing a substance past the critical point of temperature and pressure creates a supercritical fluid. A supercritical fluid is a highly compressed fluid that retains the properties of both liquid and gases.
These fluids can be used in a wide variety of industrial applications. There are many common uses that include but are not limited to:
Food industry: decaffeinating coffee, extracting natural dyes and aromas
Pharmaceuticals: to create super fine powders
Pesticides: removal from various crops
Plastics and polymers: uses supercritical carbon dioxide as a plasticizer
Textile Industry: dying textiles, tanning leather, and dry cleaning processes
Energy production: liquefaction of biomass
Waste treatment: allows for more efficient destruction of hazardous wastes
In general, supercritical fluids have densities equal to that of liquids, and viscosities and diffusion capabilities similar to gases. These behaviors allow supercritical fluids to be applied in an industrial setting where they can diffuse through solids like a gas, but dissolve substances the same as a liquid.
When viewing a phase diagram, the supercritical fluid combinations can be found in the space above and to the right of the critical point.
Importance of the Critical Point in Thermodynamics
Thermodynamics is the study of how heat, work, temperature, and energy interact in a scientific process. The thermodynamic state of a system is set by the parameters of a system. In the case of critical point thermodynamics, this means the parameters are temperature and pressure. Critical point becomes an important target in many thermodynamic systems because past the critical point gases cannot be condensed back into liquids. This will change the dynamic of the system if it relies upon that ability to vaporize and condense as needed.
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What Is a Triple Point? -----------------------
Triple point on a phase diagram is indicated by the point where all three equilibrium lines meet. This is the temperature and pressure combination that allows for all three phases of a substance to occur simultaneously. There will be solid, liquid and gas molecules all existing in a state of equilibrium.
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Critical and Triple Point Phase Diagram Examples ------------------------------------------------
A triple point phase diagram notes not only the triple point of a substance, but also the critical point. A phase diagram is drawn by plotting the points of temperature and pressure combinations that determine equilibrium between solid and liquid states, solid and gas states, and liquid and gas states. Where the three lines meet is the triple point. Where the line ends between liquid and gas becomes the critical point. Beyond that point, the substance becomes a supercritical fluid.
Water and carbon dioxide are two common substances in which phase diagrams are used.
This diagram shows the triple of point of water where each of the three equilibrium lines meet. It is represented by a yellow dot. The critical point of water is represented by the red dot at the end of the liquid-gas equilibrium line.
Supercritical oxygenated water is used to help destroy hazardous wastes in an environment that is less able to dissolve solutes and may provide for safer destruction.
The phase diagram of carbon dioxide shows the triple point and critical point of the substance. Triple point is where the three lines meet and critical point is at the end of the liquid and gas line.
The supercritical fluid of carbon dioxide is used in the food industry to remove a chemical from cork that would taint the taste of food and drinks. Supercritical carbon dioxide is also used in the pharmaceutical industry to create the fine powders found in various make up applications, and in the cosmetic industry to aid in extraction of materials from natural sources, such as plants and vegetables.
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Lesson Summary --------------
A phase diagram is a graph that has all the temperature and pressure combinations of a specific substance for each state of matter. Equilibrium lines denote places where two phases can exist in equilibrium, such as solid and gas, liquid and solid and liquid and gas. The point at which the three lines converged equals the triple point, a pressure and temperature combination in which all three states exist simultaneously. The critical point, a temperature and pressure combination in which a vapor can no longer be condensed back into a liquid is found at the end of the equilibrium line between liquid and gas.
The critical point plays a notable role in thermodynamics, the study of heat, work, temperature and energy in scientific processes. The point on a phase diagram past the critical point is where supercritical fluid exists. Any temperature and pressure combination above and the right of critical point is where the supercritical fluid can be found.
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Video Transcript
Phase Diagram
Look on the back of a package of cake mix. See where it says 'high-altitude directions?' Ever wonder why it has this section? It is because things cook differently at higher altitudes. The pressure is less the higher you are on Earth, so the chemicals, such as the baking soda in cake, behave differently. We can predict how substances will behave at different pressures and temperatures with a phase diagram.
A phase diagram is a graph of the physical state of a substance (solid, liquid or gas) and the temperature and pressure of the substance. Phase diagrams are unique to every different substance.
Let's start by looking at a phase diagram and exploring everything that's on it. The x-axis of the graph shows temperature. As temperature increases, most substances change from solid to liquid and then to gas. The y-axis shows pressure.
Example of a phase diagram
Look at the three solid lines on the graph. These are the phase equilibrium lines. They are the lines on a phase diagram where two phases are in equilibrium. One line shows the equilibrium point between solid and liquid. One shows the equilibrium point between liquid and gas. The third shows the equilibrium point between solid and gas. Yes - solid and gas. Some substances can change directly from solid to gas and gas to solid without going through the liquid phase.
Look at the solid equilibrium line. If you move to the left of this line, either by increasing pressure or lowering temperature, the substance turns solid. If you move to the right, the substance is liquid until it hits the liquid/gas equilibrium line.
Triple Point
The three-phase equilibrium lines meet at one point. This triple point is the point where the temperature and pressure conditions are right for all three states (solid, liquid and gas) to exist together at equilibrium. The triple point is like the peak of a pyramid. If you are standing on the top of the pyramid, no matter which direction you step, you are stepping onto a different side. No matter where you move from a triple point, you are entering a different phase.
You enter a different phase regardless of where you move from the triple point.
Critical Point
The three phases of a substance can sometimes be indistinguishable from each other. Look at the line on this graph separating liquid from gas:
At the critical point, the gas and liquid states are identical.
As the temperature increases, the substance reaches a critical point. At the critical point, the gas and liquid states of a substance are identical and the substance is in one phase. Above the critical point, the substance is a supercritical fluid. This is a state where the gas and liquid phase of a substance are indistinguishable.
Lesson Summary
A phase diagram is a graph of the physical state of a substance (solid, liquid or gas) and the temperature and pressure of the substance. There are three solid lines on the phase diagram that are called the phase equilibrium lines. These lines show where two phases are in equilibrium.
The critical point on the phase diagram shows where the gas and liquid states of a liquid are identical and the substance is in one phase. Above the critical point, a substance is a supercritical fluid, where the gas and liquid phase of a substance are indistinguishable.
Learning Outcomes
Following your viewing of this video lesson, you could have the ability to:
Read a phase diagram and explain its importance
Identify phase equilibrium lines, triple points and critical points on a phase diagram
Describe at what point a substance is a supercritical fluid
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Chemistry 101: General Chemistry
14 chapters | 133 lessons | 11 flashcard sets
Ch 1. Experimental Chemistry and Introduction...
Ch 2. Atom
Ch 3. The Periodic Table
Ch 4. Nuclear Chemistry
Ch 5. Chemical Bonding
Ch 6. Liquids and Solids
Kinetic Molecular Theory | Definition, Assumptions & Examples 8:07
Critical Point & Triple Point Phase Diagrams | What is a Phase Diagram? 3:29
5:38 Next Lesson Phase Changes and Heating Curves
Crystal Structures and the Unit Cell 3:59
Ch 7. Gases
Ch 8. Solutions
Ch 9. Stoichiometry
Ch 10. Chemical Reactions
Ch 11. Equilibrium
Ch 12. Kinetics
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Ch 14. Studying for Chemistry 101
Critical Point & Triple Point Phase Diagrams | What is a Phase Diagram? Related Study Materials
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16186 | https://math.stackexchange.com/questions/564194/the-automorphism-group-of-the-complete-binary-rooted-tree-of-height-3 | Stack Exchange Network
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The automorphism group of the complete binary rooted tree of height $3$
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$\begingroup$
Can someone give me some help with this problem:
How do I find the automorphism group of the complete binary rooted tree of height $3$ ($15$ vertices)? when an automorphism $F$ on a graph $G=(V,E)$ is defined as follows:
${v,u}$ is an edge in $E$ iff ${F(v),F(u)}$ is an edge in $F(E)$.
I tried to start counting the permutations, but it seems there is a different and better way to approach the problem.
combinatorics
group-theory
graph-theory
trees
automorphism-group
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edited Dec 16, 2020 at 3:26
RobPratt
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asked Nov 12, 2013 at 17:48
zivikzivik
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$\endgroup$
6
1
$\begingroup$ Well, it will be a non-abelian group of order $2^7=128,$ so that should narrow things down a bit, but there has to be a better way of going about it than that. $\endgroup$
Cameron Buie
– Cameron Buie
2013-11-12 17:52:39 +00:00
Commented Nov 12, 2013 at 17:52
$\begingroup$ Can you explain why the size is 128? $\endgroup$
zivik
– zivik
2013-11-12 17:59:56 +00:00
Commented Nov 12, 2013 at 17:59
$\begingroup$ Certainly. For clarity, the root of the tree will be called $\varnothing;$ the nodes of height one, $T,$ $F;$ of height two, $TT,$ $TF,$ $FT,$ $FF;$ of height three, $TTT,$ $TTF,$ $TFT,$ $TFF,$ $FTT,$ $FTF,$ $FFT,$ $FFF.$ $\varnothing$ has immediate successors $T,F,$ and if $N$ is the name of a node of height one or two, then its immediate successors are named $NT,NF.$ Hopefully, that's easy enough to visualize (draw a picture, if not). Now, let us construct an automorphism $f:G\to G,$ where $G$ is the complete binary rooted tree of height three. We have to have $f(\varnothing)=\varnothing.$ $\endgroup$
Cameron Buie
– Cameron Buie
2013-11-12 18:32:07 +00:00
Commented Nov 12, 2013 at 18:32
$\begingroup$ (cont'd) We must have either $f(T)=T$ or $f(T)=F$ ($2$ choices), and once we've decided on $f(T)$, we will have determined what $f(F)$ has to be. Now, if $N$ is the name of a node of height one or two (of which there are $6$), and if we have already determined $f(N),$ then either $f(NT)=f(N)T$ or $f(NT)=f(N)F$ ($2$ choices in each case), and once we have decided on $f(NT),$ we will have determined what $f(NF)$ has to be. In this way, we somewhat recursively define $f$, making $7$ choices between two possibilities, which uniquely determine the automorphism $f:G\to G$. $\endgroup$
Cameron Buie
– Cameron Buie
2013-11-12 18:37:22 +00:00
Commented Nov 12, 2013 at 18:37
$\begingroup$ (cont'd) Put another way: (1) $f(\varnothing)=\varnothing;$ (2) two choices for $f(T),$ and our choice of $f(T)$ determines $f(T)$ and $f(F);$ (3) two choices for $f(TT),$ and our choice for $f(TT)$ determines $f(TT)$ and $f(TF);$ (4) two choices for $f(FT),$ and our choice for $f(FT)$ determines $f(FT)$ and $f(FF);$ (5) two choices for $f(TTT),$ and our choice for $f(TTT)$ determines $f(TTT)$ and $f(TTF);$ (6) two choices for $f(TFT),$ and our choice for $f(TFT)$ determines $f(TFT)$ and $f(TFF);$ (7) two choices for $f(FTT),$ and our choice for $f(FTT)$ determines $f(FTT)$ and $f(FTF);$ $\endgroup$
Cameron Buie
– Cameron Buie
2013-11-12 18:41:12 +00:00
Commented Nov 12, 2013 at 18:41
| Show 1 more comment
1 Answer 1
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23
$\begingroup$
Let $G_n$ denote the automorphism group of the complete binary rooted tree of height $n$. So, $G_1 \cong C_2$, the cyclic group of order $2$.
What about height $2$? You have automorphisms switching one pair of leaves on the left and the other pair on the right (in the usual planar picture of the rooted tree). Call these $a_1$ and $a_2$. Together they generate a subgroup of $G_2$ that is isomorphic to $C_2 \times C_2$, the Klein $4$ group. There is an additional automorphism generator, say $b$ that exchanges the left and right branches at the root. $G_2 \cong \langle a_1, a_2, b \rangle$. In fact the subgroup $\langle a_1, a_2 \rangle$ is normal in $G_2$, and $$ G_2 \cong (C_2 \times C_2) \rtimes C_2. $$ This construction is called a wreath product and can be written $$ C_2 \wr C_2. $$
Now, for height $3$ (and beyond), we iterate the construction. There are two isomorphic copies of $G_2$ in $G_3$ generated by the subtrees on either side of the root. The direct product of these two generate a normal subgroup inside of $G_3$, permuted by one additional generator at the root. So, $$ \begin{align} G_3 &\cong (G_2 \times G_2) \rtimes C_2 \ &\cong \bigg( \big( (C_2 \times C_2) \rtimes C_2 \big) \times \big( (C_2 \times C_2) \rtimes C_2 \big) \bigg) \rtimes C_2, \end{align} $$ a group of order $2^7 = 128$, generated by $7$ involutions.
