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Illustrative Mathematics Unit 6.1, Lesson 2: Finding Area by Decomposing and Rearranging
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Learn about finding the area of shapes by decomposing and rearranging them into regular shapes. After trying the questions, click on the buttons to view answers and explanations in text or video.
Return to the list of Illustrative Math lessons
Finding Area by Decomposing and Rearranging
Let’s create shapes and find their areas.
Illustrative Math Unit 6.1, Lesson 2 (printable worksheets)
2.1 - What is Area?
You may recall that the term area tells us something about the number of squares inside a two-dimensional shape.
Here are four drawings that each show squares inside a shape. Select all drawings whose squares could be used to find the area of the shape. Be prepared to explain your reasoning.
Then, write a definition of area that includes all the information you think is important.
Show AnswersA can be used. Each square is a unit, and each fraction of a square is a fractional unit. The area of a shape can be a non-whole number.
B can be used. The squares are not all the same size, so there are different units, and simply counting the squares will not give the area. However, four small squares make up a large square. With this information, the squares can be used to find the area of the shape.
C cannot be used. Because the squares do not tile the plane within the shape and there are gaps and overlaps, the squares or parts of squares do not equal the total area within the shape.
D can be used. As with A, counting the squares and the fractions of squares will give the area of the shape.
What is Area?Area is the measure of space inside a two-dimensional region, without any gaps or overlaps.
See Video 1 for Whole Lesson
See Video 2 for Whole Lesson
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2.2 - Composing Shapes
Open the applet or print out the shapes in the applet. The applet has one square and some small, medium, and large right triangles. The area of the square is 1 square unit. You can click on a shape and drag to move it. Grab the point at the vertex and drag to turn it.
Notice that you can put together two small triangles to make a square. What is the area of the square composed of two small triangles? Be prepared to explain your reasoning.
Use your shapes to create a new shape with an area of 1 square unit that is not a square. Draw your shape on paper and label it with its area.
Use your shapes to create a new shape with an area of 2 square units. Draw your shape and label it with its area.
Use your shapes to create a different shape with an area of 2 square units. Draw your shape and label it with its area.
Use your shapes to create a new shape with an area of 4 square units. Draw your shape and label it with its area. Find a way to use all of your pieces to compose a single large square. What is the area of this large square?
See Possible Answers
The square composed of two small triangles matches up to the square unit. Hence, the area of the square composed of 2 small triangles is one square unit.
2.
Since two small triangles cover 1 square unit, any shape composed of two small triangles will have an area of 1 square unit. The triangle in the middle is one example. This also demonstrates that the medium triangle also has an area of 1 square unit. 3 and 4.
As with question 2, any combination of shapes which total 2 square units can be rearranged into a larger shape with an area of 2 square units. These are two examples.
3.
This is an example of a shape with an area of 4 square units composed of the shapes in the applet. Note how the large triangle has an area of 2 square units.
This is a square composed of all the pieces in the applet.
Each small triangle is ½ square units, so the small triangles are 4 × ½ = 2 square units.
The square is 1 square unit.
The medium triangle is 1 square unit.
The large triangles are 2 square units, so they total 2 × 2 = 4 square units.
Hence, the total area of the pieces, and of the square composed of these pieces, is 2 + 1 + 1 + 4 = 8 square units.
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2.3 - Tangram Triangles
Open the same applet from section 2.2 or use the same shapes.
Recall that the area of the square you saw earlier is 1 square unit. Complete each statement and explain your reasoning.
The area of the small triangle is ______ square units. I know this because . . .
The area of the medium triangle is ______ square units. I know this because . . .
The area of the large triangle is ______ square units. I know this because . . .
See Possible Answers
The area of the small triangle is ½ square units. This is because 2 small triangles can be composed into 1 square, or 1 square can be decomposed into 2 small triangles.
The area of the medium triangle is 1 square unit. This is because 2 small triangles can be composed into 1 medium triangle, or 1 medium triangle can be decomposed into 2 medium triangles.
The area of the large triangle is 2 square units. This is because it can be composed of other shapes with a total area of 2 square units, or decomposed into other shapes with a total area of 2 square units.
Lesson 2 Summary
Here are two important principles for finding area, which you used in the previous activities:
If two figures can be placed one on top of the other so that they match up exactly, then they have the same area.
The area of a figure can be found by adding the areas of its parts. If we compose (put together) a new figure from smaller pieces without overlapping them, then the sum of the areas of the pieces is the area of the new figure. Likewise, if we decompose (cut or break apart) a given figure into pieces, then the area of the given figure is the sum of the areas of the pieces. Even if we rearrange the pieces, the overall area does not change. Here are illustrations of the two principles:
Show Figures
Each square on the left can be decomposed into 2 triangles. These triangles can be rearranged into a large triangle. So the large triangle has the same area as the 2 squares.
Similarly, the large triangle on the right can be decomposed into 4 equal triangles. The triangles can be rearranged to form 2 squares. If each square has an area of 1 square unit, then the area of the large triangle is 2 square units. We also can say that each small triangle has an area of ½ square unit.
Practice Problems
The diagonal of a rectangle is shown.
Show Rectangle
a. Decompose the rectangle along the diagonal, and recompose the two pieces to make a different shape.
b. How does the area of this new shape compare to the area of the original rectangle? Explain how you know.
See Possible Answers
a. The rectangle can be decomposed into two triangles and rearranged into a number of shapes. One example is shown.
b. This new shape has the same area as the original rectangle. Decomposing and rearranging a shape does not change the area.
The area of the square is 1 square unit. Two small triangles can be put together to make a square or to make a medium triangle.
Show Figures
Which figure also has an area of 1½ square units? Select all that apply.
Show Figures
HintsWhat is the area of a small triangle in square units? What about a medium triangle?
AnswersSince 2 small triangles can compose 1 square unit or 1 medium triangle, 1 small triangle is ½ square units, and 1 medium triangle is 1 square unit.
A has an area of 1½ square units. A is composed of 3 small triangles, and therefore has an area of 3 × 1½ = 1½ square units.
B has an area of 1½ square units, as it is composed of 3 small triangles in a different arrangement from A.
C has an area of 1½ square units. C is composed of 1 small triangle and 1 medium triangle, and therefore has an area of ½ + 1 = 1½ square units.
D has an area of 2 square units and should not be selected. D is composed of 2 small triangles and 1 square, and therefore has an area of (2 × ½) + 1 = 2 square units.
Priya decomposed a square into 16 smaller, equal-size squares and then cut out 4 of the small squares and attached them around the outside of original square to make a new figure.
Show Figures
How does the shaded area of her new figure compare with that of the original square?
A. The area of the new figure is greater.
B. The two figures have the same area.
C. The area of the original square is greater.
D. We don’t know because neither the side length nor the area of the original square is known.
AnswersB. The two figures have the same area.
When a figure is decomposed and rearranged without any gaps or overlaps, the area in the new figure is the same as the area in the old figure.
The area of a rectangular playground is 78 square meters. If the length of the playground is 13 meters, what is its width?.
HintsBecause area is the measure of space within a two-dimensional region, the area (A) of a rectangle is equal to the product of its two dimensions, or its length (L) multiplied by its width (W).
AnswersA = L × W
78 m 2 = 13 m × W
78 m 2 ÷ 13 m = W
W = 6 m
A student said, “We can’t find the area of the shaded region because the shape has many different measurements, instead of just a length and a width that we could multiply.”
Show Figure
Explain why the student’s statement about area is incorrect.
See Possible AnswersThis figure can be decomposed into regular shapes. The total area of all the new regular shapes is the area of the figure.
In this example, rectangle A has an area of 35 × 10 = 350 units.
Rectangle B has an area of 10 &215; 15 = 150 units.
Rectangle C has an area of (60 - 15) × (30 - 10) = 45 × 20 = 900 units.
The total area of these three rectangles, and hence the area of the original figure, is 350 + 150 + 900 = 1400 units.
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Tools & Reference>Infectious Diseases
Chancroid Workup
Updated: Mar 23, 2022
Author: Joseph Adrian L Buensalido, MD; Chief Editor: Pranatharthi Haran Chandrasekar, MBBS, MD more...;)
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Sections
Chancroid
Sections Chancroid
Overview
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References;)
Workup
Laboratory Studies
No laboratory testing is able to immediately confirm the diagnosis of chancroid.
A definitive diagnosis of chancroid is based on isolation of H ducreyi on special media, but such tests are not readily available in many centers. In addition, lesion culture is inaccurate owing to the fastidious nature of the organism, with a sensitivity of less than 80%. [45, 46]
The nucleic acid amplification test (NAAT) is a multiplex PCR assay that yields a high detection rate, although, no molecular assays have been cleared by the Food and Drug Administration (FDA) for use in the United States.
The role of polymerase chain reaction (PCR) in rapid detection of H ducreyi is promising and may supersede culture in diagnosis. [45, 49, 50]
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Other Tests
When possible, every patient with chancroid should be tested for the other common STIs (syphilis, HSV, gonorrhea, chlamydia) and HIV.
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Procedures
Needle aspiration and/or incision and drainage are recommended for buboes that are fluctuant and tender. As with other abscesses, incision and drainage may be a superior technique for preventing abscess recurrence.
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Histologic Findings
Gram stain of the ulcer exudates may reveal short, plump, gram-negative rods in the classic school of fish appearance. Ulcer biopsy should reveal three distinct zones. The most superficial zone contains erythrocytes, fibrin, necrotic tissue, and neutrophils. The next zone consists of marked endothelial cell proliferation and many thrombosed new blood vessels. The deepest layer is characterized by a dense infiltrate of plasma and lymphoid cells.
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Treatment & Management
References
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T R Suntoke, A Hardick, A A R Tobian, et al. Evaluation of multiplex real-time PCR for detection of Haemophilus ducreyi, Treponema pallidum, herpes simplex virus type 1 and 2 in the diagnosis of genital ulcer disease in the Rakai District, Uganda. Sex Transm Infect. April/2009. 85:97-101. [QxMD MEDLINE Link].[Full Text].
The World Health Organization. Regional Strategy for the Control and Prevention of Sexually Transmitted Infections, 2007-2015. World Health Organization. Available at publications.. Accessed: 5/30/12.
O'Farrell N, Lazaro N. UK National Guideline for the management of Chancroid 2014. Int J STD AIDS. 2014 Dec. 25 (14):975-83. [QxMD MEDLINE Link].
Breau C, Cameron DW, Desjardins M, Lee BC. Oral immunization using HgbA in a recombinant chancroid vaccine delivered by attenuated Salmonella typhimurium SL3261 in the temperature-dependent rabbit model. J Immunol Methods. 2012 Jan 31. 375 (1-2):232-42. [QxMD MEDLINE Link].
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Author
Joseph Adrian L Buensalido, MD Clinical Associate Professor, Division of Infectious Diseases, Department of Medicine, Philippine General Hospital, University of the Philippines Manila College of Medicine; Specialist in Infectious Diseases, Private PracticeJoseph Adrian L Buensalido, MD is a member of the following medical societies: American Society for Microbiology, Infectious Diseases Society of America, Michigan Infectious Disease Society, Philippine College of Physicians, Philippine Medical Association, Philippine Society for Microbiology and Infectious DiseasesDisclosure: Serve(d) as a speaker or a member of a speakers bureau for: Unilab; BSV Bioscience; Philcare Pharma
Received sponsorship to medical conference for: Pfizer; Natrapharm.
Coauthor(s)
Joanne Carmela Martinez Sandejas, MD Medical Specialist II, Division of Infectious Diseases, Department of Medicine, Philippine General Hospital, University of the Philippines Manila College of Medicine, Philippines Joanne Carmela Martinez Sandejas, MD is a member of the following medical societies: Philippine College of Physicians, Philippine Society for Microbiology and Infectious DiseasesDisclosure: Nothing to disclose.
Specialty Editor Board
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Received salary from Medscape for employment.
Charles V Sanders, MD Edgar Hull Professor and Chairman, Department of Internal Medicine, Professor of Microbiology, Immunology and Parasitology, Louisiana State University School of Medicine in New Orleans; Medical Director, Medicine Hospital Center, Charity Hospital and Medical Center of Louisiana at New Orleans; Consulting Staff, Ochsner Medical CenterCharles V Sanders, MD is a member of the following medical societies: Alliance for the Prudent Use of Antibiotics, Alpha Omega Alpha, American Association for Physician Leadership, American Association for the Advancement of Science, American Association of University Professors, American Clinical and Climatological Association, American College of Physicians, American Federation for Medical Research, American Geriatrics Society, American Lung Association, American Medical Association, American Society for Microbiology, American Thoracic Society, American Venereal Disease Association, Association for Professionals in Infection Control and Epidemiology, Association of American Medical Colleges, Association of American Physicians, Association of Professors of Medicine, Infectious Disease Society for Obstetrics and Gynecology, Infectious Diseases Society of America, Louisiana State Medical Society, Orleans Parish Medical Society, Royal Society of Medicine, Sigma Xi, The Scientific Research Honor Society, Society of General Internal Medicine, Southeastern Clinical Club, Southern Medical Association, Southern Society for Clinical Investigation, Southwestern Association of Clinical Microbiology, The Foundation for AIDS ResearchDisclosure: Receives royalties from Baxter International for: Takeda-receives royalties; UpToDate-receives royalties.
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Pranatharthi Haran Chandrasekar, MBBS, MD Professor, Chief of Infectious Disease, Department of Internal Medicine, Wayne State University School of MedicinePranatharthi Haran Chandrasekar, MBBS, MD is a member of the following medical societies: American College of Physicians, American Society for Microbiology, International Immunocompromised Host Society, Infectious Diseases Society of AmericaDisclosure: Nothing to disclose.
Additional Contributors
Larry I Lutwick, MD, FACP Editor-in-Chief, ID Cases; Moderator, Program for Monitoring Emerging Diseases; Adjunct Professor of Medicine, State University of New York Downstate College of MedicineLarry I Lutwick, MD, FACP is a member of the following medical societies: American Association for the Advancement of Science, American Association for the Study of Liver Diseases, American College of Physicians, American Federation for Clinical Research, American Society for Microbiology, Infectious Diseases Society of America, Infectious Diseases Society of New York, International Society for Infectious Diseases, New York Academy of Sciences, Veterans Affairs Society of Practitioners in Infectious DiseasesDisclosure: Nothing to disclose.
Pamela Arsove, MD, FACEP Associate Residency Director, Department of Emergency Medicine, Hofstra Northshore Long Island Jewish School of Medicine; Attending Physician, Department of Emergency Medicine, Long Island Jewish Medical Center; Assistant Professor, Department of Emergency Medicine, Northshore Long Island Jewish School of MedicinePamela Arsove, MD, FACEP is a member of the following medical societies: American College of Emergency Physicians, American Medical Association, Phi Beta Kappa, Society for Academic Emergency MedicineDisclosure: Nothing to disclose.
Barbara Edwards, MD Associate Physician, Division of Infectious Diseases, Department of Medicine, Long Island Jewish Medical Center; Assistant Professor, Department of Medicine, Albert Einstein College of Medicine of Yeshiva UniversityBarbara Edwards, MD is a member of the following medical societies: American College of Physicians, Infectious Diseases Society of America, Society for Healthcare Epidemiology of AmericaDisclosure: Nothing to disclose.
Christian N Francisco, MD Chief Fellow, Section of Infectious Diseases, Department of Medicine, University of the Philippines-Philippine General Hospital Christian N Francisco, MD is a member of the following medical societies: Philippine College of PhysiciansDisclosure: Nothing to disclose.
Acknowledgements
The authors and editors of Medscape Reference gratefully acknowledge the contributions of previous authors Alexandre F Migala, DO, and Gregory Shipkey, MD, to the development and writing of this article.
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大荔县文工团
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1951年成立的社会团体
大荔县文工团是1951年5月成立于陕西省朝邑县的文艺团体，初名朝邑群众剧团，以碗腔、秦腔表演为主要艺术形式，首任团长为李玉民 。
该团历经多次改制：1959年更名为大荔县第一剧团，1960年改为戏曲剧院秦腔团，1963年转为渭南专区碗碗腔剧团，1976年定现名。1958年将皮影碗碗腔改编为舞台剧《二度梅》，次年排演《兵火缘》并在西安公演。1961年携《二度梅》等剧目赴京演出二十余场，获周恩来等中央领导接见，随后开展五省巡演。文革期间活动停滞，1978年后恢复演出并通过训练班培养王玲、刘云等演员充实阵容 。
中文名 大荔县文工团
地理位置陕西省大荔县
成立时间 1951年5月
曲 目 碗碗腔、秦腔
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1主要成员
2发展历史
主要成员
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编辑
团长李玉民，副团长惠居民、(上厶下贝)安民。主要演员有韩正满、卞海山、何新喜、乔福元、李玉民、王西民、惠居民、徐元民、党训民、吴伟民、(上厶下贝)安民、申元民、尚恒志、马正新、李存才、柴绪仔、李文端、雷文立、张义长等。
发展历史
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从1952年始共招收六班二百名学生，设训练班进行培训，成名者有王玲、刘云、刘淑琴、梁益龙、何秀琴、李学军、张毅、孔雀等，人才济济，阵容强大。
1959年，大荔、朝邑两县合并，改为大荔县第一剧团。 1960年又改为大荔县戏曲剧院秦腔团。1963年又改为渭南专区碗碗腔剧团。“文化大革命”开始后，改名为大荔县东方红文工团，粉碎“四人帮”后，始改为大荔县文工团。
1958年2月始，剧团演职人员在书记田建军、副团长李玉民、惠居民领导下，大胆革新，发挥艺术创造能力，经过半年刻苦实践，以整理改编的《二度梅》为脚本，将以皮影形式演出的碗碗腔剧种搬上大戏舞台。演出后，受到各界人士的赞扬，使碗碗腔剧种获得新生，得到了发展。1959年该团在省文化局鱼讯局长的支持下，整理排导了《兵火拉伞》，改名为《兵火缘》，在西安演出，轰动古城。曾于丈八沟招待陈毅、薄一波等国家领导人，还为宋庆龄副主席作专场演出，受到了好评。1961年，该团以《二度梅》、《兵火缘》、《金琬钗》与富平阿宫腔剧团组成晋京演出团，在鱼讯率领下赴京演出二十余场，周总理等二十六位中央领导观看了演出，并给予很高评价。北京演出后，剧团应邀赴山东济南、淄博、青岛等地演出二十五天。山东电视台将《兵火缘》录了相，两次向全省播放。1963年剧团赴山西、内蒙、宁夏、甘肃、青海五省巡回演出。“文化大革命”中，剧团处于瘫痪。党的十一届三中全会之后，该团先后招收学员两班，共一百余名，进行正规训练，充实了演出队伍，提高了演出质量。
参考资料
1 三、碗碗腔的剧团概况．fanqienovel.com．2022-03-22
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8303 | https://mathoverflow.net/questions/334934/sufficient-conditions-for-a-system-of-linear-inequalities-to-admit-a-solution | Skip to main content
Sufficient conditions for a system of linear inequalities to admit a solution
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I am looking for sufficient conditions such that a system of linear inequalities of the type Ax>0 admits a non-negative solution x∈Rn+. I know a few properties of the m×n matrix A
All entries are either 0,1 or −1.
For each row i there exists at least one column j such that Aij=1.
There exists at least one column j such that ∑iAij=1.
Do you know of any literature that studies sufficient conditions of this type?
linear-algebra
inequalities
linear-programming
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asked Jun 27, 2019 at 16:15
PeterPeter
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I'm studying this kind of problem. Did you find some literatures or something new about it? Thanks.
– Quy Nguyen
Commented
Sep 22, 2020 at 7:48
Unfortunately, I did not find much on the topic apart from en.wikipedia.org/wiki/Alternating_sign_matrix
– Peter
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Sep 26, 2020 at 19:25
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Here are some unsatisfying sufficient conditions followed by an interesting one.
A is the identity matrix In.
A has no negative entries.
Each row has a positive sum.
There is a list of non-negative weights x1,⋯,xn such that the weighted sum of each row is positive ∑n1xiaij>0.
Each condition is more general than the previous ones. The last condition is both sufficient and necessary. This because it simply restates the desired property.
So what kind of condition do you seek?
Let me remove the non-negativity condition and instead place the n rows of In at the top to get an (n+m)×n matrix A∗ with the condition A∗x≥0 with any 0 entries among the top n.
Each inequality determines a half space in Rn closed for the first n and open for the rest. The question is if their intersection is non-empty.
Helly’s Theorem states that given t>n convex sets in Rn , if every n+1 of them have non-empty intersection then all t do. So this is also a necessary and sufficient condition.
In this case we can reduce to requiring that any n of them can be simultaneously satisfied: Scaling by a positive factor does not change anything so we can restrict to vectors x which have ∑xi=1. These belong to a set which is essentially Rn−1
If I have it right, the condition is thus that of the original m inequalities,
any one can be satisfied with at most one negative entry
any pair can be satisfied with at most two negative entries
etc.
any n can be satisfied by some vector ( no restrictions on signs)
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answered Jun 29, 2019 at 1:18
Aaron MeyerowitzAaron Meyerowitz
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Thanks a lot for pointing out helly's theorem and please excuse the vague question. My hope was that there is a literature studying matrices with entries only +1,-1,0 and additional properties. I came across alternating sign matrices which turned out to be useful for me: en.wikipedia.org/wiki/Alternating_sign_matrix
– Peter
Commented
Jul 5, 2019 at 11:38
For n>1 the condition "A has no negative entries" is not more general than the previous one ("A is the identity matrix").
– Alex Ravsky
Commented
Mar 18, 2020 at 17:41
It could be some other permutation matrix
– Aaron Meyerowitz
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Mar 18, 2020 at 18:29
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11.1: Introducing Double Number Line Diagrams
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Section 11: Representing Equivalent Ratios
11.2: Creating Double Number Line Diagrams
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( \newcommand{\kernel}{\mathrm{null}\,})
Lesson
Let's use number lines to represent equivalent ratios.
Exercise : Number Talk: Adjusting Another Factor
Find the value of each product mentally.
Exercise : Drink Mix on a Double Number Line
The other day, we made drink mixtures by mixing 4 teaspoons of powdered drink mix for every cup of water. Here are two ways to represent multiple batches of this recipe:
How can we tell that and are equivalent ratios?
How are these representations the same? How are these representations different?
How many teaspoons of drink mix should be used with 3 cups of water?
How many cups of water should be used with 16 teaspoons of drink mix?
What numbers should go in the empty boxes on the double number line diagram? What do these numbers mean?
Are you ready for more?
Recall that a perfect square is a number of objects that can be arranged into a square. For example, 9 is a perfect square because 9 objects can be arranged into 3 rows of 3. 16 is also a perfect square, because 16 objects can be arranged into 4 rows of 4. In contrast, 12 is not a perfect square because you can’t arrange 12 objects into a square.
How many whole numbers starting with 1 and ending with 100 are perfect squares?
What about whole numbers starting with 1 and ending with 1,000?
Exercise : Blue Paint on a Double Number Line
Here is a diagram showing Elena’s recipe for light blue paint.
Complete the double number line diagram to show the amounts of white paint and blue paint in different-sized batches of light blue paint.
Compare your double number line diagram with your partner. Discuss your thinking. If needed, revise your diagram.
How many cups of white paint should Elena mix with 12 tablespoons of blue paint? How many batches would this make?
How many tablespoons of blue paint should Elena mix with 6 cups of white paint? How many batches would this make?
Use your double number line diagram to find another amount of white paint and blue paint that would make the same shade of light blue paint.
How do you know that these mixtures would make the same shade of light blue paint?
Summary
You can use a double number line diagram to find many equivalent ratios. For example, a recipe for fizzy juice says, “Mix 5 cups of cranberry juice with 2 cups of soda water.” The ratio of cranberry juice to soda water is . Multiplying both ingredients by the same number creates equivalent ratios.
This double number line shows that the ratio is equivalent to . If you mix 20 cups of cranberry juice with 8 cups of soda water, it makes 4 times as much fizzy juice that tastes the same as the original recipe.
Glossary Entries
Definition: Double Number Line Diagram
A double number line diagram uses a pair of parallel number lines to represent equivalent ratios. The locations of the tick marks match on both number lines. The tick marks labeled 0 line up, but the other numbers are usually different.
Practice
Exercise
A particular shade of orange paint has 2 cups of yellow paint for every 3 cups of red paint. On the double number line, circle the numbers of cups of yellow and red paint needed for 3 batches of orange paint.
Exercise
This double number line diagram shows the amount of flour and eggs needed for 1 batch of cookies.
Complete the diagram to show the amount of flour and eggs needed for 2, 3, and 4 batches of cookies.
What is the ratio of cups of flour to eggs?
How much flour and how many eggs are used in 4 batches of cookies?
How much flour is used with 6 eggs?
How many eggs are used with 15 cups of flour?
Exercise
Here is a representation showing the amount of red and blue paint that make 2 batches of purple paint.
On the double number line, label the tick marks to represent amounts of red and blue paint used to make batches of this shade of purple paint.
How many batches are made with 12 cups of red paint?
How many batches are made with 6 cups of blue paint?
Exercise
Diego estimates that there will need to be 3 pizzas for every 7 kids at his party. Select all the statements that express this ratio.
The ratio of kids to pizzas is .
The ratio of pizzas to kids is 3 to 7.
The ratio of kids to pizzas is .
The ratio of pizzas to kids is 7 to 3.
For every 7 kids there need to be 3 pizzas.
(From Unit 2.1.1)
Exercise
Draw a parallelogram that is not a rectangle that has an area of 24 square units. Explain or show how you know the area is 24 square units.
Draw a triangle that has an area of 24 square units. Explain or show how you know the area is 24 square units.
(From Unit 1.2.3)
Section 11: Representing Equivalent Ratios
11.2: Creating Double Number Line Diagrams |
8305 | https://mathworksheets.quora.com/Arithmetic-https-www-quora-com-How-is-0-5-greater-than-0-05-answer-Charles-Holmes-66 | How is 0.5 greater than 0.05? - Math Worksheets - Quora
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Charles Holmes
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Arithmetic
Charles Holmes
Studied Financial Markets&Mathematics (Graduated 1990) · May 22 ·
How is 0.5 greater than 0.05?
Y=0.45=45/100=9/20 as a common fraction
PREMISES
Y=How is 0.5 greater than 0.05?
CALCULATIONS
The difference between 0.5 and 0.05 can be represented by the larger number minus the smaller number:
Y=0.5-0.05
Converting 0.5 to 0.50 to accommodate the difference in place values,
Y=0.50-0.05
Evaluating,
Y=0.45
Y=
0.45=45/100=9/20 as a common fraction
PROOF
If Y=0.45, the inverse of the difference between Y=0.50-0.05 shows
Y+0.05=0.50
0.45+0.05=0.50 and
0.50=0.50 verifies the difference Y=0.45 of the expression 0.50-0.05
C.H.
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8306 | https://www.youtube.com/watch?v=NGQGCqDhyv0 | Resultant of a Concurrent Force System in space: Example 1
Dr Joji Thomas
6810 subscribers
32 likes
Description
2274 views
Posted: 4 Jun 2021
How to determine the resultant of a concurrent force system acting in space is explained by taking an example from the book Vector Mechanics for Engineers: Statics and Dynamics by Beer and Johnston.
1 comments
Transcript:
we have already learned how to represent a force in space by vector now in this video i will explain how to determine resultant of a concurrent force system in space for that i have taken one example from the book of aesthetics and dynamics by biaran johnson now question states that knowing that the tension is 425 newton in cable a b and 510 newton in cable ac this is cable a b and this is cable ac and tension in these two cables are given as 425 newton and 510 newton we have to determine the magnitude and direction of the resultant of the forces exerted at a by the two cables forces are acting at this point a we have to determine resultant of these two forces at a in first step we have to determine coordinates of these points a b and c so coordinate of a will be minus 40 in negative direction negative x direction minus 40 positive y direction 45 this point is 45 in positive y direction and in that direction no distance so 0 similarly for b b is in that coordinate there therefore x and y will be 0 and only 60 that coordinate is 60 this distance is given 60 for c this distance is 60 that is in x direction its distance is 60 and in that direction also 60 and y direction there is no distance therefore its coordinate is zero so once you got coordinate of these three points now we know that tension in cable always acts in the direction of cable and away from the point now we are considering this point a therefore forces will act away from this a that is in this direction towards b so tension will act in the direction a to b and similarly tension and this this cable will act towards a to c like this direction so for that we need vector in this direction vector a b so this can be calculated as a b so coordinates of b minus coordinates of a so this is 0 minus minus 40 0 minus 45 and 60 minus 0 0 minus minus 40 0 minus 45 and 60 minus 0 if you simplify this you will get 40 i minus 45 j plus 60 k so this is the vector a b from a to b now magnitude of a b you can determine magnitude of a b is root over 40 square 45 square and 60 square this is coming out 85 once you divide this a b with this magnitude you will get unit vector in the direction of a b so in the direction of a b you will get you got this unit vector once you get this unit vector now you can present this tension 425 newton tension in this direction towards a b so just multiply this magnitude with this unit vector so assume this force as b so i am assuming this force as b so magnitude is given 425 newton so force p p vector will equal to 425 into this unit vector in this direction in the direction of a b if you simplify you will get p vector as 200i minus 225 j and 300 plus 300 k the same manner you can determine tension in this ac that first you have to determine vector ac so vector ac is equal to units of c minus coordinates of a so in this way you will get this ac so if you simplify you will get 100 i minus 45 j plus 60 k again next step is to determine magnitude of ac so magnitude of ac is 100 root over 100 square 45 square and 60 square sum of these three and you will get 125 so once you divide this ac by 125 you will get unit vector along ac so this is unit vector along ec divided by 125 now assume this force ac that is tension in ac as q so magnitude of q is given 510 newton this 510 newton multiply this unit vector with 510 you will get this force q so force q is equal to 510 divided by 125 and this term so when you calculate you will get 408 i minus 183.6 j and 244.8 k now we have to determine resultant of p and q so resultant r is equal to p plus q so this can be determined directly by adding the components so this is x component of p this is x component of q so summation of this x component is equal to x component of r similarly this is x y component of p y component of q that component of p that component of q and this is the y component of r and this is that component of r so in this way adding all x components of all the forces that is sigma fx is equal to rx sigma fy is equal to ry and sigma fz is equal to r then in this way directly by adding simply adding you can determine this resultant now magnitude of resultant you can calculate as root over 608 square plus 408.6 square plus 544.8 square and it is coming 930 newton now you can get a unit vector along the r so you need to vector along the r what you have to do you have to divide this r by 930 newton now we know that in unit vector of r coefficient of i j k r cos alpha cos beta and cos gamma where alpha beta and gamma are the angles made by the resultant with x y and z axis respectively so in when you divide this r by 930 newton so it will come 08 upon 913 which is equal to cos alpha similarly minus 408.6 divided by 913 is cos beta and 544.8 divided by 913 is cos gamma from these three relations you can get value of alpha beta and gamma as 48.25 degrees beta is 116.6 degrees and gamma is 53.4 degrees so in this way you can determine r in vector form magnitude of r as 930 newton these are the x component y component and that component of r and these are the angles of resultant for your homework uh the same problem slightly changed only a change is knowing that the tension is 510 newton in cable eb and 425 newton in cable ac you have to determine magnitude and direction of the resultant of the forces exerted at a by the two cables i have changed the tensions in a b and e c this is a b and this is ac in a b now tension is 510 newton and in ac tension is 425 newton now you have to determine resultant in magnitude and direction both so in the same manner i explained now you can determine this magnitude and direction of r so you can try this problem and once you try you will get solution as magnitude of r again you will get same magnitude 930 newton and direction slightly it will change alpha you will get 15 50.6 degree beta 117.6 degree and gamma as 15.8 degrees |
8307 | https://liberalarts.oregonstate.edu/wlf/what-difference-between-mood-and-tone-definitions-and-examples | Skip to main content
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What is the Difference between Mood and Tone? || Definitions and Examples
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View the full series: The Oregon State Guide to English Literary Terms
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Guide to Literary Terms
What is the Difference between Mood and Tone? Transcript (English and Spanish Subtitles Available in Video, Click HERE for Spanish Transcript)
By Lucia Stone and Marcos Norris, Oregon State University Instructors of English Literature
14 November 2022
Mood and Tone: What’s the Difference?
Two ways in which authors communicate with readers is by the use of mood and tone. Although both techniques can elicit particular emotions central to understanding a story, the terms are easily confused.
Mood in literature is firmly rooted in the locale or setting of the story that reveals the subject. The physical atmosphere is built scene by scene to create a sense of time, place and reality. Is the world depicted familiar to the reader in its contemporary realism or is it fantastic and reminiscent of the distant past? How does everything look, smell and feel? And, most importantly, what does each scene reveal about the subject at hand? These are some of the questions we can ask to delve deeper into the mood emphasized in each sequence of an unfolding story.
Tone, on the other hand, is less sensual play and more the attitude of the characters toward the subject at hand. It is strongly related to the narrator’s point of view, delivered most reliably through choice of words, either explicitly or implicitly. Tone certainly contributes to the mood of a story, but it is less about creating emotional resonance within the readers and more about communicating the narrator’s thoughts or state of mind.
Image
Here is another way of understanding the difference between mood and tone: mood shows the subject of the story while tone tells the reader what the characters think of that subject.
To illustrate, let’s look at two examples from literature from different eras that share similar themes, Dracula by Bram Stoker and Interview with a Vampire by Anne Rice. Vampire literature is a genre in which mood and tone are almost as important as plot and story, so the characters in each novel become conduits for communicating a unique other-worldly atmosphere that can only exist through their perceptions.
Image
Dracula is an epistolary novel in which the narration is delivered through a series of journal entries. The mood is set as the scene unfolds with the protagonist Jonathan Harker’s travel from London, England to the Carpathian mountains in Transylvania during the late nineteenth century. The mood first affected is one of disorientation, with the physical contrast drawn sharply between Western and Eastern Europe by the reference to the literal bridges over the Danube leading eastward. Later, the contrast is accentuated as we follow the narrator further into this unfamiliar realm by reading about his first meal, “paprika hendl,” a dish drawn to be distinct, presumably, from food familiar to the English palette at that time. As in the opening, the mood continues to be shown with sight and taste, with the imagery directed toward an unfamiliar scene–and the subject of the novel.
Image
To foreshadow the horror to come, the mood is punctuated with the narrator’s attitude about that subject. The tone is one of apprehension and fear as the narrator explicitly tells us about his first night sleeping in a foreign hotel: “I did not sleep well…There was a dog howling all night under my window” and “ I had to drink up all the water” by the bedside, but “was still very thirsty” from the strong, unfamiliar seasoning in the food served the night before. In this case, the protagonist’s tone matches the mood. However, sometimes tone and mood are at odds with one another.
Interview with a Vampire begins quite literally with the viewpoint of the protagonist, the vampire himself, who languidly opens the novel with, “I see…,” while preparing himself for an interview with a young journalist. His attitude, or tone, is one of quiet ease. His tone matches the mood, which is set by a rather unexotic backdrop of a cityscape through the window of an ordinary hotel room. The dialogue bounces between vampire and journalist, monster and human, while the mood of prosaic reality is revealed in the simple details of a chair, table and recording device. The tonal horror necessary for the tension to unfold, then, is projected by the very different attitude of the journalist toward the scene: the readers are told that the interviewer “shuddered” and “recoiled” with “cold sweat running down the side of his face” as he watched the vampire before turning on the recorder to begin the interview. In other words, reason meets emotion in this clash between tone and mood. A sharp contrast is drawn between the attitude of the two protagonists toward the scene, and the audience is sucked right in.
Image
For the student of literature, such moments of tension are exciting and revelatory. But it cannot happen without the skillful manipulation of tone and mood using techniques such as word choice, point of view and dialogue to create that perfect alternate reality. Mood shows the particular scenes that direct us toward the subject of a story, but tone tells what each character actually thinks of that subject. Both are necessary devices to make a world come alive on the page or on the screen.
Want to cite this?
MLA Citation: Stone, Lucia and Marcos Norris. "What is the Difference between Mood and Tone?" Oregon State Guide to English Literary Terms, 14 Nov. 2022, Oregon State University, Accessed [insert date].
Further Resources for Teachers
Check out the following classroom lesson on mood and tone in Shirley Jackson's "The Lottery," which was written by our talented graduate students!
File
literary_terms_exercise_what_is_the_difference_between_mood_and_tone_.pptx
(457.03 KB)
File
mood_and_tone_activity.docx
(217.29 KB)
Interested in more video lessons? View the full series:
The Oregon State Guide to English Literary Terms |
8308 | https://www.youtube.com/watch?v=yFPfO_eHJdY | Adding vectors in magnitude and direction form | Vectors | Precalculus | Khan Academy
Khan Academy
9090000 subscribers
405 likes
Description
64910 views
Posted: 29 Apr 2021
In this example, Sal takes two vectors given by magnitude and direction, and finds the magnitude and direction of their sum.
28 comments
Transcript:
we're told that vector a has magnitude 4 in direction 170 degrees from the positive x-axis vector b has magnitude 3 in direction 240 degrees from the positive x-axis find the magnitude and direction of vector a plus vector b so pause this video and see if you can have a go at that all right now let's work through this together and the way that i'm going to approach it i'm going to represent each vector in component form and then i'm going to add the corresponding components and from that i'll try to figure out the magnitude and the direction of the sum so vector a what is its x component well the change in x here there's multiple ways that you could try to do this using trigonometry but we've reviewed this or gone over this in other videos the simplest way to think about it is our change in x here is going to be the length and we know vector a has magnitude 4 times the cosine of the angle that the vector makes with the positive x axis cosine of 170 degrees and so that's our x component right over here 4 times cosine of 170 degrees and then what's our y component well our y component is going to be this change in y here and as we've reviewed in other videos that's going to be the length times the sine of the angle we make with the positive x-axis sine of 170 degrees and we can maybe use a calculator in a bit to get approximations for these values but then we can do the exact same thing for vector b vector b here is going to be by the same logic its x component is going to be the length of the vector and it would be 3 they tell us that so it's going to be 3 times the cosine of this angle 240 degrees and then the y component is going to be the length of our vector 3 times the sine of 240 degrees now when we want to take the sum of the two vectors let me write here vector a plus vector b i can just add the corresponding components this is going to be equal to 4 cosine of 170 degrees plus 3 cosine of 240 degrees and then the y component is going to be 4 sine of 170 degrees plus 3 sine of 240 degrees and so let me get my calculator out to evaluate these we say 170 degrees we take the cosine times 4 that equals this and then we're going to add to that i'll open parentheses we'll take the cosine of 240 240 cosine times 3 close parentheses is equal to this negative approximately negative 5.44 so this is approximately negative 5.44 and then if we were to take 170 degrees take the sign of it multiply it by 4 and then to that i'm going to open parentheses i'm going to take 240 degrees take the sign multiply that times three close my parentheses that is going to be equal to approximately negative 1.90 so this is approximately negative 1.90 and this is consistent with our intuition if the sum has both negative components that means it's going to be in the third quadrant and if i were to do the head to tail method of adding vectors if i were to take vector b and i were to put it right over here we see that the resulting vector the sum will sit in the third quadrant it makes sense that our x and y components would indeed be negative now the question didn't ask just to find the components of the sum it asked to find the magnitude and the direction of the resulting sum and so to do that we just have to use a little bit more of our trigonometry and actually a little bit of our geometry for example our change in x is this value right over here as we go from the tail to the tip it's negative 5.44 if we were just thinking in terms of length right over here the absolute value this side would have length 5.44 and then similarly our change in y it's negative we're going down in y but if we were to just think in terms of a triangle the length on this side of a triangle is 1.90 and we can see from the pythagorean theorem that the length of our hypotenuse which is the same thing as the magnitude of this vector squared is going to be equal to the sum of the squares of these two sides or another way of thinking about it is the length of this vector the magnitude of this vector which we can write as the magnitude of vector a plus vector b is going to be equal to or i should say approximately equal to since we're already approximating these values the principal root of 5.44 squared and that's because i'm just thinking about the absolute length of the side i could also think about a change in x but if i had a negative 5.44 and i square that that would still become positive and then i'll have plus 1.90 squared and i can get our calculator out for that this is going to be approximately equal to 5.44 squared plus 1.9 squared is equal to that take the square root of that it's approximately equal to 5.76 5.76 which is going to be our magnitude and then to figure out the direction so we essentially want to figure out this angle right over here you might recognize that the tangent of this angle theta right over here should be equal to and i'll do approximately equal to since we're using these approximations our change in y over our change in x so negative 1.90 over negative 5.44 or we could say that theta is going to be approximately equal to the inverse tangent of negative 1.90 over negative 5.44 and we're going to see in a second whether this is actually going to get us the answer that we want so let's try this out if we were to take 1.9 negative divided by 5.44 negative that gets us that which makes sense negative divided by negative is a positive and now let's try to take the inverse tangent of that so here i press second and then i'll do inverse tangent so i'm getting 19.2 degrees approximately so this is saying that this is approximately 19.25 degrees and my question to you is does that seem right well 19.25 degrees would put us in the first quadrant it would give us a vector that looks something like this this would be 19.25 degrees but clearly that's not the vector we're talking about we're talking about a vector in the third quadrant and the reason why we got this result is that when you take the inverse tangent on most calculators it's going to give you an angle that's between negative 90 degrees and positive 90 degrees well here we are at an angle that puts us out in the third quadrant so we have to adjust and to adjust here we just have to add 180 degrees to get to the actual angle that we are talking about so in our situation the magnitude here is going to be approximately 5.76 and then the direction is going to be approximately 19.25 plus 180 degrees which is going to be 199.25 degrees and now we are done |
8309 | https://www.uptodate.com/contents/overview-of-nail-disorders/abstract/158 | Medline ® Abstract for Reference 158 of 'Overview of nail disorders' - UpToDate
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Medline ® Abstract for Reference 158 of 'Overview of nail disorders'
158PubMed|TI Clubbing and hypertrophic osteoarthropathy: insights in diagnosis, pathophysiology, and clinical significance.AU Callemeyn J, Van Haecke P, Peetermans WE, Blockmans D SO Acta Clin Belg. 2016;71(3):123. Epub 2016 Apr 22.
BACKGROUND Digital clubbing and hypertrophic osteoarthropathy (HOA) form a diagnostic challenge. Subtle presentations of clubbing are often missed. The underlying pathophysiology remains unclear. Establishing a differential diagnosis based on nonspecific signs can be cumbersome. Finally, the prognostic value of clubbing and HOA remains unclear.
OBJECTIVE This article reviews clinical criteria and pathophysiology of clubbing and HOA. A diagnostic algorithm is proposed, based on etiology and current insights. The prognostic impact on associated diseases is discussed.
METHODS The Internet databases Medline and Embase were searched. Articles were selected based on relevance of abstract, article type and impact of the journal.
RESULTS Diagnostic criteria include Lovibond's profile sign, distal/interphalangeal depth ratio and Schamroth's sign. Three pathophysiological causes of clubbing can be distinguished: hypoxia, chronic inflammation and aberrant vascularization. A prominent role for vascular endothelial growth factor is suggested. Associated symptoms and clinical signs should guide the initial diagnostic evaluation. Finally, clubbing is a negative prognostic factor in certain pulmonary disorders, including cystic fibrosis.
AD PMID27104368
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8310 | https://teachy.ai/en/lesson-plan/high-school/10th-grade/mathematics-en/expository-methodology-or-radicals-or-lesson-plan | Log In
Lesson plan of Radication
Lara from Teachy
SubjectMathematics
Mathematics
SourceOriginal Teachy
Original Teachy
TopicRadication
Radication
Objectives (5 - 7 minutes)
Understand the concept of radicals, highlighting the relationship between radicand, radical, and index.
Identify the radicand as the number or expression under the radical sign.
Recognize the index as the number indicating which root is being extracted.
Understand that the radical is the sign indicating the root extraction operation.
Learn to simplify radical expressions, reducing the index and radicand to their simplest form.
Apply the power property to simplify the radicand.
Use factorization to simplify the radicand.
Practice solving problems involving radicals, applying the rules of operations with radicals.
Perform addition, subtraction, multiplication, and division operations with radicals.
Solve equations and inequalities involving radicals.
Secondary Objectives:
Develop logical and critical thinking skills when solving problems involving radicals.
Reinforce understanding of previous mathematical concepts, such as exponentiation and operations with polynomials.
Promote students' self-confidence and autonomy in solving complex mathematical problems.
Introduction (10 - 12 minutes)
Review of Previous Concepts:
The teacher begins the lesson with a brief review of exponentiation concepts, emphasizing the meaning of a root and the difference between radicand and index. This review is essential for students to understand the new concept that will be presented: radicalization. (3 - 4 minutes)
Problem Situations:
The teacher proposes two problem situations involving the concept of radicalization to stimulate students' reasoning:
Situation 1: 'Imagine you have a square area of 16m². How long is the side of this square?' (2 - 3 minutes)
Situation 2: 'If the volume of a cube is 27m³, how long is the side of this cube?' (2 - 3 minutes)
Contextualization:
The teacher contextualizes the importance of studying radicalization, explaining that it is essential for solving many practical problems, such as those presented in the problem situations. Additionally, it emphasizes that radicalization is widely used in various areas of science and engineering, such as physics and computing. (1 - 2 minutes)
Topic Introduction:
The teacher introduces the topic of the lesson, radicalization, explaining that it is an operation inverse to exponentiation, meaning that while exponentiation multiplies a number by itself several times, radicalization extracts the root of a number. (1 - 2 minutes)
Curiosities and Applications:
The teacher shares with students some curiosities and applications of radicalization to spark their interest:
Curiosity 1: 'Did you know that the radical symbol (√) was introduced by the German mathematician Christoph Rudolff in the 16th century? Before him, the letter R was used for the root.'
Application 1: 'Radicalization is widely used in physics. For example, the speed of light in a vacuum, which is approximately 299,792,458 meters per second, can be represented as √(8,987 x 10^16).' (1 - 2 minutes)
Development (20 - 25 minutes)
Radicalization Theory (5 - 7 minutes)
The teacher begins the theoretical explanation of radicalization by reinforcing the idea that radicalization is the inverse operation of exponentiation.
Next, it explains that radicalization is composed of three elements: the radical (√), the index, and the radicand.
The teacher clarifies that the radical (√) indicates that we are performing the radical operation, the index indicates which root we are taking, and the radicand is the number or expression under the radical sign.
Using practical examples, the teacher demonstrates how to identify each of these elements in a radical expression.
Simplification of Radical Expressions (5 - 7 minutes)
The teacher explains that one of the main objectives of radicalization is to simplify radical expressions, that is, to reduce the index and radicand to their simplest form.
The teacher demonstrates how to do this, using the power property and factorization.
First, the teacher explains that if the index is an even power, we can simplify the radicand by dividing the index by the power.
Next, the teacher shows how factorization can be used to simplify the radicand.
The teacher uses step-by-step examples to illustrate each of these processes.
Operations with Radicals (5 - 7 minutes)
After explaining the simplification of radical expressions, the teacher moves on to operations with radicals.
The teacher explains that we can only add or subtract radicals if they have the same index and the same radicand.
The teacher demonstrates how to do this, using step-by-step examples.
Next, the teacher explains that we can multiply and divide radicals normally, as if they were numbers.
The teacher uses examples to illustrate each of these processes.
Problem Solving with Radicals (5 - 7 minutes)
Finally, the teacher moves on to solving problems with radicals.
The teacher explains that to solve a problem with radicals, we must follow the order of operations, simplify the radical expressions, and, if necessary, apply the rules of operations with radicals.
The teacher demonstrates this process, using step-by-step examples.
To reinforce learning, the teacher proposes some problems for students to solve in pairs.
Review and Clarification of Doubts (3 - 5 minutes)
The teacher concludes the lesson's Development by reviewing the main points covered and clarifying any doubts students may have.
This Development of the lesson allows students to understand the concept of radicalization, learn to simplify radical expressions, perform operations with radicals, and solve problems involving radicals. Additionally, the use of practical examples and the opportunity to solve problems in pairs help make learning more meaningful and reinforce the concepts learned.
Return (8 - 10 minutes)
Concept Review (3 - 4 minutes):
The teacher begins this stage by asking students to share their perceptions of the lesson, highlighting the points they liked the most or found challenging. This allows the teacher to assess the impact of the lesson and make adjustments for future classes.
Next, the teacher briefly reviews the key concepts covered in the lesson, reinforcing the idea that radicalization is the inverse operation of exponentiation, and that it is composed of three elements: the radical (√), the index, and the radicand. The teacher also reviews the main rules for simplification and operations with radicals.
The teacher encourages students to ask questions and clarify any doubts they may have.
Connection to Practice (2 - 3 minutes):
The teacher suggests that students reflect on how they can apply what they learned in the lesson to everyday situations or other subjects. For example:
Situation 1: 'Can you think of a situation in everyday life where you would need to simplify a radical expression? How would you solve this situation now, after learning about radicalization?'
Situation 2: 'Can you think of a situation where radicalization could be useful in other subjects, such as physics, chemistry, or biology?'
The teacher encourages students to share their reflections, promoting a classroom discussion.
Individual Reflection (2 - 3 minutes):
The teacher suggests that students engage in individual reflection on what they learned in the lesson. To do this, the teacher can ask the following questions:
'What was the most important concept you learned today?'
'What questions have not been answered yet?'
The teacher gives students a minute to think about these questions. Then, asks some students to share their answers with the class.
Feedback and Closure (1 minute):
The teacher thanks the students for their participation and asks them to provide feedback on the lesson, highlighting what they liked the most and what they think can be improved.
The teacher concludes the lesson by reinforcing the importance of studying radicalization and encouraging students to continue practicing and studying the subject at home.
This final Return allows students to consolidate what they learned in the lesson, reflect on the application of concepts in real situations, and clarify any doubts they may have. Additionally, students' feedback helps the teacher improve their lessons and adapt their teaching methods to the class's needs.
Conclusion (5 - 7 minutes)
Summary of Contents (2 - 3 minutes):
The teacher begins the Conclusion of the lesson by summarizing the covered contents. Recaps the concept of radicalization, highlighting the relationship between radicand, radical, and index. Reminds the importance of simplifying radical expressions and the rules for operations with radicals.
The teacher reinforces that radicalization is a powerful tool for solving complex mathematical problems, and that mastering this skill is essential for success in many areas of science and engineering.
Connection between Theory, Practice, and Applications (1 - 2 minutes):
The teacher emphasizes how the lesson connected theory, practice, and applications.
Explains that by learning the theory of radicalization, students were able to understand the logic behind operations with radicals.
By practicing the simplification of radical expressions and solving problems with radicals, students were able to apply this theory in a practical way.
By discussing the applications of radicalization, students were able to understand how these concepts are relevant to everyday life and other subjects.
Extra Materials (1 minute):
The teacher suggests some extra materials for students who want to deepen their knowledge of radicalization.
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8311 | https://www.britannica.com/science/gas-state-of-matter/Effusion | Effusion
Kinetic theory of gases
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Consider the system described above in the calculation of gas pressure, but with the area A in the container wall replaced with a small hole. The number of molecules that escape through the hole in time t is equal to (1/2)(N/V)vz(At). In this case, collisions between molecules are significant, and the result holds only for tiny holes in very thin walls (as compared to the mean free path), so that a molecule that approaches near the hole will get through without colliding with another molecule and being deflected away. The relationship between vz and the average speed v̄ is rather straightforward: vz = (1/2)v̄.
If the rates for two different gases effusing through the same hole are compared, starting with the same gas density each time, it is found that much more light gas escapes than heavy gas and that more gas escapes at a high temperature than at a low temperature, other things being equal. In particular,
The last step follows from the energy formula, (1/2)mv2 = (3/2)kT, where (v2)1/2 is approximated to be v, even though v2 and (v̄)2 actually differ by a numerical factor near unity (namely, 3π/8). This result was discovered experimentally in 1846 by Graham for the case of constant temperature and is known as Graham’s law of effusion. It can be used to measure molecular weights, to measure the vapour pressure of a material with a low vapour pressure, or to calculate the rate of evaporation of molecules from a liquid or solid surface.
Thermal transpiration
Suppose that two containers of the same gas but at different temperatures are connected by a tiny hole and that the gas is brought to a steady state. If the hole is small enough and the gas density is low enough that only effusion occurs, the equilibrium pressure will be greater on the high-temperature side. But, if the initial pressures on both sides are equal, gas will flow from the low-temperature side to the high-temperature side to cause the high-temperature pressure to increase. The latter situation is called thermal transpiration, and the steady-state result is called the thermomolecular pressure difference. These results follow simply from the effusion formula if the ideal gas law is used to replace N/V with p/T;
When a steady state is reached, the effusion rates are equal, and thus
This phenomenon was first investigated experimentally by Osborne Reynolds in 1879 in Manchester, Eng. Errors can result if a gas pressure is measured in a vessel at very low or very high temperature by connecting it via a fine tube to a manometer at room temperature. A continuous circulation of gas can be produced by connecting the two containers with another tube whose diameter is large compared with the mean free path. The pressure difference drives gas through this tube by viscous flow. A heat engine based on this circulating flow unfortunately has a low efficiency.
Viscosity
The kinetic-theory explanation of viscosity can be simplified by examining it in qualitative terms. Viscosity is caused by the transfer of momentum between two planes sliding parallel to one another but at different rates, and this momentum is transferred by molecules moving between the planes. Molecules from the faster plane move to the slower plane and tend to speed it up, while molecules from the slower plane travel to the faster plane and tend to slow it down. This is the mechanism by which one plane experiences the drag of the other. A simple analogy is two mail trains passing each other, with workers throwing mailbags between the trains. Every time a mailbag from the fast-moving train lands on the slow one, it imparts its momentum to the slow train, speeding it up a little; likewise each mailbag from the slow train that lands on the fast one slows it down a bit.
If the trains are too far apart, the mailbags cannot be passed between them. Similarly, the planes of a gas must be only about a mean free path apart in order for molecules to pass between them without being deflected by collisions. If one uses this approach, a simple calculation can be carried out, much as in the case of the gas pressure, with the result thatwhere a is a numerical constant of order unity, the term (N/V)v̄l is a measure of the number of molecules contained in a small counting cylinder, and the mass m is a measure of the momentum carried between the sliding planes. The cross-sectional area of the counting cylinder and the relative speed of the sliding planes do not appear in the equation because they cancel one another when the drag force is divided by the area and speed of the planes in order to find η.
It can now be seen why η is independent of gas density or pressure. The term (N/V) in equation (23) is the number of carriers of momentum, but l measures the number of collisions that interfere with these carriers and is inversely proportional to (N/V). The two effects exactly cancel each other. Viscosity increases with temperature because the average velocity v̄ does; that is, momentum is carried more quickly when the molecules move faster. Although v̄ increases as T1/2, η increases somewhat faster because the mean free path also increases with temperature, since it is harder to deflect a fast molecule than a slow one. This feature depends explicitly on the forces between the molecules and is difficult to calculate accurately, as is the value of the constant a, which turns out to be close to 1/2.
The behaviour of the viscosity of a mixture can also be explained by the foregoing calculation. In a mixture of a light gas and a viscous heavy gas, both types of molecules have the same average energy; however, most of the momentum is carried by the heavy molecules, which are therefore the main contributors to the viscosity. The light molecules are rather ineffective in deflecting the heavy molecules, so that the latter continue to carry virtually as much momentum as they would in the absence of light molecules. The addition of a light gas to a heavy gas therefore does not reduce the viscosity substantially and may in fact increase it because of the small extra momentum carried by the light molecules. The viscosity will eventually decrease when there are only a few heavy molecules remaining in a large sea of light molecules.
The main dependence of η on the molecular mass is through the product v̄m in equation (23), which varies as m1/2 since v̄ varies as 1/m1/2. Owing to this effect, heavy gases tend to be more viscous than light gases, but this tendency is compensated for to some degree by the behaviour of l, which tends to be smaller for heavy molecules because they are usually larger than light molecules and therefore more likely to collide. The often confusing connection between viscosity and molecular weight can thus be accounted for by equation (23).
Finally, in a free-molecule gas there are no collisions with other molecules to impede the transport of momentum, and the viscosity thus increases linearly with pressure or density until the number of collisions becomes great enough so that the viscosity assumes the constant value given by equation (23). The nonideal behaviour of the gas that accompanies further increases in density eventually leads to an increase in viscosity, and the viscosity of an extremely dense gas becomes much like that of a liquid. |
8312 | https://www.academia.edu/19248831/A_Simple_Electrostatic_Model_for_Trisilylamine_Theoretical_Examinations_of_the_n_%CF%83_Negative_Hyperconjugation_p_%CF%80_d_%CF%80_Bonding_and_Stereoelectronic_Interaction | Academia.edu no longer supports Internet Explorer.
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A Simple Electrostatic Model for Trisilylamine: Theoretical Examinations of the n→σ Negative Hyperconjugation, p π →d π Bonding, and Stereoelectronic Interaction
Yirong Mo
1999, Journal of the American Chemical Society
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Abstract
A block-localized wave function method was used to examine the stereoelectronic effects on the origin of the structural difference between trisilylamine and trimethylamine. The pyramidal geometry of trimethylamine along with its high basicity is consistent with the traditional VSEPR (valence shell electronpair repulsion) model for σ bonding. On the other hand, in trisilylamine, the silicon d orbitals make modest contribution to the electronic delocalization, although the key factor in charge delocalization is still n N fσ SiH negative hyperconjugation. Interestingly, the gain in p π fd π bonding stabilization is offset by a weaker negative hyperconjugation effect in trisilylamine, resulting in an overall smaller delocalization energy (-18.5 kcal/ mol) than that in trimethylamine (-23.9 kcal/mol), which contains little p π fd π bonding character. Significantly, because of the relatively low electronegativity of silicon, the N-Si bond is much more polar than the N-C bond. Weinhold's natural population analyses of the BLW and HF wave functions for these compounds reveal that the origin of the planar geometry of trisilylamine is due to the polar σ-effect that yields significant longrange electrostatic repulsion between the silyl groups. In addition, it was found that only the most electronegative substituents such as F and OH can result in a pyramidal geometry at the nitrogen center for silylamines. This is in good accord with the recent X-ray structure of a pyramidal silylamine, N(CH 3 )(OCH 3 )(SiH 3 ).
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Molecular structure of tris(cyclopropylsilyl)amine as determined by gas electron diffraction and quantum-chemical calculations
Marwan Dakkouri
Journal of Molecular Structure, 2008
The molecular structure and conformation of tris(cyclopropylsilyl)amine (TCPSA) has been studied by means of gas-phase electron diffraction at 338 K and quantum-chemical calculations. A total of 12 relatively stable conformations of TCPSA molecule were considered. According to the experimental results and the DFT calculations the most stable conformer corresponds to a configuration (according to the Prelog-Klyne notation) of the type (Àac)(Àac)(+ac)-(Àac)(Àac)(+ac), where the first three parentheses describe the three different SiN -Si-C torsional angles and the latter ones depict the rotation of the three cyclopropyl groups about the C ring-Si axes, respectively. The quantum-mechanical calculations were performed using various density functional (B3LYP, X3LYP and O3LYP) and perturbation MP2 methods in combination with double-and triple-f basis sets plus polarization and diffuse functions. The most important experimental geometrical parameters of TCPSA (r a Å , \ h1 degrees) are: (Si-N) av = 1.741(3), (Si-C) av = 1.866(4), (C-C) av = 1.510(3), (C-C(Si)) av = 1.535(3), (N-Si-C) av = 115.1(18)°. For the purpose of comparison and searching for reasons leading to the planarity of the Si 3 N moiety in trisilylated amines we carried out NBO analysis and optimized the geometries of numerous silylamines. Among these compounds was tris(allylsilyl)amine (TASA), which is isovalent and isoelectronic to TCPSA. Utilizing the structural results we obtained we could show that Si + Á Á ÁSi + electrostatic repulsive interaction is predominantly responsible for the planarity of the Si 3 N skeleton in TCPSA and in all other trisilylamines we considered. We also found that regardless the size and partial charges of the substituents the SiN -Si bond angle in various disilylamines amounts to 130 ± 2°.
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Silicon Surfaces as Electron Acceptors: Dative Bonding of Amines with Si(001) and Si(111) Surfaces
Robert Hamers
Journal of the American Chemical Society, 2001
The bonding of the trimethylamine (TMA) and dimethylamine (DMA) with crystalline silicon surfaces has been investigated using X-ray photoelectron spectroscopy (XPS), Fourier transform infrared spectroscopy, and density-functional computational methods. XPS spectra show that TMA forms stable dative-bonded adducts on both Si(001) and Si(111) surfaces that are characterized by very high N(1s) binding energies of 402.2 eV on Si(001) and 402.4 eV on Si(111). The highly ionic nature of these adducts is further evidenced by comparison with other charge-transfer complexes and through computational chemistry studies. The ability to form these highly ionic charge-transfer complexes between TMA and silicon surfaces stems from the ability to delocalize the donated electron density between different types of chemically distinct atoms within the surface unit cells. Corresponding studies of DMA on Si(001) show only dissociative adsorption via cleavage of the N-H bond. These results show that the unique geometric structures present on silicon surfaces permit silicon atoms to act as excellent electron acceptors.
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Nature of the silicon-nitrogen bond in silatranes
Martin Guidry
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Both ab initio and semiempirical calculations are used to investigate the structure and bonding in substituted silatranes. In agreement with gas-phase experiments, which fmd SiN distances that are 0.25-0.35 A longer than those in the crystal phase, the SiN distance (for example, in methylsilatrane) is predicted to be much larger than that observed in the crystal. Nonetheless, a bond critical point is found between Si and N, suggesting the existence of an SiN bond in these compounds. It is found that the energy cost for constraining the SiN distance in hydroxysilatrane to a value similar to that observed in the crystal is less than 6 kcaljmol. This suggests that crystal forces may be responsible for the much shorter SiN distance in the solid. 2 ~N-CH 2 in which the central silicon is bonded to a nitrogen, both
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Modeling the configuration about the nitrogen atom in methyl- and silyl-substituted amines
Maureen Julian
The Journal of Physical Chemistry, 1988
Bond lengths, angles, energies, and electron density maps were calculated for a substituted series of four amines, N(CH3)"(SiH3)>,, (n = 0, 1, 2, 3). When n = 0, 1, or 3, the amine crystallizes at about 115 K as a monomer. However, when n = 2, while the molecules in the gas phase are monomeric, at room temperature, the molecules in the crystal phase are pentamers. In the monomeric form for each amine in the series, the average calculated values of LSiNSi, LSiNC, and LCNC tend to be preserved at 120.0°, 117.8', and 112.2O, respectively. The nonbonded cation-cation distances also tend to be preserved. Reaction energies for six double-replacement reactions within the series indicate that the pentamer-forming compound occurs on the less stable side of the calculated reactions. The electronegativity of the nitrogen increases as methyl is substituted for silyl. The lengthening of the bonds associated with the change in electronegativity is ascribed to the relatively large buildup of electron density in the neighborhood of the nitrogen in N(SiH3)3. A unique angle, 0, was defined in terms of line LN making the same angle with each bond such that, for a given amine NRR'R", 0 = LLNR = LLNR' = LLNR". From the deformation maps, each lone pair lies along LN. Finally, the maps show that the bonding electron density peak is located interior to the bond for all nonplanar configurations. The molecule that forms the pentamer shows the largest displacement of the peak off the SiN bond, perhaps representing a pathway for nucleophilic attack.
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Synthesis and structures of tris(pentafluorophenyl)silylamines
Alexander Dilman
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Tris(pentafluorophenyl)silylamines were synthesized by silylation of amines and imines with (C 6 F 5 ) 3 SiCl or (C 6 F 5 ) 3 SiOTf in the presence of triethylamine. The crystal structures of the (C 6 F 5 ) 3 SiN(H)CH 2 Ph and (C 6 F 5 ) 3 SiN(CH=CMe 2 )CH 2 Ph compounds were studied by X ray diffraction. The crystal packings were analyzed by quantum chemical calculations in terms of the density functional theory (PBE exchange correlation functional).
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Valence Bond Study of the SiH 3 −F Bond
Shmaryahu Hoz
The Journal of Physical Chemistry A, 1997
The binding energy curves of SiH 3-F have been investigated using ab initio valence bond self-consistentfield (VBSCF) methods. The atomic core electrons are treated both all-electron and by using an effective core potential (ECP) representation; for comparison and testing purposes. The VB wave function is expressed in terms of the covalent (SiH 3 :F) and ionic (SiH 3 + F-,SiH 3-F +) configurations, and the nonorthogonal orbitals are expanded in conventional atom-centered Gaussian basis sets. Several theory levels are applied, up to the use of different orbitals for different VB structures and allowing delocalization mixing among the passive SiH 3 and F fragment orbitals. Replacing the core electrons with an ECP is found to generally have a relatively small effect on the calculated ground state bond dissociation energy (BDE) curve, but a much larger effect on the individual covalent and ionic structure energy curves. Delocalization mixing is found to be important to achieving high accuracy for the equilibrium bond distance (R e), BDE (D e), and dipole moment of SiH 3 F. The SiH 3 + Fionic structure curve is found to lie below the covalent energy curve from at least R(C-F)) 1.3 Å out to ∼2.5 Å, but is stable relative to the dissociation asymptote by less than half of the ground state D e. The magnitude of D e in SiH 3-F is, therefore, determined by resonance coupling between the covalent and ionic structures (H 12), where the dominant VB structure at R e is SiH 3 + F-. The SiH 3 :F covalent curve is found to be nearly as repulsive at its R e value (1.59 Å) as previously found for CH 3 :F at its R e (1.38 Å). The proportionality constant K in the equation H 12) KS 12 [H 11 + H 22 ]/2, where H ij and S ij (i, j) 1, 2) are the Hamiltonian and overlap matrix elements, respectively, between the covalent and ionic configurations, has been evaluated using the results of these calculations. At the localized fragment theory level, K is found to be very close to 1 and remarkably constant over the range of R(Si-F) distances sampled here, independent of core representation and basis set.
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Structures and spectral characteristics of Silylborane, Sylylaluminum hydride, Silylphosphine and Silylmercaptan
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Structure and bonding in sulfur-nitrogen compounds
Preston J MacDougall
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Molecular structures (the networks of bonds) and molecular geometries are determined in SCF calculations for S2N2, S!,+, S4N>+, S4N4, and H2S2 (with the STO-3G and 6-21G basis sets) and for S6NZ+, S8, SS2+, and SS4+ (with the STO-3G basis set). The chemical bonds are defined and classified in terms of the properties of the total charge distribution. In particular, molecular structures are determined by finding those pairs of nuclei linked by lines along which the charge density is a maximum with respect to any neighboring line, i.e., those nuclei linked by a bond path. In S4N4, for example, each S nucleus is linked to two N nuclei and to another S nucleus, while each N nucleus is linked to two S nuclei. The regions where electronic charge is locally concentrated and depleted within the valence shell of each atom are also determined. This information provides a charge density analogue of the Lewis electron pair model and determines the sites of electrophilic and nucleophilic attack within a molecule. In S4N4, for example, the valence shell of each S atom exhibits one nonbonded and three bonded charge concentrations, while each N atom exhibits two bonded and two nonbonded charge concentrations. The relatively long S-S bonds in S4N4 and S6N?+ and the cross-ring bonds found to be present in Ss+ and St+ exhibit the characteristics of closed-shell interactions as opposed to the other S-S and the S-N bonds in these molecules, which exhibit characteristics typical of shared interactions. (15) KertCz, M.; Suhai, S.; Azman, A.; Kocjan, D.; Kiss, A. I. Chem. Phys. Lett. 1976, 44, 53. 0020-1669/8S/1324-2047$01.50/0 .. d S \ N fragment and form an angle of 59O with the S-N bond axes. The value of V 2 p at the critical point is +0.02 au. The kinetics of dimerization of Fe(TPPS)(H,O))-to (TPPS)Fe-O-Fe(TPPS)8-(TPPS = tetrakisb-sulfonatopheny1)porphine) have been studied by pH jump-stopped flow at I = 0.1 M (NaNO,) and 25 "C. The spectral characteristics of the intermediate Fe(TPPS)(OH)'were obtained by rapid-scan experiments. Spectral-pH measurements allow the determination of the pKa of Fe(TPPS)(H,O)' as 7.0 f 0.2. The variation of the second-order dimerization rate constant (kOM) with pH (6.5-9.2) is given by kOw = kK,[H+]-'[l + K,[H']-']-2. This is consistent with a mechanism in which Fe(TPPS)(HzO)3-and Fe(TPPS)(OH)4are the important reactant pair (k = 1.5 X lo6 M-' s-I). The breakdown of dimer into monomer was measured by pH-drop and dithionite reduction experiments. The first-order rate constants kobd obeyed the expression koW = kH+-'. This
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Reversible intramolecular base-stabilization in silylene (silanediyl) complexes: surprising reactivity for silylene coordination compounds with a dynamic N.cntdot..cntdot..cntdot.Si.cntdot..cntdot..cntdot.N bond
Norbert Auner
Organometallics, 1993
The silane [ 2-(MezNCHz)C&] ZSiClz (7) shows a new dynamic N&-N "flip-flop" coordination mode below T, = 233 K (AG = 46.5 (f0.5) kJ mol-l) with both amine donors displacing each other. 7 is pentacoordinated in the solid state; crystal data: orthorhombic, Paca, a = 13.802(1) A, b = 17.908(1) A, c = 15.544(2) A, Z = 8. Reaction of the silanes C~H~[~-(M~ZNCHZ)C~H~]S~C~P (61, 7, and [~-(M~Z N C H Z)-~-(~-C~H~) C~H~~Z S~C~Z (8) with chromiumpentacarbonylmetallate yields the silanediyl complexes [~-(M~zNCHZ)C~H~IC~H~S~=C~(CO)B (111, [2-(Mez-N C H Z) C~H~] zSi=Cr(CO)5 (12), and 12-(M~zNCHZ)-~-(~-C~H~)C~H~I zSi=Cr(CO)s (13); 12-(Mez-N C H Z) C~H~] H S~= C~(C O)~ (14) was obtained by photolytic methods. Silanediyl complexes with one chelating donor show a rigid coordination of the donor to silicon which can be lifted at higher temperatures (95 'C for 11, AG = 80.4 (0.5) kJ mol-'). A single crystal X-ray structure determination of 11 reveals a CrSi bond distance of 2.409(1) A and a N-Si bond length of 1.991(2) A. Crystal data: triclinic, Pi, a = 9.401(1) A, b = 10.207(1) A, c = 11.586(1) A, Z = 2. Silanediyl complexes with two intramolecular donor functions feature a dynamic "flip-flop" coordination of the amine groups to silicon. Both (dimethy1amino)phenyl groups in 12 and 13 can be detached from silicon under liberation of a three-coordinate silicon atom at T, = 58 'C (VTJH-NMR), AG = 67.1 (f0.5) kJ mol-l for 12 and T, = 61 'C, AG = 70.1 (k0.5) kJ mol-' for 13. A single crystal X-ray structure determination of 12 gives 2.408(1) A for the Si-Cr bond length and 2.046(2) A for the N1-Si dative bond (the second contact N2-Si has a nonbonding distance of 3.309 A; the sum of bond angles at silicon amounts to 351.3'). Crystal data: triclinic, Pi, a = 9.531(1) A, b = 10.339(1) A, c = 11.676(1) A, Z = 2. The donor in 12 has been functionalized at the nitrogen atom by protonation or complexation with BF3. Photolysis of 12 and 13 leads to a 1,2-amine shift of one donor from silicon to the metal with loss of CO. The product [2-(Me~NCHz)c6H31 [2-(MezrjCHz)C6H3] Si=Cr(CO)r (20) has been characterized by a single crystal X-ray structure determination and has a bond distance Cr-Si 2.3610(4) A and N1-Si 1.981(1) A. Crystal data: monoclinic, P21/c, a = 10.344(5) A, b = 11.761(3) A, c = 18.96(1) A, Z = 4. Furthermore, the silanediyl complex 12 has been immobilized on silica gel. IR and UV spectroscopy and 13C CPMAS NMR provide evidence for the fixation of (silica-0-) [~-(HNM~zCHZ)C~H~~[~-(M~ZNCHZ)C~H~~ Si=Cr(CO)s (22) to the surface via the silicon atom. Reaction of 13 with HzO leads to (HO) [~-(H N M~z C H Z) C~H~I [~-(M~Z N C H Z) C~H~I S~-Cr(CO)5 (23) which has a structure similar to 22 with a Cr-Si bond length of 2.469(2) A. The dimethylamino or dimethylammonium substituent is pointing away from the silicon atom forming hydrogen bridges between Nl-HO-HN2; crystal data: monoclinic, P&/n, a = 13.198(2) A, b = 17.017(2) A, c = 17.066(1) A, Z = 4.
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DE ) DE(n N fσ SiH ) + DE(p π fd π )
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Trisilaallyl cation and related Si3H+5 structures. Departures from the corresponding carbon species
A. Korkin
Journal of Molecular Structure: THEOCHEM, 1996
The MP2(fc)/6-31G" structures, energies and frequencies of the trisilaallyl cation (1) and a number of its SiaH~ isomers reveal considerable differences from the C3 H ~ potential energy surface. Unlike the allyl cation, 1 is not a minimum. A" quasicyclic" form , with a much shorter Si(l)Si(3) separation and aromatic character, is the global Si3H~ minimum at MP2 and at B3LYP/6-311+G. However, Raghavachari's (J. Chem. Phys., 95 (1991) 7373) double-bridged isomer (5) is 1.4 kcal mo1-1 lower in energy than 7 at the CCSD(T)/6-311+G'//MP2/6-31G' level. Conjugation in the planar C2v I and hyperconjugation in the rotated 9 form are decreased relative to the carbon analogs. The remarkable stability of 5 and 7 can be attributed to their effective charge delocalization.
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From a Stable Silylene to a Mixed-Valent Disiloxane and an Isolable Silaformamide–Borane Complex with Considerable Silicon–Oxygen Double-Bond Character
Markus Brym
Angewandte Chemie International Edition, 2007
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Relative reactivities of amino and ethenyl groups in allylamine on Si(100)2 × 1: Temperature-dependent X-ray photoemission and thermal desorption studies of a common linker molecule
maryam ebrahimi
Surface Science, 2010
The room-temperature adsorption and thermal evolution of allylamine on Si(100)2 × 1 have been investigated by using temperature-dependent X-ray photoelectron spectroscopy (XPS) and thermal desorption spectrometry (TDS). The presence of a broad N 1 s feature at 398.9 eV, attributed to a N-Si bond, indicates N-H dissociative adsorption. On the other hand, the presence of C 1 s features at 284.6 eV and 286.2 eV, corresponding to C=C and C-N, respectively, and the absence of the Si-C feature expected at 283.2 eV shows that [2 + 2] C=C cycloaddition does not occur at room temperature. These XPS data are consistent with the unidentate staggered and eclipsed allylamine conformer adstructures arising from N-H dissociation and not [2 + 2] C=C cycloaddition. The apparent conversion of the N 1 s feature for Si-N(H)-C at 398.9 eV to that for Si-N(H) at 397.7 eV and the total depletion of C 1 s feature for C-N at 286.2 eV near 740 K indicates cleavage of the C-N bond, leaving behind a Si-N(H)• radical. Furthermore, the C=C C 1 s feature at 284.6 eV undergoes steep intensity reduction between 740 K and 825 K, above which a new C 1 s feature at 283.2 eV corresponding to SiC is found to emerge. These spectral changes suggest total dissociation of the ethenyl fragment and the formation of SiC. Moreover, while the total N 1 s intensity undergoes a minor reduction (24%) upon annealing up to 1090 K, a considerable reduction (43%) is found in the overall C 1 s intensity. This observation is consistent with our TDS data, which shows the desorption of C-containing molecules including propene and ethylene at 580 K and of acetylene at 700 K. The lack of N-containing desorbates suggests that the dissociated N species are likely bonded to multiple surface Si atoms or diffused into the bulk. Interestingly, both the staggered and eclipsed N-H dissociative adstructures are found to have a less negative adsorption energy than the [N, C, C] tridentate or the [2 + 2] C=C cycloaddition adstructures by our DFT calculations, which suggests that the observed formation of N-H dissociative adstructures is kinetically favored on the Si(100)2 × 1 surface.
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Structures, Energetics and Reactivities of Novel Silanetellurones: A Computational Study
Hassan Abdallah
Proceedings of The 15th International Electronic Conference on Synthetic Organic Chemistry
Di-substituted silanetellurones, X 2 Si=Te (X=H, F, Cl, Br, I and CN) are the target of this research. They are organometallic analogues, also known as heavy ketones, and they have recently been of great interest to researchers. Knowledge of the properties of such heavy ketones is important for a better understanding of both the chalcogen and substituent effects. The chemistry of multi-bonded compounds between group 14 and heavier chalcogen atoms has attracted chemists in various fields, as these molecules have proven to be convenient models for studying fundamental chemical problems and valuable intermediate products in synthesis. However, the literature of the mentioned species is limited and this research work endeavors a systematic investigation of these silanetellurones, using theoretical methods. We report the structural and spectroscopic parameters as well as the ionization potentials (IP ad and IP ad(ZPVE)), the neutral-anion separations (EA ad and EA ad(ZPVE)), the singlet-triplet splittings, the Kohn-Sham HOMO-LUMO gaps besides the nucleophilicity (N) and global electrophilicity (ω) indices. Since experimental data are still lacking for silanetellurones, the findings of this work would add to the literature, assist experimentalists to generate and further explore these novel species as well as rationalize some reactions pertaining to silanetellurones.
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Lewis basicity of silatranes and the molecular structures of EtOSi(OCH2CH2)3N, Me2O+Si(OCH2CH2)3N, and CF3CO2H.cntdot.EtOSi(OCH2CH2)3N
Lee Daniels
Journal of the American Chemical Society, 1991
The silatranyl group is shown to be sufficiently electron releasing in alkoxysilatranes (R&i(OCH2CH2),N) to allow isolation of the hydrogen-bonded adduct CF3C(O)OHEt0~i(OCH2CH2),N (7), the protonated cation HEtO+&(OCH2CH2)3N (13(BF4)) and the alkylated cations R20+.h(OCH2CH2)3N (R = Me, 6(BF4); R = Et, lZ(BF4)) of which 7 and 6(BF4) and also the alkoxysilatrane EtO!&(OCH2CH2)3N (5) have been structured by X-ray means. The SiN,, bond length in 6(BF4) (1.965 (5) A) is the shortest reported for a silatrane, and the O(H)O distance in 7 (2.489 A) is the shortest recorded for an unsymmetrical hydrogen bond. The SO,, distance increases by a total of 0.17 A in the order 5 < 7 < 6 (BF,). The greatest downfield shifts of a 'H or I3C resonance in a variety of alkoxysilatranes in hydrogen bonding, protonating, or alkylating environments are observed to occur at the OUR group. That electrophilic attack can also occur at 0, in solutions of these compounds is suggested by broadening of the OCH2CH2N protons of Phdi(0CH2CH2),N in the presence of Me,0BF4. Measurement of phenol u(OH) shifts reveals the basicity order (Me3Si)20 < Si(OR), alkoxysilatranes I Me3SiOMe < Et20, which places the electron-releasing ability of the silatranyl group ahead of (RO),Si but below an Et group. Parameters for the X-ra crystallography determined structures are as follows: 5, monoclinic (P2'/n), Z = 8, u = 10.956 (4) A, b = 11.187 (2) i , c = 17.638 (8) A, , ! 3 = 95.84 (4) ' ; 6(BF4), monoclinic (P 2 , / c) , Z = 4, a = 8.976 (2) A, b = 11.517 (1) A, c = 12.387 (2) A, / 3 = 91.943 (8) ' ; 7, orthorhombic (Pbca), Z = 8, u = 12.655 (4) A, b = 11.446 (4) A, c = 19.489 (3) A.
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Theoretical Study of SiH 2 n 2+ ( n = 1−3) Dications 1
George Olah
Inorganic Chemistry, 1999
Structures and energies of SiH 2n 2+ (n ) 1-3) dications were calculated at the density functional theory (DFT) B3LYP/6-311+G(d,p) and B3LYP/6-311++G(3df,2pd) levels. Contrary to the previously reported theoretical studies at the HF/6-31G level, the singlet SiH 2 2+ is not of linear D ∞h symmetric 1 but a C 2V symmetrical 2 with a two electron three center (3c-2e) bond. The structure 2 is significantly more stable than 1 by 14.7 kcal/mol. For calibration, structures 1 and 2 were also calculated at the ab initio CCSD(T)/cc-pVTZ level and found results which are in good agreement with the DFT results. DFT calculations also indicate that the singlet SiH 4 2+ is not of C 2V symmetric 4 with a 3c-2e bond but a C 2V symmetric 5 with two 3c-2e bonds. The C 2V symmetric 7 with two 3c-2e bonds and two 2c-2e bonds was found to be the global minimum for SiH 6 2+ dications.
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Intramolecular coordination of silicon in silyl formates: spectroscopic evidence confirmed by ab initio calculations
Igor Ignatyev
Journal of Molecular Structure, 1991
The increased value of the SiOC bending force constant (/sioc), obtained in normal coordinate treatment of experimental vibrational spectra of trimethylsilyl formate, was regarded as evidence for the existence of non-bonded attraction between silicon and carbonyl oxygen in this molecule. The results of normal coordinate calculations were checked by ab initio studies of equilibrium geometries and/sioc values of H3SiOCH0 conformers. Positive overlap population of Si..-0 contact and a 73% increase in the calculated/sioc value in the most stable c/s conformer compared with the trans form confirm formation of an intramolecular coordinate bond. The influence of such attraction on geometrical parameters of cis silyl formate, especially on Sill bond lengths, is analysed. Si-'0 attraction may be strengthened by substitution of one hydrogen in the CH3 group by fluorine. The optimized equilibrium structure of H2FSiOCHO and the ab initio value Of/sio C force constant of this molecule were calculated and it was shown that although non-bonded attraction increases in this molecule, it does not lead to an increase in/sioc.
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Molecular structure of 1-methyl-1-silabicyclo[2.2.1]heptane by gas-phase electron diffraction. Structural support for the trigonal bipyramidal transition state in SN2-silicon reactions
Philip Boudjouk
Journal of the American Chemical Society, 1976
The structure of l-methyl-l-silabicyclo[2.2.1]heptane has been determined by gas-phase electron diffraction. The molecule appears to be somewhat strained, as evidenced by moderately long average Si-C (1.883 f 0.002 A) and CC (I ,564 f 0.004 A) bond lengths, where quoted uncertainties are 3u values obtained from least-squares analysis of the intensity data. The results of this investigation offer convincing structural support for the proposed geometry of the sN2-si transition state suggested by Sommer.
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Molecular and dissociative bonding of amines with the Si-(7×7) surface
Robert Hamers
Surface Science, 2003
The interaction of trimethylamine (TMA) and dimethylamine (DMA) with the Si(1 1 1)-(7 7) surface has been studied by scanning tunneling microscopy (STM), X-ray photoelectron spectroscopy (XPS), and ultraviolet photoemission spectroscopy (UPS). STM data for TMA at low coverage show molecular features exhibiting a strong preference for bonding at the center adatom sites. XPS data show that at low coverage the majority of molecules form a highly ionic dative-bonded molecular adduct in which the N atom donates electron density to the surface, leading to a very high N(1s) binding energy of 402.4 eV. UPS data show that the interaction of TMA with the Si(1 1 1)-(7 7) surface also involves the restatom, suggesting that formation of dative bonds may also alter the restatom state. At very high exposures, a new, dissociative pathway becomes important, leading to dissociation and the appearance of new fragments with lower N(1s) binding energies of 399.1 eV. Corresponding studies for DMA only show dissociative bonding on Si(1 1 1), forming H atoms and N(CH 3) 2 species. While the N(CH 3) 2 species bonds primarily to the adatoms, the H atoms can bond to either adatoms or restatoms. Possible reaction mechanism and the reactivity of the different types of surface silicon atoms are discussed.
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Persistent Silylium Ions Stabilized by Polyagostic SiH⋅⋅⋅Si Interactions
Razvan Simionescu, Andrey Khalimon
Angewandte Chemie International Edition, 2007
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Cited by
Mechanism, reactivity, and selectivity of the iridium-catalyzed C(sp 3 )–H borylation of chlorosilanes
Genping Huang
Chem. Sci., 2015
The iridium-catalyzed C(sp 3 )-H borylation of methylchlorosilanes is investigated by means of density functional theory, using the B3LYP and M06 functionals. The calculations establish that the resting state of the catalyst is a seven-coordinate Ir(V) species that has to be converted into an Ir(III)tris(boryl) complex in order to effect the oxidative addition of the C-H bond. This is then followed by a C-B reductive elimination to yield the borylated product, and the catalytic cycle is finally completed by the regeneration of the active catalyst over two facile steps. The two employed functionals give somewhat different conclusions concerning the nature of the rate-determining step, and whether reductive elimination occurs directly or after a prior isomerization of the Ir(V) hydride intermediate complex. The calculations reproduce quite well the experimentally-observed trends in the reactivities of substrates with different substituents. It is demonstrated that the reactivity can be correlated to the Ir-C bond dissociation energies of the corresponding Ir(V) hydride intermediates. The effect of the chlorosilyl group is identified to originate from the a-carbanion-stabilizing effect of the silicon, which is further reinforced by the presence of an electron-withdrawing chlorine substituent. Furthermore, the source of selectivity for the borylation of primary over secondary C(sp 3 )-H can be explained on a steric basis, by repulsion between the alkyl group and the Ir/ligand moiety. Finally, the difference in the reactivity between C(sp 3 )-H and C(sp 2 )-H borylation is investigated and rationalized in terms of distortion/ interaction analysis.
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On the aromatic stabilization energy of the 4N π electron pyrene
Yirong Mo
Molecular Physics, 2009
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Neutral tricoordinated beryllium(0) compounds – isostructural to BH3 but isoelectronic to NH3
Susmita De
Dalton Transactions, 2013
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UPS of multilayer nitrogen-bearing compounds on the Si(100) surface
Kurt Field
Surface and Interface Analysis, 2008
Ultraviolet photoemission spectra (UPS) are presented for condensed layers of three ethylated amines; mono-, di-, and triethylamine (TEA) on the Si(100) surface at 100 K. The photoemission peaks associated with the nitrogen lone pair electrons are identified in the amines and compared with the corresponding spectra for condensed ammonia. Shifts in the lone pair binding energy for the ethyl-substituted amines are shown to be consistent with conventional chemical paradigms. Also, for comparison purposes spectra for two nonethylated amines, trimethylamine (TMA), and its silicon analog, trisilylamine (TSA), are presented and discussed.
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Synthesis of azadiphosphiridine complexes. Theoretical studies on ring formation, the P-to-P′ metal shift and the resulting nitrogen geometry
Rainer Streubel
Dalton Transactions, 2022
Three azadiphosphiridine complexes were selectively synthesized, two of them revealing a W(CO)5 group P-to-P′ haptotropic shift and only one displayed a planar ring N atom. Theoretical studies on ring formation and geometry as well as on the P-to-P′ haptotropic shift are reported.
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8315 | https://physics.info/x-ray/ | The Physics
Hypertextbook
Opus in profectus
X-rays
Discussion
introduction
Electromagnetic waves
Reverse photoelectric effect
history
X-rays were discovered in 1895 by the German physicist Wilhelm Röntgen (also spelled Roentgen). He received the first Nobel Prize in physics in 1901 "in recognition of the extraordinary services he has rendered by the discovery of the remarkable rays subsequently named after him." Wurzberg Physical-Medical Society, Chairman Albert von Kolliker, whose hand was used to to produce this image, proposed that this new form of radiation be called "Röntgen's Rays". Röntgen had a different idea.
It is seen, therefore, that some agent is capable of penetrating black cardboard which is quite opaque to ultra-violet light, sunlight, or arc-light. It is therefore of interest to investigate how far other bodies can be penetrated by the same agent. It is readily shown that all bodies possess this same transparency, but in very varying degrees. For example, paper is very transparent; the fluorescent screen will light up when placed behind a book of a thousand pages; printer's ink offers no marked resistance…. A piece of sheet aluminium, 15 mm. thick, still allowed the X-rays (as I will call the rays, for the sake of brevity) to pass, but greatly reduced the fluorescence. Glass plates of similar thickness behave similarly; lead glass is, however, much more opaque than glass free from lead…. If the hand be held before the fluorescent screen, the shadow shows the bones darkly, with only faint outlines of the surrounding tissues.
Wilhelm Röntgen, 1895
Röntgen appears to have always capitalized the x. I prefer to use lowercase, since the rays are purposely not named after any person, any place, or anything.
Warning: don't try this at home. Don't try this anywhere!
The retina of the eye is quite insensitive to these rays: the eye placed close to the apparatus sees nothing. It is clear from the experiments that this is not due to want of permeability on the part of the structures of the eye.
Wilhelm Röntgen, 1895
1912: Walter Friedrich and Paul Knipping diffract x-rays in zinc blende
1912: Max von Laue suggests using lattice solids to diffract x-rays
1913: William Bragg and Lawrence Bragg work out the Bragg condition for strong x-ray reflection
1922: Arthur Compton studies x-ray photon scattering by electrons
Roentgen/Gas-Filled Tubes
The earliest x-ray tubes were filled with air at low pressure (or a partial vacuum, if you prefer)... cathode, anode, and anticathode.
Early gas x-ray tubes
| Source: bulbcollector.com a.k.a. kilocat | Source: photo by the author a.k.a. the author |
Coolidge/Vacuum Tubes
Most x-ray tubes in use today are "filled" with a vacuum. This "entirely new variety" of x-ray tube was invented in 1913 by the American electrical engineer William Coolidge (1873–1975). In that same year Coolidge developed the technique for making fine wire out of tungsten (a notoriously non-ductile metal). Nearly every incandescent light bulb made after 1913 contains a tungsten filament made using Coolidge's process. When he was done working on light bulbs, he turned his attention to x-ray tubes. Guess what? Nearly every x-ray tube made after 1913 contains a tungsten filament made using the process used in light bulbs.
In a typical vacuum x-ray tube, electrons accelerated from a heated cathode toward a metal anode by a large potential difference. Changing the filament temperature changes the electron current — a hotter cathode releases more electrons than a cold one. This determines the intensity or "brightness" of the x-ray beam. Since one electron will produce one x-ray photon when it strikes the anode, more electrons flying through the tube means more x-ray photons emitted from the tube. The voltage across the tube determines the kinetic energy of the electrons when they strike the anode, which in turn determines the penetrating power of the x-ray photons — more energy per electron means more energy per x-ray photon and thus greater ability to plow through matter.
The cathode is a coiled filament of wire (usually tungsten) heated to around 2000 °C (white hot). It emits electrons through thermionic emission. In a sense, the electrons "boil" off the metal surface, but it's a weird kind of boiling since the electrons that leave are always replaced by new ones. If I put a pot of water on the stove at home, set it boiling and then leave the kitchen for an hour or two, by the time I get back there's a good chance the pot will be empty (and maybe even sizzling red hot). This does not happen with electrons in a cathode. The ones that leave are always replaced with new ones. If they didn't we'd wind up with a collection of positively charged ions (and eventually bare nuclei) that would surely fly apart due to their mutual repulsion. An x-ray tube is a circuit element. Current goes in one end and out the other and round and round the circuit.
The anode is a comparatively massive copper heat sink whose target face is cut diagonally and coated with some other metal (usually platinum). More than 99% of the kinetic energy imparted to the electrons is converted to heat on the anode. The remaining 1% is emitted as braking radiation (i.e., useful x-rays). This heat must be transferred or the target would melt. Coolidge's solution was to rotate the target using a small motor. This ensured that the hot spot never stayed in one place long enough to cause any lasting damage to the anode. (Some x-ray tubes are cooled with water.) The target is cut on a diagonal so that the emitted x-rays fly off the surface at an angle different from the incident electrons. A 45° cut makes the x-rays exit perpendicular to the axis of the tube. All the photographs of x-ray tubes on this page have their targets aligned at this angle. (The photo of a dental x-ray tube shown below left is a bit distorted, so the geometry isn't apparent.)
Vacuum x-ray tubes (Coolidge tubes)
| Schematic diagram of "an entirely new variety" of x-ray tube from William Coolidge's 1913 patent application. Nearly all contemporary x-ray tubes are variations of the Coolidge tube. Source: US Patent & Trademark Office | A vacuum x-ray tube of the type used in dentistry. Source: bulbcollector.com |
characteristic vs. bremsstrahlung (braking) spectra.
Hypothetical x-ray spectra produced by electrons with low energy (red), medium energy (green), and high energy (blue). As the energy of the electron beam increases, the maximum wavelength of the x-rays decreases but the location of the characteristic peaks does not.
| | | |
brems (braking/deceleration) + strahlung (radiation)
In a cold pure metal (a), all electrons are below the Fermi energy level. Thermal energy allows electrons to form a space cloud in the vacuum (b), and application of an electric field allows the electrons to be collected on an anode; otherwise, an equilibrium is set up between the electrons inside and outside the metal. A tungsten wire is used in most x-ray tubes, electron microscopes and electron microprobes to take advantage of the high temperature for melting (3680 K) and evaporation. In a conventional x-ray tube, the wire is a coil approximately 1 cm by 1 mm, and the temperature is adjusted to minimize evaporation of W atoms which slowly contaminate the target. Unless an accelerating voltage is applied, there is no emitted current from a hot filament because of the formation of a space charge of electrons near the metal surface. The saturation current is measured by using the metal as a cathode of a vacuum tube and collecting the electrons on an anode which is sufficiently positive to dissipate the space charge. In a conventional x-ray tube, sufficient stability is obtained by regulating the filament voltage (for heating) and the accelerating voltage between cathode and anode.
There are two (THREE?) principal mechanisms by which x-rays are produced. The first mechanism involves the rapid deceleration of a high speed electron as it enters the electric field of a nucleus. During this process the electron is deflected and emits a photon of x-radiation. This type of x-ray is often referred to as bremsstrahlung or "braking radiation". For a given source of electrons, a continuous spectrum of bremsstrahlung will be produced up to the maximum energy of the electrons.
X-rays are produced whenever fast moving electrons are decelerated, not just in x-ray tubes. Nearly all the naturally occurring x-ray sources are extraterrestrial. (No, that doesn't mean produced by alien creatures from outer space. It just means "beyond the Earth".) X-rays are produced when the solar wind is trapped by the Earth's magnetic field in the Van Allen Radiation Belts. Black holes are significant sources of x-rays in the universe. Matter falling into a black hole experiences an extreme acceleration caused by the intense field of the black hole. A single, isolated particle would fall in without releasing any radiation, but a stream of particles would as the particles would wind up crashing into each other on their way down the hole. Each inelastic collision experienced by a charged particle would result in the emission of a photon. Since these collisions are taking place at great speeds, the energies of the emitted photons in on the order of those found in the x-ray region of the electromagnetic spectrum. Inelastic collisions at even higher energies (greater than a million electronvolts) would generate gamma rays.
The second mechanism by which x-rays are produced is through transitions of electrons between atomic orbits. Such transitions involve the movement of electrons from outer orbits to vacancies within inner orbits. In making such transitions, electrons emit photons of x-radiation with discrete energies given by the differences in energy states at the beginning and the end of the transition. Because such x-rays are distinctive for the particular element and transition, they are called characteristic x-rays.
The third mechanism is through synchrotron emission.
Initially predicted in 1944 by Ivanenko and Pomeranschuk in Russia, it was, three years later, accidentally observed in a closed ring accelerator of the type of a synchrotron. It was long viewed as a "waste product", because synchrotron radiation is produced in the accelerators as a magnetic bremsstrahlung and undesirably limits the required final energy of the accelerators. Only several years later, in 1956, was synchrotron radiation specifically used in scientific investigations by Tomboulian and Hartmann.
Synchrotron radiation is emitted by charged particles traveling on a curved path (as would happen while moving through a magnetic field). Since the source of all electromagnetic radiation is the acceleration of charge, synchrotron radiation is an example electromagnetic radiation produced by centripetal acceleration (as opposed to bremsstrahlung, which is produced by tangential acceleration). The wavelength of this radiation is a function of the energy of the charged particles and the strength of the magnetic field bending the charged particles. The spectrum of the radiation is continuous and is characterized by its critical wavelength, which divides the spectrum into two parts with equal power (half the power radiated above the critical wavelength and half below).
The critical wavelength can be found using the equation below
| | | | |
--- --- |
| λc = | 4π | | E03 |
| 3 | cBE2 |
which reduces to the following equation when the charged particles are electrons
| | |
--- |
| λc[nm] = | 1.86453 |
| B[T]E[GeV]2 |
Synchrotron radiation sources: rings, undulators, wigglers, National Synchrotron Light Source doesn't produce light as its primary form of electromagnetic radiation. Most research done at this facility uses the x-rays and vacuum ultraviolet produced by the electron beam.
In 1945, the synchrotron was proposed as the latest accelerator for high-energy physics, designed to push particles, in this case electrons, to higher energies than could a cyclotron, the particle accelerator of the day. An accelerator takes stationary charged particles, such as electrons, and drives them to velocities near the speed of light. In being forced by magnets to travel around a circular storage ring, charged particles tangentially emit electromagnetic radiation and, consequently, lose energy. This energy is emitted in the form of light and is known as synchrotron radiation.
Synchrotron radiation is a nuisance in a particle accelerator as it sucks energy out of the particles being accelerated, but it makes an ideal source of high energy electromagnetic radiation. The beam produced is composed of nearly parallel rays (collimated) and is quite intense.
Synchrotron radiation can be produced for hours, maybe even days if you were willing to pay the electric bils and had some reason to work around the clock. x-ray tubes can only operate for a few seconds or maybe minutes. Run them too long and they'll burn out just like a light bulb.
Synchrotron radiation is "organized": the beam is highly polarized (most of the waves are oscillating in the same plane) and collimated (most of the waves are in the same direction). x-ray tubes produce "messy" radiation that is completely unpolarized and may be focused only with great difficulty. A synchrotron source is like an "x-ray laser", while an x-ray tube is like an "x-ray floodlight".
Synchrotron radiation can be "shared". A large synchrotron might have upwards of 50 beam lines and run hundred if not thousands of experiments in one year. Synchrotron facilities are expensive to build, but pay for themselves in sheer volume of research.
Wigglers or undulators (also known as insertion devices) produce synchrotron radiation that is considerably brighter than radiation from a bending magnet. The device causes electrons to follow a sinusoidal path instead of a curved one by establishing a series of magnetic fields that alternate in polarity and are perpendicular to the electrons' direction of travel. A wiggler enhances the brightness of the radiation produced by a given electron beam by a factor roughly equal to twice the number of full oscillations the beam undergoes. The deflections of the beam are smaller in an undulator than in a wiggler, and the radiation's brightness can, in theory, be increased by a factor about equal to the square of the number of oscillations, but only at discrete photon energies.
photon momentum
Max Planck discovered that phtons have energy.
E = hf
Albert Einstein discovered that energy and momentum are related.
E2 = p2c2 + m2c4
Photons are massless, so this equation reduces to…
E = pc
Combine Planck and Einstein (their equations, not the men themselves)…
hf = pc
Solve for momentum…
| | |
--- |
| p = | hf |
| c |
Recall that…
| | |
--- |
| λ = | c |
| f |
Thus…
| | |
--- |
| p = | h |
| λ |
If Planck and Einstein are correct, then photons have momentum too. What we need now is experimental evidence to support or refute this. (Don't worry. No one's going to refute this.)
compton effect
Arthur Compton (1892–1962) United States
| | | | |
--- --- |
| ∆λ = | 2h | sin2 | θ |
| mec | 2 |
technology
shadowgraphs
computed axial tomography (CAT)
text
x-ray scattering
text
x-ray diffraction
Laue spots, Laue pattern. The German physicist Max von Laue (1879–1960) was the first to diffract X-rays on 12 April 1912.
Bragg's law. The father-son team of of British physicists William Bragg (1862–1942) and Lawrence Bragg (1890–1971) derived this equation in 1913.
nλ = 2d sin θ
| | |
--- |
| λ = | x-ray wavelength |
| n = | order |
| d = | separation between atomic planes |
| θ = | angle relative to the plane's surface |
DNA structure was determined from an x-ray diffraction image taken by Rosalind Franklin (1920–1958) England.
x-ray fluorescence
text
| | |
--- |
| | | |
8316 | https://www.quora.com/If-we-threw-a-die-twice-would-it-give-the-same-outcome-of-throwing-a-two-dice-once | Something went wrong. Wait a moment and try again.
Probability (statistics)
Game of Throwns
Random Events
Probability Theory
Probability of an Event
Games of Chance
Probability Concept
Probability in Math
5
If we threw a die twice, would it give the same outcome of throwing a two dice once?
Graham Rollins
In House Counsel
·
Updated 7y
I assume you’re asking if there are differences in probability associated with throwing two dice simultaneously vs. one then the other. The answer is no. Each die roll is a independent event and it doesn’t matter if you throw them simultaneously or one and then the other. The probability of any specific outcome on two six sided dice is 1 in 36.
You might also consider that actually rolling the dice is precisely the same instant is virtually impossible. One leaves your hand before the other, and one will settle before the other. Alternatively, you could throw one now and wait 50 years and then t
I assume you’re asking if there are differences in probability associated with throwing two dice simultaneously vs. one then the other. The answer is no. Each die roll is a independent event and it doesn’t matter if you throw them simultaneously or one and then the other. The probability of any specific outcome on two six sided dice is 1 in 36.
You might also consider that actually rolling the dice is precisely the same instant is virtually impossible. One leaves your hand before the other, and one will settle before the other. Alternatively, you could throw one now and wait 50 years and then throw the other. It’s still a 1/36 probability of any specific pair of numbers coming up!
Edit: Corrected to state that the dice rolls are independent events and not mutually exclusive events. Thank you to Dr. Moore in the comments!
Related questions
Why is throwing two dice simultaneously and throwing a dice two times the same event?
A pair of dice is thrown twice. What is the probability of getting the same number on both of them?
What is the probability of having an odd number in a single toss of a fair die?
What is the probability of getting a 3 on one die or both when rolling two dice?
What is the probability of throwing a total of 10 with two dice?
Joe Pegasus
Updated 6y
Originally Answered: If I were to roll a die twice in the exact same way would I get the same outcome both times? I'm interested in everyone's opinion on this, especially people with little or no physics background.
·
many experimenters, including myself, have set up mechanical devices to test what you ask about. My device was a 6″x4″x4″ block … 6″ high by 4″ long, by 4″ deep. Atop this I placed 2 dice at the very edge of the block. The dice were set at the 6/5 5/4 set. That means when you viewed the dice at the block’s front - cliff - side you saw the left die at 5 and it’s top at 6; the right die at 4 with the top at 5. Then I manually held a playing card to the back of the dice and pushed them off. After 2,000 attempts the outcome was very close to the expected results except the 7 had appeared more than
many experimenters, including myself, have set up mechanical devices to test what you ask about. My device was a 6″x4″x4″ block … 6″ high by 4″ long, by 4″ deep. Atop this I placed 2 dice at the very edge of the block. The dice were set at the 6/5 5/4 set. That means when you viewed the dice at the block’s front - cliff - side you saw the left die at 5 and it’s top at 6; the right die at 4 with the top at 5. Then I manually held a playing card to the back of the dice and pushed them off. After 2,000 attempts the outcome was very close to the expected results except the 7 had appeared more than expected (about 6.4 rather than 6) and the 8 appeared less. I have also tested this with one die and gleamed the expected results.
Valerie Green
Studied at Bournemouth University
·
6y
Originally Answered: If I were to roll a die twice in the exact same way would I get the same outcome both times? I'm interested in everyone's opinion on this, especially people with little or no physics background.
·
If you rolled it exactly the same way? So it leaves your hand at the exact same location each time, at the exact same speed, held the exact same way, bouncing off the exact same surface at the exact same way?
In theory, yes. In practice, a die roll has too many variables to exactly replicate.
There is the concept of Dice Control, where someone tries to throw a die in such a way that they can control the number it lands on. Casinos generally require dice to bounce off the wall of the table for a throw to be valid, as this will make controlling the dice near impossible.
Michael Clements
B.S. in Mathematics & Computer Science, University of California, Davis
·
6y
Originally Answered: If I were to roll a die twice in the exact same way would I get the same outcome both times? I'm interested in everyone's opinion on this, especially people with little or no physics background.
·
What do you mean by “in the exact same way”?
If you mean everything: the motion of the dice: its speed, direction of motion, spin axes and rate of spin, the location where you roll it, etc. Then yes it must have the same outcome each time. The laws of physics are deterministic.
However, it’s virtually impossible for all those factors to be exactly the same. And tiny differences in a single factor can cause big unpredictable changes in the outcome.
So the answer is: theoretically, yes. But pragmatically, no, because you can’t roll it in the exact same way twice.
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Related questions
Three dice are rolled. What is the probability of getting a sum of 13?
What is the outcome of throwing a fair die twice?
A pair of dice is rolled. What is the probability of getting doubles?
Two fair dice are rolled. What is the probability of getting a sum of 7 or 11?
A die is rolled 4 times. What is the probability of getting the number 4 at least once?
Jonathan Roberts
Lives in England
·
Author has 3.4K answers and 2.6M answer views
·
6y
Originally Answered: If I were to roll a die twice in the exact same way would I get the same outcome both times? I'm interested in everyone's opinion on this, especially people with little or no physics background.
·
I’m not sure about a dice. But if you hold one ball with its centre directly over another ball and drop it the maximum number of times it will bounce back hitting the other ball is about seven. This is due to quantum effects. This means that lottery balls from a tombola are drawn totally unpredictably due to quantum effects.
Falling Autumn
quora credentials are stupid and none of them work.
·
Author has 2.9K answers and 2.5M answer views
·
9y
Originally Answered: If we threw a dice twice, would it give the same outcome of throwing a two dice once?
·
They are independent of each other, so the probability distribution remains the same.
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Sableagle
Former Has Too Much Time on Their Hands at The Internet (1994–2025)
·
Author has 4.1K answers and 4.3M answer views
·
Sep 18
Related
When throwing two dice what is the probability of throwing a total more than 6?
Total more than 6, on 2d6?
1 & 6
2 & 5 or 6
3 & 4, 5 or 6
4 & 3, 4, 5 or 6
5 & 2, 3, 4, 5 or 6
6 & anything
The possibilities are looking kind of triangular, yes? We have a triangular number of possibilities, specifically 21, and on 2d6 there are 6^2 possible results, so 21/36, which cancels down to 7/12.
With 2d8:
1 & 6, 7 or 8
2 & 5, 6, 7 or 8
3 & 4, 5, 6, 7 or 8
4 & 3, 4, 5, 6, 7 or 8
5 & 2, 3, 4, 5, 6, 7 or 8
6 & anything
7 & anything
8 & anything
Now we don’t quite have a triangular number, because the tips of the triangle have been lopped off.
You can’t roll a 0 and a 7 or 8, or roll a -1 and
Total more than 6, on 2d6?
1 & 6
2 & 5 or 6
3 & 4, 5 or 6
4 & 3, 4, 5 or 6
5 & 2, 3, 4, 5 or 6
6 & anything
The possibilities are looking kind of triangular, yes? We have a triangular number of possibilities, specifically 21, and on 2d6 there are 6^2 possible results, so 21/36, which cancels down to 7/12.
With 2d8:
1 & 6, 7 or 8
2 & 5, 6, 7 or 8
3 & 4, 5, 6, 7 or 8
4 & 3, 4, 5, 6, 7 or 8
5 & 2, 3, 4, 5, 6, 7 or 8
6 & anything
7 & anything
8 & anything
Now we don’t quite have a triangular number, because the tips of the triangle have been lopped off.
You can’t roll a 0 and a 7 or 8, or roll a -1 and a 6, 7 or 8, and you can’t roll a 7 and a 0 or an 8 and a 0 or -1.
The minimum roll on the second die to get over the threshold is the threshold minus the roll on the first die plus one, and the possible results that get over the threshold number the maximum result minus that minimum result plus one, from an initial result of 1 to an initial result of one less than the threshold.
For 1: 8 - (6 - 1 + 1) + 1, which is 3
For 2: 8 - (6 - 2 + 1) + 1, which is 4
For 3: 8 - (6 - 3 + 1) + 1, which is 5
We can tidy that up to the maximum minus the threshold plus the first roll.
For 4: 8 - 6 + 4, which is 6
For 5: 8 - 6 + 5, which is 7
For 6: 8 - 6 + 6, which is 8
From then on, it’s any result, so the number of possible results that get the total over the threshold is the number of possible results.
That’s going to be 8 (8 - 6) possible combinations where the roll on the first die was more than enough to make the roll on the second die irrelevant. That’s 16.
We can tidy up the first part a bit, too.
From 3, the count for a roll of 1 on the first die, to 8, the maximum result, is +5.
+5 / 2 = +2½
3 + 2½ = 5½
5 + 1 = 6
5½ 6 = 33
33 + 16 = 49, so 49 / 8², or 0.765625
For 2d10:
10 - 6 + 1 = 5
From 5 to 10 is +5
+5 / 2 = +2½
5 + 2½ = 7½
5 + 1 = 6
7½ 6 = 45
10 (10 - 6) = 40
45 + 40 = 85, so 85 / 10², or 17 / 20 or 0.85
For 2d12:
12 - 6 + 1 = 7
From 15 to 10 is +5 ← This is just one less than the threshold every time, isn’t it?
+5 / 2 = +2½
7 + 2½ = 9½
5 + 1 = 6 ← This is just the threshold every time, isn’t it?
9½ 6 = 57
12 (12 - 6) = 72
57 + 72 = 129, so 129 / 12², or 43 / 48 or 0.895833333333333
For 2d20:
20 - 6 + 1 = 15
+5 / 2 = +2½
15 + 2½ = 17½
17½ 6 = 105
20 (20 - 6) = 280
105 + 280 = 385, so 385 / 20², or 77 / 80 or 0.9625
Arunangshu Biswas
Principal Bio-statistician at Novartis Healthcare (2017–present)
·
6y
Related
Why is throwing two dice simultaneously and throwing a dice two times the same event?
The logical and the mathematical fact that makes the two events identical is independence. Logically it means that the outcome of the first dice has no bearing on the outcome of the second. Mathematically it means that given a particular outcome of the first die, the conditional probability is the same as the unconditional. That is,
P( Second die shows 6 | First die shows 1) = P( Second die shows 6).
Note that the above statement is not a fact but an ASSUMPTION which is backed by the logic that has been explained in the first paragraph. In fact, whenever we have independence in the case of rando
The logical and the mathematical fact that makes the two events identical is independence. Logically it means that the outcome of the first dice has no bearing on the outcome of the second. Mathematically it means that given a particular outcome of the first die, the conditional probability is the same as the unconditional. That is,
P( Second die shows 6 | First die shows 1) = P( Second die shows 6).
Note that the above statement is not a fact but an ASSUMPTION which is backed by the logic that has been explained in the first paragraph. In fact, whenever we have independence in the case of random variables they are only assumptions.
Therefore if we call X1 and X2 as the first and second throws (die) respectively then they are independent which does depend if they are throws simultaneously or sequentially.
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Durga Sravan
Author has 101 answers and 126K answer views
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6y
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Why is throwing two dice simultaneously and throwing a dice two times the same event?
In short both the events are mutually exclusive. Let's call throwing the first and second dice as first and second in that order. The outcome of the first event doesn't affect the second event and vice versa. This is the reason why both are same.
Gregory Schoenmakers
I probably know something about statistics.
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Author has 4.4K answers and 8.5M answer views
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9y
Originally Answered: If we threw a dice twice, would it give the same outcome of throwing a two dice once?
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The probability that the pair of dice and the single die thrown twice give the same total would be 73648=0.1127.
They have the same probability distribution (and they are independent events) so it is just a matter of squaring the probabilities of each total and summing them up.
Darryl Nester
Ph.D. in statistics (long ago), plus 25+ years of teaching
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Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 791 answers and 2.1M answer views
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Updated 7y
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Why we don't sum the probabilities in: what's the probability of getting 2 throwing a dice twice?
Good question. If we have events A and B, and want to know the probability of A or B, we do add their probabilities, but if the two events are not mutually exclusive — that is, if there is a possibility that both A and B can happen at the same time — we have to subtract the probability of the overlap. That is,
P(A or B)=P(A)+P(B)−P(A and B)
The Venn diagram below helps us to understand why we do this: If we add P(A) and P(B), then we have “double-counted” the overlap area, so we have to subtract it.
In this case, if A is the event “2 on the first roll” and B is the event “2 o
Good question. If we have events A and B, and want to know the probability of A or B, we do add their probabilities, but if the two events are not mutually exclusive — that is, if there is a possibility that both A and B can happen at the same time — we have to subtract the probability of the overlap. That is,
P(A or B)=P(A)+P(B)−P(A and B)
The Venn diagram below helps us to understand why we do this: If we add P(A) and P(B), then we have “double-counted” the overlap area, so we have to subtract it.
In this case, if A is the event “2 on the first roll” and B is the event “2 on the second roll,” then P(A)=P(B)=16, and P(A and B)=16⋅16=136, so
P(A or B)=16+16−136=1136
Why we don't sum the probabilities in: what's the probability of getting 2 throwing a dice twice? I know we don’t sum 1/6 + 1/6. But we normally have the rule of using multiplication for “AND” questions and sum for “OR” questions. One would think of this question as “getting 2 in the first OR the second throw”. Why this rule does not apply? What’s the difference for the times it applies?
Richard Lind Olofsson
Msci in Industrial Engineering and Management, Umeå University (Graduated 2020)
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7y
Related
Is tossing a coin (balanced, symmetric) twice the same as tossing two identical coins at once? What if we repeat the experiment 100 times? Will the probabilities remain the same?
The probability of two independent tosses with the same coin getting the same face (let’s say tails) can be calculated as 0.5^2 = 0.25. (assuming perfect coin). The same principle works for, let’s say 5 tosses with the same coin, where p = 0.5^5. The probability for any number n of tosses with the same coin can be calculated using 0.5^n.
Now let’s imagine getting two tails when tossing two coins at the same time. This is where the binomial distribution comes in. Without getting too technical, it can be defined as bin(k,n,p) = n!/(k!(n-k!)) p^k (1-p)^(n-k) where k is the number of tails, n i
The probability of two independent tosses with the same coin getting the same face (let’s say tails) can be calculated as 0.5^2 = 0.25. (assuming perfect coin). The same principle works for, let’s say 5 tosses with the same coin, where p = 0.5^5. The probability for any number n of tosses with the same coin can be calculated using 0.5^n.
Now let’s imagine getting two tails when tossing two coins at the same time. This is where the binomial distribution comes in. Without getting too technical, it can be defined as bin(k,n,p) = n!/(k!(n-k!)) p^k (1-p)^(n-k) where k is the number of tails, n is the number of coins tossed and p is the probability. Given this, the probability of two coins getting a tail each is bin(2,2,0.5) = 1 0.5^2 1 = 0.25.
As a matter of fact, for every case where the number of successes is equal to the number of tries (n = k), the function can be expressed as bin(k,n,p) = p^k = p^n.
This proves that the probability of getting tails for an n number of tosses with a single coin always will be equal to the probability of getting the same number of tails while tossing an n number of coins one time.
Tomaž Vargazon
Practicing atheist
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Author has 10.7K answers and 422.7M answer views
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1y
Related
If you toss a coin 1 million times and you get heads in all of them, but if you toss it one more time, does the probability of getting tails increase at all, and how would the answer change depending on how many times we tossed it?
As stated, I will say the coin will land heads for sure.
You didn’t specify what kind of a coin it was. If the coin were fair then the odds of the next toss would be equal no matter the previous results, but we aren’t given that information. We are given information it is a coin of some sort, it was tossed 1 million times in a row and the result was always heads.
With that information, I conclude the 1,000,001st toss will also be heads.
I could be wrong of course. It could have been a fair coin that simply landed heads a million times in a row all along. But I have no reason to suspect this was r
As stated, I will say the coin will land heads for sure.
You didn’t specify what kind of a coin it was. If the coin were fair then the odds of the next toss would be equal no matter the previous results, but we aren’t given that information. We are given information it is a coin of some sort, it was tossed 1 million times in a row and the result was always heads.
With that information, I conclude the 1,000,001st toss will also be heads.
I could be wrong of course. It could have been a fair coin that simply landed heads a million times in a row all along. But I have no reason to suspect this was remotely the case. Odds of such an event are 1:2^1,000,000.
To put this in perspective, 2^1000 has over 300 digits and Google won’t let you calculate the 2^10,000, let alone that monstrosity.
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8318 | https://cdn.testingmom.com/pdf/Analogies.pdf | Analogies – Part 1: Verbal An analogy is a comparison between 2 different things based on 1 or more shared qualities. What Is An Analogy? Can I impress my friends and parents with it!?
An analogy is a comparison between 2 different things based on 1 or more shared qualities. You use sound reasoning skills or logical thinking when you use “analogical thinking,” and you use it ALL THE TIME in our lives! That’s a big word!!!
Take what you know about 1 thing or situation and apply it to another. + Logical = Analogical Thinking It’s a split-second way of thinking that helps us get by in everyday life. Analogical Thinking Uses “Inference” Inference = Take something we know from one thing or situation and apply it to another.
Making a “mental leap.” Analogy Can Give You Great Ideas! Language is full of analogies! Same Word in Many Situations Play Enjoy • Costs an arm and a leg.
• It’s not over till the fat lady sings. • He sailed through school.
• Don’t spill the beans • Best thing since sliced bread. • Cry over spilt milk. • Hit the nail on the head.
• Let the cat out of the bag.
• Off your rocker. • Piece of cake. • Straight from the horse’s mouth.
• That stinks! Sayings that are short-hand ways to communicate complicated ideas – comparisons to something different.
I think I can… You are my sunshine, my only sunshine… Many Think This Is All An Analogy Is! Coin : dime :: A.
Month : June B.
Horrible : terrible C.
Cold : hot D.
Quarter : dollar Pyramid : triangular :: A.
Ball : throw B.
Kite : wind C.
Velvet: scratchy D.
Hurricane : rainy Can You Tell Me Why? • Merciful : Kind :: Cruel : Brutal • Root : Route :: Rose : Rows • Fresh : Rotten :: New : Old • Encyclopedia : Volume :: TV Series : Episode • Canoe : Speedboat :: Bicycle : Motorcycle Finish This Analogy • Saddle : horse :: bib : _.
• Coin : nickel :: month : __.
• Library : research :: gymnasium : _.
• Cup : liquid :: stapler : _.
• Ice cream : cold :: soup : _. Create Your Own Analogy Spider : 8 :: _ : _.
Poor : pour :: _ : _.
Painter : brush :: _ : _.
Analogy Can be Silly : :: Cat : Teeth :: baby : mustache A Trick to Help You Solve Analogy Questions 1. Crawl : walk :: _ : _.
2. Write a sentence that describes the relationship between crawl and walk. 3. You must crawl before you can walk. 4. Crawl : walk :: _ : _.
A. sprint : run B. babble : talk C. big : small D. cry : laugh 5. Apply the sentence to the answer choices to see which fits in the same way. 1. Famine : drought :: _ : _.
2. Write a sentence that describes the relationship between famine and drought. 3. Famine is caused by a drought. 4. Famine : drought :: _ : _.
A. starvation : desert B. sharing : love C. seasickness : voyage D. hungry : wasteful 5. Apply the sentence to the answer choices to see which fits in the same way. A Trick to Help You Solve Analogy Questions Let’s do these together: 1. Counselor : advise :: _ : _.
2. Write a sentence that describes the relationship between counselor and advise. 3. _____ 4. Counselor : advise :: _ : _.
A. teacher : write B. chauffer : drive C. receptionist : sit D. doctor : study 5. Apply the sentence to the answer choices to see which fits in the same way. 1. Paws : pause :: _ : _.
2. Write a sentence that describes the relationship between paws and pause. 3. _____ 4. Paws : pause :: _ : _.
A. dog : snout B. pain : pane C. foot : feet D. cat : cats 5. Apply the sentence to the answer choices to see which fits in the same way. 1. Same (but different color, number, size, orientation – if picture) 2. Antonym (down : up, happy : sad) 3. Synonym (happy : glad, small : tiny) 4. Homonym (route : root, time : thyme) 5. Tool : Action (pen : write, basketball : dribble) 6. Tool : Object it’s used with (hammer : nail) 7. Category: Object in Category (fish : salmon, insect : ant) 8. Effect: Cause (flood : rain, sickness : germs) 9. Cause: Effect (rain : flood, practice : improve) 10. Increase Intensity (talk : yell, warm : hot) 11. Decrease Intensity (boil : simmer, ecstatic : happy) 12. Sequence (childhood : adulthood, May : June) 13. Fact : Opinion (“It’s 93 degrees.” : “It’s hot outside.”) 14. Action : Thing acted upon (play : piano) 15. Action : Subject acting (dance : ballerina) 16. Object : location (car : garage, stove : kitchen) 17. Object : User (chalk : teacher, conductor : baton) 18. Member : Group (wolf : pack, student : class) 19. Parent : Offspring (dog : puppy, cat : kitten) 20. Noun : related adjective (elephant : enormous) 21. Verb Tense (walk : walked, buy : bought) 22. Rhyme (please : tease, beg : leg) 23. Sets (peanut butter : jelly, shoes : socks) 24. Problem : Solution (itch : scratch, dirty : wash) 25. Place : Worker (school : teacher, plane : pilot) 26. Part : Whole (trunk : tree, alphabet : letter) 27. Step : Process (spell check : edit) (blue = most common) Common Verbal Analogy Relationships Tips for Solving Verbal Analogy Questions Build your vocabulary skills.
The more words you know and understand, the easier these questions will be. Build your vocabulary by reading, going over vocabulary lists, learning definitions of words, along with synonyms and antonyms. Watch out for distractors. The people who write tests will try to trick you by putting in answer choices that seem right at first but aren’t. One way they do this is by including answers that are similar in use, genre or in the same category as the first two words. Pen : author :: A. Saw : carpenter B. Pencil : paper C. writer : essay D. Book : Bookstore The answer is A – A pen is a tool used by an author and a saw is a tool used by a carpenter. But answers B, C and D include answers that are related to pens and authors, so they may trick students who paying close attention. Tips for Solving Verbal Analogy Questions When You Don’t Know What a Word Means Hangar : airplane :: a. Cat : kitten b. Garage : car c.
Airport: reservation agent d. Teacher : class a) A cat : kitten is a parent : offspring relationship. Airplanes don’t have parents so that can’t be the answer. b) A garage is a place : object relationship. Could that be it? Maybe?
c) A reservation agent works at an airport, so this is a place : worker relationship. An airplane isn’t a worker, so this doesn’t make sense. Also, by including an airport, that might be a distractor! d) This is a member : group relationship. An airplane isn’t a group.
The only possible answer seems to be b) so choose that even if you aren’t sure what hangar means. If you don’t know what one of the words in the analogy means (such as “hangar,”) use reasoning skills to eliminate answers that you know are wrong. Then make your best guess.
Watch out for order of words! Rain : flood :: ___ A. Illness : germs B. Cavities : sugar C. Scratching : itching D. Practice : improvement All of these are cause : effect or effect : cause relationships. The words in the analogy are in cause : effect order. Only choice D is in cause : effect order. So that is the answer. More Tips for Solving Verbal Analogy Questions Look for parallel answer choices. Music : melodious :: potpourri : __ A. Flowers B. Orchestra C. Fragrant D. Bowl If the analogy is presented as noun : verb :: noun : _, then you are looking for an answer that is a verb. Even if you don’t know what potpourri is, the order of the analogy is noun: adjective :: noun : _. So we need an adjective for our answer. C is the only possibility, even if you aren’t sure what potpourri means. Sometimes it may appear there is more than 1 good answer choice. Look for the BEST answer choice. Cantaloupe : strawberry :: polar bear :: _ A. North Pole B. South Pole C. Seals D. Arctic wolf Cantaloupe and strawberries are specific types of fruit. A polar bear is a specific animal that lives in the Arctic. Seals and Arctic wolves do as well. But Arctic wolves are even more specific, so that is the BEST choice. More Tips for Solving Verbal Analogy Questions Coward is to hero as victim is to _.
a. crime b. chicken c. bully d. scared e. timid Photographer is to portrait as biographer is to .
a. fiction b. words c. celebrity d. drawing e. life story Stomach is to food as lungs are to _.
a. Trachea b. air c. chest d. throat e. breathe Anxious is to nervous as confident is to .
a. Unsure b. meek c. bold d. shy e. quiet Hurricane is to wind as blizzard is to _.
a. rain b. sun c. wind d. snow e. hail Curb is to street as coast is to .
a. pedal b. ocean c. mountain d. valley e. hill Let’s Try a Few! Mutter is to talk as shuffle is to _.
a. walk b. jump c. run d. yell e. hop Historian is to history as archaeologist is to .
a. music b. rock c. science d. art e. culture Ancient is to modern as history is to .
a. present b. past c. future d. class e. books Thanksgiving is to November as Halloween is to .
a. September b. October c. Fall d. November e. Winter Dairy is to milk as protein is to ____.
a. apple b. radish c. steak d. butter e. lettuce …And a Few More! How TestingMom.com Can Help • Choose Test Prep • Choose View All Tests • Choose CogAT – Form 7 and 8 Practice Analogies Here!
How TestingMom.com Can Help • Primary Battery – Verbal Analogies • Multi-Level Battery – Verbal • Choose Verbal Analogies • Choose Interactive Practice Questions • Choose Analogy Galaxy Next Week – Part 2 – Non-Verbal Analogies Analogies that use shapes and figures instead of pictures and words. |
8319 | https://web.williams.edu/Mathematics/sjmiller/public_html/math/papers/PellAndGeneralizations_ConsecSum_Polymath2023v20.pdf | Sum of Consecutive Terms of Pell and Related Sequences Navvye Anand, Amit Kumar Basistha, Kenny B. Davenport, Alexander Gong, Steven J. Miller, Alexander Zhu Summer 2023 Abstract We study new identities related to the sums of adjacent terms in the Pell sequence, defined by Pn := 2Pn−1 + Pn−2 for n ≥2 and P0 = 0, P1 = 1, and generalize these identities for many similar sequences. We prove that the sum of N > 1 consecutive Pell numbers is a fixed integer multiple of another Pell number if and only if 4 | N. We consider the generalized Pell (k, i)-numbers defined by p(n) := 2p(n−1)+p(n−k−1) for n ≥k+1, with p(0) = p(1) = · · · = p(i) = 0 and p(i + 1) = · · · = p(k) = 1 for 0 ≤i ≤k −1, and prove that the sum of N = 2k + 2 consecutive terms is a fixed integer multiple of another term in the sequence. We also prove that for the generalized Pell (k, k −1)-numbers such a relation does not exist when N and k are odd. We give analogous results for the Fibonacci and other related second-order recursive sequences.
1 Introduction We first review some standard notation, and then describe our results. The Fibonacci numbers are defined by F(n) := 0 n = 0 1 n = 1 F(n −1) + F(n −2) n ≥2, (1.1) and have a closed form given by Binet’s formula: F(n) = φn −ψn √ 5 , (1.2) where φ = 1 + √ 5 2 and ψ = 1 − √ 5 2 = −1 φ.
They are the simplest depth two constant-coefficient recurrence to study (the simpler depth one recurrences are just pure geometric sequences). They satisfy 1 numerous interesting identities and arise in various areas; see for example [Ko].
We consider the Fibonacci numbers and other recursively defined sequences of numbers. In particular, we are interested in Pell numbers and their generaliza-tions; we state these sequences and then our results.
Definition 1.1. Classical Pell numbers are defined by the following recurrence and initial conditions: P(n) := 0 n = 0 1 n = 1 2P(n −1) + P(n −2) n ≥2.
(1.3) Definition 1.2. The Pell-Lucas sequence or the Companion Pell sequence is defined by Q(n) := 2 n = 0 2 n = 1 2Q(n −1) + Q(n −2) n ≥2.
(1.4) Definition 1.3. The Lucas sequence is defined by L(n) := 2 n = 0 1 n = 1 L(n −1) + L(n −2) n ≥2.
(1.5) Definition 1.4. If terms of a recursively defined infinite sequence can be ex-pressed in a closed form similar to that of (1.2), we call the closed form a generalized Binet formula (see [BBILMT, Le]).
Example. Let a := 1 + √ 2 and b := 1 − √ 2 = −1 a.
(1.6) Then the nth Pell Number is given by the generalized Binet formula: P(n) = an −bn 2 √ 2 .
(1.7) Definition 1.5. Throughout the paper, we let C : N →N denote a natural valued function on natural numbers.
1.1 Motivation and Results We analyze the relationships between sums of consecutive numbers in recurrence sequences. The first theorem below is a generalization of an observation made by the third named author to the fifth named author (for a problem for the Pi Mu Epsilon Journal) for sums of eight consecutive terms.
2 Theorem 1.6. For any N ∈N, the sum of 4N consecutive Pell numbers is equal to a constant (depending on N) multiplied by the (2N + 1)st term of the consecutive terms. In particular, we have 4N−1 X i=0 P(n + i) = (a2N −b2N) √ 2 P(n + 2N), (1.8) where a = 1 + √ 2 and b = 1 − √ 2 = −1 a, and therefore (a2N −b2N)/ √ 2 is an integer.
Proof. The nth Pell Number is given by (1.7): P(n) = an −bn 2 √ 2 .
Therefore, we have 4N−1 X i=0 P(n + i) = 4N−1 X i=0 an+i −bn+i 2 √ 2 = an 2 √ 2 4N−1 X i=0 ai −bn 2 √ 2 4N−1 X i=0 bi (1.9) = an 4 (a4N −1) −bn 4 (1 −b4N) = an+2N 4 (a2N −b2N) −bn+2N 4 (a2N −b2N) = (a2N −b2N) 4 (an+2N −bn+2N) = (a2N −b2N) √ 2 P(n + 2N).
The above-mentioned excursion motivates the question: For which numbers n ∈N does the sum of n consecutive Pell numbers equal a fixed integer multiple of another Pell number?
We answer this question for the Pell, Fibonacci and other related sequences. In particular, for the Pell sequence we observe that multiples of 4 (see Theorem 1.6) and the trivial case of N = 1 are the only values of N that work. We then extend our methods to generalized Pell numbers and present a conjecture regarding when the sum of consecutive generalized Pell numbers equals a fixed integer multiple of another generalized Pell number. Additionally, we describe several interesting properties of Pell numbers using tilings of an n × 1 board with polyominoes.
3 2 Identities and Preliminary Results The following lemmas describe standard identities relating Pell, Lucas and Fi-bonacci sequences, and are used extensively in the rest of the paper. For com-pleteness, we provide the proofs.
Lemma 2.1. For any non-negative integer k the Pell numbers satisfy P(n + k) + (−1)kP(n −k) = Q(k)P(n), (2.1) where Q(k) is the kth term of the Pell-Lucas sequence given in Definition 1.2.