In other words, $G_3$ is an iterated wreath product: $$ G_3 \cong G_2 \wr C_2 \cong \big( C_2 \wr C_2 \big) \wr C_2. $$
Beyond $n = 3$, we have the recursive formula $$ G_{n + 1} \cong G_n \wr C_2 \cong \big( G_n \times G_n \big) \rtimes C_2, $$ which gives the recurrence relation for the orders of the automorphism groups: $$ \big| G_{n + 1} \big| = 2\big| G_n \big|^2. $$ Together with the base case $\big| G_1 \big| = 2$, the can be solved to show that $$ \big| G_n \big| = 2^{2^n - 1}. $$
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edited Dec 16, 2020 at 0:19
answered Nov 12, 2013 at 18:40
Sammy BlackSammy Black
28.8k33 gold badges3939 silver badges6565 bronze badges
$\endgroup$
5
1
$\begingroup$ This is an under-appreciated answer. +1 $\endgroup$
user123641
– user123641
2014-08-30 02:48:17 +00:00
Commented Aug 30, 2014 at 2:48
1
$\begingroup$ Also worth noting that $G_n$ is isomorphic to the sylow-$2$ subgroups of the symmetric group on $2^n$ points. $\endgroup$
Josh B.
– Josh B.
2020-12-16 04:39:47 +00:00
Commented Dec 16, 2020 at 4:39
$\begingroup$ mind commenting while we need a semidirect product here? for me it looks like the third $C_2$ (related to the generaor $b$) is also a normal subgroup and so we would have $G_2 = (C_2\times C_2)\times C_2$ etc. but i seems to miss something $\endgroup$
asgeige
– asgeige
2021-10-25 13:16:21 +00:00
Commented Oct 25, 2021 at 13:16
$\begingroup$ No, the subgroup $\langle b \rangle = {1, b}$ is not normal in $G_2$. Notice that $a_1 b a_1 \not\in \langle b \rangle$ as it has both branches swapped at the root and both pairs of leaves swapped as well. As a permutation of the leaves from left-to-right, it's $4321$ in one-line permutation notation. $\endgroup$
Sammy Black
– Sammy Black
2021-10-26 05:14:24 +00:00
Commented Oct 26, 2021 at 5:14
$\begingroup$ thanks @SammyBlack makes sense! Cool construction,... $\endgroup$
asgeige
– asgeige
2021-11-03 17:14:50 +00:00
Commented Nov 3, 2021 at 17:14
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16187 | https://www.chessandpoker.com/dice_strategy_guide.html | Dice Strategy and Probability
Dice Strategy and Probability
Understanding Probabilities When Playing Dice Games
Dice games are probably the most widely played and well-known genre ever created. The concept of Dice games have been around for literally thousands of years, with ancient peoples sometime using shells and bones as Dice to add the element of controlled chance into their otherwise lackluster games. Also known to many as "Bones", Dice have been thrown by millions of people and have given birth to other games which are derived from their concept, most notably Dominoes.
With so many Dice-based games available, we decided to create a collection of the various theory and strategy ideas associated with Dice for gamers to access whenever they needed them. Our dice probability charts are well organized and detail the probabilities that affect up to five dice when thrown simultaneously. We'll also detail the ease in which new Dice games can be created using the Chess and Poker Dot Com dice strategy guide by inventing one of our own based on the concepts of Dice Probability generated when multiple Dice interact. Let's have a look to see how dice interaction can be understood and utilized.
Alea jacta est: Dice Probability
The appeal of Dice games is rooted in the fact that when two or more Dice are thrown simultaneously, a beautiful interaction occurs. Before the initial throw the Dice are in fact separate entities. However, once multiple Dice are thrown, the totals they create (arrived at by adding all the values together) unbalance the games in which they are being used into various channels of chance and strategy. Based on the number of Dice thrown, certain totals become increasingly more likely than others. As more and more Dice are thrown together at once, the expected outcomes begin to fluctuate and alter the strategy that a player may use accordingly.
Probability is the number of times something occurs divided by the number of times it could occur.
We'll take a look at the various combinations you will arrive at when rolling two, three, four and even five dice together, as well as some of the games that these outcomes affect. When researching the Dice statistics for this page, I could not find any one grouping of charts that showed all the numbers I was looking for. Logically, I then decided to create them myself. The following charts should be read from left to right, and they detail all of the possible Dice Counts, Sums and statistical percentages (probability) that they should occur. The following charts are based on the sums of all the Dice that are thrown. I call this the "Dice Count". For example, if you roll two Fives, your Dice Count would be 10 (5+5=10) and rolling a Three, Four and a Six would give you a 13 Count (3+4+6=13).
ChessandPoker.com Dice Probability Charts
The top row of each chart (shaded black) show the various Dice Counts. In most cases, there are two Counts shown in each box. This means that the Sums and Percentages detailed below them apply to both of the counts listed. For example, 2 / 12 means that both Two Counts and Twelve Counts share the exact same probability of occurring. If only one Count is shown, then the percentages apply to only that certain Count.
The middle row (shaded in white) show the exact number of combinations the Counts listed directly above them occur out of ALL THE POSSIBLE COMBINATIONS for that number of Dice. For example, in the Two Dice chart below there is a "3" listed in white below the 4 / 10 Counts. This means that when you're rolling two Dice, a Four and also a Ten both have 3 different combinations each out of the 36 total number of combinations possible between two Dice (there's more about this below). So there are three different ways to make a Four (1-3, 2-2, 3-1) and also three ways to make a Ten (4-6, 5-5, 6-4). And finally, the bottom row (shaded in blue) shows the number of possible combinations displayed as a percentage. Using the above 4 / 10 section for Two Dice, we see that each count has three different combinations or a 8.33% chance of occuring. This method is used for all of the charts in this guide.
2 or 123 or 114 or 105 or 96 or 87
123456
2.78%5.56%8.33%11.11%13.89%16.67%
Two Dice Probability (36 combinations)
3 or 184 or 175 or 166 or 157 or 148 or 139 or 1210 or 11
1361015212527
.46%1.39%2.78%4.63%6.94%9.72%11.57%12.50%
Three Dice Probability (216 combinations)
4/245/236/227/218/209/1910/1811/1712/1613/1514
141020355680104125140146
.08%.31%.77%1.54%2.70%4.32%6.17%8.02%9.65%10.80%11.27%
Four Dice Probability (1296 combinations)
5/306/297/288/279/2610/2511/2412/2313/2214/2115/2016/1917/18
15153570126205305420540651735780
.01%.06%.19%.45%.90%1.62%2.64%3.92%5.40%6.94%8.37%9.45%10.03%
Five Dice Probability (7776 combinations)
Understanding the Probability Charts
As you can see, the Counts on the left side of the charts are much less likely to occur than those on the right hand side. The further right you go on a chart, the more likely the listed counts are to occur. For example, the right-most entry in the "Two Dice Probability" chart is the 7 Count. This means that the 7 Count is the most likely to occur, with a 16.67% chance of occurrence. So how do you utilize the raw data found in our charts? Here's an explanation with the oft-used two-dice game Craps. One of the first bets you can make in the game of Craps is the Pass Line Bet. This bet will win instantly for you when the shooter (the person throwing the dice) throws a 7 or 11 on their first throw. You will lose this bet if the shooter instead throws either a 2, 3 or 12 on the first throw. Let's use the Two Dice Probability Chart to determine the chances of each of those outcomes.
The winning throws for the Pass Line bet are 7 or 11. Using the chart, we see that a 7 count has a 16.67% chance of occurring and the 11 count has about a 5.56% chance. That totals up to approximate winning chances of 22.23% The losing throws are 2 (2.78), 3 (5.56) or 12 (2.78) counts. So the losing chances equal 11.12%. We can see that concerning the Pass Line bet the shooter is almost twice as likely to roll a winning total than they are to roll a losing one. If this were the only bet and factor in the game it would be pretty simple, wouldn't it? Actually, there is much more to the game than just this as any Craps player will surely tell you. Using the chart again we can see that while the player is more likely to throw a winning 7 or 11 compared to a losing 2, 3 or 12, a whopping 66.66% (about 2/3 of the time) one of the other counts (4, 5, 6, 8, 9 or 10) should occur.
Now we are beginning to see how the charts can be used to better understand the Dice games you are involved with. But what about the Dice games that don't already exist? Knowing the mechanics of what makes a dice game work should put you well on your way to creating games of your own. Want an easy 3 Dice game? Make a game where the players get huge rewards when they throw a 10 or 11 count. Since this is the most likely occurrence for three dice, more often than not they should come out pretty good. Need it a bit harder? Try a 4 dice game where you have to roll a 4, 5, 6, 22, 23 or 24 count to win a point. Even though that's 6 winning counts, the combined percentage chance of a winning count occurring is only 2.32%! Read through the following section that covers our own invented dice game, and note how the reward system is geared based on the likely outcomes of two dice.
Strung: Original Dice Game Invented by James Yates
Let's explore our invented game called Strung to show how a game dependent on Dice sums might be constructed. The following table is brief rundown of the few rules of the game:
Number of Players: Two
Dice Required: Two Per Player
Wagering Devices: 50 Chips Per Player
CountsActions for Each Count (The sum of both Dice)
4 thru 10The lesser roll must contribute the difference in Chips into the Pot from their own stack. The numbers in this range are known as the Give Counts.
3 or 11Player wins 5 chips out of the Pot (or all of them if the Pot has less than 5). These are both known as the Take Counts. They are both equal in strength and are very good Counts.
2 or 12Player wins all chips currently in the Pot. These are known as the Strung Counts. They are both equal in strength and the best Counts you can get in the game.
Both players roll their two dice simultaneously (but separate from each other).
They then add their two dice together to arrive at their respective Dice Counts.
Players gain or lose chips based on their own count compared to their opponents.
The first player to capture all of their opponents Chips wins the game.
In the case of both players rolling Strung Counts, regardless of whether or not they are 2 or 12 counts, both players must re-roll with no other action permitted.
In the case of both players rolling Take Counts, regardless of whether or not they are 3 or 11 Counts, both players must also re-roll with no other action permitted.
If one player rolls a Strung Count and the other a Take Count, the Strung Count "Trumps" the Take Count and wins the whole pot. The Take Count does not get to take their usual 5 chips in this situation.
Finally, if both players roll matching Give Counts, they are simply required to keep re-rolling until one of the players must contribute to the Pot.
Gameplay
To start the game, both players should each have a pair of dice and 50 wagering units each (in this example we used Poker chips). At the same time, each player rolls their two dice in front of them. They then each add up their Dice Count and compare it to the other players dice count. Chips are then won, lost or put into the Pot based on the chart above.
As we know from the Two Dice Probability chart, there are 12 different Counts (sums) possible between two dice. The least likely of all the counts are the 2 and 12, which are tied with only one occurrence each. In this game, as in almost all other games of any genre, the rarer an occurrence is the more value it receives in the game. For example, in Poker the less likely a hand is to occur the higher it ranks. The Straight Flush is the most statistically unlikely hand in the game and is therefore the best hand you can have, followed by Four of a Kind which is the second least likely combination to occur. In most games LOW FREQUENCY EQUALS HIGH VALUE.
So following this logic, the premium rolls in our game are the least likely of all the two dice combos. The two (1-1) and the twelve (6-6) counts make up only two combinations out of the total 36 combos for two dice. They are known as the Strung counts. If you roll a two or twelve and your opponent rolls anything other than a two or twelve themselves, you win all the chips in the current Pot. But if they also roll a Strung count, a stalemate occurs and both players must roll again with neither player winning any chips.
The next unlikely group would be the three (1-2, 2-1) and eleven (5-6, 6-5) counts that are known as the Take counts. Therefore we have also given these rolls a special bonus. Whenever you roll either of these counts and your opponent doesn't also roll a Take count or Strung count, you get to take out five chips from the Pot into your stack. If your opponent had rolled a Strung count, unfortunately the Strung count would have trumped your second-best Take Count and they'd have won the whole Pot (you wouldn't even get to take out your five chips). So Take counts are the second least likely counts and beat everything except for the Strung counts.
But how do you get any chips into the Pot in the first place? If neither player rolls a Strung or Take count, the player who has the lower count must contribute the difference in the counts to the Pot. So if you roll a 9 count and your opponent rolls a 5 count, they would have to put in four chips into the Pot. The difference between your higher count and their lower count is four (9 minus 5 equals 4), so that's the amount they have to cough up. Whether or not they'll be able to win it back on the next roll will depend entirely on the Dice. And this is as it should be!
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Sulfur Trioxide (SO₃): Chemistry, Structure, Properties & Reactions
Sulfur Trioxide (SO₃): Chemistry, Structure, Properties & Reactions
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What is Sulfur Trioxide (SO₃)? – Formula, Structure and Key Reactions
Sulfur Trioxide (SO₃) is essential in chemistry and helps students understand various practical and theoretical applications related to this topic. It plays a central role in the creation of sulfuric acid and is also known for its strong oxidizing and dehydrating properties, making it an important chemical in both industry and academics.
What is Sulfur Trioxide in Chemistry?
A sulfur trioxide refers to an inorganic compound with the chemical formula SO₃. This concept appears in chapters related to oxides of sulfur, acid-base behavior, and environmental chemistry, making it a foundational part of your chemistry syllabus. Sulfur trioxide is mainly recognized as the anhydride of sulfuric acid and is significant in the manufacture of fertilizers, detergents, and explosives.