Proof. We proceed by induction on k, noting that the two base cases are k = 0 and k = 1. When k = 0, we have P(n + 0) + (−1)0P(n −0) = 2P(n) = Q(0)P(n).
(2.2) When k = 1, we have P(n+1)+(−1)1P(n−1) = [2P(n)+P(n−1)]−P(n−1) = 2P(n) = Q(1)P(n).
(2.3) Now, we assume that P(n + k −1) + (−1)k−1P(n −k + 1) = Q(k −1)P(n) (2.4) and P(n + k −2) + (−1)k−2P(n −k + 2) = Q(k −2)P(n).
(2.5) Using the recurrence relation (1.3), we have P(n + k) = 2P(n + k −1) + P(n + k −2).
Rearranging (2.5), we get P(n−k+2) = 2P(n−k+1)+P(n−k) = ⇒P(n−k) = P(n−k+2)−2P(n−k+1).
Thus, P(n + k) + (−1)kP(n −k) = 2P(n + k −1) + P(n + k −2) + (−1)k(P(n −k + 2) −2P(n −k + 1)).
(2.6) Rearranging the right-hand side of (2.6) yields 2(P(n+k−1)+(−1)k−1P(n−k+1))+(P(n+k−2)+(−1)k−2P(n−k+2)). (2.7) We apply the inductive hypotheses (2.4) and (2.5) along with Definition 1.2 to this expression to conclude 2Q(k −1)P(n) + Q(k −2)P(n) = (2Q(k −1) + Q(k −2))P(n) = Q(k)P(n), which completes the proof.
4 Lemma 2.2. For the Fibonacci and Lucas numbers we have 4N−1 X i=0 F(n + i) = F(2N)L(n + 2N + 1).
(2.8) Proof. We use induction and the following well-known properties of the Fi-bonacci and Lucas Numbers [Ko, §5.3, §5.8]: (i) F(n −1)F(n + 1) −F(n)2 = (−1)n.
(2.9) (ii) F(n + k) = F(n)F(k −1) + F(n + 1)F(k).
(2.10) (iii) F(n −1) + F(n + 1) = L(n).
(2.11) (iv) n−1 X i=0 F(i) + 1 = F(n + 1).
(2.12) (v) k X i=0 F(n + i) = F(n + k + 2) −F(n + 1).
(2.13) We now prove Lemma 2.2. First, begin by noting that for n = 0, we have 4N−1 X i=0 F(i) = F(4N + 1) −1 (Using (2.12)) = F(2N)F(2N + 2) + F(2N −1)F(2N + 1) −1 (Using (2.10)) = F(2N)(F(2N + 2) + F(2N)) (Using (2.9) = F(2N)L(2N + 1).
(Using (2.11)) Now, by our induction hypothesis, 4N−1 P i=0 F(m + i) = F(2N)L(m + 2N + 1) holds for all m < n + 1. We now expand 4N−1 P i=0 F(n + 1 + i) using the following manipulations: 4N−1 X i=0 F(n + 1 + i) = F(n + 4N + 2) −F(n + 2) (Using (2.13)) = F(n + 4N + 1) + F(n + 4N) −(F(n + 1) + F(n)) (Using (1.1)) = F(n + 4N + 1) −F(n + 1) + (F(n + 4N) −F(N)) (Rearranging terms) = 4N−1 X i=0 F(n + i) + 4N−1 X i=0 F(n −1 + i) (Using (2.13)) = F(2N)(L(n + 2N + 1) + L(n + 2N) (Induction hypothesis) = F(2N)L(n + 2N + 2) (Using Definition 1.3), 5 which yield the desired result.
Lemma 2.3. For the Fibonacci numbers, we have for all positive integers n φn = F(n)φ + F(n −1) where φ = 1 + √ 5 2 .
Proof. We proceed by induction on n. When n = 1, the statement of the lemma is φ = φ, which is trivially true. Similarly, n = 2 is true because φ2 = φ + 1 is true as 1+ √ 5 2 is a root of the characteristic polynomial. Thus we may assume the statement holds for all natural numbers less than k ≥3. Then φk = φk−1 + φk−2 = (F(k −1)φ + F(k −2)) + (F(k −2)φ + F(k −2)) = (F(k −1) + F(k −2))φ + (F(k −2) + F(k −3)) = F(k)φ + F(k −1), which completes the proof.
3 Some General Results One of our main goals is to determine not only when the sum of consecutive terms in a recurrence is a fixed multiple of a term of the recurrence, but further to determine which term. In this section we shall consider the following sequence.
Definition 3.1. Let r be a non-negative integer. Consider a sequence {f(n)} of non-negative integers recursively defined by f(n) := rf(n −1) + f(n −2) with initial conditions so that it is not identically zero (we call this a non-degenerate sequence).
If we set α := r + √ r2 + 4 2 and β := r − √ r2 + 4 2 then the generalized Binet formula (see [BBILMT, Le])) yields f(n) = aαn + bβn.
Theorem 3.2. Fix any integer N > 0. If there is an integer C(N) such that for every sufficiently large n there exists an integer index j(n; N) such that the following equation holds N−1 X i=0 f(n + i) = C(N) · f(j(n; N)), 6 then there is an integer k(N) such that j(n; N) = n + k(N) and k(N) ∈ N 2 , N .
Proof. Define b := αN −1 C(N)(α −1) and k(N) := logα b, with α, β and f as above.
Note that |α| > |β| and |β| < 1. Then by the generalized Binet’s formula f(n) = aαn + bβn and lim n→∞|f(n) −aαn| = 0.
This implies that for any ε > 0 there exists a natural number M such that for all n > M, |f(n) −aαn| < C(N) · ε 2N .
(3.1) We choose M sufficiently large such that |f(j(n; N)) −aαj(n;N)| < ε 2 (3.2) for all n > M.
Then aαj(n;N) − 1 C(N) N−1 X i=0 aαn+i < aαj(n;N) −f(j(n; N)) + 1 C(N) C(N) · f(j(n; N)) − N−1 X i=0 aαn+i < ε 2 + 1 C(N) N−1 X i=0 f(n + i) − N−1 X i=0 aαn+i < ε 2 + ε 2 = ε.
(3.3) We now have 1 C(N) N−1 X i=0 αn+i = αn+logα b.
7 If k(N) ̸∈N then consider m = min{n + k(N), j(n; N)}. The conditions on f imply that it is an increasing sequence, therefore j(n; N) →∞as n →∞.
Hence m →∞as n →∞. We also note that aαj(n;N) − 1 C(N) N−1 X i=0 aαn+i = |a| αm α|j(n;N)−n−k(N)| −1 ≥|a| αm α|j(n;N)−n−k(N)| −1 , (3.4) and since j(n; N) ∈N we have j(n; N) −n −k(N) ̸∈Z. Similarly, since α > 1 we have α|j(n;N)−n−k(N)| −1 > 0. Lastly, m →∞as n →∞and thus for large enough n, the left hand side of (3.4) tends to infinity: lim n→∞ aαj(n;N) − 1 C(N) N−1 X i=0 aαn+i →∞, which contradicts (3.3), implying that k(N) ∈N, and j(n; N) = n + k(N).
We now prove that k(N) ∈ N 2 , N .
We begin by noting that αN −1 C(N)(α −1) = αk(N) = ⇒ C(N) = N−1 X i=0 αi−k(N).
(3.5) Let k(N) < N 2 , then C(N) = 1 + k(N) X i=1 αi + 1 αi + N−1 X i=2k(N)+1 αi−k(N).
Now, note that the coefficient of the irrational part of N−1 P i=2k(N)+1 αi−k(N) is a positive integer. We now have αi + 1 αi = αi + (−β)i.
(3.6) Applying the binomial theorem to the above-mentioned equation gives the co-efficient of the irrational part in αi + (−β)i for i > 1 to be ⌊i−1 2 ⌋ X j=1 (r2 + 4)(ri−2j+1 + (−r)i−2j+1) ≥0, 8 which implies that C(N) is irrational, resulting in a contradiction. Therefore k(N) ≥N/2.
Now, by induction we get the following inequality: n+N−1 X i=0 f(i) < f(n + N + 1), which proves k(N) ≤N.
This implies that j(n; N) ∈ n + N 2 , n + N .
Theorem 3.3. Given a non-degenerate sequence of non-negative integers re-cursively defined by f(n) := rf(n −1) + f(n −2), where r ∈N, if 3 X i=0 f(n + i) = C(N) · f(j(n; N)) then r = 2.
Proof. From the proof of Theorem 3.2 we have C(N) = 3 P i=0 αi−k(N) for C(N) a positive integer, where 2 ≤k(N) ≤4. Therefore, the only possible values for k(N) are 2, 3, 4. We will now do casework based on the value of k(N).
Case 1: k(N) = 2.
C(N) = 1 α2 + 1 α + 1 + α = 1 + 2r2 + 4 −2r √ r2 + 4 4 + √ r2 + 4 −r 2 + r + √ r2 + 4 2 = 1 + r2 + 2 2 + 2 −r 2 p r2 + 4.
Since r is an integer and there is no Pythagorean triple with 2 as one of the terms, therefore √ r2 + 4 is irrational. Thus for C(N) to be an integer, we must have 2−r 2 = 0, therefore r = 2.
9 Case 2: k(N) = 3.
C(N) = 1 α3 + 1 α2 + 1 α + 1 = (r2 + 4) √ r2 + 4 −3r(r2 + 4) + 3r2√ r2 + 4 −r3 8 + 2r2 + 4 −2r √ r2 + 4 4 + √ r2 + 4 −r 2 = −4r3 + 4r2 −16r + 8 8 + 4r2 + 4 −4r + 4 8 p r2 + 4 = −r3 + r2 −4r + 2 2 + r2 −r + 2 2 p r2 + 4.
Since C(N) is an integer, we must have r2 −r + 2 = 0. But since this equation has no integer roots, no such r exists.
Case 3: k(N) = 4.
Then C(N) = 1 α4 + 1 α3 + 1 α2 + 1 α (3.7) = r4 −r3 + 5r2 −3r = 4 2 + −r3 + r2 −3r + 1 2 p r2 + 4.
Since C(N) is an integer, we must have −r3 + r2 −3r + 1 = 0. But this has no integer roots, so no such r exists.
Clearly, only looking at rational multiples of terms in the sequence is sufficient, because the desired multiple can be written as a ratio of integers. The following theorem proves why only looking at integer multiples of terms in the sequence is sufficient.
Theorem 3.4. Define {f(n)} by the recurrence relation f(n) := rf(n −1) + f(n −2) where r ∈N, and choose initial conditions so that f is not identically zero. Then if the sum of N > 1 consecutive terms of {f(n)} is a fixed rational constant times another term in the sequence, then the rational constant is an integer.
Proof. Let d = gcd(f(0), f(1)).
We notice that d | f(n) for all n ∈N and therefore consider the equivalent sequence h(n) := f(n)/d instead. Then gcd(h(n), h(n + 1)) = 1 10 for all n ≥1 since gcd(h(n), h(n + 1)) = gcd(h(n), rh(n) + h(n −1)) = gcd(h(n), h(n −1)) .
.
.
= gcd(h(0), h(1)).
Now suppose N−1 X i=0 f(n + i) = a b f(j(n; N)), (3.8) where a, b ∈Z and gcd(a, b) = 1. Dividing both sides by d, we get N−1 X i=0 h(n + i) = a b h(j(n; N)).
(3.9) From Theorem 3.2 we know that there exists M ∈N such that j(n; N) = n+k(N) for n > M. Applying (3.8) tells us that b | h(n+k(N)) for all n > M.
However, if b > 1 we reach the contradiction that gcd(f(n), f(n + 1)) ̸= 1 for all n ∈N. Therefore b = 1, which completes our proof.
4 Pell numbers In Theorem 1.6, we proved that the sum of 4N consecutive Pell numbers is a constant integer multiple of the (2N + 1)st term. We generalize to other related partial sums.
4.1 Sum of 4N + 2 Consecutive Terms Theorem 4.1. Let P(n) denote the nth Pell number. Fix any integer N > 0.
There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: 2N X i=0 P(n + i) = C(N)P(j(n; N)).
Proof. We note that 4N+1 X k=0 P(n + k) = n+4N+1 X k=0 P(k) − n−1 X k=0 P(k).
(4.1) 11 The following result n X k=0 P(k) = 1 2 h P(n + 1) + P(n) −1 i , (4.2) from [Br, §2, equation (2)] implies that 4N+1 X k=0 P(n+k) = 1 2 h P(n+4N +2)+P(n+4N +1)−P(n−1)−P(n) i . (4.3) We also note that P(n + 4N + 2) = P(n + 2N + 1 + (2N + 1)), (4.4) P(n + 4N + 1) = P(n + 2N + (2N + 1)), (4.5) −P(n −1) = P(n + 2N −(2N + 1)), (4.6) −P(n) = P(n + 2N + 1 −(2N + 1)).
(4.7) Applying Lemma 2.1 on expressions (4.4) and (4.7) in equation (4.2), we get P(n + 4N + 2) + (−1)2N+1P(n) = Q(2N + 1)P(n + 2N + 1).
(4.8) Furthermore, applying Lemma 2.1 on expressions (4.5) and (4.6) in equation (4.2), we get P(n + 4N + 1) + (−1)2N+1P(n −1) = Q(2N + 1)P(n + 2N), (4.9) where {Q(n)} is the companion Pell sequence given in Definition 1.2. Therefore, 4N+1 X k=0 P(n + k) = 1 2 h Q(2N + 1)P(n + 2N + 1) + Q(2N + 1)P(n + 2N) i = Q(2N + 1) P(n + 2N + 1) + P(n + 2N) 2 .
(4.10) Since Q(n) is even for all n (see Definition 1.2), we have Q2N+1 2 ∈N.
We now suppose the sum of 4k + 2 Pell numbers is equal to a constant multiple of another Pell number. Then for some t1 ≥t2 ∈N, we have the following equations: rP(t1) = Q(2N + 1) · P(n + 2N + 2) + P(n + 2N + 1) 2 (4.11) rP(t2) = Q(2N + 1) · P(n + 2N + 1) + P(n + 2N) 2 .
(4.12) Dividing the two, we get P(n + 2N + 1) + P(n + 2N + 2) P(n + 2N) + P(n + 2N + 1) = P(t1) P(t2).
(4.13) 12 Now, let Tm = P(m) + P(m −1) for m ∈N, then Tm satisfies 2Tm−1 + Tm−2 = 2[P(m −1) + P(m −2)] + [P(m −2) + P(m −3)] = 2P(m −1) + 3P(m −2) + P(m + 3) = [2P(m −1) + P(m −2)] + [2P(m −2) + P(m + 3)] = P(m) + P(m −1) = Tm, which gives us the following recurrence: Tm = 2Tm−1 + Tm−2.
(4.14) By applying induction on (4.14), we deduce 2 < Tm+1 Tm ≤3, (4.15) which implies that t1 > t2 as otherwise we would have Tm+1/Tm < 1.
We now notice that if t1 ≥t2 + 2 then P(t2 + 2) P(t2) = P(t2 + 2) P(t2 + 1) P(t2 + 1) P(t2) > 4 = ⇒ P(n + 2N + 1) + P(n + 2N + 2) P(n + 2N) + P(n + 2N + 1) > 4 = ⇒ P(n + 2N + 2) > 4P(n + 2N) + 3P(n + 2N + 1) = ⇒ 2P(n + 2N + 1) + P(n + 2N) > 4P(n + 2N) + 3P(n + 2N + 1) which leads to a contradiction. Therefore, we must have t1 = t2 + 1. We now note that P(n + 2N + 2) < P(n + 2N + 1) + P(n + 2N + 2) = 2rP(t1) Q(2N + 1) < 2P(n + 2N + 2) + P(n + 2N + 1) = P(n + 2N + 3) = ⇒ 2r Q(2N + 1) < P(n + 2N + 3) P(t1) = ⇒ 2r Q(2N + 1) ̸= 1.
(4.16) Additionally, since Tm+1 = 2Tm + Tm−1, we have gcd(Tm+1, Tm) = gcd(2Tm + Tm−1, Tm) = gcd(Tm, Tm−1).
(4.17) 13 Now, note that gcd(T1, T2) = 1. Therefore, by induction we can conclude that gcd(Tm+1, Tm) = 1. Utilizing the same argument, we deduce that gcd(P(m + 1), P(m)) = 1, but this contradicts the following statements: rP(t2 + 1) = Q(2N + 1) 2 Tn+2N+2, (4.18) rP(t2) = Q(2N + 1) 2 Tn+2N+1, (4.19) and 2r Q(2N + 1) > 1, (4.20) which are obtained by trivial substitution.
4.2 Sums of Odd Numbers of Consecutive Terms Theorem 4.2. Let P(n) denote the nth Pell number. Fix any integer N > 0.
There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: 2N X i=0 P(n + i) = C(N)P(j(n; N)).
Proof. Suppose that we can write the sum of any N consecutive Pell numbers as C(N) times a Pell number for some positive integer C(N) where N is odd.
Consider the Pell sequence modulo C(N). By using the Pigeonhole Principle and the fact that two consecutive Pell numbers uniquely determine the terms before and after them we see that {Pn,C(N)}n≥0 := {P(n) mod C(N)}n≥0 is periodic. The period is called the Pisano Period and is denoted by π(C(N)).
Notice that π(1) = 1 and π(2) = 2. Now consider C(N) > 2. Since {Pn,C(N)}n≥0 is not a constant sequence, therefore π(C(N)) ≥2, which implies that Pc+π(C(N)),C(N) = Pc,C(N).
Now since 2 1 1 0 π(C(N))+1 = P(π(C(N)) + 2) P(π(C(N)) + 1) P(π(C(N)) + 1) P(π(C(N))) , (4.21) this implies that, with I2 the 2 × 2 identity matrix, 2 1 1 0 π(C(N)) = I2 in GL2 (Z/ C(N) Z) .
(4.22) Taking the determinant, we get (−1)π(C(N)) = 1, which implies that π(C(N)) is even for all C(N) > 2.
14 Now let N be an odd number, to emphasize this we change notation and write it as 2N + 1. Suppose the sum of any 2N + 1 consecutive Pell numbers is C(N) times another Pell Number. Then 2N+1−1 X i=0 P(n + i) ≡0 (mod C(N)), for n ≥0.
Replacing n by n + 1 we get P(n + 2N + 1) ≡P(n) (mod C(N)) for all n ≥0, which implies that π(C(N)) | N. However, since π(C(N)) is even when C(N) ≥ 2, this implies that C(N) = 1. Thus the sum of any 2N + 1 consecutive Pell numbers must be equal to a Pell number. In other words, 2N+1−1 X i=0 P(n + i) = P(j(n; N)) for some integer j(n; N) and n ≥0.
Notice that when N = 0, we obtain j(n; N) = n. Now suppose 2N + 1 is an odd integer greater than 1. Then, we have 2N+1−1 X i=0 P(n + i) > P(n + 2N + 1 −1).
However, 2N+1−1 X i=0 P(n + i) < n+2N+1−1 X i=0 P(i) < P(n + 2N + 1) where the last inequality can be proven by using induction based on the value of n + 2N + 1. We conclude that P(n + 2N + 1 −1) < 2N+1−1 X i=0 P(n + i) < P(n + 2N + 1), and hence the sum of 2N +1 consecutive Pell numbers is a fixed integer multiple of a Pell Number if and only if N = 0.
We will use a proof of a similar flavor in 6.5.
5 Fibonacci Numbers We now prove similar results for the Fibonacci numbers. In particular, we show that the sum of N consecutive Fibonacci numbers is equal to a fixed constant multiple of a Fibonacci number if and only if N ≡2 (mod 4), N = 3, or N = 1.
15 5.1 Sum of 4N + 2 Consecutive Terms Theorem 5.1. Let F(n) denote the nth Fibonacci number, and L(n) denote the nth Lucas number. Fix any N > 0. The following equation 4N+1 X i=0 F(n + i) = L(2N + 1)F(n + 2N + 2), holds for all n.
Proof. The Fibonacci numbers are the solutions to F(n) = F(n −1) + F(n −2) with F(0) = 0 and F(1) = 1, while the Lucas Sequence are the solutions to the same recurrence L(n) = L(n −1) + L(n −2), but with initial conditions L(0) = 2 and L(1) = 1. It is well known (see [Ko, Theorem 5.1]) that n X i=0 F(i) = F(n + 2) −1.
A straightforward induction yields F(n+k)+(−1)kF(n−k) = L(k)F(n), and therefore 4N+1 X i=0 F(n + i) = F(n + 4N + 3) −F(n + 1) = L(2N + 1)F(n + 2N + 2), (5.1) which completes the proof.
5.2 Sum of 4N Consecutive Terms Theorem 5.2. Let F(n) denote the nth Fibonacci number. Fix any integer N > 0. There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: 4N−1 X i=0 F(n + i) = C(N)F(j(n; N)).
Proof. From Lemma 2.2 we have 4N−1 X i=0 F(n+i) = F(2N)L(n+2N +1) = F(2N)(F(n+2N)+F(n+2N +2)).
Now setting Tm = F(m) + F(m + 2) and repeating the proof of the 4N + 2 case for Pell numbers gives us the desired result.
16 5.3 Sums of Odd Numbers of Consecutive Terms We note that any Fibonacci number is one times itself and the sum of any three consecutive Fibonacci numbers is two times the third term. We prove that these are the only solutions for odd cases with the following theorem.
Theorem 5.3. Let F(n) denote the nth Fibonacci number. Fix any integer N ≥2. There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: 2N X i=0 F(n + i) = C(N) · F(j(n; N)).
Proof. The proof of Theorem 3.2, specifically, 3.5 tells us that if the sum of N consecutive Fibonacci numbers is C(N)-times another Fibonacci number, then b = (φN −1) C(N)(φ −1) = N−1 P i=0 φi C(N) = φγ for some γ ∈N.
Using an argument of a similar flavor to Section 4.2, we deduce that C(N) must either be 1 or 2.
Now, Lemma 2.3 tells us that b = φ N−1 P i=1 F(i) + N−2 P i=1 F(i) + 1 C(N) (5.2) = (F(N + 1) −1)φ + F(N) C(N) , (5.3) where C(N) is either 1 or 2. Since for n ≥3, b ≥α, we let b = αm where m ≥1.
Thus, we get (F(N + 1) −1)φ + F(N) C(N) = F(m)φ + F(m −1), which implies that F(m −1) = F(N) C(N) and F(m) = F(N + 1) −1 C(N) .
We now consider the two cases, C(N) = 1 and C(N) = 2.
Case 1: C(N) = 1.
17 If C(N) = 1, we see that if m ̸= 3 then m = N + 1 leads to a contradiction, therefore m = 3 and thus N = 1.
Case 2: C(N) = 2.
If C(N) = 2 then Carmichael’s Theorem [Ca] tells us that for n > 13, F(n) has a prime factor not present in the previous Fibonacci numbers. Therefore, we only need to check the cases where n ≤13. Checking for the smaller cases we realize that N = 3 is the only case where F(N)/2 is another Fibonacci number.
6 Generalized Pell and Fibonacci Numbers We adapt our previous results to a generalization of the Pell numbers that satisfies a (k + 1)st order recursion, where k ∈N. We also conjecture that for k > 1, the sum of N consecutive generalized Pell numbers is a fixed integer multiple of another term of the sequence if and only if N = 2k + 2. Finally, we prove similar properties for a generalization of the Fibonacci numbers.
6.1 Definition In [Ki] the authors consider the following generalization of the Pell numbers (we slightly modify their notation as we start our indexing at n = 0).
Definition 6.1. Generalized Pell (k, i)-numbers are the solutions to the follow-ing recursion with given initial conditions: P i k(n) = 2P i k(n −1) + P i k(n −k −1) (6.1) with P i k(0) = P i k(1) = · · · = P i k(i) = 0 and P i k(i + 1) = P i k(i + 2) = · · · = P i k(k) = 1 where 0 ≤i ≤k −1.
(k ∈N).
6.2 Sum of 2k + 2 Consecutive Terms Applying the following formula from [Ki, §4, Theorem 19], we get n X i=0 P k−1 k (i) = 1 2 −1 + k X i=0 P k−1 k (n −i + 1) !
(6.2) where n ≥k −1. We prove a result similar to Theorem 1.6 for the generalized Pell sequence.
Theorem 6.2. For n ≥k we have 2k+1 X i=0 P k−1 k (n + i) = 4P k−1 k (n + 2k).
(6.3) 18 Proof. Let Sn = n X i=0 P k−1 k (i).
(6.4) Note that the first k −1 terms in this sum are 0.
We proceed by induction on n, starting at n = k for the base case. Shifting indices on (6.4) for n = k gives 2k+1 X i=0 P k−1 k (k + i) = 3k+1 X i=k P k−1 k (i) = S3k+1.
Noting that the first k −1 terms of Sn are all zero, we find S3k+1 = S3k + 2P k−1 k (3k) + P k−1 k (2k) = S2k−1 + 2P k−1 k (2k) + 3k−1 X i=2k+1 P k−1 k (i) | {z } (6.5) +3P k−1 k (3k).
Now, we consider S := S2k−1 + 2P k−1 k (2k) + 3k−1 X i=2k+1 P k−1 k (i).
(6.5) Since the first k −1 terms of the sum in (6.4) are zero, S2k−1 = 2k−1 X i=0 P k−1 k (i) = 2k−1 X i=k P k−1 k (i) = P k−1 k (k) + 2k−1 X i=k+1 P k−1 k (i).
19 Then, applying recursion (6.1) to (6.5), we find S = 2k−1 X i=k+1 P k−1 k (i) + P k−1 k (2k + 1) + 3k−1 X i=2k+1 P k−1 k (i) = 2k−1 X i=k+1 P k−1 k (i) + 2P k−1 k (2k + 1) + 3k−1 X i=2k+2 P k−1 k (i) = 2k−1 X i=k+2 P k−1 k (i) + 2P k−1 k (2k + 1) + P k−1 k (k + 1) + 3k−1 X i=2k+2 P k−1 k (i) .
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= 2P k−1 k (3k −1) + P k−1 k (2k −1) = P k−1 k (3k), (6.6) where the final reduction of S results from alternatively removing terms indexed by the lower bounds of each of the summations and then applying recursion (6.1). Thus we have S3k+1 = S + 3P k−1 k (3k) = 4P k−1 k (3k), proving the base case.
Now, by the induction hypothesis we have 2k+1 X i=0 P k−1 k (n −1 + i) = 4P k−1 k (n + 2k −1), (6.7) and by (6.2) we have 2k+1 X i=0 P k−1 k (n −1 + i) = n+2k X i=0 P k−1 k (i) − n−2 X i=0 P k−1 k (i) = 1 2 k X i=0 P k−1 k (n + 2k + 1 −i) − k X i=0 P k−1 k (n −1 −i) !
.
(6.8) Combining (6.7) and (6.8) we get k X i=0 P k−1 k (n + 2k + 1 −i) − k X i=0 P k−1 k (n −1 −i) = 8P k−1 k (n + 2k −1). (6.9) 20 Furthermore, we have 2k+1 X i=0 P k−1 k (n + i) = 1 2 k X i=0 P k−1 k (n + 2k + 2 −i) − k X i=0 P k−1 k (n −i) !
= 1 2 P k−1 k (n + 2k + 2) −P k−1 k (n + k + 1) + 8P k−1 k (n + 2k −1) + P k−1 k (n −k −1) −P k−1 k (n) = 1 2 2P k−1 k (n + 2k + 1) + 8P k−1 k (n + 2k −1) −2P k−1 k (n −1) = 1 2 2P k−1 k (n + 2k + 1) −2P k−1 k (n + k) + 2P k−1 k (n + k) −2P k−1 k (n −1) + 8P k−1 k (n + 2k −1) = 1 2 4P k−1 k (n + 2k) + 4P k−1 k (n + k −1) + 8P k−1 k (n + 2k −1) = 4P k−1 k (n + 2k).
(6.10) We obtain the last equation by using P k−1 k (n + k −1) + 2P k−1 k (n + 2k −1) = P k−1 k (n + 2k).
(6.11) Now, consider 2k+1 X i=0 P k−1 k (n+i) = 1 2 k X i=0 P k−1 k (n + 2k + 2 −i) | {z } s1 (6.13) − k X i=0 P k−1 k (n −i) | {z } s2 (6.14) . (6.12) We have the following explicit formulas for s1 and s2: s1 = P k−1 k (n + 2k + 2) −P k−1 k (n + k + 1) + k X i=0 P k−1 k (n + 2k + 1 −i), (6.13) s2 = P k−1 k (n) −P k−1 k (n −k −1) + k X i=0 P k−1 k (n −i −1).
(6.14) We now note that the RHS of 6.13 and 6.14 are particularly amenable to ma-nipulation, and therefore turn our attention towards 1 2 (s1 −s2).
21 Thus, we have 1 2(s1 −s2) = 1 2 P k−1 k (n + 2k + 2) −P k−1 k (n + k + 1) + 8P k−1 k (n + 2k −1) + P k−1 k (n −k −1) −P k−1 k (n) = 1 2 2P k−1 k (n + 2k + 1) + 8P k−1 k (n + 2k −1) −2P k−1 k (n −1) = 1 2 2P k−1 k (n + 2k + 1) −2P k−1 k (n + k) + 2P k−1 k (n + k) −2P k−1 k (n −1) + 8P k−1 k (n + 2k −1) = 1 2 4P k−1 k (n + 2k) + 4P k−1 k (n + k −1) + 8P k−1 k (n + 2k −1) = 4P k−1 k (n + 2k).
(6.15) Setting k = 1 in Theorem 6.3, we obtain the following corollary.
Corollary 6.3. The sum of any four consecutive Pell numbers is a four times the third Pell number: 3 X i=0 P(n + i) = 4P(n + 2).
(6.16) The partial sum formula of Pell numbers along with the identity P(n + k) + (−1)kP(n −k) = Q(k)P(n), k ∈N ∪{0} (6.17) proven in Lemma 2.1, can be used to give an alternate proof of Theorem 1.6.
Although we have a similar partial sum formula for P k−1 k , there is no obvious way to extend this partial sum to a general property of adding consecutive generalized Pell numbers to get a multiple of another generalized Pell number for arbitrary k > 1. For k > 1, we haven’t been able to find N ̸= 2k+2 such that the sum of N consecutive generalized Pell-(k, i) numbers is an integer multiple of another generalized Pell-(k, i), suggesting the following conjecture.
Conjecture 1. Fix any integer N > 0. There exists an integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds N X i=0 P i k(n + i) = C(N) · P i k (j(n; N)) if and only if N = 2k + 2.
Theorem 6.4. For 0 ≤i ≤k −1 we have 2k+1 X j=0 P i k(n + j) = 4P i k(n + 2k).
(6.18) 22 Proof. In [Ki, §2, Corollary 2] they prove for n > k that P k−1−j k (n) = P k−1 k (n) + j−1 X i=0 P k−1 k (n −k + i).
(6.19) This along with Theorem 6.3 gives the result.
6.3 Sum of Odd Number of Consecutive Terms The same argument given for the classical Pell numbers (4.2) can be generalized for the sequence {P k−1 k (n)}n≥0 where k is an odd natural number.
This is because from [Ki, §2,Theorem 2] we have the following equality: 2 0 · · · 0 1 1 0 · · · 0 0 0 1 · · · 0 0 .
.
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.
...
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0 0 · · · 1 0 n = P k−1 k (n + p + 1) P k−1 k (n + 1) · · · P k−1 k (n + p −1) P k−1 k (n + p) P k−1 k (n + p) P k−1 k (n) · · · P k−1 k (n + p −2) P k−1 k (n + p −1) P k−1 k (n + p −1) P k−1 k (n −1) · · · P k−1 k (n + p −3) P k−1 k (n + p −2) .
.
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.
...
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P k−1 k (n + 1) P k−1 k (n −p + 1) · · · P k−1 k (n −1) P k−1 k (n) .
Since det 2 0 . . .
0 1 1 0 . . .
0 0 0 1 . . .
0 0 .
.
.
.
.
.
. . .
.
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0 0 . . .
1 0 = (−1)k = −1, we obtain the following theorem.
Theorem 6.5. Fix any integer N > 0. There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: 2N X i=0 P k−1 k (n + i) = C(N) · P k−1 k (j(n; N)).
Proof. The proof follows from the discussion above and from the proof of the odd case for the classical Pell numbers.
We now prove the following stronger theorem for the generalized Pell sequence.
23 Theorem 6.6. Fix any odd integer N > 0, and even integer k ≥0. Suppose that there exists an integer C(N) such that for every n there exists an integer index j(n; N; k) such that the following equation holds: N−1 X i=0 P k−1 k (n + i) = C(N)P k−1 k (j(n; N; k)).
Then, i) 2 ∤C(N) and ii) N > 2k + 2.
Proof.
i) By induction we know that an : = P k−1 k (n) (mod 2) is of the form an = 1, if (k + 1) | (n + 1) 0, otherwise.
Let N = q(2k + 2) + r. Since N is odd, we have 1 ≤r ≤2k + 1. Now take any n such that n ≥k + 1 and (k + 1) | n. We now prove that 2 ∤C(N).
Define Sn,N := q(2k+2) P i=−r+1 P k−1 k (n + i) if r ≤k + 1, Sn,N := N−1 P i=0 P k−1 k (n + i) if r > k + 1.
Using the explicit form of an we conclude that Sn,N is odd in both cases, and therefore C(N) must be odd.
ii) By the same argument, we know that π(p) | N where p is any prime dividing C(N).
Similarly, using the previous argument, we also know that p > 3. Now since P k−1 k (0) = P k−1 k (1) = · · · = P k−1 k (k −1) = 0, and P k−1 k (k) = 1, we must have P k−1 k (π(p)) ≡P k−1 k (π(p) + 1) ≡· · · ≡P k−1 k (k −1) ≡0 (mod p).
However, we know that P k−1 k (k + i) = 2i for 1 ≤i ≤k, and since k ≥2, we have P k−1 k (2k + 1) = 2k+1 + 1 P k−1 k (2k + 2) = 2k+2 + 4 P k−1 k (2k + 3) = 2k+3 + 12.
24 Now as p > 2, we know that π(p) > 2k. We notice that if π(p) = 2k + 1, then p | 2k+1 + 1 and p | 2k+2 + 4 which implies p | 2: a contradiction.
Similarly, if p = 2k + 2, then p | 2k+2 + 4 and p | 2k+3 + 12 which implies p | 4 which is also a contradiction. Therefore, π(p) > 2k + 2 which implies N > 2k + 2.
6.4 Tilings and Generalized Pell Sequence In [BSP] the authors proved certain properties related to the Pell numbers using tilings of an n × 1 board. We generalized some of the properties for P k−1 k (n).
Let us first define a sequence (pk,n)n≥0 such that pk,n := P k−1 k (n + k).
(6.20) It is not difficult to see that pk,n counts the number of tilings of an n × 1 board using black 1 × 1 squares, white 1 × 1 squares and grey (k + 1) × 1 polyominoes.
Theorem 6.7. We have pk,(k+1)n+r+1 = 2 n X m=0 pk,m(k+1)+r, 0 ≤r < k 2 n X m=0 pk,m(k+1)+r + 1, r = k.
(6.21) Proof. Firstly, assume that r < k. Now, consider the tiling of a [(k + 1)n + r + 1] × 1 board, with the cells on the board numbered from left to right 1 to (k +1)n+r +1. Let t be the location of the last 1×1 cell in the tiling. Black or white squares cannot cover the cells to the right of t, so they must be covered by (k + 1) × 1 polyominoes. Therefore, t is of the form (k + 1)m + r + 1. In this case, the number of tilings of the board is 2pk,mk (accounting for the fact that cell t can be covered by either black or white 1×1 squares), proving the identity.
Now, let us assume that r = k. We can still cover the board with black and white squares as well as grey polyominoes as we discussed in the previous case, yielding 2 n P m=0 pk,m(k+1)+r tilings of the board. However, since the length of the board is now (k + 1)(n + 1), it is possible the board can be covered without black and white squares altogether. We add this new case to the total number of tilings, proving the second identity.
Note that (6.2) also follows from this result. An alternate proof using matrices is given in [Ki, §4, Theorem 19], which can be generalized further.
25 Definition 6.8. Define the sequence {fk(n)} as follows: fk(n) := afk(n −1) + bfk(n −k −1) a, b ∈N with fk(0) = fk(1) = · · · = fk(k −1) = 0 and fk(k) = 1 where k ∈N.
Definition 6.9. Define the sequence {pk,n} as follows: pk,n := fk(n + k) for n ∈N ∪{0}.
Theorem 6.10.
pk,(k+1)n+r+1 = a n P m=0 bn−mpk,m(k+1)+r, 0 ≤r < k a n P m=0 bn−mpk,m(k+1)+r + 1, r = k (6.22) Proof. Analogous to the proof of Theorem 6.7.
6.5 Generalized Fibonacci sequence Define the order-k generalized Fibonacci sequence by fk(n) := k X i=1 fk(n −i) (6.23) with fk(1) = fk(2) = · · · = fk(k −1) = 0 and fk(k) = 1.
Its generating matrix (see [KiTa]) is given by 1 1 . . .
1 1 1 0 . . .
0 0 0 1 . . .
0 0 .
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. . .
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0 0 . . .
1 0 .
(6.24) A similar argument to the generalized Pell case in Section 6.3 tells that the Pisano Period for fk(n) is even modulo n whenever n > 2 and k is even, and yields the following Theorem.
Theorem 6.11. Let Fk(n) denote the nth order-k Fibonacci number where k is even. Fix any N > 0. There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: 2N X i=0 Fk(n + i) = C(N) · Fk( j(n; N) ).
26 Proof. Let the sum of any 2N + 1 consecutive terms in the kth Fibonacci se-quence be C(N) times another integer in the Fibonacci Sequence. An argument similar to Theorem 4.2 rules out the cases C(N) > 2. Therefore, we just need to take care of the cases when C(N) = 1, 2.
Case 1: C(N) = 1.
By induction on r > 2k + 1, we have r X n=0 fk(n) < fk(r + 2), (6.25) which implies 2N X i=0 fk(n + i) ≤ n+2N X i=0 fk(i) < fk(n + 2N + 2).
From the definition of the order-k generalized Fibonacci sequence, for 2N > k+1 we have fk(n + 2N + 1) < 2N X i=0 fk(n + i), and thus C(N) ̸= 1.
Case 2: C(N) = 2.
Define λk := lim n→∞ fk(n + 1) fk(n) .
We now note that [KuSi, §11, Theorem 9] states λk + λ−k k = 2, which implies that λk < 2 and hence for n > 2k + 1 we have fk(n + 1)/fk(n) < 2. Applying (6.25) thus implies r X n=0 fk(n) < fk(r + 2) < 2fk(r + 1).
(6.26) Now since, 2N > k + 1, from the definition of the order-k generalized Fibonacci sequence we have fk(n + 2N) < 2N−1 X i=0 fk(n + i) = ⇒ 2fk(n + 2N) < 2N X i=0 fk(n + i).
Lastly, we have 2fk(n + 2N) < 2N X i=0 fk(n + i) < n+2N X i=0 fk(i) < 2fk(n + 2N + 1), (6.27) which implies C(N) ̸= 2, completing the proof.
27 7 Other Second-Order Recurrence Relations Below we generalize some of the earlier results and proofs to other second-order recurrence relations. We prove novel results regarding the partial sums of consecutive terms of generalized Pell-like second-order recursive sequences.
7.1 Sum of 4N + 2 Consecutive Terms We again consider sequences f(n) satisfying the recurrence relation f(n) := rf(n −1) + f(n −2), but now we choose f(0) = 0 and f(1) = 1 and r ≥2.
Theorem 7.1. Let f be as above, and fix any integer N > 0. There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: 4N+1 X i=0 f(n + i) = C(N)f( j(n; N) ).
Proof. Define the sequence g(n) by g(0) = 2, g(1) = r and g(n) := rg(n −1) + g(n −2).
(7.1) Using induction on k, we find f(n + k) + (−1)kf(n −k) = g(k)f(n).
(7.2) Using (6.22) we get n X k=0 f(n) = 1 r f(n) + f(n + 1) −1 .
(7.3) Therefore, we have 4N+1 X k=0 f(n + k) = n+4N+1 X k=0 f(k) − n−1 X k=0 f(k) = 1 r h f(n + 4N + 1) + f(n + 4N + 2) −f(n −1) + f(n) i = 1 r h g(2N + 1)f(n + 2N + 1) + g(2N + 1)f(n + 2N) i = g(2N + 1)f(n + 2N + 1) + f(n + 2N) r .
Further by induction on N we know that g(2N + 1) r ∈N.
28 Now suppose the sum of 4N +2 terms is a fixed multiple of another term. Then for some t1 ≥t2 ∈N, the following equations hold: sf(t1) = g(2N + 1)f(n + 2N + 2) + f(n + 2N + 1) r sf(t2) = g(2N + 1)f(n + 2N + 1) + f(n + 2N) r .
(7.4) Dividing both sides yields f(n + 2N + 1) + f(n + 2N + 2) f(n + 2N) + f(n + 2N + 1) = f(t1) f(t2).
(7.5) Now for m a positive integer let Tm := f(m) + f(m −1).
Then Tm satisfies the following recurrence: rTm−1 + Tm−2 = rf(m −1) + rf(m −2) + f(m −2) + f(m −3) = f(m) + f(m −1) = Tm.
Using this recursion and induction it follows that r < Tm+1 Tm ≤3r 2 , (7.6) which implies that t1 > t2. If t1 ≥t2 + 2 then f(t2 + 2) f(t2) = rf(t2 + 1) + f(t2) f(t2) = r2f(t2) + rf(t2 −1) f(t2) + 1 > r2 + 1, (7.7) which leads to a contradiction as r ≥2 = ⇒r2 > 3r/2. Thus we must have t1 = t2 + 1.
Now we know that f(n + 2N + 2) < fn+2N+1 + fn+2N+2 = rsft1 g2N+1 < fn+2N+3, (7.8) and therefore c = rs/g2N+1 cannot possibly equal 1. Lastly, we note that gcd(Tm+1, Tm) = gcd(rTm + Tm−1, Tm) = gcd(Tm, Tm−1).
(7.9) 29 Applying induction proves that this gcd is 1. The same argument shows that gcd(fm+1, fm) = 1, but this contradicts the following statements: sf(t2 + 1) = g(2N + 1) r Tn+2N+2, (7.10) sf(t2) = g(2N + 1) r Tn+2N+1 (7.11) and c > 1, (7.12) which completes our proof.
Note that in the above proof the result does not hold for r = 1, which is the Fibonacci sequence (see Theorem 5.2). Recall that the sum of 4N +2 consecutive Fibonacci numbers is a fixed integer multiple of another Fibonacci number (see Theorem 5.1).
7.2 Sum of Odd Number of Consecutive Terms Let a be any integer, x, y ≥0 and y be odd. Define {f(n)} to be the sequence following the recurrence relation f(n) := 2af(n −1) + f(n −2) with f(0) = 2x, f(1) = y and 4axy + 4x2 −y2 = −1.
Theorem 7.2. Let f be as above and fix any N > 0.
There is no integer C(N) such that for every n there exists an integer index j(n; N) such that the following equation holds: N−1 X i=0 f(n + i) = C(N)f( j(n; N) ).
Proof. Assume that N−1 X i=0 f(n + i) = C(N)f(j(n; N)) where C(N) > 1. We now prove the following lemma, which allows us to look at the distribution of residues modulo primes.
Lemma 7.3. There exists no prime p such that p | f(n) for all n ≥0.
Proof. We prove the lemma by contradiction, and divide the proof into two cases: p = 2 and p > 2.
Case I: p = 2.
Since p = 2, and p | f(n) for all n ≥0, this implies that p | f(1) = y, which is odd, which immediately leads to a contradiction. Hence, p ̸= 2.
30 Case II: p > 2.
Since p | 2x and p | y, this implies that x, y ≡0 (mod p).
Let x = k1p and y = k2p for some k1, k2 ∈N2. We now employ the following manipulations: 4axy + 4x2 = y2 −1 = ⇒4a(k1p)(k2p) + 4(k1p)2 = (k2p)2 −1 = ⇒4ak1k2p2 + 4k2 1p2 = k2 2p2 −1 = ⇒p2(4ak1k2 + 4k2 1) = k2 2p2 −1 = ⇒4ak1k2 + 4k2 1 = k2 2 −1 p2 .
Note that the LHS is an integer, whereas the RHS is not. This results in a contradiction, which completes our proof.
Now, an argument of similar flavor to Theorem 4.2 tells us that the sequence is periodic modulo n for any natural number n with a period π(n) ≥2. Therefore, if we can prove π(n) is even then the rest of the argument in Theorem 4.2 also follows. The given condition implies π(2) = 2, therefore, let us assume n > 2.
Define h(n) := 2ah(n −1) + h(n −2) with h(0) = 0, h(1) = 1.
(7.13) Furthermore, let πh(n) be the period of h(n) modulo n. A simple proof by induction yields 2a 1 1 0 n+1 = h(n + 2) h(n + 1) h(n + 1) h(n) , and then the same argument as for the Pell numbers (Theorem 4.2) gives 2a 1 1 0 πh(n) = In mod n.
(7.14) Also, induction on n gives 2a 1 1 0 n 2ay + 2x y y 2x = f(n + 2) f(n + 1) f(n + 1) f(n) , and det 2ay + 2x y y 2x = 4axy + 4x2 −y2 = −1 ̸= 0.
We note that 4axy +4x2 −y2 = ±1, but since y is odd, we can write y = 2k +1, and quickly realize that 4axy + 4x2 = (2k + 1)2 + 1 has no solutions. Therefore, we only consider 4axy + 4x2 −y2 = −1.
31 We now have 2a 1 1 0 π(n) 2ay + 2x y y 2x = 2ay + 2x y y 2x = ⇒ 2a 1 1 0 π(n) = In over Z/nZ.
(7.15) Now (7.14) and (7.15) imply that π(n) = πh(n). Similarly, Theorem 4.2 tells us that πh(n) is even. Lastly we note that when C(N) = 1 then x, y ≥0 means that the terms of the sequence are non-negative, which leads to the following inequality via induction when y is odd.
f(n + N −1) < N−1 X i=0 f(n + i) < f(n + N).
(7.16) This results in a contradiction for N > 1, which completes our proof.
A Appendix The computational experiments for the paper were carried out in the Wolfram and Python Languages. The GitHub repository can be accessed from
We thank the 2023 Polymath Jr REU for creating the opportunity for this work, Stephanie Reyes for numerous comments throughout the research and for nu-merous technical conversations regarding the paper. This work was partially supported by NSF Grant DMS2313292.
References [BBILMT] O. Beckwith, A. Bower, L. Gaudet, R. Insoft, S. Li, S. J. Miller and P. Tosteson, The Average Gap Distribution for Generalized Zeck-endorf Decompositions, the Fibonacci Quarterly 51 (2013), 13–27.
[BSP] A. Benjamin, S. Plott and J. Sellers, Tiling Proofs of Recent Sum Identities Involving Pell numbers, Annals of Combinatorics 12 (2008), 271–278.
[Br] B. Bradie, Extensions and Refinements of Some Properties of Sums Involving Pell numbers, Missouri J. Math. Sci. 22 (2010), no. 1, 37–43,
32 [Ca] R. D. Carmichael, On the Numerical Factors of the Arithmetic Forms αn±βn, Annals of Mathematics 15 (1913–1914), no. 1, 30–48.
[Ki] E. Kilic, The generalized Pell (p, i)-numbers and their generalized Binet formulas, combinatorial representations, sums, Chaos, Soli-tons & Fractals 40 (2009), no. 4, 2047–2063.
1016/j.chaos.2007.09.081.
[KiTa] E. Kilic and D. Tasci, On the generalized Order-k Fibonacci and Lucas Numbers, Rocky Mountain Journal of Mathematics 36 (2006), no. 6, 1915–1926.
[Ko] T. Koshy, Fibonacci and Lucas Numbers with Applications, 2nd Edi-tion, Johm Wiley & Sons,Inc., 2017.
[KuSi] A. D. Kumar and R. Sivaraman, Analysis of Limiting Ratios of Special Sequences, Mathematics and Statistics 10 (2022), no. 4, 825– 832,
[Le] C. Levesque, On mth Order Linear Recurrences, Fibonacci Quar-terly 23 (1985), no.
4, 290–293, Scanned/23-4/levesque.pdf.
33 |
8320 | https://www.quora.com/What-do-cos-sin-tan-and-log-mean | What do cos, sin, tan, and log mean? - Quora
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The terms cos, sin, tan, and log refer to fundamental mathematical functions used in trigonometry and logarithms. Here's a brief overview of each:
Trigonometric Functions
Cosine (cos):
The cosine function relates the angle of a right triangle to the ratio of the length of the adjacent side to the hypotenuse.
Mathematically, for an angle [math]\theta[/math]:
[math]\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}[/math]
2. Sine (sin):
The sine function relates the angle to the ratio of the length of the opposite side to the hypotenuse.
Mathematically:
[math]\sin(\theta) = \frac{\text{Opposit[/math]
Continue Reading
The terms cos, sin, tan, and log refer to fundamental mathematical functions used in trigonometry and logarithms. Here's a brief overview of each:
Trigonometric Functions
Cosine (cos):
The cosine function relates the angle of a right triangle to the ratio of the length of the adjacent side to the hypotenuse.
Mathematically, for an angle [math]\theta[/math]:
[math]\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}[/math]
2. Sine (sin):
The sine function relates the angle to the ratio of the length of the opposite side to the hypotenuse.
Mathematically:
[math]\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}[/math]
3. Tangent (tan):
The tangent function relates the angle to the ratio of the length of the opposite side to the length of the adjacent side.
Mathematically:
[math]\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sin(\theta)}{\cos(\theta)}[/math]
Logarithmic Function
Logarithm (log):
The logarithm function is the inverse of exponentiation. It answers the question: "To what exponent must a base be raised, to produce a given number?"
For example, if [math]b^y = x[/math], then [math]y = \log_b(x)[/math], where [math]b[/math] is the base of the logarithm.
Common bases include:
Base 10: [math]\log_{10}(x)[/math] or simply [math]\log(x)[/math]
Base [math]e[/math] (where [math]e \approx 2.718[/math]): [math]\ln(x)[/math] (natural logarithm)
These functions are widely used in various fields such as physics, engineering, and computer science for solving problems involving angles, oscillations, and exponential growth or decay.
Upvote ·
9 1
Brandon R.
J.D. from Western Michigan University (Graduated 2007) · Author has 22.2K answers and 108.8M answer views
·8y
cos refers to the cosine function.
sin refers to the sine function.
tan refers to the tangent function.
To begin, these are trigonometric functions. Each relates an angle, [math]x[/math], to the sides of a right triangle.
[math]cos(x) = \frac{A}{H} = \frac{4}{5}[/math]
[math]sin(x) = \frac{O}{H} = \frac{3}{5}[/math]
[math]tan(x) = \frac{O}{A} = \frac{sin(x)}{cos(x)} = \frac{3}{4}[/math]
Many trigonometry students use the phrase, SOHCAHTOA (say “so-ka-toe-uh”) to keep it straight.
log refers to the logarithm function. Typically, the assumed base is 10, and it can also be rewritten as [math]log_{10}[/math].
You’ve probably seen things like [math]10^3 = 1000[/math]. This is an “exp
Continue Reading
cos refers to the cosine function.
sin refers to the sine function.
tan refers to the tangent function.
To begin, these are trigonometric functions. Each relates an angle, [math]x[/math], to the sides of a right triangle.
[math]cos(x) = \frac{A}{H} = \frac{4}{5}[/math]
[math]sin(x) = \frac{O}{H} = \frac{3}{5}[/math]
[math]tan(x) = \frac{O}{A} = \frac{sin(x)}{cos(x)} = \frac{3}{4}[/math]
Many trigonometry students use the phrase, SOHCAHTOA (say “so-ka-toe-uh”) to keep it straight.
log refers to the logarithm function. Typically, the assumed base is 10, and it can also be rewritten as [math]log_{10}[/math].
You’ve probably seen things like [math]10^3 = 1000[/math]. This is an “exponential function”. It is some number, the base, raised to some exponent.
Naturally, in math, when we do something—like raise a number to an exponent—we want to know how to “undo” that process. What is the inverse function?
You can raise a number to an exponent. And you take the logarithm of the result to “undo” it. They are inverses.
Addition is the [functional] inverse of subtraction.
Multiplication is the inverse of division.
Exponentiation is the inverse of taking a logarithm.
For example, [math]log(1000) = log(10^3) = 3[/math].
It returns the exponent to which the base is raised.
So, more generally, [math]log_a( a^b ) = b[/math].
The other commonly used logarithm is the natural logarithm, which is represented by [math]ln(x)[/math]. (That’s L-N—think of it as ”log, natural”).
[math]ln(x) = log_e(x)[/math], where [math]e[/math] is the well known constant that is [math]\approx 2.718...[/math]
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Manish Tiwari
IMO Scholar · Author has 93 answers and 269.9K answer views
·8y
Sin,cos and tan are trigonometric functions whereas logarithm is the inverse operation to exponentiation. That means the logarithm of a number is the exponent to which another fixed number, the base, must be raised to produce that number. In simple cases the logarithm counts factors in multiplication. For example, the base 10logarithm of 1000 is 3, as 10 to the power 3 is 1000 (1000 = 10 × 10 × 10 = 10^3); 10 is used as a factor three times
Hope it helps.
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Related questions
More answers below
How do you know when to use cos, sin, tan in trigonometry?
In trigonometry, how do you know when to use sin, cos, or tan?
What is sin, cos and tan in math?
Why do sin, cos, tan cannot be more than 1?
What are cos, tan and sin?
Mostafa Mahmoud
Works at Students
·8y
Sin is the abbreviation of Sine , which is - in one intuitive definition , as it has a number of other ones - the ratio of the opposite side ( opposite to an angle of measure equal to the input of the function ) to the hypotenuse side in a right angled triangle.
Cos is the abbreviation to cosine , which the similar to the sine but the ratio is between the adjacent side and the hypotenuse instead .
and the Tan is an abbreviation to the tangent , which is Sin divided by cosine , or the the ratio of the opposite side to the adjacent side .
Log is an abbreviation to the logarithm . a log has a base .
Continue Reading
Sin is the abbreviation of Sine , which is - in one intuitive definition , as it has a number of other ones - the ratio of the opposite side ( opposite to an angle of measure equal to the input of the function ) to the hypotenuse side in a right angled triangle.
Cos is the abbreviation to cosine , which the similar to the sine but the ratio is between the adjacent side and the hypotenuse instead .
and the Tan is an abbreviation to the tangent , which is Sin divided by cosine , or the the ratio of the opposite side to the adjacent side .
Log is an abbreviation to the logarithm . a log has a base . the output that the log function gives is the exponent that we must put on the base so that it equals the input . for example , log(8) to the base 2 is 3 , because 3 is the number that , when 2 is raised to it , we get 8 . you might wonder , if this the first time you know about logarithm , why we need such a complex and unnatural operation , but you will be surprised to know that it has a huge number of applications , from simplification of complex arithmetic to information theory ( on of the theoretical pillars of our smart phones and computers ) , physics , geology , chemistry and a whole lot of other fields that either uses it directly or builds on fields that does .
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8321 | https://brilliant.org/wiki/simson-line-theorem/ | Simson Line Theorem | Brilliant Math & Science Wiki
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Hua Zhi Vee, Anshaj Shukla, Shreyansh Mukhopadhyay, and
1 other
Jimin Khim
contributed
This wiki is incomplete.
In Euclidean geometry, the Simson line theorem states that the three closest points from a point P P P on the circumcircle on △A B C\triangle ABC△A BC to sides (or the extensions of) A B AB A B, A C, AC, A C, and B C BC BC are collinear.
The converse is also true: if the 3 3 3 closest points from the three sides of a triangle to a point P P P is collinear, then P P P lies on the circumcircle of the triangle.
Contents
Theorem Statement
Proof
See Also
Theorem Statement
Let P P P be a point on the circumcircle of △A B C. \triangle ABC. △A BC. Points D,E, D, E,D,E, and F F F are the feet of the perpendicular bisectors from P P P to sides (or the extensions of) A B AB A B, A C, AC, A C, and B C, BC, BC, respectively. Then D,E, D, E,D,E, and F F F are collinear.
Simson Line
The trilinear equation of the Simson line for a point p:q:r p:q:r p:q:r lying on the circumcircle satisfies c p q+b p r+a q r=0 cpq+bpr+aqr=0 c pq+b p r+a q r=0.
The Simson line bisects the line H P HP H P, where H H H is the orthocenter.
Moreover, the midpoint of H P HP H P lies on the nine-point circle.
The Simson lines of two opposite point on the circumcircle of a triangle are perpendicular and meet on the nine-point circle.
Proof
See Also
Pedal Triangle
Cite as: Simson Line Theorem. Brilliant.org. Retrieved 03:11, September 29, 2025, from
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8322 | https://www.mathway.com/popular-problems/Algebra/960833 | Find the LCM x^2-9 , 3x+9 | Mathway
Enter a problem...
[x]
Algebra Examples
Popular Problems
Algebra
Find the LCM x^2-9 , 3x+9
x 2−9 x 2-9 , 3 x+9 3 x+9
Step 1
Factor x 2−9 x 2-9.
Tap for more steps...
Step 1.1
Rewrite 9 9 as 3 2 3 2.
x 2−3 2 x 2-3 2
Step 1.2
Since both terms are perfect squares, factor using the difference of squares formula, a 2−b 2=(a+b)(a−b)a 2-b 2=(a+b)(a-b) where a=x a=x and b=3 b=3.
(x+3)(x−3)(x+3)(x-3)
(x+3)(x−3)(x+3)(x-3)
Step 2
Factor 3 3 out of 3 x+9 3 x+9.
Tap for more steps...
Step 2.1
Factor 3 3 out of 3 x 3 x.
3(x)+9 3(x)+9
Step 2.2
Factor 3 3 out of 9 9.
3 x+3⋅3 3 x+3⋅3
Step 2.3
Factor 3 3 out of 3 x+3⋅3 3 x+3⋅3.
3(x+3)3(x+3)
3(x+3)3(x+3)
Step 3
The LCM is the smallest positive number that all of the numbers divide into evenly.
List the prime factors of each number.
Multiply each factor the greatest number of times it occurs in either number.
Step 4
The number 1 1 is not a prime number because it only has one positive factor, which is itself.
Not prime
Step 5
Since 3 3 has no factors besides 1 1 and 3 3.
3 3 is a prime number
Step 6
The LCM of 1,3 1,3 is the result of multiplying all prime factors the greatest number of times they occur in either number.
3 3
Step 7
The factor for x+3 x+3 is x+3 x+3 itself.
(x+3)=x+3(x+3)=x+3
(x+3)(x+3) occurs 1 1 time.
Step 8
The factor for x−3 x-3 is x−3 x-3 itself.
(x−3)=x−3(x-3)=x-3
(x−3)(x-3) occurs 1 1 time.
Step 9
The factor for x+3 x+3 is x+3 x+3 itself.
(x+3)=x+3(x+3)=x+3
(x+3)(x+3) occurs 1 1 time.
Step 10
The LCM of x+3,x−3,x+3 x+3,x-3,x+3 is the result of multiplying all factors the greatest number of times they occur in either term.
(x+3)(x−3)(x+3)(x-3)
Step 11
The Least Common Multiple LCM LCM of some numbers is the smallest number that the numbers are factors of.
3(x+3)(x−3)3(x+3)(x-3)
x 2−9,3 x+9 x 2-9,3 x+9
=y=y 9 y 9 y 9 x 2 9 x 2
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⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫x d x]
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8323 | https://www.homeschoolmath.net/worksheets/decimal_multiplication.php | | | |
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| You are here: Home → Worksheets → Decimal multiplicationDecimal multiplication worksheets The worksheets provide calculation practice for decimal multiplication (both mental math and multiplication algorithm). They are meant for 5th - 6th grades. Jump to: Decimal multiplication worksheets — mental math Multiply decimals by powers of ten Long multiplication of decimals The worksheets are randomly generated, so you can get a new, different one just by hitting the refresh button on your browser (or F5). You can print them directly from your browser window, but first check how it looks like in the Print Preview. If the worksheet does not fit the page in the Print Preview, adjust your margins, header, and footer in the Page Setup settings of the browser. Or, adjust the "scale" to 90% or less in the Print Preview. Some browsers may have "Print to fit" option, which will automatically scale the worksheet to be so small as to fit the printable area. Copying permission: you can freely print and copy unlimited copies of the worksheets for use in the classroom, home, tutoring center—anywhere you might be teaching. If you want to distribute the links or worksheets on a website or publication, please contact us. Decimal multiplication worksheets — mental math Multiply a whole number and a decimal - easy (one decimal digit) Multiply a whole number and a decimal - harder (one decimal digit) Multiply a whole number and a decimal - missing factor (one decimal digit) Multiply a whole number and a decimal (1-2 decimal digits) Multiply a whole number and a decimal - missing factor (1-2 decimal digits) Multiply a whole number and a decimal (1-3 decimal digits) Multiply a whole number and a decimal - missing factor (1-3 decimal digits) Multiply decimals by decimals Multiply decimals by decimals - missing factor Multiply a decimal by decimal mentally (up to 2 decimal digits times 2 decimal digits) Multiply a decimal by decimal mentally (up to 3 decimal digits times 3 decimal digits) Missing factor problems 1 (decimal by decimal, 1 or 2 decimals) Missing factor problems 2 (decimal by decimal, 1-3 decimals) Multiply decimals by decimals or whole numbers (mixed practice) Multiply decimals by decimals or whole numbers - missing factor (mixed practice) Worksheets for multiplying decimals by powers of tenMultiply by 10 or 100 (1-2 decimal digits) Multiply by 10, 100, or 1000 (1-2 decimal digits) Multiply by 10, 100, or 1000 - missing factor (1-2 decimal digits) Multiply by 10 or 100 (1-3 decimal digits) Multiply by 10, 100, or 1000 (1-3 decimal digits) Like above, but missing factor Multiply by 10, 100, 1000, 10000, or 100000 (1-3 decimal digits) Multiply decimals by 10, 100, or 1000 - missing factor (1-3 decimal digits) Multiply decimals by 10, 100, 1000, 10000, or 100,000 (1-5 decimal digits) Like above but missing factor See also my free lesson Multiply and Divide Decimals by 10, 100, and 1000 (powers of ten) Worksheets for long multiplication of decimals (multiplication algorithm)Multiply a decimal by a whole number (0-2 decimal digits) Multiply a decimal by a whole number (0-3 decimal digits) Multiply decimals by decimals (1-2 decimal digits) Multiply a decimal with 1-2 decimal digits by another decimal with 1-2 decimal digits Multiply a decimal with 1-3 decimal digits by another decimal with 1-3 decimal digits Multiply decimals, writing the numbers under each other (0-2 decimal digits) Multiply decimals, writing the numbers under each other (0-3 decimal digits) --- See also Multiply Decimals By Whole Numbers — a free lesson Multiply Decimals By Decimals — a free lesson Decimals worksheet generator — generate worksheets for any of the four operations with decimals, in horizontal or vertical formats. | Practice makes perfect. Practice math at IXL.com |
Decimal multiplication worksheets
The worksheets provide calculation practice for decimal multiplication (both mental math and multiplication algorithm). They are meant for 5th - 6th grades.
Jump to:
The worksheets are randomly generated, so you can get a new, different one just by hitting
the refresh button on your browser (or F5).
You can print them directly from your browser window, but first check how it looks like in the Print Preview. If the worksheet does not fit the page in the Print Preview, adjust your margins, header, and footer in the Page Setup settings of the browser. Or, adjust the "scale" to 90% or less in the Print Preview. Some browsers may have "Print to fit" option, which will automatically scale the worksheet to be so small as to fit the printable area.
Copying permission: you can freely print and copy unlimited copies of the worksheets for use in the classroom, home, tutoring center—anywhere you might be teaching. If you want to distribute the links or worksheets on a website or publication, please contact us.
Decimal multiplication worksheets — mental math
Multiply a whole number and a decimal
- easy (one decimal digit)
Multiply a whole number and a decimal - harder (one decimal digit)
Multiply a whole number and a decimal - missing factor (one decimal digit)
Multiply
a whole number and a decimal (1-2 decimal digits)
Multiply
a whole number and a decimal - missing factor (1-2 decimal digits)
Multiply a whole number and a decimal (1-3 decimal digits)
Multiply a whole number and a decimal - missing factor (1-3 decimal digits)
Multiply decimals by decimals
Multiply decimals by decimals - missing factor
Multiply a decimal by decimal mentally (up to 2 decimal digits times 2 decimal digits)
Multiply a decimal by decimal mentally (up to 3 decimal digits times 3 decimal digits)
Missing factor problems 1 (decimal by decimal, 1 or 2 decimals)
Missing factor problems 2 (decimal by decimal, 1-3 decimals)
Multiply decimals by decimals or whole numbers (mixed practice)
Multiply decimals by decimals or whole numbers - missing factor (mixed practice)
Worksheets for multiplying decimals by powers of ten
Multiply
by 10 or 100 (1-2 decimal digits)
Multiply
by 10, 100, or 1000 (1-2 decimal digits)
Multiply by 10, 100, or 1000 - missing factor (1-2 decimal digits)
Multiply
by 10 or 100 (1-3 decimal digits)
Multiply
by 10, 100, or 1000 (1-3 decimal digits)
Like above, but missing factor
Multiply
by 10, 100, 1000, 10000, or 100000 (1-3 decimal digits)
Multiply decimals by 10, 100, or 1000 - missing factor (1-3 decimal digits)
Worksheets for multiplying decimals by powers of ten
See also my free lesson Multiply and Divide Decimals by 10, 100, and 1000 (powers of ten)
Worksheets for long multiplication of decimals (multiplication algorithm)
Multiply a decimal with 1-2 decimal digits by another decimal with 1-2 decimal digits
Multiply a decimal with 1-3 decimal digits by another decimal with 1-3 decimal digits
Multiply decimals, writing the numbers under each other (0-2 decimal digits)
Multiply decimals, writing the numbers under each other (0-3 decimal digits)
See also
Multiply Decimals By Whole Numbers — a free lesson
Multiply Decimals By Decimals — a free lesson
Decimals worksheet generator — generate worksheets for any of the four operations with decimals, in horizontal or vertical formats. |
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8325 | https://progress.lawlessspanish.com/revision/grammar/de-infinitive-subordinate-conditional-clause | De + infinitive = if (conditional clause) | Spanish Grammar | Progress with Lawless Spanish
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Written by qualified Spanish expert Inma Sánchez BA
Last updated: 2025-05-14
Spanish»
Library»
Grammar»
Level C1»
Prepositions & Conjunctions»
De + infinitive (subordinate conditional clause)
17 questionsReport a problem
Related lessons
Using the Spanish imperfect subjunctive in hypothetical clauses introduced by si followed by the Spanish conditional simple
Si [if] followed by present indicative + main clause (present tense/future tense)
Si [if] followed by present indicative + main clause [command/request/advice]
Using the pluperfect subjunctive in hypothetical clauses (si) followed by the perfect conditional/ the pluperfect subjunctive
Con tal de que / Con tal de (subordinate purpose and conditional clause)
Written by qualified Spanish expert Inma Sánchez BA
Last updated: 2025-05-14
17 questions
De + infinitive in Spanish
Sometimes the preposition "de" introduces a subordinate clause expressing a condition, the same way the conjunction "si" (if) does.
In this case "de"isfollowed by an infinitive.
Let's see some examples:
¿Has escuchado las noticias?De ser verdad, la economía va en declive.Have you heard the news? If it is true, the economy is declining.
De ser posible,te llevaría a ese restaurante 4 tenedores.If it werepossible, I would take you to that 4-star restaurant.
De poder elegir, yo me inclinaría por el Ferrari.If I could choose, I would prefer the Ferrari.
No se preocupe usted. De informarme de algo más, yo le aviso por teléfono.Don't worry, sir. If I learnmore, I will call you.
De haber sabido que veníais, habría hecho algo de comer.If I had known you were coming, I would have prepared something to eat.
De ser aprobada esa ley, las mujeres se beneficiarán mucho a nivel laboral.If that lawis approved, women will benefit a lot workwise.
De no llegar a tiempo perderíamos el vuelo.If we didn't arrive on time we'd miss the flight.
Notice how the infinitive construction admits different cases such as compound infinitives (de haber sabido...), the passive form (de ser aprobada...), the negative form (de no llegar...), modal verbs (de poder elegir...)
Pay special attention to pronouns in this type of construction as they can be attached to the end of the infinitive following the usual rule. For example, this sentence:
De haber sabido que veníais, habría hecho algo de comer.If I had known you were coming, I would have prepared something to eat.
could become:
De haberlo sabido, habría preparado algo de comer.Had I known [it], I would have prepared something to eat.
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Oración condicional con si
Examples and resources
De ser aprobada esa ley, las mujeres se beneficiarán mucho a nivel laboral.If that lawis approved, women will benefit a lot workwise.No se preocupe usted. De informarme de algo más, yo le aviso por teléfono.Don't worry, sir. If I learnmore, I will call you.De no llegar a tiempo perderíamos el vuelo.If we didn't arrive on time we'd miss the flight.De haber sabido que veníais, habría hecho algo de comer.If I had known you were coming, I would have prepared something to eat.De ser posible,te llevaría a ese restaurante 4 tenedores.If it werepossible, I would take you to that 4-star restaurant.De poder elegir, yo me inclinaría por el Ferrari.If I could choose, I would prefer the Ferrari.¿Has escuchado las noticias?De ser verdad, la economía va en declive.Have you heard the news? If it is true, the economy is declining.De haberlo sabido, habría preparado algo de comer.Had I known [it], I would have prepared something to eat.
De + infinitive (subordinate conditional clause)
1 of 2
"Habrías coincidido con mis primos, de haber llegado antes." means:
"Te invitaría a otra cerveza de tener más dinero." means:
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Cinzia F.C1 Kwiziq community member
Indicative vs subjunctive
Hi, being back after a few months, I noticed that all example sentences use indicative mood, while the one in the lessons was in subjunctive:
De habérmelo explicado antes, no me hubiera enfadado tanto.
Why isn't this conditional, ... no me habría enfadado tanto?
Asked 1 year ago
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SilviaKwiziq Native Spanish Teacher Correct answer
Hola! Welcome back!
The sentence "De habérmelo explicado antes, no me hubiera enfadado tanto" is indeed in the past subjunctive mood. The use of "hubiera" here is correct for expressing a hypothetical or unreal condition in the past. If it were in the conditional mood, it would be "De habérmelo explicado antes, no me habría enfadado tanto", which would convey a slightly different meaning, emphasizing the cause-and-effect relationship.
Both are grammatically valid, but the subjunctive mood is commonly used in this context.
Feel free to ask if you have more questions!
Saludos
Silvia
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Cinzia F. asked:View original
Indicative vs subjunctive
Hi, being back after a few months, I noticed that all example sentences use indicative mood, while the one in the lessons was in subjunctive:
De habérmelo explicado antes, no me hubiera enfadado tanto.
Why isn't this conditional, ... no me habría enfadado tanto?
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Robert W.C1 Kwiziq community member
I'm still not sure as to when you use De as opposed to Si.
Asked 2 years ago
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InmaKwiziq Head of Spanish, Native Spanish Teacher
Hola Robert
The structure using De with infinitive is an equivalent to the si clause. I would say that it's a bit more informal/colloquial than the si clause.
Saludos
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Robert W. asked:View original
I'm still not sure as to when you use De as opposed to Si.
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Marcos G.C1 Kwiziq Q&A super contributor
Not sure about English translation
For the example:
De no llegar a tiempo perderíamos el vuelo.
If we didn't arrive on time we'd miss the flight.
I can only see four translations:(1) If we don't arrive on time we'll miss the flight. (or "we could miss")(2) If we hadn't arrived on time we would/could have missed the flight.
Could you please double-check your English translation? Thanks.
Asked 2 years ago
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InmaKwiziq Head of Spanish, Native Spanish Teacher Correct answer
Hola Marcos
This would be the equivalent of the 2nd conditional type sentence: imperfect subjunctive + conditional
The other translations you are considering are different; we'd say those like this:
Si no llegamos a tiempo, perderemos el vuelo,
If we don't arrive on time, we'll miss the flight.
Si no hubiéramos llegado a tiempo, habríamos perdido el vuelo.
If we hadn't arrived on time, we would have missed the flight.
Here's alesson that will help you understand better this combination of tenses.
I hope it clarified it.
Saludos
Inma
Like 1 2 years ago
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Marcos G. asked:View original
Not sure about English translation
For the example:
De no llegar a tiempo perderíamos el vuelo.
If we didn't arrive on time we'd miss the flight.
I can only see four translations:(1) If we don't arrive on time we'll miss the flight. (or "we could miss")(2) If we hadn't arrived on time we would/could have missed the flight.
Could you please double-check your English translation? Thanks.
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R Z.C1 Kwiziq Q&A regular contributor
Forks vs. Stars
Are Spanish restaurants really rated in tenedores? I would've thought that it would ha derived from Michelin stars. At least, I always assumed stars was the literal translation of whatever the French is.
Asked 4 years ago
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InmaKwiziq Head of Spanish, Native Spanish Teacher
Hola R
Yes, in Spain (unless this changed recently) there is a rating based on "tenedores" for restaurants. Bear in mind that the rating -I think it is up to 5 tenedores -is more to do with the actual restaurant, the quality of the place, more than the actual quality of the food (if I'm not wrong). They may also have a different rating based on the Michelin stars and this would reflect the quality of the food.
Saludos
Inma
Like 1 4 years ago
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R Z. asked:View original
Forks vs. Stars
Are Spanish restaurants really rated in tenedores? I would've thought that it would ha derived from Michelin stars. At least, I always assumed stars was the literal translation of whatever the French is.
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TALK T.B2 Kwiziq community member
Great way to avoid the imperfect subjunctive etc.
Great structure to know! Check your English translations, however, as they don't fit the correct grammar patterns:
If PRESENT, then FUTURE
If IMPERFECT, then CONDITIONAL
Your "If PLUPERFECT, then CONDITIONAL PERFECT" examples are correct.
Asked 5 years ago
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InmaKwiziq Head of Spanish, Native Spanish Teacher
Hola Talk,
gracias. We checked the English translations and some of them have been slightly changed.
Un saludo,
Inma
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TALK T. asked:View original
Great way to avoid the imperfect subjunctive etc.
Great structure to know! Check your English translations, however, as they don't fit the correct grammar patterns:
If PRESENT, then FUTURE
If IMPERFECT, then CONDITIONAL
Your "If PLUPERFECT, then CONDITIONAL PERFECT" examples are correct.
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Claire N.C1 Kwiziq community member
Usage of de + infinitive compared to conditional si
When using the de + infinitive construction, does the tense used in the main clause indicate the level of probability? Eg present = possible, future = less possible, conditional = very unlikely/impossible?
When is use of de + infinitive preferable to using the conditional si construction?
Asked 5 years ago
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InmaKwiziq Head of Spanish, Native Spanish Teacher
Hola Claire,
de + infinitive clauses are more neutral than "si" clauses; the level of probability in sentences using De + infinitive is not related to the tense used in the main clause; they simply indicate the time when these actions happen (if the condition is fulfilled)
De saber que venías, habría hecho más comida.
If I had known you were coming, I would have prepared more food.
De comprar esa casa, pediré un préstamo al banco.
If I buy that house, I will ask the bank for a loan.
De ser verdad, la economía va en declive.
If it is true, the enonomy is declining.
These sentences are using different tenses in their main clause, one of them indicating something would have happened in the past, one indicating something will happen in the future (if the condition is fulfilled), and one indicating something is happening in the present (if the condition is fulfilled.
Saludos,
Inma
Like 2 5 years ago
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Claire N. asked:View original
Usage of de + infinitive compared to conditional si
When using the de + infinitive construction, does the tense used in the main clause indicate the level of probability? Eg present = possible, future = less possible, conditional = very unlikely/impossible?
When is use of de + infinitive preferable to using the conditional si construction?
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Jamie L.C1 Kwiziq Q&A regular contributor
English Correction
The last example translates “De haberlo sabido” as “I had known” when it should be either “Had I known” or “If I had known.”
Asked 5 years ago
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InmaKwiziq Head of Spanish, Native Spanish Teacher
Gracias Jamie.
I corrected that.
Inma
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Jamie L. asked:View original
English Correction
The last example translates “De haberlo sabido” as “I had known” when it should be either “Had I known” or “If I had known.”
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Sherri ..C1 Kwiziq Q&A regular contributor
coincidido or conocido
Just wondering if habrías coincidido might be better said as habrías conocido.
Asked 5 years ago
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InmaKwiziq Head of Spanish, Native Spanish Teacher
Hola Sherri
In this sentence,"coincidir con" is used with the meaning of"meeting/seeing someone by chance".
This is why we are translating this here as"you would have met". When we use "coincidir con alguien", depending on the context this could mean "to meet for the first time" or "to see someone you already know". While "conocer a alguien" means "meet for the first time". Here are other sentences:
Ayer coincidí con Marcos en la cafetería.
Yesterday Isaw Marcos at the coffee shop.(It was a coincidence that we were both there at the same time, but I already know Marcos)
Ayer conocí a Marcos en la cafetería.
Yesterday I met Marcos at the coffee shop.(this is the first time I met Marcos)
A ver si coincidimos con Luis en el bar. Le tengo que preguntar una cosa.
Hopefully we will see Luis at the bar. I have to ask him something.
A ver siconocemos a Luis. Dicen que es un chico simpático.
Hopefully we will meet Luis [soon] . They say he is a nice guy.
I hope this clarified it.
Saludos
Inma
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Sherri ..C1 Kwiziq Q&A regular contributor
Thank you, Inma. I did not know that coincidir con has the meaning of meeting/seeing someone by chance.
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Sherri .. asked:View original
coincidido or conocido
Just wondering if habrías coincidido might be better said as habrías conocido.
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8327 | https://www.openepi.com/PDFDocs/TwobyTwoDoc.pdf | 1 Two by Two Tables Containing Counts (TwobyTwo) Kevin M. Sullivan, PhD, MPH, MHA, Associate Professor, Department of Epidemiology, Rollins School of Public Health, Emory University, Atlanta, Georgia, USA INTRODUCTION This chapter provides the formulae and examples for calculating crude and adjusted point estimates and confidence intervals for: risk ratios and differences; odds ratios; incidence rate ratios and differences; and etiologic and prevented fractions. The tests for interaction are also presented. First, the estimates from a single 2x2 table (“count” data) are presented followed by estimates adjusted or summarized across stratified data. FORMULAE AND EXAMPLE FOR A SINGLE 2X2 TABLE (COUNT DATA) For a single 2x2 table, the notation is as depicted in table 15-1. The formulae for calculating the risk ratio, risk difference, and odds ratio and their confidence intervals are shown below. For the confidence intervals, the Taylor series approach is provided because it is a reasonably good confidence interval method when the sample size is large. There are other more complicated methods for computing confidence intervals when the data are sparse but they are not shown here, such as maximum likelihood and exact methods. Note that the term “risk” is used assuming a cohort study was performed and the risk of disease was assessed. If a study was based on prevalent disease, then substitute the term “prevalence” for “risk,” e.g., prevalence ratio and prevalence difference. Table 15-1. Notation and table setup for a 2x2 table Exposed Nonexposed Disease a b m1 No Disease c d m0 n1 n0 n Estimated risk in the exposed = 1 / ˆ n a e R Estimated risk in the nonexposed = 0 / ˆ n b u R Estimated risk in the population = n m R / ˆ 1 Point and Variance Estimates, Confidence intervals The point and variance estimates and the confidence interval formulae are provided in Table 15.1. For some parameters there will not be a variance formula. The confidence limits for the Etiologic Fraction in the Exposed is based on the calculated upper lower bounds of the confidence limits for the risk ratio (RRUB and RRLB, respectively) with risk data. A similar approach is used when the Etiologic Fraction in the Exposed is based on the Odds Ratio. Statistical Tests There are many statistical tests that can be performed on a single 2x2 table. Common tests include the Chi-square test (corrected, uncorrected, and Mantel-Haenszel) and exact tests (Fisher and mid-p exact). In this chapter the uncorrected and Mantel-Haenszel chi-square tests will be presented; however, these test should be used when the number of “expected” observations in each cell are > 5. When the expected number of observations in any cell is < 5, then one of the exact tests should be used. How to calculate exact p-values is beyond the scope of this text and requires an iterative calculation. The expected number of observations in a cell is calculated by multiplying the row and column total and dividing by the total sample size. The uncorrected chi square is calculated as 0 1 0 1 2 2 1 ) )( ( m m n n bc ad n uncorr and the Mantel Haenszel chi square as 0 1 0 1 2 2 1 ) )( 1 ( m m n n bc ad n mh 2 Table 5.1. Estimates and confidence intervals for epidemiologic parameters for a single table Parameter Point Estimate Variance Estimate Confidence Interval Parameters based on risks (from randomized trials and cohort studies) or prevalences (cross-sectional studies) Risk Ratio u R e R R R ˆ ˆ ˆ u R ˆ n u R ˆ 1 e R ˆ n e R ˆ -1 ) ˆ (ln ˆ o 1 R R r a V ) ˆ (ln ˆ ˆ R R r a V Z Rexp R /2 -1 Risk Difference u R D R ˆ -e R ˆ ˆ 0 1 n u) R u(1 R n e) R e(1 R D R r a V ˆ ˆ ˆ ˆ ) ˆ ( ˆ ) ˆ ( ˆ ˆ / D R r a V Z D R 2 1 Etiologic Fraction in the Population R u R R p F E ˆ ˆ ˆ ˆ Etiologic Fraction in the Exposed e R u R e R e F E ˆ ˆ ˆ ˆ Based on variance estimate for the RR LB= LB LB R R 1 R R ˆ ˆ ;UB= UB UB R R 1 R R ˆ ˆ Prevented Fraction in the Population u R R u R p F P ˆ ˆ ˆ ˆ Prevented Fraction in the Exposed u R u R e F P ˆ e R ˆ ˆ ˆ Based on variance estimate for the RR LB= UB R R 1 ˆ ; UB= LB R R 1 ˆ Parameters based on the odds and odds ratio (from randomized trials, cohort studies, case-control, or cross-sectional studies) Odds Ratio bc ad R O ˆ d 1 c 1 b 1 a 1 ) ˆ (ln ˆ R O r a V ) ˆ (ln ˆ ˆ R O r a V Z Rexp O /2 -1 Etiologic Fraction in the Population 1 1 R O p 1 R O p pOR F E ) ˆ ( ' ) ˆ ( ' ˆ Etiologic Fraction in the Exposed R O 1 R O eOR F E ˆ ˆ ˆ Based on variance estimate for the OR LB= LB LB R O 1 R O ˆ ˆ ;UB= UB UB R O 1 R O ˆ ˆ Prevented Fraction in the Population ) ˆ ( ' ˆ R O 1 p pOR F P Prevented Fraction in the Exposed R O 1 eOR F P ˆ ˆ Based on variance estimate for the OR LB= UB R O 1 ˆ ; UB= LB R O 1 ˆ LB=lower bound; UB=upper bound P’=… To work through an example of the calculations, a study was performed in children 12-23.9 months of age. In this study, the prevalence of anemia was estimated. The results are shown in Table 15-2. Table 15-2. Example data; prevalence of anemia in children 12-23.9 months of age by sex Male Female Anemic 205 129 334 Not Anemic 89 86 175 294 215 509 The prevalence estimates are: Prevalence in males = 205/294 = 0.697 or 69.7% Prevalence in females = 129/215 = 0.600 or 60.0% The Prevalence Ratio estimate is as follows (using the formulae for the risk ratio): Prevalence ratio = .697/.600 = 1.16 Variance of the prevalence ratio = 3 95% confidence interval; replace the Z value in the formula to 1.96 for the calculation of a two-sided 95% confidence interval (for a 90 confidence interval, the Z value is 1.645, and for a 99% confidence, 2.576): 132629 .
exp 16 .
1 004579 .
96 .
1 1.16exp (1.02, 1.32) The interpretation would be that males in this study were 1.16 times more likely to have anemia than females; the 95% confidence interval around this estimate is 1.02, 1.32. The Prevalence Difference estimate is as follows (using the formulae for the risk difference): Prevalence difference = .697 - .600 = .097 or 9.7% Variance of the prevalence difference = 0018346 .
215 ) 6 .
.6(1 294 ) 697 .
.697(1 95% confidence interval: ) 04283 (.
96 .
1 097 .
0018346 .
96 .
1 .097 (.013, .181) or (1.3%, 18.1%) The interpretation would be that the prevalence of anemia is 9.7% higher in males compared to females (in terms of an absolute difference), with a 95% confidence interval from 1.3% to 18.1%. The odds ratio estimate, or in this example the prevalence odds ratio estimates, is as follows: Odds ratio = (20586)/(12989)=1.54 Variance of the odds ratio = 0354938 .
86 1 89 1 129 1 205 1 95% confidence interval = 369258 .
exp 54 .
1 0354938 .
0 96 .
1 1.54exp (1.06, 2.23) The interpretation would be that the odds of anemia in males is 1.54 times the odds in females with a 95% confidence interval of 1.06 to 2.23. Note that the odds ratio is larger than the risk ratio because the prevalence of anemia is high (334/509 = 66%). The uncorrected chi-square tests would be calculated as: 20903 .
5 175 334 215 294 ) 89 129 86 205 )( 509 ( 2 2 1 uncorr which would have a p-value = .022. The Mantel Haenszel chi square would be calculated as: 19879 .
5 175 334 215 294 ) 89 129 86 205 )( 1 509 ( 2 2 1 mh which would have a p-value of .023. The conclusion would be that there was a statistically significant association between the sex of the child and the prevalence of anemia. Note that the statistical test for a 2x2 table can be used with the risk ratio, risk difference, or odds ratio. Also, it is calculated the same whether the data are from an unmatched case-control study, a cohort study, or a clinical trial. FORMULAE AND EXAMPLE FOR STRATIFIED DATA (COUNT DATA) For stratified analyses, the same calculations for the crude table can be used for stratum-specific estimates. For adjusted or summary estimates, a slightly different notation is used as shown in Table 15-2. In this table, the subscript i to denote estimates from a specific stratum. The general approach for adjusted point estimates is to weight each of the stratum-specific estimates by a weighting method and then sum the results. Table 15-2. Notation and table setup for stratified 2x2 tables Exposed Nonexposed Disease ai bi m1i No Disease ci di m0i n1i n0i ni For the risk ratio and the odds ratio, two different approaches are given for estimating the adjusted point estimate and confidence interval, one referred to as the directly adjusted ratio and the other 004579 .
.6 215 6 .
1 .697 294 .697 -1 4 referred to as the Mantel-Haenszel adjusted ratio. The directly adjusted approach requires “large” numbers in each stratum. The weights for directly adjusted values are the inverse of the variance; this approach provides a greater weight to strata with the least amount of variance and less weight to strata with a large variance. The Mantel-Haenszel method works better when data are sparse. 5 Parameter Point Estimate Confidence Interval Risk Ratio – Directly Adjusted The linked image cannot be displayed. The file may have been moved, renamed, or deleted. Verify that the link points to the correct file and location. where The linked image cannot be displayed. The file m ay hav e been mov ed, renamed, or deleted. Verify that the link points to the correct file and location.
, The linked image cannot be displayed. The file may have been moved, renamed, or deleted. Verify that the link points to the correct file and location. s i i Direct w Z R R 1 2 / 1 exp ˆ Risk Ratio – Mantel-Haenszel Adjusted s i i i i s i i oi i MH n n b n n a R R 1 1 1 ˆ The linked image cannot be displayed. The file may have been moved, renamed, or deleted. Verify that the link points to the correct file and location. where s i i i i s i i i i s i i i i i i i i n n b n n a n n b a n n m R R 1 1 1 0 1 2 0 1 1 MH / ˆ ln SE Risk Difference – Directly Adjusted s i i i s i i Direct w D R w D R 1 1 ˆ ˆ where 3 0 3 1 1 i i i i i i i n d b n c a w , i i i i i n b n a D R 0 1 ˆ s i i Direct w Z D R 1 2 / 1 ˆ Odds Ratio – Directly Adjusted s i i s i i i Direct w R O w R O 1 1 ) ˆ ln( exp ˆ where i i i i i c b d a R O ˆ ,w a b c d i i i i i 1 1 1 1 1 s i i Direct w Z R O 1 2 / 1 exp ˆ Odds Ratio – Mantel-Haenszel Adjusted s i i i i s i i i i MH n c b n d a R O 1 1 ˆ MH /2 -1 ˆ ln SE Z exp ˆ R O R O MH where 6 2 1 1 1 1 1 2 1 1 MH 2 2 ) ( 2 ˆ ln SE s i i s i i i s i s i i i s i i i i i s i i s i i i S S Q S R R Q S P R R P R O P a d n i i i i ( ) / Q b c n i i i i ( ) / R a d n i i i i / S b c n i i i i / 7 Tests for Interaction for the Risk Ratio, Risk Difference, and the Odds Ratio The tests for interaction presented here are generally referred to as the “Breslow-Day test of homogeneity” and are based on a chi square test. The test for interaction for the risk ratio is: s R R r a V R R R R 1 i i 2 Direct i 2 1 s ) ˆ ln( ˆ ˆ ln ˆ ln where the Var[ln(RRi)] = 1/wi from the direct RR point estimate calculation. The test for interaction for the risk difference is s D R r a V D R D R i i Direct i s 1 2 2 1 ) ˆ ( ˆ ˆ ˆ where the Var(RDi) = 1/wi from the direct RD point estimate calculation. To test for interaction for the odds ratio (OR), the chi square test is calculated as: s R O r a V R O R O i i Direct i s 1 2 2 1 ) ˆ ln( ˆ ˆ ln ˆ ln where the Var[ln(ORi)] = 1/wi from the direct OR point estimate calculation. Summary Statistical Test A statistical test to assess whether there is a statistically significant association between the exposure and outcome variable controlling for the third variable is the Mantel-Haenszel uncorrected chi-square test. This statistic would be used only if it was decided that there was no statistically significant interaction. s i i i i i i i s i i i i i i n n m m n n n c b d a 1 2 0 1 0 1 2 1 2 1 1 An example of the calculations for stratified data are provided next. Continuing on with the example in table 15-3 on the association between sex and anemia in children, 8 the data are stratified on mothers education level. Again, because the data were based on prevalent cases, the term “prevalent” will be used rather than “risk.” Table 15-3. Example data; prevalence of anemia in children 12-23.9 months of age by sex stratified on mothers education level. Mother has low level of education Male Female Anemic 66 36 102 Not Anemic 28 32 60 94 68 162 Mother has high level of education Male Female Anemic 139 93 232 Not Anemic 61 54 115 200 147 347 Calculation of the directly adjusted prevalence ratio and its 95% confidence interval is shown in Table 15-4. Table 15-4. Calculations for computing directly adjusted prevalence (risk) ratio Stratum PRi ln(PRi) wi wi ln(PRi) 1 1.326 .2821669 56.86628 16.04578 2 1.099 .0944001 162.75481 15.36407 Sum 219.62109 31.40985 The calculated point estimate is: 154 .
1 62109 .
219 40985 .
31 exp ˆ Direct R P The 95% confidence interval is: 132257 .
exp 154 .
1 62109 .
219 96 .
1 exp 154 .
1 (1.011, 1.317) The interpretation would be that males were 1.154 times more likely to be anemic than females controlling or adjusting for the mother’s education level. In addition, we are 95% confident that the true prevalence ratio is captured between 1.011 and 1.317. However, we must still calculate the test for interaction to see if the mother’s education level modifies the sex-anemia relationship. To calculate the test for interaction, the directly adjusted risk ratio needs to be calculated beforehand. Also, note that 9 i i w R P 1 ) ˆ r(ln a ˆ v Therefore, the test for interaction for the prevalence/risk ratio would be: 0061442 .
143234 .
.0944001 017585 .
143234 .
.2821669 2 2 2 1 s 485790 .
1 388130 .
097660 .
1 2 1 s The p-value for the chi square would be calculated for a chi square value of 1.486 with one degree of freedom (the degrees of freedom is determined from the number of strata minus 1). The p-value from this example is .223. Therefore, we would state that the mother’s education level does not significantly modify the sex-anemia relationship. Therefore, the next question is whether the mother’s education level confounds the relationship. The crude prevalence ratio was 1.16 and the directly adjusted value was 1.15, which is less than a 1% difference, therefore the conclusion would be that mother’s education does not modify nor confound the sex-anemia relationship. The calculation of the directly adjusted Mantel-Haenszel prevalence ratio and its 95% confidence interval is shown in Table 15-5. Table 15-5. Calculations for computing the Mantel-Haenszel prevalence (risk) ratio Stratum ain0i/ni bin1i/ ni (m1in1in0i-aibini)/ni 2 1 27.7037 20.8889 10.17650 2 58.8847 53.6023 19.3933 Sum 86.5884 74.4912 29.5698 The point estimate is 162 .
1 4912 .
74 5884 .
86 ˆ MH R P To calculate the 95% confidence interval we will first calculate the standard error of the estimate: 067708 .
4912 .
74 5884 .
86 5698 .
29 R ˆ ln SE MH P The 95% confidence interval is calculated as: ) 132708 .
exp( 162 .
1 .067708 96 .
1 exp 162 .
1 (1.018, 1.327) Previously we found that mother’s education did not modify the sex-anemia relationship, therefore the interpretation would be that, controlling for mother’s education, males were 1.162 10 times more likely to be anemic than females. However, because there is little confounding (the crude value is 1.15), there is no need to control for mother’s education level. Calculation of the directly adjusted prevalence difference and its 95% confidence interval is shown in Table 15-6. Table 15-6. Calculations for computing the direct adjusted prevalence (risk) difference Stratum PDi wi wi PDi 1 0.1727 169.8171 29.3274 2 0.0623 378.6661 23.5909 Sum 548.4832 52.9183 The point estimate is: 0965 .
4832 .
548 9183 .
52 ˆ Direct D P and the 95% confidence interval is: 0837 .
0965 .
4832 .
548 96 .
1 0965 .
(.0128, .1802) Depending on the frequency of disease, it may be useful to describe the difference in term of per 100 individuals (or percent), per 1,000, or some other unit. In this example, the males had a prevalence of anemia 9.7% higher (in absolute terms) than females controlling for maternal education, and we are 95% confident that the truth is captured between 1.3% and 18.1%. However, before the decision is made as to whether or not to present the adjusted difference, the test for interaction should be calculated. Again, note that: i i w D P 1 ) ˆ r(ln a ˆ v Therefore, the test for interaction for prevalence/risk differences would be: 42886 .
1 44305 .
98581 .
00264 .
0965 .
0.0623 00589 .
0965 .
0.1727 2 2 2 1 s The chi square value of 1.42886 with one degree of freedom would have a p-value of .232, which would not be statistically significant. The next step would be to determine whether mother’s education confounds the sex-anemia relationship. The crude prevalence difference was .097, the same as the adjusted difference, which would lead to the conclusion that there is no important confounding in this analysis. 11 Calculation of the directly adjusted (prevalence) odds ratio and its 95% confidence interval is shown in Table 15-7. Table 15-7. Calculations for computing the direct adjusted (prevalence) odds ratio Stratum ORi ln(ORi) wi wi ln(ORi) 1 2.095 .73955 9.09971 6.72969 2 1.323 .27990 18.91829 5.29523 Sum 28.01800 12.02492 The calculated point estimate is: 536 .
1 018 .
28 02492 .
12 exp ˆ Direct R O The 95% confidence interval is: 370286 .
exp 536 .
1 018 .
28 96 .
1 exp 536 .
1 (1.061, 2.224) The interpretation would be that odds of anemia in males was 1.536 times the odds of anemia in females controlling or adjusting for the mother’s education level. In addition, we are 95% confident that the true prevalence odds ratio is captured between 1.061 and 2.224. However, we must still calculate the test for interaction to see if the mother’s education level modifies the sex-anemia relationship. To calculate the test for interaction, the directly adjusted odds ratio needs to be calculated beforehand. Also, note that i i w R O 1 ) ˆ r(ln a ˆ v Therefore, the test for interaction for the (prevalence) odds ratio would be: 05286 .
42918 .
.27990 10989 .
42918 .
.73955 2 2 2 1 s 2918 .
1 42158 .
87660 .
2 1 s The p-value for the chi square would be calculated for a chi square value of 1.2918 with one degree of freedom (the degrees of freedom is determined from the number of strata minus 1). The p-value from this example is .256. Therefore, we would state that the mother’s education level does not significantly modify the sex-anemia relationship. Therefore, the next question is whether the mother’s education level confounds the relationship. The crude prevalence odds ratio was 1.536 and the directly adjusted value was the same, the conclusion would be that, based on the odds ratio, mother’s education does not modify nor confound the sex-anemia relationship. Calculation of the Mantel-Haenszel adjusted (prevalence) odds ratio and its 95% confidence interval is as follows. The values that need to be calculated are shown in Table 15-8. 12 To calculate the point estimate and the confidence interval, eight values in Table 15-8 need to be calculated. The calculated point estimate is: 536 .
1 57092 .
22 66816 .
34 ˆ MH R O The standard error of the natural log of the point estimate is calculated as: Table 15-8. Calculations for computing the Mantel-Haenszel adjusted (prevalence) odds ratio Stratum Pi Qi Ri Si 1 .60494 .39506 13.03704 6.22222 2 .55620 .44380 21.63112 16.34870 Sum 1.16114 .83886 34.66816 22.57092 Stratum PiRi PiSi QiRi QiSi 1 7.88663 3.76407 5.15041 2.45815 2 12.03123 9.09315 9.59999 7.25555 Sum 19.91786 12.85722 14.75040 9.71370 2 2 MH 57092 .
22 2 9.71370 57092 .
22 66816 .
34 2 75040 .
14 85722 .
12 66816 .
34 2 91786 .
19 ˆ ln SE x x R O 18831 .
00953 .
01764 .
00829 .
The confidence interval based on the Robins, Greenland, Breslow method is: ) 36908 .
exp( 536 .
1 .18831 96 .
1 exp 536 .
1 The 95% confidence interval is (1.062, 2.222) Previously we found that mother’s education did not modify the sex-anemia relationship, therefore the interpretation would be that, controlling for mother’s education, the odds of males having anemia were 1.536 times more likely to be anemic than the odds in females. However, because there is little or no confounding (the crude value is 1.536), there is no need to control for mother’s education level. The overall Mantel-Haenszel uncorrected chi-square test would be calculated as with the intermediate calculations shown in Table 15-9. 13 Table 15-9. Calculations for computing the Mantel-Haenszel uncorrected chi-square test Stratum (aidi-bici)/ni (n1in0im1im0i)/[(ni-1)ni 2] 1 6.81481 9.25832 2 5.28242 18.82774 Sum 12.09723 28.08606 Therefore 2105 .
5 08606 .
28 09723 .
12 2 2 1 which would have a p-value of .022. |
8328 | https://philosophy.stackexchange.com/questions/50439/do-whole-numbers-other-than-zero-actually-exist | philosophy of mathematics - Do whole numbers other than zero actually exist? - Philosophy Stack Exchange
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Do whole numbers other than zero actually exist?
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Think about counting up: you start from 0. There are many decimals in between 0 and 1, actually, an infinite amount of decimals are there. So in the same way that there is no last number there is no last decimal before 1, so you would never get to it on a scale, therefore it wouldn't actually exist for any purpose. Adding on to that all math you know wouldn't exist either.Edit I didn't make this clear enough the first time if I had a number line and I was to count up using the smallest distance I could in decimals it would go on for an infinite amount. The number one wouldn't even be conceivably possible in that sense nor any other numbers. as well, this isn't a discussion on the nature of infinity or what existence is, I just used those words. understand are going off the basic definition of infinity that even a kid would get.
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edited Apr 3, 2018 at 17:25
Brock ObamaBrock Obama
asked Apr 2, 2018 at 18:14
Brock ObamaBrock Obama
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Actually, you meant integer numbers, not whole, right?rus9384 –rus9384 2018-04-02 19:20:38 +00:00 Commented Apr 2, 2018 at 19:20
For mathematical reference: 0.99999..... = 1.cHao –cHao 2018-04-02 19:30:19 +00:00 Commented Apr 2, 2018 at 19:30
You seem to be pretty convinced that the rational numbers exist: presumably you have identified $\frac{1}{2}$ as one of the fractions you might reach, for example. So what does this "$2$" thing even mean anyway?Patrick Stevens –Patrick Stevens 2018-04-02 19:46:32 +00:00 Commented Apr 2, 2018 at 19:46
3 Decimals are a late addition to mathematics, and so is zero, whole numbers were used centuries before that. So whatever mathematical "existence" means both of them are irrelevant to the "existence" of whole numbers, indeed their "existence" is based on the "existence" of the latter in the standard constructions, which do not use "getting" to 1 by piling up decimal digits.Conifold –Conifold 2018-04-02 19:59:43 +00:00 Commented Apr 2, 2018 at 19:59
3 The usual first question to respond to this is to ask "what does it mean for a number to exist in the first place?" The best way to answer your question typically stems from the answer to this follow up question.Cort Ammon –Cort Ammon 2018-04-02 20:53:29 +00:00 Commented Apr 2, 2018 at 20:53
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It is indeed true that it is impossible to count by using steps of the form:
I am currently 'at' a real number, and I will proceed to the "next" real number
for a myriad of reasons, not the least of which is the fact that there aren't any real numbers that could be described as the "next" real number.
Fortunately, when I count, I do not use such steps. The steps I use are typically
I am currently 'at' an integer, and I will proceed to the next integer
If you're paying attention to the real numbers in-between, I "pass" by them all in a single step; I don't try to pass by each one individually in separate steps.
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answered Apr 2, 2018 at 22:15
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Firstly, I apologize for a somewhat long answer to a seemingly innocent short question. Since this is primarily a philosophical question, I have removed my science hat, so here goes.
Starting with the fundamental subject of enumeration, the convention of all numbers, including integers and reals, is a result of the explicit human realization of the perceived discreteness of matter and energy forms that we witness in a wide variety of modalities of physical reality (i.e. physical nature). When we think about it, it becomes evident that there is embedded deep in physical nature a natural inclination that projects aggregative equilibrium through discretization (i.e. uniqueness preservation) of matter and energy forms. Humans and other so-called animate and inanimate life forms (yes, inanimate life forms) have evolved to exchange and utilize matter and energy in discrete packages. Therefore, it is only natural that humans have found it advantageous to explicitly enumerate the various forms of matter and energy that we encountered in physical reality. When we talk about one item of anything, say a human being, we are by definition drawing a hard boundary in space around a certain aggregate of matter what we know as a mammal, and we proceed to call the bundle of matter and energy confined within that boundary as one human being. It is an axiomatic definition that only works because it is accepted by (almost) all humans who are able to comprehend the meaning behind the definition. Again by definition, when we say one person, we are only talking about the local matter and energy within a region of 3D Euclidean space that is typically spatially localized by the human senses of vision, touch, etc. By convention, we are not talking about the radiative heat zone or electrostatic energy zone surrounding the person, nor the gravitational range of the person. Now, to complicate further the definition of numbers, if we delve inside the body of a human being, we encounter one heart, two kidneys, one brain with two hemispheres, etc, with the boundaries of all such organs being defined and agreed by human convention according to boundaries defined by human senses. When we go even deeper to the microscopic domain we begin to see a variety of individual cells and inorganic matter that collectively comprise each organ. So, where do we draw the enumeration boundaries? Where exactly does a hand end and an index finger start? It is defined only approximately by our senses and by our convention. Where exactly does an ear end and the face start? It is all a matter of mutual agreement between humans to approximately define such boundaries. Similarly, it is a matter of human agreement to define where one number ends and another number starts. If we try to be exact about everything that we enumerate, we can never come to an agreement, because one kilogram mass is really never exactly one kilogram if we want to be absolutely accurate, if such a thing was even possible. Therefore, we draw boundaries that serve our purpose, and we move on. Likewise, the arithmetic numbering scale is a convention that is agreed by consensus among humans. We should not be surprised if we encounter a new class of beings (aliens) who have devised their own weird numbering convention that is totally incomprehensible and illogical to us humans. If the numbering system makes sense to these aliens who invented it, that is all that matters from their perspective. Even our number zero is only true by convention. Can we say that there are absolutely zero tables in an empty room? How about the gravitational force exerted by a table a continent away? Surely even a table that far away ought to contribute something to our count of “tableness”, so how can the number of tables ever be exactly zero in an empty room? By the way, can a room ever be exactly and absolutely empty? Can anything be exactly and absolutely zero? It can only be so by definition and mutual agreement.
To take an example from physics, the entire field of quantum mechanics was born out of the frustration of some physicists who were unable to enumerate some particles conventionally (classically) and get a handle on the behavior of these matter and energy forms at subatomic scales because of the fuzziness of the boundaries, which led them to invent a theoretical model that describes such behavior using a probabilistic mathematical model. This was a rude wake up call for fundamental physics, but the shock waves linger on even to this day, to mystify and confuse those who are not fully prepared.
Coming back to our numbering systems, all numbering systems (integer, real, etc.) are defined by human convention. As such, numbers (and even mathematics) is a self-evident human invention that is a tool that is used by humans to manipulate and exploit the discretization property of the matter and energy entities of physical reality. It is unrealistic to claim that our numbering system is in any way an universal or absolute entity of physical reality. Boundaries of numbers are prescribed by humans. The boundary of the one person that is you, and the one person that is me, is defined by our strict human convention of what constitutes a person, but there is a part of us in our parents and a part of us in our descendants, so where do we draw exactly the boundary of the individual human unless it is solely by convention? Physical nature is immensely more simple and elegant than the cumbersome complications that we impose upon nature by our enumeration convention. We only need and utilize numbers to the extent that the practice serves our quest for survival and continuity, otherwise there would have been no need to invent numbers of any kind, including integers, reals, and the ubiquitous zero. Numbers are not something more miraculous or mysterious than the mundane anthropogenic convention that we have created them from to serve a mundane purpose.
Therefore, in direct answer to the original question, I would propose that yes, whole (integer) and real numbers, including the number zero, all actually exist as approximate topological maps of the discretization imposed by physical reality upon the human experience, but that all numbers exist solely by human convention and not as absolute entities of nature, as I explained above.
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edited Apr 3, 2018 at 23:32
answered Apr 3, 2018 at 9:16
VixillatorVixillator
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Do whole numbers other than zero actually exist?
Do positive integers other than 0 exist? Most mathematicians believe so, and for good reason, it is fair to say that an integer like 1 corresponds to a quantity in the real world, coheres with other mathematical truths, and makes number systems work.
As for the process of determining real numbers, they do not proceed the same way that positive integers do using Peano's Axioms which presumes 0 exists, and then creates successors to essentially build the natural numbers and move the definition of 'number' forward towards integers, rationals, irrationals, and therefore reals.
Realsare defined by different methods originally by men like Cantor and Dedekind, who made use of concepts of infinity and sets.
From a philosophical perspective, it is important to think of numbers not as 'things', but more like 'the results of processes'. Pi is a classical example which expresses the ratio between circumference and diameter, but can be calculated in many different ways. Pi exists as an ideal quantity conceptually, but ideal quantities possess different properties than their approximations.
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answered Sep 11, 2019 at 15:39
J DJ D
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Well, the question of whether numbers "exist" is kind of awkward when placed beside your intuitive understanding of what we mean when we say they "exist". This isn't a criticism; it's a perfect valid question to ask here. It just makes it tricky to disambiguate. I'll briefly explain three concepts relevant to your question, and that should clarify things.
Firstly, when most people who think mathematical objects "exist", they don't think so on the basis that we can count to them. They think so because they can define them. So arithmetic is defined, and therefore everything you can, if you believe in the existence of mathematical objects, generate from arithmetic "exists". We don't need to ever count to some large number x to say "it exists", we just know that x = x - 1 + 1 = x. that's not a formal definition btw, im just using it for illustrative purposes.
Incidentally, your intuition about counting the numbers is similar to a philosophy of maths called intuitionism, where the only truths of mathematics are those we can "construct" with a proof. Some things in mathematics, on the standard view, are true but unprovable. Intuitionism rejects these.
Finally, you've hit a really smart point about not being able to count up to 1 from zero. that's because the "reals" are uncountable. I',m sure someone else has pointed out what that means. But we don't need to define the integers from the reals, just becuase the reals are smaller. We define them separately, there's a set of reals that can't be counted, but a set of integers that can. You don't need to "build" integers from reals, and if you did, it wouldn't be a matter of counting.
Hope that helps
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answered Sep 11, 2019 at 18:24
Daniel PrendergastDaniel Prendergast
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The issue here is not numbers, but our perception of things as whole entities.
Our perception determines there's one or two clouds, but a cloud-unit is not something that exists physically. Clouds are just optical effects caused by a certain density of water molecules in a region, which is not physically determined, but it is bounded by our mind. In other words, you could see a cloud, but somebody in a different earth position could not see it, despite looking at the same region in the sky. You can see one cloud where other observer could see five or zero.
Numbers represent ideas or concepts (to understand what a concept is and how are they acquired, see Kant's Critique of Pure Reason). One thing is the concept (zero, one, 10⁸⁴ apples, an infinite number of apples) and another is the ideal of a physical existence of apples, which is not a real fact. Apples are like clouds, except that molecules are more dense and perhaps different. Your brain tells you where an apple ends and and another begins, but that's not the physical reality. Out of our minds, everything is just a fuzzy substance-like made of quarks or strings, which are not things like in the macroscopic realm, and don't behave as things.
Numbers do exist in our brains. So, all numbers (including zero, one, -2/5, PI or infinite) can exist in your brain. I'm pretty sure that the number 928754629384 hadn't existed in your brain until I've wrote it here. Numbers do not exist physically, nor the things that they represent.
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answered Sep 12, 2019 at 5:15
RodolfoAPRodolfoAP
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Your question is contradictory, if you don't believe in the existence of whole numbers other than 0, how could you count the "infinite" amount of decimals?
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answered Apr 2, 2018 at 21:51
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1 Actually, you can't count the infinite amount of decimals either way, because the infinity of the reals is larger than the infinity of the integers.Challenger5 –Challenger5 2018-04-03 01:44:02 +00:00 Commented Apr 3, 2018 at 1:44
@Challenger5 Depends what you mean by "decimals". There are only countably many terminating decimal expansions, and "only consider terminating decimal expansions" is a pretty reasonable interpretation of a naive question.Patrick Stevens –Patrick Stevens 2018-04-03 08:57:58 +00:00 Commented Apr 3, 2018 at 8:57
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I'll leave my two cents here :)
So I had a question also, much like the OP. But mine was, "IS there such a thing as a whole number? For example, the number 1 is made up of infinite small pieces/parts/decimal places/points. So in reality everything is an infinite fraction of tiny pieces of itself, so 1, is really ALL.
There would be no reason to ever leave the number one, nor may it even be possible. If we arrived at 2 somehow, 2 of what? 2 of 1? Well, now we have another problem, you can’t have 2 things that both equal infinity; this could cause all sorts of problems.
So, instead we stand at one and theorize what two might look like, for we have decided that it is imperative we arrive at 2 as a matter of fact, based purely, on supposition.
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edited Sep 11, 2019 at 17:22
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Hello, and welcome to Philosophy.SE. I think there are a number of problems with this answer, namely an unconventional understanding of infinity, the lack of a definite answer and the lack of any sources to support the position expressed here. Would you mind to edit the text to pose an answer proper instead of just expressing the context of thought that led you to a similar (yet ultimately unanswered) question?Philip Klöcking –Philip Klöcking♦ 2019-09-11 07:58:26 +00:00 Commented Sep 11, 2019 at 7:58
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8329 | https://www.sterlingtt.com/2023/11/14/examples-of-hydropower/ | Exploring examples of hydropower
Updated on November 14, 2023.
The modern history of hydropower can be traced back to the 1800s, with hundreds of hydropower plants opening globally.
In recent years, the demand for renewable energy sources has been at an all-time high, making hydroelectric energy a desirable asset for any economy. In this article, we will discuss examples of hydropower and how this can look on a large scale.
What is hydropower?
Hydropower, sometimes referred to as hydroelectric energy, is a form of renewable energy generated from the movement of water. Hydropowered machines convert the kinetic energy of the water into useful work directly or electricity to be used elsewhere.
Using water for work has roots far back in history, in the ancient civilisations of Rome, Greece and China. There, hydropower took the form of waterwheels that powered flour mills or water scoops that moved hammers in smitheries.
Today, hydropower is most likely to refer to facilities where the water turns a turbine to generate electricity. This can also be called hydroelectric power. These are often in the form of dams.
Examples of types of hydropower
There are three different examples of hydroelectric power stations that we see used on a large scale: impoundment, diversion, and pumped storage. All three types of facilities have their unique features.
Impoundment
In an impoundment facility, a large body of water is captured behind a dam (in other words, the water is impounded). When the need arises for more power to be generated, water is allowed to pass through the turbines of the dam, making them rotate and generate electricity.
Impoundment facilities are the most commonly used hydroelectric power stations.
In recent years, there has been a more significant push for impoundment facilities to include fish steps or fishways in their design in the hope of causing less of an environmental impact with their construction.
Diversion
A diversion facility, sometimes called a “run-of-river facility”, temporarily diverts a river away from its natural course. The river may be split into two: one section is left to run uninterrupted, and the kinetic energy of the other is used to generate power.
A penstock (a chamber consisting of gates, valves, and turbines) channels the split-off section of water, and then later reconnects with the original body of water.
This type of hydropower facility is typically more environmentally friendly than impoundment facilities due to reduced disruption to the river’s natural flow.
Pumped-storage
Pumped storage facilities, sometimes called PSH facilities, function as a giant battery. With PSH facilities, electricity from other sources, such as wind or nuclear power, moves water from one reservoir to another at a higher ascension. It happens during times of low energy consumption. The water is then released back into the lower reservoir, passing through the turbines, at peak times for energy consumption.
PSH facilities are generally highly economical but can cause higher levels of pollution than other hydropower options.
What makes hydropower a good option?
Hydroelectric energy is a renewable energy source, as it relies on the natural water cycle. As long as water resources are managed sustainably, it can provide a constant and reliable source of electricity.
Moreover, compared to other electrical generating facilities, hydropower facilities may be more affordable to both construct and operate.
Pumped storage hydropower facilities can generate power immediately to the power grid, making them a valuable backup power source during major electrical outages and disruptions.
Beyond this, hydropower facilities also offer benefits outside of electrical generation such as flood control, irrigation support, and providing clean drinking water.
It’s worth noting that while hydropower has significant advantages and will likely be an essential tool in a more sustainable future, it does have environmental and social impacts, such as altering river ecosystems. Any potential drawbacks of a hydropower project must be considered carefully by the designers.
Is hydropower used globally?
The capabilities of hydropower are utilised all around the world. According to the International Energy Agency, in 2020, hydropower supplied 17% of global electricity generation. It made it the third-largest contributor, behind coal and natural gas.
What is the largest hydropower in the world?
Presently, the Three Gorges Dam in the Yangtze River holds the title for the largest hydroelectric power facility globally. Boasting more than 30 turbine generators, this dam is recognised for its unmatched productivity among hydroelectric facilities worldwide.
Image source: Wikipedia Three Gorges Dam China 2009
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8330 | https://brainly.com/question/40247798 | [FREE] Which of the following is the equation of a tangent line to the parabola y^2 = 2px that passes through the - brainly.com
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Which of the following is the equation of a tangent line to the parabola y 2=2 p x that passes through the point A(7 4,−4)?
Option 1: y=2 x−9
Option 2: y=2 x+7 36
Option 3: y=2 x+7 36
Option 4: y=−2 x−9
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To find the equation of a tangent line to the parabola y^2=2px that passes through a given point, substitute the coordinates of the point into the equation of the parabola to find the point of tangency, calculate the slope of the tangent line using the point of tangency, and use the slope and the coordinates of the point of tangency in the equation y = mx + b to find the y-intercept.
Explanation
The equation of a tangent line to the parabola y2 = 2px that passes through the point A (4/7,-4) can be found using the general equation for a line, y = mx + b, where m is the slope and b is the y-intercept.
First, we need to find the slope of the tangent line. We know that the point A lies on the tangent line, so we can substitute the coordinates of point A into the equation of the parabola to find the corresponding x-coordinate of the point of tangency.
Substituting the x-coordinate back into the equation of the parabola will give us the corresponding y-coordinate of the point of tangency. Once we have the coordinates of the point of tangency, we can calculate the slope of the tangent line. Finally, we can substitute the slope and the coordinates of the point of tangency into the equation y = mx + b and solve for b to find the y-intercept.
After performing these calculations, we find that the correct equation of the tangent line is y = 2x + 36/7.
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The equation of the tangent line to the parabola y 2=2 p x that passes through the point A(7 4,−4) is given by Option 2: y=2 x+7 36. This was determined by finding the slope of the tangent line and verifying the coordinates of point A against the available options.
Explanation
To find the equation of a tangent line to the parabola given by y 2=2 p x that passes through the point A(7 4,−4), we can follow these steps:
Identify the General Equation of the Tangent Line: The equation of a line can be expressed in the form y−y 1=m(x−x 1), where m is the slope and (x 1,y 1) is a point on the line.
Calculate the General Point of Tangency: For the parabola y 2=2 p x, a point of tangency can be expressed as (x 0,y 0). Generally, the slope of the tangent at this point can be derived from the parabola's equation, showing y 0=±2 p x 0. The slope of the tangent line m=d x d y=y 0p.
Substitute the Coordinates: Substitute the coordinates of point A into the tangent line equation to express it in terms of the slope and the point of tangency. You are looking for m and the corresponding y 0 that satisfies both conditions.
Check Possible Options: Set the equation found in step 1 in the form y=m x+b. Now, compare the resulting expression from the calculations with the provided options:
Option 1: y=2 x−9
Option 2: y=2 x+7 36
Option 3: y=2 x+7 36 (note this is a repeat of option 2)
Option 4: y=−2 x−9
Test the Options by Checking the Point A: Substitute x=7 4,y=−4 into each option. The only equation that satisfies this calculation is Option 2: y=2 x+7 36.
In summary, through calculations to find the point of tangency and substituting the coordinates of point A into potential equations, we determine that the correct equation of the tangent line is Option 2: y=2 x+7 36.
Examples & Evidence
Before confirming which equation is correct, one might substitute the coordinates of point A into each option to see if it holds true for that point, which serves as an important verification step for determining the tangent line.
This approach aligns with the principles of calculus, where understanding the properties of tangents to curves, especially parabolas, relies on recognizing slopes and using point coordinates to ascertain the validity of potential equations.
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Company X tried selling widgets at various prices to see how much profit they would make. The following table shows the widget selling price, x, and the total profit earned at that price, y. Write a quadratic regression equation for this set of data, rounding all coefficients to the nearest tenth. Using this equation, find the profit, to the nearest dollar, for a selling price of 14.75 dollars. | Price (x) | Profit (y) | :---: | | 12.50 | 884 | | 16.25 | 1309 | | 23.25 | 1996 | | 33.75 | 2003 | | 39.00 | 1814 |
The data shows poll ratings of a politician over time. | month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | :---: :---: :---: :---: :---: :---: | rating | 12 | 18 | 25 | 32 | 41 | 37 | 30 | 27 | 35 | 42 | 49 | 56 | a. Is the data Linear, Quadratic, or Cubic? b. Write the equation of the regression curve. c. Use your equation to find the average rate of change from May to October.
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8331 | https://ihs-headache.org/wp-content/uploads/2020/05/ICHD-3-Pocket-version.pdf | INTERNATIONAL HEADACHE SOCIETY Company limited by guarantee, registered in England no.2988368 Registered charity no. 1042574 The International Classification of Headache Disorders 3rd Edition (ICHD-3) Abbreviated pocket version for reference by professional users only prepared by the Headache Classification Committee of the International Headache Society Jes Olesen (Chairman), Timothy J Steiner (Secretary), Lars Bendtsen, David Dodick, Anne Ducros, Stefan Evers, Michael First, Peter J Goadsby, Andrew Hershey, Zaza Katsarava, Morris Levin, Julio Pascual, Michael B Russell, Todd Schwedt, Cristina Tassorelli, Gisela M Terwindt, Maurice Vincent, Shuu-Jiun Wang International Headache Society 2017/18 2 Introduction to Abbreviated Pocket Version The International Classification of Headache Disorders, 3rd edition, is published in Cephalalgia and freely accessible at This abbreviated version includes the most common or important headache disorders (and some that are frequently over- or misdiagnosed) as an aide memoire for those familiar with the classification principles and experienced in their application. It lists the diagnostic criteria but omits explanatory introductions, descriptions, notes and comments, which in many cases are key to proper and accurate usage. Classification ICHD-3 code Diagnosis 1. Migraine 1.1 Migraine without aura 1.2 Migraine with aura 1.2.1 Migraine with typical aura 1.2.1.1 Typical aura with headache 1.2.1.2 Typical aura without headache 1.2.2 Migraine with brainstem aura 1.2.3 Hemiplegic migraine 1.2.3.1 Familial hemiplegic migraine (FHM) 1.2.3.1.1 FHM type 1 (FHM1) 1.2.3.1.2 FHM type 2 (FHM2) 1.2.3.1.3 FHM type 3 (FHM3) 1.2.3.1.4 FHM, other loci 1.2.3.2 Sporadic hemiplegic migraine 1.2.4 Retinal migraine 1.3 Chronic migraine 1.4 Complications of migraine 1.4.1 Status migrainosus 1.4.2 Persistent aura without infarction 1.4.3 Migrainous infarction 1.4.4 Migraine aura-triggered seizure 1.5 Probable migraine 1.5.1 Probable migraine without aura 1.5.2 Probable migraine with aura 1.6 Episodic syndromes that may be associated with migraine 1.6.1 Recurrent gastrointestinal disturbance 1.6.1.1 Cyclical vomiting syndrome 1.6.1.2 Abdominal migraine 1.6.2 Benign paroxysmal vertigo 1.6.3 Benign paroxysmal torticollis 2. Tension-type headache (TTH) 2.1 Infrequent episodic TTH 3 2.1.1 Infrequent episodic TTH associated with pericranial tenderness 2.1.2 Infrequent episodic TTH not associated with pericranial tenderness 2.2 Frequent episodic TTH 2.2.1 Frequent episodic TTH associated with pericranial tenderness 2.2.2 Frequent episodic TTH not associated with pericranial tenderness 2.3 Chronic TTH 2.3.1 Chronic TTH associated with pericranial tenderness 2.3.2 Chronic TTH not associated with pericranial tenderness 2.4 Probable TTH 2.4.1 Probable infrequent episodic TTH 2.4.2 Probable frequent episodic TTH 2.4.3 Probable chronic TTH 3. Trigeminal autonomic cephalalgias 3.1 Cluster headache 3.1.1 Episodic cluster headache 3.1.2 Chronic cluster headache 3.2 Paroxysmal hemicrania 3.2.1 Episodic paroxysmal hemicrania 3.2.2 Chronic paroxysmal hemicrania 3.3 Short-lasting unilateral neuralgiform headache attacks 3.3.1 Short-lasting unilateral neuralgiform headache attacks with conjunctival injection and tearing (SUNCT) 3.3.1.1 Episodic SUNCT 3.3.1.2 Chronic SUNCT 3.3.2 Short-lasting unilateral neuralgiform headache attacks with cranial autonomic symptoms (SUNA) 3.3.2.1 Episodic SUNA 3.3.2.2 Chronic SUNA 3.4 Hemicrania continua 3.5 Probable trigeminal autonomic cephalalgia 3.5.1 Probable cluster headache 3.5.2 Probable paroxysmal hemicrania 3.5.3 Probable short-lasting unilateral neuralgiform headache attacks 3.5.4 Probable hemicrania continua 4. Other primary headache disorders 4.1 Primary cough headache 4.1.1 Probable primary cough headache 4.2 Primary exercise headache 4.2.1 Probable primary exercise headache 4.3 Primary headache associated with sexual activity 4.3.1 Probable primary headache associated with sexual activity 4.4 Primary thunderclap headache 4.5 Cold-stimulus headache 4 4.5.1 Headache attributed to external application of a cold stimulus 4.5.2 Headache attributed to ingestion or inhalation of a cold stimulus 4.5.3 Probable cold-stimulus headache 4.5.3.1 Headache probably attributed to external application of a cold stimulus 4.5.3.2 Headache probably attributed to ingestion or inhalation of a cold stimulus 4.6 External-pressure headache 4.6.1 External-compression headache 4.6.2 External-traction headache 4.6.3 Probable external-pressure headache 4.6.3.1 Probable external-compression headache 4.6.3.2 Probable external-traction headache 4.7 Primary stabbing headache 4.7.1 Probable primary stabbing headache 4.8 Nummular headache 4.8.1 Probable nummular headache 4.9 Hypnic headache 4.9.1 Probable hypnic headache 4.10 New daily persistent headache (NDPH) 4.10.1 Probable NDPH 5. Headache attributed to trauma or injury to the head and/or neck 5.1 Acute headache attributed to traumatic injury to the head 5.1.1 Acute headache attributed to moderate or severe traumatic injury to the head 5.1.2 Acute headache attributed to mild traumatic injury to the head 5.2 Persistent headache attributed to traumatic injury to the head 5.2.1 Persistent headache attributed to moderate or severe traumatic injury to the head 5.2.2 Persistent headache attributed to mild traumatic injury to the head 5.3 Acute headache attributed to whiplash 5.4 Persistent headache attributed to whiplash 5.5 Acute headache attributed to craniotomy 5.6 Persistent headache attributed to craniotomy 6. Headache attributed to cranial and/or cervical vascular disorder 6.1 Headache attributed to cerebral ischaemic event 6.1.1 Headache attributed to ischaemic stroke (cerebral infarction) 6.1.1.1 Acute headache attributed to ischaemic stroke 6.1.1.2 Persistent headache attributed to past ischaemic stroke 6.1.2 Headache attributed to transient ischaemic attack 6.2 Headache attributed to non-traumatic intracranial haemorrhage 6.2.1 Acute headache attributed to non-traumatic intracerebral haemorrhage 5 6.2.2 Acute headache attributed to non-traumatic subarachnoid haemorrhage 6.2.3 Acute headache attributed to non-traumatic acute subdural haemorrhage 6.2.4 Persistent headache attributed to past non-traumatic intracranial haemorrhage 6.3 Headache attributed to unruptured vascular malformation 6.3.1 Headache attributed to unruptured saccular aneurysm 6.3.2 Headache attributed to arteriovenous malformation 6.3.3 Headache attributed to dural arteriovenous fistula 6.3.4 Headache attributed to cavernous angioma 6.3.5 Headache attributed to encephalotrigeminal or leptomeningeal angiomatosis (Sturge Weber syndrome) 6.4 Headache attributed to arteritis 6.4.1 Headache attributed to giant cell arteritis 6.4.2 Headache attributed to primary angiitis of the central nervous system 6.4.3 Headache attributed to secondary angiitis of the central nervous system 6.5 Headache attributed to cervical carotid or vertebral artery disorder 6.5.1 Headache or facial or neck pain attributed to cervical carotid or vertebral artery dissection 6.5.1.1 Acute headache or facial or neck pain attributed to cervical artery dissection 6.5.1.2 Persistent headache or facial or neck pain attributed to past cervical artery dissection 6.5.2 Post-endarterectomy headache 6.5.3 Headache attributed to carotid or vertebral angioplasty or stenting 6.6 Headache attributed to cranial venous disorder 6.6.1 Headache attributed to cerebral venous thrombosis 6.6.2 Headache attributed to cranial venous sinus stenting 6.7 Headache attributed to other acute intracranial arterial disorder 6.7.1 Headache attributed to an intracranial endarterial procedure 6.7.2 Angiography headache 6.7.3 Headache attributed to reversible cerebral vasoconstriction syndrome (RCVS) 6.7.3.1 Acute headache attributed to RCVS 6.7.3.2 Acute headache probably attributed to RCVS 6.7.3.3 Persistent headache attributed to past RCVS 6.7.4 Headache attributed to intracranial artery dissection 6.8 Headache and/or migraine-like aura attributed to chronic intracranial vasculopathy 6.8.1 Headache attributed to Cerebral Autosomal Dominant Arteriopathy with Subcortical Infarcts and Leukoencephalopathy (CADASIL) 6 6.8.2 Headache attributed to Mitochondrial Encephalopathy, Lactic Acidosis and Stroke-like episodes (MELAS) 6.8.3 Headache attributed to Moyamoya angiopathy 6.8.4 Migraine-like aura attributed to cerebral amyloid angiopathy 6.8.5 Headache attributed to syndrome of retinal vasculopathy with cerebral leukoencephalopathy and systemic manifestations 6.8.6 Headache attributed to other genetic vasculopathy 6.9 Headache attributed to pituitary apoplexy 7. Headache attributed to non-vascular intracranial disorder 7.1 Headache attributed to increased cerebrospinal fluid pressure 7.1.1 Headache attributed to idiopathic intracranial hypertension 7.1.2 Headache attributed to intracranial hypertension secondary to metabolic, toxic or hormonal cause 7.1.3 Headache attributed to intracranial hypertension secondary to chromosomal disorder 7.1.4 Headache attributed to intracranial hypertension secondary to hydrocephalus 7.2 Headache attributed to low cerebrospinal fluid (CSF) pressure 7.2.1 Post-dural puncture headache 7.2.2 CSF fistula headache 7.2.3 Headache attributed to spontaneous intracranial hypotension 7.3 Headache attributed to non-infectious inflammatory disease 7.3.1 Headache attributed to neurosarcoidosis 7.3.2 Headache attributed to aseptic (non-infectious) meningitis 7.3.3 Headache attributed to other non-infectious inflammatory disease 7.3.4 Headache attributed to lymphocytic hypophysitis 7.3.5 Syndrome of transient Headache and Neurological Deficits with cerebrospinal fluid Lymphocytosis (HaNDL) 7.4 Headache attributed to intracranial neoplasia 7.4.1 Headache attributed to intracranial neoplasm 7.4.1.1 Headache attributed to colloid cyst of the third ventricle 7.4.2 Headache attributed to carcinomatous meningitis 7.4.3 Headache attributed to hypothalamic or pituitary hyper- or hyposecretion 7.5 Headache attributed to intrathecal injection 7.6 Headache attributed to epileptic seizure 7.6.1 Ictal epileptic headache 7.6.2 Post-ictal headache 7.7 Headache attributed to Chiari malformation type I 7.8 Headache attributed to other non-vascular intracranial disorder 7 8. Headache attributed to a substance or its withdrawal 8.1 Headache attributed to use of or exposure to a substance 8.1.1 Nitric oxide (NO) donor-induced headache 8.1.1.1 Immediate NO donor-induced headache 8.1.1.2 Delayed NO donor-induced headache 8.1.2 Phosphodiesterase inhibitor-induced headache 8.1.3 Carbon monoxide-induced headache 8.1.4 Alcohol-induced headache 8.1.4.1 Immediate alcohol-induced headache 8.1.4.2 Delayed alcohol-induced headache 8.1.5 Cocaine-induced headache 8.1.6 Histamine-induced headache 8.1.6.1 Immediate histamine-induced headache 8.1.6.2 Delayed histamine-induced headache 8.1.7 Calcitonin gene-related peptide (CGRP)-induced headache 8.1.7.1 Immediate CGRP-induced headache 8.1.7.2 Delayed CGRP-induced headache 8.1.8 Headache attributed to exogenous acute pressor agent 8.1.9 Headache attributed to occasional use of non-headache medication 8.1.10 Headache attributed to long-term use of non-headache medication 8.1.11 Headache attributed to use of or exposure to other substance 8.2 Medication-overuse headache (MOH) 8.2.1 Ergotamine-overuse headache 8.2.2 Triptan-overuse headache 8.2.3 Non-opioid analgesic-overuse headache 8.2.3.1 Paracetamol (acetaminophen)-overuse headache 8.2.3.2 Non-steroidal anti-inflammatory drug-overuse headache 8.2.3.2.1 Acetylsalicylic acid-overuse headache 8.2.3.3 Other non-opioid analgesic-overuse headache 8.2.4 Opioid-overuse headache 8.2.5 Combination-analgesic-overuse headache 8.2.6 MOH attributed to multiple drug classes not individually overused 8.2.7 MOH attributed to unspecified or unverified overuse of multiple drug classes 8.2.8 MOH attributed to other medication 8.3 Headache attributed to substance withdrawal 8.3.1 Caffeine-withdrawal headache 8.3.2 Opioid-withdrawal headache 8.3.3 Estrogen-withdrawal headache 8.3.4 Headache attributed to withdrawal from chronic use of other substance 9. Headache attributed to infection 9.1 Headache attributed to intracranial infection 9.1.1 Headache attributed to bacterial meningitis or meningoencephalitis 9.1.1.1 Acute headache attributed to bacterial meningitis or meningoencephalitis 8 9.1.1.2 Chronic headache attributed to bacterial meningitis or meningoencephalitis 9.1.1.3 Persistent headache attributed to past bacterial meningitis or meningoencephalitis 9.1.2 Headache attributed to viral meningitis or encephalitis 9.1.2.1 Headache attributed to viral meningitis 9.1.2.2 Headache attributed to viral encephalitis 9.1.3 Headache attributed to intracranial fungal or other parasitic infection 9.1.3.1 Acute headache attributed to intracranial fungal or other parasitic infection 9.1.3.2 Chronic headache attributed to intracranial fungal or other parasitic infection 9.1.4 Headache attributed to localized brain infection 9.2 Headache attributed to systemic infection 9.2.1 Headache attributed to systemic bacterial infection 9.2.1.1 Acute headache attributed to systemic bacterial infection 9.2.1.2 Chronic headache attributed to systemic bacterial infection 9.2.2 Headache attributed to systemic viral infection 9.2.2.1 Acute headache attributed to systemic viral infection 9.2.2.2 Chronic headache attributed to systemic viral infection 9.2.3 Headache attributed to other systemic infection 9.2.3.1 Acute headache attributed to other systemic infection 9.2.3.2 Chronic headache attributed to other systemic infection 10. Headache attributed to disorder of homoeostasis 10.1 Headache attributed to hypoxia and/or hypercapnia 10.1.1 High-altitude headache 10.1.2 Headache attributed to aeroplane travel 10.1.3 Diving headache 10.1.4 Sleep apnoea headache 10.2 Dialysis headache 10.3 Headache attributed to arterial hypertension 10.3.1 Headache attributed to phaeochromocytoma 10.3.2 Headache attributed to hypertensive crisis without hypertensive encephalopathy 10.3.3 Headache attributed to hypertensive encephalopathy 10.3.4 Headache attributed to pre-eclampsia or eclampsia 10.3.5 Headache attributed to autonomic dysreflexia 10.4 Headache attributed to hypothyroidism 10.5 Headache attributed to fasting 10.6 Cardiac cephalalgia 10.7 Headache attributed to other disorder of homoeostasis 9 11. Headache or facial pain attributed to disorder of the cranium, neck, eyes, ears, nose, sinuses, teeth, mouth or other facial or cervical structure 11.1 Headache attributed to disorder of cranial bone 11.2 Headache attributed to disorder of the neck 11.2.1 Cervicogenic headache 11.2.2 Headache attributed to retropharyngeal tendonitis 11.2.3 Headache attributed to craniocervical dystonia 11.3 Headache attributed to disorder of the eyes 11.3.1 Headache attributed to acute angle-closure glaucoma 11.3.2 Headache attributed to refractive error 11.3.3 Headache attributed to ocular inflammatory disorder 11.3.4 Trochlear headache 11.4 Headache attributed to disorder of the ears 11.5 Headache attributed to disorder of the nose or paranasal sinuses 11.5.1 Headache attributed to acute rhinosinusitis 11.5.2 Headache attributed to chronic or recurring rhinosinusitis 11.6 Headache attributed to disorder of the teeth 11.7 Headache attributed to temporomandibular disorder 11.8 Head or facial pain attributed to inflammation of the stylohyoid ligament 11.9 Headache or facial pain attributed to other disorder of cranium, neck, eyes, ears, nose, sinuses, teeth, mouth or other facial or cervical structure 12. Headache attributed to psychiatric disorder 12.1 Headache attributed to somatization disorder 12.2 Headache attributed to psychotic disorder 13. Painful lesions of the cranial nerves and other facial pain 13.1 Pain attributed to a lesion or disease of the trigeminal nerve 13.1.1 Trigeminal neuralgia 13.1.1.1 Classical trigeminal neuralgia 13.1.1.1.1 Classical trigeminal neuralgia, purely paroxysmal 13.1.1.1.2 Classical trigeminal neuralgia with concomitant continuous pain 13.1.1.2 Secondary trigeminal neuralgia 13.1.1.2.1 Trigeminal neuralgia attributed to multiple sclerosis 13.1.1.2.2 Trigeminal neuralgia attributed to space-occupying lesion 13.1.1.2.3 Trigeminal neuralgia attributed to other cause 13.1.1.3 Idiopathic trigeminal neuralgia 13.1.1.3.1 Idiopathic trigeminal neuralgia, purely paroxysmal 13.1.1.3.2 Idiopathic trigeminal neuralgia with concomitant continuous pain 13.1.2 Painful trigeminal neuropathy 10 13.1.2.1 Painful trigeminal neuropathy attributed to herpes zoster 13.1.2.2 Trigeminal post-herpetic neuralgia 13.1.2.3 Painful post-traumatic trigeminal neuropathy 13.1.2.4 Painful trigeminal neuropathy attributed to other disorder 13.1.2.5 Idiopathic painful trigeminal neuropathy 13.2 Pain attributed to a lesion or disease of the glossopharyngeal nerve 13.2.1 Glossopharyngeal neuralgia 13.2.1.1 Classical glossopharyngeal neuralgia 13.2.1.2 Secondary glossopharyngeal neuralgia 13.2.1.3 Idiopathic glossopharyngeal neuralgia 13.2.2 Painful glossopharyngeal neuropathy 13.2.2.1 Painful glossopharyngeal neuropathy attributed to a known cause 13.2.2.2 Idiopathic painful glossopharyngeal neuropathy 13.3 Pain attributed to a lesion or disease of nervus intermedius 13.3.1 Nervus intermedius neuralgia 13.3.1.1 Classical nervus intermedius neuralgia 13.3.1.2 Secondary nervus intermedius neuralgia 13.3.1.3 Idiopathic nervus intermedius neuralgia 13.3.2 Painful nervus intermedius neuropathy 13.3.2.1 Painful nervus intermedius neuropathy attributed to herpes zoster 13.3.2.2 Post-herpetic neuralgia of nervus intermedius 13.3.2.3 Painful nervus intermedius neuropathy attributed to other disorder 13.3.2.4 Idiopathic painful nervus intermedius neuropathy 13.4 Occipital neuralgia 13.5 Neck-tongue syndrome 13.6 Painful optic neuritis 13.7 Headache attributed to ischaemic ocular motor nerve palsy 13.8 Tolosa-Hunt syndrome 13.9 Paratrigeminal oculosympathetic (Raeder’s) syndrome 13.10 Recurrent painful ophthalmoplegic neuropathy 13.11 Burning mouth syndrome 13.12 Persistent idiopathic facial pain 13.13 Central neuropathic pain 13.13.1 Central neuropathic pain attributed to multiple sclerosis 13.13.2 Central post-stroke pain 14. Other headache disorders 14.1 Headache not elsewhere classified 14.2 Headache unspecified 11 PART 1. THE PRIMARY HEADACHES 1. Migraine 1.1 Migraine without aura A. At least five attacks fulfilling criteria B-D B. Headache attacks lasting 4-72 hours (when untreated or unsuccessfully treated) C. Headache has at least two of the following four characteristics: 1. unilateral location 2. pulsating quality 3. moderate or severe pain intensity 4. aggravation by or causing avoidance of routine physical activity (eg, walking or climbing stairs) D. During headache at least one of the following: 1. nausea and/or vomiting 2. photophobia and phonophobia E. Not better accounted for by another ICHD-3 diagnosis. 1.2 Migraine with aura A. At least two attacks fulfilling criteria B and C B. One or more of the following fully reversible aura symptoms: 1. visual 2. sensory 3. speech and/or language 4. motor 5. brainstem 6. retinal C. At least three of the following six characteristics: 1. at least one aura symptom spreads gradually over ≥5 minutes 2. two or more aura symptoms occur in succession 3. each individual aura symptom lasts 5-60 minutes 4. at least one aura symptom is unilateral 5. at least one aura symptom is positive 6. the aura is accompanied, or followed within 60 minutes, by headache D. Not better accounted for by another ICHD-3 diagnosis. 1.2.1 Migraine with typical aura A. Attacks fulfilling criteria for 1.2 Migraine with aura and criterion B below B. Aura with both of the following: 1. fully reversible visual, sensory and/or speech/language symptoms 2. no motor, brainstem or retinal symptoms. 1.2.1.1 Typical aura with headache A. Attacks fulfilling criteria for 1.2.1 Migraine with typical aura and criterion B below B. Headache, with or without migraine characteristics, accompanies or follows the aura within 60 minutes. 12 1.2.1.2 Typical aura without headache A. Attacks fulfilling criteria for 1.2.1 Migraine with typical aura and criterion B below B. No headache accompanies or follows the aura within 60 minutes. 1.2.2 Migraine with brainstem aura A. Attacks fulfilling criteria for 1.2 Migraine with aura and criterion B below B. Aura with both of the following: 1. at least two of the following fully reversible brainstem symptoms: a) dysarthria b) vertigo c) tinnitus d) hypacusis e) diplopia f) ataxia not attributable to sensory deficit g) decreased level of consciousness (GCS ≤13) 2. no motor or retinal symptoms. 1.2.3 Hemiplegic migraine A. Attacks fulfilling criteria for 1.2 Migraine with aura and criterion B below B. Aura consisting of both of the following: 1. fully reversible motor weakness 2. fully reversible visual, sensory and/or speech/language symptoms. 1.2.3.1 Familial hemiplegic migraine A. Attacks fulfilling criteria for 1.2.3 Hemiplegic migraine B. At least one first- or second-degree relative has had attacks fulfilling criteria for 1.2.3 Hemiplegic migraine. 1.3 Chronic migraine A. Headache (migraine-like or tension-type-like) on ≥15 days/month for >3 months, and fulfilling criteria B and C B. Occurring in a patient who has had at least five attacks fulfilling criteria B-D for 1.1 Migraine without aura and/or criteria B and C for 1.2 Migraine with aura C. On ≥8 days/month for >3 months, fulfilling any of the following: 1. criteria C and D for 1.1 Migraine without aura 2. criteria B and C for 1.2 Migraine with aura 3. believed by the patient to be migraine at onset and relieved by a triptan or ergot derivative D. Not better accounted for by another ICHD-3 diagnosis. 13 2. Tension-type headache (TTH) 2.1 Infrequent episodic TTH A. At least 10 episodes of headache occurring on <1 day/month on average (<12 days/year) and fulfilling criteria B-D B. Lasting from 30 minutes to 7 days C. At least two of the following four characteristics: 1. bilateral location 2. pressing or tightening (non-pulsating) quality 3. mild or moderate intensity 4. not aggravated by routine physical activity such as walking or climbing stairs D. Both of the following: 1. no nausea or vomiting 2. no more than one of photophobia or phonophobia E. Not better accounted for by another ICHD-3 diagnosis. 2.2 Frequent episodic TTH As 2.1 except: A. At least 10 episodes of headache occurring on 1-14 days/month on average for >3 months (12 and <180 days/year) and fulfilling criteria B-D. 2.3 Chronic TTH As 2.1 except: A. Headache occurring on 15 days/month on average for >3 months (180 days/year), fulfilling criteria B-D B. Lasting hours to days, or unremitting D. Both of the following: 1. no more than one of photophobia, phonophobia or mild nausea 2. neither moderate or severe nausea nor vomiting 3. Trigeminal autonomic cephalalgias 3.1 Cluster headache A. At least five attacks fulfilling criteria B-D B. Severe or very severe unilateral orbital, supraorbital and/or temporal pain lasting 15-180 minutes (when untreated) C. Either or both of the following: 1. at least one of the following symptoms or signs, ipsilateral to the headache: a) conjunctival injection and/or lacrimation b) nasal congestion and/or rhinorrhoea c) eyelid oedema d) forehead and facial sweating e) miosis and/or ptosis 2. a sense of restlessness or agitation D. Occurring with a frequency between one every other day and 8 per day E. Not better accounted for by another ICHD-3 diagnosis. 14 3.1.1 Episodic cluster headache A. Attacks fulfilling criteria for 3.1 Cluster headache and occurring in bouts (cluster periods) B. At least two cluster periods lasting from 7 days to 1 year (when untreated) and separated by pain-free remission periods of ≥3 months. 3.1.2 Chronic cluster headache A. Attacks fulfilling criteria for 3.1 Cluster headache, and criterion B below B. Occurring without a remission period, or with remissions lasting <3 months, for at least 1 year. 3.4 Hemicrania continua A. Unilateral headache fulfilling criteria B-D B. Present for >3 months, with exacerbations of moderate or greater intensity C. Either or both of the following: 1. at least one of the following symptoms or signs, ipsilateral to the headache: a) conjunctival injection and/or lacrimation b) nasal congestion and/or rhinorrhoea c) eyelid oedema d) forehead and facial sweating e) miosis and/or ptosis 2. a sense of restlessness or agitation, or aggravation of the pain by movement D. Responds absolutely to therapeutic doses of indomethacin E. Not better accounted for by another ICHD-3 diagnosis. 4. Other primary headache disorders 4.3 Primary headache associated with sexual activity A. At least two episodes of pain in the head and/or neck fulfilling criteria B-D B. Brought on by and occurring only during sexual activity C. Either or both of the following: 1. increasing in intensity with increasing sexual excitement 2. abrupt explosive intensity just before or with orgasm D. Lasting from 1 minute to 24 hours with severe intensity and/or up to 72 hours with mild intensity E. Not better accounted for by another ICHD-3 diagnosis. 4.5 Cold-stimulus headache 4.5.1 Headache attributed to ingestion or inhalation of a cold stimulus A. At least two episodes of acute frontal or temporal headache fulfilling criteria B and C 15 B. Brought on by and occurring immediately after a cold stimulus to the palate and/or posterior pharyngeal wall from ingestion of cold food or drink or inhalation of cold air C. Resolving within 10 minutes after removal of the cold stimulus D. Not better accounted for by another ICHD-3 diagnosis. 4.7 Primary stabbing headache A. Head pain occurring spontaneously as a single stab or series of stabs and fulfilling criteria B and C B. Each stab lasts for up to a few seconds C. Stabs recur with irregular frequency, from one to many per day D. No cranial autonomic symptoms E. Not better accounted for by another ICHD-3 diagnosis. 4.8 Nummular headache A. Continuous or intermittent head pain fulfilling criterion B B. Felt exclusively in an area of the scalp, with all of the following four characteristics: 1. sharply-contoured 2. fixed in size and shape 3. round or elliptical 4. 1-6 cm in diameter C. Not better accounted for by another ICHD-3 diagnosis. 4.9 Hypnic headache A. Recurrent headache attacks fulfilling criteria B-D B. Developing only during sleep, and causing wakening C. Occurring on ≥10 days/month for >3 months D. Lasting from 15 minutes up to 4 hours after waking E. No cranial autonomic symptoms or restlessness F. Not better accounted for by another ICHD-3 diagnosis. 4.10 New daily persistent headache (NDPH) A. Persistent headache fulfilling criteria B and C B. Distinct and clearly-remembered onset, with pain becoming continuous and unremitting within 24 hours C. Present for >3 months D. Not better accounted for by another ICHD-3 diagnosis. PART 2. THE SECONDARY HEADACHES A new headache occurring in temporal association with another disorder recognized to be capable of causing it is diagnosed as secondary to that disorder. This remains true even when the headache has the characteristics of a primary headache (migraine, tension-type headache or one of the trigeminal autonomic cephalalgias). 16 When a pre-existing primary headache becomes chronic or is made significantly worse (usually meaning a two-fold or greater increase in frequency and/or severity) in close temporal relation to such a disorder, both the primary and the secondary headache diagnoses should be given, provided that there is good evidence that the disorder can cause headache. General diagnostic criteria for secondary headaches enact these rules. General diagnostic criteria for secondary headaches: A. Any headache fulfilling criterion C B. Another disorder scientifically documented to be able to cause headache has been diagnosed C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to the onset of the presumed causative disorder 2. either or both of the following: a) headache has significantly worsened in parallel with worsening of the presumed causative disorder b) headache has significantly improved in parallel with improvement of the presumed causative disorder 3. headache has characteristics typical for the causative disorder 4. other evidence exists of causation D. Not better accounted for by another ICHD-3 diagnosis. 5. Headache attributed to trauma or injury to the head and/or neck 5.2 Persistent headache attributed to traumatic injury to the head A. Any headache fulfilling criteria C and D B. Traumatic injury to the head has occurred C. Headache is reported to have developed within 7 days after one of the following: 1. the injury to the head 2. regaining of consciousness following the injury to the head 3. discontinuation of medication(s) impairing ability to sense or report headache following the injury to the head D. Headache persists for >3 months after its onset E. Not better accounted for by another ICHD-3 diagnosis. 5.2.1 Persistent headache attributed to moderate or severe traumatic injury to the head A. Headache fulfilling criteria for 5.2 Persistent headache attributed to traumatic injury to the head 17 B. Injury to the head associated with at least one of the following: 1. loss of consciousness for >30 minutes 2. Glasgow Coma Scale (GCS) score <13 3. post-traumatic amnesia lasting >24 hours 4. alteration in level of awareness for >24 hours 5. imaging evidence of a traumatic head injury such as skull fracture, intracranial haemorrhage and/or brain contusion. 5.4 Persistent headache attributed to whiplash A. Any headache fulfilling criteria C and D B. Whiplash, associated at the time with neck pain and/or headache, has occurred C. Headache has developed within 7 days after the whiplash D. Headache persists for >3 months after its onset E. Not better accounted for by another ICHD-3 diagnosis. 6. Headache attributed to cranial and/or cervical vascular disorder 6.2 Headache attributed to non-traumatic intracranial haemorrhage 6.2.2 Acute headache attributed to non-traumatic subarachnoid haemorrhage (SAH) A. Any new headache fulfilling criteria C and D B. SAH in the absence of head trauma has been diagnosed C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in close temporal relation to other symptoms and/or clinical signs of SAH, or has led to the diagnosis of SAH 2. headache has significantly improved in parallel with stabilization or improvement of other symptoms or clinical or radiological signs of SAH 3. headache has sudden or thunderclap onset D. Either of the following: 1. headache has resolved within 3 months 2. headache has not yet resolved but 3 months have not yet passed E. Not better accounted for by another ICHD-3 diagnosis. 6.4 Headache attributed to arteritis 6.4.1 Headache attributed to giant cell arteritis (GCA) A. Any new headache fulfilling criterion C B. GCA has been diagnosed 18 C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in close temporal relation to other symptoms and/or clinical or biological signs of onset of GCA, or has led to the diagnosis of GCA 2. either or both of the following: a) headache has significantly worsened in parallel with worsening of GCA b) headache has significantly improved or resolved within 3 days of high-dose steroid treatment 3. headache is associated with scalp tenderness and/or jaw claudication D. Not better accounted for by another ICHD-3 diagnosis. 6.5 Headache attributed to cervical carotid or vertebral artery disorder 6.5.1.1 Acute headache or facial or neck pain attributed to cervical artery dissection A. Any new headache and/or facial or neck pain fulfilling criteria C and D B. Cervical carotid or vertebral dissection has been diagnosed C. Evidence of causation demonstrated by at least two of the following: 1. pain has developed in close temporal relation to other local signs of the cervical artery dissection, or has led to its diagnosis 2. either or both of the following: a) pain has significantly worsened in parallel with other signs of the cervical artery dissection b) pain has significantly improved or resolved within 1 month of its onset 3. either or both of the following: a) pain is severe and continuous for days or longer b) pain precedes signs of acute retinal and/or cerebral ischaemia 4. pain is unilateral and ipsilateral to the affected cervical artery D. Either of the following: 1. headache has resolved within 3 months 2. headache has not yet resolved but 3 months have not yet passed E. Not better accounted for by another ICHD-3 diagnosis. 6.6 Headache attributed to cranial venous disorder 6.6.1 Headache attributed to cerebral venous thrombosis (CVT) A. Any new headache, fulfilling criterion C B. CVT has been diagnosed 19 C. Evidence of causation demonstrated by both of the following: 1. headache has developed in close temporal relation to other symptoms and/or clinical signs of CVT, or has led to the discovery of CVT 2. either or both of the following: a) headache has significantly worsened in parallel with clinical or radiological signs of extension of the CVT b) headache has significantly improved or resolved after improvement of the CVT D. Not better accounted for by another ICHD-3 diagnosis. 6.7 Headache attributed to other acute intracranial arterial disorder 6.7.3.1 Acute headache attributed to reversible cerebral vasoconstriction syndrome (RCVS) A. Any new headache fulfilling criterion C B. RCVS has been diagnosed C. Evidence of causation demonstrated by either or both of the following: 1. headache, with or without focal deficits and/or seizures, has led to angiography (with “strings and beads” appearance) and diagnosis of RCVS 2. headache has one or more of the following characteristics: a) thunderclap onset b) triggered by sexual activity, exertion, Valsalva manœuvres, emotion, bathing and/or showering c) present or recurrent during ≤1 month after onset, with no new significant headache after >1 month D. Either of the following: 1. headache has resolved within 3 months of onset 2. headache has not yet resolved but 3 months from onset have not yet passed E. Not better accounted for by another ICHD-3 diagnosis. 7. Headache attributed to non-vascular intracranial disorder 7.1 Headache attributed to increased cerebrospinal fluid (CSF) pressure A. New headache, or a significant worsening of a pre-existing headache, fulfilling criterion C B. Intracranial hypertension has been diagnosed, with both of the following: 1. CSF pressure exceeds 250 mm CSF (or 280 mm CSF in obese children) 2. normal CSF composition C. Evidence of causation demonstrated by at least two of the following: 20 1. headache has developed in temporal relation to the intracranial hypertension, or led to its discovery 2. headache is relieved by reducing the intracranial hypertension 3. papilloedema D. Not better accounted for by another ICHD-3 diagnosis. 7.1.1 Headache attributed to idiopathic intracranial hypertension (IIH) A. New headache, or a significant worsening of a pre-existing headache, fulfilling criterion C B. Both of the following: 1. IIH has been diagnosed 2. CSF pressure exceeds 250 mm CSF (or 280 mm CSF in obese children) C. Either or both of the following: 1. headache has developed or significantly worsened in temporal relation to the IIH, or led to its discovery 2. headache is accompanied by either or both of the following: a) pulsatile tinnitus b) papilloedema D. Not better accounted for by another ICHD-3 diagnosis. 7.2 Headache attributed to low cerebrospinal fluid (CSF) pressure A. Any headache fulfilling criterion C B. Either or both of the following: 1. low CSF pressure (<60 mm CSF) 2. evidence of CSF leakage on imaging C. Headache has developed in temporal relation to the low CSF pressure or CSF leakage, or led to its discovery D. Not better accounted for by another ICHD-3 diagnosis. 7.2.1 Post-dural puncture headache A. Headache fulfilling criteria for 7.2 Headache attributed to low CSF pressure, and criterion C below B. Dural puncture has been performed C. Headache has developed within 5 days of the dural puncture D. Not better accounted for by another ICHD-3 diagnosis. 7.2.3 Headache attributed to spontaneous intracranial hypotension A. Headache fulfilling criteria for 7.2 Headache attributed to low CSF pressure, and criterion C below B. Absence of a procedure or trauma known to be able to cause CSF leakage C. Headache has developed in temporal relation to occurrence of low CSF pressure or CSF leakage, or has led to its discovery D. Not better accounted for by another ICHD-3 diagnosis. 21 7.4 Headache attributed to intracranial neoplasia 7.4.1 Headache attributed to intracranial neoplasm A. Any headache fulfilling criterion C B. A space-occupying intracranial neoplasm has been demonstrated C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to development of the neoplasm, or led to its discovery 2. either or both of the following: a) headache has significantly worsened in parallel with worsening of the neoplasm b) headache has significantly improved in temporal relation to successful treatment of the neoplasm 3. headache has at least one of the following four characteristics: a) progressive b) worse in the morning and/or when lying down c) aggravated by Valsalva-like manœuvres d) accompanied by nausea and/or vomiting D. Not better accounted for by another ICHD-3 diagnosis. 8. Headache attributed to a substance or its withdrawal 8.1 Headache attributed to use of or exposure to a substance 8.1.3 Carbon monoxide (CO)-induced headache A. Bilateral headache fulfilling criterion C B. Exposure to CO has occurred C. Evidence of causation demonstrated by all of the following: 1. headache has developed within 12 hours of exposure to CO 2. headache intensity varies with the severity of CO intoxication 3. headache has resolved within 72 hours of elimination of CO D. Not better accounted for by another ICHD-3 diagnosis. 8.1.4.2 Delayed alcohol-induced headache A. Any headache fulfilling criterion C B. Alcohol has been ingested C. Evidence of causation demonstrated by all of the following: 1. headache has developed within 5-12 hours after ingestion of alcohol 2. headache has resolved within 72 hours of onset 3. headache has at least one of the following three characteristics: 22 a) bilateral b) pulsating quality c) aggravated by physical activity D. Not better accounted for by another ICHD-3 diagnosis. 8.2 Medication-overuse headache (MOH) A. Headache occurring on ≥15 days/month in a patient with a pre-existing headache disorder B. Regular overuse for >3 months of one or more drugs that can be taken for acute and/or symptomatic treatment of headache C. Not better accounted for by another ICHD-3 diagnosis. 8.2.1 Ergotamine-overuse headache A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of ergotamine on ≥10 days/month for >3 months. 8.2.2 Triptan-overuse headache A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of one or more triptans, in any formulation, on ≥10 days/month for >3 months. 8.2.3 Non-opioid analgesic-overuse headache 8.2.3.1 Paracetamol (acetaminophen)-overuse headache A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of paracetamol on ≥15 days/month for >3 months. 8.2.3.2 Non-steroidal anti-inflammatory drug (NSAID)-overuse headache A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of one or more NSAIDs (other than acetylsalicylic acid) on ≥15 days/month for >3 months. 8.2.3.2.1 Acetylsalicylic acid-overuse headache A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of acetylsalicylic acid on ≥15 days/month for >3 months. 8.2.4 Opioid-overuse headache A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of one or more opioids on ≥10 days/month for >3 months. 8.2.5 Combination analgesic-overuse headache A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of one or more combination-analgesic medications on ≥10 days/month for >3 months. 23 8.2.6 MOH attributed to multiple drug classes not individually overused A. Headache fulfilling criteria for 8.2 MOH B. Regular intake of any combination of ergotamine, triptans, non-opioid analgesics and/or opioids on a total of ≥10 days/month for >3 months without overuse of any single drug or drug class alone. 8.2.7 MOH attributed to unspecified or unverified overuse of multiple drug classes A. Headache fulfilling criteria for 8.2 MOH B. Both of the following: 1. regular intake of any combination of ergotamine, triptans, non-opioid analgesics and/or opioids on ≥10 days/month for >3 months 2. the identity, quantity and/or pattern of use or overuse of these classes of drug cannot be reliably established. 9. Headache attributed to infection 9.1 Headache attributed to intracranial infection 9.1.1 Headache attributed to bacterial meningitis or meningoencephalitis A. Headache of any duration fulfilling criterion C B. Bacterial meningitis or meningoencephalitis has been diagnosed C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to the onset of the bacterial meningitis or meningoencephalitis 2. headache has significantly worsened in parallel with worsening of the bacterial meningitis or meningoencephalitis 3. headache has significantly improved in parallel with improvement in the bacterial meningitis or meningoencephalitis 4. headache is either or both of the following: a) holocranial b) located in the nuchal area and associated with neck stiffness D. Not better accounted for by another ICHD-3 diagnosis. 9.1.2 Headache attributed to viral meningitis or encephalitis A. Any headache fulfilling criterion C B. Viral meningitis or encephalitis has been diagnosed C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to the onset of the viral meningitis or encephalitis 2. headache has significantly worsened in parallel with worsening of the viral meningitis or encephalitis 24 3. headache has significantly improved in parallel with improvement in the viral meningitis or encephalitis 4. headache is either or both of the following: a) holocranial b) located in the nuchal area and associated with neck stiffness D. Not better accounted for by another ICHD-3 diagnosis. 9.2 Headache attributed to systemic infection 9.2.2 Headache attributed to systemic viral infection A. Headache of any duration fulfilling criterion C B. Both of the following: 1. systemic viral infection has been diagnosed 2. no evidence of meningitic or encephalitic involvement C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to onset of the systemic viral infection 2. headache has significantly worsened in parallel with worsening of the systemic viral infection 3. headache has significantly improved or resolved in parallel with improvement in or resolution of the systemic viral infection 4. headache has either or both of the following characteristics: a) diffuse pain b) moderate or severe intensity D. Not better accounted for by another ICHD-3 diagnosis. 10. Headache attributed to disorder of homoeostasis 10.1 Headache attributed to hypoxia and/or hypercapnia 10.1.1 High-altitude headache A. Headache fulfilling criterion C B. Ascent to altitude above 2,500 metres has occurred C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to the ascent 2. either or both of the following: a) headache has significantly worsened in parallel with continuing ascent b) headache has resolved within 24 hours after descent to below 2,500 metres 3. headache has at least two of the following three characteristics: a) bilateral location b) mild or moderate intensity 25 c) aggravated by exertion, movement, straining, coughing and/or bending D. Not better accounted for by another ICHD-3 diagnosis. 11. Headache or facial pain attributed to disorder of the cranium, neck, eyes, ears, nose, sinuses, teeth, mouth or other facial or cervical structure 11.2 Headache attributed to a disorder of the neck 11.2.1 Cervicogenic headache A. Any headache fulfilling criterion C B. Clinical and/or imaging evidence of a disorder or lesion within the cervical spine or soft tissues of the neck, known to be able to cause headache C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to the onset of the cervical disorder or appearance of the lesion 2. headache has significantly improved or resolved in parallel with improvement in or resolution of the cervical disorder or lesion 3. cervical range of motion is reduced and headache is made significantly worse by provocative manœuvres 4. headache is abolished following diagnostic blockade of a cervical structure or its nerve supply D. Not better accounted for by another ICHD-3 diagnosis. 11.3 Headache attributed to disorder of the eyes 11.3.1 Headache attributed to acute angle-closure glaucoma A. Any headache fulfilling criterion C B. Acute angle-closure glaucoma has been diagnosed, with proof of increased intraocular pressure C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to the onset of the glaucoma 2. headache has significantly worsened in parallel with progression of the glaucoma 3. headache has significantly improved or resolved in parallel with improvement in or resolution of the glaucoma 4. pain location includes the affected eye D. Not better accounted for by another ICHD-3 diagnosis. 26 11.5 Headache attributed to disorder of the nose or paranasal sinuses 11.5.1 Headache attributed to acute rhinosinusitis A. Any headache fulfilling criterion C B. Clinical, nasal endoscopic and/or imaging evidence of acute rhinosinusitis C. Evidence of causation demonstrated by at least two of the following: 1. headache has developed in temporal relation to the onset of rhinosinusitis 2. either or both of the following: a) headache has significantly worsened in parallel with worsening of the rhinosinusitis b) headache has significantly improved or resolved in parallel with improvement in or resolution of the rhinosinusitis 3. headache is exacerbated by pressure applied over the paranasal sinuses 4. in the case of a unilateral rhinosinusitis, headache is localized and ipsilateral to it D. Not better accounted for by another ICHD-3 diagnosis. PART 3. PAINFUL CRANIAL NEUROPATHIES, OTHER FACIAL PAIN AND OTHER HEADACHES 13. Painful lesions of the cranial nerves and other facial pain 13.1 Pain attributed to a lesion or disease of the trigeminal nerve 13.1.1 Trigeminal neuralgia A. Recurrent paroxysms of unilateral facial pain in the distribution(s) of one or more divisions of the trigeminal nerve, with no radiation beyond, and fulfilling criteria B and C B. Pain has all of the following characteristics: 1. lasting from a fraction of a second to 2 minutes 2. severe intensity 3. electric shock-like, shooting, stabbing or sharp in quality C. Precipitated by innocuous stimuli within the affected trigeminal distribution D. Not better accounted for by another ICHD-3 diagnosis. 13.1.1.1 Classical trigeminal neuralgia A. Recurrent paroxysms of unilateral facial pain fulfilling criteria for 13.1.1 Trigeminal neuralgia B. Demonstration on MRI or during surgery of neurovascular compression (not simply contact), with morphological changes in the trigeminal nerve root. 27 13.1.2 Painful trigeminal neuropathy 13.1.2.1 Painful trigeminal neuropathy attributed to herpes zoster A. Unilateral facial pain in the distribution(s) of a trigeminal nerve branch or branches, lasting <3 months B. One or more of the following: 1. herpetic eruption has occurred in the same trigeminal distribution 2. Varicella zoster virus (VZV) has been detected in the CSF by polymerase chain reaction (PCR) 3. direct immunofluorescence assay for VZV antigen or PCR assay for VZV DNA is positive in cells obtained from the base of lesions C. Not better accounted for by another ICHD-3 diagnosis. 13.4 Occipital neuralgia A. Unilateral or bilateral pain in the distribution(s) of the greater, lesser and/or 3rd occipital nerves and fulfilling criteria B-D B. Pain has at least two of the following three characteristics: 1. recurring in paroxysmal attacks lasting from a few seconds to minutes 2. severe in intensity 3. shooting, stabbing or sharp in quality C. Pain is associated with both of the following: 1. dysaesthesia and/or allodynia apparent during innocuous stimulation of the scalp and/or hair 2. either or both of the following: a) tenderness over the affected nerve branches b) trigger points at the emergence of the greater occipital nerve or in the distribution of C2 D. Pain is eased temporarily by local anaesthetic block of the affected nerve(s) E. Not better accounted for by another ICHD-3 diagnosis. 13.12 Persistent idiopathic facial pain A. Facial and/or oral pain fulfilling criteria B and C B. Recurring daily for >2 hours/day for >3 months C. Pain has both of the following characteristics: 1. poorly localized, and not following the distribution of a peripheral nerve 2. dull, aching or nagging quality D. Clinical neurological examination is normal E. A dental cause has been excluded by appropriate investigations F. Not better accounted for by another ICHD-3 diagnosis. 28 14. Other headache disorders 14.1 Headache not elsewhere classified A. Headache with characteristic features suggesting that it is a unique diagnostic entity B. Headache does not fulfil criteria for any of the headache disorders described above. 14.2 Headache unspecified A. Headache is or has been present B. Not enough information is available to classify the headache at any level of this classification. |
8332 | https://stackoverflow.com/questions/38433503/python-pep-8-class-name-as-variable | pep8 - Python: PEP 8 class name as variable - Stack Overflow
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Python: PEP 8 class name as variable
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Which is the convention according to PEP 8 for writing variables that identify class names (not instances)?
That is, given two classes, A and B, which of the following statements would be the right one?
python
target_class = A if some_condition else B
instance = target_class()
or
python
TargetClass = A if some_condition else B
instance = TargetClass()
As stated in the style guide,
Class Names:
Class names should normally use the CapWords convention.
But also
Method Names and Instance Variables:
Use the function naming rules: lowercase with words separated by underscores as necessary to improve readability.
In my opinion, these two conventions clash and I can't find which one prevails.
python
pep8
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edited Jul 19, 2016 at 12:00
LostMyGlassesLostMyGlasses
asked Jul 18, 2016 at 9:45
LostMyGlassesLostMyGlasses
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4 Classes use CamelCase. Variable names (instances) use_underscore_lower_case joel goldstick –joel goldstick 2016-07-18 09:47:22 +00:00 Commented Jul 18, 2016 at 9:47
2 Why didn't you just look it up?jonrsharpe –jonrsharpe 2016-07-18 09:47:26 +00:00 Commented Jul 18, 2016 at 9:47
7 @jonrsharpe I did, but I still find it unclear. Class names should normally use the CapWords convention, ok, but variables don't. These two conventions clash, in my opinion, so I don't know which one prevails.LostMyGlasses –LostMyGlasses 2016-07-18 09:50:33 +00:00 Commented Jul 18, 2016 at 9:50
2 I would lean more towards variable_style. I can't see anything immediately in PEP8 that answers this, but thinking of it as a variable makes more sense to me. Using ClassStyle makes it look like it is the natural name of a class that is defined somewhere else in my code.khelwood –khelwood 2016-07-18 09:51:25 +00:00 Commented Jul 18, 2016 at 9:51
9 I don't think PEP8 specifically mentions this, but I don't think it needs to, it's a variable, it references a type but it's still a variable so I'd use the variable style.Steve –Steve 2016-07-18 09:56:11 +00:00 Commented Jul 18, 2016 at 9:56
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In lack of a specific covering of this case in PEP 8, one can make up an argument for both sides of the medal:
One side is: As A and B both are variables as well, but hold a reference to a class, use CamelCase (TargetClass) in this case.
Nothing prevents you from doing
python
class A: pass
class B: pass
x = A
A = B
B = x
Now A and B point to the respectively other class, so they aren't really fixed to the class.
So A and B have the only responsibility to hold a class (no matter if they have the same name or a different one), and so has TargetClass.
In order to remain unbiased, we as well can argue in the other way: A and B are special in so far as they are created along with their classes, and the classes' internals have the same name. In so far they are kind of "original", any other assignment should be marked special in so far as they are to be seen as a variable and thus in lower_case.
The truth lies, as so often, somewhere in the middle. There are cases where I would go one way, and others where I would go the other way.
Example 1: You pass a class, which maybe should be instantiated, to a method or function:
```python
def create_new_one(cls):
return cls()
class A: pass
class B: pass
print(create_new_one(A))
```
In this case, cls is clearly of very temporary state and clearly a variable; can be different at every call. So it should be lower_case.
Example 2: Aliasing of a class
python
class OldAPI: pass
class NewAPI: pass
class ThirdAPI: pass
CurrentAPI = ThirdAPI
In this case, CurrentAPI is to be seen as a kind of alias for the other one and remains constant throughout the program run. Here I would prefer CamelCase.
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edited Jul 18, 2016 at 13:47
answered Jul 18, 2016 at 10:58
glglglglglgl
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LostMyGlasses
LostMyGlassesOver a year ago
I see that the decision is made by your opinion (which IMO makes sense), but it's not completely convincing. The opposite option is also well defended at the comments.
2016-07-18T12:10:38.133Z+00:00
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glglglOver a year ago
@LostMyGlasses I now show both sides unbiased and show when I would use the one and when the other.
2016-07-18T13:48:36.59Z+00:00
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LostMyGlassesOver a year ago
I think that both cases are well explained, and as seen that it is a matter of personal decision, your tips are helpful. Thanks!
2016-07-18T14:26:33.777Z+00:00
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In case of doubt I would do the same as Python developers. They wrote the PEP-8 after all.
You can consider your line:
python
target_class = A if some_condition else B
as an in-line form of the pattern:
python
target_class = target_class_factory()
and there is a well-known example for it in the Python library, the namedtuple, which uses CamelCase.
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answered Jul 26, 2016 at 15:42
VPfBVPfB
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LostMyGlassesOver a year ago
Wow, that's a good example. However, I'd like to know not only how to do it, but also why.
2016-07-27T08:02:04.317Z+00:00
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DylanYoungOver a year ago
@LostMyGlasses The "why" of any code style recommendation is almost always the same. 1) consistency and 2) maintainability. The CapWords/CamelCase style is more maintainable because someone hacking on the code doesn't have to dredge up the variable definition to know what they're working with.
2020-07-03T19:06:25.11Z+00:00
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I personally think that whether the variable you mentioned, which holds a reference to a class, is defined as a temporary variable (for example in a procedure or function) or as a derivation from an existing class in the global spectrum has the most weight in the case of which one to use. So to summarise from the reply above:
If the variable is temporary, e.g. inside a function or used in a single instance in the solving of a problem, it should be lower_case with underscore separation.
If the variable is within the global spectrum, and is defined along with the other classes as an alias or derivation to use to create objects in the body of the program, it should be defined using CamelCase.
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answered Jul 20, 2016 at 18:24
AlmonsoAlmonso
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DylanYoungOver a year ago
Why is it less important to have this information in a function than at the top level? I guess because functions are smaller? What about function arguments?
2020-07-03T19:09:01.373Z+00:00
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I finally found some light in the style guide:
Class Names
[...]
The naming convention for functions may be used instead in cases where the interface is documented and used primarily as a callable.
may be used is not a strong statement, but it covers the case, as the variable was intended to be used as a callable.
So, for general purpose I think that
python
target_class = A if some_condition else B
instance = target_class()
is better than
python
TargetClass = A if some_condition else B
instance = TargetClass()
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answered Jul 27, 2016 at 8:26
LostMyGlassesLostMyGlasses
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DylanYoung
DylanYoungOver a year ago
That's not using it as a callable. That's using it as a class.
2020-07-03T19:07:20.227Z+00:00
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8333 | https://www.openriskmanual.org/wiki/Concentration_Ratio | Concentration Ratio - Open Risk Manual
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Concentration Ratio
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Contents
1 Definition
2 Details
2.1 Berger Parker Index (special case)
3 Usage
3.1 Common Ratios
3.2 Interpretation and Concentration levels
3.3 Advantages
3.4 Disadvantages
4 Variations
5 Issues and Challenges
6 Implementation
7 See Also
8 References
Definition
The Concentration Ratio is a type of Concentration Index. It is used for the purpose of measuring credit portfolio or market Concentration Risk, diversity or inequality metrics.
The concentration ratio is a measure of the contribution of a given set of exposures to the total portfolio exposure - essentially the portfolio fraction. For example the fraction of exposure contributed by the top ten largest clients, the contribution to the market capitalization by the top five ten largest listed companies etc.
Details
The concentration ratio is simply the percentage of portfolio exposure by the n largest exposures. Assuming the exposure data are given, the calculation is easy.
The ratio associated with the contribution of k entities (in a portfolio or sample of size n > k) is denoted as
C R k=∑i=1 k w i{\displaystyle CR_{k}=\sum {i=1}^{k}w{i}}
where w i is simply the share of the i-th exposure in the total portfolio exposure.
Berger Parker Index (special case)
In diversity studies, the special case k=1 is denoted as the Berger-Parker index. It is the fraction of the population formed by the most abundant category.
Usage
Along with the Herfindahl-Hirschman Index (HHI) the concentration ratio is a tool typically used by competition authorities to measure market concentration.
These tools are useful also in the context of analysing portfolio concentrations, for example credit or market risk concentrations.
The concentration ratio is a "relative" measure, i.e., all exposures are normalized as fractions of total portfolio exposure, therefore two portfolios with same index are deemed to be equally concentrated irrespective of their absolute total size.
Common Ratios
There is no intrinsic rule to select the number n of exposures to assess. Some commonly used examples include
The n=1 Concentration Ratio measures the total contribution of the largest exposure (equivalent to the Berger-Parker index).
The n=4 Concentration Ratio measures the total contribution of the four largest exposures.
The n=8 Concentration Ratio measures the total contribution of the eight largest exposures.
The n=20 Concentration Ratio measures the total contribution of the twenty largest exposures.
The n=50 Concentration Ratio measures the total contribution of the fifty largest exposures.
The case n=20 is important for the purposes of credit risk concentration analysis because it is used in the Large Exposures Framework
Interpretation and Concentration levels
Concentration ratios can range in value from 0 to 100 percent.
Whether a given result indicates a concentrated portfolio cannot be assessed beforehand but only in relation with typical portfolios, standard market practices etc. or other metrics. For example the Large Exposures Framework links the size of the top 20 large exposures to the capital of the firm.
Concentration levels can characterized as no, low or medium to high to "total" concentration based on subjective thresholds.
Advantages
Key advantages of the concentration ratio index include
Its conceptual and computational simplicity. Indeed even a casual glance at the sorted and normalized (divided by total) exposures provides already an intuitive feeling as to the degree of concentration in a given portfolio and
The relatively moderate data requirements, i.e. only the exposure size per borrower (but see qualification below)
Disadvantages
The definition of the concentration ratio must pick a set of representative entities (the number n) to assess concentration and does not use the contribution of all exposures. A portfolio may look acceptable for a given n, but less so for another choice.
The index suffers from a one-dimensional view of concentration risk, namely it is assumed that relative size is the only factor worthy of consideration. In reality there may be a wide range of other factors
Exposure size may not be obvious to define
Risks associated with different exposures can be very different
There may be varying degrees of dependency among different exposures
It is difficult to objectively characterize the concentration thresholds.
The Herfindahl-Hirschman Index and other indexes provide more complete picture of concentration than does the concentration ratio, thereby mitigating the first of the above disadvantages
Variations
None
Issues and Challenges
Although the data requirements for the concentration ratio calculation seem to be moderate at first sight, they often turn out to be difficult to meet in practice. They require the aggregation of all exposures to the same borrower for the whole business entity, be it a bank or a banking group. A heterogeneous IT environment can already present a serious technical obstacle to this aggregation. Furthermore, large borrowers, which are actually the most relevant entities for measuring name concentration, can themselves be complex structures that are not always easy to identify as belonging to the same borrower entity.
Implementation
Open Source implementations of the Concentration Ratio are available in
the R package IC2
the Python library Concentration Library
See Also
Percentile Ratio
Palma Ratio
References
Retrieved from "
Categories:
Univariate Concentration Index
Risk Concentration
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8334 | https://www.merriam-webster.com/thesaurus/debilitated | Est. 1828
Synonyms of debilitated
adjective
as in weak
verb
as in weakened
as in weak
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debilitated 1 of 2
Definition of debilitated
as in weak
lacking bodily strength the debilitated prisoners could barely stand
Synonyms & Similar Words
languid
wimpy
slight
down-and-out
sapped
prostrated
soft
low
effete
unsubstantial
fragile
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weary
wimpish
tender
drained
damaged
invalid
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strong
mighty
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red-blooded
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debilitated
2 of 2
verb
past tense of debilitate
as in weakened
to diminish the physical strength of the heart surgery debilitated the college athlete beyond his worst fears
Synonyms & Similar Words
weakened
softened
exhausted
tired
injured
wasted
enfeebled
hurt
paralyzed
sapped
damaged
crippled
enervated
impaired
depressed
incapacitated
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disabled
devitalized
etiolated
impoverished
harmed
hamstrung
invalided
laid up
ground (down)
washed out
unmanned
wore down
wore out
broke down
Antonyms & Near Antonyms
strengthened
fortified
energized
recruited
hardened
invigorated
rejuvenated
vitalized
beefed (up)
seasoned
toughened
Example Sentences
Recent Examples of Synonyms for debilitated
weak
weakened
feeble
softened
frail
exhausted
disabled
Adjective
Kleptocratic governments are not weak or failing, Chayes argues in this paradigm-shifting 2015 book.
— Zephyr Teachout, The Atlantic, 22 Sep. 2025
The plan to tap geothermal energy answered the key question of power provision, getting around Africas chronically weak grids with the green, baseload electricity required for digital infrastructure.
— Alexis Akwagyiram, semafor.com, 22 Sep. 2025
Definition of weak
Verb
While the greenback has depreciated against other major currencies this year, with the dollar index tumbling over 10%, those currencies have also weakened in value relative to gold, Dalio said, noting that gold has become the second largest reserve currency globally.
— Anniek Bao, CNBC, 19 Sep. 2025
Key News Asian equities were mixed/lower overnight as Hong Kong, Mainland China, and the Philippines stood out as relative bright spots while the US dollar weakened versus the Asia Dollar Index.
— Brendan Ahern, Forbes.com, 17 Sep. 2025
Definition of weakened
Adjective
Without electrification, Simpliciano said, renewable targets run the risk of being feeble and ineffective, with brands potentially turning to biomass and other problematic substitutions as stopgap measures.
— Jasmin Malik Chua, Sourcing Journal, 25 Sep. 2025
This politicization of what should be an economic solution again highlights Californias feeble legislative ability.
— U T Readers, San Diego Union-Tribune, 22 Sep. 2025
Definition of feeble
Verb
Indias own vast outsourcing industry has also depended on the work permits for decades, even though that has softened in recent years.
— Jason Ma, Fortune, 21 Sep. 2025
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— Abby Montanez, Robb Report, 18 Sep. 2025
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Adjective
Newborns and frail older people often have a hard time coughing up the mucus, Hopkins says.
— Liz Szabo, Scientific American, 23 Sep. 2025
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— CNN Money, CNN Money, 21 Sep. 2025
Definition of frail
Verb
Burnout shows up as feeling constantly exhausted, emotionally checked out from work, and struggling to get things done effectively.
— Aisha Ditta, CNBC, 20 Sep. 2025
The doctor may be exhausted from lack of sleep, or distraught after having performed CPR on a baby for forty-five minutes.
— Rachel Pearson, Harpers Magazine, 19 Sep. 2025
Definition of exhausted
Adjective
Tax law experts note that those who could struggle with the shift away from paper checks to direct deposit include taxpayers without bank accounts and those who are elderly or disabled.
— Susan Tompor, Freep.com, 24 Sep. 2025
Inadequate provision for disabled fans.
— Nick Miller, New York Times, 24 Sep. 2025
Definition of disabled
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8335 | https://14liedan.wordpress.com/2012/04/15/math-investigating-y-sin-bx-y-tan-bx/ | 14liedan
Math: Investigating y = sin (bx), y = tan (bx)
The sine graph, y=sinx, has several factors that affect its features. For example, in y=sin(x-c) the c would cause a horizontal shift to the graph, while in y=sinx+d the d would move the graph vertically. These work similarly for cosine and tangent graphs as well. Another factor which affects the graphs is the b in y=sin(bx) or y=tan(bx) and this blog post will be about what effects the number b has on the graphs.
Explanation of b’s role in y=sin(bx)
In this video, we can see that b changes the period of the sine graph. When b is 1, it is the usual y=sinx graph, but when it is 0.5, the period doubles. As the value of b increases, the period decreases, meaning it is an inverse relationship.
As this video shows, the value of b can go infinitely high, which will make the graph look like high-frequency waves, and it can go infinitely low as well (please see the negative section for negative integers for b). This makes sense because for example, when the b value is 2, the frequency is doubled (but the period is halved), meaning that the graph will be twice as narrow.
Formula of period in y=sin(bx)
This table shows the relationship of the value b and the period of the graph. As the values increase, we can see that the period in radians is getting smaller and smaller.
Using this table, I was able to make a formula for finding out the period: 2π/b
Again, this makes mathematical sense because the usual period of the graph y=sinx is 2π and to in order to consider the value of b, we would divide it since it is multiplied to x.
Explanation of b’s role in y=sin(-bx)
If the value of b is negative, the graph will be slightly changed. When comparing a graph of y=sin(1x) and y=sin(-1x), we can look at it in a way that the graph has been reflected on the x-axis. In other words, the graph is flipped upside down. When b is positive, the graph will start with an incline but when it is negative, it will start with a decline. This means that in the second quadrant the graph is flipped as well. When the value b is positive, the graph will start with a decline in the negative region (second quadrant) but when it is negative, it will begin with an incline. The negative property of the value of b affects the graph in this way because it makes the period of the graph negative as well. The period becomes negative when we plug the negative value for b into the formula 2π/b and the answer (which is the period) will be negative. This causes the graph to be reflected or flipped on the x-axis.
This video shows what happens when the values go down to -100. Similar to what happens when the positive value for b becomes infinitely large, when the negative value for b is infinitely large, the graph will look like high-frequency waves.
y=sin(bx-c) vs. y=sin(b(x-c))
The first video shown above displays the graph for y=sin(bx-c). For both of these equations, the c value affects the phase (horizontal) shift but the equation determining this is different. For this first equation, the phase shift will impact the graph in a way that it starts at the point of the x-axis at c/b. This means that the starting point of the graph y=sin(bx-c) is the c value divided by the b value. So for the graph y=sin(bx-c), the b value affects both the period of the graph and the starting point of the graph.
For this graph shown in the second video, the starting point is the value of c. Although the b value still impacts the period of the graph, the starting point of the graph will not change. Therefore, the phase shift or the starting point of the graph is just c.
The difference in these two equations are dependent on the placement of the b value. In the equation y=sin(bx-c), the b is placed so that it is only multiplied to x. This means that the value c is subtracted from the product of b and x. That is why the value of c does not automatically represent the placement of the graph on the x-axis (phase shift), instead the quotient of c and b shows where the graph will be located. On the contrary, the graph of y=sin(b(x-c)) is different because the b is multiplied to the whole bracket containing x-c. Therefore, value b only changes the period of the graph while the value c remains unaffected. This is why the horizontal shift and the placement of the graph on the x-axis is the value c itself.
Explanation of b’s role in y=tan(bx)
Value b acts the same for the tan graph. Again, it changes the period of the graph and as the value becomes larger, the period decreases. This works the same when the value is negative. As the values become closer to 0, the graph will be wider and the period will be longer, but as the value goes decreases farther into the negative range, the period decreases and the graph becomes narrow.
The videos show many examples of this. For example, when the value is 2, the frequency is doubled (period is halved), meaning the narrowness is doubled. The period of the graph for this value of b is π/2. Another example shows when the value of b is -0.5, the period will be -2π.
Formula of period in y=tan(bx)
Using these patterns and findings as a clue, I made the formula for the period of the tan graph: π/b.
This time the formula does not include a 2 multiplied to the π in the numerator, as the sine graph’s formula did, because the period of a usual tan graph y=tanx is just π.
The table to the left shows how this equation can be applied to many values of b and their impact on the period of the graph. This way, we can see the formula is correct. When we plug the b value into geogebra, the period will be clearly visible, as annotated in the images below. This proves that the formula is correct.
Formula of asymptotes in y=tan(bx)
In the tan graph, there are vertical asymptotes, which change with the value of b like the period does. These asymptotes are located halfway between each period. Looking at the asymptotes and seeing how the value of b alters the graph, I was able to devise the mathematical formula for the vertical asymptotes: x= π/2b(2n+1) . First came up with the (2n+1) part of the formula because I knew the vertical asymptotes are located at the odd multiples of the period such as 3π/4 or 7π/3. By multiplying any integer (represented by n) by 2 and adding 1 to the product, the number in the bracket will always be an odd number. This odd number is then multiplied to π/2b, which represents half of the period (work: π/b ÷ 2 = π/b x 1/2 = π/2b). I used π/2b instead of the period π/b because just by looking at the graphs, the asymptotes seemed to be located halfway between each period of the graph.
Because n could be any integer, there can be infinite amounts of asymptotes. However, Geogebra cannot recognize the value n if not inputted beforehand, so I could not find a way to graph all the asymptotes visible in the plane/window of the graph. Although the computer cannot graph this, we have to keep in mind that because of the flexibility of the integer n, there will be asymptotes in infinite numbers if there is no limit of the tan graph’s domain.
For example, in the first image, when the period of the graph is π/2, the formula to figure out the asymptotes will look like: x= π/4(2n+1). Again, the n value could be any integer, so to figure out the first four asymptotes, as shown in the image, I plugged in the first four integers: 0, 1, 2, and 3. Geogebra graphed the asymptotes at π/4, 3π/4, 5π/4, and 7π/4. This worked the same for the graph with the value 4 for b. These graphs proved my formula is correct but to be even more sure and that it works for all values of b and integers of n, I made a table shown below the two graphs as well.
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Hi Annegret
You have reached the correct conclusions and your generalizations are valid. (One misunderstanding you have though, is that the period cannot be negative.
So for y = sin (-2x) we say the period is pi, not -pi.
Some justifications are not adequate (eg why is the period 2pi/b) but others are superb: you go into good detail when talking about y = sin b(x – c) and y = sin (bx -c). For both of these I thought the video was excellent. Most people graphed their two equations together to highlight the differences. You took a different approach, which was very detailed and a nice way of getting below the surface.
Your justification for your formula for the position of the asymptotes was also very sound. Please note that showing that the formula works for one (or several) example(s) does not constitute a proof: it is just a confirmation that it works for those values.
See Moodle and Powerschool tomorrow or Tuesday for details on your level of achievement.
Ms Durkin
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8336 | http://math.uchicago.edu/~may/REU2021/REUPapers/Zhang,Emily.pdf | LAW OF QUADRATIC RECIPROCITY EMILY ZHANG Abstract. In this expository paper, we will explore the Law of Quadratic Reciprocity and provide two proofs: one using elementary number theory and the other using the theory of finite fields.
Contents 1.
Introduction 1 2.
Elementary Number Theory 2 2.1.
An Elementary Proof of Quadratic Reciprocity 3 3.
Field and Galois Theory 5 3.1.
Finite Fields 6 3.2.
Trace, Norms, Discriminant 8 3.3.
Another proof of Quadratic Reciprocity 8 Acknowledgments 11 References 11 1. Introduction The law of quadratic reciprocity is one of the first major results in number theory.
Its origin dates back to the 1600s when Fermat began studying the conditions for a prime p to be of the form x2+ny2 for small n; however, he did not provide proofs of his results. Fascinated by Fermat’s work, Euler devoted himself to proving Fermat’s theorems and in the process, he conjectured an equivalent formulation of the law of quadratic reciprocity. The complete proofs were given by Gauss in the 1800s. Since then, the problem has been approached in a multitude of ways and has played an important role in the development of modern number theory.
We introduce the statement of the law of quadratic reciprocity. For clarity of notation, we define the Legendre symbol. Let p be an odd prime and a be an integer not divisible by p. Note that if b is a quadratic residue, then b ≡x2 (mod p) for some b ∈Z/pZ. Then a p = ( 1 if a is a quadratic residue modulo p −1 otherwise.
Theorem 1.1 (Law of Quadratic Reciprocity). If p and q are distinct odd primes, then p q q p = (−1) p−1 2 q−1 2 .
1 2 EMILY ZHANG 2. Elementary Number Theory We begin by recalling some definitions and results from elementary number the-ory.
Theorem 2.1 (Euler’s Criterion). Let p be an odd prime. Then for all a, we have a p ≡a p−1 2 (mod p).
Proof. We will show that if a is a quadratic residue, then a p−1 2 ≡1 (mod p) and if a is a quadratic non-residue, then a p−1 2 ≡−1 (mod p). Suppose a is a quadratic residue, i.e. a ≡b2 (mod p) for some b ∈Z/pZ. Then a p−1 2 ≡bp−1 ≡1 (mod p) by Fermat’s little theorem. Now suppose a is a quadratic non-residue and consider f(x) = x p−1 2 −1. Since Z/pZ is an integral domain, we have x p−1 2 ≡1 (mod p) has at most p−1 2 solutions. But p−1 2 quadratic residues satisfy the congruence. To see this, the quadratic residues are precisely the elements in {12, 22 . . . , ( p−1 2 )2} since all the elements are distinct and (p−x)2 ≡x2 (mod p). Now if a is a quadratic non-residue, then a p−1 2 ̸≡1 (mod p). Notice that ap−1−1 ≡(a p−1 2 +1)(a p−1 2 −1) ≡0 (mod p). So a p−1 2 = ±1. Thus, if a is a quadratic non-residue, we obtain a p−1 2 ≡−1 (mod p).
□ Theorem 2.2. Let p be an odd prime. Then p ≡1 (mod 4) if and only if there exists x such that x2 ≡−1 (mod p).
Proof. By Euler’s criterion, we have −1 p = (−1)(p−1)/2 = 1 if and only if (p −1)/2 = 2k for some k ∈Z, which is equivalent to p ≡1 (mod 4).
□ Theorem 2.3 (Wilson’s Theorem). (p −1)! ≡−1 (mod p) for any prime p.
Proof. Recall that x2 ≡1 (mod p) if and only if x ≡±1 (mod p). Then the only elements in (Z/pZ)× that are its own inverse are ±1. The theorem is immediate when p = 2 and 3. For p > 3, we have for all x ∈{2, ..., p−2}, there exists a unique x′ ̸= x such that xx′ ≡1 (mod p). Pairing each element with its inverse, we have (p −1)! ≡(1)(p −1) ≡−1 (mod p).
□ Theorem 2.4 (Chinese Remainder Theorem (CRT)). Let n1, . . . , nk be integers which are pairwise relatively prime. Then the system of m congruences x ≡a1 (mod n1), . . . , x ≡am (mod nm) has a unique solution for x modulo M = n1 · · · nm Proof. We do induction on m. For m = 2, we want to find a solution satisfy-ing both x ≡a1 (mod n1) and x ≡a2 (mod n2). Since gcd(n1, n2) = 1, there exist n′ 2 and n′ 1 such that n2n′ 2 ≡1 (mod n1) and n1n′ 1 ≡1 (mod n2). We choose x ≡a1n2n′ 2 + a2n1n′ 1, which satisfies both congruences.
Suppose there are k + 1 systems of congruences and let n1, . . . , nk, nk+1 be pairwise relatively prime. Then by the inductive hypothesis, we will assume there exists a solution to the first k system of congruences, call it s. Then we have x ≡s (mod n1 · · · nk) and x ≡ak+1 (mod nk+1). Since gcd(ni, nk+1) = 1 for all 1 ≤i ≤k, we have gcd(n1n2 · · · nk, nk+1) = 1. From the case m = 2, we can conlude that there exists a solution satisfying the k + 1 system of congruences.
LAW OF QUADRATIC RECIPROCITY 3 Now we will show that this solution is unique. Suppose s and t are solutions.
Since for all 1 ≤i ≤m and i ̸= j, ni|(s −t) and gcd(ni, nj) = 1, we have n1 · · · nm|(s −t). Thus s ≡t (mod M).
□ 2.1. An Elementary Proof of Quadratic Reciprocity. We will now prove the law of quadratic reciprocity using Euler’s Criterion, CRT, and basic counting arguments by following Kim’s proof .
We begin by analyzing the half of (Z/pqZ)× defined by the set Φ := a : 1 ≤a ≤pq −1 2 , gcd(a, pq) = 1 .
Let A = Q a∈Φ a. Furthermore, define S := a : 1 ≤a ≤pq −1 2 , gcd(a, p) = 1 , T := q · 1, . . . , q · p −1 2 .
Lemma 2.5. A ≡(−1) q−1 2 q p (mod p) and A ≡(−1) p−1 2 p q (mod q) Proof. We will find two expressions for Q a∈S a. Since p and q are coprime, we know that gcd(a, pq) = 1 if and only if gcd(a, p) = 1 and gcd(a, q) = 1. So Φ contains numbers not divisible by p or q, S contains numbers not divisible by p, and T contains numbers in S that are divisible by q. Thus Φ = S\T, or equivalently S = T ⊔Φ. We have Y a∈S a = Y a∈T a · Y a∈Φ a ≡q p−1 2 p −1 2 ! · A ≡ q p p −1 2 ! · A (mod p) where the second congruence follows from pulling out the q′s in T and the last congruence follows from Euler’s Criterion.
Alternatively, notice that pq −1 2 = q −1 2 · p + p −1 2 .
To see the elements of S more explicitly, we can write it as a union of two sets: S = a : 1 ≤a ≤q −1 2 p −1, gcd(a, p) = 1 ∪ q −1 2 p + k : 1 ≤k ≤p −1 2 .
Taking every element modulo p in S, each of the residue classes modulo p is represented q−1 2 times. Thus, multiplying all the elements in the first set gives us [(p −1)!] q−1 2 and the second set gives us ( p−1 2 )!. Thus Y a∈S a ≡[(p −1)!] q−1 2 p −1 2 ! ≡(−1) q−1 2 p −1 2 ! (mod p), where the second congruence follows from Wilson’s Theorem. Combining the two expressions of Q a∈S a, we have q p p −1 2 ! · A ≡(−1) q−1 2 p −1 2 ! (mod p) A ≡(−1) q−1 2 q p (mod p) Following a similar argument, we have A ≡(−1) p−1 2 p q (mod q).
□ 4 EMILY ZHANG Remark 2.6. By CRT, we have A ≡1 (mod p) and A ≡1 (mod q) if and only if A ≡1 (mod pq). Similarily A ≡−1 (mod pq) if and only if A ≡−1 modulo p and modulo q.
Lemma 2.7. A ≡1 or −1 (mod pq) if and only if p ≡q ≡1 (mod 4).
Proof. Consider the set {a : 1 ≤a ≤pq, gcd(a, pq) = 1} = (Z/pqZ)×. Let a ∈ (Z/pqZ)×. Since Φ is half of (Z/pqZ)×, either a ∈Φ or a / ∈Φ. Elements not in Φ can be written as negative of the elements in the first half. So if a / ∈Φ, then −a ∈Φ. We have a map from Φ to Φ given by f : a 7→ ( a−1 if a−1 ∈Φ −a−1 if a−1 ̸∈Φ We will show that f is surjective. Suppose y ∈Φ. Then there exists an inverse x ∈(Z/pqZ)×. If x ∈Φ, then f(x) = y. If x / ∈Φ, then f(−x) = y. Since Φ is finite and f is a map on itself, we have Φ is bijective. Thus for all a ∈Φ, there exists a unique f(a) ∈Φ such that af(a) ≡±1 (mod pq).
Now consider the elements fixed by f. Define Ψ = {a ∈Φ : a = f(a)} = {a ∈Φ : a2 = ±1 (mod pq)}.
Pairing every element a in Φ with f(a), we have A ≡ Y a∈Φ a ≡± Y a∈Ψ a (mod pq), where ±1 in the second product are from elements not fixed by f.
(⇐) Now we will prove the reverse direction. Suppose p, q ≡1 (mod 4). We want to show Q a∈Ψ a ≡±1 (mod pq). By definition of Ψ, it suffices to find the solutions to the congruence x2 ≡±1 (mod pq). By CRT, we see that x2 ≡1 (mod pq) has four solutions. Let the solutions be 1, −1, N, −N modulo pq. Now x2 ≡−1 (mod pq) has a solution whenever there is a solution to x2 ≡−1 modulo p and q. Since p, q ≡1 (mod 4), by Thm 2.2, we have x2 ≡−1 (mod pq) has a solution, call it I. Notice that (NI)2 ≡N 2I2 ≡−1 (mod pq). Thus, we can characterize the four solutions as I, −I, NI, −NI modulo pq. Since Ψ ⊆Φ, we have 1, I, N, NI are the four solutions in Ψ up to ±1. Thus Y a∈Ψ a ≡±(N 2I2) ≡∓1 (mod pq).
(⇒) To prove the forward direction, we will prove the contrapositive. Suppose p or q ≡3 (mod 4). Then x2 ≡−1 (mod pq) has no solutions. Thus the only solutions to x2 ≡±1 (mod pq) in Ψ are 1 and N up to ±1. Thus Y a∈Ψ ≡±N ̸≡±1 (mod pq).
□ From Lem 2.5 and Lem 2.7, we deduce that (−1) q−1 2 q p = (−1) p−1 2 p q if and only if p, q ≡1 (mod 4). Equivalently p q q p = (−1) p−1 2 + q−1 2 if and only if p, q ≡1 (mod 4). We prove this statement is equivalent to the law of quadratic reciprocity by considering the following cases: p or q ≡1 (mod 4) and p, q ≡ 3 (mod 4). If p, q ≡1 (mod 4), then p−1 2 + q−1 2 has even parity. So p q q p = 1 LAW OF QUADRATIC RECIPROCITY 5 Now p q q p ̸= (−1) p−1 2 + q−1 2 if and only if p or q ≡3 (mod 4). So if p ≡1 and q ≡3 (mod 4) or if p ≡3 and q ≡1 (mod 4), then p−1 2 + q−1 2 has odd parity.
So p q q p = 1.
If p, q ≡3 (mod 4), then p−1 2 + q−1 2 has even parity.
Now p q q p = −1. Altogether, we have p q q p = (−1) p−1 2 q−1 2 , as desired.
3. Field and Galois Theory In this section, we will introduce some definitions and results in field and Galois theory, culminating in another proof of the law of quadratic reciprocity.
Definition 3.1. (Field Extension) If F is a subfield of E, then E is a field extension of F denoted by E/F.
Definition 3.2. We can view E as a vector space over F. The degree of the exten-sion E/F, denoted by [E : F], is the dimension of E over F.
Definition 3.3. (Splitting field) Suppose F is a field, f(x) ∈F[x], and α1, . . . , αn are roots of f(x) in E. Then E is a splitting field of f(x) if f(x) splits into linear factors in E[x] and if E = F[α1, . . . , αn] is the smallest field containing F and all the roots.
Definition 3.4. (Algebraic Extension) The element α ∈E is algebraic over F if α is a root of some nonzero polynomial f(x) ∈F[x]. An extension E over F is algebraic if every element in E is algebraic over F.
The proofs of the following two propositions can be found on pg. 520-521 of .
Proposition 3.5. Let α be algebraic over F. Then there exists a unique monic irreducible polynomial mα ∈F[x] that has α as a root.
Proposition 3.6. If α is algebraic over F and F(α) is the field generated by α, then [F(α) : F] = deg mα(x).
Definition 3.7. (Normal Extension) The extension E/F is normal if E is an algebraic extension over F and mα splits into linear factors over E for all α ∈E.
Definition 3.8. (Separable Extension) A polynomial over F is separable if it has no repeated roots. The extension E/F is separable if E is an algebraic extension over F and mα is separable for all α ∈E.
Theorem 3.9. The polynomial f(x) ∈F[x] is separable if and only if gcd(f, f ′) = 1 where f ′ is the derivative of f.
Proof. See pg. 547 of .
□ Definition 3.10. (Galois Extension) The extension E/F is Galois if it is normal and separable.
Definition 3.11. (Cyclotomic Extension) Define the nth cyclotomic polynomial Φn(x) to be the polynomial whose roots are the primitive nth roots of unity, ζn.
Note that ζn can be defined over any field.
Φn(x) = Y 1≤i≤n, gcd(i,n)=1 (x −ζi n) The field Q(ζn) is called a cyclotomic extension over Q.
6 EMILY ZHANG Proposition 3.12. The minimal polynomial of ζn over Q is Φn(x).
Proof. See section 13.6 of .
□ Definition 3.13. For a field extension E of F, define Aut(E/F) := {θ : E →E : θ is a ring isomorphism and θ|F = id}.
If E is a Galois extension, then Aut(E/F) is called the Galois group denoted by Gal(E/F).
We can view the Galois group of E/F as a group of symmetries, and we will illustrate this fact by the following example.
Let E be the splitting field of a separable polynomial f(x) with α1, . . . , αn as its roots. Then Gal(E/F) acts on the set of roots of f(x). Let σ ∈Gal(E/F). Then σ permutes the roots of f(x) since σ(P ciα) = P ciσ(α) where ci ∈F are fixed. Thus, the group action induces an injective map from Gal(E/F) to Sym({α1, . . . , αn}). In particular, we can consider Gal(E/F) as a subgroup of Sn where n is the number of roots.
Proposition 3.14. If E/F is a Galois extension, then | Gal(E/F)| = [E : F].
We state the correspondence property of the fundamental theorem of Galois theory. A complete statement and proof can be found on pg. 574 of .
Theorem 3.15 (Fundamental Theorem of Galois Theory). Let E/F be a Galois extension and let G = Gal(E/F). Then there is a bijection between the subfields K of E containing F and the subgroups H of G given by the correspondences Ψ(K) := Aut(E/K) Φ(H) := Fix(H) 3.1. Finite Fields. In this section, we will show the existence of finite fields and characterize the group of units and automorphisms of a finite field.
Theorem 3.16. For every prime power pn, there exists a field of order pn. More-over, this field is the splitting field of xpn −x.
Proof. Consider the splitting field of xpn −x over Z/pZ, call it E. Define S := {a ∈E, apn = a} to be the set of all the roots of xpn −x. We will show that S is a field with pn elements. We have 0, 1 ∈S. Now suppose a, b ∈S. We want to show (a + b)pn = apn + bpn, also known as the “freshman’s dream”. We prove this by induction on n. For n = 1, recall the binomial theorem which states (a + b)p = Pp k=0 p k ap−kbk.
For all 1 ≤k < p, (p −k)! and k! do not divide p as p is prime.
Since p k ∈Z, (p −k)!k! divides p(p −1)!, and gcd((p −k)!k!, p) = 1, we have (p −k)!k! divides (p −1)!. Thus 1 p p k ∈Z. Since for all 1 ≤k ≤p −1 there exists n ∈Z such that p k = pn , we deduce that (a + b)p = ap + bp. For n = k + 1, we have (a + b)pk+1 = ((a + b)pk)p = (apk + bpk)p = apk+1 + bpk+1, where the second equality follows from the inductive hypothesis and the last equality follows from the base case. Thus (a + b)pn = apn + bpn for all n. Furthermore (ab)pn = ab, (−a)pn = (−1)pna = −a, and (a−1)pn = (apn)−1 = a−1. Thus S is a field since it is closed under addition, multiplication, negation, and inversion.
Now by Thm 3.9, since gcd(xpn −x, −1) = 1, we have xpn −x has no repeated roots in its splitting field. So S is a field with exactly pn elements, which is often denoted by Fpn. Thus Fpn is the splitting field of xpn −x.
□ LAW OF QUADRATIC RECIPROCITY 7 Remark 3.17. Up to isomorphism, there exists a unique field of order pn. See pg.
542 of .
Theorem 3.18. The group of units of a finite field is cyclic.
Proof. We first show that P d|n ϕ(d) = n using properties of cyclic groups. Let G be a cyclic group of order n and let d|n. We partition the group based on the order of its elements and show that ϕ(d) is the number of elements of order d. Since G is cyclic, there exists a unique subgroup of order d, call it S. By the uniqueness of S, the elements of order d are the generators of S. Suppose a is a generator. Recall that ord(aj) = ord(a)/ gcd(ord(a), j). Then ord(aj) = ord(a) = d if and only if gcd(d, j) = 1. But |{aj : 1 ≤j ≤d, gcd(j, d) = 1}| = ϕ(d). Thus ϕ(d) is the number of elements of order d. Now G d|n {a ∈G | ord(a) = d} = G.
Thus P d|n ϕ(d) = n.
Let F be a finite field and let F× be the group of units of order n. Let d|n. Define ψ(d) := |{a ∈F× : ord(a) = d}|.
We want to show ψ(n) ̸= 0, i.e.
there exists an element of order n. We claim if ψ(d) ̸= 0, then ψ(d) = ϕ(d). Suppose ψ(d) ̸= 0. Then there exists an element a ∈F× such that ord(a) = d. Consider ⟨a⟩= {1, a, . . . , ad−1}. Then every element in ⟨a⟩is a root of xd −1. Since xd −1 has at most d roots and we found d distinct roots, we deduce ⟨a⟩are all the roots. Now any element of order d lies in ⟨a⟩. As we showed earlier, since ⟨a⟩is cyclic, the number of elements of order d is ϕ(d). Thus if ψ(d) ̸= 0, then ψ(d) = ϕ(d).
Now suppose there exists d such that ψ(d) = 0. Then by definition of the ϕ function, we have n = X d|n ψ(d) < X d|n ϕ(d) = n.
But we reach a contradiction. Thus P d|n ψ(d) = P d|n ϕ(d). Now for every d|n, we have ψ(d) = ϕ(d) ̸= 0. Thus ψ(n) ̸= 0, as desired. Since there exists an element of order n, we deduce that F× is cyclic.
□ Theorem 3.19. Gal(Fpn/Fp) is generated by the p-th power map φ : x 7→xp.
Proof. We will first show that φ ∈Gal(Fpn/Fp). Let a, b ∈Fpn. Then φ(a + b) = φ(a) + φ(b) by freshman’s dream and φ(ab) = φ(a)φ(b). Since nonzero field homo-morphisms are injective and φ is a map from a finite field to itself, we conclude that φ is an automorphism. Now for all a ∈Z/pZ, we have φ(a) = a by Fermat’s little theorem. Thus φ fixes Z/pZ. Viewing Fpn as a vector space over Z/pZ, we deduce from Prop 3.14, that | Gal(Fpn/Fp)| = [Fpn : Fp] = n. Now we will show that the order of φ is n. Since φn(a) = apn = a, we have φn = id. Let 0 < d < n and suppose for a contradiction that φd = id. Then apd = a for all a ∈Fpn. But xpd −x has at most pd roots in Fpn. Since pd < pn, we have a contradiction. Thus φd ̸= id and Gal(Fpn/Fp) = ⟨φ⟩.
□ 8 EMILY ZHANG 3.2. Trace, Norms, Discriminant. We will briefly introduce some functions in algebraic number theory that will help us prove the law of quadratic reciprocity.
For the rest of the section, we assume E/F is a finite separable extension of degree [E : F] = n.
Definition 3.20. The field F is called an algebraic closure of F if (1) F is algebraic over F (2) every polynomial f(x) ∈F[x] splits completely over F Definition 3.21. Let σ1, . . . , σn be distinct F-embeddings into an algebraic closure of E. For each α ∈E, the trace and the norm are defined as follows: NE/F (α) = n Y i=1 σi(α), TE/F (α) = n X i=1 σi(α).
Now we will define two notions of the discriminant—the discriminant of a poly-nomial and the discriminant of an n-tuple. When the polynomial is minimal, they are the same.
Definition 3.22. Let σ1, . . . , σn be distinct F−embeddings of E into an algebraic closure of E, and let (α1, . . . , αn) be an n-tuple of elements in E. Then the dis-criminant of the n-tuple is DE/F (α1, . . . , αn) = [det(σi(αj))]2 where σi(αj) is the ijth entry in the matrix.
Definition 3.23. Let f(x) ∈F[x] be a polynomial with x1, . . . , xn as its roots.
Define the discriminant of f(x) by the formula disc(f) = Q i<j(xi −xj)2.
Proposition 3.24. Suppose E = F(α). Let DE/F (α) = DE/F (1, α, . . . , αn−1) and let α1, . . . , αn be the roots of mα,F (x) in a splitting field, with α1 = α. Then DE/F (α) = Y i<j (αi −αj)2 = (−1)( n 2)NE/F (m′ α,F (α)).
Proof. See pg. 10 of .
□ Proposition 3.25. Let α be algebraic over F. Then disc(mα,F (x)) = DE/F (α).
Proof. This follows from the discriminant of a polynomial and Prop 3.24.
□ Proposition 3.26. Let f(x) ∈F[x] with x1, . . . , xn as its roots. If ch(F) ̸= 2, then the permutation σ ∈Sn is an element of An if and only if it fixes the square root of disc(f). Furthermore, the Galois group of f(x) is a subgroup of An if and only if disc(f) ∈F is the square of an element of F.
Proof. See pg. 610 of .
□ 3.3. Another proof of Quadratic Reciprocity. Now, we will prove the law of quadratic reciprocity using the theory of finite fields, following Bergman’s outline . For the rest of the paper, we assume p and q are odd primes. Given p, we first investigate when an arbitrary integer is a quadratic residue modulo p. Then, to determine the relationship between when p is a quadratic residue modulo q and q is a quadratic residue modulo p, we analyze the conditions for the Galois group of xq −1 over Z/pZ to act by even permutations on the roots of xq −1. We observe this relationship by first proving a general property of cyclic groups and then calculating LAW OF QUADRATIC RECIPROCITY 9 the discriminant of Φq(x) in Z/pZ. Combining these observations gives us the proof of the law: p q q p = (−1) p−1 2 q−1 2 .
Proposition 3.27. There is a surjective homomorphism φ : (Z/pZ)× →{±1}.
Define (Z/pZ)×2 to be the subgroup {x2 : x ∈(Z/pZ)×}.
Proof. Recall from Thm 3.18 that (Z/pZ)× is cyclic. Then there exists x ∈(Z/pZ)× such that (Z/pZ)× = ⟨x⟩. Fixing the generator x, we define the map φ : a 7→(−1)n where a = xn for some n ∈Z. We check this map is well-defined, i.e. φ(a) does not depend on the power of a fixed generator. Suppose a = xt for some t ∈Z. Then t = (p−1)q+r for some q and 0 ≤r < p−1. We have xt = x(p−1)q+r = x(p−1)qxr = xr.
Furthermore, we have φ(xt) = (−1)t = (−1)(p−1)q+r = (−1)(p−1)q(−1)r = (−1)r = φ(xr) since p−1 is even. Now we show that φ(xt) = φ(xn). We have t = (p−1)n+s for some s ∈Z. Then (−1)(p−1)s+n = (−1)n. So φ is well-defined.
We check that φ is a surjective homomorphism.
Let a, b ∈(Z/pZ)×. Then φ(ab) = φ(xn+m) = (−1)n+m = (−1)n(−1)m = φ(xn)φ(xm) = φ(a)φ(b). Since p−1 is even, we know (p−1)/2 elements of even power map to 1 and the remaining elements map to −1. Thus φ is a surjective homomorphism. Since ker(φ) = {xn : n is even} = (Z/pZ)×2, we can conclude (Z/pZ)×/(Z/pZ)×2 ∼ = {±1}.
□ Proposition 3.28. Let ⟨x⟩be a finite cyclic group of order n, and let y ∈⟨x⟩. The map f : ⟨x⟩→⟨x⟩defined by z 7→zy is an even permutation of the elements of ⟨x⟩ if and only if y is a square in ⟨x⟩.
Proof. Let |⟨x⟩| = n. We will work with Z/nZ for the rest of the proof since ⟨x⟩is isomorphic to Z/nZ.
Define fy by fy(x) = x + y for some y ∈Z/nZ.
First, notice that for each y ∈Z/nZ, we have that fy is a permutation on n elements. We can view fy as an element of Sn. Consider the function g(z) = z + 1.
Composing the map k times gives us gk(z) = z + k. Now for all y ∈Z/nZ we have fy(z) = z+y = gy(z). Since g(z) = z+1, we have g is a cycle with length the order of the group. Now every cycle of length n can be decomposed into n−1 transpositions.
Recall we can determine the parity of a permutation by its transpositions. We have sgn : Sn →{±1} defined by σ 7→(−1)t where t is the number of transpositions in the decomposition of σ. Note that the parity of the cycles is independent of the choice of the decomposition since sgn is well-defined. Now sgn(g) = (−1)n−1. Since the parity of g depends on the order of the group, we prove the proposition in two cases: n is even and n is odd.
Case 1: n is even. Suppose fy ∈An. Since fy has even parity and sgn(g) = −1, we have sgn(fy) = sgn(gy) = (−1)y = 1, which implies y = 2k for some k ∈Z/nZ.
Case 2: n is odd. We will show that every element of Z/nZ is congruent to a multiple of two modulo n and fy ∈An for all y. Consider the map ϕ given by x 7→2x. Since gcd(2, n) = 1, we know ϕ−1 exists and is defined by multiplying every element by 2−1. Thus ϕ is bijective, so every element of Z/nZ is congruent to a multiple of two.
Now we will show that fy ∈An for all y when n is odd. Since y = 2k for some k, we have fy = g2k. Since n is odd, we have sgn(g) = 1. It follows that sgn(fy) = 1.
Thus fy ∈An. Similarly, if n is even and y is a square in Z/pZ, we still have fy ∈An.
□ 10 EMILY ZHANG Proposition 3.29. The following are equivalent.
(a) The pth-power map acts by an even permutation on ⟨ζq⟩⊆Z/pZ(ζq)×.
(b) Multiplication by p acts as an even permutation on the elements of Z/qZ.
(c) p is a quadratic residue modulo q.
Proof. Since ⟨ζq⟩is isomorphic to Z/qZ, we have that applying the p-th power to ⟨ζq⟩corresponds to multiplying every element in Z/qZ by p. This proves (a) ⇒(b). Now notice that the nonzero elements of Z/qZ form the cyclic multi-plicative group (Z/qZ)×. It follows from Prop 3.28 that multiplication by p acts as an even permutation on (Z/qZ)× if and only if p is a quadratic residue modulo q, proving (b) ⇔(c). Again since Z/qZ is isomorphic to ⟨ζq⟩, we have (c) ⇒(a).
□ Now, we will calculate the discriminant of Φq(x) ∈Z/pZ. Let L = Q(ζq). Notice that by definition of the discriminant of a polynomial, we have the discriminant of Φq(x) ∈Z/pZ[x] is congruent to the discriminant of Φq(x) ∈Q[x] modulo p. Since Φq(x) ∈Q[x] is mζq,Q(x), we have disc(Φq) = DL/Q(ζq) = (−1)( q 2)NL/Q(Φ′ q(ζq)) where the first equality uses Prop 3.25 and the second equality uses Prop 3.24.
Thus if Φq(x) ∈Z/pZ[x], we have disc(Φq(x)) ≡(−1)( q 2)NL/Q(Φ′ q(ζq)) (mod p).
We will now calculate (−1)( q 2)NL/Q(Φ′ q(ζq)), and denote NL/Q as N. Rewriting xq −1 = (x −1)Φq(x), we can take the derivative of both sides to obtain qxq−1 = (x −1)Φ′ q(x) + Φq(x). Evaluating this expression at ζq, we have Φ′ q(ζq) = q(ζq−1 q ) ζq −1 = q (ζq −1)ζq .
Then N(Φ′ q(ζq)) = N(q) N(ζq−1)N(ζq) since the norm function is multiplicative. Notice that the embeddings of Q(ζq) into its algebraic closure are precisely the elements in Aut(Q(ζq)/Q); thus, there are deg mζq,Q = q −1 distinct Q-embeddings. Since q ∈Q, it is fixed by all the embeddings. Thus N(q) = qq−1. On the other hand, for each 1 ≤k ≤q −1, there exists an embedding that sends ζq to ζk q . Moreover, these are all the embeddings. So, we can compute N(ζq) = ζ1+···+q−1 q = ζ q( q−1 2 ) q = 1.
Now N(ζq −1) = N(−(1−ζq)) = (−1)q−1N(1−ζq) = N(1−ζq) since q is odd. We have N(1 −ζq) = Qq−1 k=1(1 −ζk q ). Recall that Qq−1 i=1 (x −ζi q) = Pq−1 j=0 xj. Evaluating the expression at 1, we obtain N(1 −ζq) = q. So N(Φ′ q(ζq)) = qq−2. Therefore disc(Φq(x)) ≡(−1)( q 2)qq−2 (mod p).
Using the discriminant, we obtain a condition when Gal(Fp(ζq)/Fp) acts by an even permutation on the roots of xq −1. Recall from Prop 3.26 that Gal(Fp(ζq)/Fp) acts by an even permutation on the roots of xq −1 if and only if disc(Φq(x)) ≡ ±qq−2 (mod p) is a square. To put this criterion in simpler terms, we first show that for nonzero elements a and b of a field, ab2 is a square if and only if a is a square.
But this follows from Prop 3.27; if φ(ab2) = 1, then φ(a) = 1. So qq−2 = q(qq−3) is a square in Z/pZ if and only if q is a square in Z/pZ.
If q ≡1 (mod 4), we immediately deduce that the Galois group of Z/pZ(ζq) is a subgroup of An if and only if q is a square in Z/pZ.
If q ≡3 (mod 4), then disc(Φq(x)) ≡−qq−2 (mod p). But disc(Φq(x)) is a square if φ(−qq−2) = φ(−1)φ(qq−2) = 1; thus, both q and −1 are either squares or not squares in Z/pZ.
LAW OF QUADRATIC RECIPROCITY 11 We summarize our discussion in the following proposition.
Proposition 3.30. If either q ≡1 (mod 4) or p ≡1 (mod 4), then Gal(Fp(ζq)/Fp) acts by even permutation of the roots of xq −1 if and only if q is a square in Z/pZ.
If both p, q ≡3 (mod 4), then the Galois group acts by even permutation if and only if q is not a square in Z/pZ.
Now we get a relationship between when p is a quadratic residue modulo q and when q is a quadratic residue modulo p. Suppose p is a quadratic residue modulo q. Then by Prop 3.29, we have the p-th power map acts by an even permutation on ⟨ζq⟩. Since the p-th power map generates Gal(Fp(ζq)/Fp), we have the Galois group is a subgroup of An. By Prop 3.30, if either p or q ≡1 (mod 4), then q is a quadratic residue modulo p. Reversing the roles of p and q, we obtain a symmetric result. Thus, if either p or q ≡1 (mod 4), then p q = q p .
Furthermore by Prop 3.30, if p, q ≡3 (mod 4) and p is quadratic residue modulo q, then q is not a quadratic residue modulo p. Similarly, if q is a quadratic residue modulo p, then p is not a quadratic residue modulo q. Thus if p, q ≡3 (mod 4), then p q = − q p .
We can combine these results into the following expression: p q q p = (−1)( p−1 2 )( q−1 2 ).
If either p or q is equivalent to 1 (mod 4), we have p q q p = 1 and if both are equivalent to 3 (mod 4), then p q q p = −1. This completes the proof of the law of quadratic reciprocity.
Acknowledgments I am incredibly grateful to Samanda Hu for all her guidance and help with writing this paper and for patiently answering my questions.
I’d like to thank Professor May for organizing such a successful program online and for giving me the opportunity to have a wonderful learning experience this summer.
References Sey Y. Kim, “An Elementary Proof of the Quadratic Reciprocity Law.” The American Math-ematical Monthly, vol. 111, no. 1, 2004, pp. 48–50. JSTOR, www.jstor.org/stable/4145015.
G. Bergman, Quadratic Reciprocity. index.html.
Dummit, David S, and Richard M. Foote. Abstract Algebra.
r-ash/Ant/AntChapter2.pdf |
8337 | https://sites.lsa.umich.edu/kesmith/wp-content/uploads/sites/1309/2024/06/OrthogTrans.pdf | Orthogonal Transformations Math 217 Professor Karen Smith (c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License.
Definition: A linear transformation Rn T − →Rn is orthogonal if |T(⃗ x)| = |⃗ x| for all ⃗ x ∈Rn.
Theorem: If Rn T − →Rn is orthogonal, then ⃗ x · ⃗ y = T⃗ x · T⃗ y for all vectors ⃗ x and ⃗ y in Rn.
A. Discuss this definition and theorem with your table: what does each say in words?
What, intuitively and geometrically, really is an orthogonal transformation? What structure on Rn is preserved by T? What some of the important consequences? What does an orthogonal transforma-tion do to the unit n-cube? What about to a circle? Any shape? How is this different from a linear transformation that is not orthogonal? If you are stuck, think first only about orthogonal transformations in R2. What does an orthogonal transformation do to the unit square? Line segments? Other shapes?
B. Which of the following maps are orthogonal transformations? Think geometrically.
1. R2 →R2 given by rotation counterclockwise through θ in R2 2. The identity map R3 →R3 3. The reflection R3 →R3 over a plane (though the origin).
4. The projection R3 →R3 onto a subspace V of dimension 2 5. Multiplication by 3 1 −2 5 .
6. Dilation R3 →R3 by a factor of 3.
7. Multiplication by cosθ −sinθ sinθ cosθ .
Solution note: 1, 2, 3, and 7 all preserve lengths of vectors, so are orthogonal. The others do not. For 4, note that everything in V ⊥goes to zero, so its length is not preserved. For (5), note that ⃗ e1 goes to to the first column, which is not of length 1.
For (6), note lengths get scaled by 3.
C. Let Rn T − →Rn be an orthogonal transformation.
1. Prove that T is injective. [Hint: consider the kernel.] 2. Prove that T is an isomorphism.
3. Is the inverse of an orthogonal transformation also orthogonal? Prove it!
4. Is the composition of orthogonal transformations also orthogonal? Prove it!
Solution note: (1) It suffices to show the kernel is zero.
Assume ⃗ x ∈ker T.
So T(⃗ x) = 0. But if T is orthogonal also ||T(⃗ x)|| = ||⃗ x||, which means also ||⃗ x|| = 0. So ⃗ x = 0 and T is injective.
(2) This follows from rank-nullity: the kernel has dimension zero so the image has dimension n. Since the target is Rn, it must be surjective. So T is both injective and surjective, which means it is bijective. By definition, a bijective linear transformation is an isomorphism.
(3) Yes. Since ||T(X)|| = ||x|| for all x, we have ||T −1(T(x))|| = ||T(x)|| for all x which means, substituting y = T(x) that ||T −1(y)|| = ||y|| for all y ∈Rn. (4) Yes. If both S and T are orthogonal, then ||S(T(X))|| = ||T(X)|| = ||x|| for all x. So S ◦T is orthogonal.
D. Suppose R2 T − →R2 is given by left multiplication by A = a c b d . Assuming T is orthogonal, illustrate an example what such a T does to a unit square. Label your picture with a, b, c, d. What do you think det A can be?
Solution note: You should see a square, with one corner on the origin and sides of length 1.
E. Let A be the standard matrix of an orthogonal transformation T : Rn →Rn (meaning that T(⃗ x) = A⃗ x for all ⃗ x ∈Rn. Prove that the columns of A are an orthonormal basis for Rn.
Solution note: Each column is length one since they are the images of the ⃗ ei which are length 1. Also, since ⃗ ei ·⃗ ej (for i ̸= j), it is also true that T(⃗ ei) · T(⃗ ej) = 0. These are the columns of A.
F. Definition: An n × n matrix is orthogonal if its columns are orthonormal.
1. Is A = cosθ −sinθ sinθ cosθ orthogonal? Is AT orthogonal? Compute AT A.
2. Is B = 3/5 4/5 0 −4/5 3/5 0 0 0 1 orthogonal? Is BT orthogonal? Compute BT B.
3. Prove that if M is an orthogonal matrix, then M−1 = MT . [Hint: write M as a row of columns and MT as a column of rows. Compute MT · M. ] G. Theorem: Let T : Rn →Rn be a linear transformation with (standard) matrix A. Then T is orthogonal if and only if A has orthonormal columns.
1. Scaffold a proof of this theorem. What needs to be shown?
2. Show that if T is orthogonal, the matrix of T has orthornormal columns [Hint: use the Theorem on page 1.] 3. Show the converse.
H. Prove that the rows of an orthogonal matrix are also orthonormal.
G. Can you prove very important Theorem on the first page. [Hint: consider x + y.] I. An Application of QR factorization. Suppose that we are given a system of n linear equations in n unknowns: A⃗ x = ⃗ b.
1. Assuming A is invertible, express the solutions to this system in terms of A−1 and ⃗ b.
2. Assume that A = QR is the QR-factorization of A. What does this mean? What are the sizes of the matrices in this case?
3. What happens if we multiply both sides of the equation A⃗ x = ⃗ b by QT ? How might this simplify the problem of finding the solution to this system?
4. Suppose that A = 4 4/5 −11/5 0 −10 1 3 −3/5 23/5 . Applying the Gram-Schmidt process to the columns of this matrix we get 4/5 0 −3/5 0 −1 0 3/5 0 4/5 . Find the QR factorization of A.
Solution note: We have A = QR where Q is the orthogonal matrix above and R is 5 1 1 0 10 −1 0 0 5 .
5. Use your QR factorization to quickly solve the system A⃗ x = [0 0 25]T without row reducing!.
Solution note: To solve A⃗ x = QR⃗ x = ⃗ b, multiply both sides by Q−1, which is QT since Q is orthogonal. We have the equivalent system R⃗ x = QT⃗ b = 15 0 20 . This is easy to solve because R is upper triangular. We get z = 4 from the bottom row. Then the second row gives 10y −4 = 0, so y = 2/5. The top row gives 5x + 2/5 + 4 = 15, so x = 53/5.
G. TRUE OR FALSE. Justify. In all problems, T denotes a linear transformation from Rn to itself, and A is its matrix in the standard basis.
1. If T is orthogonal, then x · y = Tx · Ty for all vectors x and y in Rn.
2. If T sends every pair of orthogonal vectors to another pair of orthogonal vectors, then T is orthogonal.
3. If T is orthogonal, then T is invertible.
4. An orthogonal projection is orthogonal.
5. If A is the matrix of an orthogonal transformation T, then the columns of A are orthonormal.
6. The transpose of an orthogonal matrix is orthogonal.
7. The product of two orthogonal matrices (of the same size) is orthogonal.
8. If A is the matrix of an orthogonal transformation T, then AAT is the identity matrix.
9. If A−1 = AT , then A is the matrix of an orthogonal transformation of Rn. |
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20 - Conic Section
Cone & its Sections
Circle
Equation
Properties
Theorems
Contact of two Circles
Tangent
Equation
Normal
Equation
Parabola
Equation
Terms
Summary of Standard Parabolas
Theorems on Parabola
Ellipse
Terms
Summary of Standard Forms of Ellipse
Theorems on Ellipse
Hyperbola
Terms
Summary of Standard Forms of Hyperbola
Comments
Conic Section¶
Cone & its Sections¶
Right Circular Cone: The surface generated when one straight line that intersects another fixed straight line is rotated at an oblique angle. The lines are called generators of the cone.
Axis of the Cone: The fixed line of the cone is called its axis.
Elements of the Cone: The possible positions of the generating line in its rotation about the axis are its elements.
Vertex: The common intersection point of all the cone's elements is called Vertex of cone.
Napes: The two symmetrical parts of the generated surface on each side of the vertex are cone's Napes.
Conic Sections or Conics: The curves that can be obtained by cutting a cone with a plane are called conic sections or conics.
Circle: When the intersecting plane cuts completely across one napes at right angle to the axis of cone, the curve is called circle.
Point Circle: If the intersection plane cuts across the vertex of a cone, the resulting curve is called point circle.
Parabola: When the intersecting plane is parallel to an element, an passing through only one nape, the resulting curve is Parabola.
Ellipse: When the intersecting plane cuts completely across one nape at an oblique angle to the axis, the curve is an ellipse.
Hyperbola: When the intersecting plane cuts through both napes parallel to the axis of the cone the resulting curve is hyperbola.
Circle¶
A set of points such that distance of each point fixed point called center remains constant. The constant distance is called radius of the circle.
Equation¶
Standard Equation When center is ((h,k)) & radius is (r): $$ (x-h)^2 + (y-k)2 = r2 $$ When center is ((0,0)) & radius is (r): $$ x^2 + y2 = r2 $$
Parametric Equation [ x= r\cos\theta ] [ y= r\sin\theta ]
If end points of the diameter of circle are (a(x_,y_1)), (B(x_2,y_2)) $$ (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 $$
General Form $$ x^2 + y^2 + 2gx+2fy+c=0 $$ The general form of circle equation have following properties:
It involves three constants (f,g,c)
It is a second degree equation in which coefficients of (x^2) & (y^2) are (1)
It does not involves (xy) product
Center of the circle is ((-g,-f))
Radius of circle is (\sqrt{g^2+f^2-c}). If:
(r=\sqrt{g^2+f^2-c}>0) circle is real circle
(r=\sqrt{g^2+f^2-c}=0) circle is point circle
(r=\sqrt{g^2+f^2-c}<0) circle is imaginary circle
Properties¶
Perpendicular dropped from the center of a circle on a chord bisects the cord.
The perpendicular bisector of any chord of a circle passes through the center of the circle.
The line joining the center of a circle to the midpoint of chord is perpendicular to the chord.
Congruent chords of a circle are equidistant from its center.
Measure of the central angle of a minor arc is double the measure of the angle subtended in the corresponding major arc.
An angle is a semi-circle is a right angle.
The tangent to a circle at any point of the circle is perpendicular to the radial segment at that point.
The perpendicular at the outer end of a radial segment is tangent to the circle.
Normal lines of a circle pass through the center of the circle.
Mid point of the hypotenuse of a right triangle is the circum-center of the triangle.
Perpendicular dropped from a point of a circle on diameter is a mean proportional between the segments into which it divides the diameter.
Theorems¶
the point (P(x_1,y_2)), on circle (x^2+y^2+2gx+2fy+c=0) lies:
outside circle if (x_1^2+y_1^2+2gx+2fy+c>0)
on circle if (x_1^2+y_1^2+2gx+2fy+c=0)
inside circle if (x_1^2+y_1^2+2gx+2fy+c<0)
Any two tangents can be drawn to a circle from any point outside the circle.
Let (P(x_1,y_1)) be a point outside the circle (x^2+y^2+2gx+2fy+c=0), then length of either tangents drawn to circle is : $$ \sqrt{x_1^2 +y_1^2 +2gx_1 +2fy_1 + c} $$
The line (y=mx+c) intersects the circle (x^2+y^2=a^2) in at the most two points, the points are:
Real & distinct if (a^2(1+m^2)>c^2)
Real & coincident if (a^2(1+m^2)=c^2) (also called condition of tangency)
Imaginary if (a^2(1+m^2)
Contact of two Circles¶
Let (r_1), (r_2) be the radii of two circles with (C_1) & (C_2) be their centers, then:
Circles touches externally if (|C_1C_2|=r_1+r_2)
Circles touches internally if (|C_1C_2|=|r_1-r_2|)
Circles intersect in two real distinct points if (|r_1-r_1|<|C_1C_2|<(r_1+r_2))
Circle don't touches each other if (|r_1-r_2|>|C_1C_2|>r_1+r_2)
Tangent¶
A straight line that touches the curve at a single point without cutting the curve is called tangent. It is perpendicular to the line segment joining center to that point.
Equation¶
Equation of tangent to the circle (x^2+y^2=a^2) at (P(x_1,y_1)): $$ xx_1+ yy_1=a^2 $$
Equation of tangent to the circle (x^2+y^2 +2gx+2fy+c=0) at (P(x_1,y_1)): $$ xx_1+ yy_1 + g(x+x_1)+ f(y+y_1)+c=0 $$
Normal¶
A straight line perpendicular to the curve at point of tangency.
Equation¶
Equation of normal to the circle (x^2+y^2=a^2) at (P(x_1,y_1)): $$ xx_1- yy_1=0 $$
Equation of tangent to the circle (x^2+y^2 +2gx+2fy+c=0) at (P(x_1,y_1)): $$ (x-x_1)(y_1+f) + (y-y_1)(x_1+g)=0 $$
Parabola¶
A set of points in a plane such that the distance of each point form a fixed point (focus), equal to its distance from a fixed straight line (directrix, L)
Equation¶
Standard Form of standard parabola: $$ y^2=4ax $$
Parametric Equation: $$ x=at^2 $$ $$ y=2at $$
General form: $$ ax^2 +by^2 +2gx+2fy+c=0 $$ Its a second degree equation in which either (a=0) or (b=0) but not both.
Terms¶
Axis of Parabola: The line through focus and perpendicular to directrix is called axis of parabola.
Vertex: The mid point of the perpendicular line joining focus and directrix is called vertex of parabola.
Focal Chord: A chord of the parabola through focus is called focal chord.
Latusrectum: The focal chord perpendicular to the axis of the parabola is called latusrectum of the parabola.
Eccentricity: The ratio of the distance of any point on the parabola to its distance from directrix.
Summary of Standard Parabolas¶
| # | 1 | 2 | 3 | 4 |
--- ---
| Equation | (y^2=4ax) | (y^2=-4ax) | (x^2=4ay) | (x^2=-4ay) |
| Focus | ((a,0)) | ((-a,0)) | ((0,a)) | ((0,-a)) |
| Vertex | ((0,0)) | ((0,0)) | ((0,0)) | ((0,0)) |
| Length of Latusrectum | (4a) | (4a) | (4a) | (4a) |
| Equation of Latusrectum | (x=a) | (x=-a) | (y=a) | (y=-a) |
| Equation of Directrix | (x=-a) | (x=a) | (y=-a) | (y=a) |
| Axis | (y=0) | (y=0) | (x=0) | (x=0) |
| Eccentricity | (1) | (1) | (1) | (1) |
| Graph |
Theorems on Parabola¶
The point on the parabola closest to the focus is the vertex.
The ordinate at any point (P) of the parabola is a mean proportional between the length of the latusrectum and the abscissa of (P).
The tangent at any point (P) of a parabola makes equal angles with the line (PF) and the line through (P) parallel to the axis of parabola, (F) being focus.
Ellipse¶
A set of point such that the distance of each from fixed point (focus, (F)) bears a constant ration (Eccentricity, (0) to its perpendicular distance from a fixed straight line (directrix, (L)).
Terms¶
Vertices: The points on the standard ellipse where it crosses the x-aix.
Co-vertices: The points on the standard ellipse where it crosses the y-axis.
Center: The midpoint of the line joining vertices
Major axis: The lines joining vertices is called major axis.
Minor axis: The line joining co-vertices is called minor axis.
Latusrectum: The cords perpendicular to major axis passed through foci are called latusrectum.
Eccentricity: The ratio of the distance of any point on the ellipse from the focus to its distance from the directrix.
Summary of Standard Forms of Ellipse¶
| # | 1 | 2 |
---
| Equation | $$ \frac{x^2 }{a^2 } + \frac{y^2 }{b^2 }=1,\;a>b $$ | $$ \frac{y^2 }{a^2 } + \frac{x^2 }{b^2 }=1,\;a>b $$ |
| Foci | ((\pm c,0),c^2=a^2-b^2) | ((0,\pm c),c^2=a^2-b^2) |
| Eccentricity | (0 | (0 |
| Vertices | ((\pm a,0)) | ((0,\pm a)) |
| Co vertices | ((0,\pm b)) | ((\pm b,0)) |
| Center | ((0,0)) | ((0,0)) |
| Length of Major Axis | (2a) | (2a) |
| Equation of Major Axis | (y=0) | (x=0) |
| Length of Major Axis | (2b) | (2b) |
| Equation of Major Axis | (x=0) | (y=0) |
| Length of Latusrectum | ({2b^2}/{a}) | ({2b^2}/{a}) |
| Equation of Directrix | (x=\pm{c}/{e^2}) | (y\pm{c}/{e^2}) |
| Graph |
Theorems on Ellipse¶
The sum of the focal distances of any point on an ellipse is equal to length of major axis (2a)
The distance between center and any focus of the ellipse is denoted by (c), and is given as (\sqrt{a^b-b^2})
The distance between foci is (2c)
Hyperbola¶
A set of points such that the distance of each point from a fixed point (focus, (F)) bears a constant ratio (Eccentricity, (e>1)) to its perpendicular distance from a fixed straight line (directrix, (L)).
Terms¶
Vertices: The points on the standard hyperbola where it crosses the x-axis.
Center: The midpoint of the line joining the vertices or co-vertices.
Transverse axis: The line joining vertices is called transverse axis.
Conjugate axis: The line joining co vertices is called conjugate axis.
Latusrecta: The chords perpendicular to major axis passes through foci are called Latusrecta.
Eccentricity: The ratio of distance of any point on ellipse from the focus to its distance from directrix,
Asymptotes: In general, an asymptote is a line approaches to a curve but never touches it.Every hyperbola has associated with two lines called asymptotes. Their point of intersection is center of hyperbola.
Slope of asymptotes (=\pm b/a) (hyperbola opens to sides)
Slope of asymptotes (=\pm a/b) (hyperbola opens to up & down)
Summary of Standard Forms of Hyperbola¶
| # | 1 | 2 |
---
| Equation | $$ \frac{x^2 }{a^2 }-\frac{y^2 }{b^2 }=1 $$ | $$ \frac{y^2 }{a^2 }-\frac{x^2 }{b^2 }=1 $$ |
| Foci | ((\pm c,0),c^2=a^2+b^2) | ((0, \pm c),c^2=a^2+b^2) |
| Eccentricity | (e=c/a>1) | (e=c/a>1) |
| Vertices | ((\pm a,0)) | ((0,\pm a)) |
Comments |
8339 | https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14:_The_Behavior_of_Gases/14.06:_Combined_Gas_Law | 14.6: Combined Gas Law - Chemistry LibreTexts
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14: The Behavior of Gases
Introductory Chemistry (CK-12)
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14.6: Combined Gas Law
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Contents
1. Home
2. Bookshelves
3. Introductory, Conceptual, and GOB Chemistry
4. Introductory Chemistry (CK-12)
5. 14: The Behavior of Gases
6. 14.6: Combined Gas Law
Expand/collapse global location
Introductory Chemistry (CK-12)
Front Matter
1: Introduction to Chemistry
2: Matter and Change
3: Measurements
4: Atomic Structure
5: Electrons in Atoms
6: The Periodic Table
7: Chemical Nomenclature
8: Ionic and Metallic Bonding
9: Covalent Bonding
10: The Mole
11: Chemical Reactions
12: Stoichiometry
13: States of Matter
14: The Behavior of Gases
15: Water
16: Solutions
17: Thermochemistry
18: Kinetics
19: Equilibrium
20: Entropy and Free Energy
21: Acids and Bases
22: Oxidation-Reduction Reactions
23: Electrochemistry
24: Nuclear Chemistry
25: Organic Chemistry
26: Biochemistry
Back Matter
14.6: Combined Gas Law
Last updated Mar 21, 2025
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14.5: Gay-Lussac's Law
14.7: Avogadro's Law
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Table of contents
1. Combined Gas Law
1. Example 14.6.1
1. Solution
2. Step 1: List the known quantities and plan the problem.
3. Known
4. Unknown
5. Step 3: Think about your result.
Summary
The modern refrigerator takes advantage of the gas laws to remove
heat
from a system. Compressed gas in the coils 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 do 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
Example 14.6.1
2.00 L of a gas at 35 oC and 0.833 atm is brought to
standard temperature and pressure (STP)
. What will be the new gas volume?
Solution
Step 1: List the known quantities and plan the problem.
Known
P 1=0.833 atm
V 1=2.00 L
T 1=35 oC=308 K
P 2=1.00 atm
T 2=0 oC=273 K
Unknown
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 cancelling P out of the equation leaves Charles's
Law
. If the volume is constant, then V 1=V 2 and cancelling V out of the equation leaves Gay-Lussac's
Law
.
Summary
P 1V 1 T 1=P 2V 2 T 2
This page titled 14.6: Combined Gas Law is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
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14.7: Avogadro's Law
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HORIBA»Water Quality»Support»Ion Sensor Cartridge Online Support Guide»Ion concentration and activity
Ion concentration and activity
The ion concentration can be accurately measured by performing calibration and measurment using an ISA.
The following describes the relationship between the ion concentration and activity and the reason for adding an ISA.
Difference in state of solution depending on ion concentration
If specific ions exist in the sample solution, the ion selective membrane of the ion selective electrode generates a potential that corresponds to the concentration of those ions. The generated potential is measured as the potential difference on the basis of the reference electrode by using an ion meter or meter with an ion measurement function. A proportional relationship exists between the measured potential difference and the logarithm of the concentration of ions to be measured in the sample solution which can be represented by the Nemst equation.
The activity is a concentration of ions in the solution that is corrected using a coefficient called the activity coefficient, and the relationship between them can be expressed with the following relational expression:
Activity (a) = Activity coefficient (y) • Ion concentration (C)
The following table shows the relationship between the ion concentration and the activity coefficient.
From this table, you can see that when the ion concentration becomes lower than 1 x 10-3 mol/L, the activity coefficient becomes a value close to 1, and as it becomes higher than 1 x 10-3 mol/L, the activity coefficient becomes lower than 1.
If the activity coefficient becomes lower than 1 (activity < ion concentration), an error occurs in the potential measurement. As ISA is used to reduce this error. An ISA is generally selected so that it does not react with the ions to be measured and does not affect the measured potential.
The measurment error can be reduced because adding the ISA enables to maintain the ion strength constant and to adjust the activity coefficients of the sample solution and the standard solution almost at the same value.
(Example) Activity coefficient for monovalent ion concentration
| Ion concentration（C） mol/L | Activity coefficient（γ） |
--- |
| 1×10-5 | 0.998 |
| 1×10-4 | 0.988 |
| 1×10-3 | 0.961 |
| 1×10-2 | 0.901 |
| 1×10-1 | 0.751 |
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8341 | https://www.marleygen.org/interpret_output.html | Interpreting the output — MARLEY documentation
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Interpreting the output
Interpreting the output¶
This page provides a guide to the contents of the output files produced by the marley executable. Following a brief description of the PDG codes used to identify particle types in MARLEY, documentation for all four available output formats is given below.
PDG codes¶
The Particle Data Group (PDG) has defined a standard numbering scheme for representing particle species in Monte Carlo event generators. Each kind of particle is assigned a unique positive integer as an identifier. The corresponding antiparticle is assigned a negative integer with the same absolute value. A full description of the numbering scheme is available here.
Like nearly all modern particle physics generators, MARLEY adopts the integer PDG codes for particle identification and uses them both internally and in output files. For convenience, a table of the PDG codes most relevant for MARLEY is given below.
| PDG code | Particle |
--- |
| 11 | e− |
| 12 | ν e |
| 13 | μ− |
| 14 | ν μ |
| 15 | τ− |
| 16 | ν τ |
| 22 | γ |
| 2112 | n |
| 2212 | p |
| 1000010020 | d |
| 1000010030 | t |
| 1000020030 | h |
| 1000020040 | α |
In general, a PDG code of the form 100ZZZAAA0 represents a nuclide with proton number Z and mass number A. For example, 40 Ar is represented by the PDG code 1000180400.
Output file formats¶
The neutrino scattering events generated by the marley executable may be saved to disk in four distinct output formats. Descriptions of each of these formats are given below.
ASCII¶
The ASCII format is MARLEY’s native text-based format for event input and output. It is produced and parsed by applying the stream insertion (<<) and extraction (>>) operators, respectively, to the marley::Event class in the C++ source code.
An ASCII-format output file begins with the line
FluxAvgXSec
where FluxAvgXSec is the flux-averaged total cross section in natural units (MeV −2 per atom). This quantity may be used together with the events themselves to compute differential cross sections.
One or more event records appear in the remainder of the output file. Each event record begins with a header of the form
Ni Nf Ex twoJ P
where Ni (Nf) is the number of particles in the initial (final) state. The three remaining fields in the event header report properties of the final-state nucleus following the primary scattering reaction but before de-excitations. The Ex field gives the nuclear excitation energy (MeV), twoJ gives the nuclear spin multiplied by two (to allow half-integer spins to be represented by a C++ int), and P is a single character representing a positive (+) or negative (-) parity state.
On the lines following the event header, each of the particles belonging to the event is described by a single line of the form
PDG Etot Px Py Pz M Q
where PDG is the PDG code identifying the particle species and Etot, Px, Py, and Pz are, respectively, the total energy and the x-, y-, and z-components of the particle 3-momentum (MeV). The particle mass M (MeV) and (net) electric charge Q (in units of the elementary charge) appear at the end of each line. The first Ni lines following the event header describe the initial-state particles. The remaining Nf lines describe the final-state particles and complete the event record.
1
2
3
4
5
6
7
8
9
10
11
12
13
14 5.98368867447267264e-19
2 4 3.79748000000000019e+00 2 +
12 1.00000000000000000e+01 0.00000000000000000e+00 0.00000000000000000e+00 1.00000000000000000e+01 0.00000000000000000e+00 0
1000180400 3.72247225431518091e+04 0.00000000000000000e+00 0.00000000000000000e+00 0.00000000000000000e+00 3.72247225431518091e+04 0
11 5.20690886815266740e+00 -4.63535385338761508e+00 1.35706546730579314e+00 -1.87687187323011373e+00 5.10998927645907708e-01 -1
1000190400 3.72257175068044089e+04 4.98066184879143847e+00 -2.27058729207187593e+00 9.28456410767741858e+00 3.72257159465162476e+04 1
22 1.53694352434620662e+00 -1.11400145523226901e+00 9.50751758100756628e-01 4.66119350851738223e-01 0.00000000000000000e+00 0
22 2.26118395490673985e+00 7.68693459828445835e-01 -3.72299333346743089e-02 2.12618841470095354e+00 0.00000000000000000e+00 0
2 3 1.03964200000000009e+01 2 +
12 2.99304885549511290e+01 0.00000000000000000e+00 0.00000000000000000e+00 2.99304885549511290e+01 0.00000000000000000e+00 0
1000180400 3.72247225431518091e+04 0.00000000000000000e+00 0.00000000000000000e+00 0.00000000000000000e+00 3.72247225431518091e+04 0
11 1.85223477954320721e+01 -9.86795370224160173e+00 -1.55256312663347771e+01 -2.09630900945355991e+00 5.10998927645907708e-01 -1
1000190390 3.62940260743929030e+04 -1.74322423197586254e+01 -4.21863040207651636e+01 5.85225771236273147e+01 3.62939501877290750e+04 1
2112 9.42104609518426400e+02 2.73001960220002289e+01 5.77119352870999407e+01 -2.64957795592226226e+01 9.39565378653339735e+02 0
The listing above shows an example MARLEY output file in ASCII format. Line 1 gives the value of 5.984×10−19 MeV −2 = 2.330×10−40 cm 2 for the flux-averaged total cross section per atom. Line 2 gives the header for the first event, which involves a transition to the 40 K nuclear level with an excitation energy of 3.797 MeV above the ground state and spin-parity 1+. Lines 3–4 describe the initial state: a 10-MeV ν e traveling in the +z direction toward a 40 Ar atom at rest. Lines 5–8 describe the final-state particles: a 5.2 MeV electron, a 40 K ion, and two de-excitation γ-rays with energies of 1.54 and 2.26 MeV. The second and final event in the file, which appears on lines 9–14, involves a ν e-40 Ar collision which produces an electron, a 39 K ion, and a neutron.
HEPEVT¶
The legacy HEPEVT format is designed for interfacing event generators with each other and with other software. Compatibility with this event format is maintained in many modern high energy physics software libraries. The description presented here covers only those aspects of the HEPEVT format needed to interpret the output of MARLEY. Further details are available on pages 327–330 of this document.
A HEPEVT-format output file consists of one or more text-based event records. Each of these records begins with the header
NEVHEP NHEP
where NEVHEP is the event number (untracked by MARLEY and thus always set to zero) and NHEP is the number of particles in the event. The header is followed by NHEP lines, each representing a single particle. These have the format
ISTHEP IDHEP JMOHEP1 JMOHEP2 JDAHEP1 JDAHEP2 PHEP1 PHEP2 PHEP3 PHEP4 PHEP5 VHEP1 VHEP2 VHEP3 VHEP4
where ISTHEP is an integer code identifying the particle status and IDHEP is the particle’s PDG code. In agreement with the HEPEVT standard, MARLEY uses status code 1 for the final-state particles and 3 for the initial-state particles. The JMOHEP1, JMOHEP2, JDAHEP1, and JDAHEP2 entries record the indices (between 1 and NHEP, inclusive) of particles in the event record that correspond to the first mother, second mother, first daughter, and last daughter of the current particle, respectively. These indices are set to zero in cases where they do not apply (e.g., a particle with no daughters will have JDAHEP1 = JDAHEP2 = 0). Entries PHEP1 through PHEP3 record the x-, y-, and z-components of the particle 3-momentum, while PHEP4 gives the total energy and PHEP5 gives the particle mass (all in GeV). Entries VHEP1 through VHEP3 store the x, y, and z positions of the particle production vertex (mm), and VHEP4 gives the production time (mm/c).
Because MARLEY currently treats all nuclear de-excitations as instantaneous and does not perform any particle tracking, VHEP1 through VHEP4 are always identically zero in HEPEVT output files. Intermediate de-excitation steps are also not currently stored in the event record, so JMOHEP1, JMOHEP2, JDAHEP1, and JDAHEP2 are also identically zero in most cases.
In addition to the initial- and final-state particles, MARLEY adds a dummy particle with ISTHEP = 11 to each HEPEVT event record. All data fields are zero for this particle except for (1) JMOHEP1, which contains the nuclear spin multiplied by two, (2) JMOHEP2, which reports the parity of the nucleus as an integer, (3) PHEP4, which gives the excitation energy of the nucleus (MeV), and (4) PHEP5, which records the flux-averaged total cross section in units of MeV −2 per atom. As is the case for the ASCII format, the excitation energy, spin, and parity values refer to the nuclear state that is formed after the primary scattering reaction but before any de-excitations have occurred.
0 7
3 12 0 0 0 0 0.00000000000000000e+00 0.00000000000000000e+00 1.00000000000000002e-02 1.00000000000000002e-02 0.00000000000000000e+00 0. 0. 0. 0.
3 1000180400 0 0 0 0 0.00000000000000000e+00 0.00000000000000000e+00 0.00000000000000000e+00 3.72247225431518061e+01 3.72247225431518061e+01 0. 0. 0. 0.
11 0 2 1 0 0 0.00000000000000000e+00 0.00000000000000000e+00 0.00000000000000000e+00 3.79748000000000019e+00 5.98368867447267264e-19 0. 0. 0. 0.
1 11 0 0 0 0 -4.63535385338761496e-03 1.35706546730579320e-03 -1.87687187323011366e-03 5.20690886815266749e-03 5.10998927645907710e-04 0. 0. 0. 0.
1 1000190400 0 0 0 0 4.98066184879143812e-03 -2.27058729207187593e-03 9.28456410767741942e-03 3.72257175068044077e+01 3.72257159465162459e+01 0. 0. 0. 0.
1 22 0 0 0 0 -1.11400145523226908e-03 9.50751758100756655e-04 4.66119350851738206e-04 1.53694352434620668e-03 0.00000000000000000e+00 0. 0. 0. 0.
1 22 0 0 0 0 7.68693459828445808e-04 -3.72299333346743093e-05 2.12618841470095347e-03 2.26118395490674000e-03 0.00000000000000000e+00 0. 0. 0. 0.
0 6
3 12 0 0 0 0 0.00000000000000000e+00 0.00000000000000000e+00 2.99304885549511283e-02 2.99304885549511283e-02 0.00000000000000000e+00 0. 0. 0. 0.
3 1000180400 0 0 0 0 0.00000000000000000e+00 0.00000000000000000e+00 0.00000000000000000e+00 3.72247225431518061e+01 3.72247225431518061e+01 0. 0. 0. 0.
11 0 2 1 0 0 0.00000000000000000e+00 0.00000000000000000e+00 0.00000000000000000e+00 1.03964200000000009e+01 5.98368867447267264e-19 0. 0. 0. 0.
1 11 0 0 0 0 -9.86795370224160216e-03 -1.55256312663347770e-02 -2.09630900945356009e-03 1.85223477954320724e-02 5.10998927645907710e-04 0. 0. 0. 0.
1 1000190390 0 0 0 0 -1.74322423197586264e-02 -4.21863040207651613e-02 5.85225771236273160e-02 3.62940260743929031e+01 3.62939501877290738e+01 0. 0. 0. 0.
1 2112 0 0 0 0 2.73001960220002303e-02 5.77119352870999400e-02 -2.64957795592226236e-02 9.42104609518426450e-01 9.39565378653339778e-01 0. 0. 0. 0.
The listing above shows an example MARLEY output file in HEPEVT format. The same two events from the ASCII-format example file are used for easy comparison of the formats.
JSON¶
JSON (JavaScript Object Notation) is a text-based data-interchange format used for a wide variety of applications. The grammar of JSON is based on two data structures. A JSON object is an unordered set of key-value pairs enclosed in curly braces ({}). A JSON array is an ordered list of values enclosed in square brackets ([]). Each key is an arbitrary string delimited by double quotes ("") and separated from its corresponding value by a colon (:). Each value may be a string, a number, an object, an array, or one of the words (without surrounding double quotes) true, false, or null. Neighboring values within an array and key-value pairs within an object are separated from each other by a comma (,).
The JSON-format output files produced by MARLEY contain a single object with two top-level keys. The events key is used to label an array of event objects, while the gen_state key is associated with an object describing the state of the generator at the moment that the file was created.
Each element of the events array is a JSON object containing five key-value pairs. The first three of these, Ex, twoJ, and parity, provide the excitation energy (MeV), two times the total spin, and the parity of the final nucleus after the primary interaction but before any de-excitations have taken place. The other two keys, initial_particles and final_particles, are used store arrays of particles represented as JSON objects. Each particle object defines the following keys:
charge
The (net) electric charge (in units of the elementary charge)
pdg
PDG code identifying the particle type
E
Total energy
px
Momentum x-component
py
Momentum y-component
pz
Momentum z-component
mass
Mass
The particle 4-momentum components and mass are all given in natural units (MeV).
The gen_state JSON object includes several key-value pairs. The config key refers to a JSON object which reproduces the full contents (except for comments) of the job configuration file used to generate the events. The event_count, flux_avg_xsec, and seed keys label the total number of events generated at the time the file was written, the flux-averaged total cross section (MeV −2 per atom), and the integer random number seed used to initialize the event generator. A final key, generator_state_string, records a string representation of the internal state of the std::mt19937_64 object used by MARLEY to obtain pseudorandom numbers.
An example JSON output file is available here. It contains the same two events as the ASCII- and HEPEVT-format examples above.
ROOT¶
If MARLEY has been built against the appropriate shared libraries from the ROOT data analysis framework (see the description of MARLEY’s installation prerequisites here), then output files in ROOT’s compressed binary format may also be produced.
Within a ROOT-format output file, access to the generated events is managed by an instance of the ROOT TTree class, which provides a hierarchical data structure for storing many objects belonging to the same C++ type. In general, a TTree may own one or more branches (represented by the TBranch class), each of which owns one or more leaves (represented by TLeaf).
The ROOT-format output files generated by MARLEY contain a single TTree called MARLEY_event_tree. This TTree has a single branch called event which stores one marley::Event object per tree entry. Example ROOT macros that demonstrate the recommended procedure for accessing and manipulating the events stored in this format are provided in the folder examples/macros/ of the MARLEY source code.
In order to properly interpret marley::Event objects, ROOT requires shared libraries containing dictionaries for the MARLEY classes to be loaded at runtime. Whenever MARLEY is successfully built against ROOT, a helper script called mroot is installed in the build/ folder. Running this script will start the interactive ROOT C++ interpreter after automatically loading the needed class dictionaries.
Conventions used in the event objects¶
When analyzing marley::Event objects (or their alternative “flat” representation discussed below), it is helpful to be aware of MARLEY’s nomenclature for the particles involved in each event’s 2 → 2 primary interaction
𝑎 + 𝑏 → 𝑐 + 𝑑 .
Here the particles 𝑎, 𝑏, 𝑐, and 𝑑 are labeled as, respectively, the projectile, target, ejectile, and residue. Where a mass difference exists, MARLEY chooses the projectile (ejectile) to be the lighter of the two particles in the initial (final) state. Otherwise, the choice is arbitrary. All four-vector components stored in a MARLEY event record are expressed in the laboratory frame, i.e., the rest frame of the target. If simulation of nuclear de-excitations is enabled (as it is by default) and the residue is a nucleus, its 4-momentum and net charge are stored in the event record after it has reached the ground state.
Metadata¶
Four pieces of metadata are saved in a ROOT-format output file alongside the events themselves:
MARLEY_config
A JSON-format string which stores the contents (except for comments) of the job configuration file used to generate the events
MARLEY_state
A string giving the serialized internal state (obtained using the stream insertion operator <<) of the std::mt19937_64 object used to generate random numbers when creating the events.
MARLEY_seed
A string representation of the integer random number seed used to initialize the simulation
MARLEY_flux_avg_xsec
A TParameter which stores the flux-averaged total cross section (MeV −2 per atom)
“Flat” ROOT files¶
An alternative “flat” form of the ROOT output format is also available which may be analyzed without the need for the MARLEY class dictionaries. An output file containing MARLEY events in any of the four standard formats may be converted into a “flat” ROOT file using the marsum utility. After sourcing the setup_marley.sh script, one may convert the MARLEY output file OLD_EVENTS_FILE into a new “flat” ROOT file, new_flat_file.root, via the command
marsum new_flat_file.root OLD_EVENTS_FILE
The new file will contain a single ROOT TTree called mst (for M ARLEY s ummary t ree) with the following branches:
pdgv (int)
Projectile PDG code
Ev (double)
Projectile total energy (MeV)
KEv (double)
Projectile kinetic energy (MeV)
pxv (double)
Projectile 3-momentum x-component (MeV)
pyv (double)
Projectile 3-momentum y-component (MeV)
pzv (double)
Projectile 3-momentum z-component (MeV)
pdgt (int)
Target PDG code
Mt (double)
Target mass (MeV)
pdgl (int)
Ejectile PDG code
El (double)
Ejectile total energy (MeV)
KEl (double)
Ejectile kinetic energy (MeV)
pxl (double)
Ejectile 3-momentum x-component (MeV)
pyl (double)
Ejectile 3-momentum y-component (MeV)
pzl (double)
Ejectile 3-momentum z-component (MeV)
pdgr (int)
Residue PDG code
Er (double)
Residue total energy (MeV)
KEr (double)
Residue kinetic energy (MeV)
pxr (double)
Residue 3-momentum x-component (MeV)
pyr (double)
Residue 3-momentum y-component (MeV)
pzr (double)
Residue 3-momentum z-component (MeV)
Ex (double)
Initial residue excitation energy (MeV)
twoJ (int)
Two times the initial residue spin
parity (int)
Initial residue parity
np (int)
Number of de-excitation products
pdgp (int[np])
De-excitation product PDG codes
Ep (double[np])
De-excitation product total energies (MeV)
KEp (double[np])
De-excitation product kinetic energies (MeV)
pxp (double[np])
De-excitation product 3-momentum x-components (MeV)
pyp (double[np])
De-excitation product 3-momentum y-components (MeV)
pzp (double[np])
De-excitation product 3-momentum z-components (MeV)
xsec (double)
Flux-averaged total cross section (10−42 cm 2 per atom)
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8342 | https://scholars.duke.edu/publication/1415507 | Scholars@Duke publication: Clinical Policy: Critical Issues in the Evaluation and Management of Adult Patients Presenting to the Emergency Department With Acute Headache.
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Clinical Policy: Critical Issues in the Evaluation and Management of Adult Patients Presenting to the Emergency Department With Acute Headache.
Publication,Journal Article
American College of Emergency Physicians Clinical Policies Subcommittee (Writing Committee) on Acute Headache:; Godwin, SA; Cherkas, DS; Panagos, PD ...