Molecular Formula and Composition
The molecular formula of sulfur trioxide is SO₃. It consists of one sulfur atom and three oxygen atoms bonded covalently, and is categorized under sulfur oxides. Its molar mass is 80.066 g/mol. In its gaseous state, SO₃ has a trigonal planar structure with 120° bond angles. The Lewis structure displays sulfur at the center double bonded to three oxygen atoms with possible resonance structures.
Preparation and Synthesis Methods
Sulfur trioxide can be prepared using both laboratory and industrial methods. In the laboratory, it is commonly obtained by heating sodium pyrosulfate or by dehydrating concentrated sulfuric acid using phosphorus pentoxide. The main industrial method is the Contact Process:
Sulfur or iron pyrites are burned in air to produce sulfur dioxide (SO₂). 2. SO₂ is purified and then oxidized over a vanadium(V) oxide (V₂O₅) catalyst at about 400–600°C to form SO₃. 3. SO₃ produced cannot be mixed directly with water, so it's first absorbed in concentrated sulfuric acid to form oleum, which is then diluted to make sulfuric acid.
Physical Properties of Sulfur Trioxide
Sulfur trioxide is a colorless to white crystalline solid or clear liquid (depending on temperature). It fumes in moist air and is highly soluble in water, reacting vigorously to form sulfuric acid. Its boiling point is 44.9°C and the melting point is 16.9°C. The density is about 1.92 g/cm³. SO₃ has no odor and is highly corrosive.
Chemical Properties and Reactions
Sulfur trioxide is a powerful oxidizing agent and highly reactive. The most notable chemical reaction is with water:
SO₃ + H₂O → H₂SO₄ (sulfuric acid) Even a small amount of water reacts suddenly and releases heat. SO₃ also reacts with sodium hydroxide to form sodium bisulfate (NaHSO₄). It is incompatible with organic materials due to its strong dehydrating nature and can cause fires or explosions if not handled carefully.
Frequent Related Errors
Confusing SO₃ with sulfite ion (SO₃²⁻) or sulfate ion (SO₄²⁻).
Drawing incorrect Lewis structures, or ignoring resonance and geometry.
Forgetting that SO₃ cannot be added directly to water without causing violent reactions.
Mistaking SO₃ for a neutral (non-acidic) oxide in writing reactions.
Uses of Sulfur Trioxide in Real Life
Sulfur trioxide is widely used in the chemical industry. Most importantly, it is the key raw material for producing sulfuric acid, which is essential in fertilizer, chemical, and detergent manufacturing. It is also used as a sulfonating agent in producing dyes and pharmaceuticals, and plays a role in the production of explosives and certain kinds of plastics.
Relation with Other Chemistry Concepts
Sulfur trioxide is closely related to topics such as oxides of sulfur and acid rain. Understanding SO₃ helps students link concepts from chemical bonding (like VSEPR theory and resonance) and environmental chemistry, highlighting its importance in air pollution and acid deposition. It also helps distinguish between similar ions like sulfite and sulfate: see Sulphite vs Sulphate for more details.
Step-by-Step Reaction Example
Begin with the reaction of sulfur trioxide with water. SO₃ (g) + H₂O (l) → H₂SO₄ (aq) 2. This is a combination reaction. The reaction is highly exothermic, meaning it produces a lot of heat, and can cause a fine acidic mist if uncontrolled. 3. In industry, SO₃ is first absorbed in concentrated sulfuric acid (not water directly) to form oleum, which is then carefully diluted to obtain sulfuric acid safely.
Lab or Experimental Tips
Always handle sulfur trioxide in a fume hood using appropriate safety gear, as it is extremely corrosive and fumes strongly. Vedantu educators recommend remembering SO₃ by its violent reaction with water—never add SO₃ directly to water, always add it to concentrated acid for safe laboratory practice.
Try This Yourself
Write the IUPAC name of SO₃.
Draw the Lewis structure and all possible resonance forms of sulfur trioxide.
List two industries where SO₃ plays a major role.
How can SO₃ contribute to acid rain? Explain in a sentence.
Final Wrap-Up
We explored sulfur trioxide (SO₃)—its formula, structure, properties, reactions, and uses in daily life and industry. For more in-depth chemistry examples and live topic classes, explore topic notes and live sessions on Vedantu, your trusted learning partner for science concepts.
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FAQs on Sulfur Trioxide (SO₃): Chemistry, Structure, Properties & Reactions
What is the chemical formula of sulfur trioxide?
Sulfur trioxide has the chemical formula SO₃. It consists of one sulfur atom bonded to three oxygen atoms. This molecule is an important industrial reagent and is the anhydride of sulfuric acid.
Is sulfur trioxide (SO₃) harmful to humans?
Sulfur trioxide is highly hazardous to humans if inhaled or touched. Key hazards include:
Corrosive to skin, eyes, and mucous membranes
Can cause severe burns and respiratory irritation
Reacts violently with water to produce sulfuric acid fumes
Always follow appropriate safety guidelines when handling SO₃.
What happens when sulfur trioxide reacts with water?
Sulfur trioxide reacts vigorously with water to form sulfuric acid (H₂SO₄):
SO₃ (g) + H₂O (l) → H₂SO₄ (aq)
This reaction is highly exothermic and produces dense, corrosive fumes
It is a critical step in industrial sulfuric acid production
What are the main uses of sulfur trioxide in industry?
Sulfur trioxide is primarily used for the manufacture of sulfuric acid. Other uses include:
Production of sulfonic acids and detergents
Laboratory reagent in organic synthesis
Intermediate in chemical industries
Is SO₃ a gas or a liquid at room temperature?
Sulfur trioxide (SO₃) exists as a colorless to white crystalline solid or a pungent gas at room temperature. Its exact state depends on temperature and pressure:
As a gas above 44.8°C
As a solid or liquid below this temperature
What is the Lewis structure of sulfur trioxide?
The Lewis structure of SO₃ shows the sulfur atom double-bonded to each of the three oxygen atoms, with resonance arrows between forms. Key points:
No lone pairs on sulfur
Three equivalent resonance structures
Molecule is trigonal planar
How is sulfur trioxide prepared industrially?
Sulfur trioxide is prepared by the catalytic oxidation of sulfur dioxide (SO₂) in the Contact process:
Sulfur is burned to produce SO₂
SO₂ is oxidized with oxygen over a vanadium(V) oxide catalyst at 450°C
The resulting SO₃ is collected and used in acid production
What is the resonance in SO₃ and how does it affect its stability?
SO₃ exhibits resonance because its double bonds to oxygen can be distributed among all three oxygen atoms.
This delocalizes electrons
Provides extra stability and equalizes S–O bond lengths
Explains trigonal planar geometry
What is the difference between SO₃ and sulfite (SO₃²⁻)?
SO₃ is sulfur trioxide (neutral molecule), while sulfite is SO₃²⁻ (anion):
SO₃ is a molecular oxide of sulfur; SO₃²⁻ is a polyatomic ion
SO₃ reacts with water to form sulfuric acid; SO₃²⁻ forms sulfurous acid salts
SO₃ is covalent, SO₃²⁻ is an ionic species
What role does sulfur trioxide play in acid rain formation?
Sulfur trioxide contributes to acid rain because:
It forms when sulfur dioxide (SO₂) in the atmosphere is further oxidized
SO₃ reacts with atmospheric moisture to create sulfuric acid (H₂SO₄)
This acid mixes with rainwater, leading to acid rain and environmental harm
How can you safely handle sulfur trioxide in the laboratory?
Safe handling of SO₃ involves:
Working in a well-ventilated fume hood
Using protective gloves, eye protection, and lab coats
Avoiding contact with water to prevent hazardous reactions
Storing SO₃ in properly sealed, corrosion-resistant containers
What are the physical properties of sulfur trioxide?
Physical properties of SO₃ include:
Molar mass: 80.07 g/mol
Appearance: Colorless to white solid, or pungent liquid/gas
Melting point: 16.9°C
Boiling point: 44.8°C
Highly reactive with water
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16189 | https://www.ck12.org/book/ck-12-chemistry-concepts---intermediate/section/14.6/ | Combined Gas Law | CK-12 Foundation
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14.6 Combined Gas Law
Difficulty Level: At Grade | Created by: CK-12
Last Modified: Dec 05, 2024
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[Figure 1]
What keeps things cold?
The modern refrigerator takes advantage of the gas laws to remove heat from a system. Compressed gas in the coils (see above) is allowed to expand. This expansion lowers the temperature of the gas and transfers heat energy from the material in the refrigerator to the gas. As the gas is pumped through the coils, the pressure on the gas compresses it and raises the gas temperature. This heat is then dissipated through the coils into the outside air. As the compressed gas is pumped through the system again, the process repeats itself.
Combined Gas Law
To this point, we have examined the relationships between any two of the variables of P, V, and T, while the third variable is held constant. However, situations arise where all three variables change. The combined gas law expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant.
P×V T=k and P 1×V 1 T 1=P 2×V 2 T 2
Sample Problem: Combined Gas Law
2.00 L of a gas at 35°C and 0.833 atm is brought to standard temperature and pressure (STP). What will be the new gas volume?
Step 1: List the known quantities and plan the problem.
Known
P 1=0.833 atm
V 1=2.00 L
T 1=35∘C=308 K
P 2=1.00 atm
T 2=0∘C=273 K
Unknown
V 2=?L
Use the combined gas law to solve for the unknown volume (V 2). STP is 273 K and 1 atm. The temperatures have been converted to Kelvin.
Step 2: Solve.
First, rearrange the equation algebraically to solve for V 2.
V 2=P 1×V 1×T 2 P 2×T 1
Now substitute the known quantities into the equation and solve.
V 2=0.833 atm×2.00 L×273 K 1.00 atm×308 K=1.48 L
Step 3: Think about your result.
Both the increase in pressure and the decrease in temperature cause the volume of the gas sample to decrease. Since both changes are relatively small, the volume does not decrease dramatically.
It may seem challenging to remember all the different gas laws introduced so far. Fortunately, Boyle’s, Charles’s, and Gay-Lussac’s laws can all be easily derived from the combined gas law. For example, consider a situation where a change occurs in the volume and pressure of a gas while the temperature is being held constant. In that case, it can be said that T 1=T 2. Look at the combined gas law and cancel the T variable out from both sides of the equation. What is left over is Boyle’s law:
P 1×V 1=P 2×V 2. Likewise, if the pressure is constant, then P 1=P 2 and canceling P out of the equation leaves Charles’s law. If the volume is constant, then V 1=V 2 and canceling V out of the equation leaves Gay-Lussac’s law.
Summary
The combined gas law shows the relationships among temperature, volume, and pressure.
P 1 V 1 T 1=P 2 V 2 T 2
Review
What is the only thing held constant in a combined gas law problem?
If you want to solve for the volume of a gas(V 2) and P 1 is greater than P 2, would you expect V 2 to be larger or smaller than V 1?
What would be the equation for finding P 2 given all the other parameters?
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Description
Describes the combined gas laws and gives an example of the use of this relationship in calculations.
Learning Objectives
State the combined gas law.
Use the law to calculate parameters in general gas problems.
Difficulty Level
At Grade
author
Ck12 Science
Subjects
science,Chemistry
Concept Nodes
SCI.CHE.659 (Combined Gas Law - Chemistry)
Grades
10,11,12
Standards Correlations
NGSS,CA,DE,DC,GA,HI,KY,LA,MA,MI,MN,MS,NC,OH,OK,SC,SD,TX,VT,VA,WV, (15 more)
Search Keywords
Volume,gas,Pressure,temperature,combined gas laws, (2 more)
License
CC BY NC
Language
English
Date Created
May 01, 2013
Last Modified
Dec 05, 2024
Vocabulary
combined gas law: Expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas.
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16190 | https://ijnaa.semnan.ac.ir/article_7922_51574aa752f263c375a0ecef60f187f3.pdf | Int. J. Nonlinear Anal. Appl. 15 (2024) 5, 111–120 ISSN: 2008-6822 (electronic) Routh stability criterion and Lyapunov-Routh method in control theory Ali Farhan Hashosha, Hadi Basirzadehb,∗ aShahid Chamran University of Ahvaz, Ahvaz, Iran bDepartment of Mathematics, Faculty of Mathematical Sciences and Computer, Shahid Chamran University of Ahvaz, Ahvaz, Iran (Communicated by Mohammad Rasoul Velayati) Abstract One of the most important issues in the linear control subject is to obtain eigenvalues of the system to study the stability of a system. It is needed to identify the sign of eigenvalues but not the value of it. To fulfil this, there are different methods such as Routh stability criterion, Lyapunov’s method and Nyquist stability criterion. In this research, we will present the most simple one to determine the sign of eigenvalues and we shall discuss and explain different types of stability. Also, we will discuss some special cases that cross a controversial mission, and a new method is proposed to calculate the stability of systems, which we call the Lyapunov-Routh method (composition of Lyapunov and Routh method).
Keywords: Stability, Stability types, Routh-Hurwitz stability criterion, Lyapunov’s method 2020 MSC: 93D05, 93C95 1 Introduction The stability of the system is to make the system equilibrium which does not accept the major and sudden changes that leads the system to be in a state of dispersion or may cause noise, which hinders the system from well functioning.
Before we give a mathematical definition of stability, let us explain this concept with examples from our daily life such as calmness after a big challenge. Mathematically, the system is considered stable when the output is within certain values, meaning and does not go to infinity, the researchers need to define the transfer function as shown in Fig. 1 .
Figure 1: ∗Corresponding author Email addresses: ali_fr@uomisan.edu.iq (Ali Farhan Hashosh), basirzad@scu.ac.ir (Hadi Basirzadeh) Received: February 2023 Accepted: July 2023 112 Farhan, Basirzadeh G(s) = Y (s) U(s) = bmsm + bm−1sm−1 + · · · + b1s + b0 ansn + an−1sn−1 + · · · + a1s + a0 = N(s) D(s) when nth is of the system (n ≥m) and D(s) is characteristic polynomial when the D(s). When getting close to zero (D(s) = 0) shown below D(s) = ansn + an−1sn−1 + · · · + a1s + a0 = 0 (1.1) are called the characteristic equation. The roots of the numerator polynomial i.e., N(s) = bmsm + bm−1sm−1 + · · · + b1s + b0 = 0 (1.2) are called the zeroes of the system. The roots of the denominator polynomial equation (1.1) are called poles of the system.
2 Stability types There are three types of system stability we will explain each of these types: 1- Absolutely stable system.
2- Conditionally stable system.
3- Marginally stable system.
1- Absolutely stable system: The control system is stable for all ranges of system component values. The system is stable if all the poles of the present transfer function are in left half of S-plan.
2- Conditionally stable system: If the system is stable for certain ranges of system component values.
3- Marginally stable system: if the system is stable by producing an output signal with constant amplitude and constant frequency of oscillations for bounded input .
The control system is marginally stable if any two poles of the transfer function are present on the imaginary axis.
Impulse responses for various root locations in the s-plan : 3 Routh stability criterion: The Routh stability criterion is an analytical procedure for determining if all the roots of a polynomial have negative real parts, and it is used in the stability analysis of linear time invariants systems . This stability criterion is useful in various engineering applications [3, 7].
It is important to obtain the sign of the G(s) roots not the value of them .to distinguish the sign of D(s) can be use Routh stability criterion. After reading the theory of the network structure , we can easily say that any pole of Routh stability criterion and Lyapunov-Routh method in control theory 113 the system located on the right side of the origin of the s- plane, makes the system unstable based on this condition . Routh began to verify the necessary and sufficient conditions for the stability of the system.
Also known as modified Hurwitz Criterion of stability of the system. We will study this criterion in two parts.
Part one will cover necessary conditions for stability of the system and part two will cover the sufficient conditions for the stability of the system. One-The necessary conditions: it includes two parts as follows All the coefficients of the characteristic of polynomial (equation (1)) should be positive and real.
I. All the coefficients of the characteristic polynomial should be non-zero.
Two-The sufficient condition: The helpful key to identify whether a system is stable or not is to consider that all elements of the first column of the Routh table have same sign.
This method yields stability information without the need to solve closed –loop system poles.
Using this method, we can tell the number of the closed –loop system poles in the half –plan, in the right half-plan, and on the iω-axis(notice that we say how many not where).
The method requires two steps: 1-generate a data table called a Routh table: i. Begin by labeling the rows with powers of s form the highest power of the denominator polynomial tos0.
ii. The first row will consist of all the even terms of the characteristic equation. Arrange them from first (even term) to last (even term). The first row is written below: a0, a2, a4, a6, . . .
iii. The second row will consist of all the odd terms of the characteristic equation. Arrange them from first (odd term) to last (odd term). The first row is written below:a1, a3, a5, a7, . . .
iv. The elements of third row can be calculated as follows a) First element: Multiply a0 with the diagonally opposite element of next column (i.e.a3) then subtract this from the product of a1 and a2 (where a2 is diagonally opposite element of next column) and finally divide the result so obtain with a1. Mathematically we write as first element b1 = (a1a2 −a3a0) a1 b) Second element: Multiply a0 with the diagonally opposite element of one column after the next column (i.e.a5) then subtract this from the product of a1 and a4 (where, a4 is diagonally opposite element of next to next column) and then finally divide the result then obtain with a1. Mathematically we write second element as follows: b2 = (a1a4 −a5a0) a1 114 Farhan, Basirzadeh c) Third element: Multiply a0 with the diagonally opposite element of one column after the next column (i.e.a7). then subtract this from the product of a1 and a6 (where, a6 is diagonally opposite element of next-to-next column) and then finally divide the result so obtain with a1. Mathematically we write as second element b3 = (a1a6 −a7a0) a1 v. The elements of the fourth row can be calculated through using the following procedure: a) First element: Multiply a1 with the diagonally the opposite element of next column (i.e.b2) then subtract this from the product of a3 and b1 (where, b1 is diagonally the opposite element of next column) and then finally divide the result so obtain with b1. Mathematically we write the first element c1 = (b1a3 −b2a1) b1 b) First element: Multiply a1 with the diagonally the opposite element of the next column (i.e.b3) then subtract this from the product of a5 and b3 (where, b1 is diagonally opposite element of next column) and finally divide the result to obtain b1. Mathematically we write the first element as follows: c2 = (b1a5 −b3a1) b1 c) Similarly, we can calculate all the elements of the fourth row.
vi. Similarly, we can calculate all the elements of all rows.
We will explain this on the table below. The Characteristic equation is: a0sn + a1sn−1 + a2sn−2 + · · · + an−1s1 + ans0 = 0.
We can multiply positive constant value and for the equation if not found s to power i that means it’s coefficient equal 0.
Stability criteria: if all the elements of the first column are positive then the system will be stable. However, if anyone of them is negative the system will be unstable. The number of roots of polynomial that are in the RHP is equal to the number of sign changes in first column.
Example 3.1. Generate the Routh table for the characteristic equation as below, s4 + 3s3 + 3s2 + 2s + 1 = 0 Solution: The system is stable.
Routh stability criterion and Lyapunov-Routh method in control theory 115 Example 3.2. Consider the following characteristics equation, develop Routh array and determine the stability sys-tem.
s4 + 2s3 + 3s2 + 4s + 5 = 0.
Solution: It is unstable because two poles in sign appear in the first column; we find that two roots of the characteristic equation lie on the right-hand plane of the s-plan.
4 Special cases in Routh Stability Criteria: There are some special cases related to Routh Stability Criteria, which are discussed below: (1) Case one: It is when the first term in any row of the array is zero while the rest of the row has at least one none zero term.
In this case we will assume a very small value (ϵ) which is tending to zero in place of zero. By replacing zero with (ϵ) we will calculate all the elements of the Routh array. After calculating all the elements, we will apply the limit at each element containing (ϵ). Through solving the limit at every element if we get positive limiting value, we will say the given system is stable otherwise in all other conditions the given system is not stable.
Example 4.1. Check whether the given system in Fig.2 is stable or unstable by Routh criterion: Figure 2: Solution: Consider the characteristics equation 1 + G(s)H(s) = 0. Then + 2 (s(s + 1)2) = 0. This implies that s3 + s2 + s + 2 = 0. Let’s create a Routh table We notice that the third element of the first column is zero, so we deleted the zero and put a value very close to zero ϵ(ϵ − →0), and then we completed the Routh table. The system is stable.
116 Farhan, Basirzadeh Example 4.2. Determine the stability of the system having a characteristic equation given below: D(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10 Solution: Let’s create a Routh table We notice that the third element of the first column is zero, so we deleted the zero and put a value very close to zero ϵ(ϵ − →0). We can calculate the element: c1 = 4ϵ −12 ϵ = −12 ϵ and d1 = (6c1 −10ϵ) c1 − →0 Then we complete the Routh table. Note that c1 is negative i.e., the transformation from positive to negative and negative to positive, the system is unstable and this indicates the presence of two poles in the right-hand plane of the s-plan.
Example 4.3. Make the Routh table for the system as below: G(s) = 10 s5 + 2s4 + 3s3 + 6s2 + 5s + 3 Solution: The characteristic equation for system is s5 + 2s4 + 3s3 + 6s2 + 5s + 3.
Then we complete the Routh table We notice that the third element of the first column is zero, so we deleted the zero and put a value very close to zero ϵ(ϵ − →0). We can calculate the c1 = 6ϵ −7) ϵ = −12 ϵ and d1 = 42ϵ −49 −6ϵ2 12ϵ −14 Routh stability criterion and Lyapunov-Routh method in control theory 117 The ϵ It is a very close value to zero. It can be a negative value close to zero or a positive value close to zero. To analyze the stability of this system, we can suppose the ϵ positive and negative as on the table below: If we assume that the value of ϵ is positive, then the fourth element of the first column is negative, and this indicates the presence of a change in the sign from positive to negative then to positive, and this indicates the presence of two of poles in the right-hand plane of the s-plan.
Also, if we assume that the value of ϵ is negative, we notice a change in the sign in the third element from the first column, this indicates the presence of two poles in the right - hand plane of the s-plan. This means that the system is unstable.
(2) Case second: When all the elements of any row of the Routh array are zero. In this case we can say the system has the symptoms of marginal stability. Let us first understand the physical meaning of having all the elements zero of any row. The physical meaning is that there are symmetrically located roots of the characteristic equation in the s plane. Now in order to find out the stability in this case we will first find out auxiliary equation. Auxiliary equation can be formed by using the elements of the row just above the row of zeros in the Routh array. After finding the auxiliary equation, we will differentiate the auxiliary equation to obtain elements of the zero row. If there is no sign change in the new Routh array formed by using auxiliary equation, then the given system is a limited stable. However, in all other cases we will say, the given system is unstable.
Example 4.4. Make the Routh table for the system shown and determine the pole distribution in the s-plan the auxiliary polynomial method G(s) = 10 s5 + 7s4 + 6s3 + 42s2 + 8s + 56.
Solution: The characteristic equation for system is s5 + 7s4 + 6s3 + 42s2 + 8s + 56.
Then we complete the Routh table Here to complete Routh table, We Derivative the equation for the second row d(s4 + 6s2 + 8) ds = 4s3 + 12s (4.1) We substitute the coefficients of the equation (4.1) in the third row instead of zero, and divide the third row by 4 becomes the table below: We notice that the system is stable because the first column did not change its signal.
118 Farhan, Basirzadeh 5 Lyapunov Routh stability method: Sometimes it is difficult to check the stability of a problem by Lyapunov’s method, and in this case, we can use Routh method for help. We call this method Lyapunov Routh method (composition of Lyapunov and Routh). In this method, first we use Lyapunov’s method to obtain the form of matrix in state space: ˙ X = AX + BU.
Now we use the Routh method to write a specific sentence of the matrix A and check the stability of the system.
As shown in the examples (6 and 7), we can easily check the stability of a parametric system.
Example 5.1. Determine the range of K for the system in Fig. 3 to be asymptotically stable.
Figure 3: Solution: using Lyapunov’s method, X1 = 3 s + 2X2 ⇒˙ x1 = −2x1 + 3x2 X2 = K s + 1 (−X1 + U) ⇒˙ x2 = −Kx1 −x22Ku The state equation is ˙ x1 ˙ x2 = −2 3 −K −1 x1 x2 + 0 K u A = −2 3 −K −1 and B = 0 K Solving A′P + PA = −I, −2 −K 3 −1 p11 p12 p21 p22 + p11 p12 p21 p22 −2 3 −K −1 = −1 0 0 −1 We obtain p = 1 18K + 12 K2 + 3K + 3 3 −2K 3 −2K 3K + 15 Routh stability criterion and Lyapunov-Routh method in control theory 119 Here it is difficult to calculate the value K So, we use a simpler method to known The stability of system by using Routh stability criterion: |A −λI| = 0 ∥ −2 3 −K −1 − λ 0 0 λ |= −2 −λ 3 −K −1 −λ = 2 + 2λ + λ + λ2 + 3K = λ2 + 3λ + 2 + 3K By using Routh table: The system is stable 2 + 3K > 0 = ⇒K > −2/3 Example 5.2. Determine a range of values of system parameter K for which the system in Fig. 4 is stable.
Figure 4: Solution: using Lyapunov’s method X1 = 3 s + 4X2 ⇒˙ x1 = −4x1 + Kx2 X2 = K s + 3X3 ⇒˙ x2 = −3x2 + Kx3 X3 = K s (−X1 + U) ⇒˙ x3 = −Kx1 + Ku The state equation is: ˙ x1 ˙ x2 ˙ x3 = −4 K 0 0 −3 K −K 0 0 x1 x2 x3 + 0 0 K u A = −4 K 0 0 −3 K −K 0 0 and B = 0 0 K Solving A′P + PA = −I, −4 0 −K K −3 0 0 K 0 p11 p12 p13 p21 p22 p23 p31 p32 p33 + p11 p12 p13 p21 p22 p23 p31 p32 p33 −4 K 0 0 −3 K −K 0 0 = −1 0 0 0 −1 0 0 0 −1 Here it is difficult to calculate the value K. So, we use a simpler method to known The stability of system by using 120 Farhan, Basirzadeh Routh stability criterion: |A −λI| = 0 −4 K 0 0 −3 K −K 0 0 − λ 0 0 0 λ 0 0 0 λ = 0 −4 −λ K 0 0 −3 −λ K −K 0 −λ = 0 λ3 + 7λ2 + 12λ + K3 = 0.