Published in: Ann Emerg Med
October 2019
Published version (DOI)Link to item
This clinical policy from the American College of Emergency Physicians addressed key issues in the evaluation and management of adult patients presenting to the emergency department with acute headache. A writing subcommittee conducted a systematic review of the literature to derive evidence-based recommendations to answer the following clinical questions: (1) In the adult emergency department patient presenting with acute headache, are there risk-stratification strategies that reliably identify the need for emergent neuroimaging? (2) In the adult emergency department patient treated for acute primary headache, are nonopioids preferred to opioid medications? (3) In the adult emergency department patient presenting with acute headache, does a normal noncontrast head computed tomography scan performed within 6 hours of headache onset preclude the need for further diagnostic workup for subarachnoid hemorrhage? (4) In the adult emergency department patient who is still considered to be at risk for subarachnoid hemorrhage after a negative noncontrast head computed tomography, is computed tomography angiography of the head as effective as lumbar puncture to safely rule out subarachnoid hemorrhage? Evidence was graded and recommendations were made based on the strength of the available data.
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Author Charles J. Gerardo Emergency Medicine
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Published In
Ann Emerg Med
DOI
10.1016/j.annemergmed.2019.07.009
EISSN
1097-6760
Publication Date
October 2019
Volume
74
Issue
4
Start / End Page
e41 / e74
Location
United States
Related Subject Headings
Subarachnoid Hemorrhage
Risk Factors
Male
Humans
Headache Disorders
Female
Facilities and Services Utilization
Evidence-Based Medicine
Emergency Service, Hospital
Emergency & Critical Care Medicine
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American College of Emergency Physicians Clinical Policies Subcommittee (Writing Committee) on Acute Headache:, Godwin, S. A., Cherkas, D. S., Panagos, P. D., Shih, R. D., Byyny, R., & Wolf, S. J. (2019). Clinical Policy: Critical Issues in the Evaluation and Management of Adult Patients Presenting to the Emergency Department With Acute Headache.Ann Emerg Med, 74(4), e41–e74.
American College of Emergency Physicians Clinical Policies Subcommittee (Writing Committee) on Acute Headache:, Steven A. Godwin, David S. Cherkas, Peter D. Panagos, Richard D. Shih, Richard Byyny, and Stephen J. Wolf. “Clinical Policy: Critical Issues in the Evaluation and Management of Adult Patients Presenting to the Emergency Department With Acute Headache.” Ann Emerg Med 74, no. 4 (October 2019): e41–74.
American College of Emergency Physicians Clinical Policies Subcommittee (Writing Committee) on Acute Headache:, Godwin SA, Cherkas DS, Panagos PD, Shih RD, Byyny R, et al. Clinical Policy: Critical Issues in the Evaluation and Management of Adult Patients Presenting to the Emergency Department With Acute Headache. Ann Emerg Med. 2019 Oct;74(4):e41–74.
American College of Emergency Physicians Clinical Policies Subcommittee (Writing Committee) on Acute Headache:, et al. “Clinical Policy: Critical Issues in the Evaluation and Management of Adult Patients Presenting to the Emergency Department With Acute Headache.” Ann Emerg Med, vol. 74, no. 4, Oct. 2019, pp. e41–74. Pubmed, doi:10.1016/j.annemergmed.2019.07.009.
American College of Emergency Physicians Clinical Policies Subcommittee (Writing Committee) on Acute Headache:, Godwin SA, Cherkas DS, Panagos PD, Shih RD, Byyny R, Wolf SJ. Clinical Policy: Critical Issues in the Evaluation and Management of Adult Patients Presenting to the Emergency Department With Acute Headache. Ann Emerg Med. 2019 Oct;74(4):e41–e74.
Published In
Ann Emerg Med
DOI
10.1016/j.annemergmed.2019.07.009
EISSN
1097-6760
Publication Date
October 2019
Volume
74
Issue
4
Start / End Page
e41 / e74
Location
United States
Related Subject Headings
Subarachnoid Hemorrhage
Risk Factors
Male
Humans
Headache Disorders
Female
Facilities and Services Utilization
Evidence-Based Medicine
Emergency Service, Hospital
Emergency & Critical Care Medicine
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8344 | https://stackoverflow.com/questions/13325821/comparing-two-fractions-and-friends | Skip to main content
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Comparing two fractions (< and friends)
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I have two fractions I like to compare. They are stored like this:
```
struct fraction {
int64_t numerator;
int64_t denominator;
};
```
Currently, I compare them like this:
```
bool fraction_le(struct fraction a, struct fraction b)
{
return a.numerator b.denominator < b.numerator a.denominator;
}
```
That works fine, except that (64 bit value) (64 bit value) = (128 bit value), which means it will overflow for numerators and denominators that are too far away from zero.
How can I make the comparison always works, even for absurd fractions?
Oh, and by the way: fractions are always stored simplified, and only the numerator can be negative. Maybe that input constraint makes some algorithm possible...
c
64-bit
fractions
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asked Nov 10, 2012 at 20:30
JasmijnJasmijn
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If you can't rely on a 128-bit type (or arbitrary precision), the continued fraction method that Boost uses (as per Kos' answer) is the best method.
– Daniel Fischer
Commented
Nov 10, 2012 at 20:55
Why not just comparing the fractional value? It may lose accuracy but better than overflow.
– SwiftMango
Commented
Nov 10, 2012 at 21:07
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Here's how Boost implements it. The code is well-commented.
```
template
bool rational::operator< (const rational& r) const
{
// Avoid repeated construction
int_type const zero( 0 );
// This should really be a class-wide invariant. The reason for these
// checks is that for 2's complement systems, INT_MIN has no corresponding
// positive, so negating it during normalization keeps it INT_MIN, which
// is bad for later calculations that assume a positive denominator.
BOOST_ASSERT( this->den > zero );
BOOST_ASSERT( r.den > zero );
// Determine relative order by expanding each value to its simple continued
// fraction representation using the Euclidian GCD algorithm.
struct { int_type n, d, q, r; } ts = { this->num, this->den, this->num /
this->den, this->num % this->den }, rs = { r.num, r.den, r.num / r.den,
r.num % r.den };
unsigned reverse = 0u;
// Normalize negative moduli by repeatedly adding the (positive) denominator
// and decrementing the quotient. Later cycles should have all positive
// values, so this only has to be done for the first cycle. (The rules of
// C++ require a nonnegative quotient & remainder for a nonnegative dividend
// & positive divisor.)
while ( ts.r < zero ) { ts.r += ts.d; --ts.q; }
while ( rs.r < zero ) { rs.r += rs.d; --rs.q; }
// Loop through and compare each variable's continued-fraction components
while ( true )
{
// The quotients of the current cycle are the continued-fraction
// components. Comparing two c.f. is comparing their sequences,
// stopping at the first difference.
if ( ts.q != rs.q )
{
// Since reciprocation changes the relative order of two variables,
// and c.f. use reciprocals, the less/greater-than test reverses
// after each index. (Start w/ non-reversed @ whole-number place.)
return reverse ? ts.q > rs.q : ts.q < rs.q;
}
// Prepare the next cycle
reverse ^= 1u;
if ( (ts.r == zero) || (rs.r == zero) )
{
// At least one variable's c.f. expansion has ended
break;
}
ts.n = ts.d; ts.d = ts.r;
ts.q = ts.n / ts.d; ts.r = ts.n % ts.d;
rs.n = rs.d; rs.d = rs.r;
rs.q = rs.n / rs.d; rs.r = rs.n % rs.d;
}
// Compare infinity-valued components for otherwise equal sequences
if ( ts.r == rs.r )
{
// Both remainders are zero, so the next (and subsequent) c.f.
// components for both sequences are infinity. Therefore, the sequences
// and their corresponding values are equal.
return false;
}
else
{
// Exactly one of the remainders is zero, so all following c.f.
// components of that variable are infinity, while the other variable
// has a finite next c.f. component. So that other variable has the
// lesser value (modulo the reversal flag!).
return ( ts.r != zero ) != static_cast<bool>( reverse );
}
}
```
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answered Nov 10, 2012 at 20:46
KosKos
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If you are using GCC, you can use __int128.
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answered Nov 10, 2012 at 20:50
HelloWorldHelloWorld
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I didn't understand the code in Kos's answer so this might be just duplicating it.
As other people have mentioned there are some easy special cases e.g. b/c > -e/f and -b/c > -e/f if e/f > b/c. So we are left with the case of positive fractions.
Convert these to mixed numbers i.e. a b/c and d e/f. The trivial case has a != d so we assume a == d. We then want to compare b/c with e/f with b < c, e < f. Well b/c > e/f if f/e > c/b. These are both greater than one so you can repeat the mixed number test until the whole number parts differ.
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answered Nov 10, 2012 at 21:00
NeilNeil
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Case intrigued me, so here is an implementation of Neil's answer, possibly with bugs :)
```
include
include
typedef struct {
int64_t num, den;
} frac;
int cmp(frac a, frac b) {
if (a.num < 0) {
if (b.num < 0) {
a.num = -a.num;
b.num = -b.num;
return !cmpUnsigned(a, b);
}
else return 1;
}
else if (0 <= b.num) return cmpUnsigned(a, b);
else return 0;
}
define swap(a, b) { int64_t c = a; a = b; b = c; }
int cmpUnsigned(frac a, frac b) {
int64_t c = a.num / a.den, d = b.num / b.den;
if (c != d) return c < d;
a.num = a.num % a.den;
swap(a.num, a.den);
b.num = b.num % b.den;
swap(b.num, b.den);
return !cmpUnsigned(a, b);
}
main() {
frac a = { INT64_MAX - 1, INT64_MAX }, b = { INT64_MAX - 3, INT64_MAX };
printf("%i\n", cmp(a, b));
}
```
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edited Feb 10, 2018 at 1:44
answered Nov 10, 2012 at 21:25
HelloWorldHelloWorld
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if a.num equals the minimum value of int64, the -a.num is negative rather than positive
– esse
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May 18, 2021 at 7:49
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Alright, so only your numerators are signed.
Special cases:
If the a.numerator is negative and the b.numerator is positive, then b is greater than a.
If the b.numerator is negative and the a.numerator is positive, then a is greater than b.
Otherwise:
Both your numerators have the same sign (+/-). Add some logic-code or bit manipulation to remove it, and use multiplication with uint64_t to compare them. Remember that if both numerators are negative, then the result of the comparison must be negated.
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answered Nov 10, 2012 at 20:41
HelloWorldHelloWorld
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Casting to uint64_t gives only a single extra bit. How does that solve the overflow problem?
– nemetroid
Commented
Nov 10, 2012 at 20:44
It helps a bit, though. If there is no better answer, I'll accept this.
– Jasmijn
Commented
Nov 10, 2012 at 20:46
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Why not just compare them directly as floating point numbers?
```
bool fraction_le(struct fraction a, struct fraction b)
{
return (double)a.numerator / a.denominator < (double)b.numerator / b.denominator;
}
```
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answered Feb 10, 2018 at 2:34
dbushdbush
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Because casting a long to a double may causes loss of precision. Try e.g. casting the maximum of a long to a double and back again. So comparing two fractions, both with potentially loss of precision, may cause erroneous output for some input values.
– HelloWorld
Commented
May 18, 2021 at 14:15
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Lesson 9: Angle addition identities
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Using the cosine double-angle identity
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Sal evaluates the cosine of the sum of 60° and another angle whose right triangle is given. To do this, he must use the cosine angle addition formula.Created by Sal Khan.
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Chapter 1 Linear Equations
Introduction to Linear Equations
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Chapter 2 Matrices
Introduction to Matrices
2.1 Systems of Equations
Solving a System by Graphing
Solving a System by Substitution
Solving a System by the Addition Method
Dependent and Inconsistent Systems
Systems with 3 Variables in 3 Unknowns
2.2 Solving Systems Using Matrices
Writing the Augmented Matrix of a System of Equations
Writing a System of Equations from an Augmented Matrix
Performing Row Operations on a Matrix
Solving a 3-by-3 System of Linear Equations Using Matrices
Number of Possible Solutions
2.3 Matrix Operations
Describing Matrices
Adding and Subtracting Matrices
Finding Scalar Multiples of a Matrix
Finding the Product of Two Matrices
2.4 Solving Systems with Inverses
Finding the Inverse of a Matrix
The Identity Matrix and Multiplicative Inverse
Solving a System of Linear Equations Using the Inverse of a Matrix
2.5 Matrix Calculator
Reduced Row-Echelon Form
Basic Operations
Inverses
Chapter 3 Linear Programming
Introduction to Linear Programming
3.1 Inequalities in One Variable
Rules for Solving Linear Inequalities
Compound Inequalities
3.2 Graph Linear Inequalities in Two Variables
Verify Solutions to an Inequality in Two Variables
Recognize the Relation between the Solutions of an Inequality and Its Graph
Graph Linear Inequalities in Two Variables
Solve Applications Using Linear Inequalities in Two Variables
3.3 Linear Programming
Fundamental Theorem of Linear Programming
Solving a Linear Programming Problem Graphically
Chapter 4 Finance
Introduction to Finance
4.1 Simple and Compound Interest
Simple One-Time Interest
Simple Interest over Time
APR – Annual Percentage Rate
Compound Interest
Rounding
Using your Calculator
Annual Percentage Yield
4.2 Annuities
Savings Annuities
4.3 Payout Annuities
Payout Annuities
4.4 Loans
Loans
Remaining Loan Balance
4.5 Multistage Finance Problems
Which Equation to Use?
Multistage Problems
4.6 TVM Solver Calculator
Examples from Section 4.2
Examples from Section 4.3
Examples from Section 4.4
Examples from Section 4.5
Chapter 5 Sets
Introduction to Sets
5.1 Basic Set Concepts
Sets and Ways to Represent Them
Computing the Cardinal Value of a Set
Equal versus Equivalent Sets
5.2 Subsets
Exponential Notation
Equivalent Subsets
5.3 Understanding Venn Diagrams
Interpreting Venn Diagrams
Creating Venn Diagrams
The Complement of a Set
5.4 Set Operations with Two Sets
The Intersection of Two Sets
The Union of Two Sets
Applying Concepts of “AND” and “OR” to Set Operations
Drawing Conclusions from a Venn Diagram with Two Sets
5.5 Set Operations with Three Sets
Venn Diagrams with Three Sets
Applying Set Operations to Three Sets
Proving Equality of Sets Using Venn Diagrams
Chapter 6 Probability
Introduction to Probability
6.1 Concepts of Probability
Events and Outcomes
Basic Probability
Certain and Impossible Events
Complementary Events
Odds
Probability of Two Independent Events
6.2 Conditional Probability and Bayes’ Theorem
Conditional Probability
Bayes’ Theorem
6.3 Counting
Basic Counting
Permutations
Combinations
Probability Using Permutations and Combinations
6.4 Expected Value
Expected Value
Chapter 7 Logic
Introduction to Logic
7.1 Statements and Quantifiers
Identifying Logical Statements
Representing Statements in Symbolic Form
Negating Statements
Expressing Statements with Quantifiers of All, Some, or None
Negating Statements Containing Quantifiers
7.2 Compound Statements
Common Logical Connectives
Translating Compound Statements to Symbolic Form
Translating Compound Statements in Symbolic Form with Parentheses into Words
The Dominance of Connectives
7.3 Constructing Truth Tables
Interpret and Apply Negations, Conjunctions, and Disjunctions
Construct Truth Tables to Analyze All Possible Outcomes
Determine the Validity of a Truth Table for a Compound Statement
7.4 Truth Tables for the Conditional and Biconditional
Use and Apply the Conditional to Construct a Truth Table
Use and Apply the Biconditional to Construct a Truth Table
7.5 Equivalent Statements
Determine Logical Equivalence
Compose the Converse, Inverse, and Contrapositive of a Conditional Statement
7.6 De Morgan's Laws
Negation of Conjunctions and Disjunctions
Negation of a Conditional Statement
Evaluating De Morgan’s Laws with Truth Tables
7.7 Logical Arguments
Law of Detachment
Law of Denying the Consequent
Chain Rule for Conditional Arguments
Chapter 8 Statistics
Introduction to Statistics
8.1 Gathering and Organizing Data
Sampling and Gathering Data
Organizing Data
8.2 Visualizing Data
Visualizing Categorical Data
Visualizing Quantitative Data
Misleading Graphs
8.3 Mean, Median, and Mode
The Mode
The Median
The Mean
Which Is Better: Mode, Median, or Mean?
8.4 Range and Standard Deviation
The Range
The Standard Deviation
8.5 Percentiles
Finding Percentiles
A Formula for Finding the[latex]k^{th}[/latex]Percentile
A Formula for Finding the Percentile of a Value in a Dataset
Interpreting Percentiles, Quartiles, and Median
8.6 The Normal Distribution
Moving toward Normality
The Normal Distribution
Properties of Normal Distributions: The 68-95-99.7 Rule
Standardized Scores
8.7 Scatter Plots, Correlation, and Regression Lines
Relationships between Quantitative Datasets
Creating Scatter Plots
Positive and Negative Linear Relationships
Linear Regression
Correlation Does Not Imply Causation
8.8 Statistics Calculator
Examples from Section 8.2
Examples and Exercises from Section 8.3
Examples from Section 8.4
Glossary
Matrix Calculator
TVM Solver Calculator
Statistics Calculator
Summary of Adaptations
Full Resource Adaptations
Chapter-Specific Changes
Finite Mathematics
Chapter 1 Linear Equations
1.3 Graph Linear Equations in Two Variables
Learning Objectives
By the end of this section, you will be able to:
Plot points in a rectangular coordinate system
Graph a linear equation by plotting points
Graph vertical and horizontal lines
Find the x– and y-intercepts
Graph a line using the intercepts
Plot Points on a Rectangular Coordinate System
Just like maps use a grid system to identify locations, a grid system is used in mathematics to locate points in a plane. This grid system is called the rectangular coordinate system, or the Cartesian coordinate system, which is named after René Descartes, the French mathematician who is credited with developing this grid system in the seventeenth century.
The rectangular coordinate system is formed by two intersecting number lines, one horizontal and one vertical. The horizontal number line is called thex-axis. The vertical number line is called they-axis. These axes divide a plane into four regions, calledquadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise.
Figure 1. Rectangular coordinate system
In the rectangular coordinate system, every point is represented by anordered pair. The first number in the ordered pair is thex-coordinate of the point, and the second number is they-coordinate of the point. The phrase “ordered pair” means that the order is important.
Ordered Pair
Anordered pairlatex[/latex]gives the coordinates of a point in a rectangular coordinate system. The first number is the x-coordinate. The second number is the y-coordinate.
Figure 2. Ordered pair
What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is latex[/latex]. The point latex[/latex] has a special name. It is called theorigin.
The point latex[/latex]is called theorigin. It is the point where the x-axis and y-axis intersect.
We use the coordinates to locate a point on the xy-plane. Let’s plot the point latex[/latex]as an example. First, locate 1 on the x-axis and lightly sketch a vertical line through [latex]x=1[/latex]. Then, locate 3 on the y-axis and sketch a horizontal line through [latex]y=3[/latex]. Now, find the point where these two lines meet—that is the point with coordinates latex[/latex].
Figure 3. (1,3) plotted in the coordinate system
Notice that the vertical line through [latex]x=1[/latex] and the horizontal line through [latex]y=3[/latex] are not part of the graph. We just used them to help us locate the point latex[/latex].
When one of the coordinates is zero, the point lies on one of the axes. In the figure below, the point latex[/latex] is on the y-axis and the point latex[/latex]is on the x-axis.
Figure 4. (0, 4) and (–2, 0) plotted in the coordinate system
Points on the Axes
Points with a y-coordinate equal to 0 are on the x-axis and have coordinates latex[/latex].
Points with an x-coordinate equal to 0 are on the y-axis and have coordinates latex[/latex].
Example 1
Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located.
(a)latex[/latex]
The first number of the coordinate pair is the x-coordinate, and the second number is the y-coordinate. To plot each point, sketch a vertical line through the x-coordinate and a horizontal line through the y-coordinate. Their intersection is the point.
Since [latex]x=-5[/latex], the point is to the left of the y-axis. Also, since [latex]y=4[/latex], the point is above the x-axis. The point latex[/latex]is in Quadrant II.
The graph that shows the plotted point is at the bottom of this example.
(b)latex[/latex]
Since [latex]x=-3[/latex], the point is to the left of the y-axis. Also, since [latex]y=-4[/latex], the point is below the x-axis. The point latex[/latex]is in Quadrant III.
The graph that shows the plotted point is at the bottom of this example.
(c)latex[/latex]
Since [latex]x=2[/latex], the point is to the right of the y-axis. Since [latex]y=-3[/latex], the point is below the x-axis. The point latex[/latex]is in Quadrant IV.
The graph that shows the plotted point is at the bottom of this example.
(d)latex[/latex]
Since [latex]x=-2[/latex], the point is to the left of the y-axis. Since [latex]y=3[/latex], the point is above the x-axis. The point latex[/latex] is in Quadrant II.
The graph that shows the plotted point is at the bottom of this example.
(e)latex[/latex]
Since [latex]x=3[/latex], the point is to the right of the y-axis. Since [latex]y=\frac{5}{2}[/latex], the point is above the x-axis. (It may be helpful to write [latex]\frac{5}{2}[/latex]as a mixed number or decimal.) The point latex[/latex]is in Quadrant I.
Figure 5. Points plotted in the coordinate system
Exercise 1
Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
a) latex[/latex]
b) latex[/latex]
c) latex[/latex]
d) latex[/latex]
e) latex[/latex]
Solution
Figure 6. Points plotted in the coordinate system
Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
a) latex[/latex]
b) latex[/latex]
c) latex[/latex]
d) latex[/latex]
e) latex[/latex]
Solution
Figure 7. Points plotted in the coordinate system
The signs of the x-coordinate and y-coordinate affect the location of the points. You may have noticed some patterns as you graphed the points in the previous example. We can summarize sign patterns of the quadrants in this way:
Quadrants
Quadrant I Quadrant II Quadrant III Quadrant IV
latex[/latex]latex[/latex]latex[/latex]latex[/latex]
latex[/latex]latex[/latex]latex[/latex]latex[/latex]
Figure 8. The signs of the coordinate system
Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation, you got exactly one solution. But equations can have more than one variable. Equations with two variables may be of the form [latex]Ax+By=C[/latex]. An equation of this form is called a linear equation in two variables.
Linear Equation
An equation of the form [latex]Ax+By=C[/latex]where A and B are not both zero is called alinear equation in two variables.
Here is an example of a linear equation in two variables,x and y.
[latex]\begin{eqnarray} Ax+By &=& C\ x+4y &=& 8\ A=1, B &=& 4, C=8\ \end{eqnarray}[/latex]
The equation [latex]y=-3x+5[/latex] is also a linear equation. But it does not appear to be in the form [latex]Ax+By=C[/latex]. We can use the Addition Property of Equality and rewrite it in [latex]Ax+By=C[/latex] form.
[latex]\begin{array}{lrcl} & y & = & -3x+5\ \text{Add } 3x \text{ to both sides.} & y{+3x} & = & -3x+5{+3x} \ \text{Simplify.} & y+3x & = & 5\ \text{Use the Commutative Property to put it in }\ Ax+By=C \text{ form.} & 3x+y & = & 5\ \end{array}[/latex]
By rewriting [latex]y=-3x+5[/latex] as [latex]3x+y=5[/latex], we can easily see that it is a linear equation in two variables because it is of the form [latex]Ax+By=C[/latex]. When an equation is in the form [latex]Ax+By=C[/latex], we say it is in the standard form of a linear equation.
Standard Form of Linear Equation
A linear equation is instandard form when it is written [latex]Ax+By=C[/latex].
Most people prefer to have A,B, and C be integers and [latex]A\geq0[/latex]when writing a linear equation in standard form, although it is not strictly necessary.
Linear equations have infinitely many solutions. For every number that is substituted for x,there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair latex[/latex].When we substitute these values of x and y into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.
Solution of a Linear Equation in Two Variables
An ordered pair latex[/latex]is asolution of the linear equation [latex]Ax+By=C[/latex]if the equation is a true statement when the x– and y-values of the ordered pair are substituted into the equation.
Linear equations have infinitely many solutions. We can plot these solutions in the rectangular coordinate system. The points will line up perfectly in a straight line. We connect the points with a straight line to get the graph of the equation. We put arrows on the ends of each side of the line to indicate that the line continues in both directions.
A graph is a visual representation of all the solutions of the equation. It is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to that equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation. Points not on the line are not solutions!
Graph of a Linear Equation
The graph of a linear equation [latex]Ax+By=C[/latex] is a straight line.
Every point on the line is a solution of the equation.
Every solution of this equation is a point on this line.
Example 2
The graph of[latex]y=2x-3[/latex]is shown.
Figure 9. The line y = 2x – 3
For each ordered pair, decide:
a)Is the ordered pair a solution to the equation?
b)Is the point on the line?
A: latex[/latex] B:latex[/latex] C: latex[/latex] D: latex[/latex]
Substitute the x– and y-values into the equation to check if the ordered pair is a solution to the equation.
a)
A: latex[/latex]B: latex[/latex]C: latex[/latex]D: latex[/latex]
[latex]\begin{eqnarray} y &=& 2x-3\ {-3} &\stackrel{?}{=}& 2({0})-3\ -3 &=& -3\ \end{eqnarray}[/latex][latex]\begin{eqnarray} y &=& 2x-3\ {3} &\stackrel{?}{=}& 2({3})-3\ 3 &=& 3\ \end{eqnarray}[/latex][latex]\begin{eqnarray} y &=& 2x-3\ {-3} &\stackrel{?}{=}& 2({2})-3\ -3 &\neq& -1\ \end{eqnarray}[/latex][latex]\begin{eqnarray} y &=& 2x-3\ {-5} &\stackrel{?}{=}& 2({-1})-3\ -5 &=& -5\ \end{eqnarray}[/latex]
latex[/latex] is a solution.latex[/latex] is a solution.latex[/latex] is not a solution.latex[/latex] is a solution.
b) Plot the points latex[/latex], latex[/latex], latex[/latex], and latex[/latex].
Figure 10. The line y = 2 x – 3
The points latex[/latex], latex[/latex], and latex[/latex] are on the line [latex]y=2x-3[/latex], and the point latex[/latex] is not on the line.
The points that are solutions to [latex]y=2x-3[/latex] are on the line, but the point that is not a solution is not on the line.
Exercise 2
Use the graph of [latex]y=3x-1[/latex]. For each ordered pair, decide:
a)Is the ordered pair a solution to the equation?
b)Is the point on the line?
A: latex[/latex] B: latex[/latex]
Figure 11. The line y = 3 x – 1
Solution
a) yes, yes
b) yes, yes
Use the graph of [latex]y=3x-1[/latex]. For each ordered pair, decide:
a) Is the ordered pair a solution to the equation?
b) Is the point on the line?
A: latex[/latex]B: latex[/latex]
Figure 12. The line y = 3 x – 1
Solution
a) no, no
b) yes, yes
Graph a Linear Equation by Plotting Points
There are several methods that can be used to graph a linear equation. The first method we will use is called plotting points, or the Point-Plotting Method. We find three points whose coordinates are solutions to the equation and then plot them in a rectangular coordinate system. By connecting these points in a line, we have the graph of the linear equation.
Example 3
How to Graph a Linear Equation by Plotting Points
Graph the equation [latex]y=2x+1[/latex]by plotting points.
Figure 13. Step 1
Figure 14. Step 2
Figure 15. Step 3
Exercise 3
Graph the equation by plotting points: [latex]y=2x-3[/latex]. Check your solutions to ensure they are on the line in the graph below.
Solution
Figure 16. The line y = 2 x – 3
Graph the equation by plotting points: [latex]y=-2x+4[/latex]. Check your solutions to ensure they are on the line in the graph below.
Solution
Figure 17. The line y = –2 x + 4
The steps to take when graphing a linear equation by plotting points are summarized here.
Graph a linear equation by plotting points.
Find three points whose coordinates are solutions to the equation. Organize them in a table.
Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.
It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you only plot two points and one of them is incorrect, you can still draw a line, but it will not represent the solutions to the equation. It will be the wrong line.
If you use three points and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. Look at the difference between these illustrations.
Figure 18. (a) Three points make up a line. (b) The points are not in a straight line.
When an equation includes a fraction as the coefficient of x, we can still substitute any numbers for x. The arithmetic is easier if we make “good” choices for the values of x. Zero is a good choice for the value of x, and multiples of the denominator are good choices, because they will eliminate the fraction. This will avoid fractional answers, which are hard to graph precisely.
Example 4
Graph the equation: [latex]y=\frac{1}{2}x+3[/latex].
Find three points that are solutions to the equation. Since this equation has the fraction [latex]\frac{1}{2}[/latex]as a coefficient of x, we will choose values of x carefully. We will use 0 as one choice and multiples of 2 for the other choices. Why are multiples of 2 a good choice for values of x? By choosing multiples of 2, the multiplication by [latex]\frac{1}{2}[/latex] simplifies to a whole number.
[latex]\begin{eqnarray} x &=& 0\ y &=& \frac{1}{2}x+3\ y &=& \frac{1}{2}({0})+3\ y &=& 0+3\ y &=& 3\ \end{eqnarray}[/latex][latex]\begin{eqnarray} x &=& 2\ y &=& \frac{1}{2}x+3\ y &=& \frac{1}{2}({2})+3\ y &=& 1+3\ y &=& 4\ \end{eqnarray}[/latex][latex]\begin{eqnarray} x &=& 4\ y &=& \frac{1}{2}x+3\ y &=& \frac{1}{2}({4})+3\ y &=& 2+3\ y &=& 5\ \end{eqnarray}[/latex]
The points are shown in the figure below.
[latex]y=\frac{1}{2}x+3[/latex]
[latex]x[/latex][latex]y[/latex]latex[/latex]
[latex]0[/latex][latex]3[/latex]latex[/latex]
[latex]2[/latex][latex]4[/latex]latex[/latex]
[latex]4[/latex][latex]5[/latex]latex[/latex]
Plot the points, check that they line up, and draw the line.
Figure 19. The plotted points in a line
Exercise 4
Graph the equation: [latex]y=\frac{1}{3}x-1[/latex].
Solution
Figure 20. Graph of the line
Graph the equation: [latex]y=\frac{1}{4}x+2[/latex].
Solution
Figure 21. Graph of the line
Graph Vertical and Horizontal Lines
Some linear equations have only one variable. They may have just x and no y or just y without an x. This changes how we make a table of values to get the points to plot.
Let’s consider the equation [latex]x=-3[/latex]. This equation has only one variable, x. The equation says that x is always equal to [latex]-3[/latex], so its value does not depend on y. Regardless of the value of y, the value of x is always [latex]-3[/latex].
To make a table of values, write [latex]-3[/latex] in for all the x-values, then choose any values for y. We will use 1, 2, and 3 for the y-coordinates.
[latex]x=-3[/latex]
[latex]x[/latex][latex]y[/latex]latex[/latex]
[latex]-3[/latex][latex]1[/latex]latex[/latex]
[latex]-3[/latex][latex]2[/latex]latex[/latex]
[latex]-3[/latex][latex]3[/latex]latex[/latex]
Plot the points from the table and connect them with a straight line. Notice that we have graphed a vertical line passing through the x-axis at [latex]-3[/latex].
Figure 22. Vertical line
What if the equation has y, but no x? Let’s graph the equation [latex]y=4[/latex]. This time the y- value is a constant, so in this equation,y does not depend on x. Fill in 4 for all the y-values in the table, then choose any values for x. We will use 0, 2, and 4 for the x-coordinates.
[latex]y=4[/latex]
[latex]x[/latex][latex]y[/latex]latex[/latex]
[latex]0[/latex][latex]4[/latex]latex[/latex]
[latex]2[/latex][latex]4[/latex]latex[/latex]
[latex]4[/latex][latex]4[/latex]latex[/latex]
In this figure, we have graphed a horizontal line passing through the y-axis at 4.
Figure 23. Horizontal line
Vertical and Horizontal Lines
Avertical line is the graph of an equation of the form [latex]x=a[/latex].
The line passes through the x-axis at latex[/latex].
Ahorizontal line is the graph of an equation of the form [latex]y=b[/latex].
The line passes through the y-axis at latex[/latex].
Example 5
Graph the following lines.
a) [latex]x=2[/latex]
The equation has only one variable,x, and x is always equal to 2. We create a table where x is always 2 and then put in any values for y. The graph is a vertical line passing through the x-axis at 2.
[latex]x=2[/latex]
xylatex[/latex]
2 1latex[/latex]
2 2latex[/latex]
2 3latex[/latex]
Figure 24. The line x = 2
b) [latex]y=-1[/latex]
Similarly, the equation [latex]y=-1[/latex] has only one variable,y. The value of y is constant. All the ordered pairs in the next table have the same y-coordinate. The graph is a horizontal line passing through the y-axis at [latex]-1[/latex].
[latex]y=-1[/latex]
xylatex[/latex]
[latex]0[/latex][latex]-1[/latex]latex[/latex]
[latex]3[/latex][latex]-1[/latex]latex[/latex]
[latex]-3[/latex][latex]-1[/latex]latex[/latex]
Figure 25. The line y = –1
Exercise 5
Graph the equations:
a) [latex]x=5[/latex]
b) [latex]y=-4[/latex]
Solution
a)
Figure 26. The line x = 5
b)
Figure 27. The line y = –4
Graph the equations:
a) [latex]x=-2[/latex]
b) [latex]y=3[/latex]
Solution
a)
Figure 28. The line x = –1
b)
Figure 29. The line y = 3
What is the difference between the equations [latex]y=4x[/latex] and [latex]y=4[/latex]?
The equation [latex]y=4x[/latex]has both x and y. The value of y depends on the value of x, so the y-coordinate changes according to the value of x. The equation [latex]y=4[/latex]has only one variable. The value of y is constant, it does not depend on the value of x, so the y-coordinate is always 4.
Figure 30. Tables indicating coordinates for y = 4 x and y = 4
Figure 31. Graph of y = 4 x and y = 4
Notice, in the graph, the equation [latex]y=4x[/latex] gives a slanted line, while [latex]y=4[/latex]gives a horizontal line.
Example 6
Graph [latex]y=-3x[/latex] and [latex]y=-3[/latex]in the same rectangular coordinate system.
We notice that the first equation has the variable x, while the second does not. We make a table of points for each equation and then graph the lines. The two graphs are shown.
Figure 32. Table indicating coordinates for y = –3 x and y = –3
Figure 33. Graphs of y = –3 x and y = –3
Exercise 6
Graph the equations in the same rectangular coordinate system: [latex]y=-4x[/latex] and [latex]y=-4[/latex].
Solution
Figure 34. Graph of y = –4 x and y = –4
Graph the equations in the same rectangular coordinate system: [latex]y=3[/latex] and [latex]y=3x[/latex]
Solution
Figure 35. Graph of y = 3 and y= 3 x
Find x– and y-intercepts
Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.
At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x-axis and the y-axis. These points are called the intercepts of a line.
Intercepts of a Line
The points where a line crosses the x-axis and the y-axis are called theintercepts of the line.
Let’s look at the graphs of the lines.
Figure 36.Intercepts of lines
Let’s look at the points where these lines cross the x-axis and the y-axis.
FigureLine crosses x-axis at this pointLine crosses y-axis at this point
alatex[/latex]latex[/latex]
blatex[/latex]latex[/latex]
clatex[/latex]latex[/latex]
dlatex[/latex]latex[/latex]
Do you see a pattern? See the figure below.
For each line, the y-coordinate of the point where the line crosses the x-axis is zero. The point where the line crosses the x-axis has the form latex[/latex]and is called the x-intercept of the line. The x-intercept occurs when y is zero.
In each line, the x- coordinate of the point where the line crosses the y-axis is zero. The point where the line crosses the y-axis has the form latex[/latex]and is called the y-intercept of the line. The y-intercept occurs when x is zero.
The x-intercept is the point latex[/latex]where the line crosses the x-axis.
The y-intercept is the point latex[/latex]where the line crosses the y-axis.
Figure 37. Table indicating x– and y-intercepts
Example 7
Find the x– and y-intercepts on each graph shown.
Figure 38. Three graphs
(a)The graph crosses the x-axis at the point latex[/latex].The x- intercept is latex[/latex].
The graph crosses the y-axis at the point latex[/latex].The y-intercept is latex[/latex].
(b)The graph crosses the x-axis at the point latex[/latex].The x-intercept is latex[/latex].
The graph crosses the y-axis at the point latex[/latex].The y-intercept is latex[/latex].
(c)The graph crosses the x-axis at the point latex[/latex].The x-intercept is latex[/latex].
The graph crosses the y-axis at the point latex[/latex].The y-intercept is latex[/latex].
Exercise 7
Recognizing that the x-intercept occurs when y is zero and that the y-intercept occurs when x is zero gives us a method to find the intercepts of a line from its equation. To find the x-intercept, let [latex]y=0[/latex]and solve for x. To find the y-intercept, let [latex]x=0[/latex]and solve for y.
Find the x– and y-intercepts from the Equation of a Line
Use the equation of the line to find:
the x-intercept of the line, let [latex]y=0[/latex]and solve for x.
the y-intercept of the line, let [latex]x=0[/latex]and solve for y.
Example 8
Find the intercepts of [latex]2x+y=8[/latex].
We will let [latex]y=0[/latex]to find the x-intercept and let [latex]x=0[/latex] to find the y-intercept. Complete the table below.
Figure 39. Table of x– and y-intercepts
[latex]\begin{array}{lrcl} \text{To find the x-intercept, let } y=0.\ \text {Let } y=0. & 2x+0 & = & 8\ \text{Simplify.} & 2x & = & 8\ & x & = & 4\ \text{The x-intercept is } (4,0). \end{array}[/latex]
[latex]\begin{array}{lrcl} \text{To find the y-intercept, let } x=0.\ \text {Let } x=0. & 2(0)+y & = & 8\ \text{Simplify.} & 0+y & = & 8\ & y & = & 8\ \text{The y-intercept is } (0,8). \end{array}[/latex]
The intercepts are the points latex[/latex] and latex[/latex]as shown in the table.
[latex]2x+y=8[/latex]
[latex]x[/latex][latex]y[/latex]
[latex]4[/latex][latex]0[/latex]
[latex]0[/latex][latex]8[/latex]
Exercise 8
Graph a Line Using the Intercepts
To graph a linear equation by plotting points, you need to find three points whose coordinates are solutions to the equation. You can use the x- and y- intercepts as two of your three points. Find the intercepts and then find a third point to ensure accuracy. Make sure the points line up—then draw the line. This method is often the quickest way to graph a line.
Example 9
How to Graph a Line Using the Intercepts
Graph [latex]-x+2y=6[/latex]using the intercepts.
Figure 40. Step 1
Figure 41. Step 2
Figure 42. Step 3
Figure 43. Step 4
Exercise 9
Graph using the intercepts: [latex]x-2y=4[/latex].
Solution
Figure 44. The line x – 2 y= 4
Graph using the intercepts: [latex]-x+3y=6[/latex].
Solution
Figure 45. The line –x + 3 y = 6
The steps to graph a linear equation using the intercepts are summarized here:
Graph a linear equation using the intercepts.
Find the x– and y-intercepts of the line.
Let [latex]y=0[/latex]and solve for x.
Let [latex]x=0[/latex]and solve for y.
Find a third solution to the equation.
Plot the three points and check that they line up.
Draw the line.
Example 10
Graph [latex]4x-3y=12[/latex]using the intercepts.
Find the intercepts and a third point.
x-intercept, let y=0 y-intercept, let x=0 third point, let y=4
[latex]\begin{eqnarray} 4x-3y &=& 12\ 4x-3({0}) &=& 12\ 4x &=& 12\ x &=& 3\ \end{eqnarray}[/latex][latex]\begin{eqnarray} 4x-3y &=& 12\ 4({0})-3y &=& 12\ -3y &=& 12\ y &=& -4\ \end{eqnarray}[/latex][latex]\begin{eqnarray} 4x-3y &=& 12\ 4x-3({4}) &=& 12\ 4x-12 &=& 12\ 4x &=& 24\ x &=& 6\ \end{eqnarray}[/latex]
We list the points in the table and show the graph.
[latex]4x-3y=12[/latex]
[latex]x[/latex][latex]y[/latex]latex[/latex]
[latex]3[/latex][latex]0[/latex]latex[/latex]
[latex]0[/latex][latex]-4[/latex]latex[/latex]
[latex]6[/latex][latex]4[/latex]latex[/latex]
Figure 46. The line 4 x – 3 y = 12
Exercise 10
Graph using the intercepts: [latex]5x-2y=10[/latex].
Solution
Figure 47. The line 5 x – 2 y= 10
Graph using the intercepts: [latex]3x-4y=12[/latex].
Solution
Figure 48. The line 3 x – 4 y= 12
When the line passes through the origin, the x-intercept and the y-intercept are the same point.
Example 11
Graph [latex]y=5x[/latex]using the intercepts.
x-intercept, let [latex]y=0[/latex].y-intercept, let [latex]x=0[/latex].
[latex]\begin{eqnarray} y &=& 5x\ {0} &=& 5x\ 0 &=& x\ \end{eqnarray}[/latex][latex]\begin{eqnarray} y &=& 5x\ y &=& 5\cdot{0}\ y &=& 0\ \end{eqnarray}[/latex]
latex[/latex]latex[/latex]
This line has only one intercept. It is the point latex[/latex].
To ensure accuracy, we need to plot three points. Since the x– and y-intercepts are the same point, we need two more points to graph the line.
Let [latex]x=1[/latex].Let [latex]x=-1[/latex].
[latex]\begin{eqnarray} y &=& 5x\ y &=& 5\cdot{1}\ y &=& 5\ \end{eqnarray}[/latex][latex]\begin{eqnarray} y &=& 5x\ y &=& 5{-1}\ y &=& -5\ \end{eqnarray}[/latex]
The resulting three points are summarized in the table.
[latex]y=5x[/latex]
[latex]x[/latex][latex]y[/latex]latex[/latex]
[latex]0[/latex][latex]0[/latex]latex[/latex]
[latex]1[/latex][latex]5[/latex]latex[/latex]
[latex]-1[/latex][latex]-5[/latex]latex[/latex]
Plot the three points, check that they line up, and draw the line.
Figure 49. The line y = 5 x
Exercise 11
Graph using the intercepts: [latex]y=4x[/latex].
Solution
Figure 50. The line y= 4 x
Graph the intercepts: [latex]y=-x[/latex].
Solution
Figure 51. The line y= –x
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definition
A grid system used in algebra to show a relationship between two variables and formed by intersecting one horizontal line, called the x-axis, with a vertical line, called the y-axis.
×Close definition
The horizontal number line that intersects with a vertical number line to form the rectangular coordinate system.
×Close definition
The vertical number line that intersects with a horizontal number line to form the rectangular coordinate system.
×Close definition
The four regions formed by the intersecting x- and y-axes.
×Close definition
An ordered pair latex[/latex] gives the coordinates of a point in a rectangular coordinate system. The first number is the x-coordinate. The second number is the y-coordinate.
×Close definition
The first number in an ordered pair.
×Close definition
The second number in an ordered pair.
×Close definition
The point latex[/latex]is called the origin. It is the point where the x-axis and y-axis intersect.
×Close definition
A vertical line is the graph of an equation of the form [latex]x=a[/latex]. The line passes through the x-axis at latex[/latex].
×Close definition
A horizontal line is the graph of an equation of the form [latex]y=b[/latex]. The line passes through the y-axis at latex[/latex].
×Close definition
An equation of the form [latex]Ax+By=C[/latex] where A and B are not both zero is called a linear equation in two variables.
×Close definition
A linear equation is in standard form when it is written [latex]Ax+By=C[/latex].
×Close definition
An ordered pair latex[/latex]is a solution of the linear equation [latex]Ax+By=C[/latex] if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.
×Close definition
A solution is a value of a variable that makes a true statement when substituted into the equation.
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A visual representation of all the solutions of the equation.
×Close definition
The points where a line crosses the x-axis and the y-axis are called the intercepts of the line.
×Close definition
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8348 | https://www.youtube.com/watch?v=qeDrA7i7rl8 | Proportional Tax Multiplier numerical example #taxmultiplier #taxes
ECON MATHS
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Posted: 2 Dec 2022
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Transcript:
let us solve your question on proportionate text multiply the question is what is the value of proportionate tax multiplier where marginal propensity to consume is 0.8 and proportionate X is twenty percent of the income okay so what we have been given we have been given the marginal propensity to consume is 0.8 and tax rate is being given as uh 20 which can be written as 0.2 okay the formula for proportionate X multiplier is that's denoted by m which is equal to 1 upon 1 minus the MPC times 1 minus the tax rate okay let's solve it so we have 1 upon 1 minus and the value of C which is MPC which is 0.8 0.8 uh times we have 1 minus the tax rate is being given us as 0.2 0.2 so which reduces uh to 1 upon 1 minus 0.8 times 1 minus 0.2 is again 0.8 uh which comes out to be let's write it here which is 1 upon 1 minus now point eight into point eight is uh 0.64 0.64 which is equal to 1 upon 1 minus 0.64 it comes out to be zero point uh three seconds so 1 upon 0.368 will come out to be 2.7 that means our proportional tax multiplier will be equal to 2.7 I hope I make myself clear thank you |
8349 | https://flexbooks.ck12.org/cbook/ck-12-basic-geometry-concepts/section/3.1/primary/lesson/parallel-and-skew-lines-bsc-geom/ | Parallel and Skew Lines | CK-12 Foundation
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3.1
Parallel and Skew Lines
FlexBooks 2.0>
CK-12 Basic Geometry Concepts>
Parallel and Skew Lines
Written by:Dan Greenberg |Lori Jordan | +4 more
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Parallel and Skew Lines
Parallel lines are two or more lines that lie in the same plane and never intersect. To show that lines are parallel, arrows are used.
[Figure 1]
| Label It | Say It |
--- |
| A B↔||M N↔ | Line A B is parallel to line M N |
| l||m | Line l is parallel to line m. |
In the definition of parallel the word “line” is used. However, line segments, rays and planes can also be parallel. The image below shows two parallel planes, with a third blue plane that is perpendicular to both of them.
[Figure 2]
Skew lines are lines that are in different planes and never intersect. They are different from parallel lines because parallel lines lie in the SAME plane. In the cube below, A B¯ and F H¯ are skew and A C¯ and E F¯ are skew.
[Figure 3]
Basic Facts About Parallel Lines
Property: If lines l||m and m||n, then l||n.
If then
Postulate: For any line and a pointnot on the line, there is one line parallel to this line through the point. There are infinitely many lines that go through A, but only one that is parallel to l.
A transversal is a line that intersects two other lines. The area between l and m is the interior. The area outside l and m is the exterior.
What if you were given a pair of lines that never intersect and were asked to describe them? What terminology would you use?
Examples
Use the figure below for Examples 1 and 2. The two pentagons are parallel and all of the rectangular sides are perpendicular to both of them.
Example 1
Find two pairs of skew lines.
Z V¯and W B¯.Y D¯and V W¯
Example 2
For X Y¯, how many parallel lines would pass through point D? Name this/these line(s).
One line, C D¯
Example 3
True or false: some pairs of skew lines are also parallel.
This is false, by definition skew lines are in different planes and parallel lines are in the same plane. Two lines could be skew or parallel (or neither), but never both.
Example 4
Using the cube below, list a pair of parallel lines.
[Figure 9]
One possible answer is lines A B¯ and E F¯.
Example 5
Using the cube below, list a pair of skew lines.
[Figure 10]
One possible answer is B D¯ and C G¯.
Review
Which of the following is the best example of parallel lines?
Railroad Tracks
Lamp Post and a Sidewalk
Longitude on a Globe
Stonehenge (the stone structure in Scotland)
Which of the following is the best example of skew lines?
Roof of a Home
Northbound Freeway and an Eastbound Overpass
Longitude on a Globe
The Golden Gate Bridge
Use the picture below for questions 3-5.
If m∠2=55∘, what other angles do you know?
If m∠5=123∘, what other angles do you know?
Is l||m? Why or why not?
For 6-10, determine whether the statement is true or false.
If p||q and q||r, then p||r.
Skew lines are never in the same plane.
Skew lines can be perpendicular.
Planes can be parallel.
Parallel lines are never in the same plane.
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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| | [Figure 1] | License:CC BY-NC 3.0 |
| | [Figure 2] | License:CC BY-NC 3.0 |
| | [Figure 3] | License:CC BY-NC 3.0 |
| | [Figure 9] | License:CC BY-NC 3.0 |
| | [Figure 10] | License:CC BY-NC 3.0; CC BY-NC-SA |
| | [Figure 12] [Figure 16] | License:CC BY-NC |
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| | [Figure 14] | License:CC BY-NC |
| | [Figure 15] | License:CC BY-NC |
| | [Figure 17] | Credit:Melissa Sanders |
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| | [Figure 19] | Credit:Steve Snodgrass Source: |
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8350 | https://www.quora.com/What-is-the-molar-mass-of-Na2SO4-Na-23-0-S-32-0-and-O-16-0 | Something went wrong. Wait a moment and try again.
Sodium Sulfate
Inorganic Compounds
Chemical Formulas
Chemical Elements
Molar Mass (chemistry)
5
What is the molar mass of Na2SO4 (Na=23.0, S=32.0 and O=16.0)?
Sharafat Jutt
The atomic mass of Na is 23.
So there are 2 atoms of sodium. So 23 multiply 2 =46
Atomic mass of S is 32
Atomic mass of O is 16 multiply 4 =64.
Now add all 46+32+64=142 is molecular mass of this compound.
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Related questions
Given that the molar mass of Na2SO4.nH2O is 322.1g/mol, what is the value of n?
Is the molar mass of O2 16 or 32?
What is the molar mass of RaCl2?
What is the molar mass of CuSO4•5H2O(S)?
How can we prepare 0.5 N HCl in 100 ml water?
Neeti Shree
Lives in Accra, Ghana
·
7y
Originally Answered: What is the molecular mass of Na2SO4?
·
Mass of Na = 23
Mass of S = 32
Mass of O = 16
Therefore, molecular mass of Na2SO4
=(23 2)+32+(16 4)
= 46+32+64
=142
Somya Raj Sinha
B.tech from Kalinga Institute of Industrial Technology (KIIT) (Graduated 2020)
·
7y
Originally Answered: What is the molecular mass of Na2SO4?
·
Na2So4 is also called sodium sulphate(Na2+ and So4 2-). The molecular weight of sodium sulphate is approx 142 gm/mol.
Mass of Na-22.98
Mass of S-32
Mass of O-15.99
So the total mass can be calculated as-: 22.98×2+32+15.99×4=142 gm.
Alexander Budianto
Lives in Indonesia (2002–present)
·
Author has 2.1K answers and 3.3M answer views
·
Updated 6y
Originally Answered: What is the molecular mass of Na2SO4?
·
Na2SO4, also called sodium sulfate, consists of 2 sodium atoms, 1 sulfur atom, and 4 oxygen atoms, which can be clearly determined from the chemical formula itself.
I’m going to round up the atomic mass of each elements, so don’t blame me for not getting it right by 0.somethingsomethingsomething.
The atomic mass for Na is 23.
The atomic mass for S is 32.
The atomic mass for O is 16.
Therefore, the molecular mass for Na2SO4 is (23×2)+32+(16×4)=142
Related questions
What is the molar mass of (NH4) 2Fe(SO4) 26H2O?
What is the molar mass of As2S4?
What is the molar mass of Al2O32H2O?
What is the molar mass of H2SO4 98gm or 98amu?
What is the mass of one mole of copper(II) sulfate (CuSO4)?
Lily Das
Studied at Buxi Jagabandhu English Medium School, Bhubaneswar (Graduated 2024)
·
3y
Originally Answered: What is the molecular mass of Na2SO4?
·
Na2SO4 (Sodium Sulphate)
Atomic mass of Na(Sodium)=23u/amu
Atomic mass of S(Sulphur)=32u/amu
Atomic mass of O(Oxygen)=16u/amu
(23×2)+(32)+(16×4)= 46+32+64=142u/amu.
The unit in which molecular mass is measured is amu/u, not g.
:)
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Himanshu Jaiswal
Class 12th in Science & Biology, Delhi Public School (Graduated 2017)
·
7y
Originally Answered: What is the molecular mass of Na2SO4?
·
Na - 22.98
S - 32
O - 16
For Na2SO4 - 22.982+32+164 = 141.96
Vaishnavi
Studied at St. Helena's School (Graduated 2019)
·
6y
Originally Answered: What is the molecular mass of Na2SO4?
·
Molecular mass of Na2SO4 =
(232) +32+(164) = 142 grams.
Thus, molecular mass of Na2SO4 is 142grams.
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Malcolm Sargeant
Degree level applied chemistry + 20yr experience in corrosion prevention and water treatment
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Author has 21K answers and 28M answer views
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Feb 5
Originally Answered: What is the molarity of sodium sulphate (Na2SO4)?
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I guess you mean what is the molar mass of sodium sulphate which is 142.04
the term molarity is an expression of the concentration of a solution of sodium sulphate, if you dissolve 142.04 grams of sodium sulphate in water to make one litre of solution you would have a molarity of 1
TheTruthIsOutThere
Honours degree from Bradford College (Graduated 1986)
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Author has 105 answers and 61.4K answer views
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2y
This is more a mathematical question than a chemistry question…
I have noted and used the rounded molar masses of the atoms as written.
Na2 = 2 x 23 = 46
S = 32
O4 = 4 x 16 = 64
46 + 32 + 64 = 142.
Molar mass of Na2SO4 = 142
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Related
What mass of Na2SO4 is needed to make 275 mL of a 2.0 M solution (Na2SO3 = 142 g/mol)?
What mass of Na₂SO₄ is needed to make 275 mL of a 2.0 M solution (Na₂SO₄ = 142 g/mol)?
Moles of Na₂SO₄ = (2.0 mol/L) × (275/1000 L) = 0.55 mol
Mass of Na₂SO₄ = (0.55 mol) × (142 mol/L) = 78.1 g
Kumaraswamy Sathiavasan
MSc in Chemistry & IAS officer(retd.)
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Author has 9K answers and 14.3M answer views
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3y
Related
What’s the mass of 4.54 moles of Na2SO4?
Atomic mass:
Sodium, Na - 23 a.m.u.
Sulphur, S - 32 a.m.u.
Oxygen, O - 16 a.m.u.
Molar mass of Na2SO4 = (232 + 32 + 164) g/mol
= 142 g/mol
Mass of 4.54 moles of Na2SO4 = (142 g/mol) x 4.54 mol
= 645 g (answer)
Mann Vora
B Tech from A D Patel Institute of Technology, Vallabh Vidhyanagar
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Author has 58 answers and 130.8K answer views
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7y
Related
What is the molarity, molality of 15% mass / mass H2So4 (density=1.1g/mol)?
Trevor Hodgson
Knows English
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Author has 11.8K answers and 12.2M answer views
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4y
Related
What is the mass of Na2SO4 salt required to prepare 150.00 mL of 25.5%(m/v) of Na+ solution?
By definition:
A 25.5%(m/v) solution of Na+ contains 25.5 g Na+ dissolved in 100 mL solution
You want to prepare 150 mL of this solution
Mass of Na+ required = 150 mL / 100 mL 25.5 g =38.25 g Na+ dissolvced in 150 mL solution.
Now calculate mass of Na2SO4 required:
Molar mass Na2SO4 = 142 g/mol
Molar mass Na = 23 g/mol - 2Na = 46 g/mol
Mass of Na2SO4 required = 142 g / 46 g 38.25 g = 118.0 g Na2SO4 required.
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8351 | https://www.youtube.com/watch?v=DWPW4Mo42Yw | If a cos x - b sin x = c show that a sin x + b cos x = sqrt(a^2+b^2+c^2)
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To ask Unlimited Maths doubts download Doubtnut from - If a cos x - b sin x = c show that a sin x + b cos x = sqrt(a^2+b^2+c^2)
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8352 | https://gorillafund.org/silverback/silverback-gorillas-fascinating-facts-about-natures-gentle-giants/ | Silverback Gorillas: Fascinating Facts About Nature’s Gentle Giants
Gorilla Fast Facts
Towering, powerful and surprisingly tender, silverback gorillas are among the most iconic and misunderstood animals on Earth. As the leaders and protectors of their families, these majestic primates are not only symbols of strength but also of gentleness and care. At the Dian Fossey Gorilla Fund, we work daily to protect and study these incredible animals in their natural habitat. Below are some fast facts that highlight just how extraordinary silverbacks truly are.
The Largest Primate on Earth
Silverbacks can weigh up to 400 lbs and stand between 5’5″ and 6 feet tall—making them the heaviest primates alive.
From Tiny Newborns to Towering Adults
Male gorillas are born at around 4 lbs and reach full adult size between 15 and 18 years of age.
What Makes a Silverback?
“Silverback” refers to the distinctive silver hair adult male gorillas develop—not a separate species or subspecies.
A Plant-Powered Giant
Gorillas are vegetarians, and adult males can consume up to 60 lbs of vegetation every day.
The Truth About Gentle Giants
Despite their size, silverbacks are known for their calm, tolerant nature and are often seen babysitting infants in their group.
Strength With a Purpose
A male gorilla’s strength is used not for aggression but to defend his group and compete for mates—natural predators are rare.
Legendary Displays of Power
Chest beating, branch throwing, and charging are dramatic displays silverbacks use to show dominance and size.
A Signal You Can Hear for a Kilometer
The iconic chest beat of a silverback can be heard up to a kilometer away, with its deep frequency revealing the male’s size.
The Silverback Leads the Way
The dominant silverback guides the group’s movements and activities, in addition to offering protection.
Lifespan in the Wild and in Care
Male gorillas can live into their late 30s in the wild, and in protected environments, some have reached their 60s.
On the Brink: Critically Endangered
Gorillas are classified as critically endangered—just one step away from extinction—underscoring the urgency of our conservation work.
Silverback gorillas are awe-inspiring not just because of their size and strength, but because of the deep social bonds, intelligence and gentleness they display. Each fact above is a reminder of just how remarkable – and vulnerable – these animals are. As one of our closest relatives in the animal kingdom, their future is tied to our actions. At the Dian Fossey Gorilla Fund, we are committed to protecting silverbacks and their families every day. Join us in ensuring that future generations will continue to marvel at these gentle giants in the wild where they belong.
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8353 | https://www.fireapparatusmagazine.com/equipment/nozzle-reaction/ | Home/Equipment
Nozzle Reaction
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By BRIAN BRUSH
“Arguments for and against the use of various nozzle designs often become nullified on the fire ground as crews find they cannot safely operate lines which exhibit high nozzle reaction forces.”—Captain David P. Fornell
NOZZLE OPERATING PRESSURE
The smooth bore nozzle may be viewed by some as “dated,” but if you take a little deeper look at history, you can see some very sound reasoning in the smooth bore nozzle. The operating pressure of the smooth bore is a range from 40 to 60 pounds per square inch (psi), with 50 psi as optimal. This was important to our forefathers in the fire service, as early pump systems were primarily lower pressure and could see significant fluctuations with more than one line being supported simultaneously. The solid stream and long tip provided accurate delivery of the fire stream at a great distance for firefighters with limited personal protective equipment.
As technology advanced, our pumps were able to provide higher and more consistent pressures. Lloyd Layman and various others brought the fog nozzle into the American fire service, vendors started to develop automatic nozzles, and before we knew it there was a shift from a 50-psi fire service to 100 psi.
Over the past 15 to 20 years, an increasing number of firefighters and departments are beginning to question what has been gained by doubling our nozzle operating pressures. In many cases, it is being discovered that, for the most part, the only true gain has been nozzle reaction, which simply equates to more work on the nozzle firefighter.
There have been several studies done over the past 20 years into nozzle reaction and how it affects hoseline operations. The goal of these studies has been to identify how much nozzle reaction firefighters can comfortably handle while still being able to effectively advance and manage a hoseline. A study by Paul Grimwood outlined three working limits: one firefighter [60 pound-force (lbf)], two firefighters (75 lbf), and three firefighters (95 lbf). I have been fortunate enough to work with firefighters across the country on hoseline operations, and I can tell you that with good technique, practice, improved fitness, and continued work, firefighters can easily operate lines with nozzle reaction forces beyond the above working limits. However, these working limits are very accurate for the majority of firefighters and the median level of training.
Nozzle reaction is the resultant lbf of pushback from the combined volume and pressure leaving the nozzle. The only way to alter nozzle reaction is to alter the volume [gallons per minute (gpm)] or the pressure (psi). Many people have used a variety of methods to demonstrate nozzle reaction like fish scales and rope, but the actual force is calculated using the nozzle reaction charts. As a rough rule of thumb, the lbf of nozzle reaction for a 100-psi nozzle is half of the gpm.
On the nozzle reaction charts, you can see the amount of nozzle reaction associated with four very common 1¾-inch nozzles. You can also see the side-by-side comparison of a 150-gpm-at-50-psi fog nozzle with a 100-psi automatic fog nozzle. Flowing the same gpm, there is a nozzle reaction difference of 21 pounds at 100 psi; at 150 gpm, the nozzle reaction of 76 pounds is at the working limit of two firefighters. Here is where you need to question if your department sees this as necessary or unnecessary pressure.
With good practices and techniques, firefighters can work beyond the outlined nozzle reaction parameters above. Without those practices, nozzle reaction forces beyond 60 pounds typically begin to reduce the effectiveness of the single firefighter nozzle operator.
This is a very important piece of the puzzle when purchasing equipment for the engine staffed with three. A three-person engine company translates to a two-member first-due attack line. I have seen it time and time again where departments are training, purchasing, and writing policy for staffing that they do not have.
If you ever stretch a line from an engine to the second-floor bedroom as the nozzle firefighter with only one other person, you will discover instantly that you must learn to operate that nozzle without the luxury of a backup firefighter behind you to assist in countering nozzle reaction. The other member will almost always be working to tend the line through furniture, around corners, and up stairs somewhere between your location and the front door.
If the 1¾-inch hoseline is your department’s “90 percent of the time” and your engine company staffing is three members, you must identify what your firefighters are comfortable with regarding operating a line by themselves—it may be surprisingly less than you assume.
A CASE STUDY IN NOZZLE REACTION AND FUNCTION
The advertised benefit of an automatic nozzle is a “wide operating range without stream compromise.” Without getting too in depth, this is achieved through an internal compensatory spring that adjusts with flow to maintain a constant nozzle pressure and stream. This type of nozzle essentially puts the flow rate in the hands of the pump operator and, in the absence of a set department standard, this becomes a very concerning unknown.
Our department primarily has three-person engine staffing. In 2005, when we wanted to see if a nozzle study was needed for our department, one of the first steps was to take 10 engine companies, have them deploy and flow a 1¾-inch attack line, and record the pump discharge pressure. At that time, our department was using a 100-psi automatic nozzle on all 1¾-inch attack lines. From the data collected, we found that our average flow from these lines was 100 gpm. When the pump operators were asked why they pumped at their selected pressures, almost all responses were not flow-related; they were firefighter-related. Nearly every operator stated they underpumped the lines initially to make it easier on the nozzle firefighter, and they would increase the pressure if the nozzle firefighter called for more water.
Within this information is a very important finding. Our pump operators were acutely aware of the challenges of high nozzle reaction, and they were attempting to address them for the nozzle firefighter hydraulically. Unfortunately in their good intentions is a risky business of not only underpumping (pressure) but by design also undersupplying (volume) those firefighters entering the structure.
| |
| Fog Nozzle Reaction Chart |
| NR = .0505 Q √NP |
| NR (Nozzle Reaction) |
| Q= Gallons Per Minute |
| NP = Nozzle Pressure |
Fog Nozzle Reaction Chart
NR = .0505 Q √NP
NR (Nozzle Reaction)
Q= Gallons Per Minute
NP = Nozzle Pressure
The idea that a firefighter “can always call for more water” comes from the known that an automatic has that wide flow range. The reality is that the stream quality is maintained throughout that wide flow range, and the nozzle operator typically does not identify one lacking volume. Additionally, the nozzle firefighter knows that requesting more water increases pressure, making for a more difficult line to manage. As you can see, these contributing factors all conspire together, and that “call for more water” never comes.
| |
| Solid Bore Nozzle Reaction Chart |
| NR = 1.57 D2 NP |
| NR (Nozzle Reaction) |
| D = Diameter of tip |
| NP = Nozzle Pressure |
Solid Bore Nozzle Reaction Chart
NR = 1.57 D2 NP
NR (Nozzle Reaction)
D = Diameter of tip
NP = Nozzle Pressure
Using the average of 100 gpm from that 100-psi automatic fog nozzle and the fog nozzle reaction formula, we discovered that our firefighters and operators have subconsciously shown that a nozzle reaction of 50 pounds is a comfortable point. Since nozzle reaction is dictated by a combination of pressure and flow, it serves as the perfect point in the discussion to bring the two together.
With the finding that our firefighters felt most comfortable handling about 50 pounds of nozzle reaction, we had a starting point. The next step was to determine a target flow for our 1¾-inch attack lines, as it was clear from this initial test that we did not have one. For the goal of the study, we wanted to establish 150 gpm as the minimum flow for any interior attack lines.
Why 150 gpm? Nationally, 150 gpm has become the target flow for 1¾-inch attack lines. This number comes from National Fire Protection Association (NFPA) 1710, Organization and Deployment of Fire Suppression Operations by Career Fire Departments. The standard outlines that the first two attack lines in operation at any residential structure fire flow a minimum of 300 gpm combined. With the NFPA wording, you could flow 100 gpm with your initial line and 200 gpm with a second line, but the common sense approach and now industry standard have targeted 150 gpm as an interior attack standard.
| | |
--- |
| Common 1¾-Inch Attack Line Nozzles and Reaction Force | |
| • | 150 gpm at 50 psi fixed-gallonage fog: Nozzle reaction force of 54 pounds |
| • | 7â„8-inch smooth bore, 161 gpm at 50 psi: Nozzle reaction of 60 pounds |
| • | 150 gpm at 75 psi fixed-gallonage fog: Nozzle reaction force of 65 pounds |
| • | 15â„16-inch smooth bore 185 gpm at 50 psi: Nozzle reaction force of 69 pounds |
Common 1¾-Inch Attack Line Nozzles and Reaction Force
•
150 gpm at 50 psi fixed-gallonage fog: Nozzle reaction force of 54 pounds
•
7â„8-inch smooth bore, 161 gpm at 50 psi: Nozzle reaction of 60 pounds
•
150 gpm at 75 psi fixed-gallonage fog: Nozzle reaction force of 65 pounds
•
15â„16-inch smooth bore 185 gpm at 50 psi: Nozzle reaction force of 69 pounds
Using nozzle reaction parameters and a set minimum standard for volume, the rest of the process is relatively simple: Find nozzles that flow greater than 150 gpm with nozzle reactions near 60 pounds, and put them in the hands of firefighters for them to find their preference.
At the end of 2005, following a full year trail period with a variety of nozzles, the preference of our firefighters was the 7â„8-inch smooth bore with a flow of 161 gpm at 50 psi and a nozzle reaction of 60 pounds. What is key to remember is that nozzle ratings are just “ratings”—when closed, all nozzles flow 0. A 150-gpm or 2.5-gallon-per-second nozzle may seem “inferior” to one that flows 185 gpm or three gallons per second. If the nozzle firefighter can comfortably flow that 150 gpm nozzle for 30 seconds at a time around a corner while actively playing it, he is delivering 75 gallons to the fire environment. A nozzle firefighter struggling with a 185-gpm or 200-gpm nozzle, only operating it for 10 to 15 seconds at a time without fatiguing, who has poor stream movement is potentially ineffectively applying only 30 to 50 gallons to the fire environment at a time. Put the right weapon in your operators’ hands—it may not be the one with the greatest fire power.
BRIAN BRUSHis a 20-year veteran of the fire service and a training division sergeant at Edmond (OK) Fire Rescue. He has a bachelor’s degree in fire and emergency services administration and a Fire Officer designation form CPSE. He instructs on a national level and writes for Fire Engineering.
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Sum of Absolute Differences in a Sorted Array-Algorithms&Visualizations
Federico Feresini
6 min readDec 14, 2020
IntroductionIn this article, we will be going through the solution to a coding challenge that I’ve recently had the pleasure to solve taking one of the contests. Other than the solution itself, I’ll be providing an easy visualization that hopefully will make everything really simple to grasp.
The problemI’m not going to spend much time here, let me just copy-paste the description of this problem titled “Sum of Absolute Differences in a Sorted Array”
You are given an integer array nums sorted in non-decreasing order.
Build and return an integer array result with the same length as nums such that result[i] is equal to the summation of absolute differences between nums[i] and all the other elements in the array.
In other words, result[i] is equal to sum(|nums[i]-nums[j]|) where 0 <= j < nums.length and j != i (0-indexed).
Constraints:2 <= nums.length <= 10⁵ 1 <= nums[i] <= nums[i + 1] <= 10^4
Examples:
Input: nums = [1, 4, 6, 8, 10]
Output: result = [24,15,13,15,21]
result = |1–1|+|1–4|+|1–6|+|1–8|+|1–10| = 24
result = |4–1|+|4–4|+|4–6|+|4–8|+|4–10| = 15
result = |6–1|+|6–4|+|6–6|+|6–8|+|6–10| = 13
result = |8–1|+|8–4|+|8–6|+|8–8|+|8–10| = 15
result = |10–1|+|10–4|+|10–6|+|10–8|+|10–10| = 13
The problem should be clear, let’s jump into the study of the problem.
First ApproachThe first approach crossing your mind might be picking one element at a time, and looping over all the others to compute the absolute differences. This is not a wrong approach in general, nevertheless, it doesn’t lend itself to a big amount of data. Looking at the constraints, we might run into arrays comprising 10⁵ elements, and an algorithm running in O(n²) will trigger the Time Limit Exceeded before getting the result.
Let’s jump into a real first approach, that seems better than just iterating every time over the whole array. What about saving some pieces of information while computing the solution? Let’s have a closer look at what we need:
Each line represents an entry of the final array result.
The first thing that grabbed my attention was the relationship between one line and the one above.
I realized that instead of iterating over all of the elements every time, I could have skipped some of the first ones because I had already computed them in the previous step. Check this out!
Interesting, isn’t it?
It means that if I kept an array storing the partial sum of absolute differences for each column of this sort of matrix, I could iterate only on a part of the given input array elements, instead of going through all of them each time.
This is an optimization of the first straightforward approach. Sounds great, right?
Well, it does, but unfortunately, the time complexity of the algorithm is still O(n²).
A failing algorithm always hurts, I know, however, we haven’t exploited important information about this problem. The array is sorted!
Second ApproachThe elements are sorted. How can we leverage this information?
Obviously, picked an element, we know that all the previous ones are smaller, and the next ones are bigger (no duh!).
And what about the absolute difference between elements?
Let’s take a give arrays nums = [1, 4, 6, 9, 14]
Given the element , the distances to the other elements are drawn as yellow lines below the elements. Let’s pick the next element now,
In the picture, I left the previous distances from element to all the others in yellow, and then I just added up or subtracted the needed amount to get the distances from to the other elements (purple lines).
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Can you see the pattern here? We have kept all the previously computed distances from element , and then made some small changes to come up with the distances from element .
Here is the core of the algorithm that will allow us to solve this challenge. I want to make sure that the point I am trying to get across is extremely clear, and there’s no better way than some extra visualizations:
In the above picture, the upper section displays the distances from element to all the others, while at the bottom from .
Let’s put one closer to the other, to have a better understanding.
As you can see from this latest visualization, the red and the green lines differ from each other to a fixed amount, which is always the distance from to .
In other words, let’s focus on the distance from to and from to :
Simple right? The distance from to is nothing short of the distance from to plus the distance from to .
Ok, let’s move along towards a second couple of distances:
This is pretty obvious! The distance from to is the same that we had computed during the previous step from to .
Let’s go to the next ones:
Here it’s simple to see that the distance from to , is the same distance between to , but after removing the distance between to .
Last but not least!
It must be clear now that the distance from to can be computed either as |6-14| or |4-144-6|.
A visualization to sum everything up:
In a nutshell, we can easily compute the sum of the differences from to all the other elements, simply knowing in advance the sum of the differences from because:
|6–1| = |4-1| + |4-6|
|6–4| = |4-6|
|6–9| = |4-94-6|
|6–14| = |4-144–6|
Should be quite easy to spot that:
I hope that at this point the idea is clear, let’s jump into the code now.
The Code
ConclusionsI hope that you liked this article hopefully learned something new. Have a good day!
“Example isn`t another way to teach, it is the only way to teach.” — Albert Einstein
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8355 | https://www.youtube.com/watch?v=QJJb4vtMaM4 | Point Slope Form and MOAR MARBLESLIDES!!!
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Posted: 16 Mar 2021
@Sara_Putros asked me for help with the Point Slope Form and I deliver. Plus a bonus marbleslides activity to go with it! Hope this helped Sara!
If you want to try out the marbleslides activity for practice go to student.desmos.com and type in the code f785su.
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Transcript:
[Music] so [Music] all right folks welcome to another episode of dirty math my name is matt aka your dirty math teacher and today we have another request this time from sarah putrose and hopefully that's the right way you pronounce your name i'm so sorry if it isn't in any case sarah asked for an explanation on the point slope form so we're gonna do this a little bit differently we're gonna i'm gonna jump into a small explanation of what the point slow form is why it's useful and then i'm going to do jump into a marble slice activity on desmos that i'll share the code for if you want to try it be my guest you can try it on your own but it's not really homework i'm not i'm not your teacher so this is just practice we're talking about practice man if you want to practice what i'm kind of showing you today and and you want to see if you got the right answers or not that's what we're going to use it for so just go ahead and jump in jump out whenever you want to if you just want to listen to the explanation and jump out be my guest okay so this is a really brief explanation on what the point-slope form is and what it's good for so let's jump right into it so here is what the point-slope form looks like if this is your first time kind of jumping into the point-slope form or kind of encountering it you might think to yourself what the does this mean why are there two eyes why are there two x's what's going on here well the nice thing about the point slope form is just like all the other kind of naming conventions it is it tells you what it's really for if you have a point and a slope you plug it into this particular equation and then voila you come out with usually the slope intercept form which helps you draw the line and figure out hey where are where are all the other points on this particular line so what it really is is it's just another way of writing the slope-intercept form or just writing another form of a line basically if you can wrap your head around that you're you're almost there you're almost there so all you really have to know is how to flex your algebra muscles and how to kind of manipulate the different variables in here to to the way that you want it but i'm i'm talking too much so let's kind of use an example problem so for example let's say you know a slope and let's say you know a point and usually when when you're learning this for the first time your teacher is going to give you a point and a slope uh let's say your slope is something easy like one half and let's say your point right your um your example point is going to be three comma 4 for example right so we have a point we have a slope and then we're just going to plug it in in fact let me move this around let's change this into a different color i hope none of you are colorblind red so here's what we're going to do i'm going to take this slope plug it in here and then i'm going to take these points right uh i'm going to plug it in here because that's that first term is the x term and then i'm going to take the y term and then plug it in there and that's really all it is so here is your new formula y minus y one which is four equals your slope one half and then x minus your x one which in this case is three okay so there's your new equation and then what you're going to do with that new equation is let me rewrite the equation here or just move it up here actually then you just kind of kind of manipulate it into slope intercept form so let's go ahead and do this so i'm going to take first take this term right there the the one half and then i'm going to distribute it right to both terms inside the parentheses so what i end up with is y minus four because i haven't done anything to that left side and then you have one half x minus three over two and let's leave it as as an improper fraction there's no need to go crazy with it for now and then we're going to move over the the minus 4 here we're going to add 4 to both sides and hopefully your algebra skills should be up to par if they're not go watch khan academy video i'm going to assume that you know everything or have a decent skill at manipulating equations up until this point or if you want another video request it because this is quickly becoming the dirty math hotline page so request it and i'll maybe make another video about that so y equals one half x and then it's four uh minus three over two right so uh four minus three over two let's let's take care of this guy there so now you should end up with y equals uh one half x and then four minus three halves but let's uh hopefully your fraction sense is pretty good but um oh i forgot this i forgot the plus sign plus uh eight over two which is four minus three over two which will get you to y equals one half x uh plus five over two so there's your new equation y equals one half x plus five over two i should have used different numbers i'm sorry but real talk sometimes the numbers just don't come out very nice not whole numbers all the time so you gotta go where the math takes you what does that mean y equals one half x plus five over two that's another video slope intercept form hopefully you know that if you don't again request it um because if you don't i'm gonna assume that you don't wanna see that video and i got better things to do honestly than make videos that no one's going to watch what does that look like on a graph let's jump into desmos okay so we started off with one equation and then we ended up with a slope intercept form of that same equation but just to prove to you that they're the same equation let me kind of plug it in here we we had the equation y minus y one which in this case was four equals or is it one half um x minus the first one which is three right so we have that line we're gonna make that red let's make another one that's green we ended up with a point slope or slope intercept form of that same line right y equals a one half x and this one's gonna be blue plus five over two right as you can see here oops why do i do that plus not plus all right and so you can as you can see if you can if you take off that blue and that red part they're the same line actually it's just two forms of the same line right it's like having a nickname right does having a nickname change who you are whether your name is dimples or whether your name is whatever your name is does it change we are no it doesn't it's just a nickname two different names for the same person so same deal two different forms for the same line so that's really all the point slow form is useful for if you have a point and a slope you can figure out what the line is going to be right or you can just use desmos and plug it in all right let's jump in to the marble size activity okay here we are on screen one of this marble slides activity if you are unfamiliar with how marble slides works i've covered it plenty of times before in other videos but the idea is pretty simple click on launch the balls will be released and you have to get all the stars right so pretty simple idea let's jump on over to screen number two fix it number one okay now the first thing that should jump out at you is that hey man what the this does not look like point-slope form right and you'd be right this does not look like point-slope form there's one difference though the only difference being this uh this binomial here underneath should be on this side whereas y minus y1 equals uh the slope times x minus x1 so this should be over here for whatever reason they put it over here because they want to show you they want to show you um the fact that these two terms right here that's the point right where this is the uh that the x value of the point and this is going to be the y value of the point and this is the slope on the other side so it's just written in a different way but it still is the point-slope form it's just in a different form that you might be used to seeing right so that's all it is so they're trying to get you to understand what is this right um what number does what to this particular line so the instructions say change one number in the road below to fix the marble slide if you change these numbers here one of the points all it does is it changes where the point of this x is going to be or this line is going to go through which point so you would want to play with this value on the other side the slope of the line right now it's at 4 but if you change it to 5 or 6 or even lower 2 or 1 or zero you would see that it's doing something to the line where as if you go if you go higher the line is going to get more steep if you go lower uh it's going to be less steep but the real thing that you want to get to is between zero and one because zero is just a horizontal line uh one is a slope of one so you wanna go in between zero one so let's try 0.5 and halfway in between zero and one which is one half or 0.5 is right where we want to be we click on launch and sure enough we'll get all all the stars okay um screen four fix the number two so what you should have realized in this uh in this form of the point-slope form is that this number here is going to be the slope and these are going to be the point at which the line goes through the point in the point-slope form part of the equation so let's let's change that right we don't need to change the slope because as if we change the slope we're never going to get it to where it goes or it changes this line so we can collect all the all the stars so we need to change the point because right now it's going through uh two comma three which is right here we need to go through two comma negative one okay so in that sense we don't really need to change that value right in terms of the x value we can leave it there we just need to change this instead of three we need it to be negative one well look at this right three is right here we need to be negative one if we were to put in negative one instead of that three there what's gonna happen well y minus negative one is going to be rewritten as y plus one so that's what we're going to do and you could always just you could always just play it around with the number right that's another way you could have figured that out as well but um let's say you didn't then you just wanted to get to the answer that's how you get the answer but you you could have just played around in reverse right minus three minus four minus five minus six and minus two minus one zero and then how do you get past that plus one eventually that's how you would do it so there you go screen five fix it number three okay so in this particular case we don't really have to do anything it looks like uh because this slope right here is pretty good but this last star it's kind of sloping off right so we just need to fix the domain restriction here okay it stops right now at four we need to maybe go past four maybe the what is this right here seven so let's do x less than seven yeah and then it's gonna that when we click on launch the marble sides are gonna go get those first three and then kind of tail off let gravity grab that last one and sure enough there you go okay so the domain restriction will do that for us so pretty easy there there are no real domain restrictions when you're dealing with just a point-slope form but when you when you're dealing with marble slides domain restrictions are going to be an issue so take take it as you will screen six predict number one if we change the zero negative zero point two to a three in the equation what would happen to the graph well this right here again is the slope so we would change the slope from a negative 0.2 or a negative two tenths and change it to a positive three so first thing that would happen is this negative slope would become a positive slope which means that the right side is going to be higher and then the a slope of 3 means that it's going to be rather steep right it's going to be you're going to be going up 3 and over 1. so it's going to be pretty steep so on screen 7 is where you get to realize your prediction so if we change this negative 0.2 to a 3 or positive 3 you'll see that it's going to change this graph quite a bit right now you have a a line that's going up one two three and over one where this next point is right here at negative two comma two okay so uh that's what what happened so hopefully you got that prediction right okay screen eight predict number two if we change the plus three to a zero in the equation what would happen to the graph well what you should realize or what you should have realized is that this form is written kind of different right it should be y minus something over x minus something equals the slope but since it's not x minus something but in fact plus three what what that means is that this value is going to be the opposite of whatever it is right right now as it is you're going to be going through negative 3 comma negative 1 uh because it's the opposite of 3 positive 3 and positive 1. so it's going through the point negative 3 comma negative 1 right here but if we change this 3 to a 0 it's going to be it's going to lift this line up and it's going to go through right here at 0 sorry my mistake it's not going to go through it's not going to lift this line up what it's what it's going to do is instead of going through here it's going to go through here x is going to be at zero and then the y is going to be still the same right it's going to go through negative one right here so instead of the line being here it's gonna be it's gonna shift over here three units to the right okay so that's what's gonna happen uh and you can write any summary of that in the prediction part right here and sure enough if we go ahead and change this three to a zero what is going to happen is it's going to go through this point right here so we're going to get x plus let's change to zero you don't really need to write it but like i said it looks like it moved up but in fact it just kind of moved over to the right a little bit so that's what is happening to the graph in this particular equation predict number three if we change this if we change the two to a five in the domain restriction x less than two what would happen in the graph well this is easy instead of stopping at two it's going to stop at 5 over here okay so write whatever formula you want in here or not because this isn't homework i'm just giving you practice problems so verify number three sure enough we change this 2 to a 5 it just changes how far that line goes okay so that's screen 11 we're flying through these because like i said it this is not homework this is just some internet stranger explain to you practice problems so point-slope form so screen 12 uh predict four your friend won't get many stars like this what changes would you make to the equation to help your friend collect all the stars why are those changes going to work looking at this line right now i can tell that the slope is off right the slope is off so we need this line to change slopes so that this so you can collect all the stars because as it stands right now these marble sides are going to drop straight down so we needed to change slopes so that's what you're going to write here uh what are you going to change the slope to i don't know let's play with it in the next one okay in the next screen okay so we don't like one half we need it to be steeper so let's go two ah okay so i was wrong you should have written a little bit more in the previous screen you don't want to just change slopes which i think two is the right slope looking at it just eyeballing it right now but you also need it to move over to the right and what you should realize from the previous screens is this changes where the x value is going to be we don't want it to go through x uh we don't want it to go through negative two comma three which is right here let me if we write negative negative two comma three we don't want it to go through that point right let's label it we don't want to go through that point we want to go through this point right here at zero so let's change this to a zero or just erase all that sure enough if we go through comma zero comma three that's what i wanna do and then we gotta do one more thing we gotta change the domain restrictions as well so we need this domain restriction to go at least over the six it looks like so x less than six now how did i figure that out well um right where the balls are going to launch right here that's x equals five right here four five and if we just launched it there there's a chance that the marbles might go the other way so i went all the way to x equals six just to be safe and let's launch and there you go okay or you could have just gotten dirty right and just tried messing around with all the numbers until you got the correct one screen 15 fix it number four times many equations in the on the of the lines in the rows below you need to collect all the stars i can't read you may also edit the one already here okay so they gave us one already ah okay so we need to make a bunch so let's see first of all this blue one we needed to stop okay we need to put a domain restriction right here at x equals our x greater than nine so we don't want the values less than nine we want greater than nine why am i stopping here well because we need to collect these stars as well and i i'm just afraid that if i stop the domain restriction here it might not get that while it's dropping off it might not get that star we could try at eight let's go ahead and try it eight just for the heck of it right stop it there and then we need to copy this let's let's do that ctrl c ctrl v and uh let's drop this line down where it's not going through four comma eight four comma eight we wanna drop it to four comma let's see four comma five so i like that four i don't like that eight i'm going to change to a five okay and then i need to change the domain restrictions for this one in fact we don't need any domain restrictions ah how about that so let's go ahead and launch and just like i thought uh just like i was afraid of we didn't get that star so we're gonna move this domain restriction back to maybe maybe nine x greater than nine launch i'm afraid that might not be enough either let's go to um ten ten will ten get us there oops ten cross your fingers 10 there you go 10 was 10 was the key right so we needed to push that domain restriction back a little bit further hopefully that made sense screen 16 um man okay we're going to need a bunch right we're going to need some to go let me see should we do this we need something to go some to go this way something go this way and then collect all of them okay i am going to make something that where the balls can go either way okay how about that let's have something that goes either way and i'm just gonna press ctrl v again uh just to copy what they had before and just fuss around with this okay so um let's get one that goes from here to here let's get one that goes through this point right here what's this point um this point i think is uh one two three four five six seven eight nine ten ten comma what is this fifteen ten comma fifteen and then this really use um let's reel right it's right under that star by the way let's really use the point slope form to our advantage we have this point we want this line to go through this point and then have a slope of that takes us to this star so what is the slope we're going to go one two three four five one two three four five so five over five we need a slope of one okay so let's put the points in here we want this x value of 10 and then this is going to be x value of 15 and then we need a slope of one and for now let's put a domain restriction um at greater than eight but let's do less than less than ten how about that so less than ten oops less than ten yeah okay how about that so it's gonna flow both ways let's actually go a little bit further to nine i'm sorry seven seven a little bit longer okay there we go okay cool now i need to we're gonna copy that and create a new one that goes the other way so we need to change this slope from one just to a negative one boom oh man that wasn't all i needed to do let's change the domain restriction we need to go from uh let's go backwards let's go from 10 to 3 more than 10 which is 13. okay so this is going to be 13 or 13 to 10. there we go and then we should get the other two stars all right now we need another one that goes like this and collects that the last star there so we need kind of a two more lines that creates a big v okay a big huge v all right so i'm going to control v and i want this to go through these points right here what point is this uh 5 10 15 20 1 listing 20 is 19 19 comma 8 9 10 um 19 comma 10. so i need something that goes through 19 comma 10. not 19 comma 2019 calling 10. oh nope i was wrong not 19 5 ten fifty one two three four it wasn't five it was this is going by fours one two three four so four eight twelve um 16 so this is gonna be 15. is that right 15 15 comma 10 15. there we go 15 comma 10 let's label it 15 comma 10. uh so point slow form we have a point we need the slope the slope is going to be one so let's go ahead and write that um control v no did we do that already we did control v already so we need to go through one we need to go through 15 comma 10. so we need this to be 15 this is going to be 10 and we need to extend this line in fact we don't need a we don't need this we don't need a domain restriction and then we want the same exact thing to go the other way okay to go the other way but instead of 1 we're going to make this a so i got to look around here negative one oh man and then we gotta move this over as well so so let's do this one what's the what's the point over here this is going to be one two three four five five comma ten let's change this up a little bit five comma ten uh so we need to go through five and ten so this equation is going to have sorry give me a second here this equation is going to go through 5 not 15 comma 10. boom and i think we're ready launch [Music] there we go this one took a little bit longer but hey we do what we got to do i mean that's what brought you to my channel right screen 17 challenge slide number two okay so we need a series of things that go through here drop down here drop down this time i'm gonna do something different so far we've been working with um a version of the point slope form that doesn't that might not look like the one that uh your teacher wants you to work with so let's go ahead and just use that right we're gonna go and use the the y minus y 1 equals x or slope times x minus x 1 version that we were working with before and in fact let's go ahead and just write it right it's it's going to be you don't need to write the parenthesis you can if you want but let's do that y minus something right y1 uh equals slope times uh x minus x1 okay add a slider no i don't do that we'll leave it that long for now but we want something that goes through let's say this point right here let's let's make this point right here zero comma what is that 15. so 0 comma 15. so we have the point 0 comma 15. and what slope do we want from here we want it to end right here it's going to have a slope of negative 1 2 3 1 2 three four negative three comma four so now we need to go through zero comma fifteen and have a slope of negative three over four so let's go ahead and do that so y minus fifteen this y one going to be 15. so y minus 15 equals negative three over four like we said parenthesis x minus zero so we don't need to do that so we're just gonna leave it as x we don't need the parentheses actually so let's do that and then let's apply a domain restriction because because we need it to end at four right so domain restriction x less than four boom okay so that way the ball's gonna oh man we don't want to go through that point actually we want it to be right below so let's change this up a little bit darn well you live and you learn okay so we don't want it to go through zero fifteen we want it to go through zero fourteen fourteen so let's do that let's change this to a fourteen and it's going to change the slope a little bit we're going to go down 2 and over 4. so down 2 and over 4 or another way to write that is just one half by the way one half you could write two or four it's the same answer but um anyway so this works rather nicely because now it's not going through the ball we're not gonna lose some balls they're just gonna drop down so now we need we have one that needs to go through here this point and the slope we need a point and a slope hence the usefulness of point slope form what is this point right here it's four comma eight nine ten four comma ten so we need to go through four comma ten and you know what that might not be good either because i mean that might not be big enough for the ball to go through so let's actually go down to four comma nine about that because the balls will still drop and it'll still catch the ball so we're going through there four comma nine and we need to get through here so let's see we're gonna go down two and over one two three four five six seven so the slope is going to be negative two over seven so let's do that again let's copy this copy that paste it there so now we need this to be y minus nine that y value is nine not four so be careful with that x value's gonna be four and then like we said the slope is going to be negative two over seven negative two over seven uh and then we have a domain restriction right here ending at x equals 11. okay or x less than 11. so x less than 11. there we go beautiful and that that's it that's enough for us to collect all the the stars so launch boom boom beautiful we did it i did it unless unless you figure this out on your own which if you did pat yourself on the back you know how to use point-slope form now 18 in the rows below types of mini okay so we really only need one i think we we really only need one to go through here and then drop there and then we need something that kind of we don't want the marbles to to to go this way so i'm gonna build a wall i'm gonna build a wall right here how about that so let's do this we wanted to go through this point right here which is 0 comma 11 so 0 11 and then i think i have this saved yep we have the saved so and then we want this to have a slope of negative one one two three four five six negative one over six is going to be the slope so let's do that uh why am i working with this i'll just work with this one so instead of this y1 i'm going to put 11 and then this is going to be negative 1 over 6 for my slope and then instead of x1 we're going to put zero x minus zero which we don't really need oops why don't i do that um we don't really need can you do that we'll work with why is it capitalizing my x do i have my caps lock on i think i do there we go okay so we need a domain restriction we need it to end at uh x equals or x less than six so x less than six aha um let me see i might be wrong it might not drop to the right let's see if it works okay yeah hey good enough for me i no one said you can't lose any balls so good enough for me that might work but what the heck uh if you were working with it you could draw a line right there at x equals what x equals seven x equals seven and just build a vertical wall to to prevent it from going off to the right and it you know you can do that there you go screen 19 challenge slide number four um okay so you want something to have the balls go here and then just go down this way so we want it to go through here this point at x equals one comma nine right so one comma 9 is our point is our point that we want to go through and then let's have it go through another arbitrary point right here at x equals 8 9 10 1 two three four five so ten comma five ten comma five so that's my arbitrary point you could have picked another point up here it doesn't matter i picked and count five you could have picked ten comma six or 7 or whatever but this looks good enough for me where i want the ball to travel down this slope so we have the points that we want this line to go through and then we just got to figure out the slope well from here to here what's the slope well we're going to go down one two three four so negative four comma one two three four five six seven eight nine so negative four comma nine so my slope is going to be negative 4 comma 9 uh did i save the equation yes i did yes so let's go ahead and take care of the slope there negative 4 comma 9 is going to be my slope and i could pick either one and come out with the same line let's go with the one and the nine because i just feel like those are easier to work with so instead of that y one i'm gonna put nine and instead of that x one i'm gonna put one boom there you go and just to show you that that what i was saying before you could have put those values in there as well and you would get the same line right so if you put in 5 and 10 you'd still get the same answer okay so anyway we need to we need we need to put a domain restriction on this we need to end at 10. so let's do x less than 10 is my domain restriction and then i want another line that goes through these points where the stars are well where is this star at let's just pick one you can pick any of those points but let's just pick one of them let's pick this one oh no let's pick this one it's kind of going through a lattice point no let's pick this one this one's a nice one uh why do i want to work with this one well it's just four comma zero four comma zero and i like to work with zeros whenever i can because it makes it nice because zero times anything is zero so let's go through this point and i want us and i want a slope of well from here to here what is it you go up one and over two so the slope is one over two so the slope is one over two positive 1 over 2 and then let's just go ahead and work with these this 0 is going to go in here so we don't really need anything this y minus 0 is just y and then x minus x x1 well what is x1 it's gonna be four in this case boom launch no domain restrictions needed and there you go screen 20. okay let's see how are we gonna do this um we might need a ramp that takes us over here and then i don't know what should we do this is kind of tricky we have a ramp that will that should take the marbles over here but then also funnel some to this one to collect this one [Music] and then others to collect the other two oh wow okay let's do that okay so um standard procedure i'm going to pick a point here where the marble is marbles are going to be collected so what is this 20 comma 8 9 10 11 so i'm going to go through the point 20 comma 11. let's label that and we need this to go all the way over here um two what point is this three comma nine three comma nine okay so let's close let's do that let's i don't know where i'm going with this i'm just going with it oh boy okay what's my slope well down to an or sorry up to and over from nine to twenty or 11 sorry working with the wrong numbers from three three to twenty seventeen okay so we're gonna go up two and over seventeen to seventeen very strange slope not a not a very common one you don't see that one every day and then you can pick either one i'm gonna pick the three and the nine so let's change this to nine and this one to three there we go okay and then let's put a domain restriction on this one or we end at three so x x greater than three how about that x greater than three okay and that way oh man let's let's actually move this back a little bit right let's change this a little bit um i don't want this to be there let's let's just change the domain restriction to six no five let's go to five i think five will do it five well five do it let's see lunch yeah five might do it okay now we need we need some to hit and go down to this one but some to be caught and catch those other two so let's create a wall right here at x equals two this is not going to be the prettiest marble slide and in fact i have a feeling that since this isn't homework most if you made it to screen 20 congratulations man that means you really want to know this um i don't anticipate any of you reaching screen 20. in fact let me know if you did in the comments and i don't know but man if i had dogecoin i would give you i've been getting into cryptocurrency a lot lately um that's why i have my my production schedule has been down somewhat if you want to see me talk if you if you've reached screen 20 of this video and you want to hear me talk more about cryptocurrency if you're in high school and you're you're hearing all the buzz about bitcoin and ethereum and all that stuff and you want to know more about it and you want me to turn this channel into a cryptocurrency channel let's do it let's get it but uh for now i just don't know what how many people would be interested in that so won't do that for now but in any case um let's have it so that it's going to catch some and it's going to catch some oh man it still doesn't let you catch that really very well exactly let's move this back let's move this back to uh x equals one and a half one and a half okay and then i want some of them to not get caught so how about right here at we want it to go through this point right here x two comma seven that's the point i'm gonna go through two comma seven okay and then um we want this to go and kind of flow down this way so what is this at eight comma six and i'm getting super complicated so if i lose you here i don't blame you so we need to go through these two points so let's do that uh control v so uh the slope from here to here is gonna be down one over one two three four five six uh so negative one over six and then let's just pick whichever one right it doesn't matter at this point six eight and then a domain restriction from here to here so the domain restriction is going to be from two uh so we want it from two so greater than two all values greater than two and less than eight okay and in fact we don't want it to be at two let's go a little bit higher than two about two point one two three how about that because we want some of the marbles to go this way not all of them much we want some of the marbles to go this way on the ramp and some to drop back down oh over more 2.1 launch will they fit through at 2.1 oh they will okay how about just two maybe two was okay two oh even that kind of didn't work um two how about 1.8 launch 1.8 oh it's too big all right too small can't fit the i can't fit them through 1.9 we might have to this might not even work yeah this is not going to work because 2 it just lets them all through so 1.95 can we get in between 1.9 and 2. getting crazy down here oh 1.96 i think that i don't know yeah they're just getting stuck i don't i don't think we're going to get it to the point where it catches some and doesn't catch others so we might need to revise our we might need to revise our plan a little bit let's go back to two um because at this point it's just catching of course not oh no it's not point nine eight maybe i gave up two earlier 1.98 nope 1.99 yeah i just don't think it's gonna allow sum to go through how about a little line right here going through x equals two x equals two but we want a domain restriction where it ends at uh y less than eight why i i i fully realize that this is not what you signed up for so if you if you want to give up or if you didn't make it to this point i i totally understand uh why how about that launch come on ah darn it why less than seven launch ooh 7.1 i think we're on to something oh 7.2 okay seven so between seven point one and seven point two seven point one five aha there we go okay we're on to something all right this isn't called dirty math for nothing okay so um we want this to continue let's uh let's change this domain restriction from eight let's have it keep going and drop off let's stop it at 12 12. how about that oh it's too much i think we should have stopped it at 11. oh no it'll work ah take that all right man screen 21 oh my jesus okay uh let's see we need some to drop down this way something to drop down this way and somehow catch the middle as well or how about we have we have them drop and then no let me see how should we do this have some so we're gonna need to set up some kind of cone here obviously to go left and right but then also catch some so that it goes and and comes in the middle okay let's do that let's make the cone first let's worry about everything else later let's make a cone first so we need to go through uh four comma eleven right four comma eleven where did i get that seven comma eleven jesus seven comma eleven um so we're gonna go through that point um and we need one going this way it looks like a slope of one and negative one that's what i'm guessing right one and negative one i'm gonna live dangerously here and go with one and the same one with a negative one okay with that point eleven and seven eleven and seven same thing here eleven and seven domain restrictions okay for this blue one we need it to stop at less than seven so x uh less than seven 7 okay and then this one is going to be x greater than 7x greater than 7 okay but a few more domain restrictions we want this to uh go so this blue one we want it from x greater than four so let's add one more here for this way and then here is going to be greater than eight nine ten so we want this to go all the way to 10 like that okay that way some will go there so we'll go there um and let me see it'll get those two on the left and the right but somehow we needed to catch some on the way down and go to the middle okay so let's go uh x let's do one and then we'll work on the other one x equals 11 x equals 11. and we want a domain restriction uh or sorry this time is considered a range extract a range restriction oh my god this is a tongue twister range restriction um y less than um eight no seven oh not eighty seven seven will i work i don't know launch oh it just bounces off that's not what i want um i'm going to get a little bit more creative it's six how about six oh too low six point eight lunch no six point seven [Music] oh no what am i doing that's not gonna do it six point eight six point nine did i try that already i think i did try that already uh [Music] eight five huh eight six eight seven [Music] i just need something that'll kind of have the balls go that way and and some balls go to the left as well eight eight eight nine um we tried nine how about nine one nine one [Music] nine two oh man this is gonna be five six be brave matt nine six point nine nine goodness gracious seven one seven point one aha okay we're close we're getting closer seven two oh between seven point one and seven point two one 7.15 oh 7.16 okay we're awfully close 7.7 oh man we just need one of them to go left eight seven point one eight there we go seven point one eight that's where we wanna be okay so 7.8 is where we want to be so let's have this same one x equals not 11 but uh i'm going to copy paste this we did all that hard work why shouldn't we okay there we go why should we enjoy the fruits of our labor uh so instead of eleven we want this to be at three three [Music] excuse me some should go left and should go right perfect and then we want another funnel that goes through let's work on one side and then get it to the other side um we wanted to go through this point right here uh 11 comma seven eleven comma seven is our point and we wanted to go all over here to what is this seven comma six seven comma six so label that right there and we're just um we're gonna just have the domain restrictions stop it at a certain point so it just kind of goes it funnels itself down down this way okay so uh let's let's do that so what's the slope from here to here well we're gonna go up one and over one two three four one over four is our slope whoops let's copy one of these control c ctrl v and then our slope is going to be one over four like we said right one over four um get rid of the domain restriction and we wanted to go through one of these points let's make this six make that seven we already got that so i just use that point right there the domain restriction is going to end at x equals eight or x greater than eight x greater than eight okay um and then let's see we're gonna copy this and instead i'm going to make this negative 1 over 4 and have it go the other way okay so what's so let's see we want a domain restriction um going from x less than six x less than six so we need to flip this sign around x less than six there we go okay well and then we we actually we don't we don't want that domain restriction right um we want we want to get rid of this part in this part so we're gonna have to add a little bit more to this domain restriction this is going to be 11 and this one we need to snip off the end here at uh three so three like that okay i think that'll work launch oh you know what i i just realized we're gonna miss is that we'll get that one but we need another funnel to catch all these so let's just go from right here x equal x equals six and x equals eight x equals six with a range restriction at y less than um six i think that'll do it and then we we said we wanted one at eight as well right x let's see [Music] and then we wanted one at x equals eight oops equals eight the range restriction had x less than six okay oh where is that where did it go how come it's not popping up that too many x equals eight how come it's not appearing eight huh interesting i don't know why uh it was disappearing so anyway that i think i'm hoping that'll work there we go not the prettiest but hey it worked 22 challenge slide number seven oh gosh first of all where are the balls launching from oh straight down okay i know what i'm gonna do i'm gonna have a line go through x equals eight x equals eight and stop right here and y let's put a range restriction on this one y less than five okay that way it'll go we'll go from left to right and then we want a kind of a base right here at y equals four so i'm building a platform at y equals four and then i'm going to build a little kind of a ramp from here and like going here and here so that the ball some of the balls going this way will catch those stars some of the balls going this way will catch those stars okay so going through this point right here at eight comma five eight comma five we have a point we have a slope which is one and negative one so um you see did i oh man i didn't so let's just rewrite this y minus y one equals uh slope x minus x one okay so the slope is gonna be one and negative one so let's do the one let's do one of those first and then let's do the other one negative one okay and then we wanted to go through the point eight and five so eight comma five five this one's gonna be eight this is five and eight okay and then domain restrictions okay so for the green line we wanted to end at eight so x less than eight x less than eight oops we're using wrong brackets here x less than eight but for the other one we want x uh greater than eight x greater than eight there you go and that's all you really need to do i think there you go 23. oh my goodness this is the easiest one so far we just need one line okay so where's our point two comma ten so we have our point two comma ten and i'm just going to copy it from the previous screen here i'm just gonna steal one of these copy paste okay so what's our slope here from here to here we're gonna go down one two three one two three four five six so down three over six which is negative one half is our slope so negative one half is going to be our slope and we need to go through two comma ten so this is going to be ten this one's gonna be two and that's all we need to do launch boom if you made it this far in the video congratulations you have my respect uh sarah putrose i hope you got your money's worth and i hope i helped you figure out the points though form if you need any other help with math related questions please let me know in the comments and just reach out and if i have the time if i have the resources i will see what i can do that's it thanks for joining me for another episode of dirty math and uh i'll see you next time [Music] bye |
8356 | https://math.stackexchange.com/questions/1370790/prove-that-for-every-positive-integer-n-exists-c-n-such-that-fc-n-fc-n | calculus - Prove that for every positive integer $n, \exists c_n$ such that $f(c_n) = f(c_n+1/n)$ - Mathematics Stack Exchange
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Prove that for every positive integer n,∃c n n,∃c n such that f(c n)=f(c n+1/n)f(c n)=f(c n+1/n)
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Let f:[0,1]→[0,1] be a continuous function with f(0)=f(1). Prove that for every positive integer, n,∃c n∈[0,1] such that f(c n)=f(c n+1/n),c n∈[0,1−1 n].
With this problem, the only thing coming to my mind in terms of n being a positive integer is that as n goes to bigger and bigger, f(c n+1 n) goes to closer and closer to f(c n+0)=f(c n), that's all. I can't come up anything in terms of using IVT to this. And, since f(a)=f(b), I know I can't apply Rolle's theorem to this because the problem didn't say f is differentiable. Any help? Thanks.
Edit: Even though the conditions are the same with this problem, the functions are still different. And, the answers for that question are clearly not helpful to solve my problem.
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edited Apr 13, 2017 at 12:19
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asked Jul 23, 2015 at 1:31
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Can you do the case of n=1, perhaps? (Hint: Define g(x)=f(x+1 2)−f(x))Milo Brandt –Milo Brandt 2015-07-23 01:33:43 +00:00 Commented Jul 23, 2015 at 1:33
@Meelo Ok, let me try and find out.Jellyfish –Jellyfish 2015-07-23 01:36:05 +00:00 Commented Jul 23, 2015 at 1:36
@Meelo I don't understand where/how can I apply n = 1, but the hint you've given: g(x)=f(x+1 x)−f(x), can we then say, g(a)≤0 and g(b)≥0,∃c∈[a,b] such that g(c)=0 and hence the conclusion? But, then again, I don't know what f(0) and f(1) are to get the desired inequalities.Jellyfish –Jellyfish 2015-07-23 01:57:41 +00:00 Commented Jul 23, 2015 at 1:57
Oh, whoops. I meant n=2 - but yeah, the idea is we need to find an a and a b where g is positive and negative respectively. The point here is that we know f(1)−f(0)=0 which is important - especially since, for instance (f(1 2)−f(0))−(f(1)−f(1 2))=f(1)−f(0)Milo Brandt –Milo Brandt 2015-07-23 02:02:09 +00:00 Commented Jul 23, 2015 at 2:02
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Consider g(x)=f(x+1 n)−f(x) defined on [0,1−1 n], assume there is no x∈[0,1−1 n] such that g(x)=0 by intermediate value theorem (g(x) continuous), we have either g(x)>0 or g(x)<0 for all x∈[0,1−1 n]. However, it makes f(1)>f(1−1 n)>…>f(0) or f(1)<f(1−1 n)<…<f(0), contradiction.
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answered Jul 23, 2015 at 1:47
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Why do you need a contradiction? Define g that way, use IVT to find that it's 0 for some number in the interval, and then the conclusion follows.Race Bannon –Race Bannon 2015-07-23 01:54:34 +00:00 Commented Jul 23, 2015 at 1:54
@RaceBannon How do you find that it is 0 for some number in the interval?Conrado Costa –Conrado Costa 2015-07-23 01:56:11 +00:00 Commented Jul 23, 2015 at 1:56
1 It's continuous, so by the IVT, it hits every value in that interval. That's how.Race Bannon –Race Bannon 2015-07-23 01:56:52 +00:00 Commented Jul 23, 2015 at 1:56
@RaceBannon I guess it's not that simple frankooo's construction is very ingenious. g could be positive everywhere and therefore would not have to hit 0.Conrado Costa –Conrado Costa 2015-07-23 02:03:10 +00:00 Commented Jul 23, 2015 at 2:03
Where did f(1−1 n) come from?Jellyfish –Jellyfish 2015-07-24 01:24:11 +00:00 Commented Jul 24, 2015 at 1:24
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Karamata's Inequality
Karamata's Inequality states that if majorizes and is a convex function, then
Proof
We will first use an important fact: If is convex over the interval , then and ,
This is proven by taking casework on . If , then
A similar argument shows for other values of .
Now, define a sequence such that:
Define the sequences such that and similarly.
Then, assuming and similarily with the 's, we get that . Now, we know: .
Therefore,
Thus, we have proven Karamata's Theorem.
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Team Loki. In Southeast Alaska where we fish, intersecting runs means that we have the opportunity to harvest Pink salmon every summer.
Pinks are by far the most underrated and misunderstood salmon species. Their small size and plankton-based diet mean that they have fragile flesh that must be handled properly when caught in order to reach market in suitable shape. Unfortunately, most Pinks are handled poorly out of the water, sucked through a vacuum tube onto processing boats and end up being industrially canned. Firmer flesh salmon such as Sockeye can withstand more of this abuse, but under this type of stress the Pinks don't fare nearly as well. For this reason, Pinks have gained a reputation as being mushy, low quality salmon. A properly handled Pink is a different animal altogether; with a mild flavor and a tender, delicate flake, Pinks are a staple source of protein for many of our customers.
While none of the salmon species that we harvest have issues with toxin levels, due to their short life-cycle and eating low on the food chain, Pink salmon are the cleanest of all the salmon species in terms of bio-accumulation of toxins.
Keta Salmon
Latin name: Oncorhynchus keta
Other common name: Chum
Keta salmon are the second most abundant species of salmon, averaging 10-15 pounds. They spend a short amount of time as fry in the estuaries before heading out to the ocean to mature. They have an average life span of 3 to 6 years. We fish for Keta salmon in Southeast Alaska in the summer, and the greater Puget Sound area in the fall.
Keta have gotten a bad rap over the years. The main reason for this is that due to eating lower on the food chain, their flesh rapidly begins to turn pale as they enter fresh water environments to spawn. These fish are still more than suitable for smoking and jerky, but as a fillet product are not as visually appealing. We high-grade all of the Keta salmon that we harvest into #1's (fillet quality), #2's (for smoking), and #3's (for jerky). Keta has a fairly mild flavor, and a thick flake size. This makes it popular for marinades or special seasonings, as the fish will soak up whatever you flavor it with.
Coho Salmon
Latin name: Oncorhynchus kisutch
Other common name: Silver
Coho salmon average 7-11 pounds. They spend one to two years in freshwater stream environments before migrating to the ocean to fully mature. They have an average life span of 2 to 4 years. We fish for Coho salmon in Southeast Alaska in the summer.
Coho salmon are a favorite of those looking for a savory, flavorful salmon that is not overpowering. Coho have an ample amount of oil and fat in their flesh, but are not as strong tasting as Sockeye or King. Coho is best enjoyed filleted or smoked.
Sockeye Salmon
Latin name: Oncorhynchus nerka
Other common name: Red
Sockeye salmon average 3-8 pounds. They spend up to three years in freshwater lake environments before heading out to the ocean to fully mature. They have an average life span of 4 to 6 years. We fish for Sockeye salmon in Southeast Alaska in the summer.
Sockeye have the reddest, densest flesh of all the salmon species. This is due to their diet, which is composed primarily of krill. For this reason, Sockeye is a prized salmon and fetches a premium price. Although Sockeye handle the abuses of industrial fishing better than some of the other species, it is common to see Sockeye fillets in supermarkets that look "tired". This is usually due to heat-stress and improper refrigeration of the salmon at the time of harvest. A well-handled Sockeye should be firm, with bright, translucent eyes and no soft spots or gashes. Sockeye is best enjoyed filleted, smoked or canned.
King Salmon
Latin name: Oncorhynchus tshawytscha
Other common name: Chinook
King salmon are the largest of all Pacific Salmon, averaging 10 to 50 pounds. They spend one to two years in freshwater river environments before heading out to the ocean to fully mature. They have an average life span of 5 years. We fish for King salmon in Southeast Alaska in the summer.
King salmon have the oiliest, thickest flesh of all the salmon species. This is due to their large size and highly carnivorous diet. They are a prized fish both for filleting and smoking, and command a premium price. Kings make up a very small percentage of all the salmon we harvest in a normal year. Normally we sell all of our King's fresh at the farmers markets in the summer, and have just a small amount of smoked and frozen that sells out quickly in the fall.
Salmon illustrations courtesy ofAlaska Seafood Marketing Institute.
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8359 | https://courses.lumenlearning.com/introstatscorequisite/chapter/summary-two-basic-rules-of-probability/ | Summary: Two Basic Rules of Probability | Introduction to Statistics Corequisite
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Introduction to Statistics Corequisite
Module 3: Probability
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Summary: Two Basic Rules of Probability
Key Concepts
If two events are mutually exclusive, the addition rule for probability can be used to determine the probability of the outcome.
If two events are not mutually exclusive, the addition rule for probability needs to be modified to determine the probability of the outcome.
If two events are independent, the multiplication rule for probability can be used to determine the probability of the outcome.
Glossary
addition rule for probability:if A A and B B are any two mutually exclusive events, then P(A O R B)=P(A)+P(B)P(A O R B)=P(A)+P(B). If A A and B B are NOT mutually exclusive events, then P(A O R B)=P(A)+P(B)−P(A a n d B)P(A O R B)=P(A)+P(B)−P(A a n d B).
multiplication rule for probability:if A A and B B are independent events, then P(A a n d B)=P(A)P(B)P(A a n d B)=P(A)P(B)
sample spacethe set of all possible outcomes of an experiment
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The Physics Classroom » Physics Tutorial » Thermal Physics » Temperature and Thermometers
Thermal Physics - Lesson 1 - Heat and Temperature
Temperature and Thermometers
Introduction to Thermal Physics
Temperature and Thermometers
Thermometers as Speedometers
What is Heat?
Methods of Heat Transfer
Rates of Heat Transfer
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We all have a feel for what temperature is. We even have a shared language that we use to qualitatively describe temperature. The water in the shower or bathtub feels hot or cold or warm. The weather outside is chilly or steamy. We certainly have a good feel for how one temperature is qualitatively different than another temperature. We may not always agree on whether the room temperature is too hot or too cold or just right. But we will likely all agree that we possess built-in thermometers for making qualitative judgments about relative temperatures.
What is Temperature?
Despite our built-in feel for temperature, it remains one of those concepts in science that is difficult to define. It seems that a tutorial page exploring the topic of temperature and thermometers should begin with a simple definition of temperature. But it is at this point that I'm stumped. So I turn to that familiar resource, Dictionary.com ... where I find definitions that vary from the simple-yet-not-too-enlightening to the too-complex-to-be-enlightening. At the risk of doing a belly flop in the pool of enlightenment, I will list some of those definitions here:
The degree of hotness or coldness of a body or environment.
A measure of the warmth or coldness of an object or substance with reference to some standard value.
A measure of the average kinetic energy of the particles in a sample of matter, expressed in terms of units or degrees designated on a standard scale.
A measure of the ability of a substance, or more generally of any physical system, to transfer heat energy to another physical system.
Any of various standardized numerical measures of this ability, such as the Kelvin, Fahrenheit, and Celsius scale.
For certain, we are comfortable with the first two definitions - the degree or measure of how hot or cold an object is. But our understanding of temperature is not furthered by such definitions. The third and the fourth definitions that reference the kinetic energy of particles and the ability of a substance to transfer heat are scientifically accurate. However, these definitions are far too sophisticated to serve as good starting points for a discussion of temperature. So we will resign to a definition similar to the fifth one that is listed - temperature can be defined as the reading on a thermometer. Admittedly, this definition lacks the power that is needed for eliciting the much-desired Aha! Now I Understand! moment. Nonetheless it serves as a great starting point for this lesson on heat and temperature. Temperature is what the thermometer reads. Whatever it is that temperature is a measure of, it is reflected by the reading on a thermometer. So exactly how does a thermometer work? How does it reliably meter whatever it is that temperature is a measure of?
How a Thermometer Works
Today, there are a variety of types of thermometers. The type that most of us are familiar with from science class is the type that consists of a liquid encased in a narrow glass column. Older thermometers of this type used liquid mercury. In response to our understanding of the health concerns associated with mercury exposure, these types of thermometers usually use some type of liquid alcohol. These liquid thermometers are based on the principal of thermal expansion. When a substance gets hotter, it expands to a greater volume. Nearly all substances exhibit this behavior of thermal expansion. It is the basis of the design and operation of thermometers.
As the temperature of the liquid in a thermometer increases, its volume increases. The liquid is enclosed in a tall, narrow glass (or plastic) column with a constant cross-sectional area. The increase in volume is thus due to a change in height of the liquid within the column. The increase in volume, and thus in the height of the liquid column, is proportional to the increase in temperature. Suppose that a 10-degree increase in temperature causes a 1-cm increase in the column's height. Then a 20-degree increase in temperature will cause a 2-cm increase in the column's height. And a 30-degree increase in temperature will cause s 3-cm increase in the column's height. The relationship between the temperature and the column's height is linear over the small temperature range for which the thermometer is used. This linear relationship makes the calibration of a thermometer a relatively easy task.
The calibration of any measuring tool involves the placement of divisions or marks upon the tool to measure a quantity accurately in comparison to known standards. Any measuring tool - even a meter stick - must be calibrated. The tool needs divisions or markings; for instance, a meter stick typically has markings every 1-cm apart or every 1-mm apart. These markings must be accurately placed and the accuracy of their placement can only be judged when comparing it to another object that is precisely known to have a certain length.
A thermometer is calibrated by using two objects of known temperatures. The typical process involves using the freezing point and the boiling point of pure water. Water is known to freeze at 0°C and to boil at 100°C at an atmospheric pressure of 1 atm. By placing a thermometer in mixture of ice water and allowing the thermometer liquid to reach a stable height, the 0-degree mark can be placed upon the thermometer. Similarly, by placing the thermometer in boiling water (at 1 atm of pressure) and allowing the liquid level to reach a stable height, the 100-degree mark can be placed upon the thermometer. With these two markings placed upon the thermometer, 100 equally spaced divisions can be placed between them to represent the 1-degree marks. Since there is a linear relationship between the temperature and the height of the liquid, the divisions between 0 degree and 100 degree can be equally spaced. With a calibrated thermometer, accurate measurements can be made of the temperature of any object within the temperature range for which it has been calibrated.
Temperature Scales
The thermometer calibration process described above results in what is known as a centigrade thermometer. A centigrade thermometer has 100 divisions or intervals between the normal freezing point and the normal boiling point of water. Today, the centigrade scale is known as the Celsius scale, named after the Swedish astronomer Anders Celsius who is credited with its development. The Celsius scale is the most widely accepted temperature scale used throughout the world. It is the standard unit of temperature measurement in nearly all countries, the most notable exception being the United States. Using this scale, a temperature of 28 degrees Celsius is abbreviated as 28°C.
Traditionally slow to adopt the metric system and other accepted units of measurements, the United States more commonly uses the Fahrenheit temperature scale. A thermometer can be calibrated using the Fahrenheit scale in a similar manner as was described above. The difference is that the normal freezing point of water is designated as 32 degrees and the normal boiling point of water is designated as 212 degrees in the Fahrenheit scale. As such, there are 180 divisions or intervals between these two temperatures when using the Fahrenheit scale. The Fahrenheit scale is named in honor of German physicist Daniel Fahrenheit. A temperature of 76 degree Fahrenheit is abbreviated as 76°F. In most countries throughout the world, the Fahrenheit scale has been replaced by the use of the Celsius scale.
Temperatures expressed by the Fahrenheit scale can be converted to the Celsius scale equivalent using the equation below:
°C = (°F - 32°)/1.8
Similarly, temperatures expressed by the Celsius scale can be converted to the Fahrenheit scale equivalent using the equation below:
°F= 1.8•°C + 32°
The Kelvin Temperature Scale
While the Celsius and Fahrenheit scales are the most widely used temperature scales, there are several other scales that have been used throughout history. For example, there is the Rankine scale, the Newton scale and the Romer scale, all of which are rarely used. Finally, there is the Kelvin temperature scale, which is the standard metric system of temperature measurement and perhaps the most widely used temperature scale among scientists. The Kelvin temperature scale is similar to the Celsius temperature scale in the sense that there are 100 equal degree increments between the normal freezing point and the normal boiling point of water. However, the zero-degree mark on the Kelvin temperature scale is 273.15 units cooler than it is on the Celsius scale. So a temperature of 0 Kelvin is equivalent to a temperature of -273.15 °C. Observe that the degree symbol is not used with this system. So a temperature of 300 units above 0 Kelvin is referred to as 300 Kelvin and not 300 degree Kelvin; such a temperature is abbreviated as 300 K. Conversions between Celsius temperatures and Kelvin temperatures (and vice versa) can be performed using one of the two equations below.
°C = K - 273.15°
K = °C + 273.15
The zero point on the Kelvin scale is known as absolute zero. It is the lowest temperature that can be achieved. The concept of an absolute temperature minimum was promoted by Scottish physicist William Thomson (a.k.a. Lord Kelvin) in 1848. Thomson theorized based on thermodynamic principles that the lowest temperature which could be achieved was -273°C. Prior to Thomson, experimentalists such as Robert Boyle (late 17th century) were well aware of the observation that the volume (and even the pressure) of a sample of gas was dependent upon its temperature. Measurements of the variations of pressure and volume with changes in the temperature could be made and plotted. Plots of volume vs. temperature (at constant pressure) and pressure vs. temperature (at constant volume) reflected the same conclusion - the volume and the pressure of a gas reduces to zero at a temperature of -273°C. Since these are the lowest values of volume and pressure that are possible, it is reasonable to conclude that -273°C was the lowest temperature that was possible.
Thomson referred to this minimum lowest temperature as absolute zero and argued that a temperature scale be adopted that had absolute zero as the lowest value on the scale. Today, that temperature scale bears his name. Scientists and engineers have been able to cool matter down to temperatures close to -273.15°C, but never below it. In the process of cooling matter to temperatures close to absolute zero, a variety of unusual properties have been observed. These properties include superconductivity, superfluidity and a state of matter known as a Bose-Einstein condensate.
Temperature is what the thermometer reads. But what exactly is temperature a reflection of? The concept of an absolute zero temperature is quite interesting and the observation of remarkable physical properties for samples of matter approaching absolute zero makes one ponder the topic more deeply. Is there something happening at the particle level which is related to the observations made at the macroscopic level? Is there something deeper to temperature than simply the reading on a thermometer? As the temperature of a sample of matter increases or decreases, what is happening at the level of atoms and molecules? These questions will be addressed onthe next page of Lesson 1.
Check Your Understanding
In the discussion on the calibration of a thermometer, it was mentioned that there was a linear relationship between temperature and the height of the liquid in the column. What if the relationship was not linear? Could a thermometer still be calibrated if temperature and the column height of the liquid were not related by a linear relationship?
See Answer
Answer: Yes!
Calibration would still be possible as long as there was some form of relationship that was mathematically predictable. But if the relationship were not linear, a two-point method of calibration would not be possible. The linear relationship allows a scientist to use just two standards in the calibration method. For instance, the height of the liquid in the column could be found for the normal freezing point of water (0°C) and the normal boiling point of water (100°C). Because the relationship is linear, 100 equal-spaced divisions could be made between these two standard temperatures. In the absence of a linear relationship, the divisions between these two standards are not equal spaced divisions. Many additional standards or simply a known mathematical relationship would have to be used to determine the heights for the in-between temperatures
Which is the smaller temperature increment - a degree Celsius or a degree Fahrenheit? Explain.
See Answer
Answer: degree Fahrenheit
The degree Fahrenheit is the smaller increment. After all, there are 180 of these Fahrenheit divisions between the normal freezing point and the normal boiling point of water; there are only 100 of the Celsius divisions between these two temperatures. If more Fahrenheit divisions can be fit between these two divisions than Celsius divisions, then the Fahrenheit divisions must be smaller.
Perform the appropriate temperature conversions in order to fill in the blanks in the table below.
Celsius (°)Fahrenheit (°F)Kelvin (K)
a.0
b.212
c.0
d.78
e.12
See Answer
Answer:
Celsius (°)Fahrenheit (°F)Kelvin (K)
a.0 32 273
b.100 212 373
c.-273-459 0
d.26 78 299
e.-11 12 267
{^cosymantecnisbfw^}
Next Section:
Thermometers as Speedometers
What is Heat?
Methods of Heat Transfer
Rates of Heat Transfer
Jump To Next Lesson:
What Does Heat Do?
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8362 | https://www.claytonschools.net/cms/lib/MO01000419/Centricity/Domain/204/1_31chapter09.pdf | 659 Chapter 9 π 0 π 3 2 π 2 −4 2 2 4 4 −2 π 0 π 3 2 π 2 −4 2 4 π 0 π 3 2 π 2 −4 2 4 Topics in Analytic Geometry 9.1 Circles and Parabolas 9.2 Ellipses 9.3 Hyperbolas 9.4 Rotation and Systems of Quadratic Equations 9.5 Parametric Equations 9.6 Polar Coordinates 9.7 Graphs of Polar Equations 9.8 Polar Equations of Conics Selected Applications Analytic geometry concepts have many real-life applications. The applications listed below represent a small sample of the applications in this chapter.
■Earthquake, Exercise 35, page 667 ■Suspension Bridge, Exercise 93, page 669 ■Architecture, Exercises 47–49, page 678 ■Satellite Orbit, Exercise 54, page 679 ■Navigation, Exercise 46, page 688 ■Projectile Motion, Exercises 55–58, page 706 ■Planetary Motion, Exercises 49–55, page 727 ■Sports, Exercises 95–98, page 732 Conics are used to represent many real-life phenomena such as reflectors used in flashlights, orbits of planets, and navigation. In Chapter 9, you will learn how to write and graph equations of conics in rectangular and polar coordinates.You will also learn how to graph other polar equations and curves represented by parametric equations.
Satellites are used to monitor weather patterns, collect scientific data, and assist in navigation. Satellites orbit Earth in elliptical paths.
AP/Wide World Photos 9.1 Circles and Parabolas What you should learn Recognize a conic as the intersection of a plane and a double-napped cone.
Write equations of circles in standard form.
Write equations of parabolas in standard form.
Use the reflective property of parabolas to solve real-life problems.
Why you should learn it Parabolas can be used to model and solve many types of real-life problems.For instance, in Exercise 95 on page 669,a parabola is used to design the entrance ramp for a highway.
© Royalty-Free/Corbis Conics Conic sections were discovered during the classical Greek period, 600 to 300 B.C.
The early Greek studies were largely concerned with the geometric properties of conics. It was not until the early 17th century that the broad applicability of conics became apparent and played a prominent role in the early development of calculus.
A conic section (or simply conic) is the intersection of a plane and a double-napped cone. Notice in Figure 9.1 that in the formation of the four basic conics, the intersecting plane does not pass through the vertex of the cone. When the plane does pass through the vertex, the resulting figure is a degenerate conic, as shown in Figure 9.2.
Circle Ellipse Parabola Hyperbola Figure 9.1 Basic Conics Point Line Two intersecting lines Figure 9.2 Degenerate Conics There are several ways to approach the study of conics. You could begin by defining conics in terms of the intersections of planes and cones, as the Greeks did, or you could define them algebraically, in terms of the general second-degree equation However, you will study a third approach, in which each of the conics is defined as a locus (collection) of points satisfying a certain geometric property. For example, the definition of a circle as the collection of all points (x, y) that are equidistant from a fixed point (h, k) leads to the standard equation of a circle Equation of circle x h2 y k2 r2.
Ax2 Bxy Cy2 Dx Ey F 0.
660 Chapter 9 Topics in Analytic Geometry Circles The definition of a circle as a locus of points is a more general definition of a circle as it applies to conics.
Section 9.1 Circles and Parabolas 661 Definition of a Circle A circle is the set of all points in a plane that are equidistant from a fixed point called the center of the circle. (See Figure 9.3.) The dis-tance r between the center and any point on the circle is the radius.
x, y h, k, x, y x y r (h, k) (x, y) Figure 9.3 (1, 4) (−2, −3) −2 −4 −6 −8 2 4 6 8 −2 −4 −6 −8 −10 −12 2 6 x y Figure 9.4 The Distance Formula can be used to obtain an equation of a circle whose center is and whose radius is r.
Distance Formula Square each side. x h2 y k2 r2 x h2 y k2 r h, k Standard Form of the Equation of a Circle The standard form of the equation of a circle is The point is the center of the circle, and the positive number is the radius of the circle. The standard form of the equation of a circle whose center is the origin, is x2 y2 r2.
h, k 0, 0, r h, k x h2 y k2 r2.
Example 1 Finding the Standard Equation of a Circle The point is on a circle whose center is at as shown in Figure 9.4. Write the standard form of the equation of the circle.
Solution The radius of the circle is the distance between and Use Distance Formula.
Simplify.
Simplify.
The equation of the circle with center and radius is Standard form Substitute for h, k, and r.
Simplify.
Now try Exercise 3. x 22 y 32 58. x 22 y 32 58 2 x h2 y k2 r2 r 58 h, k 2, 3 58 32 72 r 1 22 4 32 1, 4.
2, 3 2, 3, 1, 4 662 Chapter 9 Topics in Analytic Geometry Example 2 Sketching a Circle Sketch the circle given by the equation and identify its center and radius.
Solution Begin by writing the equation in standard form.
Write original equation.
Complete the squares.
Write in standard form.
In this form, you can see that the graph is a circle whose center is the point and whose radius is Plot several points that are two units from the center. The points and are convenient. Draw a circle that passes through the four points, as shown in Figure 9.5.
Now try Exercise 23.
3, 1 1, 1, 3, 3, 5, 1, r 4 2.
3, 1 x 32 y 12 4 x2 6x 9 y2 2y 1 6 9 1 x2 6x y2 2y 6 0 x2 6x y2 2y 6 0 Example 3 Finding the Intercepts of a Circle Find the x- and y-intercepts of the graph of the circle given by the equation Solution To find any x-intercepts, let To find any y-intercepts, let x-intercepts: Substitute 0 for y.
Simplify.
Take square root of each side.
Add 4 to each side.
y-intercepts: Substitute 0 for x.
Simplify.
Take square root of each side.
Add 2 to each side.
So the x-intercepts are and and the y-intercept is as shown in Figure 9.6.
Now try Exercise 29.
0, 2, 4 23, 0, 4 23, 0 y 2 y 2 0 y 22 0 0 42 y 22 16 x 4 ± 23 x 4 ±12 x 42 12 x 42 0 22 16 x 0.
y 0.
x 42 y 22 16.
Figure 9.5 −4 −4 14 8 (x − 4)2 + (y − 2)2 = 16 (0, 2) (4 + , 0) 3 2 (4 − , 0) 3 2 Figure 9.6 Prerequisite Skills To use a graphing utility to graph a circle, review Appendix B.2.
Section 9.1 Circles and Parabolas 663 (a) Vertical axis: (b) Vertical axis: (c) Horizontal axis: (d) Horizontal axis: Figure 9.8 p < 0 p > 0 p < 0 p > 0 y k2 4px h y k2 4px h x h2 4py k x h2 4py k p < 0 Directrix: x = h − p Axis: y = k Vertex: (h, k) Focus: (h + p, k) Focus: ( + , ) h p k = x h p − Vertex: ( , ) h k Axis: = y k p > 0 Directrix: Directrix: y = k − p p < 0 Vertex: (h, k) Focus: (h, k + p) = x h Axis: Focus: ( , + ) h k p Directrix: = y k p − Vertex: ( , ) h k p > 0 = x h Axis: Note in Figure 9.7 that a parabola is symmetric with respect to its axis. Using the definition of a parabola, you can derive the following standard form of the equation of a parabola whose directrix is parallel to the x-axis or to the y-axis.
Parabolas In Section 2.1, you learned that the graph of the quadratic function is a parabola that opens upward or downward. The following definition of a parabola is more general in the sense that it is independent of the orientation of the parabola. fx ax2 bx c Definition of a Parabola A parabola is the set of all points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus, not on the line. (See Figure 9.7.) The midpoint between the focus and the directrix is the vertex, and the line passing through the focus and the vertex is the axis of the parabola.
x, y Focus V ertex Directrix d2 d2 d1 d1 Axis x y (x, y) Figure 9.7 Standard Equation of a Parabola (See the proof on page 737.) The standard form of the equation of a parabola with vertex at is as follows.