By using Routh table: Note that K appears twice in the first column in the third element and the fourth through 84 −K3 7 must be 84 −K3 > 0 = ⇒K < 4.38 in the fourth element K3 must be K > 0 Thus, for the system to be stable, it must be 0 < K < 4.38 .
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D.N. Burghes and A. Graham, Control and optimal control theories with applications, Horwood Publishing, 2004.
K. Dekker, Formula manipulation in ALGOL 68 and application to Routh’s algorithm, Computing 26 (1981), 167–187.
F. Golnaraghi and B.C. Kuo, Automatic Control Systems. McGraw-Hill Education, 2017.
M. Gopal, Control Systems Principles and Design, Second edition, Tata McGraw Hill, 2000.
G. Leitmann, The Calculus of Variations and Optimal Control: An Introduction, Springer Science & Business Media, 2013.
A.P. Liyaw, Hgram patterns of Routh Stability zones in linear systems, Int. J. Math. Educ. Sci. Technol. 28 (1997), no. 2, 225–241.
J. Macki and A. Strauss, Introduction to Optimal Control Theory, Springer Science & Business Media, 2012.
T.D. Roopamala and S.K. Katti, Comments on Routh stability criterion, Int. J. Comput. Sci. Inf. Secur. 7 (2010), 77–78.
E.J. Routh, A Treatise on the Stability of Motion, Macmillan, London, U. K., 1877.
Q. Li, What can Routh table offer in addition to stability?, J. Control Theory Appl. 1 (2003), no. 1, 9–16. |
16191 | https://brainly.com/question/49043912 | [FREE] Use the sum or difference formula for cosine to rewrite \cos(x + \pi/4) in terms of \sin(x) and \cos(x). - brainly.com
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Use the sum or difference formula for cosine to rewrite cos(x+π/4) in terms of sin(x) and cos(x). Your answer should not have π/4 in it.
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The rewritten expression for cos(x + π/4) in terms of sin(x) and cos(x) is √2/2 sin(x) + √2/2 cos(x).
Explanation:
To rewrite cos(x + π/4) in terms of sin(x) and cos(x), we utilize the sum formula for cosine, which states that cos(a + b) = cos(a) cos(b) - sin(a) sin(b). Substituting π/4 for b, we have cos(x + π/4) = cos(x) cos(π/4) - sin(x) sin(π/4). The cosine and sine of π/4 are both equal to √2/2. Therefore, cos(x + π/4) = cos(x) (√2/2) - sin(x) (√2/2).
Next, we recognize that cos(π/4) and sin(π/4) can be expressed in terms of cos(x) and sin(x) using the unit circle, where the cosine and sine values are the x and y coordinates, respectively, of the point on the unit circle corresponding to the angle π/4. For π/4, the coordinates are (√2/2, √2/2). Therefore, cos(π/4) = √2/2 and sin(π/4) = √2/2.
Substituting these values into our expression, we get cos(x + π/4) = cos(x) (√2/2) - sin(x) (√2/2). Simplifying further, we obtain √2/2 cos(x) - √2/2 sin(x). This can be rewritten as √2/2 sin(x) + √2/2 cos(x), which is the final expression for cos(x + π/4) in terms of sin(x) and cos(x). Hence, the rewritten expression does not contain π/4.
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We can rewrite cos(x+4 π) using the cosine sum formula, resulting in 2 2(cos(x)−sin(x)). This expression utilizes the values of cos(4 π) and sin(4 π), both equal to 2 2. Thus, the final expression does not include 4 π.
Explanation
To rewrite cos(x+4 π) in terms of sin(x) and cos(x), we use the cosine sum formula:
cos(a+b)=cos(a)⋅cos(b)−sin(a)⋅sin(b)
Setting a=x and b=4 π, we have:
cos(x+4 π)=cos(x)⋅cos(4 π)−sin(x)⋅sin(4 π)
The values of cos(4 π) and sin(4 π) are both 2 2. Therefore, substituting these values in, we find:
cos(x+4 π)=cos(x)⋅2 2−sin(x)⋅2 2
This can be rewritten as:
cos(x+4 π)=2 2(cos(x)−sin(x))
Additionally, we can express it as:
cos(x+4 π)=21(cos(x)−sin(x))
Thus, we have successfully expressed cos(x+4 π) in terms of sin(x) and cos(x) without using 4 π in the final expression.
Examples & Evidence
For example, if x=0, then cos(0+4 π)=cos(4 π)=2 2, and substituting sin(0)=0 and cos(0)=1, we find that 2 2(1−0)=2 2, confirming our expression is correct.
The values of cos(4 π) and sin(4 π) equal to 2 2 are well-established in trigonometry based on the unit circle, where a 45-degree angle in a right triangle results in equal leg lengths, thus validating the identity used.
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### Free Mathematics solutions and answers
Community Answer Rewrite cos ( x + π 4 ) in terms of sin (x) and cos (x)
Community Answer 1 Use the sun or difference formula for cosine to rewrite cos(x + pi/6) in terms of sine(x) and cos (x). You answer should not have pi/6 in it
Community Answer Sum and Difference Identities Score: 0/201/2 answered Rewrite cos(x+ 5π\3 ) in terms of sin(x) and cos(x).
Community Answer Rewrite cos (x - 11π/6) in terms of sin(x) and cos(x)
Community Answer Rewrite cos(x-pi/6) in terms of sin(x) and cos(x)
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16192 | https://physics.stackexchange.com/questions/297411/quasistatic-and-reversible-processes | thermodynamics - Quasistatic and Reversible processes - Physics Stack Exchange
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Quasistatic and Reversible processes
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I'm having trouble understanding the quasistatic process concept. I understand that for any process we have well defined initial and final states and the problem is in specifying the path, and the path is important as we need to know it in order to calculate work or heat because they vary depending on the path.
But I don't quite understand how implementing the process slowly can make the system in equilibrium at all points during the process, in other words, how does the system change but in the same time it remains in equilibrium? Also I read in my textbook that if a piston compresses a gas very fast, that results in a higher pressure region near the piston's surface and hence the pressure is not uniform in the gas.
I just find all of that confusing and I want to understand why the concept of quasistatic processes is important and what problem would there be in the theory of thermodynamics if we don't define processes this way? ( we cant apply integration if there is not a set of points or a path for the process?
And if we want a path we have to assume that at any point the pressure(or any property) is uniform and we can only assume that if the process is done extremely slowly so that its almost not happening and the system isn't changing!).
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edited Dec 8, 2016 at 19:00
user108787
asked Dec 8, 2016 at 16:11
Khalid T. SalemKhalid T. Salem
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1 Worthy to check Chester Miller's Reversible Vs Irreversible Gas Compression/Expansion Work.user36790 –user36790 2016-12-09 04:32:59 +00:00 Commented Dec 9, 2016 at 4:32
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But I don't quite understand how implementing the process slowly can make the system in equilibrium at all points during the process, in other words, how does the system change but in the same time it remains in equilibrium?
The quasistatic process is not any more real than the ideal gas laws are. Apologies if you know this already, but to me, and I do sincerely apologise if I have misunderstood you, you seem to be asking a question that asks how it works in real life. It doesn't. It is an idealisation, so that we can take it that the gas, in theory, has time to continually equilibrate to the changing conditions. We also, for ease of calculation, ignore other aspects, such as the friction of the piston against the cylinder. This is work done when pushing on the piston, which we assume it is negligible.
As Jackl points out, integration of tiny little steps is involved.
Quasistatic ( I suppose you could say the clue is in the name), is usually good enough in practice, in that to avoid it, you would need to push the piston so fast that the gas fails to respond at the same rate. That is, the piston would exceed the speed of sound in the gas.
What problem would there be in the theory of thermodynamics if we don't define processes this way? (we can't apply integration if there is not a set of points or a path for the process?
As far as I know, if we did not make this reasonable approximation to real life, we would get caught up in having to account for factors that have no real bearing on the teaching of thermodynamics.
In other words, we would be obliged to, unlike the ideal gas law, obtain the particular function, for each case, that relates volume to pressure. In cases such as free expansion, tbere is no defined path or function and we can use the quasistatic process to replicate the initial and final states, and allow an estimate of work done, heat and entropy changes using a reversable processes
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edited Dec 9, 2016 at 3:39
answered Dec 8, 2016 at 19:34
user108787 user108787
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Good, except for the last sentence: if the process is not quasi-static, the system is unlikely to have a well-defined pressure at all (due to pressure gradients within the system), which means there is no such function that relates volume to pressure.march –march 2016-12-09 00:53:05 +00:00 Commented Dec 9, 2016 at 0:53
@march Such as path unknown free expansion processes, so quasi static allows us to do a reversible calculation, with the same initial and final states?? Sorry, I am still at the stage of one notebook dedicated solely to keep the zillions of equations and their twists and turns in. I never studied TD before, it's much broader in scope than I had assumed. If I am wrong in my question in the first line above, that's OK ,please ignore it, I will keep slogging at it. Thank you.user108787 –user108787 2016-12-09 01:53:57 +00:00 Commented Dec 9, 2016 at 1:53
1 @CountTo10, The point of march's statement is that during irreversible processes, there can be internal viscous gradient at job; so you can't define a unique pressure for the system. That's one of the reasons why reversible processes keep track of its path and the system can be returned along the same path since it doesn't lose the information of the trajectory. Irreversible processes due to the viscous gradient lose the information of the intermediate states, so it can't re-trace the same path even though the system can be returned to the initial state.user36790 –user36790 2016-12-09 04:00:32 +00:00 Commented Dec 9, 2016 at 4:00
@ MAFIA36790, can you elaborate more on the "viscous" gradient?Khalid T. Salem –Khalid T. Salem 2016-12-10 13:04:50 +00:00 Commented Dec 10, 2016 at 13:04
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The quasistatic hypothesis is what make you use the equations, such as Gas Laws, in every point of your transition (in every point in space and in every moment in time), because you know that in every point that gas is in equilibrium. This is necessary especially when you have to integrate them and therefore you need that the equation you are integrating is valid in the whole interval of integration. This idea of making changements happen slowly implies that the equations are still true at least approximately, because changements are small.
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answered Dec 8, 2016 at 16:49
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I want to understand why the concept of quasistatic processes is important and what problem would there be in the theory of thermodynamics if we don't define processes this way?
Thermodynamics (despite the name) is the study of systems in equilibrium. The set of all possible equilibrium states is called the state space. If you imagine a high-dimensional space, with a dimension for each thermodynamic variable (e.g., pressure, temperature, volume, etc), then the state space is a surface in that space determined by the constraints imposed by equation(s) of state. The ideal gas is a good example: consider an ideal gas with fixed particle number. We have a three-dimensional space with axes for pressure, temperature, and volume. But the equilibrium states of an ideal gas are only those values of (p,V,T)(p,V,T) that satisfy p V=N k T p V=N k T. The state space is therefore a two-dimensional surface. Almost all points in (p,V,T)(p,V,T) space do not represent an ideal gas. We could add another thermodynamic variable as a fourth dimension, e.g., the internal energy, U U. But there is another equation of state: U=3 2 N k T U=3 2 N k T (for a monatomic gas), so the state space remains two-dimensional.
A quasistatic process is a transition from one equibrium state to another equilibrium state where the system remains in equilibrium at all times. Therefore, a quasistatic process is represented as a curve on the state space, i.e., a continuous sequence of equilibrium states. This is obviously an idealization: such a transition would have to occur infinitely slowly. But, it is meant to represent the infinitesimal limit of making a very small change to the system, allowing it to return to equilibrium, and then repeating.
On the state space, we can apply the tools of differential geometry to calculate changes in the thermodynamic variables along a given curve (just as one does, e.g, on the curved spacetime of Einstein's general relativity). The basic tool for performing such analysis is the thermodynamic identity, which, in its simplest form, is:
d U=T d S−p d V d U=T d S−p d V
This is a relationship between covectors (or, differential forms) that is valid at every point on the state space. While the equation(s) of state determine what the state space is, the thermodynamic identity provides the fundamental relationship between thermodynamic variables on the state space. (Operationally, we can think of covectors as objects that can be integrated over a path).
All of this analysis is performed using quasistatic transitions/processes on a state space.
So, you might ask, why do we need to call them "quasistatic transitions"? Can't we just call them thermodynamic transitions, since they seem to be all there is? Well, for physical systems they are definitively not the only kind of transitions. As mentioned above, they are an idealization that is impossible to realize in practice. Most realistic thermodynamic transitions, from one equilibrium state to another equilibrium state, occur completely out of equilibrium, with only the initial and final states represented on the state space. Examples include: rapid expansion/compression of a gas, mixing of two gases, expansion of a gas into a vacuum, heating a pot of water on a stove (even slowly: the heating sets up a temperature gradient such that the water is always out of equilibrium). Unless we want to restrict the theory to only idealized, impossible processes, we must at least recognize that much of thermodynamics occurs completely "off of" the state space.
The preceding paragraph does not mean that quasistatic processes are not useful, however. Because each equilibrium state is represented on the state space, we could take a non-quasistatic transition (represented by just the distinct initial and final points on the state space) and consider a quasistatic transition (curve) that connects these two points. For any calculations involving changes in state variables (functions that depend only on the position on the state space, e.g., internal energy, entropy, and all thermodynamic variables) the result will be the same for the quasistatic and non-quasistatic process. A typical example is calculating the change in entropy when a hot cup of water and a cold cup of water are allowed to come to equilibrium with each other. The system is never in equilibrium while their temperatures are changing. But, because entropy is a state variable, we can use any quasistatic process that would connect the initial and final states, e.g., separately, and slowly, changing the temperatures of each cup using hot plates (to that final equilibrium temperature required by conservation of energy). In that way, the entropy change of each cup is can be calculated using the thermodynamic identity: Δ S=∫d U/T=∫c m d T/T Δ S=∫d U/T=∫c m d T/T.
So, in summary: (1) quasistatic processes and the thermodynamic identity allow for calculations involving the integration over a path on the state space; (2) often these calculations can be used as a proxy for a process that is not quasistatic, but has the same initial and final states; (3) there would be no "problem ... in the theory of thermodynamics" if we consider other transitions, it's just that there are not formal tools for making calculations for them.
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edited Feb 14, 2023 at 4:02
answered Feb 14, 2023 at 3:56
Ben HBen H
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Define phase transitions and phase transition temperatures
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We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.
Vaporization and Condensation
When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation. When the rate of condensation becomes equal to the rate of vaporization, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure 5.4.15.4.1, and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.
Figure 5.4.15.4.1: In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue.
The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.
Example 5.4.15.4.1: Explaining Vapor Pressure in Terms of IMFs
Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:
Solution
Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two −OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.
Exercise 5.4.15.4.1
At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:
| Compound | methanol CH3OH | ethanol C2H5OH | propanol C3H7OH | butanol C4H9OH |
--- ---
| Vapor Pressure at 20 °C | 11.9 kPa | 5.95 kPa | 2.67 kPa | 0.56 kPa |
Answer
: All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:
Pmethanol>Pethanol>Ppropanol>Pbutanol
As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure 5.4.25.4.2. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.
Figure 5.4.25.4.2: Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, KE, to escape from the liquid into the gas phase.
Boiling Points
Video 5.4.15.4.1: Mercury (Hg) boils at room pressure at 356 °C.
When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure 5.4.35.4.3 shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure.
Figure 5.4.35.4.3: The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)
Example 5.4.25.4.2: A Boiling Point at Reduced Pressure
A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure 5.4.35.4.3 to determine the boiling point of water at this elevation.
Solution
The graph of the vapor pressure of water versus temperature in Figure 5.4.35.4.3 indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.
Exercise 5.4.25.4.2
The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use Figure 5.4.35.4.3 to determine the approximate atmospheric pressure at the camp.
Answer
: Approximately 40 kPa (0.4 atm)
The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation:
P=Ae−ΔHvap/RT
P=Ae−ΔHvap/RT(5.4.1)
where
ΔHvapΔHvap is the enthalpy of vaporization for the liquid,
RR is the gas constant, and
lnAlnA is a constant whose value depends on the chemical identity of the substance.
Equation 5.4.15.4.1 is often rearranged into logarithmic form to yield the linear equation:
lnP=−ΔHvapRT+lnA
lnP=−ΔHvapRT+lnA(5.4.2)
This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the examples and exercises that follow. If at temperature T1T1, the vapor pressure is P1P1, and at temperature T2T2, the vapor pressure is T2T2, the corresponding linear equations are:
lnP1=−ΔHvapRT1+lnA
lnP1=−ΔHvapRT1+lnA(5.4.3)
and
lnP2=−ΔHvapRT2+lnA
lnP2=−ΔHvapRT2+lnA(5.4.4)
Since the constant, ln A, is the same, these two equations may be rearranged to isolate lnAlnA and then set them equal to one another:
lnP1+ΔHvapRT1=lnP2+ΔHvapRT2lnP1+ΔHvapRT1=lnP2+ΔHvapRT2
which can be combined into:
ln(P2P1)=ΔHvapR(1T1−1T2)
ln(P2P1)=ΔHvapR(1T1−1T2)(5.4.5)
Example 5.4.35.4.3: Estimating Enthalpy of Vaporization
Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.
Solution
The enthalpy of vaporization, ΔHvapΔHvap, can be determined by using the Clausius-Clapeyron equation (Equation 5.4.55.4.5):
ln(P2P1)=ΔHvapR(1T1−1T2)
ln(P2P1)=ΔHvapR(1T1−1T2)
Since we have two vapor pressure-temperature values
T1=34.0oC=307.2KT1=34.0oC=307.2K
P1=10.0kPaP1=10.0kPa and
T2=98.8oC=372.0KT2=98.8oC=372.0K
P2=100kPaP2=100kPa
we can substitute them into this equation and solve for ΔHvapΔHvap. Rearranging the Clausius-Clapeyron equation and solving for ΔHvapΔHvap yields:
ΔHvap=R⋅ln(P2P1)(1T1−1T2)=(8.3145J/mol⋅K)⋅ln(100kPa10.0kPa)(1307.2K−1372.0K)=33,800J/mol=33.8kJ/mol
ΔHvap=R⋅ln(P2P1)(1T1−1T2)=(8.3145J/mol⋅K)⋅ln(100kPa10.0kPa)(1307.2K−1372.0K)=33,800J/mol=33.8kJ/mol
Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.
Exercise 5.4.35.4.3
At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.
Answer
: 47,782 J/mol = 47.8 kJ/mol
Example 5.4.45.4.4: Estimating Temperature (or Vapor Pressure)
For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?
Solution
If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, ΔHvap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation (Equation 5.4.15.4.1) :
ln(P2P1)=ΔHvapR(1T1−1T2)
ln(P2P1)=ΔHvapR(1T1−1T2)
Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (T1T1 = 80.1 °C = 353.3 K, P1P1 = 101.3 kPa, ΔHvapΔHvap = 30.8 kJ/mol) and want to find the temperature (T2T2) that corresponds to vapor pressure P2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for T2T2. Rearranging the Clausius-Clapeyron equation and solving for T2T2 yields:
T2=(−R⋅ln(P2P1)ΔHvap+1T1)−1=(−(8.3145J/mol⋅K)⋅ln(83.4kPa101.3kPa)30,800J/mol+1353.3K)−1=346.9Kor73.8∘C
T2=⎛⎝⎜⎜⎜−R⋅ln(P2P1)ΔHvap+1T1⎞⎠⎟⎟⎟−1=⎛⎝⎜⎜⎜−(8.3145J/mol⋅K)⋅ln(83.4kPa101.3kPa)30,800J/mol+1353.3K⎞⎠⎟⎟⎟−1=346.9Kor73.8∘C
Exercise 5.4.45.4.4
For acetone (CH3)2CO(CH3)2CO, the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C?
Answer
: 30.1 kPa
Video 5.4.25.4.2: An overview of the role of vapor pressure on boiling point.
Enthalpy of Vaporization
Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, ΔHvapΔHvap. For example, the vaporization of water at standard temperature is represented by:
H2O(l)⟶H2O(g)ΔHvap=44.01kJ/mol
H2O(l)⟶H2O(g)ΔHvap=44.01kJ/mol(5.4.6)
As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:
H2O(g)⟶H2O(l)ΔHcon=−ΔHvap=−44.01kJ/mol
H2O(g)⟶H2O(l)ΔHcon=−ΔHvap=−44.01kJ/mol(5.4.7)
Example 5.4.55.4.5: Using Enthalpy of Vaporization
One way our body is cooled is by evaporation of the water in sweat (Figure 5.4.45.4.4). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); ΔHvap=43.46kJ/molΔHvap=43.46kJ/mol at 37 °C.
Figure 5.4.45.4.4: Evaporation of sweat helps cool the body. (credit: “Kullez”/Flickr)
Solution We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:
1.5L×1000g1L×1mol18g×43.46kJ1mol=3.6×103kJ
1.5L×1000g1L×1mol18g×43.46kJ1mol=3.6×103kJ
Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.
Exercise 5.4.55.4.5: Boiling Ammonia
How much heat is required to evaporate 100.0 g of liquid ammonia, NH3NH3, at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol?
Answer
: 28 kJ
Melting and Freezing
When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure 5.4.55.4.5.
Figure 5.4.55.4.5: (a) This beaker of ice has a temperature of −12.0 °C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 °C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 °C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 °C. (credit: modification of work by Mark Ott).
If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).
The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.
The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, ΔHfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process:
H2O(s)→H2O(l)ΔHfus=6.01kJ/mol
H2O(s)→H2O(l)ΔHfus=6.01kJ/mol(5.4.8)
The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C:
H2O(l)→H2O(s)ΔHfrz=−ΔHfus=−6.01kJ/mol
Sublimation and Deposition
Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid CO2) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure 5.4.6). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.
Figure 5.4.6: Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)
Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:
CO2(s)⟶CO2(g)ΔHsub=26.1kJ/mol
Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:
CO2(g)⟶CO2(s)ΔHdep=−ΔHsub=−26.1kJ/mol
Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law.
solid⟶liquidΔHfusliquid⟶gasΔHvap_solid⟶gasΔHsub=ΔHfus+ΔHvap
Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure 5.4.7. For example:
Figure 5.4.7: For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.
Summary
Video 5.4.3: An overview of phase changes with regards to kinetics.
Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.
Key Equations
P=Ae−ΔHvap/RT
lnP=−ΔHvapRT+lnA
ln(P2P1)=ΔHvapR(1T1−1T2)
Glossary
boiling point
: temperature at which the vapor pressure of a liquid equals the pressure of the gas above it
Clausius-Clapeyron equation
: mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance
condensation
: change from a gaseous to a liquid state
deposition
: change from a gaseous state directly to a solid state
dynamic equilibrium
: state of a system in which reciprocal processes are occurring at equal rates
freezing
: change from a liquid state to a solid state
freezing point
: temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point
melting
: change from a solid state to a liquid state
melting point
: temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point
normal boiling point
: temperature at which a liquid’s vapor pressure equals 1 atm (760 torr)
sublimation
: change from solid state directly to gaseous state
vapor pressure
: (also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature
vaporization
: change from liquid state to gaseous state
Contributors
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at
Adelaide Clark, Oregon Institute of Technology
Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes.
Guillotined Chemistry is a production of Mark Anticole and available for free on Youtube.
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Ciencia
¿Cuál es la diferencia entre peso y masa?
En el lenguaje cotidiano confundimos habitualmente “peso” con “masa”. Son dos magnitudes físicas relacionadas, aunque no designa la misma magnitud.
Publicado por
Eugenio M. Fernández Aguilar
Físico, escritor y divulgador científico. Director de Muy Interesante Digital
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En el lenguaje cotidiano confundimos habitualmente “peso” con “masa”. Son dos magnitudes físicas relacionadas, pero no son lo mismo. Desde el punto de vista coloquial podemos entendernos bien y no cometemos un error grave, pero desde la perspectiva de la física el error es garrafal. Aunque, siendo precisos, el propio Diccionario de la Real Academia Española le da la razón a la Física. Menos mal.
¿Cuál es la diferencia entre peso y masa? (Eugenio M. Fernández Aguilar)
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Entendiendo los conceptos de masa y peso
¿Qué es la masa y cómo se mide?
La masa es una propiedad intrínseca de un objeto que representa la cantidad de materia que contiene. En el Sistema Internacional de Unidades, la masa se mide en kilogramos (kg). Esta magnitud es constante sin importar el lugar en el que se encuentre el objeto, ya que no depende de factores externos como la gravedad. La masa también está relacionada con la inercia, que es la resistencia de un objeto a cambiar su estado de movimiento. Esto significa que un objeto con mayor masa requerirá más fuerza para acelerar o desacelerar. Las balanzas de platillos son instrumentos que miden la masa, ya que comparan la cantidad de materia de un objeto con una referencia conocida.
Aunque la masa es una cantidad escalar, lo que implica que solo tiene magnitud y no dirección, es fundamental para entender otros conceptos físicos como la densidad y la energía cinética. La masa no cambia incluso si el objeto se mueve a diferentes lugares del universo. Por ejemplo, la masa de una persona es la misma en la Tierra que en la Luna, aunque el peso varíe debido a la diferencia en la gravedad. Este concepto es crucial para entender fenómenos físicos y para realizar cálculos precisos en ingeniería y ciencia.
Desde 2019, la definición del kilogramo se basa en la constante de Planck, lo que ha permitido una medida aún más precisa de la masa. Este cambio refleja la importancia de la masa en la ciencia moderna y cómo su comprensión ha evolucionado con el tiempo. La masa no solo es un concepto teórico, sino que también tiene aplicaciones prácticas en la industria, la tecnología y la vida cotidiana.
¿Qué es el peso y cómo se calcula?