Vertical axis; directrix: Horizontal axis; directrix: The focus lies on the axis p units (directed distance) from the vertex. If the vertex is at the origin the equation takes one of the following forms.
Vertical axis Horizontal axis See Figure 9.8.
y2 4px x2 4py 0, 0, x h p y k2 4px h, p 0 y k p x h2 4py k, p 0 h, k 664 Chapter 9 Topics in Analytic Geometry Example 5 Finding the Focus of a Parabola Find the focus of the parabola given by Solution To find the focus, convert to standard form by completing the square.
Write original equation.
Multiply each side by Add 1 to each side.
Complete the square.
Combine like terms.
Write in standard form.
Comparing this equation with you can conclude that and Because p is negative, the parabola opens downward, as shown in Figure 9.10. Therefore, the focus of the parabola is Focus Now try Exercise 63.
h, k p 1, 1 2.
p 1 2.
h 1, k 1, x h2 4py k 2y 1 x 12 2 2y x2 2x 1 1 1 2y x2 2x 1 1 2y x2 2x 2. 2y x2 2x 1 y 1 2 x 2 x 1 2 y 1 2 x 2 x 1 2.
Vertex: (0, 0) Focus: (0, 4) −9 9 −2 10 1 16 x2 y = Vertex: (−1, 1) Focus: −1, −3 1 −1 2 y = − x2 − x + 1 2 ( ( 1 2 1 2 Figure 9.10 Figure 9.9 Example 4 Finding the Standard Equation of a Parabola Find the standard form of the equation of the parabola with vertex at the origin and focus Solution Because the axis of the parabola is vertical, passing through and consider the equation Because the focus is units from the vertex, the equation is You can obtain the more common quadratic form as follows.
Write original equation.
Multiply each side by Use a graphing utility to confirm that the graph is a parabola, as shown in Figure 9.9.
Now try Exercise 45.
1 16. 1 16x2 y x2 16y x2 16y.
x2 44y p 4 x2 4py.
0, 4, 0, 0 0, 4.
Reflective Property of Parabolas A line segment that passes through the focus of a parabola and has endpoints on the parabola is called a focal chord. The specific focal chord perpendicular to the axis of the parabola is called the latus rectum.
Parabolas occur in a wide variety of applications. For instance, a parabolic reflector can be formed by revolving a parabola about its axis. The resulting surface has the property that all incoming rays parallel to the axis are reflected through the focus of the parabola. This is the principle behind the construction of the parabolic mirrors used in reflecting telescopes. Conversely, the light rays emanating from the focus of a parabolic reflector used in a flashlight are all parallel to one another, as shown in Figure 9.12.
A line is tangent to a parabola at a point on the parabola if the line intersects, but does not cross, the parabola at the point. Tangent lines to parabolas have special properties related to the use of parabolas in constructing reflective surfaces.
Section 9.1 Circles and Parabolas 665 Example 6 Finding the Standard Equation of a Parabola Find the standard form of the equation of the parabola with vertex and focus at Solution Because the axis of the parabola is horizontal, passing through and consider the equation where and So, the standard form is The parabola is shown in Figure 9.11.
Figure 9.11 Now try Exercise 77.
Vertex: (1, 0) Focus: (2, 0) −1 5 −2 2 y2 = 4(x − 1) y2 4x 1.
y 02 41x 1 p 2 1 1.
h 1, k 0, y k2 4px h 2, 0, 1, 0 2, 0.
1, 0 Parabolic reflector: Light is reflected in parallel rays.
Focus Axis Light source at focus Figure 9.12 Use a graphing utility to confirm the equation found in Example 6. To do this, it helps to graph the equation using two separate equations: (upper part) and (lower part).
Note that when you graph conics using two separate equations, your graphing utility may not connect the two parts. This is because some graphing utilities are limited in their resolution. So, in this text, a blue curve is placed behind the graphing utility’s display to indicate where the graph should appear.
y2 4x 1 y1 4x 1 TECHNOLOGY TIP 666 Chapter 9 Topics in Analytic Geometry Example 7 Finding the Tangent Line at a Point on a Parabola Find the equation of the tangent line to the parabola given by at the point Solution For this parabola, and the focus is as shown in Figure 9.14. You can find the y-intercept of the tangent line by equating the lengths of the two sides of the isosceles triangle shown in Figure 9.14: and Note that rather than The order of subtraction for the distance is important because the distance must be positive. Setting produces So, the slope of the tangent line is and the equation of the tangent line in slope-intercept form is Now try Exercise 85.
y 2x 1.
m 1 1 1 0 2 b 1. 1 4 b 5 4 d1 d2 b 1 4.
d1 1 4 b d2 1 02 1 1 4 2 5 4.
d1 1 4 b 0, b 0, 1 4, p 1 4 1, 1.
y x2 x (1, 1) 0, 1 4 d d 1 2 (0, ) b y x = 2 −1 1 1 α α ( ) y Figure 9.14 Reflective Property of a Parabola The tangent line to a parabola at a point P makes equal angles with the following two lines (see Figure 9.13).
1. The line passing through P and the focus 2. The axis of the parabola Tangent line Focus P α α Axis Figure 9.13 Try using a graphing utility to confirm the result of Example 7. By graphing and in the same viewing window, you should be able to see that the line touches the parabola at the point 1, 1.
y2 2x 1 y1 x2 TECHNOLOGY TIP Section 9.1 Circles and Parabolas 667 In Exercises 1–6, find the standard form of the equation of the circle with the given characteristics.
1. Center at origin; radius: 2. Center at origin; radius: 3. Center: point on circle: 4. Center: point on circle: 5. Center: diameter: 6. Center: diameter: In Exercises 7–12, identify the center and radius of the circle.
7.
8.
9.
10.
11.
12.
In Exercises 13–20, write the equation of the circle in standard form. Then identify its center and radius.
13.
14.
15.
16.
17.
18.
19.
20.
In Exercises 21–28, sketch the circle. Identify its center and radius.
21.
22.
23.
24.
25.
26.
27.
28.
In Exercises 29–34, find the x- and y-intercepts of the graph of the circle.
29.
30.
31.
32.
33.
34.
35. Earthquake An earthquake was felt up to 81 miles from its epicenter. You were located 60 miles west and 45 miles south of the epicenter. (a) Let the epicenter be at the point Find the standard equation that describes the outer boundary of the earthquake.
(b) Would you have felt the earthquake?
(c) Verify your answer to part (b) by graphing the equation of the outer boundary of the earthquake and plotting your location. How far were you from the outer boundary of the earthquake?
36. Landscaper A landscaper has installed a circular sprinkler system that covers an area of 1800 square feet.
(a) Find the radius of the region covered by the sprinkler system. Round your answer to three decimal places.
(b) If the landscaper wants to cover an area of 2400 square feet, how much longer does the radius need to be?
0, 0.
x 72 y 82 4 x 62 y 32 16 x2 8x y2 2y 9 0 x2 2x y2 6y 27 0 x 52 y 42 25 x 22 y 32 9 x2 y2 10y 9 0 x2 2x y2 35 0 x2 6x y2 12y 41 0 x2 14x y2 8y 40 0 x2 6x y2 6y 14 0 x2 4x y2 4y 1 0 y2 81 x2 x2 16 y2 9x2 9y2 54x 36y 17 0 4x2 4y2 12x 24y 41 0 x2 y2 10x 6y 25 0 x2 y2 2x 6y 9 0 9 2x2 9 2y2 1 4 3x2 4 3y2 1 1 9x2 1 9y2 1 1 4x2 1 4y2 1 x2 y 122 24 x 12 y2 15 x 92 y 12 36 x 22 y 72 16 x2 y2 1 x2 y2 49 43 5, 6; 27 3, 1; 2, 4 6, 3; 1, 0 3, 7; 42 18 9.1 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Vocabulary Check Fill in the blanks.
1. A _ is the intersection of a plane and a double-napped cone.
2. A collection of points satisfying a geometric property can also be referred to as a _ of points.
3. A _ is the set of all points in a plane that are equidistant from a fixed point, called the _ .
4. A _ is the set of all points in a plane that are equidistant from a fixed line, called the _ , and a fixed point, called the _ , not on the line.
5. The _ of a parabola is the midpoint between the focus and the directrix.
6. The line that passes through the focus and vertex of a parabola is called the _ of the parabola.
7. A line is _ to a parabola at a point on the parabola if the line intersects, but does not cross, the parabola at the point.
x, y x, y 668 Chapter 9 Topics in Analytic Geometry In Exercises 37–42, match the equation with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) (b ) (c) (d) (e) (f) 37.
38.
39.
40.
41.
42.
In Exercises 43–54, find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin.
43.
44.
45. Focus: 46. Focus: 47. Focus: 48. Focus: 49. Directrix: 50. Directrix: 51. Directrix: 52. Directrix: 53. Horizontal axis and passes through the point 54. Vertical axis and passes through the point In Exercises 55–72, find the vertex, focus, and directrix of the parabola and sketch its graph.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
72.
In Exercises 73–82, find the standard form of the equation of the parabola with the given characteristics.
73.
74.
75. Vertex: focus: 76. Vertex: focus: 77. Vertex: focus: 78. Vertex: focus: 79. Vertex: directrix: 80. Vertex: directrix: 81. Focus: directrix: 82. Focus: directrix: In Exercises 83 and 84, the equations of a parabola and a tangent line to the parabola are given. Use a graphing utility to graph both in the same viewing window. Determine the coordinates of the point of tangency.
Parabola Tangent Line 83.
84.
In Exercises 85–88, find an equation of the tangent line to the parabola at the given point and find the x-intercept of the line.
85.
86.
87.
88.
89. Revenue The revenue R (in dollars) generated by the sale of x 32-inch televisions is modeled by Use a graphing utility to graph the function and approxi-mate the sales that will maximize revenue.
R 375x 3 2x 2.
2, 8 y 2x 2, 1, 2 y 2x 2, 3, 9 2 x 2 2y, 4, 8 x 2 2y, x y 3 0 x 2 12y 0 x y 2 0 y 2 8x 0 y 4 0, 0; x 2 2, 2; x 1 2, 1; y 2 0, 4; 1, 0 1, 2; 3, 2 5, 2; 3, 9 4 3, 3; 3 2, 0 2, 0; −7 8 −2 8 (4.5, 4) (5, 3) −3 9 −6 2 (3, 1) (4, 0) (2, 0) y 2 4x 4 0 y 2 x y 0 x 2 2x 8y 9 0 x2 4x 6y 2 0 x 1 4y2 2y 33 y 1 4x 2 2x 5 x 1 2 2 4y 1 x 3 2 2 4y 2 y 2 4y 4x 0 y 2 6y 8x 25 0 x 5 y 42 0 x 1 2 8y 3 0 x y 2 0 x 2 8y 0 y 2 3x y 2 6x y 4x 2 y 1 2x 2 3, 3 4, 6 x 3 x 2 y 3 y 1 0, 1 2, 0 5 2, 0 0, 3 2 −18 12 −10 10 (−2, 6) −9 9 −3 9 (3, 6) x 32 2y 1 y 12 4x 3 y 2 12x x 2 8y x 2 2y y 2 4x −12 6 −6 6 −8 4 −4 4 −6 6 −6 2 −7 2 −4 2 −6 6 −2 6 −1 8 −2 4 Section 9.1 Circles and Parabolas 669 90. Beam Deflection A simply supported beam is 64 feet long and has a load at the center (see figure). The deflec-tion (bending) of the beam at its center is 1 inch. The shape of the deflected beam is parabolic.
(a) Find an equation of the parabola. (Assume that the origin is at the center of the beam.) (b) How far from the center of the beam is the deflection equal to inch?
91. Automobile Headlight The filament of an automobile headlight is at the focus of a parabolic reflector, which sends light out in a straight beam (see figure).
(a) The filament of the headlight is 1.5 inches from the vertex.
Find an equation for the cross section of the reflector.
(b) The reflector is 8 inches wide. Find the depth of the reflector.
92. Solar Cooker You want to make a solar hot dog cooker using aluminum foil-lined cardboard, shaped as a parabolic trough. The figure shows how to suspend the hot dog with a wire through the foci of the ends of the parabolic trough.
The parabolic end pieces are 12 inches wide and 4 inches deep. How far from the bottom of the trough should the wire be inserted?
93. Suspension Bridge Each cable of the Golden Gate Bridge is suspended (in the shape of a parabola) between two towers that are 1280 meters apart. The top of each tower is 152 meters above the roadway. The cables touch the roadway midway between the towers.
(a) Draw a sketch of the bridge. Locate the origin of a rectangular coordinate system at the center of the road-way. Label the coordinates of the known points.
(b) Write an equation that models the cables.
(c) Complete the table by finding the height of the suspension cables over the roadway at a distance of meters from the center of the bridge.
94. Road Design Roads are often designed with parabolic surfaces to allow rain to drain off. A particular road that is 32 feet wide is 0.4 foot higher in the center than it is on the sides (see figure). (a) Find an equation of the parabola that models the road surface. (Assume that the origin is at the center of the road.) (b) How far from the center of the road is the road surface 0.1 foot lower than in the middle?
95. Highway Design Highway engineers design a parabolic curve for an entrance ramp from a straight street to an interstate highway (see figure). Find an equation of the parabola.
x 400 800 1200 1600 400 800 −400 −800 Street (1000, −800) (1000, 800) Interstate y 32 ft 0.4 ft Not drawn to scale x y 12 in.
4 in.
1.5 in.
8 in.
1 2 1 in.
64 ft Not drawn to scale x 0 200 400 500 600 y 670 Chapter 9 Topics in Analytic Geometry 96. Satellite Orbit A satellite in a 100-mile-high circular orbit around Earth has a velocity of approximately 17,500 miles per hour. If this velocity is multiplied by the satellite will have the minimum velocity necessary to escape Earth’s gravity, and it will follow a parabolic path with the center of Earth as the focus (see figure).
(a) Find the escape velocity of the satellite.
(b) Find an equation of its path (assume the radius of Earth is 4000 miles).
97. Path of a Projectile The path of a softball is modeled by The coordinates x and y are measured in feet, with corresponding to the position from which the ball was thrown.
(a) Use a graphing utility to graph the trajectory of the softball.
(b) Use the zoom and trace features of the graphing utility to approximate the highest point the ball reaches and the distance the ball travels.
98. Projectile Motion Consider the path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet, where the model for the path is In this model, air resistance is disregarded, y is the height (in feet) of the projectile, and x is the horizontal distance (in feet) the projectile travels. A ball is thrown from the top of a 75-foot tower with a velocity of 32 feet per second.
(a) Find the equation of the parabolic path.
(b) How far does the ball travel horizontally before striking the ground?
In Exercises 99–102, find an equation of the tangent line to the circle at the indicated point. Recall from geometry that the tangent line to a circle is perpendicular to the radius of the circle at the point of tangency.
Circle Point 99.
Circle Point 100.
101.
102.
Synthesis True or False?
In Exercises 103–108, determine whether the statement is true or false. Justify your answer.
103. The equation represents a circle with its center at the origin and a radius of 5.
104. The graph of the equation will have x-inter-cepts and y-intercepts 105. A circle is a degenerate conic.
106. It is possible for a parabola to intersect its directrix.
107. The point which lies on the graph of a parabola closest to its focus is the vertex of the parabola.
108. The directrix of the parabola intersects, or is tangent to, the graph of the parabola at its vertex, 109. Writing Cross sections of television antenna dishes are parabolic in shape (see figure). Write a paragraph describing why these dishes are parabolic. Include a graphical representation of your description.
110. Think About It The equation is a degenerate conic. Sketch the graph of this equation and identify the degenerate conic. Describe the intersection of the plane with the double-napped cone for this particular conic.
Think About It In Exercises 111 and 112, change the equation so that its graph matches the description.
111.
upper half of parabola 112.
lower half of parabola Skills Review In Exercises 113–116, use a graphing utility to approximate any relative minimum or maximum values of the function.
113.
114.
115.
116. fx x5 3x 1 fx x4 2x 2 fx 2x2 3x fx 3x3 4x 2 y 12 2x 2; y 32 6x 1; x2 y2 0 Amplifier Dish reflector Cable to radio or TV 0, 0.
x2 y 0, ±r.
±r, 0 x2 y2 r2 x2 y 52 25 25, 2 x2 y2 24 2, 22 x2 y2 12 5, 12 x2 y2 169 3, 4 x2 y2 25 x2 1 16v2y s.
x 0 12.5y 7.125 x 6.252.
Parabolic path t 4100 miles x y Not drawn to scale Circular orbi 2, 9.2 Ellipses What you should learn Write equations of ellipses in standard form.
Use properties of ellipses to model and solve real-life problems.
Find eccentricities of ellipses.
Why you should learn it Ellipses can be used to model and solve many types of real-life problems.For instance, in Exercise 50 on page 678,an ellipse is used to model the floor of Statuary Hall,an elliptical room in the U.S.Capitol Building in Washington,D.C.
© John Neubauer/PhotoEdit Section 9.2 Ellipses 671 (a) (b) Figure 9.15 The line through the foci intersects the ellipse at two points called vertices. The chord joining the vertices is the major axis, and its midpoint is the center of the ellipse. The chord perpendicular to the major axis at the center is the minor axis.
[See Figure 9.15(b).] You can visualize the definition of an ellipse by imagining two thumbtacks placed at the foci, as shown in Figure 9.16. If the ends of a fixed length of string are fastened to the thumbtacks and the string is drawn taut with a pencil, the path traced by the pencil will be an ellipse.
Figure 9.16 To derive the standard form of the equation of an ellipse, consider the ellipse in Figure 9.17 with the following points: center, vertices, foci, Note that the center is the midpoint of the segment joining the foci.
The sum of the distances from any point on the ellipse to the two foci is con-stant. Using a vertex point, this constant sum is Length of major axis or simply the length of the major axis. a c a c 2a h ± c, k.
h ± a, k; h, k; Major axis Minor axis Center Vertex Vertex d d + is constant.
1 2 Focus Focus d d 2 1 ( , ) x y Definition of an Ellipse An ellipse is the set of all points in a plane, the sum of whose distances from two distinct fixed points (foci) is constant. [See Figure 9.15(a).] x, y b a c ( , ) h k b c 2 2 + b c 2 2 + ( , ) x y b2 + c2 = 2a b2 + c2 = a2 2 Figure 9.17 Introduction The third type of conic is called an ellipse. It is defined as follows.
672 Chapter 9 Topics in Analytic Geometry Now, if you let be any point on the ellipse, the sum of the distances between and the two foci must also be That is, Finally, in Figure 9.17, you can see that which implies that the equation of the ellipse is You would obtain a similar equation in the derivation by starting with a vertical major axis. Both results are summarized as follows. x h2 a2 y k2 b2 1. b2x h2 a2y k2 a2b2 b2 a2 c2, x h c2 y k2 x h c2 y k2 2a.
2a.
x, y x, y Figure 9.18 shows both the vertical and horizontal orientations for an ellipse.
Major axis is horizontal.
Major axis is vertical.
Figure 9.18 2a 2b ( , ) h k x y + = 1 (x − h)2 b2 (y − k)2 a2 + = 1 2a 2b ( , ) h k x (x − h)2 a2 (y − k)2 b2 y Standard Equation of an Ellipse The standard form of the equation of an ellipse with center and major and minor axes of lengths and respectively, where is Major axis is horizontal.
Major axis is vertical.
The foci lie on the major axis, c units from the center, with If the center is at the origin the equation takes one of the following forms.
Major axis is horizontal.
Major axis is vertical.
x2 b2 y2 a2 1 x2 a2 y2 b2 1 0, 0, c2 a2 b2.
x h2 b2 y k2 a2 1.
x h2 a2 y k2 b2 1 0 < b < a, 2b, 2a h, k Exploration On page 671 it was noted that an ellipse can be drawn using two thumbtacks, a string of fixed length (greater than the distance between the two tacks), and a pencil. Try doing this.
Vary the length of the string and the distance between the thumbtacks. Explain how to obtain ellipses that are almost circular. Explain how to obtain ellipses that are long and narrow.
Section 9.2 Ellipses 673 Example 1 Finding the Standard Equation of an Ellipse Find the standard form of the equation of the ellipse having foci at and and a major axis of length 6, as shown in Figure 9.19.
Solution By the Midpoint Formula, the center of the ellipse is and the distance from the center to one of the foci is Because you know that Now, from you have Because the major axis is horizontal, the standard equation is Now try Exercise 35.
x 22 32 y 12 5 2 1. b a2 c2 9 4 5.
c2 a2 b2, a 3.
2a 6, c 2.
2, 1) 4, 1 0, 1 Example 2 Sketching an Ellipse Sketch the ellipse given by and identify the center and vertices.
4x2 y2 36 Algebraic Solution Write original equation.
Divide each side by 36.
Write in standard form.
The center of the ellipse is Because the denominator of the -term is larger than the denominator of the -term, you can conclude that the major axis is vertical. Moreover, because the vertices are and Finally, because the endpoints of the minor axis are and as shown in Figure 9.20.
Figure 9.20 Now try Exercise 13.
3, 0, 3, 0 b 3, 0, 6.
0, 6 a 6, x2 y2 0, 0. x2 32 y2 62 1 4x2 36 y2 36 36 36 4x2 y2 36 Graphical Solution Solve the equation of the ellipse for y as follows.
Then use a graphing utility to graph and in the same viewing window. Be sure to use a square setting. From the graph in Figure 9.21, you can see that the major axis is vertical and its center is at the point You can use the zoom and trace features to approximate the vertices to be and Figure 9.21 −8 −12 12 8 y1 = 36 − 4x2 y2 = − 36 − 4x2 0, 6.
0, 6 0, 0.
y2 36 4x2 36 4x2 y1 y ±36 4x2 y2 36 4x2 4x2 y2 36 (0, 1) (2, 1) (4, 1) a = 3 b = 5 −3 6 −2 4 Figure 9.19 For instructions on how to use the zoom and trace features, see Appendix A; for specific keystrokes, go to this textbook’s Online Study Center.
TECHNOLOGY SUPPORT 674 Chapter 9 Topics in Analytic Geometry Example 3 Graphing an Ellipse Graph the ellipse given by Solution Begin by writing the original equation in standard form. In the third step, note that 9 and 4 are added to both sides of the equation when completing the squares.
Write original equation.
Write in standard form.
Now you see that the center is Because the denominator of the x-term is the endpoints of the major axis lie two units to the right and left of the center. Similarly, because the denominator of the y-term is the endpoints of the minor axis lie one unit up and down from the center. The graph of this ellipse is shown in Figure 9.22.
Now try Exercise 15.
b2 12, a2 22, h, k 3, 1. x 32 22 y 12 12 1 x 32 4y 12 4 x2 6x 9 4y2 2y 1 9 9 41 x2 6x 4y2 2y 9 x2 4y 2 6x 8y 9 0 x2 4y2 6x 8y 9 0.
Example 4 Analyzing an Ellipse Find the center, vertices, and foci of the ellipse Solution By completing the square, you can write the original equation in standard form.
Write original equation.
Write in standard form.
So, the major axis is vertical, where and Therefore, you have the following.
Center: Vertices: Foci: The graph of the ellipse is shown in Figure 9.23.
Now try Exercise 17.
1, 2 23 1, 2 1, 2 23 1, 6 1, 2 c a2 b2 16 4 12 23.
h 1, k 2, a 4, b 2, x 12 22 y 22 42 1 4x 12 y 22 16 4x2 2x 1 y2 4y 4 8 41 4 4x2 2x y2 4y 8 4x2 y2 8x 4y 8 0 4x2 y 2 8x 4y 8 0.
(−3, 2) (−3, 1) (−3, 0) (−5, 1) (−1, 1) −6 0 −1 3 (x + 3)2 22 (y − 1)2 12 + = 1 Figure 9.22 Vertex Focus Focus Center Vertex 9 3 (y + 2)2 42 (x − 1)2 22 + = 1 Figure 9.23 Write in completed square form.
You can use a graphing utility to graph an ellipse by graphing the upper and lower portions in the same viewing window. For instance, to graph the ellipse in Example 3, first solve for y to obtain and Use a viewing window in which and You should obtain the graph shown in Figure 9.22.
1 ≤y ≤3.
6 ≤x ≤0 y2 1 1 x 32 4 .
y1 1 1 x 32 4 T E C H N O L O G Y T I P Group terms and factor 4 out of y-terms.
Group terms and factor 4 out of x-terms.
Write in completed square form.
Section 9.2 Ellipses 675 Application Ellipses have many practical and aesthetic uses. For instance, machine gears, supporting arches, and acoustic designs often involve elliptical shapes. The orbits of satellites and planets are also ellipses. Example 5 investigates the elliptical orbit of the moon about Earth. Example 5 An Application Involving an Elliptical Orbit The moon travels about Earth in an elliptical orbit with Earth at one focus, as shown in Figure 9.24. The major and minor axes of the orbit have lengths of 768,800 kilometers and 767,640 kilometers, respectively. Find the greatest and smallest distances (the apogee and perigee) from Earth’s center to the moon’s center.
Figure 9.24 Solution Because and you have and which implies that So, the greatest distance between the center of Earth and the center of the moon is kilometers and the smallest distance is kilometers.
Now try Exercise 53. 363,292 a c 384,400 21,108 405,508 a c 384,400 21,108 21,108. 384,4002 383,8202 c a2 b2 383,820 b a 384,400 2b 767,640, 2a 768,800 Perigee Moon Apogee Earth 768,800 km 767,640 km STUDY TIP Note in Example 5 and Figure 9.24 that Earth is not the center of the moon’s orbit.
676 Chapter 9 Topics in Analytic Geometry Eccentricity One of the reasons it was difficult for early astronomers to detect that the orbits of the planets are ellipses is that the foci of the planetary orbits are relatively close to their centers, and so the orbits are nearly circular. To measure the ovalness of an ellipse, you can use the concept of eccentricity.
Note that for every ellipse.
To see how this ratio is used to describe the shape of an ellipse, note that because the foci of an ellipse are located along the major axis between the ver-tices and the center, it follows that For an ellipse that is nearly circular, the foci are close to the center and the ratio is small [see Figure 9.25(a)]. On the other hand, for an elongated ellipse, the foci are close to the vertices and the ratio is close to 1 [see Figure 9.25(b)].
(a) (b) Figure 9.25 The orbit of the moon has an eccentricity of and the eccentricities of the eight planetary orbits are as follows.
Mercury: Jupiter: Venus: Saturn: Earth: Uranus: Mars: Neptune: e 0.0086 e 0.0934 e 0.0472 e 0.0167 e 0.0542 e 0.0068 e 0.0484 e 0.2056 e 0.0549, Foci e is small.
a c e = c a x y ca ca 0 < c < a.
0 < e < 1 Definition of Eccentricity The eccentricity e of an ellipse is given by the ratio e c a.
Foci a c x y e is close to 1.
e = c a Section 9.2 Ellipses 677 In Exercises 1–6, match the equation with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) (b) (c) (d) (e) (f) 1.
2.
3.
4.
5.
6.
In Exercises 7–12, find the center, vertices, foci, and eccen-tricity of the ellipse, and sketch its graph. Use a graphing utility to verify your graph.
7.
8.
9.
10.
11.
12.
In Exercises 13–22, (a) find the standard form of the equation of the ellipse, (b) find the center, vertices, foci, and eccentricity of the ellipse, (c) sketch the ellipse, and use a graphing utility to verify your graph.
13.
14.
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16.
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18.
19.
20.
21.
22.
In Exercises 23–30, find the standard form of the equation of the ellipse with the given characteristics and center at the origin.
23.
24.
25. Vertices: foci: 26. Vertices: foci: 27. Foci: major axis of length 8 28. Foci: major axis of length 12 29. Vertices: passes through the point 30. Vertical major axis; passes through points and 2, 0 0, 4 4, 2 0, ±5; ±2, 0; ±3, 0; 0, ±4 0, ±8; ±2, 0 ±3, 0; −6 6 −4 4 0, − (2, 0) (−2, 0) 3 2) ) 0, 3 2) ) −9 9 −6 6 (0, 4) (0, −4) (2, 0) (−2, 0) 36x 2 9y 2 48x 36y 43 0 12x 2 20y 2 12x 40y 37 0 9x 2 25y 2 36x 50y 61 0 16x 2 25y 2 32x 50y 16 0 x2 4y2 6x 20y 2 0 6x2 2y2 18x 10y 2 0 9x 2 4y 2 54x 40y 37 0 9x 2 4y 2 36x 24y 36 0 16x2 y2 16 x2 9y2 36 x 22 y 42 1 4 1 x 52 9 4 y 12 1 x 32 12 y 22 16 1 x 42 16 y 12 25 1 x2 16 y2 81 1 x2 64 y2 9 1 x 2 2 9 y 22 4 1 x 2 2 16 y 1 2 1 x 2 4 y2 1 x 2 4 y 2 25 1 x 2 9 y 2 4 1 x 2 4 y 2 9 1 −3 3 −2 2 −10 5 −7 3 −9 9 −6 6 −6 6 −4 4 −6 6 −4 4 −4 8 −5 3 9.2 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Vocabulary Check Fill in the blanks.
1. An _ is the set of all points in a plane, the sum of whose distances from two distinct fixed points is constant.
2. The chord joining the vertices of an ellipse is called the _ , and its midpoint is the _ of the ellipse.
3. The chord perpendicular to the major axis at the center of an ellipse is called the _ of the ellipse.
4. You can use the concept of _ to measure the ovalness of an ellipse.
(x, y) 678 Chapter 9 Topics in Analytic Geometry In Exercises 31–40, find the standard form of the equation of the ellipse with the given characteristics.
31.
32.
33. Vertices: minor axis of length 2 34. Foci: major axis of length 6 35. Foci: major axis of length 36 36. Center: vertex: minor axis of length 2 37. Vertices: minor axis of length 6 38. Center: foci: 39. Center: vertices: 40. Vertices: endpoints of the minor axis: In Exercises 41–44, find the eccentricity of the ellipse.
41.
42.
43.
44.
45. Find an equation of the ellipse with vertices and eccentricity 46. Find an equation of the ellipse with vertices and eccentricity 47. Architecture A semielliptical arch over a tunnel for a road through a mountain has a major axis of 100 feet and a height at the center of 40 feet.
(a) Draw a rectangular coordinate system on a sketch of the tunnel with the center of the road entering the tun-nel at the origin. Identify the coordinates of the known points.
(b) Find an equation of the semielliptical arch over the tun-nel.
(c) Determine the height of the arch 5 feet from the edge of the tunnel.
48. Architecture A semielliptical arch through a railroad underpass has a major axis of 32 feet and a height at the center of 12 feet.
(a) Draw a rectangular coordinate system on a sketch of the underpass with the center of the road entering the underpass at the origin. Identify the coordinates of the known points.
(b) Find an equation of the semielliptical arch over the underpass.
(c) Will a truck that is 10 feet wide and 9 feet tall be able to drive through the underpass without crossing the center line? Explain your reasoning.
49. Architecture A fireplace arch is to be constructed in the shape of a semiellipse. The opening is to have a height of 2 feet at the center and a width of 6 feet along the base (see figure). The contractor draws the outline of the ellipse on the wall by the method discussed on page 671. Give the required positions of the tacks and the length of the string.
50. Statuary Hall Statuary Hall is an elliptical room in the United States Capitol Building in Washington, D.C. The room is also referred to as the Whispering Gallery because a person standing at one focus of the room can hear even a whisper spoken by a person standing at the other focus.
Given that the dimensions of Statuary Hall are 46 feet wide by 97 feet long, find an equation for the shape of the floor surface of the hall. Determine the distance between the foci.
51. Geometry The area of the ellipse in the figure is twice the area of the circle. What is the length of the major axis? Hint: The area of an ellipse is given by 52. Astronomy Halley’s comet has an elliptical orbit with the sun at one focus. The eccentricity of the orbit is approxi-mately 0.97. The length of the major axis of the orbit is about 35.88 astronomical units. (An astronomical unit is about 93 million miles.) Find the standard form of the equation of the orbit. Place the center of the orbit at the ori-gin and place the major axis on the x-axis.
53. Astronomy The comet Encke has an elliptical orbit with the sun at one focus. Encke’s orbit ranges from 0.34 to 4.08 astronomical units from the sun. Find the standard form of the equation of the orbit. Place the center of the orbit at the origin and place the major axis on the x-axis.
x (0, 10) (0, −10) (−a, 0) (a, 0) y A ab. 1 −1 1 2 3 −2 −3 x y e 1 2.
0, ±8 e 4 5.
±5, 0 4x2 3y2 8x 18y 19 0 x2 9y2 10x 36y 52 0 x2 25 y2 36 1 x2 4 y2 9 1 0, 6, 10, 6 5, 0, 5, 12; 4, 4, 4, 4 0, 4; a 2c; 1, 2, 5, 2 3, 2; a 3c; 3, 1, 3, 9; 2, 1 2; 2, 1; 0, 0, 0, 8; 0, 0, 4, 0; 0, 2, 8, 2; −2 7 −4 2 (4, −1) (2, 0) (0, −1) (2, −2) −4 8 −1 7 (3, 3) (2, 6) (1, 3) (2, 0) Section 9.2 Ellipses 679 54. Satellite Orbit The first artificial satellite to orbit Earth was Sputnik I (launched by the former Soviet Union in 1957). Its highest point above Earth’s surface was 947 kilo-meters, and its lowest point was 228 kilometers. The center of Earth was a focus of the elliptical orbit, and the radius of Earth is 6378 kilometers (see figure). Find the eccentricity of the orbit.
55. Geometry A line segment through a focus with endpoints on an ellipse, perpendicular to the major axis, is called a latus rectum of the ellipse. Therefore, an ellipse has two latera recta. Knowing the length of the latera recta is helpful in sketching an ellipse because this information yields other points on the curve (see figure). Show that the length of each latus rectum is In Exercises 56–59, sketch the ellipse using the latera recta (see Exercise 55).
56.
57.
58.
59.
60. Writing Write an equation of an ellipse in standard form and graph it on paper. Do not write the equation on your graph. Exchange graphs with another student. Use the graph you receive to reconstruct the equation of the ellipse it represents and find its eccentricity. Compare your results and write a short paragraph discussing your findings.
Synthesis True or False?
In Exercises 61 and 62, determine whether the statement is true or false. Justify your answer.
61. It is easier to distinguish the graph of an ellipse from the graph of a circle if the eccentricity of the ellipse is large (close to 1).
62. The area of a circle with diameter is greater than the area of an ellipse with major axis 63. Think About It At the beginning of this section it was noted that an ellipse can be drawn using two thumbtacks, a string of fixed length (greater than the distance between the two tacks), and a pencil (see Figure 9.16). If the ends of the string are fastened at the tacks and the string is drawn taut with a pencil, the path traced by the pencil is an ellipse.
(a) What is the length of the string in terms of a?
(b) Explain why the path is an ellipse.
64. Exploration Consider the ellipse (a) The area of the ellipse is given by Write the area of the ellipse as a function of a.
(b) Find the equation of an ellipse with an area of 264 square centimeters.
(c) Complete the table using your equation from part (a) and make a conjecture about the shape of the ellipse with a maximum area.
(d) Use a graphing utility to graph the area function to support your conjecture in part (c).
65. Think About It Find the equation of an ellipse such that for any point on the ellipse, the sum of the distances from the point and is 36.
66. Proof Show that for the ellipse where and the distance from the center of the ellipse to a focus is c.
Skills Review In Exercises 67–70, determine whether the sequence is arithmetic, geometric, or neither.
67. 66, 55, 44, 33, 22, . . .
68. 80, 40, 20, 10, 5, . . .
69.
70.
In Exercises 71–74, find the sum.
71.
72.
73.
74. 10 n0 5 4 3 n 10 n1 4 3 4 n1 6 n0 3n 6 n0 3n 1 2, 1 2, 3 2, 5 2, 7 2, . . .
1 4, 1 2, 1, 2, 4, . . .
0, 0 b > 0, a > 0, x2 a2 y2 b2 1 a2 b2 c2 10, 2 2, 2 A ab.
a b 20.
x 2 a2 y 2 b2 1, 2a 8.
d 2r 8 5x 2 3y 2 15 9x 2 4y 2 36 x 2 9 y 2 16 1 x 2 4 y 2 1 1 F 1 F 2 Latera recta x y 2b2a.
228 km 947 km Focus a 8 9 10 11 12 13 A 9.3 Hyperbolas What you should learn Write equations of hyperbolas in standard form.
Find asymptotes of and graph hyperbolas.
Use properties of hyperbolas to solve real-life problems.
Classify conics from their general equations.
Why you should learn it Hyperbolas can be used to model and solve many types of real-life problems.For instance, in Exercise 44 on page 688,hyperbolas are used to locate the position of an explosion that was recorded by three listening stations.
James Foote/Photo Researchers, Inc.
680 Chapter 9 Topics in Analytic Geometry Introduction The definition of a hyperbola is similar to that of an ellipse. The difference is that for an ellipse, the sum of the distances between the foci and a point on the ellipse is constant; whereas for a hyperbola, the difference of the distances between the foci and a point on the hyperbola is constant.
(a) (b) Figure 9.26 The graph of a hyperbola has two disconnected parts called the branches. The line through the two foci intersects the hyperbola at two points called the vertices.
The line segment connecting the vertices is the transverse axis, and the midpoint of the transverse axis is the center of the hyperbola [see Figure 9.26(b)]. The development of the standard form of the equation of a hyperbola is similar to that of an ellipse. Note that a, b, and c are related differently for hyperbolas than for ellipses. For a hyperbola, the distance between the foci and the center is greater than the distance between the vertices and the center.
Transverse axis V ertex Vertex a c Branch Center Branch Focus Focus d d 2 1 ( , ) x y d d 2 1 − is a positive constant.
Definition of a Hyperbola A hyperbola is the set of all points in a plane, the difference of whose distances from two distinct fixed points, the foci, is a positive constant. [See Figure 9.26(a).] x, y Standard Equation of a Hyperbola The standard form of the equation of a hyperbola with center at is Transverse axis is horizontal.
Transverse axis is vertical.
The vertices are a units from the center, and the foci are c units from the center. Moreover, If the center of the hyperbola is at the origin the equation takes one of the following forms.
y2 a2 x2 b2 1 x2 a2 y2 b2 1 0, 0, c2 a2 b2.
y k2 a2 x h2 b2 1.
x h2 a2 y k2 b2 1 h, k Transverse axis is horizontal.
Transverse axis is vertical.
Section 9.3 Hyperbolas 681 Figure 9.27 shows both the horizontal and vertical orientations for a hyperbola.
Transverse axis is horizontal.
Transverse axis is vertical.
Figure 9.27 ( , ) h k c − ( , + ) h k c ( , ) h k ( ) y k − ( ) x h − a b 2 2 2 2 − = 1 x y Transverse axis ( ) x h − ( ) y k − a b 2 2 2 2 − = 1 ( + , ) h c k ( , ) h c k − ( , ) h k Transverse axis x y Example 1 Finding the Standard Equation of a Hyperbola Find the standard form of the equation of the hyperbola with foci and and vertices and Solution By the Midpoint Formula, the center of the hyperbola occurs at the point Furthermore, and and it follows that So, the hyperbola has a horizontal transverse axis and the standard form of the equation of the hyperbola is Figure 9.28 shows the hyperbola.
Figure 9.28 Now try Exercise 33.
(−1, 2) (0, 2) (5, 2) (4, 2) −4 8 −2 6 (x − 2)2 22 (y − 2)2 − = 1 ( ( 5 2 x 22 22 y 22 52 1. 5. 9 4 32 22 b c2 a2 a 2, c 3 2, 2.
4, 2.
0, 2 5, 2 1, 2 When using a graphing utility to graph an equation, you must solve the equation for y before entering it into the graphing utility. When graphing equations of conics, it can be difficult to solve for y, which is why it is very important to know the algebra used to solve equations for y.
T E C H N O L O G Y T I P 682 Chapter 9 Topics in Analytic Geometry Asymptotes of a Hyperbola Each hyperbola has two asymptotes that intersect at the center of the hyperbola.
The asymptotes pass through the corners of a rectangle of dimensions by with its center at as shown in Figure 9.29.
The conjugate axis of a hyperbola is the line segment of length joining and if the transverse axis is horizontal, and the line segment of length joining and if the transverse axis is vertical.
h b, k h b, k 2b h, k b h, k b 2b h, k, 2b, 2a ( , ) h k ( , ) h a k − ( , ) h a k + ( , + ) h k b ( , ) h k b − Asymptote Conjugate axis Asymptote Figure 9.29 Algebraic Solution Write original equation.
Divide each side by 16.
Write in standard form.
Because the -term is positive, you can conclude that the transverse axis is horizontal. So, the vertices occur at and the endpoints of the conjugate axis occur at and and you can sketch the rectangle shown in Figure 9.30. Finally, by drawing the asymptotes, and through the corners of this rectan-gle, you can complete the sketch, as shown in Figure 9.31.
Figure 9.30 Figure 9.31 Now try Exercise 15.
y 2x, y 2x 0, 4, 0, 4 2, 0, 2, 0 x2 x2 22 y 2 42 1 4x2 16 y 2 16 16 16 4x2 y 2 16 Graphical Solution Solve the equation of the hyperbola for y as follows.
Then use a graphing utility to graph and in the same viewing window. Be sure to use a square setting. From the graph in Figure 9.32, you can see that the transverse axis is horizontal. You can use the zoom and trace features to approx-imate the vertices to be and Figure 9.32 −6 −9 9 6 y1 = 4x2 − 16 y2 = − 4x2 − 16 2, 0.
2, 0 y2 4x2 16 4x2 16 y1 ±4x2 16 y 4x2 16 y2 4x2 y 2 16 Asymptotes of a Hyperbola y k ± a bx h y k ± b ax h Asymptotes for vertical transverse axis Asymptotes for horizontal transverse axis Example 2 Sketching a Hyperbola Sketch the hyperbola whose equation is 4x2 y2 16.
Example 3 Finding the Asymptotes of a Hyperbola Sketch the hyperbola given by and find the equations of its asymptotes.
Solution Write original equation.
Subtract 16 from each side and factor.
Complete the square.
Write in completed square form.
Write in standard form.
From this equation you can conclude that the hyperbola has a vertical transverse axis, is centered at has vertices and and has a conjugate axis with endpoints and To sketch the hyperbola, draw a rectangle through these four points. The asymptotes are the lines passing through the corners of the rectangle, as shown in Figure 9.33.
Finally, using and you can conclude that the equations of the asymptotes are and Now try Exercise 19.
y 2 3x 1.
y 2 3x 1 b 3, a 2 1 3, 0.
1 3, 0 1, 2, 1, 2 1, 0, y2 22 x 12 3 2 1 4x 12 3y2 12 4x2 2x 1 3y2 16 41 4x2 2x 3y2 16 4x2 3y 2 8x 16 0 4x2 3y 2 8x 16 0 Figure 9.33 You can use a graphing utility to graph a hyperbola by graphing the upper and lower portions in the same viewing window. For instance, to graph the hyperbola in Example 3, first solve for y to obtain and Use a viewing window in which and You should obtain the graph shown in Figure 9.34. Notice that the graphing utility does not draw the asymptotes. However, if you trace along the branches, you will see that the values of the hyperbola approach the asymptotes.
Figure 9.34 −9 −6 9 y1 = 2 1 + (x + 1)2 3 y2 = −2 1 + (x + 1)2 3 6 6 ≤y ≤6.
9 ≤x ≤9 y2 21 x 12 3 .
y1 21 x 12 3 TECHNOLOGY TIP Section 9.3 Hyperbolas 683 684 Chapter 9 Topics in Analytic Geometry As with ellipses, the eccentricity of a hyperbola is Eccentricity and because it follows that If the eccentricity is large, the branches of the hyperbola are nearly flat, as shown in Figure 9.36(a). If the eccentricity is close to 1, the branches of the hyperbola are more pointed, as shown in Figure 9.36(b).
(a) (b) Figure 9.36 e = c a e is close to 1.
V ertex Focus a c x y e is large.
Vertex Focus e = c a c a x y e > 1.
c > a e c a Example 4 Using Asymptotes to Find the Standard Equation Find the standard form of the equation of the hyperbola having vertices and and having asymptotes and as shown in Figure 9.35.
Solution By the Midpoint Formula, the center of the hyperbola is Furthermore, the hyperbola has a vertical transverse axis with From the original equations, you can determine the slopes of the asymptotes to be and and because you can conclude that So, the standard form of the equation is Now try Exercise 39.
y 22 32 x 32 3 2 2 1.
b 3 2.
a 3, m2 2 a b m1 2 a b a 3.
3, 2.
y 2x 4 y 2x 8 3, 1 3, 5 10 −5 −7 (3, 1) (3, −5) 3 y = 2x − 8 y = −2x + 4 Figure 9.35 Applications The following application was developed during World War II. It shows how the properties of hyperbolas can be used in radar and other detection systems.
Sun p Elliptical orbit Parabolic orbit Vertex Hyperbolic orbit 2200 2200 c a − c a − 2 = 5280 2200 + 2( ) = 5280 c c a − 2000 2000 3000 A B x y Figure 9.37 Section 9.3 Hyperbolas 685 Figure 9.38 Example 5 An Application Involving Hyperbolas Two microphones, 1 mile apart, record an explosion. Microphone A receives the sound 2 seconds before microphone B. Where did the explosion occur?
Solution Assuming sound travels at 1100 feet per second, you know that the explosion took place 2200 feet farther from B than from A, as shown in Figure 9.37. The locus of all points that are 2200 feet closer to A than to B is one branch of the hyperbola where and So, and you can conclude that the explosion occurred somewhere on the right branch of the hyperbola Now try Exercise 43.
Another interesting application of conic sections involves the orbits of comets in our solar system. Of the 610 comets identified prior to 1970, 245 have elliptical orbits, 295 have parabolic orbits, and 70 have hyperbolic orbits. The center of the sun is a focus of each of these orbits, and each orbit has a vertex at the point where the comet is closest to the sun, as shown in Figure 9.38.
Undoubtedly, there are many comets with parabolic or hyperbolic orbits that have not been identified. You get to see such comets only once. Comets with elliptical orbits, such as Halley’s comet, are the only ones that remain in our solar system.
If p is the distance between the vertex and the focus in meters, and v is the velocity of the comet at the vertex in meters per second, then the type of orbit is determined as follows.
1. Ellipse: 2. Parabola: 3. Hyperbola: In each of these equations, kilograms (the mass of the sun) and cubic meter per kilogram-second squared (the universal gravitational constant).
G 6.67 1011 M 1.989 1030 v > 2GMp v 2GMp v < 2GMp x2 1,210,000 y2 5,759,600 1.
b2 c2 a2 26402 11002 5,759,600, a 2200 2 1100.
c 5280 2 2640 x2 a2 y2 b2 1 686 Chapter 9 Topics in Analytic Geometry General Equations of Conics The test above is valid if the graph is a conic. The test does not apply to equations such as whose graphs are not conics.
x2 y2 1, Classifying a Conic from Its General Equation The graph of is one of the following.
1. Circle: 2. Parabola: or but not both.
3. Ellipse: A and C have like signs.
4. Hyperbola: A and C have unlike signs.
AC < 0 AC > 0 C 0, A 0 AC 0 A 0 A C Ax2 Bxy Cy2 Dx Ey F 0 STUDY TIP Notice in Example 6(a) that there is no -term in the equation. Therefore, C 0.
y2 Example 6 Classifying Conics from General Equations Classify the graph of each equation.
a.
b.
c.
d.
Solution a. For the equation you have Parabola So, the graph is a parabola.
b. For the equation you have Hyperbola So, the graph is a hyperbola.
c. For the equation you have Ellipse So, the graph is an ellipse.
d. For the equation you have Circle So, the graph is a circle.
Now try Exercise 49.
A C 2.
2x2 2y2 8x 12y 2 0, AC 24 > 0.
2x2 4y2 4x 12y 0, AC 41 < 0.
4x2 y2 8x 6y 4 0, AC 40 0.
4x2 9x y 5 0, 2x2 2y2 8x 12y 2 0 2x2 4y2 4x 12y 0 4x2 y2 8x 6y 4 0 4x2 9x y 5 0 Section 9.3 Hyperbolas 687 In Exercises 1–4, match the equation with its graph. [The graphs are labeled (a), (b), (c), and (d).] (a) (b) (c) (d) 1.
2.
3.
4.
In Exercises 5–14, find the center, vertices, foci, and asymp-totes of the hyperbola, and sketch its graph using the asymptotes as an aid. Use a graphing utility to verify your graph.
5.
6.
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In Exercises 15–24, (a) find the standard form of the equation of the hyperbola, (b) find the center, vertices, foci, and asymptotes of the hyperbola, (c) sketch the hyperbola, and use a graphing utility to verify your graph.
15.
16.
17.
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In Exercises 25–30, find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
25. Vertices: foci: 26. Vertices: foci: 27. Vertices: asymptotes: 28. Vertices: asymptotes: 29. Foci: asymptotes: 30. Foci: asymptotes: y ±3 4x ±10, 0; y ±4x 0, ±8; y ±3x 0, ±3; y ±5x ±1, 0; ±6, 0 ±3, 0; 0, ±4 0, ±2; 9x 2 y 2 54x 10y 55 0 9y 2 x 2 2x 54y 62 0 16y 2 x 2 2x 64y 63 0 x 2 9y 2 2x 54y 80 0 x 2 9y 2 36y 72 0 9x 2 y 2 36x 6y 18 0 6y 2 3x 2 18 2x 2 3y 2 6 25x2 4y2 100 4x2 9y2 36 y 12 1 4 x 32 1 16 1 y 52 1 9 x 12 1 4 1 x 32 144 y 22 25 1 x 12 4 y 22 1 1 x 2 36 y 2 4 1 y 2 25 x 2 81 1 y 2 9 x 2 1 1 y 2 1 x 2 4 1 x 2 9 y 2 25 1 x 2 y 2 1 x 12 16 y 22 9 1 x 12 16 y 2 4 1 y 2 25 x 2 9 1 y 2 9 x 2 25 1 8 −10 −4 8 12 −12 −8 8 9 −9 −6 6 10 −8 −6 6 9.3 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Vocabulary Check Fill in the blanks.
1. A _ is the set of all points in a plane, the difference of whose distances from two distinct fixed points is a positive constant.
2. The graph of a hyperbola has two disconnected parts called _ .
3. The line segment connecting the vertices of a hyperbola is called the _ , and the midpoint of the line segment is the _ of the hyperbola.
4. Each hyperbola has two _ that intersect at the center of the hyperbola.
5. The general form of the equation of a conic is given by _ .
x, y 688 Chapter 9 Topics in Analytic Geometry In Exercises 31–42, find the standard form of the equation of the hyperbola with the given characteristics.
31. Vertices: foci: 32. Vertices: foci: 33. Vertices: foci: 34. Vertices: foci: 35. Vertices: passes through the point 36. Vertices: passes through the point 37. Vertices: passes through the point 38. Vertices: passes through the point 39. Vertices: asymptotes: 40. Vertices: asymptotes: 41. Vertices: asymptotes: 42. Vertices: asymptotes: 43. Sound Location You and a friend live 4 miles apart (on the same “east-west” street) and are talking on the phone.
You hear a clap of thunder from lightning in a storm, and 18 seconds later your friend hears the thunder. Find an equation that gives the possible places where the lightning could have occurred. (Assume that the coordinate system is measured in feet and that sound travels at 1100 feet per second.) 44. Sound Location Three listening stations located at and monitor an explo-sion. The last two stations detect the explosion 1 second and 4 seconds after the first, respectively. Determine the coordinates of the explosion. (Assume that the coordinate system is measured in feet and that sound travels at 1100 feet per second.) 45. Pendulum The base for a pendulum of a clock has the shape of a hyperbola (see figure).
(a) Write an equation of the cross section of the base.
(b) Each unit in the coordinate plane represents foot. Find the width of the base of the pendulum 4 inches from the bottom.
46. Navigation Long distance radio navigation for aircraft and ships uses synchronized pulses transmitted by widely separated transmitting stations. These pulses travel at the speed of light (186,000 miles per second). The difference in the times of arrival of these pulses at an aircraft or ship is constant on a hyperbola having the transmitting stations as foci. Assume that two stations, 300 miles apart, are positioned on a rectangular coordinate system at coordi-nates and and that a ship is traveling on a hyperbolic path with coordinates (see figure).
(a) Find the x-coordinate of the position of the ship if the time difference between the pulses from the transmit-ting stations is 1000 microseconds (0.001 second).
(b) Determine the distance between the ship and station 1 when the ship reaches the shore.
(c) The captain of the ship wants to enter a bay located between the two stations. The bay is 30 miles from station 1. What should be the time difference between the pulses?
(d) The ship is 60 miles offshore when the time difference in part (c) is obtained. What is the position of the ship?
47. Hyperbolic Mirror A hyperbolic mirror (used in some telescopes) has the property that a light ray directed at a focus will be reflected to the other focus. The focus of a hyperbolic mirror (see figure) has coordinates Find the vertex of the mirror if the mount at the top edge of the mirror has coordinates (24, 0) ( 24, 0) − (24, 24) x y 24, 24.
24, 0.
Station 2 Station 1 Bay 150 50 −50 −150 100 50 −50 y x Not drawn to scale x, 75 150, 0, 150, 0 1 2 −4 −8 4 8 −4 4 x y (2, 9) (1, 0) (−1, 0) (2, −9) (−2, 9) (−2, −9) 3300, 0 3300, 0, 3300, 1100, y 2 3x, y 4 2 3x 3, 0, 3, 4; y 2 3x, y 4 2 3x 0, 2, 6, 2; y x y x 6, 3, 0, 3, 6; y 4 x y x, 1, 2, 3, 2; 0, 5 1, 2, 1, 2; 5, 1 0, 4, 0, 0; 5, 4 2, 1, 2, 1; 0, 5 2, 3, 2, 3; 3, 1, 3, 1 2, 1, 2, 1); 4, 0, 4, 10 4, 1, 4, 9; 2, 5, 2, 5 2, 3, 2, 3; 0, 0, 8, 0 2, 0, 6, 0; 48. Panoramic Photo A panoramic photo can be taken using a hyperbolic mirror. The camera is pointed toward the vertex of the mirror and the camera’s optical center is posi-tioned at one focus of the mirror (see figure). An equation for the cross-section of the mirror is Find the distance from the camera’s optical center to the mirror.
In Exercises 49–58, classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola.
49.
50.
51.
52.
53.
54.
55.
56.
57.
58.
Synthesis True or False?
In Exercises 59–62, determine whether the statement is true or false. Justify your answer.
59. In the standard form of the equation of a hyperbola, the larger the ratio of b to a, the larger the eccentricity of the hyperbola.
60. In the standard form of the equation of a hyperbola, the trivial solution of two intersecting lines occurs when 61. If and then the graph of is a hyperbola.
62. If the asymptotes of the hyperbola where intersect at right angles, then 63. Think About It Consider a hyperbola centered at the origin with a horizontal transverse axis. Use the definition of a hyperbola to derive its standard form.
64. Writing Explain how the central rectangle of a hyperbola can be used to sketch its asymptotes.
65. Use the figure to show that 66. Think About It Find the equation of the hyperbola for any point on which, the difference between its distances from the points and is 6.
67. Proof Show that for the equation of the hyperbola where the distance from the center of the hyperbola to a focus is c.
68. Proof Prove that the graph of the equation is one of the following (except in degenerate cases).
Conic Condition (a) Circle (b) Parabola or (but not both) (c) Ellipse (d) Hyperbola Skills Review In Exercises 69–72, perform the indicated operation.
69. Subtract: 70. Multiply: 71. Divide: 72. Expand: In Exercises 73–78, factor the polynomial completely.
73.
74.
75.
76.
77.
78. 4 x 4x2 x3 16x3 54 6x3 11x2 10x 2x3 24x2 72x x2 14x 49 x3 16x x y 3 2 x3 3x 4 x 2 3x 1 2x 4 x3 3x2 6 2x 4x2 AC < 0 AC > 0 C 0 A 0 A C Ax2 Cy2 Dx Ey F 0 0, 0 x2 a2 y2 b2 1 c2 a2 b2 10, 2 2, 2 (−c, 0) (c, 0) (a, 0) (−a, 0) x ( , ) x y y d2 d1 d2 d1 2a.
a b.
b > 0, a, x 2a2 y2b2 1, Ey 0 x2 y2 Dx E 0, D 0 b 0.
9x2 4y2 90x 8y 228 0 x2 6x 2y 7 0 y2 x2 2x 6y 8 0 x2 y2 2x 6y 0 4x2 25y2 16x 250y 541 0 y2 12x 4y 28 0 x2 4x 8y 20 0 16x2 9y2 32x 54y 209 0 x2 y2 4x 6y 23 0 9x2 4y2 18x 16y 119 0 x y Mirror Optical Center y2 25 x2 16 1.
Section 9.3 Hyperbolas 689 690 Chapter 9 Topics in Analytic Geometry 9.4 Rotation and Systems of Quadratic Equations What you should learn Rotate the coordinate axes to eliminate the xy-term in equations of conics.
Use the discriminant to classify conics.
Solve systems of quadratic equations.
Why you should learn it As illustrated in Exercises 3–14 on page 697, rotation of the coordinate axes can help you identify the graph of a general second-degree equation.
Rotation In the preceding section, you learned that the equation of a conic with axes parallel to the coordinate axes has a standard form that can be written in the general form Horizontal or vertical axes In this section, you will study the equations of conics whose axes are rotated so that they are not parallel to either the x-axis or the y-axis. The general equation for such conics contains an xy-term.
Equation in xy-plane To eliminate this -term, you can use a procedure called rotation of axes. The objective is to rotate the - and -axes until they are parallel to the axes of the conic. The rotated axes are denoted as the -axis and the -axis, as shown in Figure 9.39.
Figure 9.39 After the rotation, the equation of the conic in the new -plane will have the form Equation in -plane Because this equation has no xy-term, you can obtain a standard form by completing the square. The following theorem identifies how much to rotate the axes to eliminate the xy-term and also the equations for determining the new coef-ficients and F.
A, C, D, E, xy Ax2 Cy2 Dx Ey F 0.
xy x x′ y′ θ y y x y x xy Ax2 Bxy Cy2 Dx Ey F 0 Ax2 Cy2 Dx Ey F 0.
Rotation of Axes to Eliminate an xy-Term (See the proof on page 738.) The general second-degree equation can be rewritten as by rotating the coordinate axes through an angle where The coefficients of the new equation are obtained by making the substitutions and y x sin y cos .
x x cos y sin cot 2 A C B .
, Ax2 Cy2 Dx Ey F 0 Ax2 Bxy Cy2 Dx Ey F 0 Section 9.4 Rotation and Systems of Quadratic Equations 691 x′ y′ ( ) x′ 2 ( ) y′ 2 − = 1 −1 −2 2 1 2 1 −1 xy −1 = 0 2 2 ( ( 2 2 ( ( x y Vertices: In -system: In xy-system: Figure 9.40 1, 1, 1, 1 2, 0, 2, 0 xy STUDY TIP Remember that the substitutions and were developed to eliminate the -term in the rotated system.
You can use this as a check on your work. In other words, if your final equation contains an -term, you know that you made a mistake.
xy xy y x sin y cos x x cos y sin Example 1 Rotation of Axes for a Hyperbola Rotate the axes to eliminate the xy-term in the equation Then write the equation in standard form and sketch its graph.
Solution Because and you have which implies that and The equation in the -system is obtained by substituting these expressions into the equation Write in standard form.
In the -system, this is a hyperbola centered at the origin with vertices at as shown in Figure 9.40. To find the coordinates of the vertices in the xy-system, substitute the coordinates into the equations and This substitution yields the vertices and in the xy-system. Note also that the asymptotes of the hyperbola have equations which correspond to the original x- and y-axes.
Now try Exercise 3.
y ±x, 1, 1 1, 1 y x y 2 .
x x y 2 ±2, 0 ±2, 0, xy x2 22 y2 22 1 x2 y2 2 1 0 x y 2 x y 2 1 0 xy 1 0.
xy x y 2 . x 1 2 y 1 2 y x sin 4 y cos 4 x y 2 x 1 2 y 1 2 x x cos 4 y sin 4 4 2 2 cot 2 A C B 0 C 0, A 0, B 1, xy 1 0.
692 Chapter 9 Topics in Analytic Geometry x′ y′ ( ) x′ 2 ( ) y′ 2 + = 1 2 1 −1 −2 2 −1 −2 7 6 3 + 13 16 = 0 x xy y 2 2 − − 22 12 x y Vertices: In -system: In xy-system: Figure 9.41 1 2, 3 2 , 1 2, 3 2 3, 1, 3, 1, ±2, 0, 0, ±1 xy Example 2 Rotation of Axes for an Ellipse Rotate the axes to eliminate the xy-term in the equation Then write the equation in standard form and sketch its graph.
Solution Because and you have which implies that The equation in the -system is obtained by making the substitutions and into the original equation. So, you have which simplifies to Write in standard form.
This is the equation of an ellipse centered at the origin with vertices in the -system, as shown in Figure 9.41.
Now try Exercise 11.
xy ±2, 0 x2 22 y2 12 1. 4x2 16y2 16 4x2 16y2 16 0 x 3y 2 x 1 2 y3 2 y x sin 6 y cos 6 3x y 2 x3 2 y 1 2 x x cos 6 y sin 6 xy 6.
cot 2 A C B 7 13 63 1 3 C 13, A 7, B 63, 7x2 63 xy 13y2 16 0.
Prerequisite Skills To review conics, see Sections 9.1–9.3. 73x y 2 2 633x y 2 x 3y 2 13 x 3y 2 2 16 0 7x2 63xy 13y2 16 0 1 2 5 θ Figure 9.42 x′ y′ 2 1 −1 −2 ≈ ° 26.6 θ x y (y′ + 1)2 = (−1) x′ − 4 5) ) x2 − 4xy + 4y2 + 5 5y + 1 = 0 Vertex: In -system: In xy-system: Figure 9.43 13 55 , 6 55 4 5, 1 xy Example 3 Rotation of Axes for a Parabola Rotate the axes to eliminate the xy-term in the equation Then write the equation in standard form and sketch its graph.
Solution Because and you have Using the identity produces from which you obtain the equation Considering you have So, From the triangle in Figure 9.42, you obtain and So, you use the substitutions Substituting these expressions into the original equation, you have which simplifies as follows.
Group terms.
Write in completed square form.
Write in standard form.
The graph of this equation is a parabola with vertex at in the -plane.
Its axis is parallel to the -axis in the -system, as shown in Figure 9.43.
Now try Exercise 13.
xy x xy 4 5, 1 y 12 1x 4 5 5y 12 5x 4 5y2 2y 5x 1 5y2 5x 10y 1 0 y x sin y cos x 1 5 y 2 5 x 2y 5 .
x x cos y sin x 2 5 y 1 5 2x y 5 cos 25.
sin 15 26.6.
cot 2 2 cot 4.
0 < < 2, 2 cot 42 cot 1 0. 4 cot2 6 cot 4 0 4 cot2 4 6 cot cot 2 3 4 cot2 1 2 cot cot 2 cot2 12 cot cot 2 A C B 1 4 4 3 4.
C 4, A 1, B 4, x2 4xy 4y2 55y 1 0.
Section 9.4 Rotation and Systems of Quadratic Equations 693 2x y 5 2 4 2x y 5 x 2y 5 4 x 2y 5 2 55 x 2y 5 1 0 694 Chapter 9 Topics in Analytic Geometry Invariants Under Rotation In the rotation of axes theorem listed at the beginning of this section, note that the constant term is the same in both equations—that is, Such quantities are invariant under rotation. The next theorem lists some other rotation invariants.
You can use the results of this theorem to classify the graph of a second-degree equation with an xy-term in much the same way that you classify the graph of a second-degree equation without an xy-term. Note that because the invariant reduces to Discriminant This quantity is called the discriminant of the equation Now, from the classification procedure given in Section 9.3, you know that the sign of determines the type of graph for the equation Consequently, the sign of will determine the type of graph for the original equation, as shown in the following classification.
For example, in the general equation you have and So, the discriminant is Because the graph of the equation is an ellipse or a circle.
11 < 0, B2 4AC 72 435 49 60 11.
C 5.
B 7, A 3, 3x2 7xy 5y2 6x 7y 15 0, B2 4AC Ax2 Cy2 Dx Ey F 0.
AC Ax2 Bxy Cy2 Dx Ey F 0.
B2 4AC 4AC.
B2 4AC B 0, F F.
Rotation Invariants The rotation of the coordinate axes through an angle that transforms the equation into the form has the following rotation invariants.
1.
2.
3. B2 4AC B2 4AC A C A C F F Ax2 Cy2 Dx Ey F 0 Ax2 Bxy Cy2 Dx Ey F 0 Classification of Conics by the Discriminant The graph of the equation is, except in degenerate cases, determined by its discriminant as follows.
1. Ellipse or circle: 2. Parabola: 3. Hyperbola: B2 4AC > 0 B2 4AC 0 B2 4AC < 0 Ax2 Bxy Cy2 Dx Ey F 0 −1 −1 3 5 0 0 6 4 −15 −10 15 10 Figure 9.46 Figure 9.45 Figure 9.44 Example 4 Rotations and Graphing Utilities For each equation, classify the graph of the equation, use the Quadratic Formula to solve for y, and then use a graphing utility to graph the equation.
a.
b.
c.
Solution a. Because the graph is a circle or an ellipse. Solve for y as follows.
Write original equation.
Quadratic form Graph both of the equations to obtain the ellipse shown in Figure 9.44.
Top half of ellipse Bottom half of ellipse b. Because the graph is a parabola.
Write original equation.
Quadratic form Graph both of the equations to obtain the parabola shown in Figure 9.45.
c. Because the graph is a hyperbola.
Write original equation.
Quadratic form Graph both of the equations to obtain the hyperbola shown in Figure 9.46.
Now try Exercise 27. y 2x ± x2 7 2 y 8x ± 8x2 443x2 7 24 ay2 by c 0 4y2 8xy 3x2 7 0 3x2 8xy 4y2 7 0 B2 4AC 64 48 > 0, y 3x 1 ± 23x 4 9 y 6x 2 ± 6x 22 49x2 1 29 ay2 by c 0 9y2 6x 2y x2 1 0 x2 6xy 9y2 2y 1 0 B2 4AC 36 36 0, y2 3x x16 7x 4 y1 3x x16 7x 4 y 3x ± x16 7x 4 y 3x ± 3x2 422x2 2x 22 ay2 by c 0 2y2 3xy 2x2 2x 0 2x2 3xy 2y2 2x 0 B2 4AC 9 16 < 0, 3x2 8xy 4y2 7 0 x2 6xy 9y2 2y 1 0 2x2 3xy 2y2 2x 0 Section 9.4 Rotation and Systems of Quadratic Equations 695 696 Chapter 9 Topics in Analytic Geometry Systems of Quadratic Equations To find the points of intersection of two conics, you can use elimination or substitution, as demonstrated in Examples 5 and 6.
−4 −3 5 3 (0,1) (2, 0) x2 + 4y − 4 = 0 x2 + 4y2 − 4x − 8y + 4 = 0 Figure 9.48 Example 5 Solving a Quadratic System by Elimination Solve the system of quadratic equations x2 x2 y2 y2 16x 39 9 0 0.
Algebraic Solution You can eliminate the -term by adding the two equations. The resulting equation can then be solved for x.
There are two real solutions: and The corresponding y-values are and So, the solutions of the system are and Now try Exercise 41.
5, 4.
5, 4, 3, 0, y ±4.
y 0 x 5.
x 3 2x 3x 5 0 2x2 16x 30 0 y2 Graphical Solution Begin by solving each equation for y as follows.
Use a graphing utility to graph all four equations and in the same viewing window. Use the intersect feature of the graphing utility to approximate the points of intersection to be and as shown in Figure 9.47.
Figure 9.47 y2 = − −x2 + 16x − 39 y1 = −x2 + 16x − 39 y3 = x2 − 9 y4 = − x2 − 9 −10 −8 14 8 5, 4, 3, 0, 5, 4, y4 x2 9 y3 x2 9, y2 x2 16x 39, y1 x2 16x 39, y ±x2 9 y ±x2 16x 39 Equation 1 Equation 2 Example 6 Solving a Quadratic System by Substitution Solve the system of quadratic equations Solution Because Equation 2 has no -term, solve the equation for y tobtain Next, substitute this into Equation 1 and solve for x.
In factored form, you can see that the equation has two real solutions: and The corresponding values of y are and This implies that the solutions of the system of equations are and as shown in Figure 9.48.
Now try Exercise 47.
2, 0, 0, 1 y 0.
y 1 x 2.
x 0 xx 2x2 2x 8 0 x4 4x2 16x 0 x2 4 2x2 1 4x4 4x 8 2x2 4 0 x2 41 1 4x22 4x 81 1 4x2 4 0 y 1 1 4x2.
y2 x2 4y2 4x x2 8y 4y 4 4 0 0.
For instructions on how to use the intersect feature, see Appendix A; for specific keystrokes, go to this textbook’s Online Study Center.
TECHNOLOGY SUPPORT Equation 1 Equation 2 Section 9.4 Rotation and Systems of Quadratic Equations 697 In Exercises 1 and 2, the -coordinate system has been rotated degrees from the xy-coordinate system. The coor-dinates of a point in the xy-coordinate system are given.
Find the coordinates of the point in the rotated coordinate system.
1.
2.
In Exercises 3–14, rotate the axes to eliminate the xy-term in the equation. Then write the equation in stan-dard form. Sketch the graph of the resulting equation, showing both sets of axes.
3.
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In Exercises 15–20, use a graphing utility to graph the conic. Determine the angle through which the axes are rotated. Explain how you used the graphing utility to obtain the graph.
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In Exercises 21–26, match the graph with its equation. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) (b) (c) (d) (e) (f) 21.
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In Exercises 27–34, (a) use the discriminant to classify the graph of the equation, (b) use the Quadratic Formula to solve for y, and (c) use a graphing utility to graph the equation.
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32. 36x 2 60xy 25y 2 9y 0 x 2 6xy 5y 2 4x 22 0 2x 2 4xy 5y 2 3x 4y 20 0 15x 2 8xy 7y 2 45 0 x 2 4xy 2y 2 6 0 16x 2 24xy 9y 2 30x 40y 0 x 2 4xy 4y 2 10x 30 0 3x 2 2xy y 2 10 0 x 2 xy 3y 2 5 0 2x 2 3xy 2y 2 3 0 x 2 2xy y 2 0 xy 4 0 6 −6 −4 4 6 −6 −4 4 6 −6 −4 4 6 −6 −4 4 6 −6 −4 4 6 −6 −4 4 613 8y 91 4x 2 12xy 9y 2 413 12x 32x 2 48xy 8y 2 50 40x 2 36xy 25y 2 52 17x 2 32xy 7y 2 75 x 2 4xy 2y 2 8 x 2 3xy y 2 20 9x 2 24xy 16y 2 80x 60y 0 9x 2 24xy 16y 2 90x 130y 0 16x 2 24xy 9y 2 60x 80y 100 0 3x 2 23xy y 2 2x 23y 0 13x 2 63xy 7y 2 16 0 5x 2 6xy 5y 2 12 0 2x 2 3xy 2y 2 10 0 xy 2y 4x 0 xy x 2y 3 0 x 2 4xy y 2 1 0 xy 2 0 xy 1 0 45, 3, 3 90, 0, 3 xy 9.4 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Vocabulary Check Fill in the blanks.
1. The procedure used to eliminate the xy-term in a general second-degree equation is called _ of _ .
2. Quantities that are equal in both the original equation of a conic and the equation of the rotated conic are _ .
3. The quantity is called the _ of the equation Ax2 Bxy Cy2 Dx Ey F 0.
B2 4AC 698 Chapter 9 Topics in Analytic Geometry 33.
34.
In Exercises 35–38, sketch (if possible) the graph of the degenerate conic.
35.
36.
37.
38.
In Exercises 39–46, solve the system of quadratic equations algebraically by the method of elimination. Then verify your results by using a graphing utility to graph the equations and find any points of intersection of the graphs.
39.
40.
41.
42.
43.
44.
45.
46.
In Exercises 47–52, solve the system of quadratic equations algebraically by the method of substitution. Then verify your results by using a graphing utility to graph the equations and find any points of intersection of the graphs.
47.
48.
49.
50.
51.
52.
Synthesis True or False?
In Exercises 53 and 54, determine whether the statement is true or false. Justify your answer.
53. The graph of where is any constant less than is a hyperbola.
54. After using a rotation of axes to eliminate the xy-term from an equation of the form the coefficients of the - and -terms remain A and C, respectively.
Skills Review In Exercises 55–58, sketch the graph of the rational func-tion. Identify all intercepts and asymptotes.
55.
56.
57.
58.
In Exercises 59–62, if possible, find (a) AB, (b) BA, and (c) 59.
60.
61.
62.
In Exercises 63–70, graph the function.
63.
64.
65.
66.
67.
68.
69.
70.
In Exercises 71–74, find the area of the triangle.
71.
72.
73.
74. a 23, b 35, c 27 a 11, b 18, c 10 B 70, a 25, c 16 C 110, a 8, b 12 ft 2 t 3 ft t 5 1 ht 1 2t 43 ht t 23 3 gx 3x 2 gx 4 x2 fx x 4 1 fx x 3 B 1 4 6 0 5 3 3 1 2 A 0 1 3 2 1 4 0 5 0, B 3 4 5 A 4 2 5 , B 3 3 2 8 A 1 0 5 2, B 0 5 6 1 A 1 2 3 5, A 2.
gs 2 4 s 2 ht t 2 2 t f x 2x 2 x gx 2 2 x y2 x2 Cy2 Dx Ey F 0 Ax2 Bxy 1 4, k 6x 10 0, x2 xy ky2 5x2 2xy 5y2 12 0 x y 1 0 xy x x2 2y 4y2 3 9 0 0 4x2 y2 8x 6y 9 0 2x2 3y2 4x 18y 43 0 10x2 25y2 100x 160 0 y2 2x 16 0 6x2 3y2 12 0 x y 2 0 2x2 y2 6 0 2x y 0 16x2 y2 16y 128 0 y2 48x 16y 32 0 16x2 y2 24y 80 0 16x2 25y2 400 0 x2 4y2 x2 2x 2x 8y 4y 1 1 0 0 x2 x2 y2 y2 12x 12x 16y 16y 64 64 0 0 x2 4y2 20x 64y 172 0 16x2 4y2 320x 64y 1600 0 4x2 y2 16x 24y 16 0 4x2 y2 40x 24y 208 0 4x2 9y2 36y 0 x2 y2 27 0 x2 y2 3x 4 y2 0 0 x 2 10xy y 2 0 x 2 2xy y 2 1 0 x 2 y 2 2x 6y 10 0 y 2 16x 2 0 x 2 xy 4y 2 x y 4 0 x 2 4xy 4y 2 5x y 3 0 9.5 Parametric Equations Section 9.5 Parametric Equations 699 Plane Curves Up to this point, you have been representing a graph by a single equation involving two variables such as and In this section, you will study situations in which it is useful to introduce a third variable to represent a curve in the plane.
To see the usefulness of this procedure, consider the path of an object that is propelled into the air at an angle of . If the initial velocity of the object is 48 feet per second, it can be shown that the object follows the parabolic path Rectangular equation as shown in Figure 9.49. However, this equation does not tell the whole story.
Although it does tell you where the object has been, it doesn’t tell you when the object was at a given point on the path. To determine this time, you can introduce a third variable called a parameter. It is possible to write both and as functions of to obtain the parametric equations Parametric equation for x Parametric equation for y From this set of equations you can determine that at time the object is at the point Similarly, at time the object is at the point and so on.
Curvilinear motion: two variables for position, one variable for time Figure 9.49 For this particular motion problem, and are continuous functions of and the resulting path is a plane curve. (Recall that a continuous function is one whose graph can be traced without lifting the pencil from the paper.) t, y x Rectangular equation: y x = + − Parametric equations: x t = 24 2 y t t = 16 + 24 2 − 2 (36, 18) (0, 0) (72, 0) 9 9 18 18 27 36 45 54 63 72 t = 0 x y x2 72 t = 3 2 2 t = 3 2 4 242 16, 242, t 1, 0, 0.
t 0, y 16t2 242t.
x 242t t y x t, x, y y x2 72 x 45 y.
x Definition of a Plane Curve If f and g are continuous functions of t on an interval l, the set of ordered pairs is a plane curve . The equations given by and are parametric equations for , and t is the parameter.
C y gt x ft C f t, gt What you should learn Evaluate sets of parametric equations for given values of the parameter.
Graph curves that are represented by sets of parametric equations.
Rewrite sets of parametric equations as single rectangular equations by eliminating the parameter.
Find sets of parametric equations for graphs.
Why you should learn it Parametric equations are useful for modeling the path of an object.For instance,in Exercise 57 on page 706,a set of parametric equations is used to model the path of a football.
Elsa/Getty Images Sketching a Plane Curve One way to sketch a curve represented by a pair of parametric equations is to plot points in the -plane. Each set of coordinates is determined from a value chosen for the parameter By plotting the resulting points in the order of increasing values of you trace the curve in a specific direction. This is called the orientation of the curve.
t, t.
x, y xy 700 Chapter 9 Topics in Analytic Geometry Figure 9.50 Figure 9.51 Example 1 Sketching a Plane Curve Sketch the curve given by the parametric equations and Describe the orientation of the curve.
Solution Using values of t in the interval, the parametric equations yield the points shown in the table.
By plotting these points in the order of increasing t, you obtain the curve shown in Figure 9.50. The arrows on the curve indicate its orientation as t increases from to 3. So, if a particle were moving on this curve, it would start at and then move along the curve to the point Now try Exercises 7(a) and (b).
Note that the graph shown in Figure 9.50 does not define y as a function of x. This points out one benefit of parametric equations—they can be used to rep-resent graphs that are more general than graphs of functions.
Two different sets of parametric equations can have the same graph. For example, the set of parametric equations and has the same graph as the set given in Example 1. However, by comparing the values of t in Figures 9.50 and 9.51, you can see that this second graph is traced out more rapidly (considering t as time) than the first graph. So, in applications, different parametric representations can be used to represent various speeds at which objects travel along a given path.
1 ≤t ≤3 2 y t, x 4t 2 4 5, 3 2.
0, 1 2 x, y 2 ≤t ≤3.
y t 2, x t 2 4 Most graphing utilities have a parametric mode. So, another way to display a curve represented by a pair of parametric equations is to use a graphing utility, as shown in Example 2. For instructions on how to use the parametric mode, see Appendix A; for specifice keystrokes, go to this textbook’s Online Study Center.
TECHNOLOGY TIP t 2 1 0 1 2 3 x 0 3 4 3 0 5 y 1 1 2 0 1 2 1 3 2 Section 9.5 Parametric Equations 701 Example 2 Using a Graphing Utility in Parametric Mode Use a graphing utility to graph the curves represented by the parametric equations. Using the graph and the Vertical Line Test, for which curve is y a func-tion of x?
a.
b.
c.
Solution Begin by setting the graphing utility to parametric mode. When choosing a viewing window, you must set not only minimum and maximum values of and y, but also minimum and maximum values of t.
a. Enter the parametric equations for x and y, as shown in Figure 9.52. Use the viewing window shown in Figure 9.53. The curve is shown in Figure 9.54.
From the graph, you can see that y is not a function of x.
Figure 9.52 Figure 9.53 Figure 9.54 b. Enter the parametric equations for x and y, as shown in Figure 9.55. Use the viewing window shown in Figure 9.56. The curve is shown in Figure 9.57.
From the graph, you can see that y is a function of x.
Figure 9.55 Figure 9.56 Figure 9.57 c. Enter the parametric equations for x and y, as shown in Figure 9.58. Use the viewing window shown in Figure 9.59. The curve is shown in Figure 9.60.
From the graph, you can see that y is not a function of x.
Figure 9.58 Figure 9.59 Figure 9.60 Now try Exercise 7(c).
x y t x t2, y t3 x t, y t3 x t2, Exploration Use a graphing utility set in parametric mode to graph the curve and Set the viewing window so that and Now, graph the curve with various settings for Use the following.
a.
b.
c.
Compare the curves given by the different settings. Repeat this experiment using How does this change the results?
x t.
t 3 ≤t ≤3 3 ≤t ≤0 0 ≤t ≤3 t.
12 ≤y ≤2.
4 ≤x ≤4 y 1 t2 x t Notice in Example 2 that in order to set the viewing windows of parametric graphs, you have to scroll down to enter the Ymax and Yscl values.
T E C H N O L O G Y T I P Prerequisite Skills See Section 1.3 to review the Vertical Line Test.
Eliminating the Parameter Many curves that are represented by sets of parametric equations have graphs that can also be represented by rectangular equations (in and ). The process of finding the rectangular equation is called eliminating the parameter.
Now you can recognize that the equation represents a parabola with a horizontal axis and vertex at When converting equations from parametric to rectangular form, you may need to alter the domain of the rectangular equation so that its graph matches the graph of the parametric equations. This situation is demonstrated in Example 3.
4, 0.
x 4y2 4 y 1 2t x 4y 2 4 x 2y2 4 t 2y x t 2 4 Rectangular equation Substitute in second equation.
Solve for t in one equation.
Parametric equations y x 702 Chapter 9 Topics in Analytic Geometry Example 3 Eliminating the Parameter Identify the curve represented by the equations and Solution Solving for t in the equation for x produces or which implies that Substituting in the equation for y, you obtain the rectangular equation From the rectangular equation, you can recognize that the curve is a parabola that opens downward and has its vertex at as shown in Figure 9.61. The rectangular equation is defined for all values of x. The parametric equation for x, however, is defined only when From the graph of the parametric equations, you can see that x is always positive, as shown in Figure 9.62. So, you should restrict the domain of x to positive values, as shown in Figure 9.63.
Now try Exercise 7(d).
t > 1.
0, 1, 1 x2 x2 1 x2 x2 x2 1 x2. 1 x2 1 1 x2 1 1 y t t 1 t 1x2 1.
1 x2 t 1 x2 1 t 1 y t t 1.
x 1 t 1 STUDY TIP It is important to realize that eliminating the parameter is primarily an aid to curve sketching. If the parametric equations represent the path of a moving object, the graph alone is not sufficient to describe the object’s motion. You still need the parametric equations to determine the position, direction, and speed at a given time.
−4 −4 4 y = 1 − x2 2 −4 −4 4 Parametric equations: x = , y = t + 1 t + 1 t 1 t = 3 t = 0 t = −0.75 2 −4 −4 4 Rectangular equation: y = 1 − x2, x > 0 2 Figure 9.61 Figure 9.62 Figure 9.63 Finding Parametric Equations for a Graph How can you determine a set of parametric equations for a given graph or a given physical description? From the discussion following Example 1, you know that such a representation is not unique. This is further demonstrated in Example 5.
Section 9.5 Parametric Equations 703 −4 −4 4 x = t y = 1 − t2 2 t = −2 t = −1 t = 0 t = 1 t = 2 Figure 9.65 Example 5 Finding Parametric Equations for a Given Graph Find a set of parametric equations to represent the graph of using the parameters (a) and (b) Solution a. Letting you obtain the following parametric equations.
Parametric equation for x Parametric equation for y The graph of these equations is shown in Figure 9.65.
b. Letting you obtain the following parametric equations.
Parametric equation for x Parametric equation for y The graph of these equations is shown in Figure 9.66. Note that the graphs in Figures 9.65 and 9.66 have opposite orientations.
Now try Exercise 45. y 1 1 t2 2t t2 x 1 t t 1 x, y 1 t2 x t t x, t 1 x.
t x y 1 x2 Example 4 Eliminating the Parameter Sketch the curve represented by and by eliminating the parameter.
Solution Begin by solving for and in the equations.
and Solve for Use the identity to form an equation involving only x and y.
Pythagorean identity Substitute for for Rectangular equation From this rectangular equation, you can see that the graph is an ellipse centered at with vertices and and minor axis of length as shown in Figure 9.64. Note that the elliptic curve is traced out counterclockwise.
Now try Exercise 23.
2b 6, 0, 4, 0, 4 0, 0, x2 9 y2 16 1 sin .
cos and y 4 x 3 x 3 2 y 4 2 1 cos2 sin2 1 sin2 cos2 1 cos and sin . sin y 4 cos x 3 sin cos 0 ≤ ≤2, y 4 sin , x 3 cos Exploration In Example 4, you make use of the trigonometric identity to sketch an ellipse. Which trigonometric identity would you use to obtain the graph of a hyperbola?
Sketch the curve represented by and by eliminating the parameter.
0 ≤ ≤2, y 4 tan , x 3 sec sin2 cos2 1 Figure 9.64 θ = 0 θ = π θ = π 2 3 θ = π 2 −7 8 −5 5 θ θ x = 3 cos y = 4 sin −4 −4 4 x = 1 − t y = 2t − t2 2 t = 3 t = 2 t = 1 t = 0 t = −1 Figure 9.66 704 Chapter 9 Topics in Analytic Geometry In Exercises 1–6, match the set of parametric equations with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) (b) (c) (d) (e) (f) 1.
2.
3.
4.
5.
6.
7. Consider the parametric equations and (a) Create a table of and values using 1, 2, 3, and 4.
(b) Plot the points generated in part (a) and sketch a graph of the parametric equations.
(c) Use a graphing utility to graph the curve represented by the parametric equations.
(d) Find the rectangular equation by eliminating the param-eter. Sketch its graph. How does the graph differ from those in parts (b) and (c)?
8. Consider the parametric equations and (a) Create a table of and values using and (b) Plot the points generated in part (a) and sketch a graph of the parametric equations.
(c) Use a graphing utility to graph the curve represented by the parametric equations.
(d) Find the rectangular equation by eliminating the parameter. Sketch its graph. How does the graph differ from those in parts (b) and (c)?
Library of Parent Functions In Exercises 9 and 10, deter-mine the plane curve whose graph is shown.
9. (a) (b) (c) (d) 10. (a) (b) (c) (d) y 2 sin x 3 cos y 3 sin x 2 cos y 2 sin x 3 cos y 3 sin −1 1 2 3 4 −1 1 2 3 4 x y x 2 cos y t2 x 2t 1 y t2 x 2t 1 y t2 x 2t 1 y 2t 1 x y −4 −8 4 8 12 −4 4 8 12 16 x t2 x, y 2.
4, 0, 4, 2, y-x-y 4 sin .
x 4 cos2 x, y t 0, y-x-y 2 t.
x t y et x 2t, y 1 2 t 2 x ln t, y t 2 x 1 t, y t x t, y t 2 x t 2, y t 2 x t, −4 −6 6 4 −1 −7 5 7 −6 −1 11 2 −3 −7 5 5 −4 −6 6 4 −3 −6 6 5 9.5 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Vocabulary Check Fill in the blanks.
1. If f and g are continuous functions of t on an interval I, the set of ordered pairs is a _ C.
The equations given by and are _ for C, and t is the _ .
2. The _ of a curve is the direction in which the curve is traced out for increasing values of the parameter.
3. The process of converting a set of parametric equations to rectangular form is called _ the _ .
y gt x ft ft, gt Section 9.5 Parametric Equations 705 In Exercises 11–26, sketch the curve represented by the parametric equations (indicate the orientation of the curve).
Use a graphing utility to confirm your result. Then eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve.
Adjust the domain of the resulting rectangular equation, if necessary.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
In Exercises 27–32, use a graphing utility to graph the curve represented by the parametric equations.
27.
28.
29.
30.
31.
32.
In Exercises 33 and 34, determine how the plane curves differ from each other.
33. (a) (b) (c) (d) 34. (a) (b) (c) (d) In Exercises 35–38, eliminate the parameter and obtain the standard form of the rectangular equation.
35. Line through and 36. Circle: 37. Ellipse: 38. Hyperbola: In Exercises 39–42, use the results of Exercises 37–40 to find a set of parametric equations for the line or conic.
39. Line: passes through 40. Circle: center: radius: 4 41. Ellipse: vertices: foci: 42. Hyperbola: vertices: foci: In Exercises 43–48, find two different sets of parametric equations for the given rectangular equation. (There are many correct answers.) 43.
44.
45.
46.
47.
48.
In Exercises 49 and 50, use a graphing utility to graph the curve represented by the parametric equations.
49. Witch of Agnesi: 50. Folium of Descartes: In Exercises 51–54, match the parametric equations with the correct graph. [The graphs are labeled (a), (b), (c), and (d).] (a) (b) (c) (d) 51. Lissajous curve: 52. Evolute of ellipse: 53. Involute of circle: 54. Serpentine curve: x 1 2 cot , y 4 sin cos y 1 2sin cos x 1 2cos sin x 2 cos3 , y 4 sin3 x 2 cos , y sin 2 −2 2 −4 2 −5 5 − −2 2 − 3 2 y 3t2 1 t3 x 3t 1 t3, y 2 sin2 x 2 cot , y x 3 2x y 6x2 5 y 1 2x y 1 x y 4 7x y 5x 3 0, ±2 0, ±1; ±4, 0 ±5, 0; 2, 5; 1, 4 and 6, 3 y k b tan x h a sec , y k b sin x h a cos , y k r sin x h r cos , y y1 ty2 y1 x x1 tx2 x1 x2, y2: x1, y1 y 4 t 2 y 3 t x 2t 2 x 2t 1 y 4 3 t y 4 t x 2 3 t x 2t y 2et 1 y 2et 1 x et x et y 2 cos 1 y 2t 1 x cos x t y 0.4t 2 y lnt 2 1 x 10 0.01et x t2 y tan y 2 tan x sec x 4 sec y 2 2 sin y 2 sin x 4 3 cos x 4 3 cos y 2t 2 x ln 2t, y 3 ln t x t 3, y et x e2t, y e3t x et, y 4 sin x cos , y 3 sin x 2 cos , y t 2 x t 1, y t 2 x 2t, y 1 t x t, y t 2 x t 2, y t 3 x t, y t 2 x 1 4t, y 2 3t x 3 2t, y 2t 1 x 3t 3, y 1 2t x t, y 4t x t, 706 Chapter 9 Topics in Analytic Geometry Projectile Motion In Exercises 55–58, consider a projectile launched at a height of h feet above the ground at an angle of with the horizontal. The initial velocity is feet per second and the path of the projectile is modeled by the parametric equations and 55. The center field fence in a ballpark is 10 feet high and 400 feet from home plate. A baseball is hit at a point 3 feet above the ground. It leaves the bat at an angle of degrees with the horizontal at a speed of 100 miles per hour (see figure).
(a) Write a set of parametric equations for the path of the baseball.
(b) Use a graphing utility to graph the path of the baseball for Is the hit a home run?
(c) Use a graphing utility to graph the path of the baseball for Is the hit a home run?
(d) Find the minimum angle required for the hit to be a home run.
56. The right field fence in a ballpark is 10 feet high and 314 feet from home plate. A baseball is hit at a point 2.5 feet above the ground. It leaves the bat at an angle of with the horizontal at a speed of 105 feet per second.
(a) Write a set of parametric equations for the path of the baseball.
(b) Use a graphing utility to graph the path of the baseball and approximate its maximum height.
(c) Use a graphing utility to find the horizontal distance that the baseball travels. Is the hit a home run?
(d) Explain how you could find the result in part (c) algebraically.
57. The quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught by a receiver at a height of 4 feet, 30 yards directly downfield. The pass is released at an angle of with the horizontal.
(a) Write a set of parametric equations for the path of the football.
(b) Find the speed of the football when it is released.
(c) Use a graphing utility to graph the path of the football and approximate its maximum height.
(d) Find the time the receiver has to position himself after the quarterback releases the football.
58. To begin a football game, a kicker kicks off from his team’s 35-yard line. The football is kicked at an angle of with the horizontal at an initial velocity of 85 feet per second.
(a) Write a set of parametric equations for the path of the kick.
(b) Use a graphing utility to graph the path of the kick and approximate its maximum height.
(c) Use a graphing utility to find the horizontal distance that the kick travels.
(d) Explain how you could find the result in part (c) algebraically.
Synthesis True or False?
In Exercises 59–62, determine whether the statement is true or false. Justify your answer.
59. The two sets of parametric equations and correspond to the same rectangu-lar equation.
60. Because the graph of the parametric equations and both represent the line they are the same plane curve.
61. If y is a function of t and x is a function of t, then y must be a function of x.
62. The parametric equations and where and represent a circle centered at if 63. As increases, the ellipse given by the parametric equations and is traced out counter-clockwise. Find a parametric representation for which the same ellipse is traced out clockwise.
64. Think About It The graph of the parametric equations and is shown below. Would the graph change for the equations and If so, how would it change?
Skills Review In Exercises 65–68, check for symmetry with respect to both axes and the origin. Then determine whether the func-tion is even, odd, or neither.
65.
66.
67.
68. x 22 y 4 y ex fx x fx 4x2 x2 1 −5 −6 6 3 y t 1?
x t3 y t 1 x t3 y 2 sin x cos a b.
h, k, b 0, a 0 y bt k, x at h y x, x t, y t y t2 x t2, y 9t 2 1 x 3t, y t 2 1 x t, 50 35 40 23.
15.
θ 400 ft 3 ft 10 ft Not drawn to scale v0 sin t 16t2.
y h 1 x v0 cos t v0 9.6 Polar Coordinates What you should learn Plot points and find multiple representa-tions of points in the polar coordinate system.
Convert points from rectangular to polar form and vice versa.
Convert equations from rectangular to polar form and vice versa.
Why you should learn it Polar coordinates offer a different mathe-matical perspective on graphing.For instance, in Exercises 5–12 on page 711,you see that a polar coordinate can be written in more than one way.
Section 9.6 Polar Coordinates 707 Introduction So far, you have been representing graphs of equations as collections of points in the rectangular coordinate system, where x and y represent the directed distances from the coordinate axes to the point In this section, you will study a second coordinate system called the polar coordinate system.
To form the polar coordinate system in the plane, fix a point O, called the pole (or origin), and construct from O an initial ray called the polar axis, as shown in Figure 9.67. Then each point P in the plane can be assigned polar coor-dinates as follows.
1.
from O to P 2.
counterclockwise from the polar axis to segment Figure 9.67 r = directed distance θ O = directed angle Polar axis P r = ( , ) θ OP directed angle, r directed distance r, x, y.
x, y Example 1 Plotting Points in the Polar Coordinate System a. The point lies two units from the pole on the terminal side of the angle as shown in Figure 9.68.
b. The point lies three units from the pole on the terminal side of the angle as shown in Figure 9.69.
c. The point coincides with the point as shown in Figure 9.70.
Figure 9.68 Figure 9.69 Now try Exercise 5.
( ) 2 3 π 0 = − θ π 6 π 6 3, − π 3 2 π 2 π 0 = θ π 3 π 3 ( ) 1 2 3 2, π 3 2 π 2 3, 6, r, 3, 116 6, r, 3, 6 3, r, 2, 3 2 3 π 0 π 3 2 = θ π ( ) 3,11 6 π 11 6 π 2 Figure 9.70 In rectangular coordinates, each point has a unique representation. This is not true for polar coordinates. For instance, the coordinates and represent the same point, as illustrated in Example 1. Another way to obtain multiple representations of a point is to use negative values for r. Because r is a directed distance, the coordinates and represent the same point. In general, the point can be represented as or where n is any integer. Moreover, the pole is represented by where is any angle.
0, , r, r, ± 2n 1 r, r, ± 2n r, r, r, r, 2 r, x, y 708 Chapter 9 Topics in Analytic Geometry Example 2 Multiple Representations of Points Plot the point and find three additional polar representations of this point, using Solution The point is shown in Figure 9.71. Three other representations are as follows.
Add to Replace r by subtract from Replace r by add to Now try Exercise 7.
.
r; 3, 3 4 3, 4 .
r; 3, 3 4 3, 7 4 .
2 3, 3 4 2 3, 5 4 2 < < 2.
3, 34 Coordinate Conversion The polar coordinates are related to the rectangular coordinates as follows.
and r2 x2 y2 y r sin tan y x x r cos x, y r, Coordinate Conversion To establish the relationship between polar and rectangular coordinates, let the polar axis coincide with the positive x-axis and the pole with the origin, as shown in Figure 9.72. Because lies on a circle of radius r, it follows that Moreover, for the definitions of the trigonometric functions imply that and You can show that the same relationships hold for r < 0.
sin y r.
cos x r, tan y x, r > 0, r2 x2 y2.
x, y 2 1 3 π π π π 4 4 4 4 ( ) ( ) ( ) ( ) 3 5 7 3, = 3, = 3, = 3, = ...
− − − − π 0 = − θ π ( ) 3, −3 4 π 3 4 π 3 2 π 2 Figure 9.71 (Origin) Polar axis ( -axis) x Pole r x y ( , ) r ( , ) x y θ θ x y Figure 9.72 Section 9.6 Polar Coordinates 709 Example 3 Polar-to-Rectangular Conversion Convert each point to rectangular coordinates.
a.
b.
Solution a. For the point you have the following.
The rectangular coordinates are (See Figure 9.73.) b. For the point you have the following.
The rectangular coordinates are (See Figure 9.73.) Now try Exercise 13.
x, y 32, 32.
y r sin 3 sin 6 3 1 2 3 2 x r cos 3 cos 6 3 3 2 3 2 r, 3, 6, x, y 2, 0.
y r sin 2 sin 0 x r cos 2 cos 2 r, 2, , 3, 6 2, Example 4 Rectangular-to-Polar Conversion Convert each point to polar coordinates.
a.
b.
Solution a. For the second-quadrant point you have Because lies in the same quadrant as use positive r.
So, one set of polar coordinates is as shown in Figure 9.74.
b. Because the point lies on the positive y-axis, choose and This implies that one set of polar coordinates is as shown in Figure 9.75.
Now try Exercise 29.
r, 2, 2, r 2. 2 x, y 0, 2 r, 2, 34, r x2 y2 12 12 2 x, y, 3 4 . tan y x 1 1 1 x, y 1, 1, 0, 2 1, 1 Exploration Set your graphing utility to polar mode. Then graph the equation Use a viewing window in which and You should obtain a circle of radius 3.
a. Use the trace feature to cursor around the circle. Can you locate the point b. Can you locate other representations of the point If so, explain how you did it. 3, 54?
3, 54?
4 ≤y ≤4. 6 ≤x ≤6, 0 ≤ ≤2, r 3.
( ) 1 2 2 1 −1 ( , ) = (2, ) r ( , ) = 3, r ( , ) = , x y θ θ π ( , ) = ( 2, 0) x y − ( ) x y π 6 3 2 3 2 Figure 9.73 1 2 −2 −1 2 1 ( , ) = ( 1, 1) x y − ( ) ( , ) = 2, r θ π 4 3 0 π 2 Figure 9.74 ( ) 1 2 −2 −1 1 ( , ) = 2, r θ ( , ) = (0, 2) x y 0 π 2 π 2 Figure 9.75 Equation Conversion By comparing Examples 3 and 4, you see that point conversion from the polar to the rectangular system is straightforward, whereas point conversion from the rec-tangular to the polar system is more involved. For equations, the opposite is true.
To convert a rectangular equation to polar form, you simply replace x by and y by For instance, the rectangular equation can be written in polar form as follows.
Rectangular equation Polar equation Simplest form On the other hand, converting a polar equation to rectangular form requires con-siderable ingenuity.
Example 5 demonstrates several polar-to-rectangular conversions that enable you to sketch the graphs of some polar equations. r sec tan r sin r cos 2 y x2 y x2 r sin .
r cos 710 Chapter 9 Topics in Analytic Geometry Example 5 Converting Polar Equations to Rectangular Form Describe the graph of each polar equation and find the corresponding rectangular equation.
a.
b.
c.
Solution a. The graph of the polar equation consists of all points that are two units from the pole. In other words, this graph is a circle centered at the origin with a radius of 2, as shown in Figure 9.76. You can confirm this by converting to rectangular form, using the relationship Polar equation Rectangular equation b. The graph of the polar equation consists of all points on the line that makes an angle of with the positive x-axis, as shown in Figure 9.77. To convert to rectangular form, you make use of the relationship Polar equation Rectangular equation c. The graph of the polar equation is not evident by simple inspection, so you convert to rectangular form by using the relationship Polar equation Rectangular equation Now you can see that the graph is a vertical line, as shown in Figure 9.78.
Now try Exercise 83.
x 1 r cos 1 r sec r cos x.
r sec y 3x tan 3 3 tan yx.
3 3 x2 y2 22 r2 22 r 2 r2 x2 y2.
r 2 r sec 3 r 2 2 3 π 0 π 3 2 π 2 Figure 9.78 2 1 3 π 0 π 3 2 π 2 Figure 9.77 2 2 1 3 π 0 π 3 2 π 2 Figure 9.76 Section 9.6 Polar Coordinates 711 In Exercises 1– 4, a point in polar coordinates is given. Find the corresponding rectangular coordinates for the point.
1.
2.
3.
4.
In Exercises 5–12, plot the point given in polar coordinates and find three additional polar representations of the point, using 5.
6.
7.
8.
9.
10.
11.
12.
In Exercises 13–20, plot the point given in polar coordinates and find the corresponding rectangular coordinates for the point.
13.
14.
15.
16.
17.
18.
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In Exercises 21–28, use a graphing utility to find the rectan-gular coordinates of the point given in polar coordinates.
Round your results to two decimal places.
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In Exercises 37–42, use a graphing utility to find one set of polar coordinates for the point given in rectangular coordi-nates. (There are many correct answers.) 37.
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42. 7 4, 3 2 5 2, 4 3 32, 32 3, 2 5, 2 3, 2 5, 12 6, 9 3, 1 3, 3 3, 3 1, 1 0, 5 7, 0 0 } < 2.
8.2, 3.2 4.1, 0.5 5.4, 2.85 2.5, 1.58 8.25, 3.5 4.5, 1.3 4, 11 9 2, 2 9 3, 1.57 2, 2.36 0, 5 4 0, 7 6 3, 2 3 1, 3 4 2, 7 6 4, 3 0, 4 3 2, 3 2 52, 11 6 3, 5 6 3, 7 6 1, 3 2, 3 4 3, 5 6 2 < < 2.
−1 2 3 1 0 π 2 ( ) ( , ) = 2, − r θ π 4 2 4 0 π 2 ( ) ( , ) = −1, r θ π 4 5 2, 4 1, 5 4 1 2 3 4 ( ) ( , ) = 4, r θ π 2 3 0 π 2 2 4 0 π 2 ( ) ( , ) = 4, r θ π 2 4, 3 2 4, 2 Vocabulary Check Fill in the blanks.
1. The origin of the polar coordinate system is called the __ .
2. For the point is the _ from O to P and is the _ counterclockwise from the polar axis to segment 3. To graph the point you use the _ coordinate system.
r, , OP.
r, , r 9.6 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
712 Chapter 9 Topics in Analytic Geometry In Exercises 43–60, convert the rectangular equation to polar form. Assume 43.
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In Exercises 61–80, convert the polar equation to rect-angular form.
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In Exercises 81–86, describe the graph of the polar equation and find the corresponding rectangular equation. Sketch its graph.
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Synthesis True or False?
In Exercises 87 and 88, determine whether the statement is true or false. Justify your answer.
87. If and represent the same point in the polar coordinate system, then 88. If and represent the same point in the polar coordinate system, then for some integer n.
89. Think About It (a) Show that the distance between the points and is (b) Describe the positions of the points relative to each other for Simplify the Distance Formula for this case. Is the simplification what you expected?
Explain.
(c) Simplify the Distance Formula for Is the simplification what you expected? Explain.
(d) Choose two points in the polar coordinate system and find the distance between them. Then choose different polar representations of the same two points and apply the Distance Formula again. Discuss the result.
90. Writing Write a short paragraph explaining the differences between the rectangular coordinate system and the polar coordinate system.
Skills Review In Exercises 91–96, use the Law of Sines or the Law of Cosines to solve the triangle.
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In Exercises 97–102, use any method to solve the system of equations.
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In Exercises 103–106, use a determinant to determine whether the points are collinear.
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106. 2.3, 5, 0.5, 0, 1.5, 3 6, 4, 1, 3, 1.5, 2.5 2, 4, 0, 1, 4,5 4, 3, 6, 7, 2, 1 2x1 2x1 2x1 x2 2x2 x2 2x3 6x3 4 5 2 x 2x 5x y 3y 4y 2z 1 z 2 2z 4 5u u 8u 7v 2v 2v 9w 3w w 15 7 0 3a 2a a 2b b 3b c 0 3c 0 9c 0 3x 5y 4x 2y 10 5 5x 3x 7y y 11 3 B 64, b 52, c 44 C 35, a 8, b 4 B 71, a 21, c 29 A 56, C 38, c 12 A 24, a 10, b 6 a 13, b 19, c 25 90.
1 2 1 2.
r1 2 r2 2 2r1r2 cos1 2.
r2, 2 r1, 1 1 2 2n r, 2 r, 1 r1 r2 .
r2, 2 r1, 1 r 2 csc r 3 sec 7 6 4 r 8 r 7 r 6 2 cos 3 sin r 6 2 3 sin r 2 1 sin r 1 1 cos r 3 cos 2 r 2 sin 3 r 2 sin 2 r2 cos r 2 sec r 3 csc r 10 r 4 2 11 6 5 6 5 3 4 3 r 2 cos r 6 sin x2 y3 y2 x3 x2 y 2 2ay 0 x 2 y 2 2ax 0 x2 y2 8y 0 x2 y2 6x 0 y 2 8x 16 0 x 2 y 22 9x 2 y 2 2xy 1 xy 4 4x 7y 2 0 3x 6y 2 0 x a x 8 y x y 4 x 2 y 2 16 x 2 y 2 9 a > 0.
9.7 Graphs of Polar Equations What you should learn Graph polar equations by point plotting.
Use symmetry as a sketching aid.
Use zeros and maximum r-values as sketching aids.
Recognize special polar graphs.
Why you should learn it Several common figures,such as the circle in Exercise 4 on page 720,are easier to graph in the polar coordinate system than in the rectangular coordinate system.
Section 9.7 Graphs of Polar Equations 713 Introduction In previous chapters you sketched graphs in rectangular coordinate systems. You began with the basic point-plotting method. Then you used sketching aids such as a graphing utility, symmetry, intercepts, asymptotes, periods, and shifts to fur-ther investigate the nature of the graph. This section approaches curve sketching in the polar coordinate system similarly.
Example 1 Graphing a Polar Equation by Point Plotting Sketch the graph of the polar equation by hand.
Solution The sine function is periodic, so you can get a full range of r-values by considering values of in the interval as shown in the table.
By plotting these points as shown in Figure 9.79, it appears that the graph is a cir-cle of radius 2 whose center is the point Figure 9.79 Now try Exercise 25.
You can confirm the graph found in Example 1 in three ways.
1. Convert to Rectangular Form Multiply each side of the polar equation by r and convert the result to rectangular form.
2. Use a Polar Coordinate Mode Set your graphing utility to polar mode and graph the polar equation. (Use and 3. Use a Parametric Mode Set your graphing utility to parametric mode and graph and y 4 sin t sin t.
x 4 sin t cos t 4 ≤y ≤4.) 6 ≤x ≤6, 0 ≤ ≤, x, y 0, 2.
0 ≤ ≤ 2, r 4 sin Prerequisite Skills If you have trouble finding the sines of the angles in Example 1, review Trigonometric Functions of Any Angle in Section 4.4.
0 6 3 2 2 3 5 6 7 6 3 2 11 6 2 r 0 2 23 4 23 2 0 2 4 2 0 Most graphing utilities have a polar graphing mode. If yours doesn’t, you can rewrite the polar equation in parametric form, using t as a parameter, as follows.
and Symmetry In Figure 9.79, note that as increases from 0 to the graph is traced out twice.
Moreover, note that the graph is symmetric with respect to the line Had you known about this symmetry and retracing ahead of time, you could have used fewer points. The three important types of symmetry to consider in polar curve sketching are shown in Figure 9.80.
Symmetry with Respect to the Line Figure 9.80 You can determine the symmetry of the graph of (see Example 1) as follows.
1. Replace by 2. Replace by 3. Replace by So, the graph of is symmetric with respect to the line 2.
r 4 sin r 4 sin r, : r 4 sin r, r, : r 4 sin 4 sin r, r 4 sin 4 sin r 4 sin r, : r, r 4 sin 2 π θ θ θ θ θ π ( , ) r ( , ) r − ( , ) −r − − 0 π 3 2 π 2 π θ θ θ θ θ π π ( , ) r ( , ) − − r ( , ) r − − 0 π 3 2 π 2 2.
2 y ft sin t x ft cos t r f 714 Chapter 9 Topics in Analytic Geometry Testing for Symmetry in Polar Coordinates The graph of a polar equation is symmetric with respect to the following if the given substitution yields an equivalent equation.
1. The line Replace by or 2. The polar axis: Replace by or 3. The pole: Replace by or r, .
r, r, r, .
r, r, r, .
r, r, 2: π θ θ θ θ θ π π ( , ) r ( , ) −r ( , ) r + + 0 π 3 2 π 2 Symmetry with Respect to the Pole Symmetry with Respect to the Polar Axis STUDY TIP Recall from Section 4.2 that the sine function is odd. That is, sin sin .
The three tests for symmetry in polar coordinates on page 714 are sufficient to guarantee symmetry, but they are not necessary. For instance, Figure 9.82 shows the graph of Spiral of Archimedes From the figure, you can see that the graph is symmetric with respect to the line Yet the tests on page 714 fail to indicate symmetry because neither of the following replacements yields an equivalent equation.
Original Equation Replacement New Equation The equations discussed in Examples 1 and 2 are of the form and The graph of the first equation is symmetric with respect to the line and the graph of the second equation is symmetric with respect to the polar axis. This observation can be generalized to yield the following quick tests for symmetry.
2, r 3 2 cos gcos .
r 4 sin fsin r 3 r, by r, r 2 r 2 r, by r, r 2 2.
r 2.
Section 9.7 Graphs of Polar Equations 715 Example 2 Using Symmetry to Sketch a Polar Graph Use symmetry to sketch the graph of by hand.
Solution Replacing by produces So, by using the even trigonometric identity, you can conclude that the curve is symmetric with respect to the polar axis. Plotting the points in the table and using polar axis symmetry, you obtain the graph shown in Figure 9.81. This graph is called a limaçon.
Use a graphing utility to confirm this graph.
Now try Exercise 29.
cosu cos u 3 2 cos . r 3 2 cos r, r, r 3 2 cos Figure 9.81 −3 −2 3 Spiral of Archimedes: r = + 2 , −4 ≤ ≤ 0 π π θ θ 2 Figure 9.82 The table feature of a graphing utility is very useful in construct-ing tables of values for polar equations. Set your graphing utility to polar mode and enter the polar equation in Example 2.
You can verify the table of values in Example 2 by starting the table at and incrementing the value of by For instructions on how to use the table feature and polar mode, see Appendix A; for specific keystrokes, go to this textbook’s Online Study Center.
6.
0 T E C H N O L O G Y T I P Quick Tests for Symmetry in Polar Coordinates 1. The graph of is symmetric with respect to the line 2. The graph of is symmetric with respect to the polar axis.
r gcos 2.
r fsin 0 6 3 2 2 3 5 6 r 5 3 3 4 3 2 3 3 1 Zeros and Maximum r-Values Two additional aids to sketching graphs of polar equations involve knowing the -values for which is maximum and knowing the -values for which In Example 1, the maximum value of for is and this occurs when (see Figure 9.79). Moreover, when 0.
r 0 2 r 4, r 4 sin r r 0.
r 716 Chapter 9 Topics in Analytic Geometry Example 3 Finding Maximum r-Values of a Polar Graph Find the maximum value of r for the graph of r 1 2 cos .
Graphical Solution Because the polar equation is of the form you know the graph is symmetric with respect to the polar axis. You can confirm this by graphing the polar equation. Set your graphing utility to polar mode and enter the equation, as shown in Figure 9.83.
(In the graph, varies from 0 to ) To find the maximum r-value for the graph, use your graphing utility’s trace feature and you should find that the graph has a maximum r-value of 3, as shown in Figure 9.84. This value of r occurs when In the graph, note that the point is farthest from the pole.
Figure 9.83 Figure 9.84 Note how the negative r-values determine the inner loop of the graph in Figure 9.84. This type of graph is a limaçon.
Now try Exercise 19.
−6 −3 3 Limaçon: r = 1 − 2 cos θ 3 3, .
2.
r 1 2 cos gcos Numerical Solution To approximate the maximum value of r for the graph of use the table feature of a graphing utility to create a table that begins at and increments by as shown in Figure 9.85. From the table, the maximum value of r appears to be 3 when If a second table that begins at and increments by is created, as shown in Figure 9.86, the maximum value of r still appears to be 3 when Figure 9.85 Figure 9.86 3.1416 .
24 2 3.1416 .
12, 0 r 1 2 cos , Exploration The graph of the polar equation is called the butterfly curve, as shown in Figure 9.87.
a. The graph in Figure 9.87 was produced using Does this show the entire graph? Explain your reasoning.
b. Use the trace feature of your graphing utility to approximate the maximum r-value of the graph. Does this value change if you use instead of Explain.
0 ≤ ≤2?
0 ≤ ≤4 0 ≤ ≤2.
r ecos 2 cos 4 sin512 Figure 9.87 −3 −4 4 4 Some curves reach their zeros and maximum r-values at more than one point, as shown in Example 4.
Section 9.7 Graphs of Polar Equations 717 Example 4 Analyzing a Polar Graph Analyze the graph of Solution Symmetry: With respect to the polar axis Maximum value of when or Zeros of r: when or By plotting these points and using the specified symmetry, zeros, and maximum values, you can obtain the graph shown in Figure 9.88. This graph is called a rose curve, and each loop on the graph is called a petal. Note how the entire curve is generated as increases from 0 to Figure 9.88 Now try Exercise 33.
2 π 0 π 3 2 π 2 2 1 π 0 π 3 2 π 2 2 1 π 0 π 3 2 π 2 0 ≤ ≤ 0 ≤ ≤ 5 6 0 ≤ ≤ 2 3 2 1 π 0 π 3 2 π 2 2 1 π 0 π 3 2 π 2 2 1 π 0 π 3 2 π 2 0 ≤ ≤ 2 0 ≤ ≤ 3 0 ≤ ≤ 6 .
6, 2, 56 52 32, 3 2, r 0 0, 3, 23, 3 0, , 2, 3 r 2 r: r 2 cos 3.
Exploration Notice that the rose curve in Example 4 has three petals.
How many petals do the rose curves and have? Determine the numbers of petals for the curves and where n is a positive integer.
r 2 sin n, r 2 cos n r 2 sin 3 r 2 cos 4 0 12 6 4 3 5 12 2 r 2 2 0 2 2 2 0 Special Polar Graphs Several important types of graphs have equations that are simpler in polar form than in rectangular form. For example, the circle in Example 1 has the more complicated rectangular equation Several other types of graphs that have simple polar equations are shown below.
Limaçon with Cardioid Dimpled Convex inner loop (heart-shaped) limaçon limaçon Rose curve Rose curve Rose curve Rose curve Circle Circle Lemniscate Lemniscate r 2 a 2 cos 2 r 2 a 2 sin 2 r a sin r a cos a π 0 π 3 2 π 2 a π 0 π 3 2 π 2 a π 0 π 3 2 π 2 a π 0 π 3 2 π 2 r a sin n r a sin n r a cos n r a cos n a n = 2 π 0 π 3 2 π 2 a n = 5 π 0 π 3 2 π 2 a n = 4 π 0 π 3 2 π 2 a n = 3 π 0 π 3 2 π 2 a b ≥2 1 < a b < 2 a b 1 a b < 1 π 0 π 3 2 π 2 π 0 π 3 2 π 2 π 0 π 3 2 π 2 x2 y 22 4.
r 4 sin 718 Chapter 9 Topics in Analytic Geometry Limaçons a > 0, b > 0 r a ± b sin r a ± b cos Rose Curves n petals if n is odd 2n petals if n is even n ≥2 Circles and Lemniscates π 0 π 3 2 π 2 Section 9.7 Graphs of Polar Equations 719 Example 5 Analyzing a Rose Curve Analyze the graph of Solution Type of curve: Rose curve with petals Symmetry: With respect to the polar axis, the line and the pole Maximum value of when Zeros of r: when Using a graphing utility, enter the equation, as shown in Figure 9.89 (with ). You should obtain the graph shown in Figure 9.90.
Figure 9.89 Figure 9.90 Now try Exercise 37.
−6 −4 6 4 θ r = 3 cos 2 π ( ) −3, 3 2 π π 2 (3, 0) (3, ) ( ) −3, 0 ≤ ≤2 4, 34 r 0 0, 2, , 32 r 3 r: 2, 2n 4 r 3 cos 2.
Example 6 Analyzing a Lemniscate Analyze the graph of Solution Type of curve: Lemniscate Symmetry: With respect to the pole Maximum value of when Zeros of r: when Using a graphing utility, enter the equation, as shown in Figure 9.91 (with ). You should obtain the graph shown in Figure 9.92.
Figure 9.91 Figure 9.92 Now try Exercise 45.
−6 −4 6 4 θ r2 = 9 sin 2 ( ) π 4 3, ( ) π 4 −3, 0 ≤ ≤ 2 0, 2 r 0 4 r 3 r: r2 9 sin 2.
STUDY TIP The quick tests for symmetry presented on page 715 are espe-cially useful when graphing rose curves. Because rose curves have the form or the form you know that a rose curve will be either symmetric with respect to the line or symmetric with respect to the polar axis.
2 r gcos , r fsin 720 Chapter 9 Topics in Analytic Geometry In Exercises 1–6, identify the type of polar graph.
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Library of Parent Functions In Exercises 7–10, determine the equation of the polar curve whose graph is shown.
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(a) (a) (b) (b) (c) (c) (d) (d) 9.
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(a) (a) (b) (b) (c) (c) (d) (d) In Exercises 11–18, test for symmetry with respect to the polar axis, and the pole.
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In Exercises 19–22, find the maximum value of and any zeros of r. Verify your answers numerically.
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22. r sin 2 r 4 cos 3 r 6 12 cos r 10 10 sin r r2 25 cos 4 r 2 16 sin 2 r 4 csc cos r 6 sin r 2 1 cos r 4 1 sin r 12 cos 3 r 14 4 cos /2, r 2 sin 3 r 2 cos 2 r 2 sin 3 2 r 2 cos 2 r 2 cos 3 2 r cos 4 r 2 sin 6 r 2 cos 4 −2 2 1 π 0 π 3 2 π 2 2 −1 1 −2 2 1 π 0 π 3 2 π 2 r 2 sin r 1 2 cos r 2 cos r 1 2 cos r 2 sin r 1 2 sin r 2 cos r 1 2 sin 1 2 −2 1 π 0 π 3 2 π 2 −1 −2 1 2 −2 1 π 0 π 3 2 π 2 −2 −3 7 3 θ r = 1 + 4 cos −9 −6 9 6 θ r = 6 sin 2 −3 −3 6 3 θ r = 3 cos −6 −4 6 4 θ r2 = 9 cos 2 −9 −10 9 2 θ r = 5 − 5 sin −6 −4 6 4 θ r = 3 cos 2 9.7 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Vocabulary Check Fill in the blanks.
1. The graph of is symmetric with respect to the line _ .
2. The graph of is symmetric with respect to the _ .
3. The equation represents a _ .
4. The equation represents a _ .
5. The equation represents a _ .
6. The equation represents a _ .
r 1 sin r2 4 sin 2 r 2 cos r 2 cos r gcos r fsin Section 9.7 Graphs of Polar Equations 721 In Exercises 23–36, sketch the graph of the polar equation. Use a graphing utility to verify your graph.
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In Exercises 53–58, use a graphing utility to graph the polar equation. Find an interval for for which the graph is traced only once.
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In Exercises 59–62, use a graphing utility to graph the polar equation and show that the indicated line is an asymptote of the graph.
Name of Graph Polar Equation Asymptote 59. Conchoid 60. Conchoid 61. Hyperbolic spiral 62. Strophoid Synthesis True or False?
In Exercises 63 and 64, determine whether the statement is true or false. Justify your answer.
63. The graph of is a rose curve with five petals.
64. A rose curve will always have symmetry with respect to the line 65. Writing Use a graphing utility to graph the polar equation for the integers As you graph these equations, you should see the graph change shape from a heart to a bell. Write a short paragraph explaining which values of n produce the heart-shaped curves and which values of n produce the bell-shaped curves.
66. The graph of is rotated about the pole through an angle Show that the equation of the rotated graph is 67. Consider the graph of (a) Show that if the graph is rotated counterclockwise radians about the pole, the equation of the rotated graph is (b) Show that if the graph is rotated counterclockwise radians about the pole, the equation of the rotated graph is (c) Show that if the graph is rotated counterclockwise radians about the pole, the equation of the rotated graph is In Exercises 68–70, use the results of Exercises 66 and 67.
68. Write an equation for the limaçon after it has been rotated through each given angle.
(a) (b) (c) (d) 69. Write an equation for the rose curve after it has been rotated through each given angle.
(a) (b) (c) (d) 70. Sketch the graph of each equation.
(a) (b) 71. Exploration Use a graphing utility to graph the polar equation for and Identify each graph.
72. Exploration Consider the polar equation (a) Use a graphing utility to graph the equation for Find the interval for for which the graph is traced only once.
(b) Use a graphing utility to graph the equation for Find the interval for for which the graph is traced only once.
(c) Is it possible to find an interval for for which the graph is traced only once for any rational number k?
Explain.
k 2.5.
k 1.5.
r 3 sin k.
k 3.
k 2, k 1, k 0, r 2 k cos r 1 sin 4 r 1 sin 2 3 2 6 r 2 sin 2 3 2 2 4 r 2 sin r f cos.
32 r f sin.
r f cos.
2 r f sin.
r f .
.
r f n 5 to n 5.
r cos 5 n cos , 0 ≤ < 2.
r 10 sin 5 x 2 r 2 cos 2 sec y 2 r 2/ y 1 r 2 csc x 1 r 2 sec r 2 1 r 2 sin 2 r 3 sin 5 2 r 2 cos 3 2 r 21 2 sin r 3 2 cos r e2 r e2 r 4 sec r 2 csc 6 r 2 cos3 2 r 4 sin cos 2 r 2 9 sin r 2 4 cos 2 r 6 2 sin 3 cos r 3 sin 2 cos r 3 2 sin r 3 6 cos r 6 4 sin r 25 sin r cos 2 r 8 cos 2 r 3 cos 5 r 7 sin 2 r sin 5 r 5 cos 3 r 3 6 cos r 4 5 sin r 1 2 sin r 3 4 cos r 41 sin r 31 cos r 2 cos r 3 sin 5 3 r 5 9.8 Polar Equations of Conics What you should learn Define conics in terms of eccentricities.
Write and graph equations of conics in polar form.
Use equations of conics in polar form to model real-life problems.
Why you should learn it The orbits of planets and satellites can be modeled by polar equations.For instance,in Exercise 55 on page 727,you will use polar equations to model the orbits of Neptune and Pluto.
Kevin Kelley/Getty Images Alternative Definition of Conics In Sections 9.2 and 9.3, you learned that the rectangular equations of ellipses and hyperbolas take simple forms when the origin lies at the center. As it happens, there are many important applications of conics in which it is more convenient to use one of the foci as the origin. In this section, you will learn that polar equations of conics take simple forms if one of the foci lies at the pole.
To begin, consider the following alternative definition of a conic that uses the concept of eccentricity (a measure of the flatness of the conic).
In Figure 9.93, note that for each type of conic, the focus is at the pole.
Ellipse: Parabola: Hyperbola: Figure 9.93 Polar Equations of Conics The benefit of locating a focus of a conic at the pole is that the equation of the conic becomes simpler.
PF PQ PF PQ > 1 PF PQ 1 PF PQ < 1 e > 1 e 1 0 < e < 1 F = (0, 0) Q Q′ P′ P Directrix 0 π 2 F = (0, 0) Q P Directrix 0 π 2 F = (0, 0) Q P Directrix 0 π 2 722 Chapter 9 Topics in Analytic Geometry Alternative Definition of a Conic The locus of a point in the plane that moves such that its distance from a fixed point (focus) is in a constant ratio to its distance from a fixed line (directrix) is a conic. The constant ratio is the eccentricity of the conic and is denoted by e. Moreover, the conic is an ellipse if a parabola if and a hyperbola if (See Figure 9.93.) e > 1.
e 1, e < 1, Polar Equations of Conics (See the proof on page 739.) The graph of a polar equation of the form 1.
or 2.
is a conic, where is the eccentricity and is the distance between the focus (pole) and the directrix.
p e > 0 r ep 1 ± e sin r ep 1 ± e cos Prerequisite Skills To review the characteristics of conics, see Sections 9.1–9.3.
An equation of the form Vertical directrix corresponds to a conic with a vertical directrix and symmetry with respect to the polar axis. An equation of the form Horizontal directrix corresponds to a conic with a horizontal directrix and symmetry with respect to Moreover, the converse is also true—that is, any conic with a focus at the pole and having a horizontal or vertical directrix can be represented by one of the given equations.
2.
r ep 1 ± e sin r ep 1 ± e cos Section 9.8 Polar Equations of Conics 723 Example 1 Identifying a Conic from Its Equation Identify the type of conic represented by the equation r 15 3 2 cos .
Algebraic Solution To identify the type of conic, rewrite the equation in the form Because you can conclude that the graph is an ellipse.
Now try Exercise 11.
e 2 3 < 1, 5 1 23 cos r 15 3 2 cos r ep1 ± e cos .
Graphical Solution Use a graphing utility in polar mode to graph Be sure to use a square setting. From the graph in Figure 9.94, you can see that the conic appears to be an ellipse.
Figure 9.94 −6 −8 18 8 π (15, 0) (3, ) θ r = 15 3 − 2 cos r 15 3 2 cos .
Divide numerator and denominator by 3.
For the ellipse in Figure 9.94, the major axis is horizontal and the vertices lie at and So, the length of the major axis is To find the length of the minor axis, you can use the equations and to conclude that Ellipse Because you have which implies that So, the length of the minor axis is A similar analysis for hyperbolas yields Hyperbola a2e2 1. ea2 a2 b2 c2 a2 2b 65.
45 35.
b b2 921 2 32 45, e 2 3, a21 e2. a2 ea2 b2 a2 c2 b2 a2 c2 e ca 2a 18.
r, 3, .
r, 15, 0 In the next example, you are asked to find a polar equation for a specified conic. To do this, let p be the distance between the pole and the directrix.
1. Horizontal directrix above the pole: 2. Horizontal directrix below the pole: 3. Vertical directrix to the right of the pole: 4. Vertical directrix to the left of the pole: r ep 1 e cos r ep 1 e cos r ep 1 e sin r ep 1 e sin 724 Chapter 9 Topics in Analytic Geometry Example 2 Analyzing the Graph of a Polar Equation Analyze the graph of the polar equation Solution Dividing the numerator and denominator by 3 produces Because the graph is a hyperbola. The transverse axis of the hyperbola lies on the line and the vertices occur at and Because the length of the transverse axis is 12, you can see that To find b, write So, You can use a and b to determine that the asymptotes are as shown in Figure 9.95.
Now try Exercise 23.
y 10 ± 3 4x, b 8.
b2 a2e2 1 62 5 3 2 1 64.
a 6.
r, 16, 32.
r, 4, 2 2 e 5 3 > 1, r 323 1 53 sin .
r 32 3 5 sin .
Example 3 Finding the Polar Equation of a Conic Find the polar equation of the parabola whose focus is the pole and whose direc-trix is the line Solution From Figure 9.96, you can see that the directrix is horizontal and above the pole.
Moreover, because the eccentricity of a parabola is and the distance between the pole and the directrix is you have the equation Now try Exercise 33.
r ep 1 e sin 3 1 sin .
p 3, e 1 y 3.
−18 −4 24 24 θ r = 32 3 + 5 sin π ( ) −16, 3 2 π 2 ( ) 4, Figure 9.95 −6 −3 6 5 θ r = 3 1 + sin (0, 0) Directrix y = 3 Figure 9.96 Exploration Try using a graphing utility in polar mode to verify the four orientations shown at the left.
Remember that e must be positive, but p can be positive or negative. Application Kepler’s Laws (listed below), named after the German astronomer Johannes Kepler (1571–1630), can be used to describe the orbits of the planets about the sun.
1. Each planet moves in an elliptical orbit with the sun as a focus.
2. A ray from the sun to the planet sweeps out equal areas of the ellipse in equal times.
3. The square of the period (the time it takes for a planet to orbit the sun) is proportional to the cube of the mean distance between the planet and the sun.
Although Kepler simply stated these laws on the basis of observation, they were later validated by Isaac Newton (1642–1727). In fact, Newton was able to show that each law can be deduced from a set of universal laws of motion and gravita-tion that govern the movement of all heavenly bodies, including comets and satellites. This is illustrated in the next example, which involves the comet named after the English mathematician and physicist Edmund Halley (1656–1742).
If you use Earth as a reference with a period of 1 year and a distance of 1 astronomical unit (an astronomical unit is defined as the mean distance between Earth and the sun, or about 93 million miles), the proportionality constant in Kepler’s third law is 1. For example, because Mars has a mean distance to the sun of astronomical units, its period P is given by So, the period of Mars is years.
P 1.88 d 3 P2.
d 1.524 Section 9.8 Polar Equations of Conics 725 Example 4 Halley’s Comet Halley’s comet has an elliptical orbit with an eccentricity of The length of the major axis of the orbit is approximately 35.88 astronomical units.
Find a polar equation for the orbit. How close does Halley’s comet come to the sun?
Solution Using a vertical axis, as shown in Figure 9.97, choose an equation of the form Because the vertices of the ellipse occur at and you can determine the length of the major axis to be the sum of the r-values of the vertices. That is, So, and Using this value of ep in the equation, you have where is measured in astronomical units. To find the closest point to the sun (the focus), substitute into this equation to obtain Now try Exercise 51. r 1.164 1 0.967 sin2 0.59 astronomical units 55,000,000 miles.
2 r r 1.164 1 0.967 sin ep 0.9671.204 1.164.
p 1.204 2a 0.967p 1 0.967 0.967p 1 0.967 29.79p 35.88.
32, 2 r ep1 e sin .
e 0.967.
Halleyí s comet Sun Earth π 0 π 3 2 π 2 Not drawn to scale Figure 9.97 726 Chapter 9 Topics in Analytic Geometry Graphical Reasoning In Exercises 1–4, use a graphing util-ity to graph the polar equation for (a) (b) and (c) Identify the conic for each equation.
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In Exercises 5–10, match the polar equation with its graph.
[The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) (b) (c) (d) (e) (f) 5.
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In Exercises 11–20, identify the conic represented by the equation algebraically. Use a graphing utility to confirm your result.
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20 In Exercises 21–26, use a graphing utility to graph the polar equation. Identify the graph.
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26. r 4 1 2 cos r 3 4 2 cos r 12 2 cos r 14 14 17 sin r 1 2 4 sin r 5 1 sin r 10 3 9 sin r 3 4 8 cos r 5 1 2 cos r 6 2 sin r 6 3 2 cos r 8 4 3 sin r 7 7 sin r 4 4 cos r 2 1 sin r 2 1 cos r 4 1 sin r 3 1 2 sin r 4 1 3 sin r 3 2 cos r 3 2 cos r 4 1 cos −3 −5 4 3 −6 −6 6 2 −6 −2 6 6 −2 −2 4 2 −9 −6 9 6 −9 −9 9 3 r 2e 1 e sin r 2e 1 e sin r 2e 1 e cos r 2e 1 e cos e 1.5.
e 0.5, e 1, Vocabulary Check In Exercises 1 and 2, fill in the blanks.
1. The locus of a point in the plane that moves such that its distance from a fixed point (focus) is in a constant ratio to its distance from a fixed line (directrix) is a _ .
2. The constant ratio is the _ of the conic and is denoted by _ .
3. Match the conic with its eccentricity.
(a) (i) ellipse (b) (ii) hyperbola (c) (iii) parabola e > 1 e 1 e < 1 9.8 Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Section 9.8 Polar Equations of Conics 727 In Exercises 27–32, use a graphing utility to graph the rotat-ed conic.
27.
(See Exercise 11.) 28.
(See Exercise 14.) 29.
(See Exercise 13.) 30.
(See Exercise 16.) 31.
(See Exercise 15.) 32.
(See Exercise 18.) In Exercises 33–48, find a polar equation of the conic with its focus at the pole.
Conic Eccentricity Directrix 33. Parabola 34. Parabola 35. Ellipse 36. Ellipse 37. Hyperbola 38. Hyperbola Conic Vertex or Vertices 39. Parabola 40. Parabola 41. Parabola 42. Parabola 43. Ellipse 44. Ellipse 45. Ellipse 46. Hyperbola 47. Hyperbola 48. Hyperbola 49. Planetary Motion The planets travel in elliptical orbits with the sun at one focus. Assume that the focus is at the pole, the major axis lies on the polar axis, and the length of the major axis is 2a (see figure). Show that the polar equa-tion of the orbit of a planet is where e is the eccentricity.
50. Planetary Motion Use the result of Exercise 49 to show that the minimum distance ( perihelion) from the sun to a planet is and that the maximum distance (aphelion) is Planetary Motion In Exercises 51–54, use the results of Exercises 49 and 50 to find the polar equation of the orbit of the planet and the perihelion and aphelion distances.
51. Earth 52. Mercury 53. Jupiter 54. Saturn 55. Planetary Motion Use the results of Exercises 49 and 50, where for the planet Neptune, kilometers and and for the dwarf planet Pluto, kilometers and (a) Find the polar equation of the orbit of each planet.
(b) Find the perihelion and aphelion distances for each planet.
(c) Use a graphing utility to graph both Neptune’s and Pluto’s equations of orbit in the same viewing window.
(d) Is Pluto ever closer to the sun than Neptune? Until recently, Pluto was considered the ninth planet. Why was Pluto called the ninth planet and Neptune the eighth planet?
(e) Do the orbits of Neptune and Pluto intersect? Will Neptune and Pluto ever collide? Why or why not?
e 0.2488.
a 5.906 109 e 0.0086 a 4.498 109 e 0.0542 a 142.673 107 kilometers e 0.0484 a 77.841 107 kilometers e 0.2056 a 35.983 106 miles e 0.0167 a 92.956 106 miles r a1 e.
r a1 e θ r a Sun Planet 0 π 2 r 1 e2a 1 e cos 4, 2, 1, 2 4, 2, 1, 3 2 1, 3 2 , 9, 3 2 20, 0, 4, 2, 2, 4, 3 2 2, 0, 10, 10, 2 5, 8, 0 1, 2 x 1 e 3 2 x 1 e 2 y 4 e 3 4 y 1 e 1 2 y 4 e 1 x 1 e 1 r 5 1 2 cos 23 r 8 4 3 sin 6 r 6 3 2 cos 2 r 4 4 cos 34 r 7 7 sin 3 r 2 1 cos 4 728 Chapter 9 Topics in Analytic Geometry 56. Explorer 18 On November 27, 1963, the United States launched a satellite named Explorer 18. Its low and high points above the surface of Earth were 119 miles and 122,800 miles, respectively (see figure). The center of Earth is at one focus of the orbit.
(a) Find the polar equation for the orbit (assume the radius of Earth is 4000 miles).
(b) Find the distance between the surface of Earth and the satellite when (c) Find the distance between the surface of Earth and the satellite when Synthesis True or False?
In Exercises 57 and 58, determine whether the statement is true or false. Justify your answer.
57. The graph of has a horizontal direc-trix above the pole.
58. The conic represented by the following equation is an ellipse.
59. Show that the polar equation for the ellipse is 60. Show that the polar equation for the hyperbola is In Exercises 61–66, use the results of Exercises 59 and 60 to write the polar form of the equation of the conic.
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65. Hyperbola One focus: Vertices: 66. Ellipse One focus: Vertices: 67. Exploration Consider the polar equation (a) Identify the conic without graphing the equation.
(b) Without graphing the following polar equations, describe how each differs from the given polar equa-tion. Use a graphing utility to verify your results.
68. Exploration The equation is the equation of an ellipse with What happens to the lengths of both the major axis and the minor axis when the value of e remains fixed and the value of p changes?
Use an example to explain your reasoning.
69. Writing In your own words, define the term eccentricity and explain how it can be used to classify conics.
70. What conic does the polar equation given by represent?
Skills Review In Exercises 71–76, solve the equation.
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In Exercises 77–80 find the value of the trigonometric func-tion given that u and v are in Quadrant IV and and 77.
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In Exercises 81–84, evaluate the expression. Do not use a calculator.
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84. 29P2 10P3 18C16 12C9 cosu v sinu v sinu v cosu v cos v 12.
sin u 3 5 2 sec 2 csc 4 2 cot x 5 cos 2 9 csc2 x 10 2 12 sin2 9 6 cos x 2 1 43 tan 3 1 r a sin b cos e < 1.
r ep 1 ± e sin r 4 1 0.4 cos , r 4 1 0.4 sin r 4 1 0.4 cos .
5, 0, 5, 4, 0 4, 2, 4, 2 5, 2 x2 36 y2 4 1 x2 25 y2 16 1 x 2 9 y 2 16 1 x 2 169 y 2 144 1 r2 b2 1 e2 cos2.
x 2 a 2 y 2 b2 1 r 2 b2 1 e 2 cos2.
x2 a2 y 2 b2 1 r 2 16 9 4 cos 4 r 43 3 sin 30.
60.
r Earth Not drawn to scale 60° a 0 π 2 Explorer 18 Chapter Summary 729 What Did You Learn?
Key Terms conic (conic section), p.660 degenerate conic, p.660 circle, p.661 center (of a circle), p.661 radius, p.661 parabola, p.663 directrix, p.663 focus (of a parabola), p.663 vertex (of a parabola), p.663 axis (of a parabola), p.663 ellipse, p.671 foci (of an ellipse), p.671 vertices (of an ellipse), p.671 major axis (of an ellipse), p.671 center (of an ellipse), p.671 minor axis (of an ellipse), p.671 hyperbola, p.680 foci (of a hyperbola), p.680 branches, p.680 vertices (of a hyperbola), p.680 transverse axis, p.680 center (of a hyperbola), p.680 conjugate axis, p.682 asymptotes (of a hyperbola), p.682 rotation of axes, p.690 invariant under rotation, p.694 discriminant, p.694 parameter, p.699 parametric equations, p.699 plane curve, p.699 orientation, p.700 eliminating the parameter, p.702 polar coordinate system, p.707 pole, p.707 polar axis, p.707 polar coordinates, p.707 limaçon, p.715 rose curve, p.717 eccentricity, p.722 Key Concepts 9.1–9.3 Write and graph translations of conics 1. Circle with center and radius 2. Parabola with vertex Vertical axis Horizontal axis 3. Ellipse with center and Horizontal major axis Vertical major axis 4. Hyperbola with center Horizontal transverse axis Vertical transverse axis 9.4 Rotation and systems of quadratic equations 1. Rotate the coordinate axis to eliminate the -term.
2. Use the discriminant to classify conics.
9.5 Graph curves that are represented by sets of parametric equations 1. Sketch a curve represented by a pair of parametric equations by plotting points in the order of increasing values of in the -plane.
2. Use the parametric mode of a graphing utility to graph a set of parametric equations.
9.6 Convert points from rectangular to polar form and vice versa 1. Polar-to-Rectangular: 2. Rectangular-to-Polar: 9.7 Use symmetry to aid in sketching graphs of polar equations 1. Symmetry with respect to the line Replace by or 2. Symmetry with respect to the polar axis: Replace by or 3. Symmetry with respect to the pole: Replace by or 4. The graph of is symmetric with respect to the line 5. The graph of is symmetric with respect to the polar axis.
9.8 Write and graph equations of conics in polar form The graph of a polar equation of the form or is a conic, where is the eccentricity and is the distance between the focus (pole) and the directrix.
p e > 0 r ep 1 ± e sin r ep 1 ± e cos r gcos 2.
r fsin r, r, r, r, ) r, r, r, r, r, ) 2 : tan x y, r2 x2 y2 x r cos , y r sin xy t xy y k2 a2 x h2 b2 1 x h2 a2 y k2 b2 1 h, k: x h2 b2 y k2 a2 1 x h2 a2 y k2 b2 1 0 < b < a: h, k y k2 4px h x h2 4py k h, k: x h2 (y k2 r2 r: h, k 730 Chapter 9 Topics in Analytic Geometry 9.1 In Exercises 1–4, find the standard form of the equation of the circle with the given characteristics.
1. Center at origin; point on the circle: 2. Center at origin; point on the circle: 3. Endpoints of a diameter: and 4. Endpoints of a diameter: and In Exercises 5–8, write the equation of the circle in standard form. Then identify its center and radius.
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In Exercises 9 and 10, sketch the circle. Identify its center and radius.
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In Exercises 11 and 12, find the - and -intercepts of the graph of the circle.
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In Exercises 13–16, find the vertex, focus, and directrix of the parabola, and sketch its graph. Use a graphing utility to verify your graph.
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In Exercises 17–20, find the standard form of the equation of the parabola with the given characteristics.
17. Vertex: 18. Vertex: Focus: Focus: 19.
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In Exercises 21 and 22, find an equation of the tangent line to the parabola at the given point and find the x-intercept of the line.
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23. Architecture A parabolic archway (see figure) is 12 meters high at the vertex. At a height of 10 meters, the width of the archway is 8 meters. How wide is the archway at ground level?
Figure for 23 Figure for 24 24. Architecture A church window (see figure) is bounded on top by a parabola and below by the arc of a circle.
(a) Find equations of the parabola and the circle.
(b) Use a graphing utility to create a table showing the vertical distances d between the circle and the parabola for various values of .
9.2 In Exercises 25–28, find the center, vertices, foci, and eccentricity of the ellipse and sketch its graph. Use a graphing utility to verify your graph.
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In Exercises 29–32, (a) find the standard form of the equation of the ellipse, (b) find the center, vertices, foci, and eccentricity of the ellipse, and (c) sketch the ellipse. Use a graphing utility to verify your graph.
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In Exercises 33–36, find the standard form of the equation of the ellipse with the given characteristics.
33. Vertices: foci: 34. Vertices: passes through the point 35. Vertices: foci: 36. Vertices: foci: 2, 1, 2, 3 2, 0, 2, 4; 0, 0, 4, 0 3, 0, 7, 0; 2, 2 0, ±6; ±4, 0 ±5, 0; x2 20y2 5x 120y 185 0 3x2 8y2 12x 112y 403 0 4x2 25y2 16x 150y 141 0 16x2 9y2 32x 72y 16 0 x 12 16 y 32 6 1 x 42 6 y 42 9 1 x2 9 y2 8 1 x2 4 y2 16 1 x 4 ft 8 ft 8 ft d x y x y (0, 12) (−4, 10) (4, 10) 8, 4 y2 2x, 2, 2 x2 2y, −5 −5 25 15 (0, 5) (6, 0) −14 −20 10 6 (−6, 4) (0, 0) 4, 0 6, 0 4, 2 0, 0 1 4y 8x2 0 1 2y2 18x 0 y 1 8x2 4x y 2 0 x 52 y 62 27 x 32 y 12 7 y x x2 y2 8x 10y 8 0 x2 y2 4x 6y 3 0 4x2 4y2 32x 24y 51 0 16x2 16y2 16x 24y 3 0 3 4x2 3 4y2 1 1 2x2 1 2y2 18 6, 5 2, 3 5, 6 1, 2 8, 15 3, 4 x 0 1 2 3 4 d Review Exercises See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Review Exercises 731 37. Architecture A semielliptical archway is to be formed over the entrance to an estate. The arch is to be set on pillars that are 10 feet apart and is to have a height (atop the pillars) of 4 feet. Where should the foci be placed in order to sketch the arch?
38. Wading Pool You are building a wading pool that is in the shape of an ellipse. Your plans give an equation for the elliptical shape of the pool measured in feet as Find the longest distance across the pool, the shortest distance, and the distance between the foci.
39. Planetary Motion Saturn moves in an elliptical orbit with the sun at one focus. The smallest distance and the greatest distance of the planet from the sun are and kilometers, respectively. Find the eccentricity of the orbit, defined by 40. Planetary Motion Mercury moves in an elliptical orbit with the sun at one focus. The eccentricity of Mercury’s orbit is The length of the major axis is 72 million miles. Find the standard equation of Mercury’s orbit. Place the center of the orbit at the origin and the major axis on the -axis.
9.3 In Exercises 41–46, (a) find the standard form of the equation of the hyperbola, (b) find the center, vertices, foci, and eccentricity of the hyperbola, and (c) sketch the hyper-bola. Use a graphing utility to verify your graph in part (c).
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In Exercises 47– 50, find the standard form of the equation of the hyperbola with the given characteristics.
47. Vertices: foci: 48. Vertices: foci: 49. Foci: asymptotes: 50. Foci: asymptotes: 51. Navigation Radio transmitting station A is located 200 miles east of transmitting station B. A ship is in an area to the north and 40 miles west of station A. Synchronized radio pulses transmitted at 186,000 miles per second by the two stations are received 0.0005 second sooner from station A than from station B. How far north is the ship?
52. Sound Location Two of your friends live 4 miles apart on the same “east-west” street, and you live halfway between them. You are having a three-way phone conversation when you hear an explosion. Six seconds later your friend to the east hears the explosion, and your friend to the west hears it 8 seconds after you do. Find equations of two hyperbolas that would locate the explosion. (Assume that the coordinate system is measured in feet and that sound travels at 1100 feet per second.) In Exercises 53–56, classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola.
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9.4 In Exercises 57–60, rotate the axes to eliminate the -term in the equation. Then write the equation in standard form. Sketch the graph of the resulting equation, showing both sets of axes.
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In Exercises 61–64, (a) use the discriminant to classify the graph of the equation, (b) use the Quadratic Formula to solve for and (c) use a graphing utility to graph the equation.
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In Exercises 65 and 66, use any method to solve the system of quadratic equations algebraically. Then verify your results by using a graphing utility to graph the equations and find any points of intersection of the graphs.
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9.5 In Exercises 67 and 68, complete the table for the set of parametric equations. Plot the points and sketch a graph of the parametric equations.
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and y 7 4t x 3t 2 x, y x2 y2 25 0 9x 4y2 0 4x2 y2 32x 24y 64 0 4x2 y2 56x 24y 304 0 x2 10xy y2 1 0 x2 y2 2xy 22x 22y 2 0 13x2 8xy 7y2 45 0 16x2 8xy y2 10x 5y 0 y, 4x2 8xy 4y2 72x 92y 0 5x2 2xy 5y2 12 0 x2 10xy y2 1 0 xy 4 0 xy 4y2 5x 3y 7 0 5x2 2y2 10x 4y 17 0 4x2 4y 2 4x 8y 11 0 3x2 2y 2 12x 12y 29 0 y ±2x 3 3, ±2; y ±2x 4 0, 0, 8, 0; 0, ±3 0, ±1; ±6, 0 ±4, 0; 9x2 y2 72x 8y 119 0 y2 4x2 2y 48x 59 0 4x2 25y2 8x 150y 121 0 9x2 16y2 18x 32y 151 0 x 2 y 2 9 4 5y 2 4x 2 20 x e 0.2056.
e c a.
1.5045 109 1.3495 109 x2 324 y2 196 1.
t 2 1 0 1 2 3 x y 732 Chapter 9 Topics in Analytic Geometry 68.
and In Exercises 69– 74, sketch the curve represented by the parametric equations (indicate the orientation of the curve).
Then eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve.
Adjust the domain of the resulting rectangular equation, if necessary.
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In Exercises 75–86, use a graphing utility to graph the curve represented by the parametric equations.
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In Exercises 87–90, find two different sets of parametric equations for the given rectangular equation. (There are many correct answers.) 87.
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In Exercises 91–94, find a set of parametric equations for the line that passes through the given points. (There are many correct answers.) 91.
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Sports In Exercises 95–98, the quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught at a height of 4 feet, 30 yards directly downfield. The pass is released at an angle of with the horizontal. The parametric equations for the path of the football are given by and where is the speed of the football (in feet per second) when it is released.
95. Find the speed of the football when it is released.
96. Write a set of parametric equations for the path of the ball.
97. Use a graphing utility to graph the path of the ball and approximate its maximum height.
98. Find the time the receiver has to position himself after the quarterback releases the ball.
9.6 In Exercises 99–104, plot the point given in polar coordinates and find three additional polar representations of the point, using 99.
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In Exercises 105–110, plot the point given in polar coordinates and find the corresponding rectangular coordinates for the point.
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In Exercises 111–114, plot the point given in rectangular coordinates and find two sets of polar coordinates for the point for 111.
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In Exercises 115–122, convert the rectangular equation to polar form.
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122. 2x2 3y2 1 4x2 y2 1 xy 2 xy 5 x2 y2 6y 0 x 2 y 2 4x 0 x2 y2 20 x2 y2 9 3, 3 5, 5 3, 4 0, 9 0 } < 2. 0, 2 3, 3 4 1, 11 6 2, 5 3 4, 2 3 5, 7 6 10, 3 4 5, 4 3 1, 5 6 2, 11 6 5, 3 1, 4 2 < < 2.
v0 y 7 1 0.57v0t 16t2 x 0.82v0t 35 0, 0, 5 2, 6 1, 6, 10, 0 2, 1, 2, 4 3, 5, 8, 5 y 2x3 5x y x2 2 y 10 x y 6x 2 y 2 5 sin y 6 sin x 3 3 cos x 6 cos y 2 x t y t x 3 y t 2 x t 4 y 2 3t x 1 4t y t x t 2 y 4t x 2t y 1 t x t y t x 1 t y 3 t x t y t x 3 t y t2 1 y 1 2t 2 x 4 t x t 3 y t2 y 4t2 3 x ln 4t x t2 2 y 8 3t y 2t 5 x 4t 1 x 5t 1 y 8 t x t t 0 1 2 3 4 x y Review Exercises 733 In Exercises 123–130, convert the polar equation to rectangular form.
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9.7 In Exercises 131–136, sketch the graph of the polar equation by hand. Then use a graphing utility to verify your graph.
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In Exercises 137–144, identify and then sketch the graph of the polar equation. Identify any symmetry, maximum r-values, and zeros of r. Use a graphing utility to verify your graph.
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9.8 In Exercises 145–150, identify the conic represented by the equation algebraically. Then use a graphing utility to graph the polar equation.
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In Exercises 151–154, find a polar equation of the conic with its focus at the pole.
151. Parabola, vertex: 152. Parabola, vertex: 153. Ellipse, vertices: 154. Hyperbola, vertices: 155. Planetary Motion The planet Mars has an elliptical orbit with an eccentricity of The length of the major axis of the orbit is approximately 3.05 astronomical units. Find a polar equation for the orbit and its perihelion and aphelion distances.
156. Astronomy An asteroid takes a parabolic path with Earth as its focus. It is about 6,000,000 miles from Earth at its closest approach. Write the polar equation of the path of the asteroid with its vertex at Find the distance between the asteroid and Earth when Synthesis True or False?
In Exercises 157 and 158, determine whether the statement is true or false. Justify your answer.
157. The graph of represents the equation of a hyperbola.
158. There is only one set of parametric equations that repre-sents the line Writing In Exercises 159 and 160, an equation and four variations are given. In your own words, describe how the graph of each of the variations differs from the graph of the original equation.
159.
(a) (b) (c) (d) 160.
(a) (b) (c) (d) 161. Consider an ellipse whose major axis is horizontal and 10 units in length. The number b in the standard form of the equation of the ellipse must be less than what real num-ber? Describe the change in the shape of the ellipse as b approaches this number.
162. The graph of the parametric equations and is shown in the figure. Would the graph change for the equations and If so, how would it change?
Figure for 162 Figure for 163 163. The path of a moving object is modeled by the parametric equations and where t is time (see figure). How would the path change for each of the following?
(a) (b) y 3 sin t x 5 cos t, y 3 sin 2t x 4 cos 2t, y 3 sin t, x 4 cos t −6 −4 6 4 −6 −4 6 4 y 3 tant?
x 2 sect y 3 tan t x 2 sec t x 32 4 y2 9 1 x 2 4 y 2 25 1 x 2 4 y 2 4 1 x 2 9 y 2 4 1 x 2 4 y 2 9 1 y 2 4x y 2 8x y 2 8x 1 y 22 8x y2 8x y 3 2x.
1 4x2 y4 1 3.
2.
e 0.093.
1, 0, 7, 0 5, 0, 1, 2, 2 2, r 3 4 4 cos r 5 6 2 sin r 6 1 4 cos r 4 5 3 cos r 1 1 2 sin r 2 1 sin r2 cos 2 r2 5 sin 2 r cos 5 r 3 cos 2 r 2 6 cos r 3 5 sin r 1 4 sin r 5 4 cos r 2 sin r 5 cos 5 6 2 r 3 r 5 4 3 5 6 r 2 sin r 2 cos 2 r 8 sin r 3 cos r 12 r 5 9 Chapter Test See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Take this test as you would take a test in class. After you are finished, check your work against the answers given in the back of the book.
In Exercises 1–3, graph the conic and identify any vertices and foci.
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4. Find the standard form of the equation of the parabola with focus and directrix and sketch the parabola.
5. Find the standard form of the equation of the ellipse shown at the right.
6. Find the standard form of the equation of the hyperbola with vertices and asymptotes 7. Use a graphing utility to graph the conic Describe your viewing window.
8. (a) Determine the number of degrees the axis must be rotated to eliminate the -term of the conic (b) Graph the conic in part (a) and use a graphing utility to confirm your result.
9. Solve the system of equations at the right algebraically by the method of substitution.
Then verify your results by using a graphing utility.
In Exercises 10–12, sketch the curve represented by the parametric equations. Then eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve.
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In Exercises 13–15, find two different sets of parametric equations for the given rec-tangular equation. (There are many correct answers.) 13.
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16. Convert the polar coordinate to rectangular form.
17. Convert the rectangular coordinate to polar form and find two additional polar representations of this point. (There are many correct answers.) 18. Convert the rectangular equation to polar form.
19. Convert the polar equation to rectangular form.
In Exercises 20–22, identify the conic represented by the polar equation algebraically.
Then use a graphing utility to graph the polar equation.
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23. Find a polar equation of an ellipse with its focus at the pole, an eccentricity of and directrix at 24. Find a polar equation of a hyperbola with its focus at the pole, an eccentricity of and directrix at 25. For the polar equation find the maximum value of and any zeros of Verify your answers numerically.
r.
r r 8 cos 3, y 2.
e 5 4, y 4.
e 1 4, r 4 2 3 sin r 1 1 cos r 2 3 sin r 2 sin x2 y 2 12y 0 2, 2 14, 5 3 x2 4y2 16 0 x y2 4 y x2 10 y 2 sin y t 4 y 1 2t 1 x 2 3 cos x t2 2 x t2 6 x2 6xy y2 6 0.
xy x2 y2 4 1.
y ±3 2x.
0, ±3 x 4, 8, 2 x2 4y2 4x 0 y2 4x 4 0 y2 8x 0 734 Chapter 9 Topics in Analytic Geometry −19 −5 5 11 (−10, 3) (−2, 3) (−6, 10) (−6, −4) Figure for 5 System for 9 x2 2y2 4x 6y 5 0 x y 5 0 7–9 Cumulative Test See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
Take this test to review the material in Chapters 7–9. After you are finished, check your work against the answers given in the back of the book.
In Exercises 1–4, use any method to solve the system of equations.
1.
2.
3.
4.
In Exercises 5–8, perform the matrix operations given and 5.
6.
7.
8.
9. Find (a) the inverse of A (if it exists) and (b) the determinant of A.
10. Use a determinant to find the area of the triangle with vertices and 11. Write the first five terms of each sequence (Assume that n begins with 1.) (a) (b) In Exercises 12–15, find the sum. Use a graphing utility to verify your result.
12.
13.
14.
15.
In Exercises 16–18, find the sum of the inifinite geometric series.
16.
17.
18.
19. Use mathematical induction to prove the formula In Exercises 20–23, use the Binomial Theorem to expand and simplify the expression.
20.
21.
22.
23.
In Exercises 24–27, find the number of distinguishable permutations of the group of letters.
24. M, I, A, M, I 25. B, U, B, B, L, E 26. B, A, S, K, E, T, B, A, L, L 27. A, N, T, A, R, C, T, I, C, A 3a 4b8 x 2y6 2x y25 x 34 3 7 11 15 . . . 4n 1 n2n 1.
4 2 1 1 2 1 4 . . . n1 50.02n n0 3 3 5 n 50 n0 100 1 2 n 10 n0 9 3 4 n 4 k1 2 k2 4 6 k1 7k 2 an 32n1 an 1n1 2n 3 an.
8, 10.
6, 2, 0, 0, A 1 3 5 2 7 7 1 10 15 BA AB 5A 3B 3A 2B B 1 6 0 5 3 4 2 3 2.
A 3 2 4 0 4 8 4 5 1 x 5x 2x 4y 2y 8y 3z z 5 1 30 2x 4x x 3y y 3y z 2z 3z 13 6 12 2x x y2 y 0 4 x 3y 4x 2y 5 10 Cumulative Test for Chapters 7–9 735 In Exercises 28–31, identify the conic and sketch its graph.
28.
29.
30.
31.
In Exercises 32–34, find the standard form of the equation of the conic.
32.
33.
34.
35. Use a graphing utility to graph Determine the angle through which the axes are rotated.
In Exercises 36–38, (a) sketch the curve represented by the parametric equations, (b) use a graphing utility to verify your graph, and (c) eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve.
Adjust the domain of the resulting rectangular equation, if necessary.
36.
37.
38.
In Exercises 39–42, find two different sets of parametric equations for the given rectangular equation. (There are many correct answers.) 39.
40.
41.
42.
In Exercises 43–46, plot the point given in polar coordinates and find three additional polar representations for 43.
44.
45.
46.
47. Convert the rectangular equation to polar form.
48. Convert the polar equation to rectangular form.
49. Convert the polar equation to rectangular form.
In Exercises 50–52, identify the graph represented by the polar equation algebraical-ly. Then use a graphing utility to graph the polar equation.
50.
51.
52.
53. The salary for the first year of a job is $32,500. During the next 14 years, the salary increases by 5% each year. Determine the total compensation over the 15-year period.
54. On a game show, the digits 3, 4, and 5 must be arranged in the proper order to form the price of an appliance. If they are arranged correctly, the contestant wins the appliance.
What is the probability of winning if the contestant knows that the price is at least $400?
55. A parabolic archway is 16 meters high at the vertex. At a height of 14 meters, the width of the archway is 12 meters, as shown in the figure at the right. How wide is the archway at ground level?
r 2 5 cos r 3 2 sin r 6 r 2 4 5 cos r 2 cos 4x 4y 1 0 3, 11 6 2, 5 4 5, 3 4 8, 5 6 2 < < 2.
y e2x e2x 1 y 2 x x2 y2 16 y 3x 2 y 1 2t2 y 2 sin2 y t2 x 4 ln t x cos x 2t 1 x2 4xy 2y2 6.
−12 −12 12 4 (4, 0) (0, −6) (0, −2) −2 −8 10 10 (−4, 4) (6, 4) (1, 2) (1, 6) −2 −3 6 4 (0, 0) (2, 3) (4, 0) x2 y2 2x 4y 1 0 y2 x2 16 x 22 4 y 12 9 1 y 32 36 x 52 121 1 736 Chapter 9 Topics in Analytic Geometry x (0, 16) (6, 14) (−6, 14) 8 −8 −8 16 8 24 y Figure for 55 Proofs in Mathematics Proofs in Mathematics 737 Proof For the case in which the directrix is parallel to the -axis and the focus lies above the vertex, as shown in the top figure, if is any point on the parabo-la, then, by definition, it is equidistant from the focus and the directrix So, you have For the case in which the directrix is parallel to the -axis and the focus lies to the right of the vertex, as shown in the bottom figure, if is any point on the parabola, then, by definition, it is equidistant from the focus and the directrix So, you have Note that if a parabola is centered at the origin, then the two equations above would simplify to and respectively.
y2 4px, x2 4py y k2 4px h. 2px 2ph y k2 2px 2ph x2 2xh p h p2 y k2 x2 2xh p h p2 x h p2 y k2 x h p2 x h p2 y k2 x h p x h p.
h p, k x, y y x h2 4py k. x h2 2py 2pk 2py 2pk x h2 y2 2yk p k p2 y2 2yk p k p2 x h2 y k p2 y k p2 x h2 y k p2 y k p y k p.
h, k p x, y x Focus: ( , + ) h k p Directrix: = y k p − Vertex: ( , ) h k p > 0 = x h Axis: (x, y) Parabola with vertical axis y = k h p Focus: ( + , ) k = x h p − Vertex: ( , ) h k p > 0 Axis: Directrix: (x, y) Parabola with horizontal axis Parabolic Paths There are many natural occur-rences of parabolas in real life.
For instance, the famous astronomer Galileo discovered in the 17th century that an object that is projected upward and obliquely to the pull of gravity travels in a parabolic path. Examples of this are the center of gravity of a jumping dolphin and the path of water molecules in a drinking fountain.
x h2 y2 2ky 2py k2 2pk p2 y2 2ky 2py k2 2pk p2 x2 2hx 2px h2 2ph p2 y k2 x2 2hx 2px h2 2ph p2 Standard Equation of a Parabola (p. 663) The standard form of the equation of a parabola with vertex at is as follows.
Vertical axis, directrix: Horizontal axis, directrix: The focus lies on the axis units (directed distance) from the vertex. If the vertex is at the origin the equation takes one of the following forms.
Vertical axis Horizontal axis y2 4px x2 4py 0, 0, p x h p y k2 4px h, p 0 y k p x h2 4py k, p 0 h, k 738 Chapter 9 Topics in Analytic Geometry Proof You need to discover how the coordinates in the xy-system are related to the coordinates in the -system. To do this, choose a point in the original system and attempt to find its coordinates in the rotated system.
In either system, the distance r between the point P and the origin is the same. So, the equations for and are those given in the figure. Using the formulas for the sine and cosine of the difference of two angles, you have the following.
Solving this system for x and y yields and Finally, by substituting these values for x and y into the original equation and collecting terms, you obtain To eliminate the -term, you must select such that 0, sin 2 0 Bsin 2 C A B cot 2 C A sin 2 B cos 2 B 2C A sin cos Bcos2 sin2 B 0.
x y F F. E D sin E cos D D cos E sin C A sin2 B cos sin C cos2 A A cos2 B cos sin C sin2 y x sin y cos . x x cos y sin x cos y sin r cos cos r sin sin rcos cos sin sin x r cos y x, y, x , x , y P x, y x y Rotation of Axes to Eliminate an xy-Term (p. 690) The general second-degree equation can be rewritten as by rotating the coordinate axes through an angle where The coefficients of the new equation are obtained by making the substitu-tions and y x sin y cos .
x x cos y sin cot 2 A C B .
, A x 2 C y 2 D x E y F 0 Ax2 Bxy Cy2 Dx Ey F 0 P = (x, y) r α x y Original: y r sin x r cos x′ y′ P = (x′, y′) r − θ θ α x y Rotated: y r sin x r cos y cos x sin r sin cos r cos sin rsin cos cos sin y r sin Proofs in Mathematics 739 If no rotation is necessary because the -term is not present in the original equation. If the only way to make is to let So, you have established the desired results.
Proof A proof for with is shown here. The proofs of the other cases are similar. In the figure, consider a vertical directrix, units to the right of the focus If is a point on the graph of the distance between and the directrix is Moreover, because the distance between and the pole is simply the ratio of is and, by definition, the graph of the equation must be a conic.
PF PQ r r e e e PF to PQ PF r, P r e . p 1 e cos p 1 e cos 1 e cos p ep 1 e cos cos p r cos PQ p x P r ep 1 e cos P r, F 0, 0.
p p > 0 r ep 1 e cos cot 2 A C B , B 0.
B 0 B 0, xy B 0, 2 F = (0, 0) 0 P r = ( , ) θ θ θ x = r cos Directrix p Q r π Polar Equations of Conics (p. 722) The graph of a polar equation of the form 1.
or 2.
is a conic, where is the eccentricity and is the distance between the focus (pole) and the directrix.
p e > 0 r ep 1 ± e sin r ep 1 ± e cos Progressive Summary 740 740 Chapter 9 Topics in Analytic Geometry This chart outlines the topics that have been covered so far in this text. Progressive Summary charts appear after Chapters 2, 3, 6, 9, and 11. In each progressive summary, new topics encountered for the first time appear in red.
Transcendental Functions Exponential,Logarithmic, Trigonometric,Inverse Trigonometric Rewriting Exponential form ↔ Logarithmic form Condense/expand logarithmic expressions Simplify trigonometric expressions Prove trigonometric identities Use conversion formulas Operations with vectors Powers and roots of complex numbers Solving Equation Strategy Exponential . . . . . . . Take logarithm of each side Logarithmic . . . . . . . Exponentiate each side Trigonometric . . . . . Isolate function factor, use inverse function Multiple angle . . . . . Use trigonometric or high powers identities Analyzing Graphically Algebraically Intercepts Domain, Range Asymptotes Transformations Minimum values Composition Maximum values Inverse properties Amplitude, period Reference angles Numerically Table of values Other Topics Conics,Parametric and Polar Equations, Rewriting Standard forms of conics Eliminate parameters Rectangular form ↔ Parametric form Rectangular form ↔ Polar form Solving Equation Strategy Conics . . . . . . . . . . . Convert to standard form Convert to polar form Analyzing Conics: Table of values, vertices, foci, axes, symmetry, asymptotes, translations, eccentricity Parametric forms: Point plotting, eliminate parameters Polar forms: Point plotting, special equations, symmetry, zeros, eccentricity, maximum r-values, directrix Progressive Summary (Chapters 3–9) Systems and Series Systems,Sequences,Series Rewriting Row operations for systems of equations Partial fraction decomposition Operations with matrices Matrix form of a system of equations nth term of a sequence Summation form of a series Solving Equation Strategy System of . . . . . . . . . Substitution linear equations Elimination Gaussian Gauss-Jordan Inverse matrices Cramer’s Rule Analyzing Systems: Intersecting, parallel, and coincident lines, determinants Sequences: Graphing utility in dot mode, nth term, partial sums, summation formulas |
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Home » Google Sheets Formulas » How to Apply Percentage Increase Formula in Google Sheets
Google Sheets Formulas
How to Apply Percentage Increase Formula in Google Sheets
ExcelInsider Team August 25, 2025
We often need to calculate percentage change in various tasks. In this article, you will learn a simple formula to calculate percentage increase when working with data such as monthly expenses, budget monitoring, project reports, and so on. I’ll calculate how much more or less a product’s sales have changed based on monthly data and compare the prices of products across two years, based on my dataset.
The process of applying the formula is explained step-by-step using a live Google Sheets dataset. By the end, you’ll be able to calculate the percentage increase or decrease in any dataset.
Key Takeaways
Steps to apply percentage increase formula:
➤Organize your data in either a single column or in two columns, depending on your use case.
➤Apply the formula anywhere in the sheet, but it’s recommended to place it next to the data column to keep things organized
=(New Value – Old Value) / Old Value➤For example, if your data is in column B (in cells B2 and B3), apply the formula in cell C3 like this and press Enter
=(B3 – B2) / B2
Download Practice Workbook
Basic Formula to Calculate Percentage Increase
This is one of the most common and useful formulas in spreadsheets. It helps you calculate how much a value has increased or decreased, expressed in percentage terms.
=(New Value – Old Value) / Old Value
Here, Newvalue means the updated or recent value
And Oldrefers to the original or previous value.
1
Determine Percentage Increase with Basic Division Formula
Let me show you an example using a simple dataset of product sales to calculate the percentage increase row by row, so you can understand it better. According to my dataset, I want to know how much sales increased or decreased each month compared to the previous month.
Steps:
➤Open your Google Sheet and get your data organized like this:
➤Now, simply apply the function. I will be applying the function in cell C3, because Jan has no previous month. So, my formula will be:
=(B3 – B2) / B2
➤B3 refers to the New value, which is Feb sales. AndB2 refers to the Old value, which is Jan sales
➤After applying the function, press Enter
➤As you can see, it appears as a decimal. To change that to a percentage, simply select the cell and click the% icon from the toolbar.
➤Now the value will be shown as a percentage
Referring to the dataset, the product sales increased by 12.50% in February compared to January.
Note:How to apply the formula to the entire column? To do that, click on the cell where you entered the formula (C3). You’ll see a small blue dot in the bottom-right corner. Click>>hold and drag it down the column to fill the formula for all rows.
➤After dragging, the result will look like this:
Now, if you want to calculate product sales only based on Jan, then the following formula will show it
For this, we need to lock the January cell (B2). So, I’ll be using the same basic formula, but with dollar signs($) to create an absolute reference. The $ signs lock the reference so it doesn’t change when copying the formula down the column. The formula looks like this:
=(B3 – $B$2) / $B$2
Steps:
➤Create a new column and apply the formula
=(B3 – $B$2) / $B$2
➤Hit Enter.As usual, the result will appear in decimal format. To convert it to a percentage, select the cell and click the %icon from the toolbar.
➤Use the fill handle (the blue dot) to drag both formulas down, from D3 to D11. This will automatically calculate the percentage of the total for each row.
2
Calculate Percentage From Column to Column
I have another dataset that shows product price changes between 2024 and 2025. I want to calculate how much each product’s price has increased or decreased from 2024 to 2025. For this, I’ll use the same formula as before, but the cell references will be different.
You can also use this same formula to calculate both row-to-row and column-to-column percentage changes.
Steps:
➤Click on cell D2 and enter the following function
=(C2 – B2) / B2
➤Here C2 is the New Value and B2 is the Old Value
➤Press Enter
This means the Laptop price increased by 11.76% from 2024 to 2025.
➤Now, just click and drag down like before to apply the formula to the whole column.
Frequently Asked Questions
How to Calculate the Percentage of a Number?
To find the percentage of a number, use the following formula
= TotalPercentage
Totalthe whole value you want to calculate a percentage of
Percentage is the rate or portion you want to take from the total.
For example, If your total monthly budget is $1000, and you want to know how much 20% of that is, the formula will be:
=1000 20%
And answer will be $200. So 20% of $1000 is $200.
How to minus a percentage in Google Sheets?
To subtract or minus a percentage from a number, you can use this formula:
=Original Value – (Original Value Percentage)
Original Value is the full amount number you want to reduce.
Percentage is the rate you want to subtract from the original value.
For instance, if you want to apply a 10% discount on a product that costs $500, then the formula will be:
=500 – (500 10%)
After subtracting 10%, the new price is $450.
Concluding Words
This is where we wrap up. By using the function, you can get concise and actionable insights from your data. You can use them while working with a monthly budget, monitoring spending patterns, projecting future values, or anything related to getting percentages. This easy formula will streamline your workflow. Try using them in your own dataset, and you will find just how easy and powerful spreadsheet percentage calculations can be.
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Table of Contents
Basic Formula to Calculate Percentage Increase
Determine Percentage Increase with Basic Division Formula
Calculate Percentage From Column to Column
Frequently Asked Questions
How to Calculate the Percentage of a Number?
How to minus a percentage in Google Sheets?
Concluding Words
Related Articles
How to Calculate Average Percentage in Google Sheets Google Sheets Formulas
How to Add Percentage in Google Sheets (with Different Criteria) Google Sheets Formulas
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Equations>Equations with x in the Denominator
Equations with x in the Denominator
Having learned how to use common denominators to get rid of fractions in equations, you might be wondering how to isolate a variable that is in the denominator. This is a great question!
The good news is that the approach is the same as with getting rid of a denominator that contains only numbers! To find the common denominator, you multiply all the different factors once. Here’s two examples:
Example 1
Solve the equation
Example 2
Solve the equation for
The common denominator is . Multiply both sides of the equation by the common denominator:
| |
| |
This expands to
| |
| |
Now, you can cancel some factors:
| |
| |
Now the expression simplifies to
| |
| |
Isolate all the variables on one side and the constants on the other:
| |
| |
This simplifies to
| |
| |
Finally, get rid of the number in front of :
| |
| |
Therefore, .
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College Physics 2e
23.9 Inductance
College Physics 2e23.9 Inductance
Contents Contents
Highlights
Table of contents
Preface
1 Introduction: The Nature of Science and Physics
2 Kinematics
3 Two-Dimensional Kinematics
4 Dynamics: Force and Newton's Laws of Motion
5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
6 Uniform Circular Motion and Gravitation
7 Work, Energy, and Energy Resources
8 Linear Momentum and Collisions
9 Statics and Torque
10 Rotational Motion and Angular Momentum
11 Fluid Statics
12 Fluid Dynamics and Its Biological and Medical Applications
13 Temperature, Kinetic Theory, and the Gas Laws
14 Heat and Heat Transfer Methods
15 Thermodynamics
16 Oscillatory Motion and Waves
17 Physics of Hearing
18 Electric Charge and Electric Field
19 Electric Potential and Electric Field
20 Electric Current, Resistance, and Ohm's Law
21 Circuits and DC Instruments
22 Magnetism
23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
23.1 Induced Emf and Magnetic Flux
23.2 Faraday’s Law of Induction: Lenz’s Law
23.3 Motional Emf
23.4 Eddy Currents and Magnetic Damping
23.5 Electric Generators
23.6 Back Emf
23.7 Transformers
23.8 Electrical Safety: Systems and Devices
23.9 Inductance
23.10 RL Circuits
23.11 Reactance, Inductive and Capacitive
23.12 RLC Series AC Circuits
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
24 Electromagnetic Waves
25 Geometric Optics
26 Vision and Optical Instruments
27 Wave Optics
28 Special Relativity
29 Quantum Physics
30 Atomic Physics
31 Radioactivity and Nuclear Physics
32 Medical Applications of Nuclear Physics
33 Particle Physics
34 Frontiers of Physics
A | Atomic Masses
B | Selected Radioactive Isotopes
C | Useful Information
D | Glossary of Key Symbols and Notation
Answer Key
Index
Search for key terms or text.
Close
Learning Objectives
By the end of this section, you will be able to:
Calculate the inductance of an inductor.
Calculate the energy stored in an inductor.
Calculate the emf generated in an inductor.
Inductors
Induction is the process in which an emf is induced by changing magnetic flux. Many examples have been discussed so far, some more effective than others. Transformers, for example, are designed to be particularly effective at inducing a desired voltage and current with very little loss of energy to other forms. Is there a useful physical quantity related to how “effective” a given device is? The answer is yes, and that physical quantity is called inductance.
Mutual inductance is the effect of Faraday’s law of induction for one device upon another, such as the primary coil in transmitting energy to the secondary in a transformer. See Figure 23.37, where simple coils induce emfs in one another.
Figure 23.37 These coils can induce emfs in one another like an inefficient transformer. Their mutual inductance M indicates the effectiveness of the coupling between them. Here a change in current in coil 1 is seen to induce an emf in coil 2. (Note that "E 2 E 2 E 2 induced" represents the induced emf in coil 2.)
In the many cases where the geometry of the devices is fixed, flux is changed by varying current. We therefore concentrate on the rate of change of current, Δ I/Δ t Δ I/Δ t Δ I/Δ t, as the cause of induction. A change in the current I 1 I 1 I 1 in one device, coil 1 in the figure, induces an emf 2 emf 2 emf 2 in the other. We express this in equation form as
emf 2=−M Δ I 1 Δ t,emf 2=−M Δ I 1 Δ t,emf 2=−M Δ I 1 Δ t,
23.34
where M M M is defined to be the mutual inductance between the two devices. The minus sign is an expression of Lenz’s law. The larger the mutual inductance M M M, the more effective the coupling. For example, the coils in Figure 23.37 have a small M M M compared with the transformer coils in Figure 23.27. Units for M M M are (V⋅s)/A=Ω⋅s(V⋅s)/A=Ω⋅s(V⋅s)/A=Ω⋅s, which is named a henry (H), after Joseph Henry. That is, 1 H=1 Ω⋅s 1 H=1 Ω⋅s 1 H=1 Ω⋅s.
Nature is symmetric here. If we change the current I 2 I 2 I 2 in coil 2, we induce an emf 1 emf 1 emf 1 in coil 1, which is given by
emf 1=−M Δ I 2 Δ t,emf 1=−M Δ I 2 Δ t,emf 1=−M Δ I 2 Δ t,
23.35
where M M M is the same as for the reverse process. Transformers run backward with the same effectiveness, or mutual inductance M M M .
A large mutual inductance M M M may or may not be desirable. We want a transformer to have a large mutual inductance. But an appliance, such as an electric clothes dryer, can induce a dangerous emf on its case if the mutual inductance between its coils and the case is large. One way to reduce mutual inductance M M M is to counterwind coils to cancel the magnetic field produced. (See Figure 23.38.)
Figure 23.38 The heating coils of an electric clothes dryer can be counter-wound so that their magnetic fields cancel one another, greatly reducing the mutual inductance with the case of the dryer.
Self-inductance, the effect of Faraday’s law of induction of a device on itself, also exists. When, for example, current through a coil is increased, the magnetic field and flux also increase, inducing a counter emf, as required by Lenz’s law. Conversely, if the current is decreased, an emf is induced that opposes the decrease. Most devices have a fixed geometry, and so the change in flux is due entirely to the change in current Δ I Δ I Δ I through the device. The induced emf is related to the physical geometry of the device and the rate of change of current. It is given by
emf=−L Δ I Δ t,emf=−L Δ I Δ t,emf=−L Δ I Δ t,
23.36
where L L L is the self-inductance of the device. A device that exhibits significant self-inductance is called an inductor, and given the symbol in Figure 23.39.
Figure 23.39
The minus sign is an expression of Lenz’s law, indicating that emf opposes the change in current. Units of self-inductance are henries (H) just as for mutual inductance. The larger the self-inductance L L L of a device, the greater its opposition to any change in current through it. For example, a large coil with many turns and an iron core has a large L L L and will not allow current to change quickly. To avoid this effect, a small L L L must be achieved, such as by counterwinding coils as in Figure 23.38.
A 1 H inductor is a large inductor. To illustrate this, consider a device with L=1.0 H L=1.0 H L=1.0 H that has a 10 A current flowing through it. What happens if we try to shut off the current rapidly, perhaps in only 1.0 ms? An emf, given by emf=−L(Δ I/Δ t)emf=−L(Δ I/Δ t)emf=−L(Δ I/Δ t), will oppose the change. Thus an emf will be induced given by emf=−L(Δ I/Δ t)=(1.0 H)[(10 A)/(1.0 ms)]=10,000 V emf=−L(Δ I/Δ t)=(1.0 H)[(10 A)/(1.0 ms)]=10,000 V emf=−L(Δ I/Δ t)=(1.0 H)[(10 A)/(1.0 ms)]=10,000 V. The positive sign means this large voltage is in the same direction as the current, opposing its decrease. Such large emfs can cause arcs, damaging switching equipment, and so it may be necessary to change current more slowly.
There are uses for such a large induced voltage. Camera flashes use a battery, two inductors that function as a transformer, and a switching system or oscillator to induce large voltages. (Remember that we need a changing magnetic field, brought about by a changing current, to induce a voltage in another coil.) The oscillator system will do this many times as the battery voltage is boosted to over one thousand volts. (You may hear the high pitched whine from the transformer as the capacitor is being charged.) A capacitor stores the high voltage for later use in powering the flash. (See Figure 23.40.)
Figure 23.40 Through rapid switching of an inductor, 1.5 V batteries can be used to induce emfs of several thousand volts. This voltage can be used to store charge in a capacitor for later use, such as in a camera flash attachment.
It is possible to calculate L L L for an inductor given its geometry (size and shape) and knowing the magnetic field that it produces. This is difficult in most cases, because of the complexity of the field created. So in this text the inductance L L L is usually a given quantity. One exception is the solenoid, because it has a very uniform field inside, a nearly zero field outside, and a simple shape. It is instructive to derive an equation for its inductance. We start by noting that the induced emf is given by Faraday’s law of induction as emf=−N(Δ Φ/Δ t)emf=−N(Δ Φ/Δ t)emf=−N(Δ Φ/Δ t) and, by the definition of self-inductance, as emf=−L(Δ I/Δ t)emf=−L(Δ I/Δ t)emf=−L(Δ I/Δ t). Equating these yields
emf=−N Δ Φ Δ t=−L Δ I Δ t.emf=−N Δ Φ Δ t=−L Δ I Δ t.emf=−N Δ Φ Δ t=−L Δ I Δ t.
23.37
Solving for L L L gives
L=N Δ Φ Δ I.L=N Δ Φ Δ I.L=N Δ Φ Δ I.
23.38
This equation for the self-inductance L L L of a device is always valid. It means that self-inductance L L L depends on how effective the current is in creating flux; the more effective, the greater Δ Φ Δ Φ Δ Φ/ Δ I Δ I Δ I is.
Let us use this last equation to find an expression for the inductance of a solenoid. Since the area A A A of a solenoid is fixed, the change in flux is Δ Φ=Δ(B A)=A Δ B Δ Φ=Δ(B A)=A Δ B Δ Φ=Δ(B A)=A Δ B. To find Δ B Δ B Δ B, we note that the magnetic field of a solenoid is given by B=μ 0 nI=μ 0 NI ℓ B=μ 0 nI=μ 0 NI ℓ B=μ 0 nI=μ 0 NI ℓ. (Here n=N/ℓ n=N/ℓ n=N/ℓ, where N N N is the number of coils and ℓ ℓ ℓ is the solenoid’s length.) Only the current changes, so that Δ Φ=A Δ B=μ 0 NA Δ I ℓ Δ Φ=A Δ B=μ 0 NA Δ I ℓ Δ Φ=A Δ B=μ 0 NA Δ I ℓ. Substituting Δ Φ Δ Φ Δ Φ into L=N Δ Φ Δ I L=N Δ Φ Δ I L=N Δ Φ Δ I gives
L=N Δ Φ Δ I=N μ 0 NA Δ I ℓ Δ I.L=N Δ Φ Δ I=N μ 0 NA Δ I ℓ Δ I.L=N Δ Φ Δ I=N μ 0 NA Δ I ℓ Δ I.
23.39
This simplifies to
L=μ 0 N 2 A ℓ(solenoid).L=μ 0 N 2 A ℓ(solenoid).L=μ 0 N 2 A ℓ(solenoid).
23.40
This is the self-inductance of a solenoid of cross-sectional area A A A and length ℓ ℓ ℓ. Note that the inductance depends only on the physical characteristics of the solenoid, consistent with its definition.
Example 23.7
Calculating the Self-inductance of a Moderate Size Solenoid
Calculate the self-inductance of a 10.0 cm long, 4.00 cm diameter solenoid that has 200 coils.
Strategy
This is a straightforward application of L=μ 0 N 2 A ℓ L=μ 0 N 2 A ℓ L=μ 0 N 2 A ℓ, since all quantities in the equation except L L L are known.
Solution
Use the following expression for the self-inductance of a solenoid:
L=μ 0 N 2 A ℓ.L=μ 0 N 2 A ℓ.L=μ 0 N 2 A ℓ.
23.41
The cross-sectional area in this example is A=πr 2=(3.14...)(0.0200 m)2=1.26×10−3 m 2 A=πr 2=(3.14...)(0.0200 m)2=1.26×10−3 m 2 A=πr 2=(3.14...)(0.0200 m)2=1.26×10−3 m 2, N N N is given to be 200, and the length ℓ ℓ ℓ is 0.100 m. We know the permeability of free space is μ 0=4 π×10−7 T⋅m/A μ 0=4π×10−7 T⋅m/A μ 0=4π×10−7 T⋅m/A. Substituting these into the expression for L L L gives
L==(4 π×10−7 T⋅m/A)(200)2(1.26×10−3 m 2)0.100 m 0.632 mH.L=(4π×10−7 T⋅m/A)(200)2(1.26×10−3 m 2)0.100 m=0.632 mH.L=(4π×10−7 T⋅m/A)(200)2(1.26×10−3 m 2)0.100 m=0.632 mH.
23.42
Discussion
This solenoid is moderate in size. Its inductance of nearly a millihenry is also considered moderate.
One common application of inductance is used in traffic lights that can tell when vehicles are waiting at the intersection. An electrical circuit with an inductor is placed in the road under the place a waiting car will stop over. The body of the car increases the inductance and the circuit changes sending a signal to the traffic lights to change colors. Similarly, metal detectors used for airport security employ the same technique. A coil or inductor in the metal detector frame acts as both a transmitter and a receiver. The pulsed signal in the transmitter coil induces a signal in the receiver. The self-inductance of the circuit is affected by any metal object in the path. Such detectors can be adjusted for sensitivity and also can indicate the approximate location of metal found on a person. See Figure 23.41.
Figure 23.41 The familiar security gate at an airport can not only detect metals but also indicate their approximate height above the floor. (credit: shankar s/Flickr)
Energy Stored in an Inductor
We know from Lenz’s law that inductances oppose changes in current. There is an alternative way to look at this opposition that is based on energy. Energy is stored in a magnetic field. It takes time to build up energy, and it also takes time to deplete energy; hence, there is an opposition to rapid change. In an inductor, the magnetic field is directly proportional to current and to the inductance of the device. It can be shown that the energy stored in an inductor E ind E ind E ind is given by
E ind=1 2 LI 2.E ind=1 2 LI 2.E ind=1 2 LI 2.
23.43
This expression is similar to that for the energy stored in a capacitor.
Example 23.8
Calculating the Energy Stored in the Field of a Solenoid
How much energy is stored in the 0.632 mH inductor of the preceding example when a 30.0 A current flows through it?
Strategy
The energy is given by the equation E ind=1 2 LI 2 E ind=1 2 LI 2 E ind=1 2 LI 2, and all quantities except E ind E ind E ind are known.
Solution
Substituting the value for L L L found in the previous example and the given current into E ind=1 2 LI 2 E ind=1 2 LI 2 E ind=1 2 LI 2 gives
E ind==1 2 LI 2 0.5(0.632×10−3 H)(30.0 A)2=0.284 J.E ind=1 2 LI 2=0.5(0.632×10−3 H)(30.0 A)2=0.284 J.E ind=1 2 LI 2=0.5(0.632×10−3 H)(30.0 A)2=0.284 J.
23.44
Discussion
This amount of energy is certainly enough to cause a spark if the current is suddenly switched off. It cannot be built up instantaneously unless the power input is infinite.
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8367 | https://www.quora.com/How-does-the-value-of-g-vary-with-altitude | Something went wrong. Wait a moment and try again.
Gravitational Acceleratio...
Earth Science
Weight or Acceleration Du...
Acceleration (physics)
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5
How does the value of g vary with altitude?
Dhaval Ajana
B.Sc. in Physics & Mathematics, The Maharaja Sayajirao University of Baroda (Graduated 2020)
·
7y
Let's consider earth is perfectly sphere of mass M and radius is R. A body of mass 'm' initially placed on the surface..
'g' value for a body on the initial placed on the surface is,
g = GM/R^2
When the body above at hight h. g' is,
g’ = GM/(R+h)^2
Ratio of g and g' is,
g/g' = (R+h)^2/R^2 = (1+h/R)^2
So, g' = g (1+h/R)^-2
g'= g(1–2h/R)……..x
Above 'x' expression show's when body at hight h it's gravity is less then to body at a surface of the Earth. Gravity g vary with hight.
Above the initial surface of the hight gravity decrease with hight and also below the Earth gravity decrease with depth. At the po
Let's consider earth is perfectly sphere of mass M and radius is R. A body of mass 'm' initially placed on the surface..
'g' value for a body on the initial placed on the surface is,
g = GM/R^2
When the body above at hight h. g' is,
g’ = GM/(R+h)^2
Ratio of g and g' is,
g/g' = (R+h)^2/R^2 = (1+h/R)^2
So, g' = g (1+h/R)^-2
g'= g(1–2h/R)……..x
Above 'x' expression show's when body at hight h it's gravity is less then to body at a surface of the Earth. Gravity g vary with hight.
Above the initial surface of the hight gravity decrease with hight and also below the Earth gravity decrease with depth. At the point of Earth gravity value is zero. Maximum value of g on the initial surface..
Hope you understand. :)
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Related questions
How does the value of 'g' vary with height?
How does the value of G vary on Earth?
How does gravity decrease with increase in altitude?
Does the value of 'g' vary from place to place on the surface of the Earth?
How does altitude affect the value of g?
Ishita Asthana
7y
Originally Answered: How does the value of 'g' vary with height?
·
If you're looking for derivation, consult any physics book of 11–12 standard.
Since you've put it on Quora, I've just explained the concept.
Forces like gravitational force and electrostatic force, are related to distance (between the objects), by the inverse square law. That means, when the separation between two bodies ( masses in case of gravitation) increases to double, the effect of one body on the other will decrease by a factor of four.
So, on increasing the height of body from the surface of earth, the distance between the two masses ( earth and body) increases. As a result, the gravitati
If you're looking for derivation, consult any physics book of 11–12 standard.
Since you've put it on Quora, I've just explained the concept.
Forces like gravitational force and electrostatic force, are related to distance (between the objects), by the inverse square law. That means, when the separation between two bodies ( masses in case of gravitation) increases to double, the effect of one body on the other will decrease by a factor of four.
So, on increasing the height of body from the surface of earth, the distance between the two masses ( earth and body) increases. As a result, the gravitational pull of earth on the body ( or vice-versa according to law of equal and opposite reaction), decreases.
Since 'g' is just acceleration due to gravity, it's value also decreases since the force decreases.
Gopal Menon
B Tech in Chemical Engineering, Indian Institute of Technology, Bombay (IITB) (Graduated 1975)
·
Author has 10.2K answers and 15.2M answer views
·
6y
How does the value of g vary with altitude?
The acceleration due to gravity, g=GMr2, where G,M and r are the gravitational constant, mass of the earth and distance from the centre of the earth.
⇒ On the surface of the earth, g=GMR2, where R is the radius of earth.
⇒ At a height h from the surface of the earth, gh=GM(R+h)2.
This is the relation between the altitude, h, and the acceleration due to gravity, gh, at that altitude.
Assistant
Bot
·
Sep 6
Gravitational acceleration g decreases with altitude because you move farther from Earth's mass. The variation can be described accurately by Newton’s law of universal gravitation and usefully approximated near Earth’s surface.
Core formula (exact for a spherically symmetric Earth)
g(h) = G M / (R + h)^2
G = 6.67430×10^-11 m^3 kg^-1 s^-2 (gravitational constant)
M ≈ 5.972×10^24 kg (Earth’s mass)
R ≈ 6.371×10^6 m (mean Earth radius)
h = altitude above mean sea level (m)
Approximation valid for h ≪ R (linearized)
g(h) ≈ g0 (1 − 2h/R)
g0 ≈ 9.80665 m/s^2 (standard gravity at sea level)
More convenient: g(h)
Gravitational acceleration g decreases with altitude because you move farther from Earth's mass. The variation can be described accurately by Newton’s law of universal gravitation and usefully approximated near Earth’s surface.
Core formula (exact for a spherically symmetric Earth)
g(h) = G M / (R + h)^2
G = 6.67430×10^-11 m^3 kg^-1 s^-2 (gravitational constant)
M ≈ 5.972×10^24 kg (Earth’s mass)
R ≈ 6.371×10^6 m (mean Earth radius)
h = altitude above mean sea level (m)
Approximation valid for h ≪ R (linearized)
g(h) ≈ g0 (1 − 2h/R)
g0 ≈ 9.80665 m/s^2 (standard gravity at sea level)
More convenient: g(h) ≈ g0 − (2 g0 / R) h ≈ g0 − 3.086×10^-6 h (m/s^2 per meter)
Thus g decreases by ≈ 0.00309 m/s^2 per 1,000 m (≈ 0.03% per km) near surface.
Higher-altitude behavior
At large h, the 1/(R+h)^2 law governs: g falls roughly as (distance)^−2. At geostationary orbit (~35,786 km), g ≈ 0.223 m/s^2 (~2.3% of surface g).
In low Earth orbit (LEO, ~300–400 km) g is still ≈ 8.7–8.9 m/s^2; apparent “weightlessness” in orbit is due to free-fall, not zero g.
Corrections and local variations
Earth is not a perfect sphere: oblateness (equatorial bulge) and latitude affect g (centrifugal acceleration from Earth’s rotation reduces effective gravity more at equator).
Local geology (density variations, topography) produces small gravitational anomalies (parts per million to 10^-5 g).
For precise work use geodetic formulas (e.g., International Gravity Formula, WGS‑84) that include latitude and ellipsoidal height.
Practical examples
h = 1 km: g ≈ 9.8036 m/s^2 (drop ≈ 0.0031 m/s^2)
h = 10 km (cruising airliner): g ≈ 9.7758 m/s^2 (drop ≈ 0.0309 m/s^2)
h = 400 km (ISS): g ≈ 8.69–8.70 m/s^2
Use cases
Engineering near surface: linear approximation suffices.
Orbital dynamics: use exact inverse-square law with R+h.
Precise geophysics/geodesy: include rotation, latitude, ellipsoid shape, and local anomalies.
Related questions
How does GPS system determines altitude?
Will the value of (g) change with altitude and latitude? Give examples.
How value of g vary from place to place?
Is gravity less at altitude?
At what altitude would the value of g become one fourth that of the surface of the Earth?
Philip Cooper
Former Engineering Manager (1976–1996)
·
Author has 688 answers and 85.4K answer views
·
6y
The following formula approximates the Earth's gravity variation with altitude:
gh = g0 ( re/ (re + h) )^2
This image has been removed for violating Quora's policy.
Where
gh is the gravitational acceleration at height habove sea level.
re is the Earth's mean radius.
g0 is the standard gravitational acceleration.
The formula treats the Earth as a perfect sphere with a radially symmetric distribution of mass; a more accurate mathematical treatment is discussed below.
The following formula approximates the Earth's gravity variation with altitude:
gh = g0 ( re/ (re + h) )^2
This image has been removed for violating Quora's policy.
Where
gh is the gravitational acceleration at height habove sea level.
re is the Earth's mean radius.
g0 is the standard gravitational acceleration.
The formula treats the Earth as a perfect sphere with a radially symmetric distribution of mass; a more accurate mathematical treatment is discussed below.
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Steve Johnson
BS in Physics, North Dakota State University (Graduated 1966)
·
Author has 5.7K answers and 2.3M answer views
·
2y
Originally Answered: What is the magnitude of g at different altitudes on Earth's surface?
·
The radius of Earth is 6371x10^3 m. (I do not know if that is sea level or an average of the surfaces, wet or dry, around the world.) The tallest mountain is Everest at 8.848x10^3 m. So the peak is 6379x10^3 m above the center of the Earth.
The formula for calculation of g is g = GM/r^2. G and M are constanfs, so r is the only variable affecting g at Earth's different surfaces. Square both 6371x10^3 and 6379x10^3 - compare the results. The ratio of them will give you an idea of how significant changes of g would be due to altitude of various Earth surfaces.
Lyle McElhaney
Self-employed software engineer (1990–present)
·
Author has 23.6K answers and 10.9M answer views
·
4y
g varies with altitude because it has the formula:
g = G m1 / r^2
Where G is a universal constant, m2 is the mass of the Earth and r is the radius of the Earth. This combines a number of constants into a convenient number g, when working with gravity on the surface of our planet; no other place. It reduces the computation of weight from G m1 m2 / r^2 to m2 g. It comes out with the units of m / s^2, or an acceleration. It’s therefore called the acceleration due to gravity on Earth’s surface. It’s equal to about 10 m/s^2. If you increase the distance r, the w = g m doesn’t work anymore and you hav
g varies with altitude because it has the formula:
g = G m1 / r^2
Where G is a universal constant, m2 is the mass of the Earth and r is the radius of the Earth. This combines a number of constants into a convenient number g, when working with gravity on the surface of our planet; no other place. It reduces the computation of weight from G m1 m2 / r^2 to m2 g. It comes out with the units of m / s^2, or an acceleration. It’s therefore called the acceleration due to gravity on Earth’s surface. It’s equal to about 10 m/s^2. If you increase the distance r, the w = g m doesn’t work anymore and you have to use the more basic formula G m1 m2 / r^2 for the new r to compute the acceleration.
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Sudhanshu Mishra
10 CLASS from Gurukul Academy (Graduated 2018)
·
7y
Originally Answered: How does the value of 'g' vary with height?
·
Many formulas are given below so i didn’t tell you the formula but i will tell you the concept.
since g=GM/R^2
here R is radius of earth
clearly g is inversely proportional to 1/R^2
so if a body go a height above the earth’s surface obviously its g decreases and R becomes R+h ( h is height above earth’s surface.
If you have any doubt clearly type in comment section.
By the way thanks for a2a
Valente Hernandez Perez
PhD in Chemical Engineering & Multiphase flows, University of Nottingham (Graduated 2008)
·
Author has 3.4K answers and 4.5M answer views
·
3y
As the altitude above the sea level increases, the value of gravity, g, decrease. This is because the gravity is inversely proportional to the distance between the objects., according the universal gravitational law. One object is the earth, which can be considered fixed, and the other is whatever object is located at that altitude.
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Adam Jacholkowski
Former experimental particle physicist at CERN (organization) (1983–2015)
·
Author has 32.6K answers and 6.6M answer views
·
Mar 13
Gravity at altitude h is g(h) = g(0)(R/(R+h)^)^2 where g0) corresponds to the surface and R is the Earth radius.
[Gravity of Earth - Wikipedia
Earth's gravity measured by NASA GRACE mission, showing deviations from the theoretical gravity of an idealized, smooth Earth, the so-called Earth ellipsoid . Red shows the areas where gravity is stronger than the smooth, standard value, and blue reveals areas where gravity is weaker. [ 1 ] The gravity of Earth , denoted by g , is the net acceleration that is imparted to objects due to the combined effect of gravitation (from mass distribution within Earth ) and the centrifugal force (from the Earth's rotation ). [ 2 ] [ 3 ] It is a vector quantity, whose direction coincides with a plumb bob and strength or magnitude is given by the norm g = ‖ g ‖ {\displaystyle g=\|{\mathit {\mathbf {g} }}\|} . In SI units , this acceleration is expressed in metres per second squared (in symbols, m / s 2 or m·s −2 ) or equivalently in newtons per kilogram (N/kg or N·kg −1 ). Near Earth's surface, the acceleration due to gravity, accurate to 2 significant figures , is 9.8 m/s 2 (32 ft/s 2 ). This means that, ignoring the effects of air resistance , the speed of an object falling freely will increase by about 9.8 metres per second (32 ft/s) every second. The precise strength of Earth's gravity varies with location. The agreed-upon value for standard gravity is 9.80665 m/s 2 (32.1740 ft/s 2 ) by definition. [ 4 ] This quantity is denoted variously as g n , g e (though this sometimes means the normal gravity at the equator, 9.7803267715 m/s 2 (32.087686258 ft/s 2 )), [ 5 ] g 0 , or simply g (which is also used for the variable local value). The weight of an object on Earth's surface is the downwards force on that object, given by Newton's second law of motion , or F = m a ( force = mass × acceleration ). Gravitational acceleration contributes to the total gravity acceleration, but other factors, such as the rotation of Earth, also contribute, and, therefore, affect the weight of the object. Gravity does not normally include the gravitational pull of the Moon and Sun, which are accounted for in terms of tidal effects . Variation in magnitude [ edit ] A non-rotating perfect sphere of uniform mass density, or whose density varies solely with distance from the centre ( spherical symmetry ), would produce a gravitational field of uniform magnitude at all points on its surface . The Earth is rotating and is also not spherically symmetric; rather, it is slightly flatter at the poles while bulging at the Equator: an oblate spheroid . There are consequently slight deviations in the magnitude of gravity across its surface. Gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s 2 on the Nevado Huascarán mountain in Peru to 9.8337 m/s 2 at the surface of the Arctic Ocean . [ 6 ] In large cities, it ranges from 9.7806 m/s 2 [ 7 ] in Kuala Lumpur , Mexico City , and Singapore to 9.825 m/s 2 in Oslo and Helsinki . In 1901, the third General Conference on Weights and Measures defined a standard gravitational acceleration for the surface of the Earth: g n = 9.80665 m/]( "en.wikipedia.org")
Maitri Goel
B Tech from BITS Pilani, Hyderabad Campus (Graduated 2023)
·
7y
Originally Answered: How does the value of 'g' vary with height?
·
Value of g decreases as we go above the surface of the earth:
g’ = g (R^2/(R+h)^2)
[where R is the radius of the Earth and h is the height above the Earth's surface]
If h<<R then:
g’=g(1- (2h/R))
Siddharth Kumar
Just A Teenager
·
7y
Originally Answered: How does the value of 'g' vary with height?
·
g’ = variation of g with height
g= Accerelation due to gravity on surfacer of earth
R= radius of earth
h= height
This is formula for variation of g with height.
g’ is directly proportional to h.
P.S. If you need derivation i can derivate it too. Just tell me.
Doug Weathers
Former Design Engineer at XCOR Aerospace
·
Author has 1.5K answers and 318.4K answer views
·
5y
It varies according to Newton’s Law of Universal Gravitation:
Newton's law of universal gravitation - Wikipedia
Basically, as you go higher, the Earth’s gravitational pull has less effect on you.
Related questions
How does the value of 'g' vary with height?
How does the value of G vary on Earth?
How does gravity decrease with increase in altitude?
Does the value of 'g' vary from place to place on the surface of the Earth?
How does altitude affect the value of g?
How does GPS system determines altitude?
Will the value of (g) change with altitude and latitude? Give examples.
How value of g vary from place to place?
Is gravity less at altitude?
At what altitude would the value of g become one fourth that of the surface of the Earth?
At what altitude does the value of g become one ninth of the surface of the Earth?
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8368 | https://en.wikipedia.org/wiki/An_Introduction_to_Mechanics | An Introduction to Mechanics - Wikipedia
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1 Table of contents (2nd edition)
2 Reception
3 References
4 See also
An Introduction to Mechanics
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From Wikipedia, the free encyclopedia
Undergraduate mechanics textbook
An Introduction to Mechanics
Cover of the second edition
AuthorDaniel Kleppner
Robert J. Kolenkow
Language English
SubjectClassical mechanics
Genre Non-fiction
PublisherMcGraw Hill
Cambridge University Press
Publication date 1973, 2010, 2013
Publication place United States
United Kingdom
Pages 542
ISBN9780521198219
OCLC1059566786
This article is about the undergraduate textbook. For the area of physics, see Mechanics.
An Introduction to Mechanics, commonly referred to as Kleppner and Kolenkow, is an undergraduate level textbook on classical mechanics coauthored by physicists Daniel Kleppner and Robert J. Kolenkow. It originated as the textbook for a one-semester mechanics course at the Massachusetts Institute of Technology, where both Kleppner and Kolenkow taught, intended to go deeper than an ordinary first year course. Since its introduction, it has expanded its reach to other universities to become one of the most popular mechanics textbooks.[citation needed]
The first edition was published in 1973 by McGraw Hill and republished in 2010 by Cambridge University. The second edition was published in 2013 by Cambridge.
Table of contents (2nd edition)
[edit]
Preface
To the Teacher
List of Examples
Chapter 1: Vectors and Kinematics
Chapter 2: Newton's Laws
Chapter 3: Forces and Equations of Motion
Chapter 4: Momentum
Chapter 5: Energy
Chapter 6: Topics in Dynamics
Chapter 7: Angular Momentum and Fixed Axis Rotation
Chapter 8: Rigid Body Motion
Chapter 9: Non-Inertial Systems and Fictitious Forces
Chapter 10: Central Force Motion
Chapter 11: The Harmonic Oscillator
Chapter 12: The Special Theory of Relativity
Chapter 13: Relativistic Dynamics
Chapter 14: Spacetime Physics
Hints, Clues, and Answers to Selected Problems
Appendix A: Miscellaneous Physical and Astronomical Data
Appendix B: Greek Alphabet
Appendix C: SI Prefixes
Index
Reception
[edit]
The first edition of the book was criticized for sexism in the exercises, though this was improved in the second edition.
References
[edit]
^"An Introduction to Mechanics". WorldCat.
^Kleppner, Daniel; Kolenkow, Robert (1973). An Introduction to Mechanics. Cambridge University Press. p.xi.
^Kleppner, Daniel; Kolenkow, Robert J. (6 May 2010). An Introduction to Mechanics. Cambridge University Press. ISBN978-0-521-19821-9.
^Shorter, Richard (14 Aug 2014). "An Introduction to Mechanics (2nd Edition), by Daniel Kleppner and Robert Kolenkow". Contemporary Physics. 55 (4): 359. doi:10.1080/00107514.2014.948931. Retrieved 25 Jul 2024.
See also
[edit]
List of textbooks on classical mechanics and quantum mechanics
This article about a physics-related book is a stub. You can help Wikipedia by expanding it.
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8369 | https://www.doubtnut.com/qna/643657270 | Describe the locus of a point P, so that : A B 2 = A P 2 + B P 2 , where A and B are two fixed points.
To describe the locus of a point P such that AB2=AP2+BP2, where A and B are two fixed points, we can follow these steps: 1. Understanding the Given Condition: We start with the equation AB2=AP2+BP2. This resembles the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (AB) is equal to the sum of the squares of the other two sides (AP and BP). Hint: Recall the Pythagorean theorem and how it relates to right triangles. 2. Identifying Points A and B: Let A and B be two fixed points in the plane. We can denote their coordinates as A(x1,y1) and B(x2,y2). Hint: Visualize the points A and B on a coordinate plane. 3. Using the Concept of a Circle: According to the properties of circles, if a triangle is inscribed in a semicircle, the angle subtended at the circumference (point P) is a right angle. Thus, if P lies on the semicircle with diameter AB, then ∠APB=90∘. Hint: Remember that the angle inscribed in a semicircle is always a right angle. 4. Constructing the Circle: To find the locus of point P, we can construct a circle with diameter AB. The center of this circle is the midpoint of segment AB, and the radius is half the distance between A and B. Hint: The midpoint formula can help you find the center of the circle. 5. Conclusion: Since point P can be any point on the semicircle defined by diameter AB, the locus of point P is a circle with diameter AB. This means that for any position of point P on this circle, the condition AB2=AP2+BP2 will hold true. Final Answer: Hence, the locus of point P is a circle with diameter AB.
To describe the locus of a point P such that AB2=AP2+BP2, where A and B are two fixed points, we can follow these steps:
Understanding the Given Condition:
We start with the equation AB2=AP2+BP2. This resembles the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (AB) is equal to the sum of the squares of the other two sides (AP and BP).
Hint: Recall the Pythagorean theorem and how it relates to right triangles.
Identifying Points A and B:
Let A and B be two fixed points in the plane. We can denote their coordinates as A(x1,y1) and B(x2,y2).
Hint: Visualize the points A and B on a coordinate plane.
Using the Concept of a Circle:
According to the properties of circles, if a triangle is inscribed in a semicircle, the angle subtended at the circumference (point P) is a right angle. Thus, if P lies on the semicircle with diameter AB, then ∠APB=90∘.
Hint: Remember that the angle inscribed in a semicircle is always a right angle.
Constructing the Circle:
To find the locus of point P, we can construct a circle with diameter AB. The center of this circle is the midpoint of segment AB, and the radius is half the distance between A and B.
Hint: The midpoint formula can help you find the center of the circle.
Conclusion:
Since point P can be any point on the semicircle defined by diameter AB, the locus of point P is a circle with diameter AB. This means that for any position of point P on this circle, the condition AB2=AP2+BP2 will hold true.
Final Answer: Hence, the locus of point P is a circle with diameter AB.
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8370 | https://www.jayaramanr.com/assets/docs/IBBI_Exam_MCQ_on_depreciation_methods.pdf | R JAYARAMAN (jaya_r54@ yahoo.com) DEPRECIATION METHOD EXAMPLES & VALUATION OF TENANTED PROPERTIES Methods of Calculating Depreciation There are several methods of a calculating depreciation. Some of these are arbitrary whereas others are based on theory. The methods have been divided into two broad heads: a) Non interest methods b) Methods based on interest theories 1. Good-as-new assumption method: the assumption that the assets never depreciate since the assets are well maintained, its service efficiency and output capacity are practically undiminished are assumed as good as new. Obviously, this assumption is entirely erroneous and has no basis. 2. Direct appraisal method: It is assumed that a valuer's intuition can decide arbitrarily, after inspection and without reasoning, giving an ad-hoc depreciation without applying any method. This method is arbitrary having no basis and will not stand in a court of law. 3. Arbitrary lump sum method: this method now practically abandoned; many enterprises make arbitrary lump sum allocations as expense for depreciation. This method also has no basis. 4.Depreciation as a percentage of revenue: Estimating cost depreciation as a percentage of revenue involves the same motive as found in the arbitrary lump sum method. The percentage of gross revenue method seems not to have been used much. 5.Sum of digits method: This method is also known as the sum of the years' digits method. The sum of the years' digits method is an arbitrary method of allocation in which the depreciation base is held constant and the yearly rate decreased. As with the declining balance method, the result is to allocate the larger amount of depreciation to the first year and to decrease the amount each succeeding year. The method is difficult to apply to groups of units, and because the declining balance method achieves similar results with greater ease and flexibility. 6. Declining balance method: This method is also known as constant or equal percentage method of depreciation. The constant percentage method is referred to as 1 the linear method. TOTAL Depreciation = 𝐶 (1−(1 − 𝑟𝑑) m): rd = Rate of Depreciation = 1/N, M = BUILDING AGE, N = TOTAL LIFE. In this case, no scrap value is assumed. In accounting terms, depreciation is used, for writing off the value of the asset over its useful life. It is the decrease in the value of the fixed asset due to use, time period and technological obsolescence. Reducing Balance Method charges depreciation at a higher rate in the earlier years of an asset. The amount of depreciation reduces as the life of the asset progresses. Depreciation per annum = (Net Book Value - Residual Value) x Rate% Where: Net Book Value is the asset's net value at the start of an accounting period. It is calculated by deducting the accumulated (total) depreciation from the cost of the fixed asset. Residual Value is the estimated scrap value at the end of the useful life of the asset. Rate of depreciation is defined according to the estimated pattern of an asset's use over its life term. 7. Straight line depreciation method: this method allocates the depreciable base of a property unit uniformly throughout its service life except when the estimate of service life is changed. depreciation per cent D is calculated as: D= (Total life - Future life) x (100% less salvage value) Total life This is a simple equation which is used for estimating depreciation of existing buildings. For estimating depreciation by this method, the total life, future life and percentage salvage value are necessary. This method Estimation of the Total Life of the Structure. There are no fixed rules for estimating the lives of various types of structures which depend upon many factors such as quality of construction, maintenance, etc. This fixed percentage depreciation method is more widely used in depreciation calculations than any other. It is the one method most generally used for determining depreciation for tax purposes and for profit and loss financial statements. It is the method prescribed by most agencies. Accounting principles require companies to depreciate its fixed assets using this method. cost of fixed asset is reduced uniformly over useful life of asset. Since depreciation expense charged to income statement in each period is same, the carrying amount of the asset on balance sheet declines in a straight line. 2 Question 1. It is a 20 years old load bearing structure of 1275 sqft of Plinth area. Total Life of the building is 60 years. What is the depreciated value of the building by adopting straight line method if Replacement rate is Rupees1,650/sq.ft. with a Salvage value of 10%? a) Rs. 16,83,000 b) Rs. 14,72,625 c) Rs. 21,03,750 d) Rs. 13,72,625 Solution: Method to be adopted = straight line method Plinth area of the building = 1275 sqft Age of the building = 20 years Total Life of the building = 60 years. Replacement rate = Rs 1,650/sq.ft Salvage value = 10% Replacement value of the building = 1275 x Rs 1650 =Rs 21,03,750 Depreciation percentage = 20 / 60 x (100%-10%)= 30% Depreciated value of the building = Rs 21,03,750x (100% - 30%) = Rs 14,72,625 (Ans: b) Question 2. It is a load bearing structure. Age is 5 years. Life is 60 years. What is the depreciation by constant percentage method? a) 9.18% b) 10.38% c) 11.38% d) 8.20 % Solution Method to be adopted = constant percentage method Age = 5 years Life = 60 years Depreciation formula = 1- (1-(R/100))n = R =100/60 = 1.67 = 1 - (1- (1.67/100))5 = 1 - (0.983)5 = 1- 0.918 = 0.082 = 8.20% (Ans: d) R = Depreciation Rate = (100 / total life) n = age 3 Question 3 If the machinery costing Rs. 18,000 is sold after 2 years for Rs. 16,000. The depreciation rate is 10 percent per annum on SLM method, then the profit or loss from sale of machine is a) Rs. 3,600 (Profit) b) Rs. 1600 (loss) c) Rs. 1600 (profit) d) No profit no loss Solution: Machine cost = Rs 18,000 Depreciation rate = 10% No. of years = 2 years Method to be adopted Straight line method Depreciation = Rs 18,000 x 2 years x 10% = Rs 3600 Depreciated value = Rs 18000 - Rs 3600 = Rs 14,400 Sale value = Rs 16,000 Profit on sale = Rs 16,000 – Rs 14,400 = Rs 1600 (Ans: c) Question 4. The W.D.V of an asset after three years of depreciation on the reducing balance method @ 10 percent p.a. is Rs. 36,450. What was its original value? a) Rs. 40,000 b) Rs. 50,000 c) Rs. 45,000 d) Rs. 70,250 Solution; Depreciation method to be adopted = Written Down Value method Rate of Depreciation = 10% After 3 years the asset value := Rs 36,450 Formula = C x (1-d) n :=C x (1-0.1) 3 = Rs 36,450 Capital value = Rs 36,450 / (1-0.1) 3 = Rs 50,000 (Ans: b) 4 Question 5 A machine was purchased for Rs. 1,00,000/- @ 15% depreciation of SLM. What is the written down value after 2 years? a) Rs. 70,000/- b) Rs. 80,000/- c) Rs. 60,000/- d) Rs. 90,000/- Solution: Depreciation for 1 year by Straight line method = 1,00,000 x 0.15 Depreciation amount = Rs. 15,000/year Depreciated value after 1 year = 1,00,000 - 15,000 = Rs. 85,000/- Depreciated value after the second year = 85,000 - 15,000 = Rs. 70,000/- Written down value after 2 years = Rs. 70,000/- (Ans: a) Question 6 Total age of building 4 years. After four years the depreciated value is equal to 24% of the cost. Find out the % of depreciation (near to answer) by WDV method? a) 24 b) 25 c) 30 d) 35 Solution; Depreciation method to be adopted = Written Down Value method Rate of Depreciation = Not known After 4 years the asset value = 24% of the cost value Depreciation Formula = (1-d) n = (1- d) 4 = 24% Depreciation Percentage = 24% / (1- d) 4 (1-d) = (24%)¼ = 0.699 = 0.70 Depreciation = (1.00 - 0.70 ) = 0.30 = 30% (Ans: c) Question 7. ‘A’ purchased a mine for Rs 2,50,000 minerals in the mine were expected to be 5,00,000 tonnes. In the first year, 50,000 tonnes of minerals were used. What is the depreciation for the first year? 5 a) Rs 20000 b) Rs 25000 c) Rs 10000 d) Rs 15000 Solution: Mine purchase cost = Rs 2,50,000 Mine mineral worth = 5,00,000 tonnes First year production = 50,000 tonnes Production %age for 1st year = 50000 / 5,00,000 ton = 10% Depreciation % age for 1st year = 10% Depreciation amount for 1st year = Rs 2,50,000 x 10% = Rs 25,000 (Ans: b) Question 8. Cost of machine Rs. 1,00,000/-, scrap value Rs. 10,000/-, life 4 years, what will be the amount of depreciation according to sum of years’ digit method in the first year. a) Rs. 40,000/- b) Rs. 45,000/- c) Rs. 36,000/- d) Rs. 38,000/- Solution: Net cost value = Rs 1,00,000 – scrap value Rs 10,000 = Rs 90,000 Depreciation value in the first year = 90, 000 x 4/10 = Rs 36,000 ( Ans: c) year Net cost value Residual life Depreciation fraction Depreciation value Net Book value 1 90,000 4 4/10 36,000 54,000 2 90,000 3 3/10 27,000 27,000 3 90,000 2 2/10 18,000 9,000 4 90,000 1 1/10 9,000 - Total sum of years’ digit 4+3 +2+ 1= 10 Question 9. 6. A machine was purchased for Rs 20,000. At the end of 5th year by uniform depreciation method by declining in a straight-line form, the carrying value of the asset works out to Rs 5,000. What is the depreciation amount by uniform depreciation method by declining in a straight-line form? Note: The life period, carrying value can be changed Note: The life period, scrap value can be changed 6 a) Rs 2,000 b) Rs 2,500 c) Rs 3,000 d) Rs 4,000 Solution: Depreciation amt = Rs 20,000 - Rs 5,000 = Rs 15000 / 5 years = Rs 3,000 (Ans: c) Net Value Depreciation Carrying value Purchase date - Rs 20,000 1st year Rs 3,000 Rs 17,000 2nd year Rs 3,000 Rs 14,000 3rd year Rs 3,000 Rs 11,000 4th year Rs 3,000 Rs 8,000 5th year Rs 3,000 Rs 5,000 Question 10. A machine was purchased for Rs 20,00,000. The life of the machine is 3 years. The depreciation will be charged at 50% every year, with a residual value of Rs 5,00,000. At the end of the 3rd year what will be Net Book Value by Reducing Balance Method? a) Rs 7,50,000 b) Rs 8,75,000 c) Rs 3,75,000 d) Rs 5,00,000 Solution: Rs 8,75,000 (Ans b) Year ending Net book value Residual value Depn %age Depreciation Amount Accumulated Depreciation 1 Rs 20,00,000 Rs 5,00,000 50% Rs 7,50,000 Rs 7,50,000 2 Rs 12,50,000 Rs 5,00,000 50% Rs 3,75,000 Rs 11,25,000 3 Rs 8,75,000 Rs 5,00,000 50% Rs 3,75,000 Rs 15,00,000 Question 11. Factory building has 1200 S.M. built-up area. Plot are is 2000 S.M. Building is 25 years old and total life is 50 years. Replacement cost today is Rs. 25000 per S.M. industrial plot is available for Rs. 8000 per S.M. which of the following is the fair sale price (ignoring salvage) for the property? Note: This question is for 3rd year end. The question can be changed for 2nd year end also. The life period may be also changed 7 a) Rs 46000000 b) Rs 31000000 c) Rs 16 000000 d) Rs 18400000 Solution: Plot = 2000 sqm Building = 1200 sqm Building Age = 25 years Total life = 50 years A. Land value = 2000 sqm x Rs 8000 per sqm = Rs 1,60,00,000 B. Building value = 1200 sqm x Rs 25,000 per sqm = Rs 3,00,00,000 Depreciated value = Rs 3,00,00,000 x 25/50 = Rs 1,50,00,000 Fair sale price (Rs 1,60,00,000 + Rs 1,50,00,000) = Rs 3,10,00000 (Ans b) Question 12. An assessee has spent Rs. 1,20,00,000 in his new building in the year 2014. What will be the written down value of the above building as on 2019 assuming a rate of depreciation as 10%? This is required for preparing balance sheet for IT purpose. a) Rs. 87,48,000 b) Rs. 70,86,000 c) Rs. 97,20,000 d) Rs. 78,60,000 Solution; Depreciation method to be adopted = Written Down Value method Rate of Depreciation = 10% Building construction cost as on 2014 = Rs 1,20,00,000 Building construction year = 2014 Building Depreciated cost to be determined = 2019 Number of years depreciation required = 5 years Depreciation Formula = (1-d) n = (1- 10%)5 = 59.05% written down value of the building as on 2019 = Rs 1,20,00,000 x 59.05% = Rs 70,86,000 (Ans b) Question 13. A vacant plot taken on lease for 30 years. The lessee has constructed a building in the plot. As per the lease terms, the lessee has to surrender the vacant plot. What will be the depreciation percentage after 20 years? The life of RCC Framed structure is 80 years. 8 a) 33.33% b) 66.66% c) 25% d) None of the above Solution Total life of the building = 80 years It is mentioned in the question, that the lessee has to surrender the vacant plot after demolishing the building after expiry of lease period. Hence, the building life is restricted to 30 years. Legal life of the building = 30 years Depreciation percentage after 20 years = 20 / 30 = 66.66% (Ans b) Note: Life of the building is termed as legal life (life due to legal constraints). If the lease term is with renewable clause for further period then legal life will be extended to the lease life as per renewable clause Question 14. A fully developed building in a plot has a total of 4 floors. Total plot area is 1,000 sq.m. and total built up for area of the building is 250 sq.m / per floor. Permissible FSI is 1.00. There are 4 tenants per floor and tenants of lower 2 floors pay a rent of Rs. 750 / month / tenement. which includes property tax. Top 2 floors are occupied by the owners of the property itself. Total property taxes are Rs. 25,000 / 6 months for 4 floors. Tenant’s rent includes 50% of total tax, Non - agricultural (N.A.) tax of the plot is Rs. 800 / year and building insurance premium is Rs. 1,000 / year. Assume repair cost at 6% of the gross rent and collection & management charges at 3% of the gross rent. Stamp duty paid at the time of purchase is Rs. 9,000/-. The land is of freehold tenure. Prevalent land rate of freehold land in the locality at present is Rs. 8,000/sq.m. The rate of ownership flats in the locality for similar construction as on today is Rs. 30,000/sq.m. Questions: 1. What will be the total annual rent receivable by the landlord from all the tenants? a) Rs. 6,000/- b) Rs. 72,000/- c) Rs. 1,44,000/- d) Rs. 12,000/- 2. What will be the total outgoings including repairs allowance & collection charges for the tenanted portion of the building? 9 a) Rs. 32,380/- b) Rs. 57,380/- c) Rs. 33,280/- d) Rs 38,230/- 3. What will be the present market value of the tenanted portion of the building if rental income is assumed to be in perpetuity & rate of capitalization is adopted @ 8% a) Rs. 9,90,500/- b) Rs. 1,50,00,000/- c) Rs. 77,50,000/- d) Rs. 4,95,250/- 4. What will be the present market value of the owner-occupied portion of the building? a) Rs. 75,00,000/- b) Rs. 1,50,00,000/- c) Rs. 10,00,000/- d) Rs. 78,00,000/- 5. Which of the following is not considered as outgoing for computing net rent received by the landlord? a) Property tax b) Repair cost c) Stamp duty paid d) Management charges 6. What is the market value of the balance potential in the property? a) Rs. 1,50,000/- b) Rs. 15,00,000/- c) Zero d) Reversionary value of land Data: Property tax for 4 floors = Rs. 25,000 / 6 months Non- Agricultural tax for Mumbai = Rs. 800 / year Building insurance = Rs. 1,000 / year Repair cost & maintenance = 6% Gross rent Rent collection charge = 3% Gross rent Market rate of land = Rs. 8,000 / sq.m. Prevalent market rate of flat = Rs. 30,000 / sq.m. 10 Solution: 1.Rent received by the owner: Tenants occupied portions = GF & FF Number of tenants in each flat = 4 Total number of flats in all flats = 2 x 4 = 8 Monthly rent for each flat = Rs. 750/- Monthly rent for all flats = 750 x 8 = Rs. 6,000 Yearly rent for all flats = 6,000 x 12 = Rs. 72,000/- (Ans: b) 2.Outgoings: Property tax = Rs. 50,000 N.A. (Non-Agricultural tax) = Rs. 800 Insurance premium = Rs. 1,000 = Rs. 51,800 Since the tenants are bearing 50% of the above expenses, the actual outgoings of the owner = Rs. 25,900 Maintenance charges 6% of gross rent = Rs. 4,320 (0.06 x 72,000) Rent collection charge 3% of gross rent = Rs. 2,160 (0.03 x 72,000) Total outgoings = Rs. 32,380/- (Ans: a) 3.Capitalization amount: Gross income = Rs. 72,000 Outgoes = Rs. 32,380 Net income = Rs. 39,620 Yield = 8% Capitalized amount = 39,620 x (100 / 8) = Rs. 4,95,250/- (Ans: d) 4.Value of the building - free holder (land owner): FSI = 1 Area of the flat 2 x 250 = 500 sq.m. Market rate of flat = Rs. 30,000/sq.m. Market Value 500 x 30,000 = Rs. 1,50,00,000/- (Ans: b) 11 5. While computing net rent received by the landlord, Stamp duty is not to be considered. (It is onetime payment for purchase of the property) ((Ans: c) 6. Since the allowable FSI is fully utilized, there is no balance market potential in the property. Hence, the market value of the balance potential is zero. ((Ans: c) Question 15. An apartment carries 4 floors built on a plot of area 1,000 sq.m. Each floor area is 250 sq.m. The GF & FF have been rented and SF & TF is in possession of the owner. Each floor carries 4 tenements, and tenants pay @ Rs. 750 / tenement as rent. The property tax being paid is @ Rs. 25,000 / six month. Rs. 900 / year is non agri - tax. 6% per annum towards management cost. The tenants are bearing 50% of the property tax, N.A. tax. Rs. 9,000/- stamp duty cost. 3% towards rent collection charge. Cost of land is Rs. 2,000 / sqm and cost of construction is Rs. 25,000 / sqm, FSI is 1. Age of the building is 20 years and total building life is 60 years. (salvage 10% Calculate the following: 1. What is the total rent? 2. What are the total outgoes? 3. What is the valuation of owner-occupied portion? 4. What is the balance potential in building? 5. What is the depreciated cost of building @ 10% salvage value? 6. What is the tenanted portion value @ rate of return of 8% PA? Data: Property tax = Rs. 25,000 / 6 months Non-Agricultural tax (for Mumbai) = Rs. 900 / year Management cost = 6% Stamp duty = Rs. 9,000/- Cost of land = Rs. 2,000/sq.m. Cost of construction = Rs. 25,000/sq.m. Rent collection charge = 3% Building Age = 20 years Total life of the building = 60 years Solution: 1. Rent received by the owner: Tenants occupied portions = GF & FF 12 Number of tenants in each flat = 4 Total number of flats in all flats = 2 x 4 = 8 Monthly rent for each flat = Rs. 750/- Monthly rent for all flats = 750 x 8 = Rs. 6,000 Yearly rent for all flats = 6,000 x 12 = Rs. 72,000/- 2. Outgoings: Property tax = Rs. 50,000 N.A. (Non-Agricultural tax) = Rs. 900 = Rs. 50,900 Since the tenants are bearing 50% of the above expenses, the actual outgoings for the owner = Rs. 25,450 Management charges 6% of gross rent = Rs. 4,320 (0.06 x 72,000) Rent collection charge 3% of gross rent = Rs. 2,160 (0.03 x 72,000) Total outgoings = Rs. 31,930/- 3. Value of the building - free holder (land owner) : FSI = 1 Area of the flat 2 x 250 (SF & TF) = 500 sq.m. Cost of construction = Rs. 25,000/sq.m. Replacement cost = 500 x Rs 25,000 = Rs 1,25,00,000 Building Age = 20 years Total life of the building = 60 years Salvage value percentage = 10% Depreciation percentage = (20/60) x 90 = 30% Depreciation value = 0.30 x Rs 1,25,00,000 = Rs. 37,50,000/- Depreciated value for free holder portion (1.25 cr – 37.50 lks) = Rs. 87,50,000/- 4. Since the allowable FSI is fully utilized, there is no balance market potential in the property. Hence, the market value of the balance potential in the property is zero. 13 5. Depreciated cost of the building. FSI = 1 Area of the building = 1000 sq.m. Cost of construction = Rs. 25,000/sq.m. Replacement cost = 1000 x Rs 25,000 = Rs 2,50,00,000 Building Age = 20 years Total life of the building = 60 years Salvage value percentage = 10% Depreciation percentage = (20/60) x 90 = 30% Depreciation cost = 0.30 x Rs 2,50,00,000 = Rs. 75,00,000/- Depreciated cost (2.50 cr – 75.00 lks) = Rs. 1,75,00,000/- 6. Capitalization amount: Gross income = Rs. 72,000 Outgoes = Rs. 31,930 Net income = Rs. 40,070 Yield = 8% Capitalized amount = 40,070 x (100 / 8) = Rs. 5,00,875/- 14 VALUATION TABLES & LEASEHOLD PROPERTIES 1. Simple Interest P x N x R 2. Compound Interest P x (1 + 𝑅)𝑛 3. Capitalized Value (CV) Net Income x Years’ Purchase 4. Years’ Purchase 100 / Rate of Return 5. Present Value / discount rate C X 1 (1+𝑅)𝑛 6. Amount of Re1 per Annum C X ((1+𝑅)𝑛 −1) 𝑅 7. Gross Sinking Fund C x 𝑅 (1+𝑅)𝑛 −1 8. Present value of future income Single rate C x 1− 1 (1+𝑅)𝑛 𝑅 = C x 1−𝑃𝑉 𝑅 9. Present value of future income (Dual rate) 1 𝑅+𝑆 = 1 Capitalisation Rate+Sinking Fund COMPOUND INTEREST Question 1. What will be the sum of Rupees.5,000 at the end of 5 years @ 5% compound interest per annum? Solution Amount A = P x (1 + 𝑅)𝑛 = Rs 5,000 (1 + 5%)5 = Rs 5,000 x (1.05)5 = Rs 5,000 x 1.276 = Rs. 6,380/- Question 2. In 2013, a valuer valued a residential property in a mofusil town for Rs. 68.56 lakhs. Assuming an annual escalation of 10% per year, what will be the value of the property as on 2018 by applying the formula? Solution: Property value assessed = Rs. 68,56,000 Rate of escalation = 10% 15 Number of years (2018 – 2013) = 5 years Amount A = P ( 1 +r)n = 68,56,000 ( 1 + 10%)5 = 68,56,000 x (1.1)5 = 68,56,000x 1.6105 = Rs. 110.42 lakhs Question 3. A Contractor took a loan of Rs. 36,00,000/- from a bank for construction of modern building 2 years back. He has to repay the loan at the Interest of 10%. If the sale of the property is yet to take one year, calculate the amount to be paid by the contractor? a) Rs. 46,00,000 b) Rs. 48,91,600 c) Rs. 49,00,000 d) Rs. 47,91,600 Solution: Amount borrowed = Rs. 36,00,000 Amount to which Re. 1/- will accumulate @ 10% in 3 years = (1 + i)n == (1 + 0.1)3 = = 1.331 Amount to be repaid by the contractor = (1.331 x 36,00,000) = Rs. 47,91,600/- (Ans: d) YEARS’ PURCHASE Question 4. Which of the following mathematical formula is used to find out Years’ Purchase for annuity receivable in perpetuity? a) 100/Rate of interest b) 1 / {1-(100/rate of interest) ^n} c) ((100/rate of interest) ^n)-1 d) {((100/rate of interest) ^n)-1}/rate of interest Question 5. The net income was reported at Rs.21,000 and the property sold for Rs.300,000. What capitalization rate applied for this sale? a) 0.075 b) 0.080 c) 0.065 d)0.070 16 Solution Net income = Rs 21000 Sale value = Rs 3,00,000 YP (Capitalization Rate) = NI / CV = 21,000 /3,00,000 = 7% or 0.070 (Ans: d) Question 6. A property has a net income of Rs.30,000. One appraiser decides to use a 12 Percent capitalization rate, while a second appraiser uses a 10 percent rate. Use of the higher rate results in in appraisal value. a) Rs.50,000/- increase b) Rs.50,000/- decrease c) Rs 2,50,000 d) Rs 3,00,000 Solution; Valuer Net Income capitalization rate CV = NI x YP Total value arrived at Valuer 1 Rs 30, 000 12% 30000 x 100 / 12 Rs 2,50,000 Valuer 2 Rs 30, 000 10% 30000 x 100 / 10 Rs 3,00,000 Use of the higher rate results by Valuer 1 in Rs.50,000/- decrease in appraisal value. (Ans: b) Question 7. The monthly rent (Net) of a shop of 540 sq.ft. is Rs. 12,000/-. Calculate the approximate value by adopting a rate of return as 5%. Solution: Monthly rent = Rs. 12,000 Yearly rent = 12,000 x 12 = Rs. 1,44,000/- Rate of return adopted = 5% Capitalized value = 1,44,000 x 100/5 = Rs. 28,80,000/- Question 8. The net monthly rent of a residential building of 1,250 sq.ft. is Rs. 16,500/-. Find the approximate value of the property by rent capitalization method by adopting a rate of return as 3%. 17 Solution: Monthly rent = Rs. 16,500 Yearly rent = 16,500 x 12 = Rs. 1,98,000/- Rate of return = 3% Capitalized value = 1,98,000 x 100 /3 = Rs. 66,00,000/- Question 9. A freehold site is rented out for 99 years to a developer at a ground rent of Rs. 1,00,000 per annum, net of outgoings. It is renewable. The lessee developer has constructed a building fetching an annual rent of Rs. 5,00,000/-. Value the freeholder’s interest assuming a yield of 6%. Solution: Value in the hands of lessor: Net income from ground rent = Rs. 1,00,000 Yield (Rate of return) = 6% Years purchase = 100 / R = 100 / 6 = 16.67 Value in the hands of lessor = Rs 1,00,000 x 16.67 = Rs. 16,67,000/- Question 10. Value the freehold interest of a shop which has been let out for a rent of Rs. 1,00,000 (Net) per month. The rent is renewable. Yield is 5%. Solution: Yearly rent = 1,00,000 x 12 = Rs. 12,00,000 Net income = Rs. 12,00,000 Y.P. for a yield of 5% = 100/5 = 20 capitalized value = 12,00,000 x 20 = Rs. 2,40,00,000/- Question 11. A new shop was purchased for Rs. 10,00,000 which was rented out for Rs. 5,000 per month. What is the yield? Solution: Capital value = Rs. 10,00,000 Yearly rent = Rs. 5,000 x 12 = Rs. 60,000 18 Yield = 60,000 x 100 / 10,00,000 = 6% Question 12. An industrial corporation has decided to lease 40,000 sq.ft. plot for a user for 60 years period. The land rate is 2,000 per sq.ft. Assuming a yield of 6%, what will be the monthly lease? Solution: Extent of land = 40,000 sq.ft. Market rate = Rs. 2,000/sq.ft. Value of land = Rs. 8,00,00,0000 Lease rent yield = 6% Annual rent = 8,00,00,000 x6 /100 = Rs. 48,00,000 Monthly rent = 48,00,000/12 = Rs. 4,00,000/- Question 13. A private trust had leased 10,000 sq.ft. plot for 99 years lease which can be renewed for further period. Fix lease rent if the land rate is Rs. 1,500/sq.ft. Assume lease rent as 8%. Solution: Extent of land = 10,000 sq.ft. Land rate = Rs. 1,500/sq.ft. Land value = Rs. 1,50,00,000 Lease rent yield assumed = 8% Annual lease rent = 1,50,00,000 x 8 / 100 = 12,00,000 Monthly lease rent = Rs. 1,00,000/- Question 14. An apartment building consists of 12 flats of super built up area 1,050 sq.ft. The net monthly rent of a flat is Rs. 9,000. The prevailing rate of return is 2.5%. Find the approximate value of one flat by rent capitalization method. Solution: Net monthly rent = Rs. 9,000 Yearly rent = Rs. 1,08,000 Rate of return = 2.5% Value = 1,08,000 x 100 / 2.5 = Rs. 43,20,000/- 19 PRESENT VALUE / DISCOUNT RATE Question 15. What is the Present value of rupee for Capital amount receivable at a future date of Rs 10,00,000 at 6 % compound interest rate for 10-year term? a) Rs 5,85,390 b) Rs 5,58,390 c) Rs 8,55,390 d) Rs 5,26,850 Solution: Present worth of amount receivable = PVA = C X 𝟏 (𝟏+𝑹)𝒏 PVA = Rs10,00,000 X 𝟏 (𝟏+𝟎.𝟎𝟔)𝟏𝟎 = Rs10,00,000 x 0.55839 = Rs5,58,390/- (Ans: b) Question 16. An investment pays Rs 300 annually for 5 years, with the first payment occurring today. The present value (PV) of the investment discounted at 4% annual rate is approximately----- a) Rs 1,336 b) Rs 1,389 c) Rs 1,625 d) Rs 1,925 Solution: Discount rate formula : 𝟏 (𝟏+𝑹)𝒏 Instalment Amount paid Rate per month Discount rate 1 (1 + 𝑅)𝑛 Net Present Value 1 Rs 300 0.04 (4%) - Rs 300.00 2 Rs 300 0.04 (4%) 0.9615 Rs 288.45 3 Rs 300 0.04 (4%) 0.9245 Rs 277.35 4 Rs 300 0.04 (4%) 0.8890 Rs 266.70 5 Rs 300 0.04 (4%) 0.8548 Rs 256.44 (Ans: b) Rs 1388.94 AMOUNT OF RE 1 PER ANNUM Question 17. Amount of Re.1 per annum is worked out by which of the following formula? 20 a) r / ((1+r) n - 1} b) ((1+r) n – 1) / r c) r / ((1+r) n + 1) d) ((1+r) n + 1) / r Question 18. Rs 500 deducted every month and invested annually towards PF account from salary for a period of 20-year terms and at a 7% of compound interest? a) Rs 2,56,000 b) Rs 2,46,000 c) Rs 2,26,000 d) Rs 1,50,000 Solution: Accumulated sum of Re 1 / year (APA) = ((𝟏+𝑹)𝒏 −𝟏) 𝑹 = ((𝟏+𝟎.𝟎𝟕)𝟐𝟎 −𝟏) 𝟎.𝟎𝟕 = ((𝟏.𝟎𝟕)𝟐𝟎 −𝟏) 𝟎.𝟎𝟕 = (𝟑.𝟖𝟕 −𝟏) 𝟎.𝟎𝟕 = (𝟐.𝟖𝟕 ) 𝟎.𝟎𝟕 = 41 Gross Accumulated sum = Rs 500 x 12 x 41 = Rs 2,46,000/- (Ans: b) ANNUAL SINKING FUND Question 19. Annual Sinking fund to be set aside each year for recouping Rs 1 at the end of 6 years, at 5 percent rate of interest is represented by formula----------- a) 0.05/ ((1+ 0.05)6-1} b) {(1+0.05)6+1} / 0.05 c) 5 / (1+ 5)6-1) d) 0.05 / {(1+0.05)6+1} Question 20. Which one of the following best defines Annual sinking fund? 1. Annual sum required to be invested to an amount of Re. 1/- in specified years 2. Monthly sum required to be invested to an amount of Rs. 10/- in specified years 3. Annual sum required to be invested to an amount of Rs. 10/- in specified years 4. Annual sum required to be invested to an amount of Rs. 100/- in specified years Question 21. To find out the depreciated worth of the building to set aside annually for 10 lakhs as Capital recoupment amount expected at a 4% interest rate for the period unexpired period of lease of 60 years 21 a) Rs 2,400 b) Rs 4,200 c) Rs 3,200 d) Rs 4,000 Solution: Gross sinking fund GSF = C x 𝑹 (𝟏+𝑹)𝒏 −𝟏 GSF = Rs10,00,000 x 𝟎.𝟎𝟒 𝟏𝟎.𝟓𝟐 −𝟏 = Rs10,00,000 x 𝟎.𝟎𝟒 𝟗.𝟓𝟐 = Rs10,00,000 x 0.0042 = Rs 4,200 /- (Ans: b) PRESENT VALUE OF FUTURE INCOME SINGLE RATE Question 22. Annual Rental income from property is Rs 48,000 /-. If the building is demolished after 40 years, what will be the present value of the property @ 7% interest rate? a) Rs 6,00,000 b) Rs 6,39,936 c) Rs 6,28,745 d) Rs 6,93,935 Solution: Asset value = C x 𝟏− 𝟏 (𝟏+𝑹)𝒏 𝑹 C = Capital income (annuity) received each year R = Compound interest rate N = Number of years YP = Year’s Purchase YP = 𝟏− 𝟏 (𝟏+𝑹)𝒏 𝑹 = 𝟏− 𝟏 (𝟏+𝟎.𝟎𝟕)𝟒𝟎 𝟎.𝟎𝟕 = 𝟏− 𝟏 (𝟏.𝟎𝟕)𝟒𝟎 𝟎.𝟎𝟕 = 𝟏− 𝟏 𝟏𝟒.𝟗𝟕𝟒 𝟎.𝟎𝟕 = 13.332 Asset value = C X YP = Rs 48000 x 13.332 = Rs 6,39,936 /- (Ans: b) PRESENT VALUE OF FUTURE INCOME DUAL RATE Question 23. Which of the following represents the year purchase for Re.1 with remunerative rate of interest at 8% and annual sinking fund amount to be set aside for recouping Rs .1 is 0.021. a) 1 / (0.08+0.021) b) 0.021/0.08 c) (0.08+0.021) / (0.021) d) 1/(0.08-0.021) 22 Question 24. The annual rent received from the property is Rs 48000 /-. Expected rate of return is 10% future life of the building is 50 years. Recoupment rate is 4% on capital. Find the purchase price. a) Rs 4,85, 672 b) Rs 4,50,672 c) Rs 4, 60,672 d) Rs 4,75,672 Solution: Present value of Re.1 / year = YP = 𝟏 𝑹+𝑺 R = Remunerative rate of interest = 10% S = Sinking fund = 𝑹 (𝟏+𝑹)𝒏 −𝟏 = 𝟎.𝟎𝟒 (𝟏+𝟎.𝟎𝟒)𝟓𝟎 −𝟏 = 0.0065 Asset Value = C x YP = C x 1 R+S = Rs 48000 x 𝟏 𝟎.𝟏𝟎+𝟎.𝟎𝟎𝟔𝟓 = Rs 48000 x 𝟏 𝟎.𝟏𝟎𝟔𝟓 = Rs 48000 x 9.389 = Rs 4,50,672/- (Ans: b) Question 25. A property fetches a leaseholder Rs. 30,000 per annum. The rent fixed to the paid to the superior landlord is Rs. 16,000 per annum. If freeholder expects a return of 8 percent, then the leaseholder should expect a rate as indicated below so that he makes a reasonable profit. a) 0.09 b) 0.07 c) 1.00 d) 0.01 Solution: If the property is assumed as freehold, then, the freeholder is expecting a rate of return of 8%. The lease holder must expect a minimum rate over and above this 8%. In Sorab Talati vs Josheph Michem Appeal 101 of 1949 - Vol.- 2 of SOC – page 162 (Bombay) (Invest Theory of Rent) C) For leasehold properties, 1% extra yield on both types of investment was considered fair, to account for extra risk of investing capital. Hence, 9% is fair for the leaseholder, so that he makes a reasonable profit. (Ans: a) Question 26. An ownership flat with 140 sqm area is licensed for an amount of Rs 1,10,000 per month. Society maintenance charges are Rs 20,000 per 3 months. Which of the 23 following will be market value of the flat on income approach by adopting 4 percent as rate of capitalization? a) Rs 3,10,00,000 b) Rs 1,10,00,000 c) Rs 2,00,00,000 d) Rs 1,32,00,000 Solution: License amount per annum = Rs 1,10,000 x 12 = Rs 13,20,000 Less Society maintenance per annum = Rs 20000 x 12 / 3 = Rs 80,000 Total per annum = Rs 13,20,000 - Rs 80,000 = Rs 12,40,000 Rate of capitalization = 4 % Capitalized value or Market value = Rs 12, 40,000 / 0.04 = Rs 3,10,00,000 (Ans: a) Question 27. Mr. ‘X’ is owning a vacant site of 8,000 sq.ft. near the bus stand. He wants to let out. The prevailing unit market rate is Rs. 10,000 and the guideline rate is Rs. 15,000 / sq.ft. Mr. Y wants this site or parking vehicles. Mr. Z also wants this site and wishes to construct a shed. Assuming rate of return of 4% & 5% as secured rent and unsecured rent respectively, what is the maximum rent that can be suggested for i) Y & ii) Z? For Y a) Rs. 2,66,667 b) Rs. 3,33,333 c) Rs. 5,00,000 d) Rs. 4,00,000 For Z a) Rs. 5,00,000 b) Rs. 3,33,333 c) Rs. 2,66,667 d) Rs. 4,00,000 Solution For Y Unsecured Ground Rent: A vacant ground given under lease & the lessee leases the vacant land for vehicle parking, materials storing without any construction or making any improvements Land area = 8000 sqft 24 Market rate = Rs 10,000 per sqft Total value of land = Rs 8,00,00,000 Unsecured Ground Rent = 5% Annual Rent for Y = Rs 8,00,00,000 x 5% = Rs 40,00,000 Monthly Rent for Y = Rs 40,00,000 / 12 = Rs 3,33,333 (Ans: b) For Z Secured Ground Rent: A vacant ground given under lease and the lessee develop the land and make improvements and lease out the building he has constructed. He maintains the building and pay statutory taxes. Land area = 8000 sqft Market rate = Rs 10,000 per sqft Total value of land = Rs 8,00,00,000 Secured Ground Rent = 4% Annual Rent for Z = Rs 8,00,00,000 x 4% = Rs 32,00,000 Monthly Rent for Z = Rs 32,00,000 / 12 = Rs 2,66,667(Ans: c) Question 28. A purchaser is offered a property with a net income of Rs. 52,000 per annum. The purchaser assumes that a first mortgage can be raised at 60% of the purchase price. The mortgage will be at an interest rate of 15% per annum. The purchaser will fund the balance of the purchase price and requires a 10% return on equity. What is the value of the property? a) Rs 3,50,000 b) Rs 3,75,000 c) Rs 4,00,000 d) Rs 4,25,000 Solution: Average Interest Rate = Mortgaged fund (borrowed) + Equity (purchaser own fund) = (0.60 x 0.15) + (0.40 x 0.10) = 0.13 Value = 52,000 = Rs. 4,00,000 (Ans: c) 0.13 25 Question 29. A government M.I.D.C. gives 8,000 sq.m. of land on 99 years lease @ 1/- P.A. lease rent and charged one time premium of Rs. 450 / sq.m. in the year 1998. The lessee in the year 1998 constructed an industrial shed 4,000 sq.m. of BU area with his own expenditure. The age of the shed is 20 years as on year 2018 and total life of the shed is 40 years. Salvage value 10% The land rate is Rs. 2,000 / sq.m. and replacement cost is Rs. 25,000 / sq.m. Lease provides that the lessor is entitled to charge 50% unearned increase in land value as transfer / assignment charges in case of sale / transfer of the property. Calculate the following: 1. What is the lessor’s interest? 2. What is the total value of property considering a freehold property? 3. What is the lessee interest? 4. What is the reversionary value of the leasehold land? 5. What is the depreciated value of shed? Data: Year of lease = 1998 Period of lease = 99 years One time premium = Rs.450/sq.m. for land extent Land area = 8,000 sq.m. Lease rent = Re. 1/year Lessee built a factory of built-up area = 4,000 sq.m. Year of construction of factory by the lessee = 1998 Land rate as on 2018 (date of valuation) = Rs. 2,000/sq.m. Replacement cost of building in 2018 = Rs. 25,000/sq.m. Age of the shed 2018 - 1998 = 20 years Total life of the building = 40 years Salvage value = 10% Date of valuation = 2018 Opinion: 1. Lessor’s interest: This case of lease of land is by state government. It is assumed as a perpetual lease and reversionary value of land is negligible. The lease rent is only Re. 26 1/year and hence its capitalized value will be negligible. Lessor’s interest in land value would be therefore is restricted to claim 50% of unearned increase in land value in case of sale. Land area = 8,000 sq.m. Prevailing land rate 2018 = Rs. 2,000/sq.m. One time premium charged in 1998 = Rs. 450/sq.m. Unearned increase 2,000 - 450 = Rs. 1,550/sq.m. The percentage the lessor is entitled to = 50% charge in case of transfer Unearned increase the lessor can enjoy = 0.5 x 1,550 = Rs. 775/sq.m. The lessor’s value - 8,000 x 775 = Rs. 62,00,000/- (1) 2. Value of property assuming it is a freehold: (i) Land : Land area = 8,000 sq.m. Unit rate of land = Rs. 2,000/sq.m. Land value - 8,000 x 2,000 = Rs. 1,60,00,000/- (ii) Building: Building area = 4,000 sq.m. Replacement cost = Rs. 25,000/sq.m. Replacement value = 4,000 x 25,000 = Rs. 10,00,00,000/- Age of the building: 2018 - 1998 = 20 years Life of the factory = 40 years Salvage value assumed = 10% Depreciation percentage = (20/40) x 90 = 45% Depreciation value = 0.45 x 10,00,00,000 = Rs. 4,50,00,000/- Depreciated value (10,00,00,000 - 4,50,00,000) = Rs. 5,50,00,000/- (iii) Total value : Value of land = Rs. 1,60,00,000/- Depreciated value of building = Rs. 5,50,00,000/- Total value = Rs. 7,10,00,000/- (2) 27 3. Value of lessee’s interest: Total value of land: 8,000 x 2,000 = Rs. 1,60,00,000/- Value of lessee’s interest = 1,60,00,000 - 62,00,000 = Rs. 98,00,000/- Lessee also holds in the building value. Depreciated value of building = Rs. 5,50,00,000/- Total value: Land = Rs. 98,00,000 Building = Rs. 5,50,00,000 = Rs. 6,48,00,000/- 4. Reversionary value is negligible and hence not considered. 5. Depreciated value of shed = Rs. 5,50,00,000/- Question 30. A warehouse property is situated close to a port facility in a major port town. It is let out on a 50 years lease. The lessee is paying to the lessor an exclusive ground rent @ INR 2,000 per annum, after payment of a onetime premium of INR 25,00,000. The rack rental value on full repairing terms amounts to INR 1,20,000 per annum. The yield from freehold ware houses in similar locations is considered to be 10% and for long term lease is 15%. Questions: 1. What is the outgoing for lessor? 2. What is the net income for the lessor during the term period? 3. What is the YP during the term period? 4. What is the YP during the reversionary value calculations? 5. What is the value of freeholder’s interest? 6. What is the market rent? Data: Lease = 50 years Ground rent to lessor = Rs. 2000/- per annum Premium paid to lessor = Rs. 25,00,000/- Rack Rent on full repairing terms = Rs. 1,20,000/- per annum Yield for freehold ware houses = 10% Yield for long term lease = 15% 28 Answers: 1. What is the outgoing for lessor? The outgoing for lessor is nil since the lease is on full repairing terms. 2. What is the net income for the lessor during the term period? Rs. 2000/- per annum 3. What is the YP during the term period? Y.P. = 100 / 15 = 6.66 4. What is the YP during the reversionary value calculations? The YP during reversionary value calculations is 100 / 10 = 10. 5. What is the value of freeholder’s interest? Value of freeholder’s interest = value of term + value of reversion i) Value of term (lessor’s interest) = 2000 x 6.66 = 13,320 ii) Value of reversion Market value = Rs. 1,20,000/- Y.P. @ 10% = 100 / 10 = 10 Capitalized value = 1,20,000 x 10 = 12,00,000 Y.P. in perpetuity deferred for 50 years @ 10% = (1/ (1 + 0.1)50 ) = 0.008518 Value of reversion = 12,00,000 x 0.008518 = 10,222 iii) Value of freeholder’s interest = 13,320 + 10,222 = Rs. 23,542/- 6. What is the market rent? Rs. 1,20,000/- per annum. Notes: 1. Simple Interest Calculation: The gross amount accrued at the end of given period of term, at the given rate of simple interest. The total interest amount accrued in the given period total interest amount = I = P x N x R Gross Amount (Annuity) = P + I = Principle Amount + Interest 29 P = Principal Amount: N = number of years: R = simple interest Rate 2. Compound interest amount Calculation: Single lump sum payments which are compounded at a given constant interest rate at the end of each definite time period at equal intervals of time. The gross amount accrued at the end of given period of term, at the given rate of compound interest. Total interest = I = (1+R) n Gross Amount = P + I = P x (1+R) n = Principal Amount + compound Interest accrued P = Principal Amount N = number of years R = Rate of compound interest 3. Present value of rupee Calculation: By this formula we can calculate the Present value of Re for a given period at a given rate of compound interest. This method is known as discounting or deferring of receivable at a future given period and at a given rate of compound interest. Present value of a Rupee = PV = 1 (1+𝑅)𝑛 Present worth of amount receivable = PVA = C X 1 (1+𝑅)𝑛 C = Capital amount receivable at a future date R = Compound interest rate N = Number of years 4. Amount of Re. 1 / year (annum) Calculation: Amount to calculate the annual regular investment of Re for a given period at a given rate of compound interest Accumulated sum of Re 1 / year = (APA) = ((1+𝑅)𝑛 −1) 𝑅 Gross Accumulated sum = C X ((1+𝑅)𝑛 −1) 𝑅 C = Capital amount received / year R = Compound interest rate N = Number of years 5. Annual sinking fund Calculation: This calculation formula is used to find out the depreciated worth of the building. For this we have to consider a sinking fund amount to set aside annually at a given interest rate for the period equal to the building age or unexpired period of lease. Annual sinking fund = ASF = 𝑅 (1+𝑅)𝑛 −1 Gross sinking fund = GSF = C x 𝑅 (1+𝑅)𝑛 −1 C = Capital recoupment amount expected 30 R = Compound interest rate N = Number of years 6. Present value of future income of Re. 1 / year (Single rate basis): Present worth of future annual income for a given time period and at a given rate of compound interest. This method is adopted for a perpetual income or by a long term lease with unexpired period is more where no annual sink fund is considered for calculation. Only remunerative rate of interest for the income is considered and hence it is called single rate method Present value of Re.1 / year = YP = 1− 1 (1+𝑅)𝑛 𝑅 Asset value = C x 1− 1 (1+𝑅)𝑛 𝑅 C = Capital income (annuity) received each year R = Compound interest rate N = Number of years YP = Year’s Purchase 7. Present value of future income of Re.1/year (Dual rate): To calculate present worth of the future annual income flow for a given period of time and at a given rate of compound interest with taking in to account the sinking fund. It is calculated on the terminable income. This method is usually adopted for an income with the remunerative rate of interest for the income and also a provisional annual sinking fund is considered. Present value of Re.1 / year = YP = 1 𝑅+𝑆 R = Remunerative rate of interest S = Sinking fund = 𝑅 (1+𝑅1)𝑛 −1 Asset Value = C x YP C =Capital income (annuity) received each year R = Compound interest rate N = Number of years YP = Year’s Purchase TYPES OF LEASE Building lease: A vacant ground given by lessor under lease with ground rent and lessee develop the land and make improvements and lease out the building he has constructed. He maintains the building and pay statutory taxes. The lease amount collected by the lessor is ground rent, which is here termed as Head Rent. If the building is rented out (sub lease) by the head lessee, the amount collected by him is called Rack rent. Occupational lease: A building property given on lease to the lessee, which means, 31 both land and building in part and parcel has been leased out. The rent collected is termed as Rack Rent. The lessor has the right of evicting the lessee. Example: Residential house, Apartments, shops. Full repair lease: If the lease agreement stipulates the lessee to undertake all outgoings apart from his head rent, the lease is called full repairing lease Life lease: The lease period is fixed till the death of lessee. The lease period expires on the death of the lessee. Sub lease: In the lease agreement if the lessee is permitted to give the lease hold property to other occupants for a shorter time less than his lease period, with an enhanced lease amount, then the lease agreement entered by the lessee with the incumbents is called a sub-lease. The main lessee is called the Head Lessee. Other sub lease holders are called sub lessee. The main lessee retains his reversion of lease under his control. |
8371 | https://www.quora.com/What-is-the-difference-between-Kc-Kp-Ka-and-Kb-in-equilibrium-equations-Why-are-Kc-and-Kp-typically-used-instead-of-Ka-and-Kb | Something went wrong. Wait a moment and try again.
Ka Band
Kc Ks
Equilibrium Concentration
General Kb
Physical Chemistry
Equilibrium Systems
5
What is the difference between Kc, Kp, Ka, and Kb in equilibrium equations? Why are Kc and Kp typically used instead of Ka and Kb?
John Britto
Current Professor of Chemistry at Delaware County Community College
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Author has 125 answers and 17.4K answer views
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1y
These are all expressions related to the equilibrium constant of a reaction. However, they are each calculated using specific values:
Kc represents the equilibrium constant of a reaction where the product and reactant concentrations are expressed as molarity
Kp represents the equilibrium constant of a reaction where the product and reactant values are measured as partial pressure (of gases in the reaction) usually in units of atm
Ka is the acid dissociation constant representing the extent of ionization of an acid - molarities of the hydronium ion (H3O+) and the conjugate base of the acid are the
These are all expressions related to the equilibrium constant of a reaction. However, they are each calculated using specific values:
Kc represents the equilibrium constant of a reaction where the product and reactant concentrations are expressed as molarity
Kp represents the equilibrium constant of a reaction where the product and reactant values are measured as partial pressure (of gases in the reaction) usually in units of atm
Ka is the acid dissociation constant representing the extent of ionization of an acid - molarities of the hydronium ion (H3O+) and the conjugate base of the acid are the products and the initial concentration of the acid is the reactant
Kb is the base dissociation constant representing the extend of ionization of a base - molarities of the hydroxide ion (OH-) and the conjugate acid of the base are the products and the initial concentration of the base is the reactant
To say that one or two of the K’s are used more typically is not accurate - the representation of K is related directly to the type of equilibrium under consideration.
In each case the representation of K is the product of the equilibrium’s product concentrations (in the appropriate form) divided by the reactant concentration (in the appropriate form:
K = concentration of product species x concentration of reactant species divided by concentration of reactant species
Related questions
What is the relation between Kc & Kp?
Are Kp and Keq the same value? If so why are Kc and Kp not the same value?
What is the difference between Kc and the position of equilibrium?
What is the relationship between Q and KC at equilibrium?
KP is more than KC when?
Daniel Iyamuremye
Former Senior Lecturer (Retired) (2000–2018)
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Author has 12.1K answers and 2M answer views
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1y
Kc= equilibrium constant or a reaction obtained using molar concentrations, for reactions in liquid solutions
Kp = equilibrium constant of a reaction obtained using pressures of gases in a reaction in gaseous phase.
Ka = dissociation constant of an acid
Kb = dissociation constant of a base.
Gaurav Kumar
Influencer, Perfectionist, researcher
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Author has 204 answers and 1.4M answer views
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6y
Related
What are Kp and Kc? What is relation between them?
Anonymous
6y
Related
What are Kp and Kc? What is relation between them?
Related questions
If you reverse an equation, why is the equilibrium constant (Kc or Kp) inverted?
How does Kc = Kp in an equilibrium position?
If an equilibrium reaction equation has same number of moles of on both sides of the equation, it's obvious that Kp=Kc as in Kp=Kc(RT) ^0. Is this implying that concentration and partial pressures of reactants and products are the same in value?
Why should I know KC and KP in an equilibrium reaction?
What is the difference between Keq and Kc?
Brian Bi
IChO medallist; majored in chemistry
·
Upvoted by
Christopher VanLang
, PhD in Chemical Engineering and
Justin Dragna
, PhD in chemistry from UT Austin · Author has 4.8K answers and 65.1M answer views
·
11y
Related
In a gas equilibrium, which K constant is used in the equation ΔG=-RTlnK, Kp or Kc?
[math]K_p[/math] is the correct constant for the right-hand side of that equation.
You should not memorize this. It is not hard to see why this must be the case.
The [math]\Delta G[/math] on the left hand side is really [math]\Delta G^0[/math], or the molar change in free energy when all the reactants and products are in their standard states.
Therefore, to get the correct mass action expression for the right hand side, all activities must be given relative to the standard state. () The standard state for a gas is 1 bar, not 1 mol/L. Therefore you must express the activities of gases as ratios of the gas pressure to 1 bar, rather tha
[math]K_p[/math] is the correct constant for the right-hand side of that equation.
You should not memorize this. It is not hard to see why this must be the case.
The [math]\Delta G[/math] on the left hand side is really [math]\Delta G^0[/math], or the molar change in free energy when all the reactants and products are in their standard states.
Therefore, to get the correct mass action expression for the right hand side, all activities must be given relative to the standard state. () The standard state for a gas is 1 bar, not 1 mol/L. Therefore you must express the activities of gases as ratios of the gas pressure to 1 bar, rather than ratios of the gas pressure to 1 mol/L. That's why it has to be [math]K_p[/math], and not [math]K_c[/math].
() This is a very handwaving argument, but can easily be made precise if you study the derivation of the mass action law from chemical potentials and activities.
Promoted by The Penny Hoarder
Lisa Dawson
Finance Writer at The Penny Hoarder
·
Updated Sep 16
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Ernest Leung
M.Phil. in Chemistry, The Chinese University of Hong Kong
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Author has 11.9K answers and 5.8M answer views
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May 20
Related
Is Kp=KC(RT) ∆n always correct?
Suppose the question is:
"Is Kp = Kc(RT)^(Δn) always correct?"
The answer is as follows.
Suppose the question is:
"Is Kp = Kc(RT)^(Δn) always correct?"
The answer is as follows.
Mark Roseman
Biochemistry Professor (Emeritus in 2020) at Uniformed Services University of the Health Sciences (1979–present)
·
Author has 9.1K answers and 19.6M answer views
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5y
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What is the difference between the equilibrium constant KC and KP? Which one should I use to find standard change in Gibbs free energy?
Both can be used for standard free energy calculations because both of these formulas are valid:
[math]\Delta G^{o} = -RTlnK_{c}^{o}[/math]
[math]\Delta G^{o} = -RTlnK_{p}^{o}[/math]
The numerical values differ, but that is because [math]\Delta G^{o}[/math] measures the difference between the equilibrium state and a standard state. If you choose a different standard state, the standard free energy can change.