El peso es una fuerza que resulta de la atracción gravitatoria que un cuerpo celeste, como la Tierra, ejerce sobre un objeto. A diferencia de la masa, el peso no es constante y varía dependiendo de la gravedad del lugar en el que se encuentre el objeto. Se mide en newtons (N), que es la unidad de fuerza en el Sistema Internacional. Para calcular el peso de un objeto, se utiliza la fórmula P = m · g, donde "P" es el peso, "m" es la masa y "g" es la aceleración debida a la gravedad. En la Tierra, esta aceleración es aproximadamente 9,8 m/s².
El peso es una cantidad vectorial, lo que significa que tiene tanto magnitud como dirección. La dirección del peso siempre apunta hacia el centro del cuerpo celeste que ejerce la fuerza gravitatoria. Por ejemplo, el peso de un objeto en la Tierra está dirigido hacia el centro del planeta. Esta característica es fundamental para comprender cómo funcionan las fuerzas en la física y cómo interactúan los objetos en el espacio.
La variabilidad del peso según la ubicación es un concepto que tiene aplicaciones prácticas en la vida cotidiana y en la ciencia. Por ejemplo, los astronautas experimentan una reducción significativa de su peso cuando están en la Luna, debido a que la gravedad lunar es aproximadamente una sexta parte de la gravedad terrestre. Este fenómeno no afecta a la masa del astronauta, que permanece constante. Entender esta diferencia es esencial para diseñar misiones espaciales y para calcular el consumo de energía en diferentes entornos gravitacionales.
Masa y peso: diferencias claves
Constancia de la masa vs. variabilidad del peso
Una de las diferencias más notables entre masa y peso es que la masa es constante, mientras que el peso puede variar dependiendo de la gravedad del lugar. La masa de un objeto no cambia, independientemente de si está en la Tierra, la Luna o en el espacio exterior. Esto se debe a que la masa es una propiedad intrínseca de la materia. En cambio, el peso es una propiedad extrínseca, ya que depende de la fuerza gravitatoria que actúa sobre el objeto. Por ejemplo, un astronauta en la Tierra puede pesar 700 N, pero en la Luna, donde la gravedad es menor, su peso se reducirá significativamente.
Esta distinción es crucial no solo en física teórica, sino también en aplicaciones prácticas. En ingeniería, por ejemplo, el diseño de estructuras y vehículos debe tener en cuenta cómo varía el peso en diferentes entornos gravitacionales. En la vida cotidiana, aunque no seamos conscientes, esta diferencia afecta cómo percibimos nuestro entorno. La sensación de peso es lo que nos ancla a la superficie de la Tierra y nos permite realizar actividades diarias con normalidad.
La constancia de la masa y la variabilidad del peso también tienen implicaciones en la educación y la comunicación científica. Es común que en el lenguaje cotidiano se confundan estos términos, lo que puede llevar a malentendidos. Por ello, es importante educar sobre estas diferencias para fomentar una comprensión más precisa de los conceptos científicos básicos.
Unidades de medida: kilogramos y newtons
Las unidades de medida para masa y peso son diferentes y reflejan la naturaleza de cada una de estas magnitudes. La masa se mide en kilogramos (kg), que es una unidad de medida del Sistema Internacional. Por otro lado, el peso se mide en newtons (N), que es la unidad de fuerza. Esta distinción es fundamental para evitar confusiones en la comunicación científica y en la vida cotidiana. Cuando decimos que algo pesa "10 kilogramos", en realidad estamos refiriéndonos a su masa, no a su peso.
El uso incorrecto de estas unidades es común y puede llevar a errores en cálculos y experimentos. En el ámbito científico, es esencial utilizar las unidades correctas para garantizar la precisión y la coherencia de los resultados. Por ejemplo, en física, la Segunda Ley de Newton, que relaciona fuerza, masa y aceleración, utiliza newtons para medir la fuerza, lo que destaca la importancia de esta unidad en el estudio de las fuerzas.
Además, comprender las unidades de medida es crucial en el contexto de la educación y la divulgación científica. Enseñar a utilizar correctamente kilogramos y newtons no solo ayuda a evitar errores, sino que también promueve una comprensión más profunda de los conceptos físicos. En última instancia, esta comprensión puede inspirar a futuras generaciones de científicos e ingenieros.
La relación entre masa, peso y gravedad
La relación entre masa, peso y gravedad es un concepto central en la física que ayuda a explicar cómo interactúan los objetos en el universo. La masa es la cantidad de materia en un objeto, mientras que el peso es la fuerza que resulta de la interacción de esa masa con un campo gravitatorio. La gravedad es la fuerza que atrae a los objetos hacia el centro de un cuerpo celeste, como la Tierra. Esta relación se expresa matemáticamente en la fórmula P = m · g, donde "P" es el peso, "m" es la masa y "g" es la aceleración gravitatoria.
La gravedad no solo afecta al peso, sino que también influye en otros fenómenos físicos, como las órbitas de los planetas y el comportamiento de los fluidos. La comprensión de esta relación es fundamental para campos como la astronomía, la ingeniería aeroespacial y la geofísica. Por ejemplo, en la exploración espacial, es crucial calcular cómo la gravedad de diferentes cuerpos celestes afectará a las naves y a los astronautas.
Además, la relación entre masa, peso y gravedad tiene aplicaciones prácticas en la vida diaria. Desde la forma en que experimentamos el peso en diferentes altitudes hasta cómo se diseñan los edificios para soportar cargas, la gravedad desempeña un papel esencial en nuestra percepción del mundo. Comprender esta relación no solo es importante para la ciencia, sino que también nos ayuda a apreciar la complejidad y la belleza del universo.
Errores comunes y su impacto en el lenguaje cotidiano
Confusión entre "kilos" y "kilogramos"
En el lenguaje cotidiano, es común referirse a la masa de un objeto utilizando el término "kilos", cuando en realidad deberíamos decir "kilogramos". Esta confusión se debe a la simplificación del lenguaje, pero puede llevar a malentendidos, especialmente en contextos científicos. El prefijo "kilo" significa mil, y no es una unidad de medida por sí mismo. Por lo tanto, cuando decimos "65 kilos", estamos omitiendo la unidad completa, que es "kilogramos".
La precisión en el uso del lenguaje es importante no solo en la ciencia, sino también en la educación y la comunicación diaria. Utilizar los términos correctos ayuda a evitar confusiones y a fomentar una comprensión más precisa de los conceptos. En la enseñanza de la ciencia, es fundamental educar a los estudiantes sobre la importancia de utilizar las unidades de medida adecuadas para promover una comprensión más profunda de los temas.
Además, en la industria y la tecnología, la precisión en el uso de las unidades de medida es crucial para garantizar la seguridad y la eficacia de los productos y procesos. Por ejemplo, en la fabricación de dispositivos electrónicos, las medidas precisas son esenciales para el funcionamiento correcto de los componentes. Por lo tanto, es importante que tanto los profesionales como el público en general sean conscientes de la importancia de utilizar correctamente los términos "kilogramos" y "kilos".
El uso incorrecto del término "peso" en lugar de "masa"
Otro error común en el lenguaje cotidiano es el uso del término "peso" para referirse a la "masa". Esta confusión se debe a la relación entre ambas magnitudes en la Tierra, donde la gravedad es constante y nos hace percibir el peso como una medida directa de la masa. Sin embargo, en física, es crucial distinguir entre estos términos para evitar errores en cálculos y experimentos. La masa es una propiedad constante de un objeto, mientras que el peso varía según la gravedad.
Este error puede tener implicaciones prácticas en la educación, la ciencia y la ingeniería. En la enseñanza de la física, es esencial educar a los estudiantes sobre la diferencia entre masa y peso para fomentar una comprensión más precisa de los conceptos científicos. En la industria, el uso incorrecto de estos términos puede llevar a errores en el diseño y la producción de productos, lo que puede afectar la seguridad y la eficacia.
Además, en la comunicación científica y la divulgación, es importante utilizar los términos correctos para garantizar la claridad y la precisión de la información. Al educar al público sobre la diferencia entre masa y peso, podemos promover una mayor comprensión de la ciencia y fomentar un mayor interés en el estudio de los fenómenos físicos.
Aplicaciones prácticas y ejemplos
¿Pesas lo mismo en la Tierra que en la Luna?
La diferencia entre peso y masa se hace evidente cuando consideramos el efecto de la gravedad en diferentes cuerpos celestes. En la Tierra, una persona puede pesar 700 N, pero en la Luna, donde la gravedad es aproximadamente una sexta parte de la terrestre, su peso se reduce a unos 116 N. Sin embargo, su masa sigue siendo la misma en ambos lugares. Este fenómeno es un ejemplo claro de cómo el peso varía según la gravedad, mientras que la masa permanece constante.
Este concepto tiene aplicaciones prácticas en la exploración espacial y la ingeniería aeroespacial. Los ingenieros deben tener en cuenta cómo la gravedad de diferentes cuerpos celestes afectará a las naves y a los astronautas. Además, en la educación, este ejemplo se utiliza a menudo para enseñar a los estudiantes sobre la diferencia entre masa y peso y cómo la gravedad influye en estas magnitudes.
En la vida cotidiana, aunque no experimentemos cambios tan drásticos en el peso, la gravedad afecta a nuestra percepción del entorno. Por ejemplo, en altitudes más altas, donde la gravedad es ligeramente menor, el peso de los objetos es un poco menor, aunque la diferencia es casi imperceptible. Comprender cómo la gravedad afecta al peso nos ayuda a apreciar la complejidad de nuestro entorno y a desarrollar una comprensión más profunda de la física.
Instrumentos de medición: balanzas y dinamómetros
Para medir la masa y el peso, se utilizan diferentes instrumentos. Las balanzas son dispositivos que miden la masa de un objeto, generalmente utilizando un sistema de comparación con una referencia conocida. Estas balanzas suelen mostrar el resultado en kilogramos, lo que puede llevar a confusiones si se utiliza incorrectamente el término "peso". Por otro lado, los dinamómetros son instrumentos diseñados para medir la fuerza, y por lo tanto, el peso. Estos dispositivos utilizan la ley de Hooke, que relaciona la fuerza ejercida por un resorte con su alargamiento o compresión.
La precisión en la medición de la masa y el peso es crucial en diversas aplicaciones, desde la investigación científica hasta la industria alimentaria y farmacéutica. En la ciencia, las medidas precisas son esenciales para garantizar la validez de los experimentos y los resultados. En la industria, la precisión es fundamental para garantizar la calidad y la seguridad de los productos.
Además, en la educación, es importante enseñar a los estudiantes a utilizar correctamente estos instrumentos para fomentar una comprensión más profunda de los conceptos científicos. Aprender a medir la masa y el peso de manera precisa no solo es una habilidad útil, sino que también ayuda a desarrollar un pensamiento crítico y analítico.
Ley de Hooke y su relación con el peso
La ley de Hooke es un principio fundamental en la física que describe cómo un resorte u objeto elástico se deforma en respuesta a una fuerza aplicada. Esta ley establece que la fuerza ejercida por un resorte es proporcional a su alargamiento o compresión. En el contexto de la medición del peso, la ley de Hooke se aplica en los dinamómetros, que utilizan resortes para medir la fuerza que representa el peso de un objeto.
La comprensión de la ley de Hooke es esencial para el diseño y la fabricación de dispositivos de medición y otros productos que utilizan principios de elasticidad. En la ingeniería, esta ley se aplica en el diseño de estructuras y materiales que deben soportar fuerzas y deformaciones. Además, en la ciencia, la ley de Hooke es fundamental para el estudio de la mecánica de sólidos y la elasticidad.
En la educación, la ley de Hooke se utiliza a menudo para enseñar a los estudiantes sobre la relación entre fuerza, masa y peso. Al comprender cómo funciona esta ley, los estudiantes pueden desarrollar una comprensión más profunda de los conceptos físicos y cómo se aplican en el mundo real. Además, la ley de Hooke es un ejemplo de cómo las leyes físicas pueden tener aplicaciones prácticas en la vida cotidiana y en la tecnología moderna.
Referencias
Judson L. Ahern. «International Gravity formula». School of Geology & Geophysics, University of Oklahoma.
National General Conference on Weights and Measures, Specifications, Tolerances, and Other Technical Requirements for Weighing and Measuring Devices, NIST Handbook 44.
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Model car queueing with two gates
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Say a school has two gates at which parents can pick up their kids. One gate is only for primary school students and the other is only for secondary school students. The school has a primary school to secondary school population ratio of p:s p:s.
Now, say I have cars queueing behind the two gates, with each cars assigned to be whether for picking up primary school students or secondary school students based on the population ratio (labeled "p" and "s" in the image). Each car takes w w seconds to pick up their kids (equal time at both primary gate and secondary gate). And s s cars have to wait and queue behind p p cars to pick up their kids.
I would like to model this queueing system and find out after time t t, what is the total number of students that are picked up from the school.
Should I use the Poisson process? How should I approach this problem? Any advice would be appreciated.