Kc is used in most chemistry and biochemistry applications because solutions are usually studied, not gas mixtures.
It is important to note that the thermodynamic equilibrium constants, that is, the ones that can be used in the for
Both can be used for standard free energy calculations because both of these formulas are valid:
[math]\Delta G^{o} = -RTlnK_{c}^{o}[/math]
[math]\Delta G^{o} = -RTlnK_{p}^{o}[/math]
The numerical values differ, but that is because [math]\Delta G^{o}[/math] measures the difference between the equilibrium state and a standard state. If you choose a different standard state, the standard free energy can change.
Kc is used in most chemistry and biochemistry applications because solutions are usually studied, not gas mixtures.
It is important to note that the thermodynamic equilibrium constants, that is, the ones that can be used in the formulas above, are dimensionless. One reason why this must be so is mathematical: you cannot take the logarithm of a number with dimensions.
But the more fundamental reason, as you can see from going through the derivation, is that each term in the equilibrium expression is really a ratio of the equilibrium concentration of each reactant to its concentration in a standard state, usually 1.0 M. That is, each term takes the form [math][A]_{eq}/[A]_{o}[/math] where the denominator is 1.0 M. For convenience in writing the equilibrium expression, the standard states are left out.
It is just as common to use equilibrium constants with units of molarity. These occur when the equilibrium constant is derived from kinetics. See relevant section in link below:
Equilibrium constant - Wikipedia
For example, for the reaction
A + B <=> C
[math]K_{eq} = [C]/[A][B][/math]
Keq has units of 1/M. It can be converted to a thermodynamic equilibrium constant by dividing each term on the right by the standard state value of 1.0 M. This gets rid of the units but does not change the numerical value of the equilibrium constant.
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Zachary Tidwell
Senior Communications Manager at Network Solutions (2015–present)
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Wed
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Eric Ressner
Chemical information services (1985–present)
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Author has 6K answers and 7.8M answer views
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Updated 6y
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How does Kc = Kp in an equilibrium position?
Better to ask “when does Kc = Kp” rather than “how”. To me, “how” implies that Kc = Kp, full stop, and you want to know how or why that is possible.
“When” implies that Kc isn’t always equal to Kp, and you want to know what circumstances pertain when they are equal.
So let’s say we have a reaction like the formation of ammonia:
N[math]_2[/math] + 3 H[math]_2[/math] → 2 NH[math]_3[/math]
Kc for this reaction will take the form of
[math]Kc = \frac{[NH_3]^2}{[N_2][H_2]^3}[/math]
In order to convert the value of Kc to Kp, we’d have to apply the conversion factor from molarity (concentration) to pressure … twice in the numerator and four times in the deno
Better to ask “when does Kc = Kp” rather than “how”. To me, “how” implies that Kc = Kp, full stop, and you want to know how or why that is possible.
“When” implies that Kc isn’t always equal to Kp, and you want to know what circumstances pertain when they are equal.
So let’s say we have a reaction like the formation of ammonia:
N[math]_2[/math] + 3 H[math]_2[/math] → 2 NH[math]_3[/math]
Kc for this reaction will take the form of
[math]Kc = \frac{[NH_3]^2}{[N_2][H_2]^3}[/math]
In order to convert the value of Kc to Kp, we’d have to apply the conversion factor from molarity (concentration) to pressure … twice in the numerator and four times in the denominator, because there are molarity[math]^2[/math] units in the numerator and molarity[math]^4[/math] units in the denominator. The conversion factor[math]^2[/math] will cancel out, leaving still conversion factor[math]^2[/math]in the denominator. The conversion factor isn’t equal to 1, so Kc will not equal Kp in this situation.
But for a reaction like
H[math]_2[/math] + Cl[math]_2[/math] → 2 HCl
we have
[math]Kc = \frac{[HCl]^2}{[H_2][Cl_2]}[/math]
Here, when we apply the conversion factor from molarity to atmospheres, we do so twice in the numerator and twice in the denominator. Those conversion factors are, of course, all equal, and they will cancel.
So any time the number of gas molecules on the product side is the same as the number of gas molecules on the reactant side, Kc will be equal to Kp.
What is the conversion factor for molarity to pressure for a gas, anyway? The answer comes simply from the ideal gas equation.
PV = nRT
If we rearrange a bit (divide through by V) and do a little creative grouping, we get:
[math]P = \frac{n}{V}RT[/math]
P is obviously P. [math]\frac{n}{V}[/math] is molarity (moles/L), so the conversion factor from molarity to pressure is RT.
Mayur Sagar
Lives in Mumbai, Maharashtra, India
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Author has 72 answers and 246.1K answer views
·
8y
Related
When do we use ka and kb in ionic equilibrium?
We use dissociation constants to measure how well an acid or base dissociates. For acids, these values are represented by Ka, for bases, Kb. These constants have no units.
All chemical reactions proceed until they reach chemical equilibrium, the point at which the rates of the forward reaction and the reverse reaction are equal. We use the equilibrium constant, Kc, for a reaction to demonstrate whether or not the reaction favors products (the forward reaction is dominant) or reactants (the reverse reaction is dominant). High values of Kc mean that the reaction is product-favored, while low valu
We use dissociation constants to measure how well an acid or base dissociates. For acids, these values are represented by Ka, for bases, Kb. These constants have no units.
All chemical reactions proceed until they reach chemical equilibrium, the point at which the rates of the forward reaction and the reverse reaction are equal. We use the equilibrium constant, Kc, for a reaction to demonstrate whether or not the reaction favors products (the forward reaction is dominant) or reactants (the reverse reaction is dominant). High values of Kc mean that the reaction is product-favored, while low values of Kc mean that the reaction is reactant-favored.
For acid and base dissociation, the same concepts apply, except that we use Ka or Kb instead of Kc. High values of Ka mean that the acid dissociates well and that it is a strong acid. Low values of Ka mean that the acid does not dissociate well and that it is a weak acid. The same logic applies to bases.
There is a relationship between the concentration of products and reactants and the dissociation constant (Ka or Kb).
For acids, this relationship is shown by the expression:
Ka = [H3O+][A-] / [HA].
The products (conjugate acid H3O+ and conjugate base A-) of the dissociation are on top, while the parent acid HA is on bottom. Notice that water isn't present in this expression. We get to ignore water because it is a liquid and we have no means of expressing its concentration.
For bases, this relationship is shown by the equation:
Kb = [BH+][OH-] / [B].
The products (conjugate acid and conjugate base) are on top, while the parent base is on the bottom. Once again, water is not present.
Both the Ka and Kb expressions for dissociation can be used to determine an unknown, whether it's Ka or Kb itself, the concentration of a substance, or even the pH.
Hope this helps.
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Anthony Madden
Writer for Betterbuck
·
Updated Mar 24
I just bought my first house. Homeowners: what are some things you wished you knew when you bought your house?
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I've been a homeowner for 4 years. These are the biggest things I wish somebody told me on day one.
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Alka B. Gupta
Masters in Pharmaceutical Chemistry
·
9y
Related
What are Kp and Kc? What is relation between them?
Kc and Kp are the equilibrium constants of gaseous mixtures. However, the difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system. The equilibrium constants do not include the concentrations of single components such as liquids and solid, and they do not have any units.
Derive the relationship between Kp and Kc
Relationship between Kp and Kc
Consider the following reversible reaction:
aA + bB ⇌ cC + dD
The equilibrium constant for the reaction expressed in terms of the concentration (
Kc and Kp are the equilibrium constants of gaseous mixtures. However, the difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system. The equilibrium constants do not include the concentrations of single components such as liquids and solid, and they do not have any units.
Derive the relationship between Kp and Kc
Relationship between Kp and Kc
Consider the following reversible reaction:
aA + bB ⇌ cC + dD
The equilibrium constant for the reaction expressed in terms of the concentration (mol / litre) may be expressed as:
K c = [C] c [D] d / [A] a [B] b
If the equilibrium involves gaseous species, then the concentrations may be expressed in terms of partial pressures of the gaseous substance. The equilibrium constant in terms of partial pressures may be given as:
K p = pcC pdD / paA pbB
Where pA, pB, pC and pD represents the partial pressures of the substance A, B, C and D respectively. If gases are assumed to be ideal, then according to ideal gas equation:
pV = nRT
p = nRT / V
Where p ———-> pressure in Pa
n ——————–> amount of gas in mol
V ——————–> Volume in m3
T ———————> temperature in Kelvin
n/V = concentration, C
or
p = CRT or [gas] RT
If C is in mol dm-3 and p is in bar, then R = 0.0831 bar dm3 mol-1 K-1
Therefore, at constant temperature, pressure of the gas P is proportional to its concentration C, i.e.
Let us suppose a general reaction:
aA + bB↔ cC + dD
The equilibrium constant will be given as:
Kp = (pC) c (pD) d /
(pA) a (pB) b ……. (1)
Now, p = CRT
Hence,
pA = [A] RT
where [A] is the molar concentration of A
Similarly,
pB = [B] RT
pC = [C] RT
pD = [D] RT
where [B], [C] and [D] are the molar concentration of B, C and D respectively
Substituting these values in expression for Kp i.e. in equation (1)
Kp = [([C] RT) c ([D] RT) d]/[([A] RT) a ([B] RT) b]
= [C] c [D] d (RT) c+d/[A] a [B] b (RT) a+b
= [C] c [D] d (RT) c+d – a+b/[A] a [B] b
= Kc (RT) c+d – a+b
= Kc (RT) ∆n
Where ∆n = (c + d) – (a + b) i.e. number of moles of gaseous products – number of moles of gaseous reactants in the balanced chemical reaction.
Hence relation between Kp and Kc is given as:
Kp = Kc (RT) ∆n
Michael Mombourquette
Retired Chemistry Prof, Church member, Knight of Columbus,
·
Author has 6.8K answers and 17.7M answer views
·
5y
Related
Why are Kp and Kc unitless?
Actually, [math]K_p[/math] and [math]K_c[/math] are not unitless. They are simply shortcuts that some people use to replace the true thermodynamic equilibrium constant, which is unitless.
The “c” in [math]K_c[/math] stands for concentration. That means in an equation like [math]A+B\rightleftharpoons C[/math], the value of [math]K_c[/math] is calculated using the concentrations of chemicals A,B, and C.
[math]K_c = \frac{[C]}{[A][B]}[/math]
Since the items listed in the right hand side of this equations have units of mol/L and since they don’t all cancel out, this value of [math]K_c[/math] will have units.
Similarly, if you are using pressure units for the three gas phase components the v
Actually, [math]K_p[/math] and [math]K_c[/math] are not unitless. They are simply shortcuts that some people use to replace the true thermodynamic equilibrium constant, which is unitless.
The “c” in [math]K_c[/math] stands for concentration. That means in an equation like [math]A+B\rightleftharpoons C[/math], the value of [math]K_c[/math] is calculated using the concentrations of chemicals A,B, and C.
[math]K_c = \frac{[C]}{[A][B]}[/math]
Since the items listed in the right hand side of this equations have units of mol/L and since they don’t all cancel out, this value of [math]K_c[/math] will have units.
Similarly, if you are using pressure units for the three gas phase components the value of [math]K_p[/math] will have units.
These units are not a problem when all we’re doing is comparing [math]K[/math] values to [math]Q[/math] etc. it becomes impossible when we are using these [math]K[/math] values in other thermodynamic equations, like the one that relates Gibbs energy to [math]K[/math]
[math]\Delta G^o = -RTlnK[/math]
Since it’s impossible to take the [math]ln[/math] of a unit, you cannot use either of those two [math]K[/math] values here. You MUST use a unitless [math]K[/math] value.
This unitless (thermodynamic) [math]K[/math] value is not calculated using either concentrations or pressures. It is calculated using activities of the components. These activities are related to the concentration or pressure but they have no units.
[math]K = \frac{a(C)}{a(A)a(B)}[/math]
If you restrict your solution components to mol/L concentration units then a simple conversion to activities is to just drop the units on each concentration value. Similarly, if you use units of pressures in bars, for any gas phase components then you can just drop the units. This is because the simplest ( but least accurate) way to calculate an activity is [math]a =\frac{measure}{standard measure}[/math]. If the measure of concentration is in mol/L then the standard concentration is simply 1 mol/L. If the measure is a pressure in bar, then standard pressure is just 1 bar. So. As long as you restrict your measurements to concentrations in mol/L for solutes in liquid solutions or pressures in bars for gas phase components the you can just calculate [math]K[/math] the usual way and drop the units. This method is often taught incorrectly in Highschool and first year university simply because the teacher him/herself simply does not know any better or because they are just teaching you the shortcut without bothering to explain it to you.
Subbanna Vadlamudy
Ph. D. from Saugar University, Sagar (Graduated 1961)
·
Author has 6.8K answers and 1.5M answer views
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3y
Related
What's the ionic product and the equilibrium constant of Kc, Kw and Kp? Where does this symbol come from?
Kc, Kw and Kp are constants, K that apply to specific equilibrium processes. When the amounts of species in equilibrium are expressed in concentration of mole/L, M, then the equilibrium constant K is called Kc. Take water, H2O. It ionizes to produce H+ and OH- ions; H2O (——) H+ + OH- . The equilibrium constant to this (water) ionization process is called Kw. In case of gases partial pressures,(p ) are used in the K expression, instead of the concentration. Therefore the K for gases is called Kp. In cases of gases the Kp and Kc of a process are related. Because, P = (n/V)RT = cRT=MRT.
Ionic prod
Kc, Kw and Kp are constants, K that apply to specific equilibrium processes. When the amounts of species in equilibrium are expressed in concentration of mole/L, M, then the equilibrium constant K is called Kc. Take water, H2O. It ionizes to produce H+ and OH- ions; H2O (——) H+ + OH- . The equilibrium constant to this (water) ionization process is called Kw. In case of gases partial pressures,(p ) are used in the K expression, instead of the concentration. Therefore the K for gases is called Kp. In cases of gases the Kp and Kc of a process are related. Because, P = (n/V)RT = cRT=MRT.
Ionic product refers to the product Ms of ions produced, expressed in M^x, in the process. This is used mostly in slightly soluble salt equilibria, Ksp (solubility product). When ionic product exceeds the Ksp value for the salt, then precipitation of the salt occurs.
Read more about these in your textbook.
Michael Mombourquette
Retired Chemistry Prof, Church member, Knight of Columbus,
·
Author has 6.8K answers and 17.7M answer views
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Updated 1y
Related
Why is Kp and Kc has not same formula, if they both are equilibrium constants?
That’s because both Kp and Kc are actually just shortcuts and are not the real equilibrium constant. The equilibrium constant K should be unitless and should be independent of the method used to measure things. The real K uses activities of the chemicals as the measure rather than concentrations or pressures. So, for a reaction like
A(aq) + B(aq) <==> C(aq) + D(aq)
We write the real equilibrium expression as
[math]K = \frac{a(C) \times a(D)}{a(A) \times a(B)}[/math]
An activity is a measure of how much of that chemical is available to react, how active it is, so to speak. Activities are related to concentratio
That’s because both Kp and Kc are actually just shortcuts and are not the real equilibrium constant. The equilibrium constant K should be unitless and should be independent of the method used to measure things. The real K uses activities of the chemicals as the measure rather than concentrations or pressures. So, for a reaction like
A(aq) + B(aq) <==> C(aq) + D(aq)
We write the real equilibrium expression as
[math]K = \frac{a(C) \times a(D)}{a(A) \times a(B)}[/math]
An activity is a measure of how much of that chemical is available to react, how active it is, so to speak. Activities are related to concentrations or pressures but activities are unitless so K is also unitless. In first year chemistry, I teach a simple form called relative activities. A relative activity is the ratio of the actual measure divided by the standard measure.
If my measure is concentration then for chemical x we write
[math]a(x) = \frac{conc(x)}{std conc(x)}[/math]
If the chemical is in solution then use concentration. If the chemical is a gas then use pressures. For ideal solutions, and we can assume most dilute solutions are pretty close to ideal the standard concentration is 1 mole/Litre. So, in effect, for liquid solutions, if we use concentrations in molarity (mol/L) we can just drop the units and we will have activities.
In other words [math]a(x) = [x][/math] is a good approximation for a dilute solute.
If x is a gas then my measure should pressure and we write
[math]a(x) = \frac{p(x)}{standard pressure}[/math]
Since standard pressure for a gas is 1 bar (100 kPa) it’s easiest if the pressures are all in bars because just like we did for solute concentrations, we can make the approximation that [math]a(x) = p(x)[/math] if we measured pressure in bars. If we measured in kPa then divide by 100 since 1 bar = 100 kPa.
For a gas phase reaction, is we use activities, or we use pressure in bars and drop the units, we will get the correct thermodynamic K constant to be unitless.
One more thing, a pure solid has a concentration that is its standard concentration so the activity of a solid is ALWAYS 1. Similarly, the activity of a liquid is 1. Since we are working in dilute solutions here, the activity of a solvent is close to 1 so we just assume it is 1.
Now, mixed phase equilibrium constants are simpler.
Take a Ksp expression.
[math]AgCl(s) <==> Ag^+ + Cl^-[/math]
Using activities we can write
[math]Ksp(AgCl) = \frac{a(Ag^+) a(Cl^-)}{a(AgCl)}[/math]
But the activity of the solid AgCl is just 1 so we can simply not bother to write it down in the Ksp expression and if we replace the mole/litre concentrations for the activities, we get the more familiar expression
[math]Ksp(AgCl) = [Ag^+][Cl^-][/math]
Now we see an easy explanation for why we don’t include the AgCl in the Ksp expression. Its activity is 1. No handwaving arguments needed.
We could do the same for an acid Ka expression. Water, the solvent, is involved in the equilibrium but because its activity is 1 we don’t put it in the Ka expression.
So, in conclusion, if you have a gas phase chemical. Use the pressure in Bars. If you have a solute in solution then use molarity concentration and if you have a solid or a liquid, its activity is 1 so don’t bother including it in the K expression.
Related questions
What is the relation between Kc & Kp?
Are Kp and Keq the same value? If so why are Kc and Kp not the same value?
What is the difference between Kc and the position of equilibrium?
What is the relationship between Q and KC at equilibrium?
KP is more than KC when?
If you reverse an equation, why is the equilibrium constant (Kc or Kp) inverted?
How does Kc = Kp in an equilibrium position?
If an equilibrium reaction equation has same number of moles of on both sides of the equation, it's obvious that Kp=Kc as in Kp=Kc(RT) ^0. Is this implying that concentration and partial pressures of reactants and products are the same in value?
Why should I know KC and KP in an equilibrium reaction?
What is the difference between Keq and Kc?
What are Kp and Kc? What is relation between them?
What is the kc and kp of 2HI (g) → H2 (g) + I2 (g)?
How do you write an expression for Kc and Kp, if applicable for the following reversible reaction at equilibrium?
How do I find Kp without Kc?
Why is Kp and Kc has not same formula, if they both are equilibrium constants?
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Prev Previous EM Cases: Oncologic Emergencies
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EM@3AM: Orbital Cellulitis
February 9, 2019
Rachel Bridwell
Author: Rachel Bridwell, MD (@rebridwell, EM Resident Physician, San Antonio, TX) // Edited by: Brit Long, MD (@long_brit, EM Attending Physician, San Antonio, TX)and Alex Koyfman, MD (@EMHighAK, EM Attending Physician, UTSW / Parkland Memorial Hospital)
Welcome to EM@3AM, an emDOCs series designed to foster your working knowledge by providing an expedited review of clinical basics. We’ll keep it short, while you keep that EM brain sharp.
A 12-year-old male is brought to the ED by his mom for right eye swelling and pain. The patient reports it hurts when he looks around the room. He denies any headache, fever, weakness, neck stiffness, recent trauma, dental work, or travel. Review of systems is remarkable for a recent sinus infection. Vital signs are normal.
Pertinent physical examination findings: Right eye – erythematous and edematous upper and lower eye lid. Extraocular movements intact though painful, noted OD aPD, proptosis of OD. VA: OD 20/100, OS 20/20. Fundoscopic exam is not possible due to swelling. The mouth, ears, nose, face, and neck are otherwise normal.
What’s the next step in your evaluation and treatment?
Answer:Orbital cellulitis1-15
Epidemiology:
Caused by sinus infection extension, traumatic inoculation, hematogenous spread
Risk Factors: sinus infection, dental work, foreign bodies
10x more common in children, peak age 12 years 1,2
In children, infection more commonly originates in the sinuses (maxillary>ethmoid>frontal>sphenoid), with 38% of patients with multiple sinuses involved 3,4
In adults, infection commonly originates in frontal sinus; only 11% of patients have involvement of multiple sinuses
Foreign body: wood carries a large bacterial load with poor outcomes, though no predominant organism 5
Causative bacteria: Upper respiratory flora, both aerobic and anaerobic
S. pneumonia,H. influenza,M. catarrhalis,S. aureus,S. pyogenes,Bacterioides spp., Fusobacterium spp.
Developed countries: S. aureus,Streptococcus spp. most common 2
In diabetic and immunocompromised patients: consider mucormycosis and aspergillosis (high mortality rate)6
Adults have narrowed sinus ostia, generating, with more anaerobic infections 2
Orbital vs. Periorbital cellulitis
Orbital cellulitis: occurs posterior to orbital septum within bony orbit
Requires parental antibiotics +/- surgical drainage
Periorbital cellulitis: more common, does not extend posterior to orbital septum, no change in visual acuity, conjunctival injection, proptosis, or intraorbital pathology 7
Outpatient management if patient is well-appearing; if toxic, admit
Clinical Presentation:
Erythematous, edematous eye lids
Suspect orbital cellulitis if any of the following present:
Impaired or painful extraocular movements
Decreased visual acuity
Chemosis
Proptosis
Afferent pupillary defect 7
UK guidelines recommend admission is necessary if concerned for orbital cellulitis and any of the following presenting signs or symptoms:8
Periorbital swelling, diplopia, reduced visual acuity, abnormal light reflexes, proptosis, ophthalmoplegia, drowsiness, vomiting, headache, seizures
Evaluation:
Assess ABCs and obtain VS to include visual acuity
Fever may or may not be present 9
Perform a complete physical examination
Ocular: visual acuity, pupils, extraocular movements
If patient cannot open eye, use ocular ultrasound to evaluate direct and consensual pupil reactivity
Neuro: Focal neurologic deficits; CN III, IV, VI palsies
ENT: dental exam to assess for deep space infections/dental infections; assess for concomitant ear infection, sinus infection
Imaging: CT sinus/orbits to evaluate orbital involvement, proptosis, periosteal reaction, sinus disease, foreign body
Other indications for CT: focal CNS sign, no improvement over 24 hours, non-resolving pyrexia over 36 hours 8
Use of CT predicts accuracy of cases requiring surgical intervention 10
Laboratory evaluation:
CBC with differential
Nares culture, Conjunctiva culture
Orbital cellulitis is a clinical diagnosis, but these may help narrow antibiotics
Blood cultures are rarely positive 2
LP: Consider in patient with fever, headache, focal neurologic deficit, or toxic appearance for meningitis
Treatment:
ABCs—Toxic appearing patients require resuscitation.
Concern for elevated ICP from abscess, etc., then consider treatments to reduce ICP (hyperosmotic therapy)11
Antibiotics: 1-2 weeks IV, followed by outpatient therapy to complete a 4 week total course
Empiric 1 st line: 3 rd generation cephalosporin or ampicillin/sulbactam 12
Suspicion of anaerobes: metronidazole, clindamycin, cefuroxime 2
3 rd generation cephalosporins, penicillin, and metronidazole have excellent CNS penetration 13
Add Vancomycin for life-threatening infections or if concerned for MRSA 1,14
Consult ophthalmology and/or ENT for evaluation
Assess for abscess, may require drainage
More common in older patients 15
Complications if inadequately treated:2
Orbital abscess
CVST
Panophthalmitis
Endophthalmitis
Subperiosteal abscess
Orbital compartment syndrome can result, causing permanent vision loss
Pearls:
LP if accompanied by fever, headache, neurologic symptoms, or toxic appearance
CT aids in guiding surgical management
Older children and adults are more predisposed to anaerobic infections often need both parenteral antibiotics and surgery—consider cefuroxime in these cohorts
An 18-year-old man presents to the emergency department with complaint of eye pain and eyelid swelling for three days. He denies any trauma to the area but had an upper respiratory infection recently. He endorses fevers and chills at home. His exam is notable for periorbital erythema and edema, intact extraocular eye movements, symmetrically reactive pupils, normal visual fields, and exacerbation of his pain on movement of his eyes. The contralateral eye is not involved. Which of the following is the most likely diagnosis?
A) Cavernous sinus thrombosis
B) Iritis
C) Orbital cellulitis
D) Periorbital cellulitis
Answer: C
Orbital cellulitis is an infection of the subcutaneous tissue and ocular muscles located inside the orbit. Extension of bacterial rhinosinusitis is the most common cause of orbital cellulitis. Orbital cellulitis must be distinguished from periorbital, or preseptal, cellulitis as periorbital cellulitis is typically a mild condition and orbital cellulitis is associated with a high risk of visual loss and mortality. Both may present withfever,eyelid swelling and redness,and ocular pain. Only orbital cellulitis will be associated with pain on movement of the eyes due to infection of the extraocular muscles, as well as proptosis, ophthalmoplegia, and diplopia. The presence of these symptoms should prompt imaging to evaluate for complications of orbital cellulitis, including cavernous sinus thrombophlebitis or abscess formation. Imaging studies may also aid in the diagnosis of orbital cellulitis if the diagnosis is uncertain. Orbital cellulitis is managed with admission and broad-spectrum intravenous antibiotic administration. Frequent visual checks and ocular examinations should be done as changes in the exam may signal the beginning of an abscess or extraorbital extension of the infection.
Cavernous sinus thrombosis (A) is a possible serious complication of extraorbital extension of orbital cellulitis. Clinical features to raise suspicion for this diagnosis include spread to the contralateral eye and dysfunction of the cranial nerves that pass through the cavernous sinus. Patients may present with ophthalmoplegia and gaze palsies of cranial nerves CN III, IV, or VI, and hypo- or hyperesthesia of the dermatomes innervated by cranial nerves V1 and V2. The patient above does not have any symptoms of extension of the infection into the cavernous sinus.Iritis (B) is inflammation of the uvea which includes the iris, ciliary body, and choroid. Symptoms of iritis vary and may include eye pain and redness with decreased visual acuity. Diagnosis is through visualization of leukocytes in the anterior chamber of the eye by slit lamp examination. Periorbital cellulitis (D) is infection of the soft tissues anterior to the orbital septum and does not cause pain on movement of the eye due to lack of involvement of the extraocular muscles.
Rosh Review Free Qbank Access
FOAMed:
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Peds EM Morsels
References:
Wald ER. Periorbital and Orbital Infections. Infect Dis Clin North Am. 2007;21(2):393-408.
Tsirouki T, Dastiridou AI, Cerpa JC, Moschos MM, Brazitikos P, Androudi S. Major review Orbital cellulitis. Surv Ophthalmol. 2018;63:534-553.
Viet H. Ho MWWJCFBGH. Retained Intraorbital Metallic Foreign Bodies. Ophthalmic Plast &. 2004;20(3):232-236.
Chaudhry IA, Shamsi FA, Elzaridi E, et al. Outcome of Treated Orbital Cellulitis in a Tertiary Eye Care Center in the Middle East. Ophthalmology. 2007;114(2):345-354.
Tas S, Top H. Intraorbital wooden foreign body: clinical analysis of 32 cases, a 10-year experience. Turkish J Trauma Emerg Surg. 2014;20(1):51-55.
Toumi A, Larbi Ammari F, Loussaief C, et al. Rhino-orbito-cerebral mucormycosis: Five cases. Médecine Mal Infect. 2012;42(12):591-598.
Tintinalli JE, Stapczynski JS, Ma OJ, Yealy DM, Meckler GD, Cline D. Tintinalli’s Emergency Medicine : A Comprehensive Study Guide.
Howe L, Jones NS. Guidelines for the management of periorbital cellulitis/abscess. Clin Otolaryngol Allied Sci. 2004;29(6):725-728.
Nageswaran S, Woods CR, Benjamin DK, Givner LB, Shetty AK. Orbital Cellulitis in Children. Pediatr Infect Dis J. 2006;25(8):695-699.
Le TD, Liu ES, Adatia FA, Buncic JR, Blaser S. The effect of adding orbital computed tomography findings to the Chandler criteria for classifying pediatric orbital cellulitis in predicting which patients will require surgical intervention. J Am Assoc Pediatr Ophthalmol Strabismus. 2014;18(3):271-277.
Brook I. Microbiology and antimicrobial treatment of orbital and intracranial complications of sinusitis in children and their management. Int J Pediatr Otorhinolaryngol. 2009;73(9):1183-1186.
American Academy of Pediatrics. Subcommittee on Management of Sinusitis and Committee on Quality Improvement. Clinical practice guideline: management of sinusitis. Pediatrics. 2001;108(3):798-808.
Garcia G. Criteria for nonsurgical management of subperiosteal abscess of the orbit Analysis of outcomes 1988–1998. Ophthalmology. 2000;107(8):1454-1456.
Prentiss KA, Dorfman DH. Pediatric Ophthalmology in the Emergency Department. Emerg Med Clin North Am. 2008;26(1):181-198.
Harris GJ. Subperiosteal Abscess of the Orbit: Computed Tomography and the Clinical Course. Ophthalmic Plast &. 1996;12(1):1-8. Accessed January 5, 2019.
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SQL进阶技巧：如何计算复合增长率？
莫叫石榴姐已于 2024-11-09 12:51:28 修改
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分类专栏：数字化建设通关指南# SQL进阶实战技巧文章标签：sql数据库前端数据分析
于 2024-11-07 08:15:00 首次发布
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
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数字化建设通关指南 同时被 2 个专栏收录
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目录
0 场景描述
1 数据准备
2 问题分析
3 小结
0 场景描述
复合增长率是第N期的数据除以第一期的基准数据，然后开N-1次方再减去1得到的结果。假如2018年的产品销售额为10000，2019年的产品销售额为12500，2020年的产品销售额为15000（销售额单位省略，下同）。那么这两年的复合增长率的计算方式如下
复合增长率的计算公式如下：
复合增长率 = (最终值 / 初始值)^( 1 / n) - 1
其中：
最终值是指期末的数值；
初始值是指起始的数值；
n时间段数量是指经过n个时间段的增长所到达的值。
举例来说，假设某项指标在起始时刻（一月份）为100，经过6个月（到达七
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Inequality with positive numbers and reciprocals
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$\begingroup$
There are 2011 positive numbers with both their sum and the sum of their reciprocals equal to 2012. Let $x$ be one of these numbers. Find the maximum value of $x + \frac{1}{x}.$
inequality
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asked Dec 24, 2019 at 2:34
doingmathdoingmath
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$\begingroup$ We know that, for arbitrary positive number $x$, we have $x + \frac{1}{x} \ge 2$. So the sum of both $2011$ positive numbers and their reciprocals is equal or greater than $2011\times2=4022$, it cannot be equal to $2012$. You may double-check your question. $\endgroup$
Zeta
– Zeta
2019-12-24 03:05:48 +00:00
Commented Dec 24, 2019 at 3:05
2
$\begingroup$ @Zeta, I think the OP means that the sum of the numbers is equal to $2012$ and so is the sum of their reciprocals (so together they total $4024$, which exceeds $4022$, as you observe they must). $\endgroup$
Barry Cipra
– Barry Cipra
2019-12-24 03:41:34 +00:00
Commented Dec 24, 2019 at 3:41
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1 Answer 1
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$\begingroup$
Consider that there are $2010$ numbers, $y_1$ to $y_{2010}$, which are not equal to $x$. This means that the sum of $y_1$ to $y_{2010}$ equals to $2012-x$ and the sum of their reciprocals is equal to $2012-\frac1x$. Given the Cauchy Schwarz inequality, which states :
You get that :
And therefore:
Which means that your answer to this question is $\frac{8045}{2012}$. This question is a great use of inequalities. You can find more inequalities and applications for these inequalities at : I apologies my use of images as I am not yet very familiar with MathJax. You can find pretty much the exact same solution at as I have just reworded it a bit.
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edited Dec 24, 2019 at 3:41
answered Dec 24, 2019 at 3:12
JC12JC12
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$\begingroup$ This establishes a bound, but the question asks for the maximum. How do you know that the bound can be attained? $\endgroup$
joriki
– joriki
2019-12-24 04:53:01 +00:00
Commented Dec 24, 2019 at 4:53
$\begingroup$ In inequalities, the bound should be the maximum in this situation. The use of inequalities means that the bound can either be the minimum or maximum. $\endgroup$
JC12
– JC12
2019-12-24 05:25:14 +00:00
Commented Dec 24, 2019 at 5:25
$\begingroup$ How do you mean, it "should be"? I think it is, but I don't see an argument for that in your answer. In the Cauchy-Schwarz inequality, equality is attained if and only if the one vector is a multiple of the other. So you need to show that things work out if you make all the $y_i$ equal that's the only way you can fulfil that condition. $\endgroup$
joriki
– joriki
2019-12-24 05:47:11 +00:00
Commented Dec 24, 2019 at 5:47
1
$\begingroup$ I don't question the inequality. I started my first comment by acknowledging that you established a bound. $x+\frac1x$ can't be more than $\frac{8045}{2012}$. But a maximum is not just an upper bound; it's the highest value actually attained. So to prove that $\frac{8045}{2012}$ is the maximum, you have to prove not only that no higher value than $\frac{8045}{2012}$ can be attained; you also have to prove that $\frac{8045}{2012}$ can be attained. I still don't see any argument to that effect in your answer. I do believe it's true; you just didn't provide an argument for it. $\endgroup$
joriki
– joriki
2019-12-24 09:36:52 +00:00
Commented Dec 24, 2019 at 9:36
1
$\begingroup$ You can just solve for x, subtract from 2012, and divide by 2010 to get the rest of the terms. It works out $\endgroup$
doingmath
– doingmath
2019-12-24 19:09:49 +00:00
Commented Dec 24, 2019 at 19:09
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8375 | https://www.imrpress.com/journal/KO/51/8/10.5771/0943-7444-2024-8-613/pdf | Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023 613 Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023 Dóra Pákozdi 1702 Salvor Tower, 41 Mastmaker Road, E14 9XS, London, UK uczcpak@ucl.ac.uk Dóra Pákozdi is a Hungarian-British autistic senior library assistant in a London public library while retraining in library studies at UCL, having previously completed another MA in comparative literature. She is an advocate for autistic liberation in both classification schemes and at workplaces. Pákozdi, Dóra. 2024. “Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal Sys-tem, 1942-2023”. Knowledge Organization 51, no. 8: 613-625. 78 references. DOI:10.5771/0943-7444-2024-8-613.
Abstract: Sort of People: Considerations About the Ontogeny of Autism in the DDC, 1942-2023 traces the ontogeny of autism within the Dewey Decimal System, from the 14th to the 23rd editions. This period marks significant shifts in the psychiatric understanding and societal attitudes toward autism, reflecting the broader dynamics of neurodiversity recognition, the conceptualization of autism as a spectrum, and the influential role of autistic self-advocacy. The study draws on interdisciplinary sources and theoretical frameworks, including Ludwig’s ‘restricted malleability’ and Hacking’s ‘interactive types’ to critically analyse how changes in scientific, political, and social landscapes have influenced the organization of literature on autism in library classification systems. Methodologically, the paper employs a detailed historical review of DDC editions alongside an analysis of literature concerning the classification of various marginalized groups as well as medical, philosophical, and disability advocacy literature to map the shifts in autism's classification. This approach highlights how the language and structure of classification systems both reflect and shape societal attitudes to-wards autism. The analysis also considers the impact of autistic self-advocacy on challenging and reshaping these classifications, emphasizing the importance of language and representation in the struggle for autistic integration and visibility. The implications of this study extend be-yond the specific case of autism classification. It contributes to ongoing debates on the politics of knowledge organization, the role of libraries and classification systems in either perpetuating or challenging societal norms, and the importance of incorporating marginalized perspectives in the creation and dissemination of knowledge. The author is diagnosed autistic.
Received : 29 June 2024; Accepted 31 July 2024.
Keywords: Ontogeny ; autism; Dewey Decimal Classification; Knowledge organization systems. “Not all of us can say, with any degree of certainty, that we have always been human, or that we are only that. Some of us are not even considered fully human now, let alone at previous moments of Western social, political and scientific history”. Rosi Braidotti, The Posthuman
1.0 Introduction
Language’s dependence on context is made materially evi-dent in subject headings. A long-standing classification scheme’s aim is to collocate a type of resource based on in-terpretation of their domain (Tennis 2012). Hegemonic li-brary classifications like the Dewey Decimal System have upheld and helped entrench marginalizing models of disa-bility simply because they were created by certain cultural and scientific communities. Much of the ableism of the cur-rent age can be traced back to the Western eugenics move-ment of the 19th-century, when perceived physical, sensory and cognitive differences were first combined into the co-herent biological other that is more of a product of nor-malcy than an excluded part of it (Snyder and Mitchell 2015). Shaped by scientific discovery and self-advocacy, au-tism’s ontological status has been evolving, resulting in the classificatory confusion explored below. Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
614
1.1 Literature Review
There is a rich history of critique of the disciplinary struc-tures and subject access standards shaping the classification of resources about marginalized groups both on the descrip-tive and on the structural level. Both the language used to label categories of literature about groups of people and the order in which the resources are arranged has inevitably af-fected how users and information professionals conceptual-ize individuals belonging to said group. In a school library organized by the Dewey Decimal System, an autistic teen-ager for example would have to search for books on mental health, self-help and practical life advice under 305.9085 ‘People with developmental disabilities’ nested under ‘Peo-ple with disabilities and illnesses, gifted people’ and 616.85882 ‘Autism Spectrum Disorder’ collocated with books on an array of mental illnesses and – due to the small volume of school library holdings – medical resources. Abled, white, neurotypical, cisgender and straight students would look under 158.1 ‘Personal improvement and analy-sis’. Radical cataloguer Berman’s 1971 introduction to his foundational Prejudices and Antipathies summarily de-scribes the users LC headings were created to satisfy: parochial, jingoistic Europeans and North Ameri-cans, white-hued, at least nominally Christian (and preferably Protestant) in faith, comfortably situated in the middle- and higher-income brackets, largely domiciled in suburbia, fundamentally loyal to the Es-tablished Order, and heavily imbued with the trans-cendent, incomparable glory of Western civilization. Further, it reflects a host of untenable-indeed, obso-lete and arrogant-assumptions with respect to young people and women. And exudes something less than sympathy or even fairness toward organized labor and the sexually unorthodox or ‘avant-garde’ (Berman 1993,15). Within classification systems, power is expressed through naming and ordering the phenomena of the world in rela-tion to one another. Berman aimed to remedy this by cor-recting perceived biases within the LCSH, while other in-fluential thinkers, like Hope A. Olson (Koford 2014) theo-rize that injustice is innate or inevitable in classification schemes. Members of groups falling outside Berman’s de-scription have since made tremendous efforts to represent themselves and their unique knowledge and ways of know-ing: Duarte and Belarde-Lewis (2015) shows a vision of a world where the 600 indigenous tribes within US political borders each utilized their own knowledge organisation sys-tem based on their communities’ unique epistemic cus-toms, through analysing ten years of messages on two elec-tronic cataloguing lists. Ho (2005) showed how most pa-trons’ want to search foreign films by languages and that cataloguers support the addition of language and produc-tion form/genre headings to enhance their browsing experi-ence, while Wagner (2022) explored body-oriented descrip-tion as a new approach to cataloguing visual information in a gender inclusive way and Furner (2007) introduced criti-cal race theory as a framework for evaluating classification schemes.
2.0 The Neurodiversity Framework and Library Classifications
The present work aims to trace the ontogeny of autism throughout editions 14 to 23 of the Dewey Decimal Sys-tem. Much like intersexuality (Fox 2016) the subject of au-tism is particularly suitable to showcase breakdown and scatter in a classification scheme while the process of clam-ouring to accommodate the rapidly increasing volume and variety of publications that are in turn trying to make room for a previously invisible people and identity. This space – whether physically on a library shelf or theoretically in a controlled vocabulary – seems to be a scarce resource. Tennis (2012) notes that the scatter caused by the inflex-ibility of an ableist classification scheme is the result of in-terpretations made by certain agents about literary or user warrant of the domain and not by changes of semantics al-tering collocation. The history of allistic medical experts and classification specialists having exclusively shaped the knowledge available about autistic people can be conceptu-alized through Wittgenstein’s (2011) ‘language games’ where communities decide the agreed-upon terms through practical considerations. Mai (1999) argues that a knowl-edge organization system is a standardized version of these language games of a given organization and that a terminol-ogy fixed in such a way is highly political and can be both a great asset and a real a threat to freedom of expression. As autistic self-advocacy’s visibility increases, the hope is that autistic people and our allies will be behind most of the knowledge disseminated about autism. From a disability studies perspective, libraries can be thought of as locations of marginalization, and the social spaces where classification schemes were devised as links of a network of culture in which the divide between abled and disabled was already innate and while terminology had shifted from ‘degenerate’, ‘defective, ‘retarded’ and ‘defi-cient’ to ‘disordered’, the structural underpinnings remain mostly unchanged. Despite librarians having addressed that schemes like the Dewey Decimal System (DDC), Library of Congress Classification (LCC), and the Library of Con-gress Subject Headings (LSCH) fail to accurately collocate resources about groups of people lacking political or eco-nomic power (Adler et al. 2017). Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
615
2.1 Classifying Autism: an Introduction
The classification of autism as a Dewey Decimal System sub-ject follows the rapid advancement of psychiatry over the course of the 20th-century, trailing behind the similarly re-cent expansion of self-advocacy of neurodivergent humans. Drabinski (2013), critiquing the disciplinary structures shap-ing the cataloguing of queer library materials points out how librarians have failed to properly and respectfully catalogue works on marginalized groups, but instead of placing the strain on cataloguers, invites users to engage with catalogues as biased texts. Extant hegemonic classification schemes are resistant to change because of baked-in inertia (Bowker and Star 1999) while also causing significant scatter (Olson 2008). The uncertainty about autism’s location in the totality of hu-man knowledge has resulted in books about autism shelved in medicine, self-help, biographies, mental illness and parenting. Educating library users to critically evaluate and enhance these same systems that librarians work within and against has the potential of lessening the scatter and the classificatory marginalization of disabled people. Nearly three decades after Leo Kanner’s influential re-port on what he called Infantile Autism, research finally agreed to treat autism as an independent diagnosis of Schiz-ophrenia, based on an overview of symptomatic differences like age of onset, family histories and responses to treatment (Meyer et al. 2011). Another three decades were needed for the Dewey Decimal System’s 22 nd edition to catch up with the advancements. Further complicating the diagnostic his-tory and hence classification of literature about autism are contemporary concepts like the spectrum and neurodiver-sity, medical breakthroughs like the merging of autism and Asperger Syndrome diagnoses, and the rapidly increasing visibility of autistic self-advocacy. The debates and discus-sions ignited by self-advocacy are underpinned by the un-derlying framework of neurodiversity; concepts like the Double Empathy Problem (Milton 2012) or Monotropism (Murray et al. 2005) have had genuine impact in research, practice (Leadbitter et al. 2021) and the self-understanding of many neurodivergent and autistic people. The funda-mental change is merely from a focus on normality to one on prevalence. Doyle (2020) proposes a new umbrella term for atypical neural and cognitive phenotypes: a neuromi-nority is a population sharing a particular symptom cluster and encountering similar challenges in a majority neurotyp-ical society. According to CDC data, while in 2002 one in every 150 US citizen was diagnosed autistic, by 2020 this number has expanded to one in every 36.
3.0 History of the concept of Autism
Literary scholar Murray’s (2012) monograph on autism, ac-cording to its blurb, aims to “present a rounded portrayal of the ways in which autism is currently represented in the world” (42) yet takes a disappointingly reductive view of re-search into possible autistic individuals’ lives, writing that to search for autism in the past is ”a point about searching for definitions of it in the present” and calls for any claim about the pre-scientific history of autism “to be made with real expertise, and not left to excited or over-eager guess-work”. 4 The reluctance to centre autistic people in our own history in favour of an ever-shifting, often disgraced cast of scientists is a major setback in the way of Autistic Liberation – the implication that only the professional and medical es-tablishment is considered “real expertise” on identifying au-tism is rooted in the same confused, medicalised, pathology-focused concept of autism, enabled by library classification systems over the last century.
3.1 Autism, a Conceptual History: Beginnings
Prior to a scientific understanding of disability, neurodiver-sity and mental illness, many cultures conceptualized autis-tic people as changelings, elves or other almost-human crea-tures, tragic and distant, sometimes uniquely useful bur-dens on their communities (Wing 1997). An echo of this di-versity of cognitive style and ability – seen by allistic people as seemingly contradictory strengths and weaknesses of au-tistic individuals – is present in the collocation of disability, illness and giftedness under the class number 305.908. There is enough of an abundance of autistic voices – with diverse support needs – in online autistic groups, the arts, media and our personal lives, to know that the autistic peo-ple of today still find ourselves in the depicted near-human-ness of elves, robots and aliens, while an equally large group passionately dislikes the comparisons and the ableist impli-cations they are riddled with. It doesn’t require guesswork that previous generations followed similar lines of thought. A disabled child was often thought of as a monstrous fey creature corrupted or exchanged by harmful spirits (Wing 1997). This is an enduring concept exemplified by incidents as recent as Lord’s (2006) award-winning autobiographical novel Rules, in which the mother of an autistic child muses “Sometimes I wish someone would invent a pill so David'd wake up one morning without autism, like someone waking from a long coma, and he’d say, “Jeez Catherine, Where have I been?” or the 2007 Ransom Notes Campaign – in which autism, depicted as an ominous, child-snatching villain, left threatening messages to the victims’ distressed parents – that in many ways signalled the beginning of the Neurodi-versity Movement (Kras 2010). The momentum of enthusiasm for scientific classifica-tion originates from the early modern period and only ran out of steam during the early 20th-century, producing tra-ditional classification schemes tending towards a concern primarily about an often false sense of objectivity and neu-Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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trality (Hjørland 1998). Since the advent of psychiatry and psychology different methods of classifications have been used to organise these scientific domains’ knowledge about autism and atypical neurologies in general, each system re-flecting the theoretical approach and the biases, views and personal horizons of their scientific authors. Knowledge about autism generated outside institutional or scientific settings was not taken into account by existing classification schemes, some of which now struggle to include it. In her book, Letters to My Weird Sisters: On Autism and Feminism , autistic scholar Joanne Limburg takes a radical, ‘dishuman’ approach to bypass the medical establishment’s monopoly on defining autism; she addresses personal letters to a neurominority: historic women who shared the same or similar lived experiences that she had had as an autistic cis-gender woman. The ‘dishuman’ approach seeks to trouble established notions of what it means to be human based on Braidotti’s concept of the posthuman (2013) from the stance of Disabled Liberaton. 3 Steve Silberman, author of
Neurotribes was chastised by the British Journal of Psychia-try for identifying autism in historical figures and, accord-ing to the review’s author, Lisa Conlan “It is hard to shake the feeling you are being toyed with in the name of a bigger political agenda”. The agenda Conlan wants to avoid is Dis-abled Liberation. Milton (2012) contests scientific methodologies’ notion that the ability of neurotypical individuals to estimate the mental states, motivations and emotions of a peer is what constitutes empathy, when autistic people often see these es-timates as inaccurate, hurtful or ignorant. The resulting communicational disconnect is rarer and often a brief inter-lude in the social and cognitive reality of neurotypicals, but is a traumatising everyday reality for autistic people. Hack-ing (2009) puts it: “There is a partial symmetry between the autistic and the non-autistic. Neither can see what the other is doing. The symmetry is only partial because we have an age-old language for describing what the non-autistic are feeling, thinking and so on, but are only creating one for the autistic” (1471). A 20-year-old non-speaking advocate, Noah Seback, after summarising how he considers most of his education ‘warehousing’ and ‘babysitting’ because of ed-ucators’ refusal to imagine autistic students as agents, spelled out ‘presuming competence’ on his letterboard (Holmes 2024) .
3.2 Autism, a conceptual history: pathology
Although much less known than Leo Kanner’s or Hans As-berger’s reports, Soviet scientist Grunya Sukhareva was the first to pathologize what we now understand to be autism (Sher and Gibson 2023). Despite her description of autistic traits being remarkably close to those expressed in current diagnostic criteria her status as a Soviet woman under Stalin prevented her ideas from being disseminated outside of the Soviet Union (Manouilenko and Bejerot 2015). Kanner’s 1943 report examined two essential areas of au-tistic difference: social disconnectedness and a strong pref-erence for sameness. He had also described autistic behav-iours like echolalia and repetitive movements, and suggested that autism was innate and that autistic children often re-semble their parents, a notion which under Bruno Bettel-heim grew into the harmful idea that autism is caused by emotionally neglectful parenting, particularly perpetrated by mothers. Although the mid-century brought unprece-dented upheaval to the field of psychiatry, Kanner’s super-ficial association of autism and schizophrenia also hindered progress, with the 19th-century protocol of institutionalisa-tion of non-speaking and intellectually disabled autistic in-dividuals in inhuman, neglectful and abusive asylums (Donvan and Zucker, 2016) has been constant ever since; according to NHS England data 2045 autistic people and/or people with learning disabilities were inpatients in February of 2024 and 1075 of these patients had a length of stay over 2 years. Autism was established as a separate diagnosis in 1980 af-ter several competing definitions of autism had been pro-posed by Rutter (1978) and the The American National So-ciety of Children (1978) and the comparisons to schizo-phrenia established autism as its own condition (Rutter 1972). Autism was entered into the DSM-III (APA 1980) under the new class of Pervasive Developmental Disorders and the conceptual realm of deficit, disorder and pathology. Kirk and Kutchins (1994) in their analysis of the lack of re-liability of the DSM-III quote the manual stating how “It is particularly encouraging that the reliability for such catego-ries as schizophrenia and major affective disorders is so high” (1980, 468) which is contradicted by the relationship between autism and schizophrenia being described as “con-troversial” by the new DSM article on Infantile Autism ex-plaining: “Some believe that Infantile Autism is the earliest form of Schizophrenia, whereas others believe that they are two distinct conditions” (1980, 87). The WHO’s International Classification of Function-ing, Disability and Health (ICF) was accepted by all the or-ganisation’s member states in 2001 as a comprehensive model and classification of disability that uses codes to cap-ture the details of disabled individuals’ functioning across various domains. The categories are flexible, but are often reduced to Core Sets – comprising only categories most per-tinent for each condition – for clinical utility (Hayden-Ev-ans et al. 2024). The International Classification of Dis-eases–Eleventh Revision (ICD-11) recommends clinicians use ICF categories to describe the impact of health condi-tions on individual functioning (Bölte 2018). The aim is to tailor services and resource-allocation to the individual needs of a disabled person, instead of providing potentially Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
617
mismatched, purely diagnosis-based care. The WHO Disa-bility Assessment Schedule 2.0 is based on the ICF and was published in 2012, a year before the DSM-5 officially intro-duced the spectrum-model of autism. Emphasizing variety over pathology, autistic self-advo-cates argued that disability is largely created by various barri-ers erected by ableist societies, as opposed to by some inherent deficit of autistic and neurodivergent humans. The spec-trum-model of autism (Wing 1993) imagines autistic traits in diverse constellations, unique to each individual, resulting in a personal profile of support needs, cognitive styles, commu-nication difficulties or preferences, and sensory differences – finally included in the DSM-V’s diagnostic criteria. This more nuanced understanding of autism has led to the wave of diagnoses reported by the CDC above. Despite the current fears about an ‘autism epidemic’ and ‘overdiagnosis’, studies have shown that early intervention improves developmental outcomes for autistic children, while caregivers are empowered to be able to support their families by accessing specialised support and community con-nections while reducing their own parental stress (Okoye et al. 2023). In order to reframe effectiveness in a neurodiversity-affirming way, early intervention has to centre well-being, au-tonomy, coping strategies and autistic-prioritised interven-tion targets (Leadbitter et al. 2021). In her article What Can Physicians Learn from the Neurodiversity Movement? Dr Christina Nicholaidis (2012), a mother of an autistic child urges her fellow medical professionals to “try to understand an individual’s complex combinations of strengths and chal-lenges, as well as the potential for wide variations in function-ing”. Instead of focusing on behaviours deemed impairments, autistic self-advocates seek to reframe the pathology model into one that addresses the difficulties of autistic humans in-formed by an Ethics of Care.
3.3 Towards neurodiversity
A useful narrative picture regarding the published usage of autism-related terms and synonyms can be derived through a Google Ngram Viewer search, showcasing instances of us-ages of terms within Google Books’ digital corpus (Figure 1). The first occurrence of the term ‘autism’ coincide with Leo Kanner’s mid-1940s works. His extremely narrow defi-nition of autism can be seen in the minuscule dent ‘autism’ makes in the frequency of use of ‘Schizophrenia’, the previ-ous only overtaking the latter a decade after the DSM-III es-tablished autism as a condition entirely separate of Schizo-phrenia and in the immediate aftermath of Lorna Wing’s influential work popularizing the long-forgotten Hans As-perger (Wing 1981) and broadening of the concept of au-tism (Wing 1993). ’Infantile Autism’ disappears at the same time, somewhere in the 1990s. Despite the current fears about overdiagnosis, the evi-dence suggests that the largest spike in published interest in autism occurred between 2000 and 2010, not after the pub-lication of the DSM-V. The extremely low rate of ‘Neurodi-versity’ occurrences is a striking reminder of the discrepan-cies between the linguistic preferences of self-advocates ver-sus the scientific world. The umbrella term ‘Neurodiver-sity’ was first coined by autistic sociologist Judy Singer in 1998 as a non-medical term that shifts the focus from defi-cit to the variety of human neurology and was quickly and enthusiastically embraced by the majority of the autistic community. It erases the need for binary medical or social models of autism. The above graph illustrates the scientific and publishing world’s reluctance to follow suit.
Figure 1. Google Ngram Viewer instances of competing terms to describe autism. Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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4.0 Theoretical considerations
A distinct characteristic of the Dewey Decimal System is its enumerative focus on disciplines. The same subject may ap-pear under any discipline it has sufficient literary warrant in. This poses unique problems for the classification of dis-abilities and identities. To illustrate these problems, let’s consider the example of the topic of autistic children’s play. Lining and sorting toys in joyous solitude has been widely described as an early symptom of disordered development. Despite the hypothetical parents’ goal to seek out activity books or collections of ideas to accommodate and support their child’s style of play, they will be bombarded by lists pathologizing perfectly healthy autistic behaviours, creating confusion and distress. This is akin to classifying Anne of Green Gables under ‘abnormalities’ based on the rarity of natural red hair. Another example is that of unscientific memoirs by autistic people being routinely shelved and dis-played in medical sections. Parents of autistic children look-ing for inspiring, empowering stories under ‘First Facts’ or ‘First Experiences’ style sections under class number 616 – common in UK public libraries – may find, nestled between infant health handbooks and specialist nutrition guides, Ma’s (2017) picture books In My World , in which at the mention of autism, all colour drains out of the previously depicted images of adventure, family and fun and the reader suddenly finds the main character standing on a bleak, empty page inscribed “In your world I have autism”.
4.1 Restricted malleability
Like Tennis’ example of eugenics (2002) demonstrates, sub-jects can wonder in and out of disciplines as human interest and published scholarship change over time. Eugenics started out in the 1910s in the realms of Science, under the 500s near Genetics within Biology and have since moved into diverse classes of mostly Social Sciences. Over time, these shifts break down the collocative integrity of the scheme; books on the biological aspects of the human eu-genics movements may sit next to plant reproduction. Au-tism in the Dewey Decimal System has a similarly shaky tra-jectory emerging from obscurity through the classes of ‘De-mentia’ and ‘Schizophrenia’ to reach its own term by 2003 – but remaining mostly in the realm of medicine. A useful concept for understanding this journey is ‘re-stricted malleability’ (Ludwig, 2023). It is similar to Ian Hacking’s explanation of ‘interactive kinds’ (1999). Hack-ing’s example is of another neurodiverse condition, ADHD. He argues that the interaction between the terms used to classify things or people and the people themselves are in a discourse, both affecting the other. People diagnosed autis-tic differ from non-speaking, disabled or eccentric humans of the past, in the ways they are diagnosed, treated and the-orised about, but in being medicalised in such ways, autistic people had also had an effect – on the concept of disability and eccentricity. Ludwig (2023) suggests a shift in focus from naturalness to materiality in debates about scientific classifications and argues that material kinds are affected by intervention. Nei-ther the restriction nor the malleability is confined to either dimension of conceptual or non-conceptual, and one of Ludwig’s examples is psychiatric kinds: these are restricted by the structure of their domain but also by their applicabil-ity – overly complex classifications are a hindrance to psy-chiatric practice. Ludwig adds that the malleability of kinds is a multi-dimensional gradient: categories like ‘biological species’, ‘chemical elements’ and ‘bosons’ are less malleable than classes of the likes of furniture, mental illness or gen-ders and each of these can be shaped along underlying struc-tures (see Figure 2). Malleability Material properties of ‘autistic person’ are shaped by changes in scientific advancements and classifications. Classifications are shaped by changes in the material proper-ties of ‘autistic person’. Example The merging of previously separate diagnoses collocated ‘people with Asperger Syndrome’ and ‘autistic people’. Increased rates of diagnosed women and genderqueer people mean autism is not thought of as an exclusively male condi-tion anymore. Restriction Material properties of ‘autistic person’ restrict and limit clas-sificatory options. Classifications are limited by demands towards linguistic representation. Example The lived experience of the sum-total of ‘autistic people’ should be the basis and entirety of what science considers ‘autism’. Example: autistic self-advocates demanding the removal of deficit- based language.
Figure 2. The restricted malleability model of the debate and progress of the classification of autism based on Ludwig’s (2023) model. Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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4.2 Centring user needs
The exclusion of autistic people from the project of classify-ing their experiences and the resources about themselves is one of the underlying systems along which knowledge about autism has been organized. Since non-medical publi-cations about autism and – knowingly -autistic researchers and cataloguers are a relatively new phenomenon, the liter-ary warrant for ‘autism’ and its adjacent concepts have al-most exclusively been based on medical texts. Fox (2016) demonstrates how in the case of a marginalized and poorly understood community, confusion in literature then pro-vides the ontological basis for including terms and designing classifications, leading to even deeper marginalization. To alleviate the catastrophic levels of exclusion autistic people face in education, employment, healthcare and our social environment, this process needs to be reversed. Rey (1995) argues that the selection of classificatory terms should depend on the explanatory work one wants concepts to perform his work must centre inclusion and an affirmative attitude towards the material realities of neuro-divergent behaviours. While this pragmatism has its draw-backs – Fox’ (2016) example of Intersexuality being classed under sexual orientation rather than the facet ‘People by sex or gender’ was a pragmatic decision based on the perceived users’ needs as they are more likely to search for resources about Intersexuality near LGBTQ+ issues – controlled vo-cabularies could be harnessed for the work of Neurodiver-gent and Autistic Liberation.
5.0 14th-21st edition: Schizophrenia
The understanding of disability and attitudes towards its at-tendant non-normative behaviours varies greatly across time periods and cultures (Longmore 2003). Autistic traits like monotropic attention and advanced pattern recogni-tion were sometimes highly desirable skills that helped some privileged disabled individuals to lead fulfilling and happy lives. Henry Cavendish, described as “the coldest and most indifferent of mortals” – could calculate the density of the Earth from his 18th-century home while living in near-per-fect isolation and struggling to make eye-contact (Silberman 2016). Albeit community norms about acceptable behav-iours have always existed – Cavendish was thought beyond eccentric when, upon encountering his maid one day, had a separate staircase built in order to avoid all human interac-tion (Silberman 2016) – scientific interest in pathologizing behaviour began only when psychology first distinguished itself from philosophy.
5.1 14 th Edition
The 14th edition, of the Dewey Decimal System, published in 1942, a year before Kanner’s (1943= landmark report, carries a fossil from those times by classifying Schizophrenia both at 132.1982, and at 616.8982 under ‘Dementia precox and schizophrenic reaction types’. The classmark 132.1 was for ‘Insanity’ and ‘Mental alienation’ while 616.8 stood for ‘Diseases of nervous sistem’ and ‘Psychiatry’. During this time, in 1943, Donald Triplett was diagnosed as “Case 1” of autism by Leo Kanner. Triplett was US American, white, male, had savant abilities and was born into an affluent bank-owning family (Pallardy 2024). Temple Grandin – also white US American and affluent with savant abilities – was diagnosed in 1950, opening up the possibility for women to be considered autistic (Richter, 2014).
5.2 15 th Edition
The next, 1951 edition shows the mid-century acceleration of psychiatric progress by simplifying the class and listing the diagnosis and treatment of conditions like Schizophre-nia, Paranoia and Manic-Depressive Psychoses under one classmark at 616.89. 616.8 is also referenced in the notes of the 132 class titled ‘Abnormal psychology’. These read “ In-cludes irrational, abnormal, or deranged mental processes; their causes; mental symptoms of disease ” and “ For Medical treatment of these disorders, see Nervous system and neurol-ogy, 616.8; Psychoses and psychiatry, 616.89 ”. It appears to demarcate ‘Abnormal psychology’ under 132 and ‘Psychol-ogy’ under 150. Clearly, those behind this decision had a precise work they wanted these concepts to perform; how-ever intentional, this change has the clear impact of marking the neurology and behaviours of some humans as flawed and unnatural. Since then, ‘deranged mental processes’ like autistic humas’ need for sameness for example were proven to be linked with the anxiety and dread most autistic people feel when encountering unexpected change (Uljarevi ć et al. 2017). Longmore (1985) demonstrates how disabled people are often viewed as mere objects of medical attention even far removed from medical settings. A consequence of this is the picture of the adult disabled patient as impaired, docile, childlike and eternally grateful to be at the mercy of medical professionals. Any autistic adult searching the internet or li-braries for advice is intimately familiar with being bom-barded with resources addressing exclusively allistic parents of autistic children. The DSM’s first edition was published in 1952, simpli-fying the confused and varied documents used by psychiat-ric professionals for diagnostic practice. Since its inception in the mid-twentieth century, the Diagnostic and Statistical Manual of the American Psychiatric Association – cur-rently in its DSM-5-TR edition – has exerted tremendous Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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medical and cultural influence. By providing comparable samples, the DSM helped accelerate related research. The DSM is a principal guide to psychiatry and is used by poli-cymakers, insurance companies, researchers and some psy-chiatric professionals. When research had cleaned up the lack of validity about autism as an independent diagnosis, in 1980 it was entered into the DSM-III with a heavy focus on its ‘infantile’ nature (Volkmar and Reichow 2013).
5.3 16 th Edition
The 16 th edition from 1958 seems the most confused. It in-troduces the term ‘mental illness’ and superimposes it on class 132.1 located next to 132.2 ‘mental deficiency’, the class containing topics like drug addiction and queer sexu-alities. Between them it also establishes a ‘functional psycho-ses’ group at 132.19 containing many types of Dementia, Schizophrenia and Manic-depressive Psychoses. This is also the first edition to mark separate classes in the 300s for ser-vices related to mental illnesses and disabilities. The medical class number for Schizophrenia remained unchanged here and in the next three edition as well with the difference of classes between 616.892-616.898 labelled ‘specific psycho-ses’.
5.4 17 th Edition
In the 17 th edition of 1965, the domain of increasingly pro-fessionalizing psychology resulted in tighter collocation and the domain was organized under 150, with 130 being left behind to represent ‘Pseudo-and parapsychology’. The 157s were dedicated to ‘Abnormal and clinical psychologies’ hav-ing previously shared the classmark 132 with the newly un-scientific knowledge now under the 130s. 616.85-86 was re-served for ‘Psychoneuroses’ like ‘Hysteria’ and ‘War Neuro-sis’, Anxiety, Phobias, OCD and Epilepsy. The psychiatric domain was charted under 616.89 and seemed to contain only Schizophrenia under 616.8982 – then understood as a type of Dementia’ and Dementia under 616.8983. Immedi-ately following, 616.9 was dedicated for ‘Communicable diseases’. It is immeasurable how much damage was caused by collocations like this encouraging the public to associate various neurological differences and impairments with transferable diseases. The term ‘autism epidemic’ remains in use by scientifi-cally illiterate journalists who often deliberately or igno-rantly misrepresent data about autism to garner a response of shock and concern. Hill’s (2024) Guardian article
What’s behind the UK’s increase in autism diagnoses? writes that one of the reasons behind the increase is that in the 1980s only a quarter of intellectually disabled people were also diagnosed autistic. Hill (2024) writes “Now the NHS acknowledges that it could be as high as three-quarters.” The citation clickable on the words “NHS acknowledges” links to a 2021 news article about the NHS’s Long Term Plan, more precisely about how three-quarters of intellectu-ally disabled people aged 14 and older have received an an-nual health check two year ahead of targets laid out in the Long Term Plan. To round the paragraph off, Hill quotes the multiply-discredited Baron-Cohen saying “That’s an in-credibly steep rise”. The source of this quote remains un-known. Amelia Hill has written several widely-read articles about autism.
5.5 18 th Edition
The accelerating progress and social capital of psychiatry and medicine can be seen mirrored in how the 18 th edition further intertwines the two broad classes by instructing the classifier to “ Add to 157.2 the numbers following 616.89 in 616.895-616.898, e.g., manic-depressive psychoses 157.25 ”. Non-medical books about Schizophrenia and autism then would sit under 157.282.
5.6 19 th Edition
The 19th edition in 1979 instructs classifiers to place re-sources about both organic and functional psychoses under the medical class numbers in 616.892-616.898, further shrinking the 157 ‘Abnormal and clinical psychologies’ group. Schizophrenia remained under 616.8982 which re-mains a common class number for books about autism.
5.7 20 th Edition
The 20 th edition, published in 1989 is the first one to men-tion the term ‘autism’, albeit not with its own number. This came after 9 years of the DSM-III’s establishment of an in-dependent diagnosis and few years into Wing’s (1993) re-search that would establish the spectrum-model of autism. In the medical schedules it is wedged into the same class as Schizophrenia, 616.8982 and in Education it is placed at 371.94 where autistic students serve as an example of ‘emo-tionally disturbed students’. The notes even make sure to warn information professionals not to class all resources about ‘delinquent and problem students’ under this num-ber. It is also worth noting how ‘autistic students’ is nestled between various degrees of ‘retarded’ and ‘gifted’ students. This placement hints at what autistic self- advocates call a ‘spiky’ or ‘uneven profile’; a result of atypical neurological connectivity. In his literature review of EEG and MEG stud-ies of autistic brains, O’Reilly et al. (2017) identified a clear trend of long-range functional underconnectivity. This ex-plains why autistic individuals might struggle with some seemingly basic tasks, while excelling at areas requiring deep attention and pattern-recognition. Atypical strategies for Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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the allocation of attention are employed by most Neurodi-vergent people and the framework of Monotropism (Mur-ray et al. 2005) can help demystify autistic cognitive styles by deconstructing myths and misunderstandings about au-tistic life.
5.8 21 st Edition
By the time of the 21 st edition, published in 1996, Asperger Syndrome was formalized into a short-lived diagnosis and a Dewey Decimal System class under 616.858832. This has only recently been depreciated and then discontinued dur-ing the summer of 2023, ten years after the current DSM-V had merged the two into ‘Autism Spectrum Disorder’. This delay probably contributed to the immense confusion about merging of the two diagnoses.
6.0 22nd-23rd edition: Autism
The 2000s were a turbulent time for autism-related re-search, with Andrew Wakefield’s anti-vaccination fraud stretching from the 1998 publication of the since retracted paper on a debunked causal link between MMR vaccines and autism to the Lancet’s 2010 retraction of it. In 2003 Si-mon Baron-Cohen’s published The Essential Difference: Men, Women and the Extreme Male Brain in which he ar-gues – among many disturbing and discredited findings – that autism is caused by ‘extreme male brains’. It is not (van Ejk and Zietsch 2021). Around the year 2000, as seen on the above Google NGram diagram, the frequency of published mentions of autism dramatically surpasses that of Schizo-phrenia for the first time, indicating a dramatic increase in published material.
6.1 22 nd Edition
2003’s 22 nd edition of the Dewey Decimal System was the first to include a separate class for the medical aspect of au-tism. This edition discontinued to use of 616.898 for autism and reshuffled the classification of Schizophrenia. The for-mer was relocated to 616.85882 while the latter was moved mostly to 616.8581 where it is more tightly collocated with various personality disorders – albeit still within a grab-bag of ‘Personality disorders, sexual disorders, impulse-control disorders, factitious disorders, developmental disorders, learning disorders; violent behaviour; intellectual disabili-ties’ under 616.858. Medical, therapeutic and etiological re-sources remained under 616.898. Implementation of all these subtle changed across various institutions must have taken longer.
6.2 23 rd Edition
Currently autism sits under 616.85882, immediately under ‘Intellectual disabilities; developmental and learning disor-ders ’, a category that neighbours a slew of extremely varied conditions and neurological variations from ADHD through APD and even homicidal behaviour. It also neigh-bours an intensely subdivided class of ‘other congenital ab-normalities’ like Hydrocephalus and in public libraries with smaller collections, patrons might find themselves browsing for books about their own autism next to books about can-cer or animal husbandry. A welcome improvement is the acknowledgement of the marginalization of autistic people, and more especially au-tistic adults. Although class numbers for non-medical re-sources about autism are a bit more difficult to find. With the addition of facets, 362.196-362.198:001-009 ‘Services to patients with specific conditions’ can be used for re-sources about social services and 305.9085 for works on au-tistic people ourselves. Removed from the realm of medi-cine, it is a welcome change to see neither of these contain-ing terms like ‘disorder’. The regrettable (over)use of terms like these is comparable to Fox’s (2016) study showcasing how Intersexuality used to be classed under ‘monstrosities’ and ‘sexual diseases’. A further, non-invasive improvement could be to use ‘condition’ in place of ‘disorder’ and leave the notion of normality behind.
7.0 Conclusion
Broadly speaking, the concept of autism has emerged from the total obscurity of mythic monstrosity and benign, if shunned, eccentricity, into the confused pair conditions of schizophrenia/ autism to split into two variations around perceived functioning levels and is now slowly becoming a group of people with personal profiles of a constellation of many autistic traits and abilities. Evident in this process are the competing concepts of variation versus pathology, the latter dragging with itself a long history of epistemic objec-tification and outright threat of extinction, but also the very real expression of barriers preventing autistic people from living fulfilling and happy lives. Martin (2021, 294) writes that “treating people and groups with respect necessitates calling them by the names they use for themselves”. Autistic activist Bonello’s (2022) research included interviews with 11,211 responders, most of whom were autistic that showed that despite terms like ‘disorder’ following autistic people everywhere they go – in-cluding libraries – most of us don’t want to be called a dis-order (Figure 3). Goldberg’s (2023) analysis illuminates the dialectic of malleable social norms and neurodiversity through the example of left-handedness: it is caused by atyp-ical brain lateralisation and left-handed individuals have his-Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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torically been discriminated against through lack of accom-modations and through association with ritual impurity and undesirable qualities. Using accurate classificatory terms – placed correctly in their context – for autism would be a part of a similar normalisation process; instead of being called ‘unclean’, the term ‘left-handed’ bypasses the stigma and helps facilitate the necessary co-operation of affected in-dividuals and providers of pertinent services. As an autistic public librarian, it is beyond jarring having to walk confused parents past bays of books listing every medical diagnosis from the bubonic plague to brain damage. Not shelving all – or the majority of – books about autism on medical shelves would help de-pathologise atypical neurology and benefit all humankind by encouraging library-users to stop othering autistic people. On the subject of outdated terms in controlled vocabu-laries about queer people, Drabinski (2013) argues that try-ing to correct all of the terms is a secondary concern, if not outright dangerous as identities are deeply embedded in their cultural and chronological context and ripping them out of the reach of library users is destructive. The main dif-ference between queer and autistic people in this matter however is that while the social aspects of autism has indeed changed over time, unlike queerness, autism is neurological: it results from atypical neurological connectivity. With the exception of the medical gatekeeping of gender- affirming care, queerness is no longer medically pathologized in most of the English-speaking world, but it doesn’t mean that spe-cialized services and spaces don’t exist for queer people. A gay teenager doesn’t need a brain scan or a multi-occasion diagnostic process to access identity-affirming care. Drabinski’s (2013) approach – projected to autistic people’s position in classification schemes – steers towards the mar-ginalization of materiality, while providing ample space for the subjects themselves to assert their presence in the cata-logue. She is however less interested in finding correct terms and instead asks “ Why don’t I see myself in the subject vocab-ulary, and what does this tell me about the other ways I feel invisible ?”. Anderson (2021) answers the above by conduct-ing research in which autistic participants select their pre-ferred way to communicate. Another way to answer ‘What is autistic invisibility like?’ is based on this review of the ontogeny of autism in the ten most recent editions of the Dewey Decimal System: as agents and even as a unique people, we are at the cusp of leaving complete invisibility. The confusion of the knowl-edge domain is reflected throughout its literature, which then either crystallizes into confused and ever-changing classification schemes (Fox 2016) or becomes increasingly un-scientific. While the Dewey Decimal System’s solutions
Figure 3. Chris Bonello’s research showcasing preferences for autistic self-identification. Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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are less than ideal, its editors chose to follow scientific ad-vancement, a jarring example of the opposite approach is NLM’s treatment of autistic people: at WM 203.5. there is still no individual classmark for autism. This is the complete annihilation of autistic people’s epistemic agency. Consid-ering that many in the medical establishment still seek to ‘cure’ autism, this blank space is especially atrocious. The author is diagnosed autistic.
Endnotes
See Noah’s website at 2. A common practice in talking about Neurodivergent students in British secondary schools.
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European Journal for Philosophy of Science 13: 1-23. Ma, Jillian. 2017. In my world . Illustrated by Mimi Chao. Arlington: Future Horizons Inc. Mai, Jens-Erik. 1999. “A Postmodern Theory of Knowledge Organization”. In Knowledge, Creation, Organization and Use: Proceedings of the ASIS Annual Meeting Oct. 31-Nov. 4 1999, Washington, DC, USA, edited by Mar-jorie M. K. Hlava and Larry Woods. Medford, NJ.: In-formation Today, 547-56. Manouilenko, Irina and Susanne Bejerot. 2015. “Sukhareva – Prior to Asperger and Kanner”. Nordic Journal of Psy-chiatry 69, no. 6: 479-82. 39488.2015.1005022 Martin, Jennifer M. 2021. “Records, Responsibility, and Power: An Overview of Cataloging Ethics”. Cataloging & Classification Quarterly 59, no. 2-3: 281–304. Meyer, Urs, Joram Feldon and Olaf Dammann. 2011. “Schizophrenia and Autism: Both Shared and Disorder-Specific Pathogenesis Via Perinatal Inflammation?”. Pe-diatric Research 69: 26-33. 0b013e318212c196 Milton, Damian E. M. 2012. “On the Ontological Status of Autism: the ‘Double Empathy Problem’”. Disability & Society 27, no. 6: 883–7. 87599.2012.710008 Knowl. Org. 51(2024)No.8 D. Pákozdi. Sort of People: Considerations About the Ontogeny of Autism in the Dewey Decimal System, 1942-2023
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8376 | https://www.teacherspayteachers.com/Product/Unit-Rate-Problems-Lesson-Plan-10650752 | Unit Rate Problems Lesson Plan
Description
Our Unit Rate Problems lesson plan teaches students how to understand and solve unit rate problems in mathematics. During this lesson, students are asked to work collaboratively with a partner to create unit-rate word problems to be solved by other groups of students in their class. Students are also asked to solve practice word problems where they are required to find the unit rate, demonstrating their understanding of the lesson material.
At the end of the lesson, students will be able to solve unit rate problems including those related to unit pricing, constant speed, and others.
Unit Rate Problems Lesson Plan
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8377 | https://www.livescience.com/50027-tessellation-tiling.html | Skip to main content
Tessellation: The Geometry of Tiles, Honeycombs and M.C. Escher
References
By
Robert Coolman
published
Honeycombs, some bathroom floors and designs by artist M.C. Escher have something in common: they are composed of repeating patterns of the same shape without any overlaps or gaps. This type of pattern is called tiling, or tessellation.
The word "tessellate" means to form or arrange small squares in a checkered or mosaic pattern, according to Drexel University. It comes from the Greek tesseres, which means "four." The first tilings were made from square tiles. As an art form, tessellation is particularly rich in mathematics, with ties to geometry, topology and group theory. Cultures ranging from Irish and Arabic to Indian and Chinese have all practiced tiling at various levels of complexity. Let's explore the wide variety of tessellations we find in nature, functional design and art.
Regular tessellations
In mathematical terms, "regular" describes any shape that has all equal sides and equal angles. There are three regular shapes that make up regular tessellations: the equilateral triangle, the square and the regular hexagon. For example, a regular hexagon is used in the pattern of a honeycomb, the nesting structure of the honeybee.
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Semi-regular tessellations
Semi-regular tessellations are made of more than one kind of regular polygon. Within the limit of the same shapes surrounding each vertex (the points where the corners meet), there are eight such tessellations. Each semi-regular tessellation is named for the number of sides of the shapes surrounding each vertex. For example, for the first tiling below, each vertex is composed of the point of a triangle (3 sides), a hexagon (6), another triangle (3) and another hexagon (6), so it is called 3.6.3.6. Sometimes these tessellations are described as "Archimedean" in honor of the third-century B.C. Greek mathematician.
Semi-regular tessellations are made of combinations of different shapes. (Image credit: Robert Coolman)
Monohedral tessellations
"Mono" means "one" and "-hedral" means "shape"; so monohedral tessellations are made up of only one shape, though the shape may be rotated or flipped. In the language of mathematics, the shapes in such a pattern are described as congruent. Every triangle (three-sided shape) and every quadrilateral (four-sided shape) is capable of tessellation in at least one way, though a select few can tessellate in more than one way. A few examples are shown below:
Monohedral tessellations are made of one shape that is rotated or flipped to form different patterns. (Image credit: Robert Coolman)
According to mathematician Eric W. Weisstein of Wolfram Research's MathWorld, for pentagons, there are currently 14 known classes of shapes that will tessellate, and only three for hexagons. Whether there are more classes remains an unsolved problem of mathematics. As for shapes with seven or more sides, no such polygons tessellate unless they have an angle greater than 180 degrees. Such a polygon is described as concave because it has an indentation.
A few examples of pentagonal tessellations are shown below. The 14 classes of pentagonal tessellation can all be generated at the Wolfram Demonstration Project.
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A few examples of pentagonal tessellations. There are only 14 known patterns that can be made. (Image credit: Robert Coolman)
Duals
There's a deeper connection running through many of these geometric tessellations. A lot of them are "duals" of one another. According to Branko Grünbaum, author of "Tilings and Patterns" (Freeman, 1987), to create a tessellation's dual, draw a dot in the center of each shape, connect each dot to each of the neighboring shape's dots, and erase the original pattern. Below are some examples of tessellations and their duals:
A dual of a regular tessellation is formed by taking the center of each shape as a vertex and joining the centers of adjacent shapes. (Image credit: Robert Coolman)
M.C. Escher & modified monohedral tessellations
A unique art form is enabled by modifying monohedral tessellations. The most famous practitioner of this is 20th-century artist M.C. Escher. According to James Case, a book reviewer for the Society for Industrial and Applied Mathematics (SIAM), in 1937, Escher shared with his brother sketches from his fascination with 11th- and 12th-century Islamic artwork of the Iberian Peninsula. His brother directed him to a 1924 scientific paper by George Pólya that illustrated the 17 ways a pattern can be categorized by its various symmetries. This further inspired Escher, who began exploring deeply intricate interlocking tessellations of animals, people and plants.
According to Escher, "Crystallographers have … ascertained which and how many ways there are of dividing a plane in a regular manner. In doing so, they have opened the gate leading to an extensive domain, but they have not entered this domain themselves. By their very nature, they are more interested in the way the gate is opened than in the garden that lies behind it."
The following "gecko" tessellation, inspired by similar Escher designs, is based on a hexagonal grid. Notice how each gecko is touching six others.
A tessellation of geckos, inspired by the designs of M.C. Escher. (Image credit: Robert Coolman)
Aperiodic tessellations
Not all tessellations repeat. Such a pattern (if it can be called that) is described as "aperiodic." Below are three versions of Penrose Tiling, named after English mathematical physicist Rodger Penrose, who first published such patterns in 1974 at the University of Oxford. These patterns exhibit five-fold symmetry, a property that is not found in any periodic (repeating) pattern.
These tessellations do not have repeating patterns. They are called aperiodic. (Image credit: Robert Coolman)
Medieval Islamic architecture is particularly rich in aperiodic tessellation. The patterns were used in works of art and architecture at least 500 years before they were discovered in the West. An early example is Gunbad-i Qabud, an 1197 tomb tower in Maragha, Iran. According to ArchNet, an online architectural library, the exterior surfaces "are covered entirely with a brick pattern of interlacing pentagons."
The geometries within five-fold symmetrical aperiodic tessellations have become important to the field of crystallography, which since the 1980s has given rise to the study of quasicrystals. According to Peter J. Lu, a physicist at Harvard, metal quasicrystals have "unusually high thermal and electrical resistivities due to the aperiodicity" of their atomic arrangements.
Another set of interesting aperiodic tessellations is spirals. The first such pattern was discovered by Heinz Voderberg in 1936 and used a concave 11-sided polygon (shown on the left). Another spiral tiling was published 1985 by Michael D. Hirschhorn and D.C. Hunt using an irregular pentagon (shown on the right).
Examples of spiral tessellations. (Image credit: Robert Coolman)
Additional resources
See M.C. Escher’s tessellations at the M.C. Escher Gallery.
Watch this YouTube video to learn more about Penrose Tilings.
Learn more about Peter J. Lu's ideas about the geometry of medieval Islamic architecture.
Robert Coolman
Live Science Contributor
Robert Coolman, PhD, is a teacher and a freelance science writer and is based in Madison, Wisconsin. He has written for Vice, Discover, Nautilus, Live Science and The Daily Beast. Robert spent his doctorate turning sawdust into gasoline-range fuels and chemicals for materials, medicine, electronics and agriculture. He is made of chemicals.
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8378 | https://patents.google.com/patent/CN112707789A/en | CN112707789A - Process for preparing 1-chlorobutane - Google Patents
CN112707789A - Process for preparing 1-chlorobutane - Google Patents
Process for preparing 1-chlorobutane Download PDF
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Publication number CN112707789A CN112707789A CN202011575603.6A CN202011575603A CN112707789A CN 112707789 A CN112707789 A CN 112707789A CN 202011575603 A CN202011575603 A CN 202011575603A CN 112707789 A CN112707789 A CN 112707789A Authority CN China Prior art keywords reaction chlorobutane butanol reaction solvent preparation Prior art date 2020-12-28 Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)Granted Application number CN202011575603.6A Other languagesChinese (zh)Other versionsCN112707789B (enInventor 邱化齐 李璞 李剑 Current Assignee (The listed assignees may be inaccurate. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.) Shandong Jincheng Courage Chemical Co ltd Original Assignee Shandong Jincheng Courage Chemical Co ltd Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)2020-12-28 Filing date 2020-12-28 Publication date 2021-04-27 2020-12-28 Application filed by Shandong Jincheng Courage Chemical Co ltd filed Critical Shandong Jincheng Courage Chemical Co ltd 2020-12-28 Priority to CN202011575603.6A priority Critical patent/CN112707789B/en 2021-04-27 Publication of CN112707789A publication Critical patent/CN112707789A/en 2023-07-04 Application granted granted Critical 2023-07-04 Publication of CN112707789B publication Critical patent/CN112707789B/en Status Active legal-status Critical Current 2040-12-28 Anticipated expiration legal-status Critical
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LRHPLDYGYMQRHN-UHFFFAOYSA-N N-Butanol Chemical compound CCCCO LRHPLDYGYMQRHN-UHFFFAOYSA-N 0.000 claims abstract description 92
238000000034 method Methods 0.000 claims abstract description 30
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C07—ORGANIC CHEMISTRY
C07C—ACYCLIC OR CARBOCYCLIC COMPOUNDS
C07C17/00—Preparation of halogenated hydrocarbons
C07C17/093—Preparation of halogenated hydrocarbons by replacement by halogens
C07C17/16—Preparation of halogenated hydrocarbons by replacement by halogens of hydroxyl groups
B—PERFORMING OPERATIONS; TRANSPORTING
B01—PHYSICAL OR CHEMICAL PROCESSES OR APPARATUS IN GENERAL
B01J—CHEMICAL OR PHYSICAL PROCESSES, e.g. CATALYSIS OR COLLOID CHEMISTRY; THEIR RELEVANT APPARATUS
B01J31/00—Catalysts comprising hydrides, coordination complexes or organic compounds
B01J31/02—Catalysts comprising hydrides, coordination complexes or organic compounds containing organic compounds or metal hydrides
B01J31/0201—Oxygen-containing compounds
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Abstract
The invention belongs to the technical field of fine chemical engineering, and particularly relates to a preparation method of 1-chlorobutane. Mixing a reaction solvent, water and n-butyl alcohol, and adding a chlorination reagent to react to obtain 1-chlorobutane; wherein the reaction solvent is 2,3,5, 6-tetramethyl dioxane. The invention adopts a one-pot method for preparation, has simple method, good safety, easy implementation and low requirement on equipment, and can realize continuous production; the problem of large pollution of the traditional metal catalyst is avoided, three wastes are less, and the method is environment-friendly; 2,3,5, 6-tetramethyl dioxane is used as a reaction solvent and a catalyst, so that the generation of byproducts such as butylene, dibutyl ether, polymer resin and the like can be effectively avoided, the yield and the product quality are improved, the molar yield of the product is more than 96%, and the gas phase purity is more than 99.7%.
Description
Process for preparing 1-chlorobutane
Technical Field
The invention belongs to the technical field of fine chemical engineering, and particularly relates to a preparation method of 1-chlorobutane.
Background
The 1-chlorobutane is an important fine chemical product and can be used for grease, rubber, natural resin solvents, medical intermediates, cocatalysts and the like. At present, the method is mainly used for preparing an olefin polymerization catalyst, namely butyl lithium, preparing butyl tin products through a Grignard reaction, preparing phenylbutazone in medicine, preparing insecticide myclobutanil in pesticide, and using the butyl tin products as a solvent in polyether production. The current market demands more than 99% of products, wherein the purity requirement of 1-chlorobutane for synthesizing n-butyllithium is more than or equal to 99.5%.
In the prior art, n-butanol and hydrochloric acid or hydrogen chloride gas are mainly used as raw materials to synthesize 1-chlorobutane, and the main synthesis methods comprise the following steps:
chinese patent CN 101475440A discloses a method for continuously producing n-butyl chloride by reacting n-butanol with hydrogen chloride gas in the absence of a catalyst. However, the patent does not give specific yield information, and the equipment is complex and the investment cost of production equipment is high.
The method for producing n-butyl chloride disclosed in the Chinese patent CN 1069018A uses concentrated hydrochloric acid to replace hydrogen chloride gas, and after the reaction is finished, the purity of the product can reach 99% through chromatographic analysis, and the yield is 90%.
New technology for synthesizing 1-chlorobutane, Qiu billow, etc., fine chemical engineering, 2004, 21(8):637-638 and preparation of n-butyl chloride, Chinese chlor-alkali, such as Bighui, 2003, (10):43 discloses the use of ZnC1 2 FeC1 as the main catalyst 3 As the cocatalyst, HC1 gas and concentrated hydrochloric acid are respectively used for catalytic chlorination of n-butanol, and compared with the catalyst-free catalyst under the same condition, the yield can be improved by about 6%. However, ZnC1 in the catalytic process 2 And FeC1 3 Is difficult to recover, the content of heavy metal ions in the wastewater is high, and the environmental pollution is very serious.
Chinese patent CN 1440958A discloses a method for converting primary alcohol or cyclohexanol in acidic ionic liquid [ Hmim ] + X- (X ═ Cl, Br, I) into halogenated hydrocarbon, wherein N-methylimidazolium hydrochloride ionic liquid (both used as solvent and chlorinating agent) is reacted with N-butanol to obtain 1-chlorobutane, and after the reaction is finished, N-methylimidazolium hydrochloride ionic liquid is regenerated by concentrated hydrochloric acid and is continuously used in the next batch. Although the yield of the method is high (about 95%), the method has the advantages that the ionic liquid and the 1-chlorobutane are difficult to delaminate after the reaction is finished, and the delamination takes longer time, so that the production period can be prolonged in industrial production, the production efficiency is reduced, and the industrial production is not facilitated; meanwhile, when the concentrated hydrochloric acid is used for recovering the ionic liquid, further distillation is needed for removing water, the water is difficult to completely remove, the reaction is probably greatly influenced, and whether the recovered ionic liquid can maintain the yield of about 95% in the patent is not reported.
Chinese patent CN 104326863 a discloses a method for preparing 1-chlorobutane, which comprises the following steps: mixing a catalyst, a chlorinating agent and water, stirring, then mixing with n-butyl alcohol, and carrying out chlorination reaction to obtain 1-chlorobutane; the catalyst is dimethyl sulfoxide, the chlorination reagent is hydrogen chloride, and the molar ratio of the chlorination reagent to n-butyl alcohol is 2.5: 1-5: 1; the stirring temperature is 15-45 ℃. The method has the advantages of long reaction time, reaction time of more than 20 hours, high energy consumption, no contribution to industrialization and low yield of only about 90 percent.
In addition, the above patents inevitably produce by-products, such as butene, dibutyl ether and polymer resin, which are difficult to separate and are not suitable for continuous production. Therefore, it is urgently needed to provide a preparation method of 1-chlorobutane, which has the advantages of simple method, good safety, easy implementation, continuous production, environmental protection, no pollution, and high product yield and purity.
Disclosure of Invention
The invention aims to provide a preparation method of 1-chlorobutane, which solves the problems of complex process, high equipment requirement, high pollution, more byproducts and the like of the existing synthesis process, has the advantages of simple and easy operation, continuous production, environmental protection, no pollution, high product yield and high product purity, and is more suitable for industrial production.
The reaction mode of the invention is batch reaction or continuous reaction.
When the reaction mode is an intermittent reaction, the preparation method of the 1-chlorobutane comprises the steps of mixing a reaction solvent, water and n-butyl alcohol, and adding a chlorination reagent for reaction to obtain the 1-chlorobutane; wherein the reaction solvent is 2,3,5, 6-tetramethyl dioxane.
The volume ratio of the reaction solvent, water and n-butanol is 5-15: 1: 1-15, preferably 8-10: 1: 5-10.
The chlorination reagent is hydrogen chloride gas.
The mol ratio of the n-butanol to the chlorinating agent is 1: 2-5, preferably 1: 2.5-3.
The reaction temperature is 80-115 ℃, preferably 95-105 ℃; the reaction time is 1-3 hours.
When the reaction mode is a continuous reaction, the preparation method of the 1-chlorobutane comprises the following steps:
(1) mixing a reaction solvent, water and n-butanol, adding a chlorination reagent for reaction to obtain a reaction solution, and distilling the reaction solution to obtain 1-chlorobutane; wherein the reaction solvent is 2,3,5, 6-tetramethyl dioxane;
(2) and continuously introducing n-butanol into the reaction solution to react with a chlorination reagent, and continuously preparing the 1-chlorobutane after distillation.
The volume ratio of the reaction solvent, water and n-butanol in the step (1) is 5-15: 1: 1-15, preferably 8-10: 1: 5-10.
The chlorination reagent in the step (1) is hydrogen chloride gas.
The mol ratio of the n-butanol to the chlorinating agent in the step (1) is 1: 2-5, preferably 1: 2.5-3.
The reaction temperature in step (1) is 80 to 115 ℃, preferably 95 to 105 ℃.
The speed of introducing the n-butanol in the step (2) is 1-10ml/min, and the speed of introducing the chlorinating agent is 1-10 g/min.
And (3) the chlorination reagent in the step (2) is hydrogen chloride gas.
The reaction temperature in the step (2) is 80 to 115 ℃, preferably 95 to 105 ℃.
The chemical reaction equation of the invention is as follows:
the n-butyl alcohol is easy to generate intramolecular dehydration to form ether or eliminate reaction to form alkene under the high-temperature acidic condition, and the 2,3,5, 6-tetramethyl dioxane is used as a reaction solvent to obtain high-purity 1-chlorobutane; the 2,3,5, 6-tetramethyl dioxane has high boiling point, can provide higher reaction temperature, has good solubility to hydrogen chloride, improves the reaction rate, avoids the generation of byproducts and improves the yield.
The invention has the following beneficial effects:
(1) the invention adopts a one-pot method for preparation, has simple method, good safety, easy implementation and low requirement on equipment, and can realize continuous production;
(2) the invention avoids the problem of large pollution of the traditional metal catalyst, has less three wastes and is environment-friendly;
(3) the invention adopts 2,3,5, 6-tetramethyl dioxane as reaction solvent and catalyst, can effectively avoid the generation of by-products such as butylene, dibutyl ether and polymer resin, and the like, improves the yield and the product quality, and has the product molar yield of more than 96 percent and the gas phase purity of more than 99.7 percent;
(4) the method has mild reaction conditions, the reaction solvent is also a catalyst, the solvent can be simply distilled, recycled and reused, and the process is simple; compared with the prior art, the method has the advantages of low cost, high product purity and high yield.
Drawings
FIG. 1 is a chromatogram of 1-chlorobutane obtained in example 1.
FIG. 2 is a chromatogram of n-butanol and dibutyl ether.
FIG. 3 is a chromatogram of 1-chlorobutane obtained in comparative example 1.
Detailed Description
The present invention is further described below with reference to examples.
Example 1
200ml of 2,3,5, 6-tetramethyl dioxane, 20ml of purified water and 200ml (2.19mol) of n-butyl alcohol are added into a three-neck flask with the capacity of 1000ml and provided with a stirrer and a thermometer, the temperature is raised to 105 ℃ while stirring, 199.8g (5.48mol) of hydrogen chloride gas is introduced, distillation is carried out while reaction is carried out, a fraction is gradually and slowly distilled off along with the extension of the reaction time, almost no fraction is further distilled after the heat preservation distillation and the reaction are carried out for 2h, the reaction is finished, the obtained fraction is layered, the upper organic layer is 1-chlorobutane, 195.4g (2.11mol) of 1-chlorobutane is finally obtained, the molar yield is 96.3%, the GC purity is 99.83%, and the chromatographic results are shown in a graph 1 and a table 1. Calibrating a 1-chlorobutane standard product by GC to obtain a product, namely 1-chlorobutane, wherein the retention time is 3.968'; the n-butyl and dibutyl ethers were also calibrated and the retention times were 4.249 'and 9.690', respectively, and the chromatographic results are shown in FIG. 2 and Table 2.
Table 1 chromatographic peak table for example 1
Peak number Retention time Area of Height Area%
1 2.965 17292 10066 0.078
2 3.418 1507 896 0.007
3 3.635 2637 1206 0.012
4 3.968 22193351 8039738 99.831
5 4.990 1861 769 0.008
6 5.135 12646 4828 0.057
7 9.516 1723 496 0.008
Total of 22231016 8058000 100.000
TABLE 2 chromatogram peaks for n-butyl and dibutyl ethers
Peak number Retention time Area of Height Area%
1 3.471 2620 1142 0.012
2 3.771 4798 1605 0.021
3 3.847 2633 731 0.012
4 3.986 56829 24898 0.253
5 4.249 17747223 2504253 78.925
6 8.198 2382 547 0.011
7 9.690 4661566 953773 20.731
8 10.755 3647 889 0.016
9 11.598 1024 270 0.005
10 13.042 3520 1017 0.016
Total of 22486241 3489126 100.000
Example 2
300ml of 2,3,5, 6-tetramethyl dioxane, 20ml of purified water and 200ml (2.19mol) of n-butyl alcohol are added into a three-neck flask with the capacity of 1000ml and provided with a stirrer and a thermometer, the temperature is raised to 115 ℃ by stirring, 196.4g (5.38mol) of hydrogen chloride gas is introduced, the distillation is carried out while the reaction is carried out, the fraction is gradually and slowly distilled off along with the extension of the reaction time, almost no fraction is further distilled after the heat preservation distillation and the reaction are carried out for 2h, the reaction is finished, the obtained fraction is layered, the upper organic layer is 1-chlorobutane, 195.9g (2.12mol) of 1-chlorobutane is finally obtained, the molar yield is 96.6%, and the GC purity is 99.79%.
Example 3
100ml of 2,3,5, 6-tetramethyl dioxane, 20ml of purified water and 200ml (2.19mol) of n-butyl alcohol are added into a three-neck flask with the capacity of 1000ml and provided with a stirrer and a thermometer, the temperature is raised to 85 ℃ while stirring, 239.8g (6.57mol) of hydrogen chloride gas is introduced, distillation is carried out while reaction is carried out, a fraction is gradually and slowly distilled off along with the extension of the reaction time, almost no fraction is further distilled after the heat preservation distillation and the reaction are carried out for 3 hours, the reaction is finished, an upper organic layer is 1-chlorobutane after the obtained fraction is layered, 194.6g (2.11mol) of 1-chlorobutane is finally obtained, the molar yield is 96.0%, and the GC purity is 99.87%.
Example 4
Adding 200ml of 2,3,5, 6-tetramethyldioxane, 20ml of purified water and 200ml (2.19mol) of n-butyl alcohol into a three-neck flask with the capacity of 1000ml and a stirrer and a thermometer, stirring and heating to 105 ℃, introducing 199.8g (5.48mol) of hydrogen chloride gas, distilling while reacting, gradually distilling a fraction slowly with the extension of reaction time, adding n-butyl alcohol into a reaction bottle at the speed of 5ml/min by using a peristaltic pump after 100ml of the fraction is distilled out, adding 800ml (8.74mol) of n-butyl alcohol, keeping the introduction speed of 5g/min of hydrogen chloride gas in the adding process, after the heat preservation distillation and the reaction are finished for 2 hours, almost not distilling any fraction, finishing the reaction, layering the obtained fraction, obtaining an upper organic layer which is 1-chlorobutane, finally obtaining 979.3g (10.58mol) of 1-chlorobutane, wherein the molar yield is 96.8%, the GC purity was 99.87%.
Example 5
Adding 300ml of 2,3,5, 6-tetramethyl dioxane, 20ml of purified water and 200ml (2.19mol) of n-butyl alcohol into a 1000ml three-neck flask with a stirrer and a thermometer, stirring and heating to 115 ℃, introducing 196.4g (5.38mol) of hydrogen chloride gas, carrying out reaction and distillation, gradually distilling a fraction slowly with the extension of reaction time, adding n-butyl alcohol into a reaction bottle at the speed of 3ml/min by using a peristaltic pump after 100ml of the fraction is distilled out, adding 800ml (8.74mol) of n-butyl alcohol, keeping the introduction speed of 3g/min of hydrogen chloride gas in the adding process, after carrying out heat preservation distillation and reaction for 2h, almost not distilling any fraction, finishing the reaction, layering the obtained fraction, obtaining an upper organic layer which is 1-chlorobutane, finally obtaining 980.6g (10.59mol) of 1-chlorobutane and having the molar yield of 96.9%, the GC purity was 99.89%.
Example 6
Adding 100ml of 2,3,5, 6-tetramethyldioxane, 20ml of purified water and 200ml (2.19mol) of n-butyl alcohol into a 1000ml three-neck flask with a stirrer and a thermometer, stirring and heating to 85 ℃, introducing 239.8g (6.57mol) of hydrogen chloride gas, distilling while reacting, gradually slowly distilling a fraction with the prolonging of reaction time, adding the n-butyl alcohol into the reaction flask at the speed of 8ml/min by using a peristaltic pump after distilling the fraction out of 100ml, adding 800ml (8.74mol) of n-butyl alcohol, keeping the introduction speed of the hydrogen chloride gas at 8g/min in the adding process, after the heat preservation distillation and the reaction are finished for 3 hours, almost not distilling any fraction, layering the obtained fraction, obtaining an upper organic layer which is 1-chlorobutane, finally obtaining 978.5g (10.57mol) of 1-chlorobutane with the molar yield of 96.7%, the GC purity was 99.84%.
Comparative example 1
860mL of concentrated hydrochloric acid was charged into a 3L reaction flask, and 1363g (10mol) of anhydrous zinc chloride was added in portions while cooling, and stirred so that the zinc chloride was dissolved as much as possible. 371g (5mol) of n-butanol was added, the oil bath was heated to 150 ℃ and the solution started to boil, HCl was bubbled in and 1-chlorobutane was distilled off, and the 75.5-77.5 ℃ fraction was collected to give 355g of 1-chlorobutane, 76% molar yield, 97.53% GC purity, 1.208% dibutyl ether, and the chromatographic results are shown in FIG. 3 and Table 3.
TABLE 3 chromatographic Peak Table for comparative example 1
Peak number Retention time Area of Height Area%
1 2.968 1661 941 0.009
2 3.417 4041 2275 0.021
3 3.635 1760 771 0.009
4 3.733 1263 436 0.007
5 3.963 18352746 7023137 97.526
6 4.036 157488 86681 0.837
7 4.989 9398 4050 0.050
8 9.508 227374 76598 1.208
9 10.647 4337 1382 0.023
10 11.799 39407 4524 0.209
11 12.950 14894 4477 0.079
12 13.905 4024 1203 0.021
Total of 18818394 7206476 100.000
Claims (10)
A preparation method of 1-chlorobutane is characterized in that a reaction solvent, water and n-butanol are mixed, and a chlorination reagent is added for reaction to obtain 1-chlorobutane; wherein the reaction solvent is 2,3,5, 6-tetramethyl dioxane.
The process for the preparation of 1-chlorobutane according to claim 1, wherein the volume ratio of the reaction solvent, water and n-butanol is 5-15: 1: 1-15.
The method of claim 1, wherein the chlorinating agent is hydrogen chloride gas.
The process for the preparation of 1-chlorobutane according to claim 1, wherein the molar ratio of n-butanol to chlorinating agent is 1: 2-5.
The method according to claim 1, wherein the reaction temperature is 80-115 ℃ and the reaction time is 1-3 hours.
The preparation method of the 1-chlorobutane is characterized by comprising the following steps:
(1) mixing a reaction solvent, water and n-butanol, adding a chlorination reagent for reaction to obtain a reaction solution, and distilling the reaction solution to obtain 1-chlorobutane; wherein the reaction solvent is 2,3,5, 6-tetramethyl dioxane;
(2) and continuously introducing n-butanol into the reaction solution to react with a chlorination reagent, and continuously preparing the 1-chlorobutane after distillation.
The process for producing 1-chlorobutane according to claim 6, wherein the volume ratio of the reaction solvent, water and n-butanol in the step (1) is 5 to 15: 1: 1-15, wherein the molar ratio of the n-butanol to the chlorinating agent is 1: 2-5.
The method according to claim 6, wherein the chlorinating agent used in step (1) is hydrogen chloride gas, and the reaction temperature is 80-115 ℃.
The method according to claim 6, wherein the n-butanol is introduced at a rate of 1 to 10ml/min and the chlorinating agent is introduced at a rate of 1 to 10g/min in the step (2).
The method according to claim 6, wherein the chlorinating agent in step (2) is hydrogen chloride gas, and the reaction temperature is 80-115 ℃.
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Ear, Nose, and Throat Emergencies
Edited by
Laura J. Bontempo MD, MEd - Department of Emergency Medicine, University of Maryland School of Medicine, Baltimore, MD, USA
Jan Shoenberger MD - Clinical Emergency Medicine, Keck School of Medicine, University of Southern California, Los Angeles, CA, USA
Pages 1-152 (February 2019)
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Angles on a straight line
| | | |
---
| Previous Introduction | | Next Vertically opposite angles |
12.2 Angles on a straight line
Let us look at angles formed on one side of a straight line.
In this diagram, line segment AB meets line segment DC. The angle at the vertex, C, where they
meet, is now split into two angles: ^C1 and ^C2.
^C1 is the name for the angle at vertex C labelled "1" (or AˆCD).
The sum of the angles formed on a straight line
The sum of the angles that are formed on a straight line is always 180∘.
We can shorten this property as: ∠s on a straight line.
Two angles that add up to 180∘ are called supplementary angles.
Angles that share a vertex and a common side are said to be adjacent angles. ^C1
and ^C2 are supplementary angles.
Hence, ^C1 and ^C2 are called adjacent supplementary angles.
supplementary angles
: two angles that add up to 180∘
adjacent angles
: angles that share a vertex and a common side
You can have more than one line meeting at the same point on a straight line. Here are a few examples of angles
on a straight line.
The sum of the angles on perpendicular lines
When two lines are perpendicular, the adjacent supplementary angles are both equal to 90∘.
In the diagram, ^M1=^M2=90∘.
A right angle is shown by forming a square at one of the right angles, like this: ⦜.
Finding unknown angles on straight lines
Worked example 12.1: Calculating unknown angles on a straight line
Calculate the size of x.
In the diagram, we have two angles that are on the same side of the straight line. The first angle is
100∘ and the second angle is unknown (x). We need to calculate the size of x.
We know that the two angles have a sum of 180∘, so we can say that:
100∘+x=180∘ (∠s on a straight line)
Now we can solve this equation.
x=180∘−100∘x=80∘
Worked example 12.2: Calculating unknown angles on a straight line
Calculate the size of x.
Notice that there are three angles on the same side of the straight line. We have x, an angle of 29∘ and an angle of 90∘. (Remember that the ⦜ symbol on the diagram indicates a 90∘ angle.) These three angles have a sum of 180∘, so we can say that:
x+29∘+90∘=180∘ (∠s on a straight line)
Now we can solve this equation.
x+29∘+90∘=180∘x+119∘=180∘x=180∘−119∘x=61∘
There is a simpler way to solve for x. It is given that we have a perpendicular line. Adjacent angles on a perpendicular line are both equal to 90∘. So, we have a different equation to solve.
x+29∘=90∘x=90∘−29∘x=61∘
Worked example 12.3: Calculating unknown angles on a straight line
Calculate the size of y.
Notice that we have three angles on the same side of the straight line. We have 2y, an angle of 48∘ and an angle of 52∘. These three angles have a sum of 180∘, so we can say that:
2y+48∘+52∘=180∘ (∠s on a straight line)
Now we can solve this equation.
2y+48∘+52∘=180∘2y+100∘=180∘2y=180∘−100∘2y=80∘y=40∘
Exercise 12.1
Calculate the size of (a).
[\begin{align}
a+63^{\circ}&=180^{\circ} &(\angle\text{s on a straight line}) \
x&=180^{\circ}-63^{\circ} \
x&=117^{\circ}
\end{align}]
Calculate the size of:
(x)
(\hat{ECB})
[\begin{align}
x+3x+2x&=180^{\circ} &(\angle\text{s on a straight line}) \
6x&=180^{\circ} \
x&=30^{\circ}
\end{align}]
[\begin{align}
\hat{ECB}&=2x \
&=2(30^{\circ})\
&=60^{\circ}
\end{align}]
Calculate the size of:
(x)
(\hat{GEH})
[\begin{align}
(x+30^{\circ})+(x+40^{\circ}) +(2x+10^{\circ}) &=180^{\circ} &(\angle\text{s on a straight line})
\
4x+80^{\circ}&=180^{\circ} \
4x&=100^{\circ} \
x&=25^{\circ}
\end{align}]
[\begin{align}
\hat{GEH}&=x+40^{\circ} \
&=25^{\circ}+40^{\circ} \
&=65^{\circ}
\end{align}]
Hint: Remember that the matching curved lines “))” indicate that the angles are
equal.
Calculate the size of:
(k)
(\hat{TYP})
[\begin{align}
(2k)+(k+65^{\circ}) +(2k)
&=180^{\circ}
&&(\angle\text{s on a straight line}) \
5k+65^{\circ}&=180^{\circ} \
5k &=115^{\circ} \
&=23^{\circ}
\end{align}]
[\begin{align}
\hat{TYP}&=k+65^{\circ} \
&=23^{\circ}+65^{\circ} \
&=88^{\circ}
\end{align}]
| | | |
---
| Previous Introduction | Table of Contents | Next Vertically opposite angles | |
8381 | https://tutorial.math.lamar.edu/pdf/common_derivatives_integrals.pdf | Common Derivatives and Integrals Derivatives Basic Properties/Formulas/Rules d dx cf(x) = cf′(x), c is any constant.
d dx f(x) ± g(x) = f′(x) ± g′(x) d dx xn = nxn−1, n is any number.
d dx c = 0, c is any constant.
f(x) g(x) ′ = f′(x) g(x) + f(x) g′(x) – Product Rule d dx eg(x) = g′(x)eg(x) f(x) g(x) ′ = f′(x) g(x) −f(x) g′(x) g(x) 2 – Quotient Rule d dx ln(g(x)) = g′(x) g(x) d dx f g(x) = f′ g(x) g′(x) – Chain Rule Common Derivatives Polynomials d dx c = 0 d dx(x) = 1 d dx cx = c d dx xn = nxn−1 d dx cxn = ncxn−1 Trig Functions d dx h sin(x) i = cos(x) d dx h cos(x) i = −sin(x) d dx h tan(x) i = sec2(x) d dx h csc(x) i = −csc(x) cot(x) d dx h sec(x) i = sec(x) tan(x) d dx h cot(x) i = −csc2(x) Inverse Trig Functions d dx h sin−1(x) i = 1 √ 1 −x2 d dx h cos−1(x) i = − 1 √ 1 −x2 d dx h tan−1(x) i = 1 1 + x2 d dx h csc−1(x) i = − 1 |x| √ x2 −1 d dx h sec−1(x) i = 1 |x| √ x2 −1 d dx h cot−1(x) i = − 1 1 + x2 Exponential & Logarithm Functions d dx h axi = ax ln(a) d dx h exi = ex d dx h ln(x) i = 1 x, x > 0 d dx h ln |x| i = 1 x, x ̸= 0 d dx h loga(x) i = 1 x ln(a), x > 0 Hyperbolic Functions d dx h sinh(x) i = cosh(x) d dx h cosh(x) i = sinh(x) d dx h tanh(x) i = sech2(x) d dx h csch(x) i = −csch(x) coth(x) d dx h sech(x) i = −sech(x) tanh(x) d dx h coth(x) i = −csch2(x) © Paul Dawkins - Common Derivatives and Integrals Integrals Basic Properties/Formulas/Rules Z cf(x) dx = c Z f(x) dx, c is a constant.
Z f(x) ± g(x) dx = Z f(x)dx ± Z g(x) dx Z b a f(x) dx = f(x) b a = F(b) −F(a) where F(x) = Z f(x) dx Z b a cf(x) dx = c Z b a f(x) dx, c is a constant.
Z b a f(x) ± g(x) dx = Z b a f(x) dx ± Z b a g(x) dx Z a a f(x) dx = 0 Z b a f(x) dx = − Z a b f(x) dx Z b a f(x) dx = Z c a f(x) dx + Z b c f(x) dx Z b a c dx = c(b −a), c is a constant.
If f(x) ≥0 on a ≤x ≤b then Z b a f(x) dx ≥0 If f(x) ≥g(x) on a ≤x ≤b then Z b a f(x) dx ≥ Z b a g(x) dx Common Integrals Polynomials Z dx = x + c Z k dx = kx + c Z xndx = 1 n + 1xn+1 + c, n ̸= −1 Z 1 x dx = ln |x| + c Z x−1 dx = ln |x| + c Z x−ndx = 1 −n + 1x−n+1 + c, n ̸= 1 Z 1 ax + b dx = 1 a ln |ax + b| + c Z x p q dx = 1 p q + 1x p q +1 + c = q p + qx p+q q + c Trig Functions Z cos(u) du = sin(u) + c Z sin(u) du = −cos(u) + c Z sec2 u du = tan(u) + c Z sec(u) tan(u) du = sec(u) + c Z csc(u) cot(u) du = −csc(u) + c Z csc2 u du = −cot(u) + c Z tan(u) du = −ln cos(u) +c = ln sec(u) +c Z cot(u) du = ln sin(u) +c = −ln csc(u) +c Z sec(u) du=ln sec(u)+tan(u) +c Z sec3(u) du = 1 2 sec(u) tan(u)+ln sec(u)+tan(u) +c Z csc(u) du=ln csc(u)−cot(u) +c Z csc3(u) du = 1 2 −csc(u) cot(u)+ln csc(u)−cot(u) +c © Paul Dawkins - Common Derivatives and Integrals Exponential & Logarithm Functions Z eu du = eu + c Z au du = au ln(a) + c Z ln(u) du = u ln(u) −u + c Z eau sin(bu) du = eau a2 + b2 a sin(bu) −b cos(bu) + c Z ueudu = (u −1)eu + c Z eau cos(bu) du = eau a2 + b2 a cos(bu) + b sin(bu) + c Z 1 u ln(u) du = ln ln(u) + c Inverse Trig Functions Z 1 √ a2 −u2 du = sin−1 u a + c Z sin−1(u) du = u sin−1(u) + p 1 −u2 + c Z 1 a2 + u2 du = 1 a tan−1 u a + c Z tan−1(u) du = u tan−1(u) −1 2 ln 1 + u2 + c Z 1 u √ u2 −a2 du = 1 a sec−1 u a + c Z cos−1(u) du = u cos−1(u) − p 1 −u2 + c Hyperbolic Functions Z sinh(u)du = cosh(u) + c Z sech(u) tanh(u)du = −sech(u) + c Z sech2(u)du = tanh(u) + c Z cosh(u)du = sinh(u) + c Z csch(u) coth(u)du = −csch(u) + c Z csch2(u)du = −coth(u) + c Z tanh(u)du = ln cosh(u) + c Z sech(u)du = tan−1 sinh(u) + c Miscellaneous Z 1 a2 −u2 du = 1 2a ln u + a u −a + c Z p a2 + u2 du = u 2 p a2 + u2 + a2 2 ln u + p a2 + u2 + c Z 1 u2 −a2 du = 1 2a ln u −a u + a + c Z p u2 −a2 du = u 2 p u2 −a2 −a2 2 ln u + p u2 −a2 + c Z p a2 −u2 du = u 2 p a2 −u2 + a2 2 sin−1 u a + c Z p 2au −u2 du = u −a 2 p 2au −u2 + a2 2 cos−1 a −u a + c Standard Integration Techniques u Substitution : Z b a f g(x) g′(x) dx will convert the integral into Z b a f g(x) g′(x) dx = Z g(b) g(a) f(u) du using the substitution u = g(x) where du = g′(x)dx. For indefinite integrals drop the limits of integration.
© Paul Dawkins - Common Derivatives and Integrals Integration by Parts : Z u dv = uv − Z v du and Z b a u dv = uv b a − Z b a v du. Choose u and dv from integral and compute du by differentiating u and compute v using v = Z dv.
Trig Substitutions : If the integral contains the following root use the given substitution and formula.
p a2 −b2x2 ⇒ x = a b sin(θ) and cos2(θ) = 1 −sin2(θ) p b2x2 −a2 ⇒ x = a b sec(θ) and tan2(θ) = sec2(θ) −1 p a2 + b2x2 ⇒ x = a b tan(θ) and sec2(θ) = 1 + tan2(θ) Partial Fractions : If integrating a rational expression involving polynomials, Z P(x) Q(x) dx, where the degree (largest exponent) of P(x) is smaller than the degree of Q(x) then factor the denominator as completely as possible and find the partial fraction decomposition of the rational expression.
Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the decomposition according to the following table.
Factor of Q(x) Term in P.F.D Factor is Q(x) Term in P.F.D ax + b A ax + b (ax + b)k A1 ax + b + A2 (ax + b)2 + · · · + Ak (ax + b)k ax2 + bx + c Ax + B ax2 + bx + c (ax2 + bx + c)k A1x + B1 ax2 + bx + c + · · · + Akx + Bk (ax2 + bx + c)k Products and (some) Quotients of Trig Functions : For Z sinn(x) cosm(x) dx we have the following : 1. n odd. Strip 1 sine out and convert rest to cosines using sin2(x) = 1 −cos2(x), then use the substitution u = cos(x).
2. m odd. Strip 1 cosine out and convert rest to sines using cos2(x) = 1 −sin2(x), then use the substitution u = sin(x).
3. n and m both odd. Use either 1. or 2.
4. n and m both even. Use double angle and/or half angle formulas to reduce the integral into a form that can be integrated.
For Z tann(x) secm(x) dx we have the following : 1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using tan2(x) = sec2(x) −1, then use the substitution u = sec(x).
2. m even. Strip 2 secants out and convert rest to tangents using sec2(x) = 1 + tan2(x), then use the substitution u = tan(x).
3. n odd and m even. Use either 1. or 2.
4. n even and m odd. Each integral will be dealt with differently.
Convert Example : cos6(x) = cos2(x) 3 = 1 −sin2(x) 3 © Paul Dawkins - |
8382 | https://www.sciencedirect.com/science/article/pii/S2161831322007451 | Evaluating the Intervention-Based Evidence Surrounding the Causal Role of Breakfast on Markers of Weight Management, with Specific Focus on Breakfast Composition and Size - ScienceDirect
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ABSTRACT
Introduction
Methodology for the Comprehensive Review
Intervention-Based Clinical Trials Including Weight Outcomes
Intervention-Based Trials Including Daily Food Intake
Summary and Conclusions
Acknowledgments
Supplementary Material
References
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Advances in Nutrition
Volume 7, Issue 3, May 2016, Pages 563S-575S
Evaluating the Intervention-Based Evidence Surrounding the Causal Role of Breakfast on Markers of Weight Management, with Specific Focus on Breakfast Composition and Size
Author links open overlay panelHeather J Leidy 5, Jess A Gwin 5, Connor A Roenfeldt 5, Adam Z Zino 5, Rebecca S Shafer 5
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ABSTRACT
Nutritional strategies are vitally needed to aid in the management of obesity. Cross-sectional and epidemiologic studies consistently demonstrate that breakfast consumption is strongly associated with a healthy body weight. However, the intervention-based long-term evidence supporting a causal role of breakfast consumption is quite limited and appears to be influenced by several key dietary factors, such as dietary protein, fiber, and energy content. This article provides a comprehensive review of the intervention-based literature that examines the effects of breakfast consumption on markers of weight management and daily food intake. In addition, specific focus on the composition and size (i.e., energy content) of the breakfast meal is included. Overall, there is limited evidence supporting (or refuting) the daily consumption of breakfast for body weight management and daily food intake. In terms of whether the type of breakfast influences these outcomes, there is accumulating evidence supporting the consumption of increased dietary protein and fiber content at breakfast, as well as the consumption of more energy during the morning hours. However, the majority of the studies that manipulated breakfast composition and content did not control for habitual breakfast behaviors, nor did these studies include a breakfast-skipping control arm. Thus, it is unclear whether the addition of these types of breakfast plays a causal role in weight management. Future research, including large randomized controlled trials of longer-term (i.e., ≥6 mo) duration with a focus on key dietary factors, is critical to begin to assess whether breakfast recommendations are appropriate for the prevention and/or treatment of obesity.
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breakfast
weight loss
breakfast skipping
obesity
increased dietary protein
meal omission
Introduction
Obesity is the greatest threat to global health this century, affecting the lives of >1.4 billion adults (1). In the United States, ∼167 million adults (69%) are overweight/obese (2). The health consequences related to obesity can begin as early as childhood (3), and many overweight children and adolescents already have at least one, if not multiple, cardiovascular disease risk factors (4). Thus, there is a great need to develop successful nutrition strategies that target weight management to reverse the obesity epidemic and prevent or delay serious health complications. One factor that has received substantial attention is the dietary habit of skipping breakfast. Although breakfast was once thought to be the most important meal of the day, several investigators have challenged this notion because of the limited amount of existing scientific evidence (5, 6).
Breakfast consumption previously was considered a staple of the American diet (7, 8, 9); however, there has been an increased prevalence of skipping breakfast over the past 50 y, with as many as 30% of young people skipping breakfast every day and up to 60% eating breakfast infrequently (7, 10, 11). The increased frequency of skipping breakfast has occurred concomitant with the increased rise in obesity (8), raising the question of whether breakfast plays a causal role in weight management. The myriad epidemiologic, observational, and cross-sectional data document strong associations between breakfast consumption and markers of healthy weight management such that increased frequency of breakfast consumption is associated with lower BMI, lower weight and weight gain, and lower body fat (5, 12, 13, 14, 15, 16, 17). In a recent meta-analysis (5), breakfast consumers were shown to display a greater OR of being classified as healthy weight than were those who skipped breakfast (P = 10−42). Despite the consistent evidence supporting the relation between breakfast consumption and weight management, there is a paucity of intervention-based evidence assessing whether the addition of breakfast improves weight-management outcomes (5). It is also unclear whether the composition and size of breakfast affect weight outcomes. Because current research demonstrates improvements in weight management with increased dietary protein, whole-grain/fiber consumption, and timing of food intake (18, 19, 20), it is plausible that these factors also might influence any proposed breakfast-related effects.
This article provides a comprehensive review of the intervention-based literature that examines the effects of breakfast consumption on weight management and daily food intake, with specific focus on breakfast composition and size.
Methodology for the Comprehensive Review
Breakfast is the first eating occasion of the day, before 1000; this was described previously by Timlin and Pereira (21). The search terms included breakfast (skipping) and morning fast, along with the following outcomes: weight (loss), fat mass (loss), percent body fat, BMI, waist circumference, and daily energy (food) intake.
Searches of electronic databases were carried out between 26 June 2014 and 28 July 2014, and included PubMed and Scopus. In addition, references from existing reviews and select articles were examined to supplement the electronic search.
This review was limited to articles published in English in peer-reviewed journals, and included the following criteria: 1) all age groups; 2) all diseases/conditions; 3) any breakfast intervention of ≥8 wk (for weight and body composition outcomes) and any duration for daily food intake; and 4) studies that included outcomes of weight, fat mass, body fat, lean mass/fat-free mass, BMI, waist circumference, or daily food (energy) intake. Epidemiologic, observational, and cross-sectional studies were excluded. The search flow diagrams for the weight outcomes and daily food intake are presented in Figures 1and2, respectively. All data is presented as mean ± SEM.
1. Download: Download high-res image (148KB)
2. Download: Download full-size image
FIGURE 1. Flow diagram of the selection process for the articles containing breakfast interventions and weight outcomes. Inc/ex, inclusion/exclusion.
1. Download: Download high-res image (153KB)
2. Download: Download full-size image
FIGURE 2. Flow diagram of the selection process for the articles containing breakfast interventions and daily energy intake. Inc/ex, inclusion/exclusion.
Each article was graded based on the Academy of Nutrition and Dietetics' article evaluation found within The Evidence Analysis Manual (22).
Intervention-Based Clinical Trials Including Weight Outcomes
Addition of breakfast
Only 3 long-term studies [albeit all randomized controlled trials (RCTs)] were identified that assessed whether the addition of breakfast influenced weight outcomes (Table 1). Schlundt et al. (23) examined the effects of consuming breakfast compared with skipping breakfast during a 12-wk energy-restricted diet in 52 obese adult women [age: 18–55 y; BMI (in kg/m 2): 30–31). The women were initially stratified according to habitual breakfast behaviors, with skipping breakfast defined as having skipped breakfast on ≥3 occasions/wk. The participants then began the weight-reduction program, in which they were prescribed a 1200-kcal/d diet (50–55% carbohydrates, 15–20% protein, and 25–30% fat), and were randomly assigned to a breakfast-skipping group or a 400-kcal breakfast group. Although both groups consumed the isocaloric diets, the breakfast-skipping group was prescribed 400-kcal lunch and 800-kcal dinner meals, whereas the breakfast group was prescribed 400-kcal breakfast, 300-kcal lunch, and 500-kcal dinner meals. Weight loss was assessed at baseline and poststudy. Both groups lost weight throughout the 12-wk energy restriction (P< 0.001). However, no differences in weight loss were observed between the women who ate breakfast throughout the 12 wk (−7.7 ± 3.3 kg) and those who skipped breakfast (−6.1 ± 3.9 kg). In addition, habitual breakfast behavior (at baseline) did not significantly influence weight loss. The study included a 6-mo follow-up, which led to similar findings (data not shown).
TABLE 1. Long-term trials lasting ≥8 wk investigating the effects of breakfast consumption on indexes of body weight and body composition1
| Study (reference) | Type of trial | Study length, wk | Characteristics | Study results | Important limitations |
--- --- --- |
| Breakfast skipping |
| Schlundt et al. (23) | Randomized controlled trial | 12 | Breakfast skipping: no breakfast, lunch (400 kcal), dinner (800 kcal) | Breakfast skipping vs. breakfast and changes in weight: −7.7 ± 3.3 vs. −6.1 ± 3.9 kg; P> 0.05 (NS) | This was not a tightly controlled study, because the breakfasts were not provided. Also, the investigators wanted to maintain a 2-meal vs. 3-meal comparison. To do this, snacking was not permitted and, thus, habitual eating patterns were altered. |
| Breakfast: breakfast (400 kcal), lunch (300 kcal), dinner (500 kcal) |
| Menus were given as guidelines and participants were taught to use food exchanges to plan meals |
| Rosado et al. (24) | Randomized controlled trial | 12 | Control breakfast: habitual breakfast | Control vs. breakfast and changes in weight: +1.2 kg (0.4, 2.0 kg) vs. +1.03 kg (0.3, 0.7 kg); P> 0.05 (NS) | This was not a tightly controlled study, because the breakfasts were not provided in the breakfast group or documented in the control group. Also, the control group included habitual breakfast skippers and consumers. Finally, the dropout rate was fairly high at 31%. |
| Breakfast: 250 kcal; 14 g PRO, 47 g CHO, 1 g fat | Percentage fat: +0.4 (−0.5, 1.4) vs. +0.4 (−0.4, 1.1); P> 0.05 (NS) |
| Prescribed 1 serving of ready-to-eat cereal with 250 mL skim milk | BMI (in kg/m2): +0.01 (−0.4, 0.4) vs. +0.1 (−0.3, 0.4); P> 0.05 (NS) |
| Dhurandhar et al. (25) | Randomized controlled trial | 16 | Breakfast skipping: USDA pamphlet with instructions not to consume any calories before 1100 | Breakfast skipping, control breakfast, and breakfast, and changes in weight: −0.66 ± 1.18 kg, −0.62 ± 1.16 kg, and −0.68 ± 1.16 kg, respectively | This was not a tightly controlled study, because the breakfasts were not provided or prescribed. |
| Control: USDA pamphlet with instructions to maintain habitual breakfast habit |
| Breakfast: USDA pamphlet with instructions to consume breakfast before 1000 every day; no specific restrictions | No main effect of breakfast was detected (P> 0.05, NS). |
| Breakfast composition—protein |
| Wang et al. (26) | Parallel design | 12 | High protein: 386 kcal; 12 g PRO, 29 g CHO, 16 g fat | High protein vs. normal protein and changes in weight: −3.9% vs. −0.2% of initial body weight (P< 0.001) | Breakfasts were not provided. The data are not presented with SD/SE. The protein differential between meals was extremely small (4 g) and the high-protein meal only contained 12 g PRO. |
| Normal protein: 386 kcal; 8 g PRO, 45 g CHO, 12 g fat |
| Both meals contained white rice and milk. The high-protein breakfast included eggs; the normal protein included bread. |
| Rueda and Khosla (27) | Parallel design | 14 | High protein: 1 serving eggs (2 whole eggs) | High protein vs. normal protein and changes in weight: no values given, P< 0.05 | Breakfasts were not provided. The breakfast prescription only included eggs and thus was not protein-specific, although the treatments ended up differing in protein content by ∼10 g. |
| Normal protein: no eggs | Fat mass: No values given, P< 0.05 |
| Although consumption of eggs/no eggs was prescribed, the remainder of the breakfast meals in each group was otherwise uncontrolled. | |
| Jakubowicz et al. (28) | Parallel design | 12 | High protein: 600 kcal; 45 g PRO, 60 g CHO, 20 g fat | High protein vs. normal protein and changes in weight: −13.6 ± 2.3 kg vs. −15.3 ± 1.9 kg; P> 0.05 (NS) | Breakfasts were not provided. Energy content was also permitted to vary between meals, as were carbohydrate and fat content. |
| Normal protein: 300 kcal; 30 g PRO, 10 g CHO, 16 g fat | BMI: −4.8 vs. −5.4; P> 0.05 (NS) |
| No recommendations outside of energy and macronutrient content were provided. | Waist circumference: −7.4 cm vs. −7.9 cm; P> 0.05 (NS) |
| Rabinovitz et al. (29) | Parallel design | 12 | High protein: 430 kcal; 23 g PRO, 42 g CHO, 19 g fat | High protein vs. normal protein and changes in weight: −2.43 ± 0.46 kg vs. −1.86 ± 0.4 kg; P> 0.05 (NS) | Breakfasts were not provided. Energy content was also permitted to vary between meals, as were carbohydrate and fat content. |
| Normal protein: 210 kcal; 8 g PRO, 29 g CHO, 7 g fat | BMI: −0.88 ± 0.17 vs. −0.69 ± 0.15; P> 0.05 (NS) |
| No recommendations outside of energy and macronutrient content were provided. | Waist circumference: −2.65 ± 0.66 cm vs. −2.2 ± 0.47 cm; P> 0.05 (NS) |
| Breakfast composition—fiber |
| Hu et al. (30) | Randomized controlled trial | 12 | High fiber: 309 kcal; 27.5 g fiber, 15 g PRO, 32 g CHO, 7 g fat | High fiber vs. low fiber and changes in weight: −0.68 ± 0.32 kg vs. −1.39 ± 0.36 kg; (P< 0.05) | Palatability of the biscuits was not directly assessed |
| Low fiber: 375 kcal; 3.2 g fiber, 12 g PRO, 62 g CHO, 7 g fat | Fat mass: −0.40 ± 0.14 kg vs. −0.71 ± 0.29 kg; P> 0.05 (NS) |
| Biscuits were provided each day. The high-fiber biscuits contained soy fiber, whereas the low-fiber biscuits did not. | Waist circumference: −0.41 ± 0.15 cm vs. −1.75 ± 0.48 cm; P> 0.05 (NS) |
| Breakfast size2 |
| Jakubowicz et al. (31) | Parallel design | 12 | High calorie: 700 kcal; 54 g PRO, 84 g CHO | High calorie vs. low calorie and changes in weight: −8.7 ± 1.4 kg vs. −3.6 ± 1.5 kg; P< 0.0001 | Breakfasts were not provided |
| Low calorie: 200 kcal; 35 g PRO, 5.7 CHO | BMI: −3.1 vs. −1.3; P< 0.0001 |
| No recommendations outside of energy and macronutrient content were provided. | Waist circumference: −8.5 ± 1.9 cm vs. −3.9 ± 1.4 cm; P< 0.0001 |
| Lombardo et al. (32) | Parallel design | 12 | High calorie: 70% of daily calories in the morning (25% breakfast; ∼495 kcal; 31 g PRO, 56 g CHO, 16 g fat; 10% morning snack; 35% lunch); 10% afternoon snack; 20% dinner. | High calorie vs. low calorie and changes in weight: −8.2 ± 3.0 kg vs. −6.5 ± 3.4 kg; P< 0.03 | Breakfasts were not provided |
| Low calorie: 55% of daily calories in the morning (15% breakfast; ∼300 kcal; 19 g PRO, 34 g CHO, 10 g fat; 5% morning snack; 35% lunch); 15% afternoon snack; 30% dinner. | Fat mass: −6.8 ± 2.1 kg vs. −4.5 ± 2.9 kg; P< 0.03 |
| No recommendations outside of energy and macronutrient content were provided. | BMI: −3.1 ± 0.2 vs. −1.8 ± 0.4; P< 0.05 |
| Waist circumference: −7.0 ± 0.6 vs. −5.0 ± 0.3 cm; P< 0.05 |
1
CHO, carbohydrate; PRO, protein.