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edited Sep 20, 2021 at 2:43
Steven OhSteven Oh
asked Sep 20, 2021 at 1:57
Steven OhSteven Oh
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1 How long does it take to pick up a student at P and S? Do s s cars wait if there is a p p parent picking up a student? One problem is that the state of the system is not just the length of the queue, but it's exact composition (like you illustrate). You could set up the balance equations for steady state, assuming e.g. Poisson arrivals for s s and p p, and exponentially distributed pick-up times at S and P, but it won't be a classic queue.vonbrand –vonbrand 2021-09-20 02:05:48 +00:00 Commented Sep 20, 2021 at 2:05
1 @vonbrand okay! so let's say it takes w w seconds to pick up a student at P and S (equal time at both gates). And yes, s s cars have to wait for p p parent to pick up their kids. I will add these information to my question. Regarding the balance equation, can you give me some pointers on how I can approach the problem and construct the balance equations assuming Poisson arrivals? Thank you so much.Steven Oh –Steven Oh 2021-09-20 02:16:36 +00:00 Commented Sep 20, 2021 at 2:16
1 Sorry if I misunderstand, but if the time for pickup is equal and s cars have to wait for p cars (p cars also have to wait for s cars anyway), then there is no practical difference between s cars and p cars, so the total time is just number of students times average pickup time, no?uUnwY –uUnwY 2021-09-20 06:42:21 +00:00 Commented Sep 20, 2021 at 6:42
1 Ah ok you mean an s car has to wait if it's behind a p car. But if a p car is behind a s car, they can both go forward at the same time to their respective gate, but in all other cases they have to wait time w. So you only have to find the number of sp pairs.uUnwY –uUnwY 2021-09-20 06:49:50 +00:00 Commented Sep 20, 2021 at 6:49
1 I haven't really thought this through, but is Poisson suitable, considering the cars are not independent events but the total number of p and s are fixed? I wonder if there is an easier way that you just have to calculate expected number m of sp pairs in a sequence of n cars, which is a bit of combinatorics. Then the time is (n-m)w, so you have time as function of car numbers which you invert to get cars as function of time. Or am I missing something?uUnwY –uUnwY 2021-09-20 13:01:39 +00:00 Commented Sep 20, 2021 at 13:01
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I am not sure I understand correctly your problem - please add remarks if needed
let's say that each w w time you will have one of following events:
car p p waits for student and there is no s s car in front- so one car is waiting
s s car is waiting and behind him is car s s so also one car is waiting
there was two cars in queue p p and s s so two cars are waiting now let's count probability that we will have pair p s p s in sequence
in general it's markov process from here we will create transition matrix m m, you can derive p(S)p(P)p(P S)p(S)p(P)p(P S) (here it will be difficult) and assign weights 1 1 2 1 1 2 and compute E(X)E(X)
from\to P S S P S P p p+s s 2(p+s)2 s p(p+s)2 S S 0 s p+s p p+s P S p p+s s 2(p+s)2 s p(p+s)2
from\to P S S P S P p p+s 0 p p+s S S s 2(p+s)2 s p+s s 2(p+s)2 P S s p(p+s)2 p p+s s p(p+s)2
now create vector initial vector v 1=(p p+s,s 2(p+s)2,s p(p+s)2)v 1=(p p+s,s 2(p+s)2,s p(p+s)2) (fill exact values s and p into these structures) and count in for loop v n+1=v n∗m v n+1=v n∗m at each step vector v v will have probabilities of events (p(P),p(S S),p(P S))(p(P),p(S S),p(P S)) and you can calculate E X n=(1,1,2)∗v n E X n=(1,1,2)∗v n which will be expected value of students picked up at step n n
```
given m, v, n
sum_x = 0
for i in range(n):
v = v m
sum_x += [1,1,2] v.T
return sum_x
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edited Sep 25, 2021 at 15:33
answered Sep 25, 2021 at 14:53
questerquester
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Thank you so much for the solution! I am just curious on why markov process is applicable here. I understand that in markov proceess an event is depent upon the previous event, but the arrival of the cars are not dependent upon any previous events. Can this scenario also be modeled using the Poisson process?Steven Oh –Steven Oh 2021-09-27 11:49:55 +00:00 Commented Sep 27, 2021 at 11:49
1 @StevenOh I assumed that you have endless stream of cars and you would like to know how quick this stream will proceed, in general you could model car arrival times with Poisson process (like calls in call center, although amount of calls in call center is much more uniform throughout the day than car arrivals to school), but there are would be two approaches: 1. cars arrive independently and "type" is chosen at random, or two poisson processes with given intensities p p and s s, if you would give more details of your system I could give more ideas quester –quester 2021-09-27 16:07:55 +00:00 Commented Sep 27, 2021 at 16:07
Thank you so much for your thoughtful answer! I think the first approach might work better with what I have been thinking. So, the cars arrive independently and their type is decided by a binomial probability. Can you give me more ideas regarding this? Thank you so much for your input.Steven Oh –Steven Oh 2021-09-28 02:44:26 +00:00 Commented Sep 28, 2021 at 2:44
1 @StevenOh here is link to paper that solves easier problem with two spots with no distinction hindawi.com/journals/ijsa/2012/145867, also systems that you want to model are M/M/c queues en.wikipedia.org/wiki/M/M/c_queue, but your system looks more complicated than these in links, probably there is something that fits your needs. This looks most promising en.wikipedia.org/wiki/Markovian_arrival_processquester –quester 2021-09-28 16:15:53 +00:00 Commented Sep 28, 2021 at 16:15
thank you so much for the resources. I have been reading them and they are super helpful. Lastly, do you think it would also be an option to model my problem in a case-by-case manner. So for each cases of the combination between car p p and car &s& (eg. ps, sp, pp, ss), I model using a poisson process, and then later I sum these up somehow. Again, thank you for your input.Steven Oh –Steven Oh 2021-09-29 08:30:43 +00:00 Commented Sep 29, 2021 at 8:30
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16197 | https://stackoverflow.com/questions/2533243/partition-a-rectangle-into-near-squares-of-given-areas | Stack Overflow
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Partition a rectangle into near-squares of given areas
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I have a set of N positive numbers, and a rectangle of dimensions X and Y that I need to partition into N smaller rectangles such that:
the surface area of each smaller rectangle is proportional to its corresponding number in the given set
all space of big rectangle is occupied and there is no leftover space between smaller rectangles
each small rectangle should be shaped as close to square as feasible
the execution time should be reasonably small
I need directions on this. Do you know of such an algorithm described on the web? Do you have any ideas (pseudo-code is fine)?
Thanks.
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edited Mar 28, 2010 at 19:01
Svante
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asked Mar 28, 2010 at 14:32
Marko DumicMarko Dumic
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What you describe sounds like a treemap:
Treemaps display hierarchical (tree-structured) data as a set of nested rectangles. Each branch of the tree is given a rectangle, which is then tiled with smaller rectangles representing sub-branches. A leaf node's rectangle has an area proportional to a specified dimension on the data.
That Wikipedia page links to a page by Ben Shneiderman, which gives a nice overview and links to Java implementations:
Then while puzzling about this in the faculty lounge, I had the Aha! experience of splitting the screen into rectangles in alternating horizontal and vertical directions as you traverse down the levels. This recursive algorithm seemed attractive, but it took me a few days to convince myself that it would always work and to write a six line algorithm.
Wikipedia also to "Squarified Treemaps" by Mark Bruls, Kees Huizing and Jarke J. van Wijk (PDF) that presents one possible algorithm:
How can we tesselate a rectangle recursively into rectangles, such that their aspect-ratios (e.g. max(height/width, width/height)) approach 1 as close as possible? The number of all possible tesselations is very large. This problem falls in the category of NP-hard problems. However, for our application we do not need the optimal solution, a good solution that can be computed in short time is required.
You do not mention any recursion in the question, so your situation might be just one level of the treemap; but since the algorithms work on one level at a time, this should be no problem.
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edited Mar 28, 2010 at 15:26
answered Mar 28, 2010 at 15:20
ThomasThomas
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Marko Dumic
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I'll take a closer look at the links you provided but I think it's not hierarchical tree map, although you're right in that it should look like one. I saw these tree maps several times (e.g. graphical measure of how much an API has changed from version to version or representation of disk usage per folder/file). I have no hierarchy in my data. E.g. suppose I want to graph 100 market shares' trade for last 24 hours where area is proportional to the volume of the trade (and change in price is represented by color).
Thomas
Thomas
From Bruls et al.: "First, we do not consider the subdivision for all levels simultaneously. This leads to an explosion in computation time. Instead, we strive to produce square-like rectangles for a set of siblings, given the rectangle where they have to fit in, and apply the same method recursively." So it sounds like that should work for you. The example in section 3.1 should already give you a good idea of how it works; pseudocode is in section 3.2.
1
I have been working on something similar. I'm prioritizing simplicity over getting as similar aspect ratios as possible. This should (in theory) work. Tested it on paper for some values of N between 1 and 10.
N = total number of rects to create, Q = max(width, height) / min(width, height), R = N / Q
If Q > N/2, split the rect in N parts along its longest side. If Q <= N/2, split the rect in R (rounded int) parts along its shortest side. Then split the subrects in N/R (rounded down int) parts along its shortest side. Subtract the rounded down value from the result of the next subrects division. Repeat for all subrects or until the required number of rects are created.
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answered Jun 16, 2010 at 22:34
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16198 | https://math.stackexchange.com/questions/1216547/find-all-primes-p-11-such-that-11-is-a-quadratic-residue-pmod-p | Find all primes $p > 11$ such that $11$ is a quadratic residue $\pmod p$. - Mathematics Stack Exchange
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Find all primes p>11 p>11 such that 11 11 is a quadratic residue (mod p)(mod p).
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Find all primes p>11 p>11 such that 11 11 is a quadratic residue (mod p)(mod p). (The answer should be in terms of congruence conditions mod a certain number.)
I believe that we need to find all primes p>11 p>11 such that (11 p)=1(11 p)=1.
or (11 p)=(−1)(p−1)/2(p 11)(11 p)=(−1)(p−1)/2(p 11) by the reciprocity law.
How do I proceed from this? Any help is appreciated.
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asked Apr 1, 2015 at 23:27
mike russelmike russel
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By quadratic reciprocity:
(11 p)=(−1)(p−1)/2(p 11)(11 p)=(−1)(p−1)/2(p 11)
We now need to work with two cases:
p≡1(mod 4)p≡1(mod 4)
This means that:
(11 p)=(p 11)(11 p)=(p 11)
The quadratic residues modulo 11 11 are 1,3,4,5 1,3,4,5 and 9 9.
p≡3(mod 4)p≡3(mod 4)
This means that:
(11 p)=−(p 11)(11 p)=−(p 11)
The quadratic non-residues modulo 11 11 are 2,6,7,8 2,6,7,8 and 10 10.
With this information, you should be able to finish the problem yourself (hint: use the Chinese Remainder Theorem).
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answered Apr 1, 2015 at 23:50
George V. WilliamsGeorge V. Williams
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Would the Chinese remainder theorem, say that there's a unique number (mod 4∗11)(mod 4∗11) satisfying the congruences: x=1,3,4,5,9(mod 11)x=1,3,4,5,9(mod 11) and x=1(mod 4)x=1(mod 4)? do we need to solve this system of 6 6 congruences?mike russel –mike russel 2015-04-02 00:13:12 +00:00 Commented Apr 2, 2015 at 0:13
@mikerussel, take them two at a time. You want x≡1(mod 11)x≡1(mod 11) and x≡1(mod 4)x≡1(mod 4), then you want x≡3(mod 11)x≡3(mod 11) and (still) x≡1(mod 4)x≡1(mod 4), then you want x≡4(mod 11)x≡4(mod 11) and x≡1(mod 4)x≡1(mod 4), etc. etc. This will give you a bunch of congruences, and if p p satisfies any of them it will be a quadratic residue.George V. Williams –George V. Williams 2015-04-02 00:15:22 +00:00 Commented Apr 2, 2015 at 0:15
Alright, Thank a lot!mike russel –mike russel 2015-04-02 00:16:52 +00:00 Commented Apr 2, 2015 at 0:16
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Treat the two factors (−1)(p−1)/2(−1)(p−1)/2 and (p 11)(p 11) separately. The first is 1 1 if p p is 1 mod 4 and it's −1−1 if p p is 3 mod 4. The second is 1 1 if p p is 1,3,4,5, or 9 mod 11 and −1−1 otherwise. Combining the two, you can determine (11 p)(11 p) in terms of the congruence class of p p mod 44 44.
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answered Apr 1, 2015 at 23:52
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16199 | https://byjus.com/binomial-expansion-calculator/ | Binomial Expansion Calculator is a free online tool that displays the expansion of the given binomial term BYJU’S online binomial expansion calculator tool makes the calculation faster, and it displays the expanded form in a fraction of seconds.
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What is Meant by Binomial Expansion?
In Algebra, a polynomial with two terms is called a binomial. The two terms are separated by either plus or minus symbol. The binomial theorem defines the binomial expansion of a given term. A binomial theorem is a mathematical theorem which gives the expansion of a binomial when it is raised to the positive integral power. Thus, the formula for the expansion of a binomial defined by binomial theorem is given as:
(\begin{array}{l}(a+b)^{n}=\sum_{k=0}^{n}\begin{pmatrix} n\ k \end{pmatrix}a^{n-k}b^{k}\end{array} )
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