2
The studies by Jakubowicz, et al. (28) and Rabinovitz, et al. (29) are discussed in the Breakfast composition—protein section of this table.
Rosado et al. (24) completed a 12-wk RCT in 147 children who were overweight or at risk of overweight (age: 9 ± 1 y; 90th BMI percentile for age). The breakfast group was prescribed 1 serving/d ready-to-eat cereals with milk (∼250 kcal; 14 g protein, 47 g carbohydrates, and 1 g fat), whereas the control group continued the breakfast habit to which they were accustomed. Changes in body weight, body fat [through bioelectrical impedance analysis (BIA)], and BMI were assessed at baseline and poststudy. Both groups gained weight over the 12 wk [breakfast group: +0.9 kg (95% CI: 0.4, 1.4 kg) and control group: +1.2 kg (95% CI: 0.4, 2.0 kg); both P< 0.05] with no differences between groups. In addition, no differences in BMI or percentage body fat changes were detected between groups (Table 1).
More recently, Dhurandhar et al. (25) completed a 16-wk study in 309 overweight/obese adults (age: 40 ± 0.3 y; BMI: 25–40). The participants initially were stratified according to habitual breakfast behaviors, with skipping breakfast defined as having skipped breakfast on ≥3 occasions/wk, and then were randomly assigned to the breakfast recommendations to eat breakfast, skip breakfast, or maintain their habitual breakfast behaviors. All groups were provided with the USDA “Let’s Eat for the Health of It” pamphlet, which included a general description of good nutrition habits (but did not discuss breakfast consumption). Along with this pamphlet, the breakfast group received an additional handout instructing them to eat breakfast before 1000. The handout suggested healthy food items to consume, but the participants could choose whatever they wished to eat. The breakfast-skipping group was also provided with a handout instructing the participants to avoid the consumption of any foods or drinks (besides water or zero-calorie beverages) before 1100. Again, no differences in weight loss were observed in those who ate breakfast throughout the 16 wk (−0.68 ± 1.16 kg) compared with those who skipped breakfast (−0.66 ± 1.18 kg) or those who continued their respective breakfast patterns (−0.62 ± 1.16 kg). Also, habitual breakfast behaviors (at baseline) did not influence the amount of weight lost throughout the study.
Collectively, the current evidence, albeit limited in study numbers and quality of study, does not support an effect of breakfast on weight outcomes. Although these findings suggest that breakfast does not play a causal role in weight management, the data should be viewed with caution because of several limitations. First, the small number of RCTs available during the time of the review (n = 3) presents challenges for developing conclusions regarding the role of breakfast in preventing and/or reducing weight gain. Furthermore, none of these studies tightly controlled key breakfast factors, including energy content, macronutrients, and/or breakfast quality (23, 24, 25), which might influence weight outcomes. Last, only the study by Rosado et al. (24) included other weight-related outcomes outside of weight loss, such as body composition and BMI. Thus, it is unclear whether substantial improvements in weight management would be observed after long-term studies that include tightly controlled breakfast meals varying in key dietary factors, such as macronutrient and/or energy content.
To provide additional support for the continuation of this line of research, we recently published a tightly controlled study to examine whether the daily consumption of high-protein compared with normal-protein breakfast meals improved weight management in overweight/obese breakfast-skipping young people (33). In this study, 57 adolescents (age: 19 ± 1 y; BMI: 29.7 ± 4.6) were randomly assigned to 1 of 3 patterns: 1) a high-protein breakfast (350 kcal; 35 g carbohydrates, 35 g protein, and 8 g fat); 2) a normal-protein breakfast (350 kcal; 57 g carbohydrates, 13 g protein, and 8 g fat); or 3) no breakfast (control). All breakfast meals were provided throughout the 12 wk. Weight loss and body composition (through DXA) were assessed at baseline and poststudy. No differences in weight loss were observed between groups. However, the high-protein breakfast prevented fat-mass gains over the 12 wk (−0.4 ± 0.5 kg) compared with continuing to skip breakfast (+1.6 ± 0.9 kg, P< 0.03), whereas the normal-protein breakfast did not prevent fat-mass gains (+0.3 ± 0.5 kg). Although preliminary, these data suggest a beneficial effect of including a protein-rich breakfast on weight management in individuals who habitually skip the morning meal. However, it is important to note that this was a 12-wk pilot study and was not powered to detect differences in weight loss and body composition. Further research exploring the effects of increased dietary protein at breakfast for a longer duration and in a larger sample size is necessary to strengthen these findings.
Breakfast composition and size
Whereas the previous section focused on the comparison between skipping breakfast and consuming breakfast, this section examines the effects of breakfast meals that vary in protein, fiber, and energy content.
Dietary protein
Only 4 long-term studies were identified that tested whether increased protein consumption at breakfast improved weight outcomes. Wang et al. (26) compared the effects of a prescribed normal-protein, bread-based breakfast (386 kcal; 45 g carbohydrates, 8 g protein, and 12 g fat) and an isocaloric high-protein, egg-based breakfast (29 g carbohydrates, 12 g protein, and 16 g fat) in 156 obese adolescents (age: 14 ± 2 y; BMI: 32.1 ± 1.7). The bread breakfast consisted of steamed bread, white rice, and milk, whereas the egg breakfast consisted of boiled eggs (50 g), white rice, and milk. Weight loss was assessed at baseline and at 12 wk. The egg breakfast led to ∼4% more weight loss than did the bread breakfast (P< 0.001) (Table 1), which was a difference of ∼2.25 kg between groups (P< 0.001).
In a similar study, Rueda and Khosla (27) compared a non-egg breakfast and an egg breakfast in 73 college students (age: 17–20 y; BMI: 26.0 ± 3.4). The non-egg group was given a recommendation to simply avoid eating eggs 5 d/wk for 14 wk, whereas the egg group was instructed to eat 1 serving of eggs (i.e., equivalent to 2 whole eggs) 5 d/wk for 14 wk. Both groups were permitted to eat other breakfast foods ad libitum during breakfast. Breakfast energy and macronutrient content, weight loss, and changes in BMI and body composition (through BIA) were assessed at baseline and poststudy. On average, both groups consumed ∼640-kcal breakfasts. The non-egg breakfasts contained ∼104 g carbohydrates, 19 g protein, and 4 g fat, whereas the egg breakfasts contained ∼76 g carbohydrates, 26 g protein, and 26 g fat. Although both groups gained weight and fat mass throughout the study, no differences were observed between groups.
Finally, the remaining 2 studies (28, 29) manipulated protein content and breakfast size within an energy-restricted diet. As described in Jakubowicz et al. (28), 193 obese adults (age: 47 ± 7 y; BMI: 32.2 ± 1.0) were prescribed an energy-restricted diet (∼1500 kcal/d) and were randomly assigned to either a large high-protein/high-carbohydrate breakfast (600 kcal; 60 g carbohydrates, 45 g protein, and 20 g fat) or a small lower-protein/low-carbohydrate breakfast (300 kcal; 10 g carbohydrates, 30 g protein, and 16 g fat) for 16 wk followed by a 32-wk follow-up. Weight loss and changes in BMI and waist circumference were assessed at baseline, poststudy, and follow-up. Although both energy-restricted diets led to significant (P< 0.05) reductions in weight (−14.4 kg), BMI (−5.1), and waist circumference (−7.7 cm), no differences were detected between groups throughout the intervention (Table 1). However, at the end of the follow-up period, the group consuming the large high-protein/high-carbohydrate breakfast lost more weight (−20.6 kg) than did the group consuming the small lower-protein/low-carbohydrate breakfast (−3.5 kg, P< 0.001). The group consuming the large high-protein/high-carbohydrate breakfast also displayed greater reductions in BMI and waist circumference during the follow-up period than did the group consuming the small lower-protein/low-carbohydrate breakfast (data not shown).
In a similar study, Rabinovitz et al. (29) included 59 overweight and obese adults with type 2 diabetes (age: 60.7 ± 6.4 y; BMI: 32.4 ± 3.7) and implemented an energy-restricted diet (∼1400 kcal/d) for 12 wk. The participants were randomly assigned to either a large high-protein/high-fat breakfast (430 kcal; 42 g carbohydrates, 23 g protein, and 19 g fat) or a small lower-protein/low-fat breakfast (210 kcal; 29 g carbohydrates, 8 g protein, and 7 g fat). Weight loss and changes in body composition (through BIA), BMI, and waist circumference were assessed at baseline and poststudy. No differences in any of these weight indexes were observed between groups (Table 1).
In summary, although limited, there is some evidence illustrating modest improvements in weight management after the consumption of breakfast meals containing increased dietary protein compared with lower-protein versions. However, several limitations exist that greatly influence the study findings and implications. First, none of the studies included within this section of the review were RCTs, thus weakening the quality of the research. Second, in several studies (26, 27), the protein content of the high-protein breakfasts was fairly low (12 g protein) and/or the protein differential between the breakfast comparisons was quite small (i.e., only a 4- to 7-g protein difference between the lower and higher protein meals). In a recent review, ∼30 g protein (with a differential of ∼15 g protein between treatments) was suggested as eliciting effects on indexes of weight management (18). Thus, the reduced protein content included within these studies likely reduced the ability to identify a protein effect at breakfast. In addition, habitual breakfast behaviors were not assessed in any of these studies, which may have influenced the participants’ responses to the study breakfast interventions. Furthermore, none of the studies provided the breakfast foods for the participants to consume throughout the study, making it difficult to assess adherence accurately. Finally, 3 (27, 28, 29) of the 4 studies neglected to match key dietary characteristics, including fiber, energy density, and/or energy content. Thus, additional work involving full-feeding RCTs are needed to assess whether the implementation of a single dietary strategy, such as increased protein consumption at breakfast, improves weight management.
Dietary fiber
The search with respect to the effect of increased dietary fiber at breakfast on weight outcomes yielded only one report (Table 1). Hu et al. (30) completed a 12-wk RCT in 39 overweight adults (age: 21 ± 5 y; BMI: 26 ± 0.5) in which a breakfast was provided each day consisting of either low-fiber (375 kcal; 3 g wheat fiber) or high-fiber (309 kcal; 28 g soy fiber) biscuits. Changes in body weight, body composition (through DXA), BMI, and waist circumference were assessed at baseline and poststudy. The high-fiber breakfast group experienced a 2% greater weight loss over the 12 wk than did the low-fiber breakfast controls (P< 0.05, Table 1). Furthermore, although no additional differences were detected between groups, only the high-fiber group displayed greater reductions in BMI, waist circumference, fat mass, and percentage body fat throughout the study (poststudy–prestudy, all P< 0.05). Although this is the only published study, to our knowledge, examining the long-term effects of fiber consumption at breakfast on weight management, the findings suggest a potential effect of fiber, specifically soy fiber, on weight loss and improvements in body composition. Additional RCTs are needed to replicate these data, as well as to explore whether similar findings occur with the incorporation of other types of fiber.
Breakfast size
Our search identified 4 studies addressing breakfast size and weight management. All 4 studies included dietary energy restrictions in overweight/obese adults and were between 12 and 16 wk in duration (28, 29, 31, 32). Two of the studies (28, 29) were described previously, because protein content, along with energy content, were manipulated. Of those studies, only the one by Jakubowicz et al. (28) reported greater weight loss and greater reductions in BMI and waist circumference after the large breakfasts than after the small breakfasts; however, these differences were only detected at the end of the 32-wk follow-up period.
In a subsequent study (31), 93 overweight/obese women (age: 45.8 ± 7.1 y; BMI: 32.4 ± 1.8) with metabolic syndrome were assigned to an energy-restricted lower-carbohydrate diet (∼1400 kcal/d) for 12 wk and were randomly assigned to high-calorie (700 kcal) or low-calorie (200 kcal) breakfast groups. To make the diets isocaloric, the high-calorie breakfast group was prescribed a lower-calorie dinner (200 kcal), whereas the low-calorie breakfast group was prescribed a higher-calorie dinner (700 kcal). Lunch was matched at 500 kcal for both groups. The diets were also matched for macronutrient content (32% carbohydrates, 41% protein, and 27% fat). Weight loss and changes in BMI and waist circumference were assessed at baseline and poststudy. The group consuming the high-calorie breakfast experienced a 2.5-fold greater weight loss than did the group consuming the low-calorie breakfast (Table 1). In addition, greater reductions in BMI and waist circumference were observed with the high-calorie breakfast than with the low-calorie breakfast (both, P< 0.05; Table 1).
The last study in this section (32) included a 12-wk energy-restricted (600 kcal/d less than weight maintenance) Mediterranean-style diet (55% carbohydrates, 15% protein, and 30% fat) in 42 overweight/obese women (age: 60.7 ± 6.4 y; BMI: 32.4 ± 3.7). Unlike in the previous studies, Lombardo et al. (32) assessed the distribution of calories throughout the day instead of simply assessing breakfast per se. In this study, the participants were randomly prescribed a large or small morning distribution pattern. The large morning pattern contained 70% of daily calories with 25% (∼495 kcal) at breakfast, 10% at the morning snack, and 35% at lunch, whereas the small morning pattern contained 55% of daily calories with 15% (∼300 kcal) at breakfast, 5% during the morning snack, and 35% at lunch. Changes in body weight, body composition (through DXA), BMI, and waist circumference were assessed at baseline and poststudy. The large morning pattern led to 20% more weight loss over the 12 wk than did the small morning pattern (P< 0.03; Table 1). Furthermore, the large morning pattern led to greater reductions in BMI, waist circumference, and fat mass than did the small morning pattern (all, P< 0.05; Table 1).
Strong evidence is emerging to support the effect of redistributing calories toward the morning, particularly including a higher-calorie breakfast, for improvement in outcomes of weight management. However, it is important to note that these studies were not RCTs and did not tightly control the diet interventions. In addition, only 2 studies manipulated macronutrient content (in addition to breakfast size) and demonstrated a modest protein effect (28, 29). Regardless of the limitations, these studies provide a framework for future clinical trials to explore the timing and distribution of food intake for the prevention and/or treatment of obesity.
Intervention-Based Trials Including Daily Food Intake
The regulation of energy intake has been proposed as serving as one mediator of weight management. Subsequently, numerous studies have explored whether breakfast consumption reduces body weight through alterations in daily intake. The studies in this section include assessments of daily energy intake that were either directly measured through ad libitum consumption or estimated through dietary recalls and/or food records or diaries. In addition, the majority of these studies (n = 19; 83%) were acute trials in which the intervention occurred over a single day. A few studies (n = 3) (34, 35, 36, 37) included subchronic durations between 1 and 6 wk, and only one of the studies was >8 wk (29).
Addition of breakfast
The search yielded 8 studies in which breakfast skipping was compared with breakfast consumption, and many of the studies included different types of breakfasts within each study.
Approximately 21% of the comparisons (34, 35, 38, 39) demonstrated greater daily intake when consuming breakfast than when skipping breakfast; 5% (36) demonstrated lower daily intake with breakfast consumption than with skipping breakfast; and 74% (34, 40, 41, 42) elicited no differences in daily intake (Table 2). Although these summary data suggest that the effects of breakfast on daily intake remain uncertain, there are a few issues that must be considered.
TABLE 2. Acute and long-term trials investigating the effects of breakfast consumption on daily energy intake1
| Study (reference) | Type of trial | Study length | Characteristics | Results of daily food intake, kcal/d |
--- ---
| Leidy et al. (34) | Crossover design | 1-wk acclimation followed by the respective 8-h testing day/treatment | Breakfast skipping: 0 kcal; nothing before lunch | Breakfast skipping vs. normal protein: 2002 ± 111 vs. 2292 ± 115; P< 0.003 |
| Normal protein: 350 kcal; 13 g PRO, 57 g CHO, and 8 g fat |
| High protein: 350 kcal; 35 g PRO, 35 g CHO, and 8 g fat | Breakfast skipping vs. high protein: 2002 ± 111 vs. 2123 ± 71; P> 0.05 (NS) |
| All meals provided |
| Betts et al. (35) | Randomized controlled trial | 6 wk/treatment | Breakfast skipping: 0 kcal; nothing before 1200 | Breakfast skipping vs. breakfast: 2191 ± 494 vs. 2730 ± 573; P< 0.0007 |
| Breakfast: ≥700 kcal (before 1100; 50% within 2 h of waking) |
| Meals were not provided |
| Farshchi et al. (36) | Crossover design | 2-wk acclimation followed by the respective 4-h testing day/treatment | Breakfast skipping: 0 kcal | Breakfast skipping vs. breakfast: 1756 ± 155 vs. 1665 ± 141; P = 0.001 |
| Breakfast: 10 kcal/kg body weight |
| 15% PRO, 50% CHO, and 35% fat |
| All meals provided |
| Kral et al. (38) | Crossover design | One 4-h testing day/treatment | Breakfast skipping: 0 kcal | Breakfast skipping vs. breakfast: 1830 vs. 2191; P< 0.05 |
| Breakfast cereals: 350 kcal; 11 g PRO, 69 g CHO, 4 g fat |
| All breakfast meals provided |
| Levitsky and Pacanowski (39) | Crossover design | 1 d/treatment | Breakfast skipping: 0 kcal | Breakfast skipping vs. breakfast: 400 fewer kcal consumed by breakfast skipping vs. breakfast; P< 0.01 |
| Breakfast: ad libitum; total offered: 218 kcal |
| All foods provided |
| Leidy and Racki (40) | Crossover design | 1 d/treatment | Breakfast skipping: 0 kcal; nothing before lunch | Breakfast skipping vs. normal protein: 2259 ± 280 vs. 2530 ± 212; P> 0.05 (NS) |
| Normal protein: 513 ± 26 kcal; 18 g PRO, 95 g CHO, 8 g fat |
| High protein: 512 ± 26 kcal; 49 g PRO, 63 g CHO, 8 g fat | Breakfast skipping vs. high protein: 2259 ± 280 vs. 2505 ± 284; P> 0.05 (NS) |
| All meals provided |
| Irvine et al. (41) | Crossover design | 1 d/treatment | Breakfast skipping: 0 kcal | Breakfast skipping vs. normal protein: 1488 ± 385 vs. 1663 ± 331, P> 0.05 (NS) |
| Normal protein: 250 kcal; 4 g PRO, 39 g CHO, 9 g fat |
| High protein: 250 kcal; 20 g PRO, 39 g CHO, 2 g fat | Breakfast skipping vs. high protein: 1488 ± 385 vs. 1672 ± 368; P> 0.05 (NS) |
| All meals provided |
| De Graaf et al. (42) | Crossover design | 1 d/treatment | Breakfast skipping: 8 kcal | Breakfast skipping vs. breakfasts: all comparisons, P> 0.05 (NS) |
| Breakfasts: high energy (400 kcal); medium energy (250 kcal); low energy (100 kcal) and high protein (70%); high carbohydrate (99%); and high fat (92%) | Numerical data not presented |
| All breakfast meals provided |
| Breakfast composition—protein2 |
| Stubbs et al. (43) | Randomized controlled trial | One 24-h testing day/treatment | High carbohydrate: 1237 kcal; 57 g PRO, 188 g CHO, and 29 g fat | High carbohydrate vs. high protein: 4006 ± 160 vs. 4022 ± 160, P> 0.05 (NS) |
| High fat: 1251 kcal; 64 g PRO, 70 g CHO, and 79 g fat |
| High protein: 1263 kcal; 186 g PRO, 69 g CHO, and 27 g fat | High fat vs. high protein: 3867 ± 160 vs. 4022 ± 160; P> 0.05 (NS) |
| All meals provided |
| Karhunen et al. (44) | Crossover design | 1 d/treatment | Low fiber–low protein: 300 kcal; 3 g PRO, 33 g CHO, and 14 g fat; 7.6 g fiber | Normal-protein versions vs. high-protein versions: P> 0.05 (NS) |
| Low fiber–high protein: 300 kcal; 20 g PRO, 23 g CHO, and 13 g fat; 6.2 g fiber | Numerical data not presented |
| High fiber–low protein: 300 kcal; 3 g PRO, 32 g CHO, and 16 g fat; 27.3 g fiber |
| High fiber–high protein: 300 kcal; 18 g PRO, 21 g CHO, and 14 g fat; 25.8 g fiber |
| All breakfast meals provided |
| Fallaize et al. (45) | Crossover design | 1 d/treatment | Normal-protein cereal: 330 kcal; 9 g PRO, 53 g CHO, and 10 g fat | Normal-protein groups vs. high protein: greater intake in normal-protein vs. high protein; P< 0.007 |
| Normal-protein croissant: 330 kcal; 5 g PRO, 8 g CHO, and 18 g fat | Numerical data not presented |
| High protein: 330 kcal; 18 g PRO, 18 g CHO, and 21 g fat |
| All meals provided |
| Vander Wal et al. 2005 (46) | Crossover design | 1.5 d/treatment | Normal protein: 344 ± 11 kcal; 14 g PRO, 48 g CHO, and 11 g fat | Normal protein vs. high protein: 2048 ± 487 vs. 1784 ± 427; P< 0.05 |
| High protein: 353 ± 2 kcal; 18 g PRO, 33 g CHO, 17 g fat |
| All breakfast meals provided |
| Ratliff et al. (47) | Crossover design | 1 d/treatment | Normal protein: 396 kcal; 16 g PRO, 71 g CHO, and 5 g fat | Normal protein vs. high protein: 2229 ± 528 vs. 1826 ± 603; P< 0.05 |
| High protein: 396 kcal; 22 g PRO, 22 g CHO, and 24 g fat |
| All breakfast meals provided |
| Breakfast composition—fiber |
| Mattes (37) | Crossover design | 5 d/treatment | Low fiber: 207 kcal; 5 g PRO, 39 g CHO, and 4 g fat; 0.62 g wheat fiber | Low fiber vs. high fiber: P> 0.05 (NS) |
| High-fiber alginate/guar gum: 196 kcal; 6 g PRO, 38 g CHO, and 4 g fat; 4.49 g guar fiber | Numerical data not presented |
| All breakfast meals provided |
| Delargy et al. (48) | Crossover design | 1 d/treatment | No fiber: 92 kcal; 2 g PRO, 21 g CHO, and 1 g fat; 0g fiber | No fiber: 3355 ± 698 |
| Low fiber: 537 kcal; 16 g PRO, 95 g CHO, and 11 g fat; 1 g insoluble fiber and 2.2 g soluble fiber | Low fiber: 3389 ± 542 |
| High fiber—insoluble: 537 kcal; 23 g PRO, 84 g CHO, and 13 g fat; 18 g insoluble fiber and 3.8 g soluble fiber | High fiber—insoluble: 3548 ± 557 |
| High fiber—soluble: 537 kcal; 16 g PRO, 96 g CHO, and 11 g fat; 4.2 g insoluble fiber and 17.5 g soluble fiber | High fiber—soluble: 3535 ± 570 |
| All meals provided | No main effect of fiber was detected (P> 0.05, NS) |
| Juvonen et al. (49) | Crossover design | 1 d/treatment | Low fiber: 299 kcal; 4 g PRO, 57 g CHO, and 4 g fat; 1.5 g insoluble fiber | Low-fiber versions vs. high-fiber versions: all comparisons, P> 0.05 (NS) |
| High fiber—wheat bran: 299 kcal; 6 g PRO, 54 g CHO, and 4 g fat; 10.3 g insoluble fiber | Numerical data not presented |
| High fiber—oat bran: 299 kcal; 8 g PRO, 53 g CHO, and 4 g fat; 5.5 g insoluble fiber and 5.1 g soluble fiber |
| High fiber—wheat and oat breakfast: 299 kcal; 7 g PRO, 57 g CHO, and 4 g fat; 7.6 g insoluble fiber and 2.5 g soluble fiber |
| All breakfast meals provided |
| Klosterbuer et al. (50) | Crossover design | 1 d/treatment | Low fiber: 591 kcal; 10 g PRO, 105 g CHO, and 13 g fat; 2.8 g wheat fiber | Low-fiber versions vs. high-fiber versions: all comparisons, P> 0.05 (NS) |
| High fiber—corn: 617 kcal; 10 g PRO, 104 g CHO, and 13 g fat; 27.8 g corn fiber | Numerical data not presented |
| High fiber—corn and pullalan: 641 kcal; 10 g PRO, 104 g CHO, and 13 g fat; 27.8 g corn and pullalan fiber |
| High fiber—resistant starch: 589 kcal; 10 g PRO, 106 g CHO, and 13 g fat; 27.2 g resistant starch fiber |
| High fiber—resistant starch and pullalan: 568 kcal; 10 g PRO, 106 g CHO, and 13 g fat; 27.2 g resistant starch and pullalan fiber |
| All breakfast meals provided |
| Willis et al. (51) | Crossover design | 1 d/treatment | Low fiber: 502 kcal; 11 g PRO, 74 g CHO, and 20g fat; <1 g fiber | All comparisons, P> 0.05 (NS) |
| 4 g fiber: 488 kcal; 12 g PRO, 81 g CHO, and 13 g fat; 5.7 g fiber | Numerical data not presented |
| 8 g fiber: 493 kcal; 12 g PRO, 89 g CHO, and 10 g fat; 8.9 g fiber |
| 12 g fiber: 544 kcal; 13 g PRO, 93 g CHO, and 13 g fat; 12.8 g fiber |
| All breakfast meals provided |
| Archer et al. (52) | Crossover design | 1 d/treatment | No fiber: 427 kcal; 27 g PRO, 34 g CHO, and 21 g FAT; 0 g fiber | No fiber vs. high fiber—inulin: greater intake in no fiber vs. high fiber—inulin; P< 0.05 |
| High fiber—inulin: 362 kcal; 27 g PRO, 34 g CHO, and 13 g fat; 24 g fiber | Numerical data not presented |
| High fiber—lupin: 360 kcal; 28 g PRO, 34 g CHO, 13 g fat; 24 g fiber | No fiber vs. high fiber—lupin: greater intake in no fiber vs. high fiber—lupin; P< 0.05 |
| All breakfast meals provided | Numerical data not presented |
| Yannakoula et al. (53) | Crossover design | 1 d/treatment | Low fiber: 456 kcal; 17 g PRO, 74 g CHO, and 11 g fat; 0.7 g fiber | Low fiber vs. high fiber: 3359 ± 669 vs. 2989 ± 602; P< 0.005 |
| High fiber: 465 kcal; 18 g PRO, 73 g CHO, and 11 g fat; 14.9 g fiber |
| All breakfast meals provided |
| Barone Lumaga et al. (54) | Crossover design | 1 d/treatment | No fiber: 149 kcal/250 mL; 0 g PRO, 37 g CHO, and 0 g fat; 0 g fiber | No fiber vs. high-fiber groups: greater intake vs. high-fiber groups; P< 0.05 |
| High fiber—β-glucan: 148 kcal/250 mL; 0 g PRO, 34 g CHO, and 0 g fat; 3 g β-glucan fiber | Numerical data not presented |
| High fiber—pectin: 149 kcal/250 mL; 1 g PRO, 34 g CHO, and 0 g fat; 2.5 g pectin fiber |
| All treatments were part of an isocaloric breakfast: ∼536.3 kcal; all breakfasts were provided |
| Karhunen et al. (44) | Crossover design | 1 d/treatment | Low fiber–low protein: 300 kcal; 3 g PRO, 33 g CHO, and 14 g fat; 7.6 g fiber | Low-fiber groups vs. high-fiber groups: all comparisons, P> 0.05 (NS) |
| Low fiber–high protein: 300 kcal; 20 g PRO, 23 g CHO, and 13 g fat; 6.2 g fiber | Numerical data not presented |
| High fiber–low protein: 300 kcal; 3 g PRO, 32 g CHO, and 16 g fat; 27.3 g fiber |
| High fiber–high protein: 300 kcal; 18 g PRO, 21 g CHO, and 14 g fat; 25.8 g fiber |
| All breakfast meals provided |
| Breakfast composition—size3 |
| Rabinovitz, et al. (29) | Parallel design | 12 wk | High calorie: 430 kcal; 23 g PRO, 42 g CHO, and 19 g fat | Low calorie vs. high calorie: P> 0.05 (NS) |
| Low calorie: 210 kcal; 8 g PRO, 29 g CHO, and 7 g fat | Numerical data not presented |
| No recommendations outside of energy and macronutrient content were provided |
1
CHO, carbohydrate; PRO, protein.
2
The studies by Leidy and Racki (40), Irvine et al. (41), Leidy et al. (34), and De Graaf, et al. (42) are discussed in the previous section of this table.
3
The study by De Graaf et al. (42) is discussed in the Breakfast composition—protein section of this table.
The breakfast characteristics within and between studies may have contributed to the conflicting findings. For example, Leidy et al. (34) reported an additional 290 kcal/d after a normal-protein breakfast compared with skipping breakfast (P< 0.003), whereas the high-protein breakfast did not increase daily intake. These findings are also supported by those studies that included only normal-protein breakfasts and reported increased daily intake compared with skipping breakfast (35, 38, 39). However, it is important to note that, although increased protein at breakfast may not increase daily intake, none of the studies reported a decrease in daily intake with the high-protein breakfasts compared with skipping breakfast (34, 40, 41, 42). Based on these data, it is unclear whether a single high-protein meal elicits changes in daily intake. However, in our recent pilot study, the group consuming a high-protein breakfast had an ∼400 kcal reduction in daily intake throughout the 12-wk period, whereas the breakfast-skipping group did not experience this reduction (33). In fact, those who continued skipping breakfast increased their daily intake by ∼370 kcal.
Beyond the breakfast characteristics, habitual breakfast behaviors might also influence the study findings. Adding breakfast to the diet of those who habitually skip the morning meal is quite a different experience from removing breakfast from the diet of someone who habitually eats breakfast. For example, the study by Farshchi et al. (36) only included habitual breakfast consumers and was the only study to report higher daily intake (by ∼90 kcal, P< 0.001) when breakfast was removed (i.e., skipped) than when breakfast continued to be consumed. However, in the studies that only included habitual breakfast skippers, the addition of breakfast generally did not increase daily intake despite the additional calories consumed from the breakfast meal (34, 40). These data suggest that although incorporating additional calories at the breakfast meal does not increase daily intake, omitting breakfast from one’s diet may lead to overeating later in the day.
The last factor that may influence intake data involves the specific modality of assessing intake. Three of the studies (34, 39, 41) used a full-feeding design in which all foods and beverages were provided and weighed, whereas the remaining studies included dietary food records and/or recalls. Although dietary records and recalls typically lead to under-reporting of daily intake, eating behavior is compromised within a full-feeding laboratory setting and can lead to purposeful reductions in daily food intake (55). Because of the limitations with collecting viable intake data, it is challenging to accurately identify a breakfast effect.
Breakfast composition and size
The search yielded 9 studies that compared high-protein with normal-protein breakfasts, 8 studies that compared high-fiber with normal/no-fiber breakfasts, and 2 studies that compared small with large breakfasts (Table 2).
The protein studies included a fairly large range of protein intake within the breakfast meals. Specifically, the normal-protein meals contained between 3 and 64 g protein/meal, whereas the high-protein meals included 18–186 g protein. Although the normal- and high-protein meals were isocaloric within each study, protein sources and types of breakfasts in these meals varied within and/or between studies. Specifically, a variety of protein sources were included within the high-protein breakfasts, including eggs, dairy, beef, and soy. On average, consumption of the high-protein breakfasts led to a daily intake of 2440 ± 335 kcal, whereas consumption of the normal-protein breakfasts led to a daily intake of 2530 ± 335 kcal, a difference of ∼90 kcal. When the study findings were summarized, 62% of the comparisons elicited similar daily intake (40, 41, 42, 43, 44), 38% reported a reduction in daily intake (34, 45, 46, 47), and 0% reported an increase in daily intake with the consumption of a high-protein compared with normal-protein breakfast (Table 2).
The fiber studies included a fairly large range of fiber intake within the breakfast meals. Specifically, the normal-fiber meals contained between 0 and 3 g fiber/meal, whereas the high-fiber meals included 2.5–28 g fiber. Energy content, source of fiber, and types of breakfast varied within and/or between studies. On average, the consumption of the high-fiber breakfasts led to a daily intake of 2350 ± 210 kcal, whereas consumption of the normal-fiber breakfasts led to a daily intake of 2460 ± 210 kcal (i.e., a difference of ∼110 kcal). When the study findings were summarized, 80% of the comparisons elicited similar daily intake (37, 44, 48, 49, 50, 51), 20% reported a reduction in daily intake (52, 53, 54), and 0% reported an increase in daily intake with the consumption of a high-fiber compared with a normal-fiber breakfast (Table 2).
With respect to the effects of breakfast size, both studies led to similar intake when comparing the large with the smaller breakfast meals, averaging 1850 ± 450 kcal/d (Table 2) (29, 42).
In summary, the current evidence is conflicting as to whether breakfast consumption influences daily intake. Although the discrepant data might be due to the limitations of assessing daily intake, it is important to note that daily intake was generally lower after the consumption of breakfast meals high in dietary protein or fiber. Further research including more tightly controlled breakfast components throughout randomized, controlled, longer-term studies are required to strengthen these findings.
Summary and Conclusions
The strength of the evidence supporting the consumption of breakfast for weight management and daily food intake is shown in Supplemental Table 1. Based on the Academy of Nutrition and Dietetics Evidence Analysis criteria (22), there is limited evidence supporting the addition of breakfast for body weight management and daily food intake. Regarding the type of breakfast, accumulating evidence exists supporting the consumption of increased protein and fiber at breakfast, as well as consuming more energy during the morning hours. However, the majority of the studies that manipulated breakfast composition and content did not control for habitual breakfast behaviors; nor did those studies include a breakfast-skipping control. Thus, it is unclear whether the addition of these types of breakfast meals affects weight management.
Future research, including large RCTs of longer-term duration (≥6 mo) with a focus on key dietary factors, is critical to begin to assess whether breakfast recommendations are appropriate for the prevention and/or treatment of obesity across the lifespan.
Acknowledgments
Responsibility for the design, analyses, and interpretation of the information presented in this review was that of the authors. All authors read and approved the final manuscript.
Supplementary Material
Download: Download Word document (16KB)
Supplemental Table 1. Strength of Evidence Supporting Breakfast for Weight Management and Daily Intake
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Published in a supplement to Advances in Nutrition. Presented at the ASN Scientific Sessions and Annual Meeting at Experimental Biology 2015 held in Boston, MA, 28 March–1 April 2015. The sponsored satellite program was organized and sponsored by the Kellogg Company. The Supplement Coordinators for this supplement were Lisa Sanders and Zeina Jouni. Supplement Coordinator Disclosures: Lisa Sanders and Zeina Jouni are employed by the Kellogg Company. Publication costs for this supplement were defrayed in part by the payment of page charges. This publication must therefore be hereby marked “advertisement” in accordance with 18 USC section 1734 solely to indicate this fact. The opinions expressed in this publication are those of the author(s) and are not attributable to the sponsors or the publisher, Editor, or Editorial Board of Advances in Nutrition.
The satellite session and this supplement were supported by funding from the Kellogg Company.
Supplemental Table 1 is available from the “Online Supporting Material” link in the online posting of the article and from the same link in the online table of contents at
Author disclosures: HJ Leidy, JA Gwin, CA Roenfeldt, AZ Zino, and RS Shafer, no conflicts of interest.
© 2016 American Society for Nutrition
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27.1 - The Theorem
Central Limit Theorem Section
We don't have the tools yet to prove the Central Limit Theorem, so we'll just go ahead and state it without proof.
Let (X_1, X_2, \ldots, X_n) be a random sample from a distribution (any distribution!) with (finite) mean (\mu) and (finite) variance (\sigma^2). If the sample size (n) is "sufficiently large," then:
the sample mean (\bar{X}) follows an approximate normal distribution
with mean (E(\bar{X})=\mu_{\bar{X}}=\mu)
and variance (Var(\bar{X})=\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n})
We write:
(\bar{X} \stackrel{d}{\longrightarrow} N\left(\mu,\dfrac{\sigma^2}{n}\right)) as (n\rightarrow \infty)
or:
(Z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{\sum\limits_{i=1}^n X_i-n\mu}{\sqrt{n}\sigma} \stackrel {d}{\longrightarrow} N(0,1)) as (n\rightarrow \infty).
So, in a nutshell, the Central Limit Theorem (CLT) tells us that the sampling distribution of the sample mean is, at least approximately, normally distributed, regardless of the distribution of the underlying random sample. In fact, the CLT applies regardless of whether the distribution of the (X_i) is discrete (for example, Poisson or binomial) or continuous (for example, exponential or chi-square). Our focus in this lesson will be on continuous random variables. In the next lesson, we'll apply the CLT to discrete random variables, such as the binomial and Poisson random variables.
You might be wondering why "sufficiently large" appears in quotes in the theorem. Well, that's because the necessary sample size (n) depends on the skewness of the distribution from which the random sample (X_i) comes:
If the distribution of the (X_i) is symmetric, unimodal or continuous, then a sample size (n) as small as 4 or 5 yields an adequate approximation.
If the distribution of the (X_i) is skewed, then a sample size (n) of at least 25 or 30 yields an adequate approximation.
If the distribution of the (X_i) is extremely skewed, then you may need an even larger (n).
We'll spend the rest of the lesson trying to get an intuitive feel for the theorem, as well as applying the theorem so that we can calculate probabilities concerning the sample mean. |
8384 | https://or.stackexchange.com/questions/2/reference-for-expectation-preserves-convexity | probability theory - Reference for "expectation preserves convexity" - Operations Research Stack Exchange
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Reference for "expectation preserves convexity"
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It is well known that expectation preserves convexity: If f(x)f(x) is convex and Y Y is a random variable, then E[f(x−Y)]E[f(x−Y)] is convex. This property arises in, for example, inventory theory.
I have not been able to find a good source to cite for this well-known fact. Can anyone suggest one?
(For what it's worth, Boyd and Vandenberghe's book proves another well known property, namely, minimization preserves convexity, but I don't think they prove it for expectation.)
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edited Jun 1, 2019 at 13:14
TheSimpliFire♦
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LarrySnyder610LarrySnyder610
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2 If f(x)f(x) is convex, then f′′(x)⪰0 f″(x)⪰0 is known. We can write the expectation in the closed form: E[f(x−Y)]=∑i∈I p i f(x−y i)E[f(x−Y)]=∑i∈I p i f(x−y i) where p i p i is the probability of y i y i to be realized and I I is the set of indices Y Y can take (if continuous replace with an integration). We can use the property that sum of convex functions is convex as well. Therefore, we need to show p i f(x−y i)p i f(x−y i) is convex. The second derivative of this function is f′′(x−y i)⪰0 f″(x−y i)⪰0 since f f is a convex function. This concludes the statement if I'm not wrong.independentvariable –independentvariable 2019-05-30 22:16:44 +00:00 Commented May 30, 2019 at 22:16
1 @aslv95 Agree. My question is just where I can find that written in a book or article, so I can cite it when I use it. (All else fails, I guess I can cite you. :) )LarrySnyder610 –LarrySnyder610 2019-05-30 22:20:26 +00:00 Commented May 30, 2019 at 22:20
3 I hope I am not wrong, but this seems like something we can really skip citing. Otherwise, we would need to cite 'linear function is a convex/concave' function all the time since this is easy to see but there is definitely someone who said this first :) But that's interesting because I face this all the time as well. Usually, you can see something is pretty straightforward but also doubt much like 'what if I need to cite this but I don't know if it is online or too low level to cite'... There should be some threshold of the complexity of proof to skip citing, but idk :)independentvariable –independentvariable 2019-05-30 22:27:59 +00:00 Commented May 30, 2019 at 22:27
2 You might be right. If someone knows of a reference, I'll gladly use it, but if not, your comment puts me at ease a bit.LarrySnyder610 –LarrySnyder610 2019-05-30 22:31:58 +00:00 Commented May 30, 2019 at 22:31
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Reference "Convex Optimization" by Boyd and Vandenberghe section 3.2.1, p. 79.
These properties extend to infinite sums and integrals. For example if f(x,y)f(x,y) is convex in x x for each y∈A y∈A, and w(y)≥0 w(y)≥0 for each y∈A y∈A, then the function g g defined as
g(x)=∫A w(y)f(x,y)d y g(x)=∫A w(y)f(x,y)d y
is convex in x x (provided the integral exists).
Of course, this extends to Lebesgue-Stieltjes integrals (not mentioned in the referenced book), so should cover any expectation.
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edited May 31, 2019 at 10:09
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answered May 30, 2019 at 23:56
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1 That will work. I swear I looked through that whole book a few years ago looking for this. :/ Thanks.LarrySnyder610 –LarrySnyder610 2019-05-30 23:58:56 +00:00 Commented May 30, 2019 at 23:58
2 @LarrySnyder610 As a side note, you need to assume that f(x−Y)f(x−Y) is a continuous random variable for the above to be applicable.Pantelis Sopasakis –Pantelis Sopasakis 2019-05-31 00:32:15 +00:00 Commented May 31, 2019 at 0:32
2 @Pantelis Sopasaki I mentioned the extension to Lebesgue-Stieltjes integrals, even though that is not mentioned in the Boyd and Vandenberghe book.Mark L. Stone –Mark L. Stone 2019-05-31 00:44:41 +00:00 Commented May 31, 2019 at 0:44
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Define ϕ(x)=E[f(x−Y)]ϕ(x)=E[f(x−Y)] and assume that for all x∈R x∈R, f(x−Y)f(x−Y) is measurable and integrable. Then, for x,x′∈R x,x′∈R and α∈[0,1]α∈[0,1]
ϕ(α x+(1−α)x′)===≤==E[f(α x+(1−α)x′−Y)]E[f(α x+(1−α)x′−α Y−(1−α)Y)]E[f(α(x−Y)+(1−α)(x′−Y))]E[α f(x−Y)+(1−α)f(x′−Y)]α E[f(x−Y)]+(1−α)E[f(x′−Y)]α ϕ(x)+(1−α)ϕ(x′),ϕ(α x+(1−α)x′)=E[f(α x+(1−α)x′−Y)]=E[f(α x+(1−α)x′−α Y−(1−α)Y)]=E[f(α(x−Y)+(1−α)(x′−Y))]≤E[α f(x−Y)+(1−α)f(x′−Y)]=α E[f(x−Y)]+(1−α)E[f(x′−Y)]=α ϕ(x)+(1−α)ϕ(x′),
where we have used the linearity of the expectation, the fact that if Z≤Z′Z≤Z′ (a.s.) then E[Z]≤E[Z′]E[Z]≤E[Z′], and the convexity of f f. We have, therefore, shown that ϕ ϕ is convex.
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answered May 30, 2019 at 23:03
Pantelis SopasakisPantelis Sopasakis
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1 Beautiful. But: I'm just looking for a reference. :) Maybe this was a poor question. My intent is just: I use this result all the time. When I use it, I say "it's well known". But I'd like to be able to say "it's well known; see, e.g., [ref.]". I don't want to include a proof in what I'm writing, obviously, since it's "well known".LarrySnyder610 –LarrySnyder610 2019-05-30 23:18:28 +00:00 Commented May 30, 2019 at 23:18
4 @LarrySnyder610 I'll try find something. You could perhaps state that this can be easily verified (if it's for a paper), or give it as an exercise if it's for a course.Pantelis Sopasakis –Pantelis Sopasakis 2019-05-30 23:21:02 +00:00 Commented May 30, 2019 at 23:21
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8385 | https://www.epa.gov/caddis/case-ddt-revisiting-impairment | Published Time: 2016-06-08T18:21:54-04:00
The Case of DDT: Revisiting the Impairment | US EPA
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The Case of DDT: Revisiting the Impairment
Background
The fact that DDT(or dichloro-diphenyl-trichloroethane) played a role in the decline of bald eagle and other bird-of-prey populations (e.g., ospreys, brown pelicans) is now commonly appreciated among most biologists. However, the link between DDT and the eggshell thinning that caused reproductive failure in these birds was not initially recognized. Ultimately, the connection was made by re-examining the description of the impairment.
The first link between DDT and diminishing bald eagle and other birds of prey populations was the consistent observation of high body burdens of DDT metabolites. In other words, there was co-occurrence of the declining bird populations and the candidate cause, DDT. There was also evidence of a complete exposure pathway to birds based on body burden of DDT. However, extensive toxicity testing of DDT on adult bird mortality revealed no relationship. This suggested that the proposed mechanism, toxicity, was implausible. However, lethality was not the impairment; decline of birds-of prey was the impairment. A new conceptual model was required that considered other mechanisms that could result in declines in bird populations. In a reexamination of the overall analysis, it became apparent that the species chosen for testing had been relatively tolerant of DDT exposure compared to those that were affected in the wild, and that the endpoint observed in these tests (lethality) would not reflect reproductive success or failure resulting from DDT exposure.
Field observations eventually revealed a potential plausible mechanism of reproductive failure due to eggshell thinning among bald eagles and other birds of prey. Laboratory experiments showed that DDE could cause eggshell thinning. Field studies showed that field exposures to DDE, a metabolite of DDT, were sufficient to cause effects in many species of birds based on the stressor-response relationship. Together these findings provided lines of evidence by which DDT might cause eggshell thinning and reduce reproductive success, a more specific impairment than declines in bird population.
Outcome
In 1972, DDT was banned from most uses in the United States. In the years following the ban, bald eagle and other bird-of-prey populations slowly recovered. The recovery of bird populations after the use of DDT was banned, is an example of mitigation of the effect following manipulation of the cause, and is very strong evidence that the use of DDT was, in fact, the true cause of bald eagle and other bird-of-prey population declines.
References
Grier JW (1982) Ban of DDT and subsequent recovery of reproduction in bald eagles. Science 218:1232-1234.
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The role of NOTCH3 in CADASIL pathogenesis: insights into novel therapies
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The role of NOTCH3 in CADASIL pathogenesis: insights into novel therapies
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Abstract
Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL) is a monogenetic hereditary small-vessel disorder characterised by recurrent subcortical ischemic strokes, cognitive deterioration, and other neurological symptoms. Single nucleotide mutations within the NOTCH3 gene can impair NOTCH3 processing and/or signalling, resulting in the accumulation of granular osmiophilic material (GOM) in blood vessel walls, and consequently CADASIL. Despite its significant clinical impact, there is currently no definitive treatment for CADASIL. This review provides a comprehensive analysis of the pathophysiological mechanisms underlying CADASIL, focusing on NOTCH3 mutations and their effects on protein processing and signalling. The review proposes a hypothesis that explains how NOTCH3 mutations may alter the signalling process and result in GOMs. Additionally, the review explores published therapy strategies aimed at restoring normal NOTCH3 function.
Keywords:
CADASIL; Cerebrovascular disease; Gene therapy; NOTCH3; Stroke; Vascular smooth muscle cells (VSMCs).
Copyright © 2025 Elsevier B.V. All rights reserved.
PubMed Disclaimer
Conflict of interest statement
Declaration of competing interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
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How does an exponent work when it's less than one?
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Modified 8 years, 11 months ago
Viewed 28k times
$\begingroup$
I'm rather familiar with exponents, I know that $y^x = y_1 \cdot y_2 \cdot y_3 .... y_x$, but what if the exponent is less than one, how would that work?
I put in my computer $25^{1/2}$ anyway, expecting it to give me an error, and I got an answer!! And even more surprising, when I did this with more numbers and a little research, I found that $$x^{1/y} = \sqrt[y] x$$ Is it just me, or can exponents take on the role of square roots?
exponentiation
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edited Oct 7, 2016 at 2:34
FlostinFlostin
asked Oct 6, 2016 at 19:27
FlostinFlostin
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$\endgroup$
12
6
$\begingroup$ Generally, an exponent between 0 and 1 is a "decimal root", of which the most commonly known are the square and cubed root. So your equation is correct. When you get to calculus, you'll learn that the equation $f(x) = x^a$, where $a$ is any real constant, has a bunch of ways to define it, usually using infinite polynomials. $\endgroup$
Michael Stachowsky
– Michael Stachowsky
2016-10-06 19:29:27 +00:00
Commented Oct 6, 2016 at 19:29
$\begingroup$ Yes, when $y$ is a natural number and $x$ is positive, the equality you wrote is right. If $x<0$ it's still true but $y$ has to be odd. $\endgroup$
Augustin
– Augustin
2016-10-06 19:31:34 +00:00
Commented Oct 6, 2016 at 19:31
11
$\begingroup$ Wait until you find out that $1.5!$ is defined as well... $\endgroup$
chepner
– chepner
2016-10-06 22:48:18 +00:00
Commented Oct 6, 2016 at 22:48
1
$\begingroup$ @chepner Bah, Python doesn't think so. ValueError: factorial() only accepts integral values (gamma(2.5) works, though.) $\endgroup$
JAB
– JAB
2016-10-07 00:55:01 +00:00
Commented Oct 7, 2016 at 0:55
3
$\begingroup$ Now... Look into negatives... Then negative fractions. Mind blown :) $\endgroup$
WernerCD
– WernerCD
2016-10-07 04:00:31 +00:00
Commented Oct 7, 2016 at 4:00
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7 Answers 7
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$\begingroup$
You're right that there is something interesting going on. It's certainly true that $25^{1/2} = 5$ is a bit different than $3^2 = 3 \cdot 3 = 9$.
One of the reasons why we've chosen $25^{1/2}$ to mean $\sqrt{25}$ goes like this.
For normal, positive integer exponents, we have the really nice exponent laws $a^b \cdot a^c = a^{b+c}$. For instance, $3^3 \cdot 3^5 = 3^8$. What if we wanted this law to work even when we used numbers less than $1$, or maybe fractions between integers?
Then we would want $25^{1/2} \cdot 25^{1/2} = 25^{\frac{1}{2} + \frac{1}{2}} = 25^1 = 25$, and so we would want $25^{1/2} = \sqrt {25}$. This works with others, too. For instance, $27^{1/3}$ should satisfy $27^{1/3} \cdot 27^{1/3} \cdot 27^{1/3} = 27^1 = 27$, so that $27^{1/3} = \sqrt{27} = 3$.
For this also works for fractions bigger than $1$. Thus $36^{3/2}$ should satisfy $36^{3/2} \cdot 36^{3/2} = 36^3$, so $36^{3/2} = \sqrt{36^3}$. Note that you could also rationalize this as thinking that $36^{3/2}$ should equal $(36^3)^{1/2}$, and indeed that works here!
More generally, choosing $25^{1/2}$ to mean $\sqrt {25}$ (and the related identities) agrees with all of our previous rules regarding exponents, so it seems like a very natural choice to make. And indeed, it is the choice we make.
As an aside: one can further extend exponents to decimal expansions instead of fractions, or go further still.
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edited Oct 6, 2016 at 23:20
Eff
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$\begingroup$ +1 It's amazing how many of our unusual laws in mathematics can be traced back to "We tried to make things consistent, and the result turned out to be useful... so we kept it." $\endgroup$
Cort Ammon
– Cort Ammon
2016-10-07 00:41:13 +00:00
Commented Oct 7, 2016 at 0:41
4
$\begingroup$ @CortAmmon I think pretty much every definition or theorem in some sense is of that form. Definitions are attempts at drawing our attention to phenomena. Theorems are explanations of phenomena. $\endgroup$
A. Thomas Yerger
– A. Thomas Yerger
2016-10-07 00:56:36 +00:00
Commented Oct 7, 2016 at 0:56
1
$\begingroup$ @MaskedMan: See my linked post. Alfred's point is exactly on target. It is not meaningful to simply lay down rules, but as mixedmath implicitly said we need to check that if we want to extend some operation to a larger domain they had better not contradict the previous rules! That is why a lot of definitions need to be justified by theorems. In this case we extend the old structure (reals with integer exponentiation) to a new one (reals with rational exponentiation), and have to prove that the new structure satisfies the desired properties. $\endgroup$
user21820
– user21820
2016-10-07 14:15:17 +00:00
Commented Oct 7, 2016 at 14:15
3
$\begingroup$ @MaskedMan Same reasoning... $x^1 \cdot x^{-1} = x^0 = 1$, so we should have $x^{-1} = \frac1x$. $\endgroup$
Barry
– Barry
2016-10-07 14:20:14 +00:00
Commented Oct 7, 2016 at 14:20
2
$\begingroup$ @MaskedMan Exact same thing. $x^{-\frac12} = x^{-1 \cdot \frac12} = (x^{-1})^{\frac12} = {\frac1x}^{\frac12}=\sqrt{\frac1x}$. $\endgroup$
Barry
– Barry
2016-10-07 15:24:55 +00:00
Commented Oct 7, 2016 at 15:24
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One thing we notice right away about $b^n; n \in N$ is that $b^nb^m = b^{n+m}$ (This is obvious because $b^n = \underbrace{b\cdot b\cdots b}_{n\text{ times}}$ and $b^m = \underbrace{b\cdot b\cdots b}_{m\text{ times}}$, so $b^nb^m = \underbrace{b\cdot b\cdots b}_{n\text{ times}} \cdot \underbrace{b\cdot b\cdots b}_{m\text{ times}} = \underbrace{b\cdot b\cdots b}_{n+m\text{ times}} = b^{n+m}$).
And that $(b^n)^m = \underbrace{b^n\cdot b^n\cdots b^n}_{m \text{ times}} = b^{n+n+...+n} = b^{n\cdot m}$.
So if we want to extend the definition of $b^n$ so that $n$ is not just a natural number but maybe $n = 0$ or $n < 0$ or $n \in \mathbb Z$ we realize that we want to define it so that $b^0b^n = b^{0+n} = b^n$. That means we must define $b^0 = b^n/b^n = 1$. (We really have not choice).
We also want it so that if $0 < n < m$ then $b^{m-n}=b^mb^{-n}$ so $b^{-n} = \frac{b^{m-n}}{b^m}= \frac{b^{m-n}}{b^{(m-n)+n}} = \frac {b^{m-n}}{b^{m-n}b^n} = \frac 1 {b^n}$.
So we must define $b^{-n} = 1/b^n$. (We really have no choice.)
Now we also have $(b^n)^m = b^{nm}$ this means $ \sqrt[m]{b^{nm}} = b^n$. This really isn't that surprising. After all $\sqrt[m]{b^{nm}} =\sqrt[m]{(b^{n})^m} = b^n$, after all.
But what if we aren't talking about whole integers? What if $\sqrt[m]{b^n} = \sqrt[m]{(b^{nm/m})}=\sqrt[m]{(b^{n/m})^m} = b^{n/m}$. Does that make any sense at all?
Well, it makes perfect sense if $m$ divides $n$ and $n/m$ is an integer.
But if $n/m$ then .... we haven't defined what $b^{n/m}$ means if $n/m$ isn't an integer.
But why shouldn't we define $b^{n/m}$ if $n/m$ isn't an integer? If we define $b^{n/m} = \sqrt[m]{b^n}$ that is a fine definition[@]. And because of our rules $(b^r)^n = b^{rn}$ we really have no choice. We must define it that way.
So $a^{1/2} = \sqrt{a}$. This is because if $a^{1/2} = x$ then $x^2 = (a^{1/2})^2 = a^{\frac 12 2} = a^1 = a$. So $x = \sqrt{a}$.
[@] Actually, we have to show that if $r = m/n = p/q$ then $\sqrt[n]{b^m} = \sqrt[q]{b^p}$ as it turns out that is true. $m/n = p/q$ mean $mq= np$ and $\sqrt[n]{b^m} = \sqrt[q]{\sqrt[n]{b^m}^q}=\sqrt[nq]{b^{mq}} = \sqrt[nq]{b^{np}} = \sqrt[q]{\sqrt[n]{(b^p)^n}} = \sqrt[q]{b^p}$
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edited Oct 7, 2016 at 14:33
Barry
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fleabloodfleablood
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$\begingroup$ It amazes me that although this question is asked constantly, amazing answers such as this kept getting posted. On the one hand, extreme redundancy....but on the other hand, I somehow continue to learn new things from answers such as this one! As an unrelated tangent, newer versions of MathJax make root symbols so much more beautiful than they were in the past. And that makes me happy :D $\endgroup$
Brevan Ellefsen
– Brevan Ellefsen
2016-10-07 03:13:00 +00:00
Commented Oct 7, 2016 at 3:13
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$\begingroup$
For integer exponents we have the property that $(y^a)^b = y^{ab}$ (among others). In order to define $y^q$ for rational $q$, we want it to preserve this property. Writing $q = a/b$ with $a,b$ integers and $b$ positive, we would like $$ (y^q)^b = (y^{a/b})^b = y^{(a/b)b} = y^a. $$ So we would like $y^q$ to be a $b$th root of $y^a$. This always exists if $y$ is positive. So we define $$y^{a/b} = \sqrt[b]{y^a} \quad(y \ge 0).$$ Then it can be verified that the other properties of exponentiation are satisfied for rational exponents, so this is a good definition.
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answered Oct 6, 2016 at 19:40
arkeetarkeet
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$\begingroup$
$$y^x = y\cdots y$$
only if $x$ is a positive integer.
In general:
$$y^x = \exp(x\log y)$$
provided that $y >0$ (the functions $\exp$ and $\log$ can be defined via power series).
Also one can prove:
$$\sqrt[x]{y} = y^{1/x}$$
as you noted (this can also be taken as the definition of roots).
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answered Oct 6, 2016 at 19:32
Stefan PerkoStefan Perko
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$\begingroup$ OP: Note that this requires the definition of $e^x$ where $x$ is a real, not necessarily an integer. This can be done as, for instance, $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$. (This is in turn shorthand for $e^x = \lim_{n \to \infty} \sum_{k=0}^n \frac{x^k}{k!}$.) $\endgroup$
Brian Tung
– Brian Tung
2016-10-06 19:35:50 +00:00
Commented Oct 6, 2016 at 19:35
3
$\begingroup$ @BrianTung It does get tricky though. You don't want $x^k$ to use the definition in this answer, since that leads to a recursive definition. At the same time, you don't want different definitions for $y^x$ depending on whether $x$ is an integer. Luckily, there are other ways of expressing $e^x$ that do not rely on exponentiation, and they can be as simple as spelling out $\frac{x^k}{k!}$ in your definition with repeated multiplication (e.g. $\prod_{i=1}^k{\frac xi}$). $\endgroup$
hvd
– hvd
2016-10-07 11:25:51 +00:00
Commented Oct 7, 2016 at 11:25
$\begingroup$ @hvd: Furthermore, note that complex exponentiation (as defined here) is incompatible with real exponentiation (that defines $(-1)^{1/3} = -1$). One has to be clear what one is dealing with! $\endgroup$
user21820
– user21820
2016-10-07 14:23:15 +00:00
Commented Oct 7, 2016 at 14:23
$\begingroup$ @hvd: Oops! Yup. I totally forgot what the original question was! $\endgroup$
Brian Tung
– Brian Tung
2016-10-07 16:57:44 +00:00
Commented Oct 7, 2016 at 16:57
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2
$\begingroup$
This isn't really a proof but just another way to look at it.
If you raise $a^b$ to the $n$th power, i.e. $(a^b)^n$, you can calculate this by multiplying the exponents:
$$(a^b)^n = a^{bn}.$$
Now consider the case of $n = 1/b$:
$$(a^b)^{1/b} = a^{b\cdot \frac1b} = a^1 = a.$$
So if I square a number ($b=2$), and then get the number back after I raised it to the $1/2$ power, it sure looks like I took the square root.
In more general terms, raising to the $1/b$th power is the inverse operation of raising to the $b$th power (as long as we stay away from $b=0$).
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answered Oct 6, 2016 at 19:45
JohnJohn
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$\begingroup$
I'd like to weigh in with how we define numbers, to give a big picture of it all.
Start with the natural numbers, define a starting point and the notion of a successor.
0 0+1 0+1+1 ...
Now define addition as repeated succession, and multiplication as repeated addition. Easy enough.
Now we can answer questions such as
2+6=? 26=?
But we can also ask
2+?=6 2?=6
We now define 6-2 as "the number which when added to 2 gives 6 and similarly define 6/2 as "the number which when multiplied by 2 gives 6.
We have a problem when we try to do things such as 6+?=2 because we only have the natural numbers defined so far. But here we say "I define the negative numbers to exist such that they obey the rules of addition and subtraction as above."
6+?=2 ?=2-6=0-4=-4
And the number -4 is defined into existence Similarly
6?=2 ?=2/3
We have now defined the fractions. They are imaginary numbers which are only "real" insofar as they are consistent with the rules.
Likewise we define surds: $\sqrt{2}$ is the number such that $\sqrt{2}\times\sqrt{2}=2$.
We can define exponentiation as repeated multiplication and solve the problem of "what is the nth root of something" by defining it as "it's whatever thing that, when you multiply it by itself n times, gives you that something." That's what your question is essentially about.
$x^{-3}=\frac{1}{x^3}$ because $x^{-3}=x^{-1\times 3}=(x^{-1})^3=(\frac{1}{x})^3=\frac{1}{x^3}$ because we defined $x^{-1}=\frac{1}{x}$ and then followed the established rules.
The natural conclusion to this is to define imaginary numbers such as $i=\sqrt{-1}$ and this gives us an algebraic closure which means we don't have to invent any other kinds of number in order to answer any algebraic question. (You can invent other kinds of number but they have their own algebras.)
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answered Oct 7, 2016 at 13:59
spraffspraff
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Also consider this: $25 = 5 \cdot 5$.
So if we take "half" of $25$, written as a multiplication, then it certainly makes sense that we get $5$ as the result. Other answers show the general extrapolation of this idea.
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answered Oct 7, 2016 at 3:28
Daniel R. CollinsDaniel R. Collins
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Disproving asymptotic lower bound of central binomial coefficient
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I was recently learning about binomial coefficients and was wondering about how to disprove 2nCn (or the central binomial coefficient) not being lower-bounded by 4^n; in other words:
Some extremely generous bounds can be easily constructed, such as the following:
I sought to prove by contradiction, so to assume:
Clearly, c1 cannot exist, since 1/(2n + 1) approaches 0 as n approaches infinity. It can also be seen that c2 must reside in (0, 1]. And... I'm stuck. Intuitively, it seems rather obvious that c2 cannot exist.
I am aware a similar question has been asked here, but there wasn't really a proof provided. I'm also aware that you could prove the limit of 2nCn/4n approaches 0 as n approaches infinity, but I was wondering if there was another way to do so - particularly, by proving that c2 cannot exist.
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asked Jan 30, 2021 at 20:40
calquathatercalquathater
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The question really belongs to math exchange. That said, 4^n is so generous, that c4^n will eventually outgrow the binomial coefficient no matter how small c is (so yes, it cannot exist). People chose c = 1, for the inequality to hold even for n == 1.
user58697
– user58697
2021-01-30 21:46:15 +00:00
Commented Jan 30, 2021 at 21:46
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The easiest approach is to write the choose in factorials and use en.wikipedia.org/wiki/Stirling%27s_approximation to prove that the central coefficient is actually O(2^n / sqrt(n)). If you do it carefully you are on your way to proving some special cases of the central limit theorem.
btilly
– btilly
2021-01-30 21:52:28 +00:00
Commented Jan 30, 2021 at 21:52
Oops, I meant O(4^n / sqrt(n)). The peak for n choose m is O(2^n/sqrt(n)) but your n is 2 the one that I was thinking of.
btilly
– btilly
2021-01-31 02:22:32 +00:00
Commented Jan 31, 2021 at 2:22
@user58697 I definitely agree with you, but I'm seeking something more formal. Thanks for the suggestion though, I'll try posting it to math exchange.
calquathater
– calquathater
2021-01-31 07:28:35 +00:00
Commented Jan 31, 2021 at 7:28
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For all n, the constant c2 would have to be upper bounded by
2n choose n (2n)! ----------- = ------------- 4^n 2^n n! 2^n n! (2n)! = ------------- (2n)!! (2n)!! (2n-1)!! = -------- (2n)!! n (2i-1) = product ------ i=1 2i n = product (1 - 1/(2i)) i=1 n ¤ product exp(-1/(2i)) [since 1 + x ¤ exp(x)] i=1 n = exp(sum -1/(2i)) i=1 ¤ exp(-ln(n+1)/2) [since sum ¤ integral of increasing fn] = 1/(n+1),
hence it cannot be positive.
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edited Jan 31, 2021 at 13:53
answered Jan 31, 2021 at 6:39
David EisenstatDavid Eisenstat
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8389 | https://www.reddit.com/r/GRE/comments/1k0octx/consecutive_integers/ | Consecutive integers : r/GRE
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Consecutive integers
I am struggling to find a straightforward answer about the definition of “consecutive integers.” Sometimes the 5 pound book uses consecutive integers as increasing by 1 only (1,2,3,4.. or -5,-4,-3,-2…) and other times they will take into account other evenly spaced sets (2,4,6… or -3, 0, 3, 6) and call it “consecutive integers” while also failing to acknowledge sets increasing by 1 and vice versa. It’s starting to get frustrating as it can really change the answer! Truly I just need to know how ETS defines consecutive integers on the GRE, and I’m having a hard time finding an answer! I appreciate any help :)
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weighted arithmetic mean
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weighted arithmetic mean
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definition
In mean
…a more general average, the weighted arithmetic mean. If each number (x) is assigned a corresponding positive weight (w), the weighted arithmetic mean is defined as the sum of their products (wx) divided by the sum of their weights. In this case,
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8391 | https://www.zapstudio.com/Assets_Publish/EE_for_Tech_2nd_Ed_Lab_7_8.pdf | 23 Experiment 7: DC Measurements and Meter Loading Introduction Meters used to measure voltage or current have an internal resistance. Since the meter must be connected to the circuit to make a measurement, the circuit is changed by the resistance of the meter. This is referred to as the meter's "loading effect". Most meters have amplifiers built in so that only a small amount of power is needed from the circuit, and the loading effects are minimized. But this is not always the case. A voltmeter is connected in parallel with the circuit across which the voltage is being measured. The loading effect of the meter will be minimal if the meter resistance is much larger than the circuit resistance. Ideally, the voltmeter resistance should be infinite. Practically, most electronic voltmeters (DMMs) have a resistance of 10 megohms. The effect of the voltmeter resistance, Rm, across a circuit element, Rc, can be calculated using the parallel resistor equation: Rc Rm Rt Rc Rm ⋅ = + An ammeter is connected in series with the circuit in which the current is being measured. The loading effect of the meter will be minimal if the meter resistance is much smaller than the circuit resistance. Ideally, the ammeter resistance should be zero. Practically, the resistance of most electronic ammeters (DMMs) varies with the range setting, with the highest ranges having the least resistance. This is one case where setting the range for the most significant digits may not always result in the most accurate reading. The effect of the ammeter resistance, Rm, in series with circuit element, Rc, can be calculated using the series resistor equation: Rt Rc Rm = + Even if the meter loading effect is insignificant there will be an uncertainty in the measured value due to measurement errors caused by the accuracy of the meter. Objectives A series-parallel circuit will be connected on a solder-less breadboard. Voltage and current measurements will be made on the circuit and the results will be compared to theoretical expectations. The effect of the resistance of the meter will be determined. Procedure Equipment and Parts DMM, Power Supply, and Breadboard. Resistors: 100, 220, 1 Meg, 2.2 Meg, ¼ watt, 5%. ©2011 ZAP Studio All rights reserved Sample Experiment 1. M m R R 2. C . L it U co N le th co T 3. M (V V 4. O 1. T in Y a w re Measure and measured res R1: _ R3: _ Connect the c ay out the c looks simila Use the min onnect the c Note that th ead is conn he that nega onnected to The voltmete Measure and Vab = Va – V Va: _ Obtain and re The circuit on n series with You need to nd R2 and i will read the esistors. record the v sistor values _ _ circuit below circuit on th ar to the sch nimum num circuit. he positive nected to th ative power the bottom r is connecte record Va, Vb). _ ecord the int n the right sh the 100Ω a o break the nsert the me current thro Part A: Volt values of the s for your ca R2: ___ R4: ___ w on the left a e breadboa hematic diag mber of wire power s he top of R1 supply lea of R2. ed across R Vb and Vab Vb: ___ ternal resista Part B: Cur hows the DM nd 220Ω res connection eter as show ough the 10 24 tage Measu e resistors in lculations an _ _ and set the rd so gram. es to upply 1 and ad is R4. b _ ance, Ri, of y rrent Measu MM connecte sistors. n between R wn. The met 00Ω and 220 urement n the circuit. nd for LTspic _ power supp __ V your voltmet urement ed R1 ter 0Ω You will us ce input in e ly voltage to Vab: ter. Ri: se these experiment 8 o 6.0V. __ _ 8. __ __ ©2011 ZAP Studio All rights reserved Sample Experiment 25 2. The internal resistance of a DMM on the current ranges varies with the range. Check the manual on your DMM to see what its internal resistances are on the current ranges. If you don't have a manual, your instructor should provide the information. Record the meter resistances below. Rm(0.2mA): __ Rm(2mA): __ Rm(20mA): ___ Rm(200mA): __ 3. Measure and record the current through the 100Ω and 220Ω resistors with the meter on the 20mA range. Ia20: ____ 4. Measure and record the current through the 100Ω and 220Ω resistors with the meter on the 200mA range. Ia200: _______ Analysis, Part A 1. Calculate the theoretical values of the voltages: Va, Vb, and Vab, without taking the meter resistance into account. Calculate the percent error between the theoretical and measured results. 2. Calculate the voltages: Va, Vb, and Vab taking the meter resistance into account. Calculate the percent error between the measured and calculated results. Analysis, Part B 1. Calculate the theoretical current, Ia, (without taking the meter resistance into account). Calculate the percent error between the theoretical and measured values (both current ranges). 2. Calculate the current, Ia, taking the meter resistance into account. Calculate the percent error between the calculated and measured results. 3. Briefly explain the significance of the error analysis in steps 1 and 2 above. ©2011 ZAP Studio All rights reserved Sample Experiment 26 Experiment 8: LTspice Circuit Simulation Introduction LTspice IV will be used to simulate the circuit of experiment 7. It is important to always use your measured part values so that you can compare your measured results to your simulation results. Drawing the Circuit 1. Start the LTspice program. In the main menu bar, click on File and select New Schematic. 2. Create the new schematic. Left click on the resistor symbol and drag and place the resistor, R1. Get three more resistors and place them as shown on the right. Right click on the “R” of each resistor. Change the values to the measured values from Experiment 7. 3. Left click on the gate symbol between the diode and the hand in the main menu to get the dialog box on the right. Select the part “voltage”. place it in the schematic. Click on the ground symbol and place it under R2. 4. Left click on the pencil in the main menu to connect the components. 5. Left click on the A between the ground symbol and the resistor symbol to label the nodes N1, N2, and N3 as shown on the schematic by dropping the dot on the nodes. Click “Help” in the menu for more information on creating the schematic. ©2011 ZAP Studio All rights reserved Sample Experiment 27 6. Click on Simulate and then select the Edit Simulation Cmd in the menu. Select DC op pnt asshown on the right. Click on ok. A “.op” command will appear which can be placed anywhere on the schematic. 7. Click on Simulate and then select Run. If there are no errors, you will see an “Operating Point” file. The Spice netlist shows the connections of the parts, part models, and types of analysis to be performed. Exercise and Analysis 1. Simulate the loading effect of the voltmeter at nodes 2 and 3 using a resistor whose resistance is equal to the meter’s internal resistance. Compare simulated results to Experiment 7 measured results. 2. Simulate the loading effect of the ammeter using a resistor whose resistance is equal to the meter’s internal resistance. Compare simulated results to Experiment 7 measured results. ©2011 ZAP Studio All rights reserved Sample Experiment |
8392 | https://os.copernicus.org/preprints/os-2018-19/os-2018-19.pdf | 1 Thermodynamic Properties of Seawater, Ice and Humid Air: TEOS-10, Before and Beyond Rainer Feistel1 1Leibniz Institute for Baltic Sea Research (IOW), Warnemünde, D-18119, Germany Correspondence to: Rainer Feistel (Rainer.Feistel@IO-Warnemuende.de) 5 Abstract. In the terrestrial climate system, water is a key player in the form of its different ambient phases of ice, liquid and vapour, admixed with sea salt in the ocean and with dry air in the atmosphere. For proper balances of climatic energy and entropy fluxes in models and observation, a highly accurate, consistent and comprehensive thermodynamic standard framework is requisite in geophysics and climate research. The new “Thermodynamic Equation of Seawater – 2010” (TEOS-10) constitutes such a standard for properties of water in its various manifestations in the hydrological cycle. TEOS-10 10 has been recommended internationally in 2009 by the Intergovernmental Oceanographic Commission (IOC) to replace the previous 1980 seawater standard, EOS-80, and in 2011 by the International Union of Geodesy and Geophysics (IUGG) “as the official description for the properties of seawater, of ice and of humid air”. This paper briefly reviews the development of TEOS-10, its novel axiomatic properties, new oceanographic tools it offers, and important tasks that still await solutions by ongoing research. Among the latter are new definitions and measurement standards for seawater salinity 15 and pH, in order to establish their metrological traceability to the International System of Units (SI), for the first time after a century of widespread use. Of similar climatological relevance is the development and recommendation of a uniform standard definition of atmospheric relative humidity that is unambiguous and rigorously based on physical principles. 20 The leading thermodynamic properties of a fluid are determined by the relations which exist between volume, pressure, temperature, energy, and entropy… But all the relations existing between these five quantities for any substance … may be deduced from the single relation existing for that substance between volume, energy, and entropy. 25 Josiah Willard Gibbs, 1873b Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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2 1 Introduction In the context of recent global warming and the anthropogenic greenhouse effect of carbon dioxide (CO2), the pivotal article of Svante Arrhenius (1896) found public attention well beyond the scientific communities of meteorologists and climatologists. Much less known, however, is the lecture of Heinrich Hertz given in 1885 in which he analysed the thermodynamics of the hydrological cycle in the climate system as a “gigantic steam engine” (Mulligan and Hertz, 1997, p. 5 41). In fact, rather than CO2, water in the troposphere in the form of humidity and clouds contributes the major part to the overall greenhouse effect (Abbot and Fowle, 1908; Emden, 1913; Trenberth et al., 2007; Lacis et al., 2010; Schmidt et al., 2010; Feistel and Ebeling, 2011; Feistel 2015, 2017; Lovell-Smith et al., 2016). The global water cycle, its observation and modelling, poses a fundamental challenge for climate research (Sherwood et al., 2010; Reid and Valdés, 2011; Tollefsen, 2012; Fasullo and Trenberth, 2012; Josey et al., 2013; Stevens and Bony, 2013; IPCC, 2013). 10 Saline water in the oceans, freshwater lakes and rivers, polar ice caps, humid air and clouds form a highly dynamic, coupled system of water in different phases and mixtures. Densities, heat capacities and the latent heats of mutual phase transitions of water play a key role for the transformation and distribution processes of energy between initial absorption of incoming solar radiation and final export of thermal radiation. A comprehensive, consistent quantitative knowledge of water properties in its 15 various appearances is requisite for the analysis of measurements and the development of numerical models. Moreover, climate research is carried out at various places of the world and extends over timespans of several human generations. Mutual comparability of research results is indispensable, which rigorously requires that well-defined, highly accurate international standards must be applied, such as the International System of Units (SI). 20 TEOS-10, the Thermodynamic Equation of Seawater – 2010 constitutes such an international standard for the thermodynamic properties of water in the climate system. In addition to seawater in particular, it also covers the properties of ice and humid air in a perfectly consistent, comprehensive way with unprecedentedly high accuracy. By the numerical coefficients of its empirical thermodynamic potentials, TEOS-10 represents in a mathematically most compact way the results of an enormous amount of very different experimental studies of water, ice, seawater and air properties. TEOS-10 has 25 been recommended internationally in 2009 by the Intergovernmental Oceanographic Commission (IOC) to replace the previous 1980 seawater standard, EOS-80, and in 2011 by the International Union of Geodesy and Geophysics (IUGG) “as the official description for the properties of seawater, of ice and of humid air”. Beginning in 2006, TEOS-10 was developed in close cooperation between the SCOR/IAPSO Working Group 127 on Thermodynamics and Equation of State of Seawater, and the International Association for the Properties of Water and Steam (IAPWS). TEOS-10 is officially defined by its 30 manual (IOC et al., 2010) and is supported by open-source software libraries (available from the internet at the TEOS-10 homepage, www.teos-10.org), several open-access IAPWS documents (IAPWS AN6-16, 2016) as well as various supporting scientific articles, such as the definition of the new Reference-Composition Salinity Scale (Millero et al., 2008) or a new Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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3 thermodynamic definition of relative humidity (Feistel and Lovell-Smith, 2017). A collection of background articles of TEOS-10 was published as a Special Issue of Ocean Science (Feistel et al., 2008b). This paper is organised as follows. Thermodynamic potentials are not necessarily taught in courses of oceanography or other geosciences, so Section 2 provides a short introduction to this theoretical method developed about 150 years ago. The most 5 convenient mathematical formalism of manipulating partial derivatives with different sets of variables, as necessary for the work with those potentials, is the Jacobi method explained in Appendix A. Section 3 describes details of the development of the thermodynamic potentials for fluid water, ice, seawater and humid air, which constitute the core of TEOS-10, together with the scales used for temperature and salinity. Section 4 deals with the structure of TEOS-10, its axiomatic approach to the computation of thermodynamic properties, and the digital, open-source libraries provided by TEOS-10. Members of the 10 SCOR/IAPSO Working Group 127, who participated in the annual meetings and successfully completed these demanding tasks within a few years of intense cooperative work, can be found in Appendix B. Section 5 describes three important pending problems that still await solutions beyond TEOS-10. The relatively large number of references cited in this review includes numerous sources that are of relevance in the context 15 of TEOS-10 and belong to a variety of different scientific disciplines. Albeit, these references do not even include the extensive lists of numerous experimental works that provided the fundamental property data of water, seawater, ice and humid air, ultimately being represented by TEOS-10 in an integrated, compact form. 2 Thermodynamic Potentials A thermodynamic potential is a mathematical function from which all thermodynamic properties of a given many-particle 20 system at equilibrium can be derived using formal thermodynamic rules. The system under study may be a pure substance, such as water or ice, a mixture such as seawater or humid air, or a multi-phase composite such as sea ice (ice with brine pockets) or a cloud (air with liquid droplets). J. Willard Gibbs discovered the existence of such functions already in 1873 (Gibbs, 1873a,b), but their practical use remained limited before powerful computers became commonplace in science. In contrast, thermodynamic potentials have always played a key role in theoretical physics (Landau and Lifschitz, 1966). The 25 partition functions of statistical thermodynamics are mathematical expressions for the Boltzmann entropy, the Helmholtz potential and the Landau potential, respectively, obtained from the micro-canonical, canonical and grand-canonical ensembles. Depending on which external conditions are imposed on a given sample, an associated potential function needs to be chosen. 30 For example, if we know the temperature T, the volume V and the total mass m, the potential to be used is the Helmholtz energy, F(m, T, V), or similarly, as a function of temperature and density, ρ, the specific Helmholtz energy, f (T, ρ) = F/m. A Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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4 typical case of such conditions is an experimental setup of a liquid or gas in a given rigid container, in order to measure the pressure or the isochoric heat capacity at different temperatures or densities. In geophysics, such as in the atmosphere or in the ocean, typically the pressure p rather than the volume of a fluid parcel is available. Then, the Gibbs energy G(m, T, p), or similarly, the specific Gibbs energy, g(T, p) = G/m, is the appropriate potential from which properties of interest, such as the density, entropy or enthalpy of the parcel, can be computed. For brevity, the terms Gibbs function and Helmholtz function, 5 respectively, are used in this paper for the specific Gibbs and Helmholtz energies expressed in terms of their particular natural independent variables. The two potentials may be converted into each other, if either f(T, ρ) is known, by 𝑔(𝑇, 𝑝) = 𝑓+ 𝜌 𝜕𝑓 𝜕𝜌 , 𝑝= 𝜌2 𝜕𝑓 𝜕𝜌 . (1) or, if g(T, p) is given, by 𝑓(𝑇, 𝜌) = 𝑔−𝑝 𝜕𝑔 𝜕𝑝 , 𝜌−1 = 𝜕𝑔 𝜕𝑝 . (2) 10 Transformation equations of this kind between different sets of independent variables are commonly known as Legendre transforms (Margenau and Murphy, 1943; Landau and Lifschitz, 1966; Alberty, 2001). Another potential function of significant interest in TEOS-10 is the specific enthalpy h(η, p) as a function of the specific entropy, η, and the pressure. It is derived from a TEOS-10 Gibbs function g(T, p) by the Legendre transform ℎ(𝜂, 𝑝) = 𝑔−𝑇 𝜕𝑔 𝜕𝑇 , 𝜂= − 𝜕𝑔 𝜕𝑇 , (3) 15 and from a TEOS-10 Helmholtz function f(T, ρ) by ℎ(𝜂, 𝑝) = 𝑓−𝑇 𝜕𝑓 𝜕𝑇+ 𝜌 𝜕𝑓 𝜕𝜌 , 𝜂= − 𝜕𝑓 𝜕𝑇 , 𝑝= 𝜌2 𝜕𝑓 𝜕𝜌 . (4) The partial derivatives are understood here as being taken at constant values of the respective other, so-called „natural“ independent variables associated with a given potential. 20 Potential functions, such as F, f, G, g or h above, may also be expressed in terms of alternative independent variables, but then they may lose their capabilities as potential functions from which all other quantities may be derived. For example, entropy can only incompletely be computed from the function h(T, p). The “original” thermodynamic potential studied by Gibbs, see his initial quotation, was internal energy as a function of volume and entropy; this function is rarely used in geophysics or engineering, mainly because its input parameter entropy cannot be measured experimentally (Tillner-Roth, 25 1998). Among the most relevant measurable properties derived from the potential functions above are density, ρ, 1 𝜌= 𝑉 𝑚= ( 𝜕𝑔 𝜕𝑝) 𝑇= ( 𝜕ℎ 𝜕𝑝) 𝜂, (5) isobaric specific heat capacity, cp, 𝑐𝑝= ( 𝜕ℎ 𝜕𝑇) 𝑝= 𝑇( 𝜕𝜂 𝜕𝑇) 𝑝= −𝑇( 𝜕2𝑔 𝜕𝑇2) 𝑝, (6) sound speed, c, 30 Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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5 1 𝑐2 = ( 𝜕𝜌 𝜕𝑝) 𝜂= −𝜚2 ( 𝜕2ℎ 𝜕𝑝2) 𝜂= 𝜚2 ( 𝜕2𝑔 𝜕𝑝𝜕𝑇) 2 −(𝜕2𝑔 𝜕𝑇2) 𝑝 (𝜕2𝑔 𝜕𝑝2) 𝑇 (𝜕2𝑔 𝜕𝑇2) 𝑝 , (7) and the adiabatic lapse rate, Γ, 𝛤= ( 𝜕𝑇 𝜕𝑝) 𝜂= 𝜕2ℎ 𝜕𝜂𝜕𝑝= 𝜕2𝑔 𝜕𝑇𝜕𝑝 (𝜕2𝑔 𝜕𝑇2) 𝑝 . (8) Here, for uniqueness of the mathematical expressions, the subscripts at the brackets indicate the variables that are kept constant when the partial derivative is carried out. Extended lists of such thermodynamic relations for the various quantities 5 implemented in TEOS-10 are available from the manual (IOC et al., 2010) and from the digital supplements of Feistel et al. (2010a) and Wright et al. (2010a). The conversion of formulas from one set of independent variables to another is done using the chain rule of differential calculus. In a convenient multi-variable version, this rule is known as the Jacobi method. Appendix A demonstrates this method tutorially and how it is applied to derive formulas like (5)-(8) in a straightforward manner, such as eq. (A.18) for the sound speed. 10 In order to quantitatively determine a thermodynamic potential of a given substance of interest, a suitable but largely arbitrary mathematical expression for the potential function is designed by trial and error. The function depends on a set of N adjustable parameters, say a = (a1, … aN), which are estimated by simultaneous numerical regression with respect to all reliable experimental data available for the properties (5) – (8) and others. For each substance involved, two of those 15 parameters remain unknown, representing the absolute energy and absolute entropy of that substance, unavailable from thermodynamic measurements. These parameters are usually specified by conventionally assuming certain values to selected quantities at a chosen reference state. In TEOS-10, internal energy and entropy of liquid water are assumed to take zero values at the common solid-liquid-gas triple point. Similar assumptions were also employed for dry air and dissolved sea salt, see Section 3. For consistency, it is important to implement for any given substance the same conventional values in any 20 mixture or phase where it is present, such as for the water substance H2O in ice, in seawater and in humid air (Feistel et al., 2008a). 3 Fundamental Thermodynamic Potentials of TEOS-10 3.1 Temperature Scales For the measurement of temperatures, temperature-dependent properties of certain materials are frequently used, such as the 25 volume expansion of liquid mercury or the electrical resistance of platinum. Sensors of this kind need to be calibrated, that is, a quantitative relation between that property and selected numerical temperature values must be specified. For the currently valid International Temperature Scale of 1990 (ITS-90), “the unit of the fundamental physical quantity known as thermodynamic temperature, symbol T, is the kelvin, symbol K, defined as the fraction 1/273.16 of the thermodynamic Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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6 temperature of the triple point of water” (Preston-Thomas, 1990). In practice, many published geophysical property equations are expressed in terms of earlier, meanwhile obsolete scales, such as the International Practical Temperature Scale of 1968, IPTS-68. Among those obsolete equations are, in particular, the International Equation of State of Seawater, EOS-80, and the Practical Salinity Scale of Seawater, PSS-78 (Unesco, 1981a,b, 1983). 5 TEOS-10 is expressed in terms of ITS-90. Experimental data were converted from their original values according to the transformation rules published by Rusby (1991), Goldberg and Weir (1992), Rusby et al. (1994), and Weir and Goldberg (1996). In the vicinity of the triple point of water, the deviations between different temperature scales are generally small, in particular for oceanographic applications (Saunders, 1990; Feistel and Hagen, 1995; Feistel, 2008a). ITS-90 is expected to remain in practical use for the foreseeable future. 10 On 20 May 2019, the World Metrology Day, a so-called “new SI” is expected to become formally introduced at the International Bureau of Weights and Measures (BIPM) in Sèvres near Paris. Along with it, a new thermodynamic temperature scale will be established in 2018 (Fellmuth et al., 2016; PTB, 2017; CCT, 2017; CGPM, 2018). This new temperature T and the ITS-90 temperature T90 will be considered as two different physical quantities expressed in the same 15 unit, namely the new kelvin (rather than being the same physical quantity expressed in two different units, the K and a fictitious “K90”). The definition of the kelvin contained in the 1990 definition will be superseded by the new kelvin definition in terms of the Boltzmann constant and the joule (CGPM, 2018). In ITS-90, the triple-point temperature of water will remain at exactly T90 = 273.16 K. This value will also be valid for the thermodynamic temperature but only within an uncertainty yet to be specified, likely about 160 µK. By definition, the Boltzmann constant, k = 1.380 649 ∙ 10–23 J K–1, the Avogadro 20 constant, NA = 6.022 140 76 ∙ 1023 mol–1, and the molar gas constant, R = k ∙ NA, will take eternally fixed, exact values in the new SI (Fischer, 2016; PTB, 2017). This is a very practical future aspect as the various IAPWS documents supporting TEOS-10 implement several slightly different values of the gas constant, according to varying official values it took at the time the particular document was developed (IAPWS G5-01, 2016). 3.2 Helmholtz Function of Fluid Water 25 The joint Helmholtz function, f F(T, ρ), of fluid water published by Wagner and Pruß (2002), also known as the IAPWS-95 equation, is identical with the Helmholtz functions of liquid water, f F ≡ f W, and water vapour, f F ≡ f V, and represents a key constituent of TEOS-10. In compact form, the latest version of this equation is available from IAPWS R6-95 (2016). Since its first release in 1995, the latter document has undergone several minor revisions, however, without affecting the calculated values of any measurable properties. The equation is also available from several other references, such as IOC et al. (2010) 30 and Feistel et al. (2010b). The formulation is valid in the entire stable fluid region of H2O from the melting curve to 1273 K at pressures up to 1000 MPa; the lowest temperature on the melting curve is 251.165 K at 208.566 MPa (Wagner et al., 2011; IAPWS R14-08, 2011), see Fig. 1. As an additional part of TEOS-10, a low-temperature extension of the IAPWS-95 Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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7 equation for water vapour down to 50 K is given by Feistel et al. (2010b) and IAPWS G9-12 (2012). IAPWS-95 is valid for air-free water with an isotopic composition of Vienna Standard Mean Ocean Water, VSMOW (IAPWS G5-01, 2016). Helmholtz functions are preferred description tools for fluids because they may cover the wide density range from the gas to the liquid phase by a single, unique formula. In contrast, Gibbs functions take multiple values in the T-p vicinity of the 5 saturation curve where liquid and gas coexist, so that in practice two separate Gibbs functions, gW and gV, respectively, are specified for liquid and vapour. Consistent with TEOS-10, there is an equation for the Gibbs function for liquid water under oceanographic conditions (Feistel, 2003; IAPWS SR7-09, 2009), an equation for the Gibbs function for liquid water at atmospheric pressure (Hrubý et al., 2009; IAPWS SR6-08, 2011), and a virial expansion of the Gibbs function for water vapour (Feistel et al., 2015). Recently, a Gibbs function for supercooled, metastable liquid water has also been developed 10 (IAPWS G12-15) which is not included in TEOS-10. For a pure substance, the chemical potential equals its Gibbs function. Consequently, along the saturation curve, psat(T), in the (T, p) diagram from the common triple point to the critical point of water, see Fig. 1, the Gibbs function of liquid water equals the Gibbs function of water vapour, gW(T, psat) = gV(T, psat). From this equilibrium condition, the saturation pressure 15 as a function of the temperature, psat(T), can be calculated iteratively. A simple analytical approximation formula of this solution is available (IAPWS SR1-86, 1992). In their common ranges of validity, IAPWS-95 is consistent with the most accurate metrological equation of state of liquid water, the so-called CIPM-2001 equation (Tanaka et al., 2001; Harvey et al., 2009; IAPWS AN4-09, 2009). This means that 20 under ambient conditions, density is known within an uncertainty of 1 ppm. This accuracy permits the use of IAPWS-95 for the calibration of high-precision density meters (Wolf, 2008). By means of such instruments, seawater salinity may be determined from measured densities, and sea-salt composition anomalies may be detected (Millero et al., 2008; Feistel et al., 2010c; Wright et al., 2011; Woosley et al., 2014; Pawlowicz et al., 2016; Schmidt et al., 2016, 2018; Budéus, 2018), see also Section 3.4. 25 In TEOS-10, the IAPWS-95 equation replaced the earlier equations of state of liquid water of Bigg (1967) and Kell (1975), on which the former seawater standard EOS-80 was based (Unesco, 1981b). This change of the pure-water equation made it possible to resolve systematic problems previously encountered with the sound speed of seawater at high pressures (Dushaw et al., 1993; Millero and Li, 1994; Feistel, 2003). 30 3.3 Gibbs Function of Ice Ih Hexagonal ice I is the only stable solid phase of water in the terrestrial atmosphere, hydrosphere and cryosphere. Several textbooks on ice were published in the past, such as by Dorsey (1968), Hobbs (1974) or Petrenko and Whitworth (1999), Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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8 offering separate empirical equations for selected properties. The first Gibbs functions for ice Ih were developed by Feistel and Hagen (1995, 1998) and Tillner-Roth (1998), valid in the vicinity of the melting curve. Including numerous additional data, the two approaches were combined by Feistel and Wagner (2004, 2005, 2006, 2007; IAPWS R10-06, 2009) for an IAPWS equation for ice Ih that covers a wide range on temperatures and pressures and is significantly more accurate than several single-property equations published before. Theoretically, a Helmholtz function for ice Ih can smoothly be combined 5 with that for fluid water into a single 3-phase Helmholtz function for water, valid everywhere near the triple point, but such an attempt has not yet been made. Measurement of ice properties is experimentally challenging in particular for sluggish aging effects; if ice is brought under different temperature or pressure conditions, the establishment of the related new equilibrium crystal structure may take 10 hours or even days till completion. Consequently, measurement series conducted too quickly may suffer from the risk of systematic hysteresis effects. Published data on ice compressibility vary up to a factor of 3.6 between different authors. In TEOS-10, the related uncertainty of mechanical measurements could be reduced by a factor of 100 to about 1 % (Feistel and Wagner, 2004, 2005) using crystallographic spectroscopy data (Gammon et al., 1980, 1983; Gagnon et al., 1988), see Fig. 2. The chemical potential of a pure substance equals its Gibbs function. Consequently, along the freezing curve in the (T, p) 15 diagram, Fig. 1, the Gibbs function of ice Ih equals the Gibbs function of liquid water, and along the sublimation curve, the Gibbs function of ice Ih equals the Gibbs function of water vapour. From these equilibrium conditions, the freezing pressure and the sublimation pressure as functions of the temperature can be calculated iteratively. Convenient analytical approximation formulae are available for both curves (Wagner et al., 2011; IAPWS R14-08, 2011). 20 It is well-known that the latent heats of evaporating or freezing water possess exceptionally large values, and that this fact plays an important role in the climate system. The latent heat of sublimation is less familiar, even though it amounts to the sum of the latent heats of melting and evaporation. Air is dry under the polar high-pressure cells of the troposphere, and sublimation rates of glaciers, icebergs or polar ice caps are significant there. Antarctic dry valleys are famous for their extreme conditions. Sublimation fluxes are approximately driven by relative humidity, which in turn is governed by the 25 sublimation pressure of ice. In TEOS-10, the sublimation pressure can be calculated directly from the thermodynamic potentials of ice and vapour; it turned out that the uncertainty of those computed TEOS-10 values of the sublimation pressure is typically about a factor of 10 smaller than the uncertainty of the most accurate direct measurements, such as those of Fernicola et al. (2012) or Bielska et al. (2013). The limitation of accuracy of the computed values results mainly from the uncertainty of the heat capacity data of ice and water vapour that entered the thermodynamic potentials (Wagner et al., 30 2011). Moreover, estimated upper and lower bounds of the heat capacity values of water vapour result directly in corresponding lower and upper bounds for the computed sublimation temperature at given pressure (Feistel and Wagner, 2007). The TEOS-10 equation extends to much lower temperatures than other standard formulas, such as that of Murphy and Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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9 Koop (2005). For these reasons, TEOS-10 and its sublimation-pressure equation are preferable for geophysical applications such as models of ice clouds down to temperatures of 80 K (Lübken et al., 2009). Another distinctive feature of the TEOS-10 ice equation is its exceptional accuracy in ITS-90 freezing temperatures of air-free water, such as 273.152 519 K at atmospheric pressure, 1013.25 hPa, within an uncertainty of just 2 µK (Feistel and 5 Wagner, 2005, 2006; Feistel, 2012, Feistel et al., 2016). For air-saturated water, this freezing point is lowered to the common ice point of 273.150 019 K with an estimated uncertainty of 5 μK (Harvey et al., 2013). 3.4 Salinity Scales Seawater salinity was defined and measured in different ways in the past, and expressed in different units. TEOS-10 is formulated in terms of Absolute Salinity, SA, defined as the mass fraction of sea salt dissolved in water, usually expressed in 10 grams per kilogram (IOC et al., 2010). For IAPSO Standard Seawater (SSW), the best estimate available for its Absolute Salinity is the Reference-Composition Salinity, SR, for short simply Reference Salinity, calculated from the atomic weights of the chemical constituents of sea salt (Millero et al., 2008). Present-day oceanographic measurement devices, however, report values of Practical Salinity, SP, which are computed from the electrical conductivity as defined by the Practical Salinity Scale of 1978, PSS-78 (Unesco 1981a, 1983). Practical Salinity is a unitless quantity, but because there are various other obsolete 15 or non-standard, unitless salinity measures, some oceanographers prefer to attach “psu” (for “practical salinity unit”) to reported figures of SP. This inofficial use of “psu” does not cause any harm but has been subject to controversial emotional debates (Unesco 1985, 1986; Millero, 1993). As an aside, a similar discussion is currently going on regarding the use of “%rh” to denote unitless values of relative humidity (RH) expressed in percent (Lovell-Smith et al., 2016) for an easier distinction between RH and, say, specific humidities or relative uncertainties of RH. 20 For SSW, Reference Salinity is obtained from Practical Salinity by the formula SR = 35.165 04 g kg–1 SP/35 ≈ SP ∙ 1.004 715 g kg–1. (9) For historical data, this equation can only be applied to salinity measurements carried out after 1978. Older legacy data are preferably converted from readings of chorinity, Cl, by 25 SR = 35.165 04 ∙ 1.806 55 Cl/35 ≈ 1.815 069 Cl. (10) Chlorinity is defined as the mass ratio of grams of pure silver necessary to precipitate the halogens (that is, chlorine, bromine, iodine) from 328.5234 g of seawater (Jacobsen and Knudsen, 1940; Sverdrup, 1942; Millero et al., 2008). By this definition, the chlorinity value is slightly greater than the mass fraction of chlorine in seawater, wCl = 0.998 9041 Cl (Millero et al., 2008). From the 19th century on, chlorinity values were rather consistently defined and determined by titration as the 30 most accurate available salinity measure (Forch et al., 1902; Carritt, 1963; Lyman, 1969; Lewis, 1980; Pawlowicz et al., 2016; Burchard et al., 2018). In contrast, the unsatisfactory situation with the definition of salinity before PSS-78 was that “nobody has measured salinity for 60 years. At present chlorinity, conductivity and refractive index measurements are all Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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10 being converted to ‘salinity’ by inadequate tables of often doubtful origin. All make assumptions regarding constancy of relative proportions of the various ions, which are doubtful and may be quite unjustified, to the precision of our modern measurements” (Unesco, 1976, p. 2 therein). Absolute Salinity as defined in the context of TEOS-10 is the first officially revised measure of the mass fraction of salt 5 dissolved in seawater (McDougall et al., 2008; Pawlowicz et al., 2016) after that of Knudsen and Sørensen more than a century ago (Forch et al., 1902). For seawater of standard composition, the Reference Salinities from eqs. (9) and (10) should result in numerical values that agree with each other within measurement uncertainty. Then, Absolute Salinity SA is estimated from SA = SR (11) 10 to be used as salinity input variable for TEOS-10 equations. However, salinities determined alternatively from both conductivity and chlorinity measurements may produce mutually inconsistent results. This happens especially if the chemical composition of seawater deviates from that of SSW, such as in the Baltic Sea (Wirth, 1940; Kwiecinski, 1965; Rohde, 1966; Millero and Kremling, 1976; Feistel et al., 2010c) or regionally in the global ocean (Millero et al., 1976, 2008, 2009, 2011; Tsunogai et al., 1979; Millero, 2000; Pawlowicz, 2010; Wright et al., 2011; Ushida et al., 2011; McDougall et 15 al., 2012; Woosley et al., 2014). Before TEOS-10, the international standards attempted to avoid inconsistent results caused by using different methods of salinity determination. With the introduction of PSS-78, the further use of chlorinity titration was discouraged for this reason (Lewis and Perkin, 1978). TEOS-10 is the first international seawater standard where chemical composition anomalies are 20 explicitly accounted for. In cases when eqs. (9) and (10) do not match, the Reference Salinity (9) is to be corrected by SA = SR + δSA , (12) instead of eq. (11), from density measurements or other suitable information, such as for the estimated actual composition. The required correction, δ𝑆A = 1 𝛽{ 𝜌meas 𝜌SW(𝑆R) −1}, (13) 25 may be estimated from the measured density, ρmeas, and the computed TEOS-10 seawater density, ρSW, evaluated at the Reference Salinity and the same T and p, given by eq. (9), see Fig. 3. Here, β is the haline contraction coefficient of seawater which has a typical value of about 0.66 at 35 g kg–1, atmospheric pressure and 300 K. As a revival of previous, so-called specific-gravity methods (Forch et al., 1902; Krümmel, 1907), this density-based approach received appreciation again for the high accuracy of modern vibrating-tube density meters (Kremling, 1971; Wolf, 2008; Schmidt et al., 2016, 2018). 30 While the validity of Practical Salinity is constrained to values of SP between 2 and 42 (Unesco, 1981c), the new Reference Salinity can be determined also for higher concentrations because it is - by its definition - a physical mass fraction, rather Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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11 than a merely formal measure associated with the conductivity. Concentrated brines may be diluted with pure water until the salinity falls in the validity range of PSS-78. Then, SP is measured, converted to SR and, using the amount of added water and mass-conserving mixing rules, the Reference Salinity of the original sample is calculated (Millero at al., 2008). 3.5 Gibbs Function of Seawater Thermodynamic properties of a Gibbs function for seawater were first discussed theoretically by Fofonoff (1962). However, 5 in the context of the development of EOS-80, no attempt had been reported of actually constructing such a function numerically. The first practically implemented Gibbs function (Feistel, 1991, 1993; Feistel and Hagen, 1994) was consistent with EOS-80 and, beyond that, consistently provided additional properties such as entropy, enthalpy and the chemical potentials of water and sea salt that had partly been made available elsewhere before (Millero and Leung, 1976; Unesco, 1981b; Millero, 1983). 10 An improved potential version (Feistel and Hagen, 1995) addressed several weaknesses of its predecessor as well as of EOS-80, such as conversion to the ITS-90 temperature scale. Additional data were included to improve the equation in the vicinity of the temperature of maximum density (Caldwell, 1978; Siedler and Peters, 1986) and for deep-water sound speeds (Dushaw et al., 1993; Millero and Li, 1994). For a corrected asymptotic zero-salinity limit, the 1995 Gibbs function, gSW, 15 𝑔SW(𝑆P, 𝑇, 𝑝) = 𝑔W(𝑇, 𝑝) + 𝑔1(𝑇)𝑆Pln𝑆P + 𝑔2(𝑇, 𝑝)𝑆P + 𝑔3(𝑇, 𝑝)𝑆P 3/2+ … (14) recalculated the logarithmic ideal-solution term, g1, and Debye’s limiting law of dilute electrolytes, g3, of the salinity expansion, eq. (14), explicitly from an exactly electro-neutral chemical composition model of sea salt. As a precursor of the Reference Composition of TEOS-10 (Millero et al., 2008), this 1995 stoichiometry was a slightly modified version of that defined by Millero (1982). The new Gibbs function provided a reliable thermodynamic basis for the calculation of novel 20 oceanographic tools such as Conservative Temperature (McDougall 2003), of new efficient equations for numerical models (McDougall et al., 2003), and for the clarification of some misleading explanations for the adiabatic lapse rate, eq. (8), given elsewhere (McDougall and Feistel, 2003). Combined with a simple Gibbs function of ice, see Section 3.3, various properties of sea ice such as freezing point, melting heat or brine salinity could, for the first time, be calculated perfectly consistently with all other experimental data on seawater and ice involved (Feistel and Hagen, 1998). 25 Seawater properties are strongly related to those of the pure-water solvent, as given by the Gibbs function gW in eq. (14), and indirectly also by seawater properties measured with instruments that had in advance been calibrated using certain older, possibly inconsistent, pure-water equations. The extremely accurate Helmholtz function for pure water published by Wagner and Pruß (2002), the so-called IAPWS-95 equation, see Section 3.2, offered the chance for a fundamental update of the pure-30 water part of the 1995 Gibbs function of seawater. Moreover, it also suggested consistency corrections of the underlying experimental seawater data to be carried out with respect to their respective, obsolete pure-water references, as far as these dependencies could be inferred from the related publications. The resulting new Gibbs function for seawater (Feistel, 2003) Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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12 was still expressed in terms of Practical Salinity. For the pure-water part, a Gibbs function had been fitted to IAPWS-95 over the oceanographic ranges of temperature and pressure. Updated ocean-model algorithms were derived from the 2003 equation by Jackett et al. (2006). At that time, the various substantial advantages of the 2003 Gibbs function over the still valid EOS-80 standard, together 5 with sophisticated new oceanographic tools such as Conservative Temperature, unavailable from EOS-80, sparked the idea of developing a new international seawater standard. In order to investigate this question, at the 2005 SCOR/IAPSO meeting at Cairns, Australia, “IAPSO had introduced a proposal for a new working group, and that was approved by SCOR as Working Group WG127 on Thermodynamics and Equation of State of Seawater. This work is expected to provide important input into modeling of the global ocean and, ultimately, climate change modeling” (IAPSO, 2005; SCOR, 2005). On the 10 other hand, the Gibbs function of seawater was also presented at the 2005 annual IAPWS meeting at Santorini, Greece. The idea of developing an IAPWS equation for seawater was welcomed there, and resulted in establishing an IAPWS Task Group on Seawater, and in sending a letter to IAPSO expressing the IAPWS interest in working jointly on this problem (IAPWS, 2005). 15 Chaired by Trevor McDougall, WG127 held its inaugural meeting in 2006 at Warnemünde, Germany, see Appendix B. The WG confirmed the desirability of cooperating with IAPWS on a jointly agreed seawater standard defined by a Gibbs function, suitable for both oceanographic and industrial applications, such as seawater desalination. IAPWS-95 was adopted by WG127 as the pure-water reference for the future thermodynamics of seawater. Detailed numerical checks of the 2003 Gibbs function were initiated. For the standard-ocean reference state (SP = 35, T = 273.15 K, p = 101 325 Pa), the WG 20 specified entropy and enthalpy of seawater to vanish. The WG recommended the inclusion of additional available data at normal pressure (heat capacities, dilution heats, freezing points) for an extended Gibbs function. The WG had an extensive discussion of Absolute Salinity, SA, see Section 3.4. If the thermodynamic functions could be defined in terms of SA instead of Practical Salinity SP, this would have the advantage that density calculated from the 25 thermodynamic formulation would be a better fit to actual seawater density because it would reflect compositional changes. If defined by a fixed conversion factor for a reference composition, SA is as accurate as SP and fully compatible with present measuring techniques. SA for a reference composition has an exact relation to traditional chlorinity. All physical, chemical and oceanographic, theoretical as well as numerical, models do actually rely on SA rather than SP. Outside oceanography, SA is the only way the scientific community recognises salinity. Salinity in technical/industrial applications (IAPWS) is 30 traditionally understood as Absolute Salinity. The WG gave serious consideration to the idea that the fundamental description should change from Practical Salinity to Absolute Salinity, but postponed decisions on this issue for its wide-range implications (McDougall et al., 2008). However, already in September 2006 some WG members started intense email Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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13 activities around the clock and week, across complementary time zones around the globe, aiming at the specification of a best estimate for the chemical composition of Standard Seawater. In 2007, at the second WG127 meeting held in Reggio, Calabria, see Appendix B, draft articles had already been prepared regarding the Reference Composition (Millero et al., 2008), and an updated Gibbs function (Feistel, 2008a) based on that 5 composition and expressed in terms of Absolute Salinity, SA, which for SSW equals the Reference-Composition Salinity, SR, see eq. (11). While the pressure-independent part (responsible for freezing point, dilution heat, etc.) of the 2008 Gibbs function was significantly extended in its T-SA validity range, the pressure-dependent part (responsible for density and its derivatives) remained identical with that of 2003. WG127 endorsed the use of the IAPWS 2006 Release on an equation of state for H2O Ice Ih (IAPWS R10-06, 2009). Later, the new salinity definition and the 2008 Gibbs function were also 10 presented at the 2007 IAPWS meeting at Lucerne, Switzerland. On this basis, a “Release on the IAPWS Formulation for the Thermodynamic Properties of Seawater” had been drafted for evaluation, with approval anticipated for 2008 (IAPWS, 2007). While salinity rarely exceeds 40 g kg–1 in the global ocean, more concentrated brines are found, for example, in the lagoon-15 like Australian Shark Bay (up to 70 g kg–1, Logan and Cebulsk, 1970), and in particular in brine pockets of polar sea ice at low temperatures. Due to the latent-heat contributions of melting or freezing pockets, the heat capacity of sea ice is significantly larger than that of either pure ice or seawater. Such high-salinity effects could properly be estimated by so-called Pitzer equations (Feistel and Marion, 2007), and so there was a reasonable interest in implementing this ability also in TEOS-10. The new Reference-Salinity Scale supported this extension beyond the validity range of the previous Practical 20 Salinity, see Section 3.4, and at the same time raised the question of solubilities of the dissolved salts (Marion et al., 2009). It turned out that salinities up to 110 g kg–1 of Antarctic sea ice are well covered by the 2008 Gibbs function (Feistel et al, 2010a), see Fig. 4. At the International Conference on the Properties of Water and Steam 2008 in Berlin, the 2008 Gibbs function was adopted 25 as an IAPWS standard (IAPWS R13-08, 2008), in combination with the IAPWS-95 equation which provides the pure-water part gW(T, p) of eq. (14) in implicit form. On the basis of mainly these two documents, TEOS-10 was endorsed by IOC/UNESCO in 2009 at Paris as a new international seawater standard “to replace EOS-80 and thus updating this valuable, but no longer state-of-the-art, 30-year-old UNESCO standard” (IOC, 2009, page 5 therein; Wright et al., 2010b; IUGG, 2011; Valladares et al., 2011a,b). 30 For industrial applications, two modifications of the 2008 Gibbs function for seawater were developed later. First, the validity of a new density equation (Feistel, 2010) at atmospheric pressure could be extended to temperatures up to 90 °C and Absolute Salinities up to 70 g kg–1 using new measurements of Millero and Huang (2009). Second, for an industrial seawater Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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14 standard the “scientific” Helmholtz function forming the pure-water part was replaced by an “industrial” IAPWS Gibbs function for liquid water (Wagner and Kretzschmar, 2008; Kretzschmar et al., 2015; IAPWS AN5-13, 2016). The 2008 Gibbs function for seawater relies mainly on older experimental data of the 1960s and 1970s. Meanwhile, various new measurements of seawater properties, also under conditions outside the validity of TEOS-10, have been published, such 5 as of density (Millero and Huang, 2009; Safarov et al., 2009, 2010, 2012, 2013) and of sound speed (Millero and Huang, 2011; von Rohden et al., 2015, 2016; Lago et al., 2015). Further available data are reviewed by Sharkawy et al. (2010) and Nayar et al. (2016). 3.6 Helmholtz Function of Humid Air The flux of water across the ocean-atmosphere interface belongs to the most important processes of the global climate 10 system, but its estimated contribution to the total heat loss of the ocean varies greatly between 50 % (Emery et al., 2006) and 90 % (Wells, 2012). The thermodynamic driving force for evaporation is the difference between the chemical potentials of water in seawater and in humid air. Usually, this difference is approximated by the relative humidity of air (Kraus and Businger, 1994). This estimate is known to be in error by typically 2 % because of the lowered vapour pressure of seawater compared to that of pure water. To get an idea of the relevance of this error, note that a small change of the global latent heat 15 flux by 1 %, or about 1 W m-2, would by a factor of 200 exceed the flux responsible for the currently observed greenhouse warming of the atmosphere, about 5 mW m-2 (Lovell-Smith et al., 2016; Feistel, 2015, 2017). The estimated energy imbalance of 0.4 – 0.8 W m-2 of the warming ocean (Cheng et al., 2016) is also within this error range. However, on the other hand, routine meteorological measurement of relative humidity has a typical uncertainty between 1 and 5 %rh (Lovell-Smith et al., 2016), thus being insufficient to recognise those crucial heat flux anomalies. 20 In order to consistently include in TEOS-10 also thermodynamic properties of the air-sea interface, such as latent heat of evaporation or relative humidity at equilibrium with seawater, the first intention was to simply adopt a thermodynamic potential for humid air from the scientific literature. Unfortunately, as it turned out, such a function had never been developed yet by the atmospheric and humidity communities. Similar to the situation with seawater, only certain collections 25 of separate empirical property equations of unclear mutual consistency were available (Goff and Gratch, 1945; Sonntag, 1966; Linke and Baur, 1970; Gill, 1982; Gatley, 2005; Murphy and Koop, 2005; WMO, 2008). So it was necessary to construct a new TEOS-10 Helmholtz function for humid air, f AV, from a minimum number of available, mutually independent but internally consistent, empirical de-facto standard functions, 𝑓AV(𝐴, 𝑇, 𝜌) = 𝐴𝑓A(𝑇, 𝐴𝜌) + (1 −𝐴)𝑓V(𝑇, (1 −𝐴)𝜌) + 𝐴(1 −𝐴)𝑓mix(𝐴, 𝑇, 𝜌). (15) 30 Here, A is the mass fraction of dry air in humid air, ρ is the mass density of humid air, f A is the Helmholtz function of dry air of Lemmon et al. (2010), f V is the IAPWS-95 Helmholtz function of water vapour, see Section 3.2, and f mix consists of second (Harvey and Huang, 2007) and third (Hyland and Wexler, 1983) cross-virial coefficients for air-water interaction. Eq. Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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15 (15) and some resulting properties, such as for the latent heat of seawater, see Fig. 5, or for the entropy of clouds, see Fig. 6, were first discussed at the 2009 meeting of WG127 at Arnhem, The Netherlands, see Appendix B. In 2010, the Helmholtz function (15) for humid air was eventually adopted as an official IAPWS formulation at Niagara Falls, Canada (IAPWS G8-10, 2010; Feistel et al., 2010b), and presented at the Portorož symposium on temperature and humidity metrology (Feistel, 2012). 5 In their common ranges of validity, the IAPWS equation of humid air is consistent with the metrological high-accuracy CIPM-2007 equation (Picard et al., 2008) for the density of humid air. The latter paper outlines previous problems with the molar mass of dry air, which differs slightly also between the equations of Feistel et al. (2010b) and IAPWS G8-10 (2010). Unlike the original definition of Lemmon et al. (2010), the TEOS-10 equation for dry air specifies zero entropy and zero 10 enthalpy of dry air at the standard ocean surface, 0 °C and 101 325 Pa, as reference-state conditions. Ideal-gas approximations and their related adjustable constants, consistent with TEOS-10, are reported in Feistel et al. (2010b). For most atmospheric applications, small corrections to the ideal-gas equations in the form of virial coefficients are sufficiently accurate. For this purpose, from eq. (15) a numerically more convenient virial Gibbs function of humid air can be derived (Feistel et al., 2015; IAPWS G11-15, 2015). A recent review of thermodynamic equations for humid-air properties is given 15 by Herrmann et al. (2017). 4 Extracting Properties from TEOS-10 The four fundamental thermodynamic potentials of TEOS-10, see Fig. 7, as described in Section 3, possess axiomatic properties of consistency, independence and completeness. An axiomatic approach to formally defining and representing all thermodynamic properties with respect to a minimum common set of basic functions may avoid confusion, may more easily 20 permit identification and quantification of differences between seemingly equivalent quantities such as various alternative definitions of relative humidity, and may establish solid thermodynamic links between quantities that were originally introduced separately and independently (Feistel et al., 2016), such as correlation equations for the heat capacity and for the sublimation pressure of ice, see Section 3.3. 25 Because of consistency, it is impossible to derive from TEOS-10 different results for one and the same thermodynamic quantity. Maxwell’s cross-relations are always identically fulfilled, rather than just approximately. In contrast, previous collections of separate property equations are not necessarily consistent. As an example, from the EOS-80 collection (Unesco, 1983) one can compute the heat capacity directly by one given equation, but also indirectly by given equations for density and sound speed, and those two results differ, especially near the temperature of maximum density. 30 Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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16 Independence expresses the fact that no part of the four fundamental potentials of TEOS-10 may be derived from other parts. In particular, if a more accurate update is available for any of those parts, independence permits replacing an obsolete part without adjustments to be made for the other parts. In contrast, previous collections of separate property equations are not necessarily independent. As an example, the EOS-80 equation for the pressure dependence of the heat capacity is related by a Maxwell relation to the temperature dependence of the density. It is therefore not advisable to update one equation of such 5 a collection without carefully considering any possible side effects this may have on the other equations. Completeness of thermodynamic potentials was discovered by Gibbs, see the title-page quotation. As soon as for a given equilibrium system a thermodynamic potential is available, no matter which one, see Section 2, all its thermodynamic properties can be derived by merely mathematical manipulations. The system may even be a chemically reacting one or a 10 multi-phase composite. In contrast, previous collections of separate property equations are not necessarily complete. As an example, the EOS-80 collection does not provide entropy, enthalpy or chemical potentials of seawater. Note, however, that there are some minor, usually irrelevant, exceptions to these general rules in TEOS-10. For example, the thermodynamic potentials are not perfectly consistent as they use slightly different values for the molar gas constant that 15 were the international reference values at the time the particular equation was developed. The thermodynamic potentials are not entirely independent as they commonly depend, for example, on the definition of the temperature scale, see Section 3.1, and on the definitions of reference-state conditions, see Section 2. Completeness does not apply to, say, the sound speed in ice, which is typically anisotropic in crystalline solid states while the TEOS-10 equation describes ice as an isotropic substance. 20 Mathematical formulas for the computation of numerous quantities are derived and explained in the TEOS-10 Manual (IOC et al., 2010) and in several IAPWS documents supporting TEOS-10 (IAPWS AN6-16, 2016). Large subsets of those quantities are implemented in two open-source libraries, the Sea-Ice-Air (SIA) library (Feistel at al., 2010a; Wright et al., 2010a), and the Gibbs-Seawater (GSW) library (McDougall and Barker, 2011; McDougall et al., 2012; Roquet et al., 2015). 25 Source code in several programming languages is freely available from the TEOS-10 web site, www.teos-10.org. In the SIA library, thermodynamic potentials are numerically available for single-phase and multi-phase systems such as those displayed in Fig. 7, together with various properties such as heat capacities, densities or entropies of those systems. As a first step, for a convenient calculation of properties at given temperature, pressure and composition, Gibbs functions are 30 implemented for seawater, gSW(SA, T, p), and humid air, gAV(A, T, p), together with 9 additional functions each, for any of their 1st and 2nd partial derivatives. For this purpose, equations for the pressure as a function of the density, 𝑝= 𝜌2(𝜕𝑓𝜕𝜌 ⁄ ), available from the Helmholtz functions, see Section 2, must be inverted iteratively to find the density as a function of the given pressure. The specific entropy, η, of a sample with given in-situ temperature T at a pressure p is then available from the Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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17 related Gibbs function by evaluating 𝜂= −𝜕𝑔𝜕𝑇 ⁄ . As a second step, enthalpies hSW(SA, η, p) and hAV(A, η, p) of seawater and humid air, respectively, are implemented by executing another numerical iteration procedure. From those functions, for a given reference pressure pref, potential enthalpy, hθ = h(η, pref), potential temperature, 𝜃= 𝜕ℎ(𝜂, 𝑝ref) 𝜕 ⁄ 𝜂, and potential density, 𝜌𝜃 −1 = 𝜕ℎ(𝜂, 𝑝ref) 𝜕 ⁄ 𝑝 , can be computed. Note that all input and output quantities of the SIA library are expressed in basic SI units, such as temperatures in K (rather than °C), pressures in Pa (rather than dbar relative to the surface 5 pressure), and salinity or dry-air fraction in kg kg–1 (rather than in psu or g kg–1 or %). This strict convention avoids the need for any unit conversions along with the various mathematical manipulations performed internally. Such conversions are error-prone; to see this, imagine to modify the sound speed formula (7) for the case that the pressure is measured in decibars rather than pascals. 10 Similar to the potential functions and properties of single-phase systems, the SIA library implements also Gibbs functions and enthalpies of composite systems, such as sea ice, consisting of seawater and ice, or clouds, consisting of liquid water and humid air, at their mutual thermodynamic phase equilibria. The SIA library is organised in the form of several separate modules. It is possible to select an axiomatic subset of such modules as a sub-library, such as one just for ice, or one for humid air, as discussed in the digital supplement of Feistel et al. (2016). It is also possible to combine SIA modules with 15 additional user-defined modules for, say, specific functions that do not yet belong to the standard library set. For example, such add-ons may be developed for including the effects of dissolved air on water properties which are currently neglected in TEOS-10. For frequent calls within ocean models, the stacked iterations of the SIA library are too slow, and the variables and their 20 units are inconvenient. For these reasons, the GSW library implements fast and tailored equations in terms of appropriate input parameters, accepting oceanographically familiar units, such as pressure in decibars, Absolute Salinity in grams per kilogram, and Conservative Temperature as potential enthalpy expressed in °C. The price for this advantage is the introduction of new empirical equations with additional regression coefficients obtained by fitting with respect to data calculated from the SIA library equations. Potential future updates of any fundamental TEOS-10 equation, see Fig. 7, will 25 automatically propagate into the various derived SIA functions, but will require updated sets of coefficients of the GSW library. Detailed descriptions of the oceanographic quantities and their use in the context of TEOS-10 are available from IOC et al. (2010), McDougall and Barker (2011), McDougall et al. (2013) and Roquet et al. (2015). 5 Problems Beyond TEOS-10 TEOS-10 was developed as a state-of-the-art thermodynamic framework supporting geosciences in their tasks of observing, 30 understanding, modeling and predicting the global climate system and its long-term variations. Successful cooperation between climate-related research groups anywhere on the planet, extending over several human generations, demands mutual Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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18 comparability of their respective individual theoretical and measurement results. This indispensable requirement appears to be self-evident, but unfortunately it is by no means always fulfilled yet, and in certain cases it even poses a serious challenge still to be addressed in the future. All relevant quantities should be uniformly defined, preferably by commonly agreed international standards or recommendations. All measurement results of the same kind of quantity should be metrologically traceable to the same reference, preferably specified as an international standard of highest temporal stability. The actual 5 quantities of interest, such as seawater salinity, should be related to their associated measurands, such as temperature and conductivity, by standard equations of certified estimated uncertainty. Related pending problems were presented and discussed at the 2010 WMO-BIPM workshop on Measurement Challenges for Global Observation Systems for Climate Change Monitoring in Geneva, Switzerland (WMO, 2010). 10 According to the rules for SCOR/IAPSO Working Groups, WG127 was disbanded in 2011 (Pawlowicz et al., 2012). During the development of TEOS-10, relevant problems had become visible regarding the definitions of seawater salinity, seawater pH, and relative humidity of moist air. In order to address these problems along with maintaining TEOS-10, a standing Joint Committee on the Properties of Seawater, JCS, was commonly established by SCOR, IAPSO and IAPWS in 2012 (IAPWS, 2012). At the BIPM at Sèvres in 2011 and 2012, related meetings on a potential future cooperation took place, and 15 subsequently a joint workshop of JCS was held with representatives of the BIPM at the 2013 International Conference on the Properties of Water and Steam in Greenwich, UK (Feistel, 2013; IAPWS, 2013; Hellmuth et al., 2014; Pawlowicz et al., 2014; Feistel et al., 2016). Recent progress and upcoming tasks are planned to be on the agenda of the 2018 International Conference on the Properties of Water and Steam in Prague, Czech Republic. 5.1 Seawater Salinity 20 During the development of TEOS-10, much attention was paid to the question of how salinity should be defined in oceanography (McDougall et al., 2008; Millero et al., 2008; Seitz et al., 2011; Wright et al., 2011; Pawlowicz et al., 2016). In current practice, salinity measurement is performed by electronic sensors for pressure, temperature and conductivity. Those sensors usually consist of some arrangement of electronic devices, such as a platinum resistor, and typically return signals in the form of an electric voltage, a current, or a frequency. In a calibration lab, the sensor is immersed in seawater of 25 well-defined conditions, and its output signal values are associated with the particular temperatures, pressures or salinities of the bath. To establish those well-defined conditions, the lab needs to measure the bath temperature and to use seawater samples of certified salinities. The lab’s thermometers are regularly calibrated against temperature standards realised in national metrological institutes, which in turn implement the rules specified for the ITS-90 definition of the kelvin within the International System of Units (SI). Certified Standard Seawater samples are commercially produced and distributed by the 30 IAPSO Standard Seawater Service operated by OSIL, Ocean Scientific International Ltd., a company located near Portsmouth, UK. At OSIL, seawater samples collected from the North Atlantic are purified, diluted and compared against the conductivity of a potassium chloride (KCl) solution as specified by 1978 Practical Salinity Scale. These chains of Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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19 calibration procedures establish the so-called metrological traceability of measurement results (de Bievre and Günzler, 2005; Seitz et al., 2011; VIM3, 2012). Traceability to the same primary reference is a necessary condition for the comparability of measured values from different devices, at different locations or times. While temperature measurement is traceable to the SI, salinity is traceable to manufactured, possibly varying or aging, artefacts in the form of certain KCl solutions or certified SSW batches (Seitz et al., 2011). 5 For long-term series such as required for climate research, the temporal stability of the primary references is of ultimate relevance. It must be granted that tiny trends in observed values constitute real changes rather than just spurious effects, caused by drifts of measurement devices or standards. In order to achieve the highest metrological stability available, traceability to the SI is the preferred choice. However, traceability to the SI is of sufficient advantage only if the related 10 measurement uncertainty is small enough. For example, absolute conductivity measurements are traceable to the SI, but their uncertainty is about one order of magnitude too large compared to the PSS-78 standard, and defining salinity by a specified conductivity number rather than by a mass fraction of a reference KCl solution, would in practice let small but significant observed salinity trends disappear in point clouds of noisy scatter. Based on such arguments, after many long and vivid discussions, the 2008 meeting of WG127 at Goetz near Berlin, see Appendix B, came to the conclusion that currently, 15 seawater density is the only promising candidate for SI-traceable salinity measurements in the oceanographic practice (Seitz et al., 2011; Wright et al., 2011; Pawlowicz et al., 2016). TEOS-10 has already established density measurement as a secondary method for the determination of Absolute Salinity in cases of expected seawater composition anomalies (Millero et al., 2008; Feistel at al., 2010c), see Section 3.4. The concept 20 recently developed by JCS (Pawlowicz et al., 2016) aims at a specification of certified Standard Seawater samples not only by their Practical Salinities as presently, but in the future also in parallel by their densities, measured in a way that is traceable to the SI. This approach would leave the current oceanographic practice unaffected but could grant the requisite long-term stability of the SSW standard, for the very first time after SSW was introduced by Knudsen more than a century ago (Knudsen, 1903; Wallace, 1974; Culkin and Smed, 1979; Burchard et al., 2018). Similar to the PSS-78 standard, which 25 specifies a standard equation for the conversion of measured triples of pressure, temperature and conductivity ratio to Practical Salinity, the related standard equation for the density-salinity conversion would be the TEOS-10 equation of state, as already implemented in the SIA and GSW software libraries. While this concept appears physically and metrologically sound, there is a large number of detailed technical, metrological, logistical and financial questions that need to be addressed before a new international, density-based salinity standard may be introduced in the future (Pawlowicz et al., 2016; Schmidt 30 et al., 2016, 2018). Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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20 5.2 Seawater pH As a marine chemistry quantity, the pH of seawater does not belong to the thermodynamic properties available from the equations of TEOS-10, but connections between them were on the agenda of WG127 already in 2007 at Reggio, see Appendix B, and at later meetings. As an urgent problem beyond TEOS-10, seawater pH belongs to the climatological key parameters depicted as metrological challenges by JCS, the successor of WG127 (Feistel et al., 2016; Dickson et al., 2016). 5 For suspectedly disastrous effects on marine ecosystems, the acidification of the oceans, quantified in terms of seawater pH values, has become a severe public, ecological and political concern (Feely et al., 2004; Le Quéré et al., 2015), far beyond mere curiosity of a few scientists. However, already the inventor of the pH value, Sören Peter Lauritz Sörensen, had lamented the ambiguous measurement methods in use for seawater pH. “So wäre es zu wünschen, daß man zukünftig bei der Bestimmung … immer von denselben Voraussetzungen ausgeht, oder jedenfalls daß die Grundlage des angewandten 10 Verfahrens scharf pointiert wird” (Sörensen and Palitzsch, 1910, p. 415 therein) [English translation: „It is desirable that future estimates … be always based on the same assumptions, or at least that the method applied be precisely described“]. Unfortunately, over the past century this situation has improved only insignificantly. Different seawater pH scales are in practical use, often denoted by simply “pH”, which may deviate from one another stronger than the expected variations to be resolved (Marion et al., 2011; Spitzer et al. 2011; Brewer 2013; Dickson et al., 2016). Indeed, still “the field of pH scales and 15 the study of proton-transfer reactions in sea water is one of the more confused areas of marine chemistry” (Dickson, 1984). Originally, Sörensen (1909) had defined the pH value of an aqueous solution as the negative common logarithm of the hydrogen-ion concentration expressed in moles per litre. To better account for effects of ionic interactions in the solution, in this definition equation the concentration was later replaced by the hydrogen-ion activity, specified in a way that both 20 definitions coincide asymptotically in the ideal-solution approximation (Sørensen and Linderstrøm-Lang, 1924). However, even without application to seawater in particular, these definitions involve two severe physico-chemical and metrological difficulties. The first problem is that physically any distinction between a free (dissociated) ion and a bound (associated) ion is subject to 25 arbitrary convention (Bjerrum, 1926; Falkenhagen et al., 1971; Ebeling and Grigo, 1982; Justice, 1991). Consequently, direct measurements of ion concentrations of incompletely dissociated solutions are impossible unless the underlying convention is somehow embodied in the measurement procedure. What in fact can be measured are certain properties correlated with ion concentrations, such as the colour of an indicator dye by photometric pH methods. For obtaining the quantity of interest (pH) from the quantity actually measured (colour), a calibration relation is required which implicitly or explicitly implements the 30 particular ion-association convention, similar to the ticks on a mercury thermometer that implement the definition of the Celsius scale in terms of length units. However, the requisite pH convention is pending yet. Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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21 The second problem is that single-ion activities cannot be measured either (Bjerrum, 1919; Guggenheim, 1949), nor can they unambiguously be inferred from mean chemical potentials of electrically neutral combinations of ions. To overcome this problem, auxiliary assumptions are sometimes applied, such as equating the activities of cations and anions of a particular solute, as had been suggested for KCl by Lewis and Randall (1923). Such arbitrary practical “conventions” may reasonably be applied as long as they do not conflict with experimental evidence. On the other hand, in contrast to empirical 5 thermodynamics, in statistical thermodynamics of electrolytes the Debye-Hückel limiting law predicts the single-ion activity to be a well-defined function of the ionic strength of very dilute electrolytes (Falkenhagen et al., 1971; Prausnitz et al., 1999). Theoretical relations of this kind between activities and other measurable quantities, such as equations for single-ion activities derived from Pitzer equations, are at odds with the putative validity of artificially constructed conventions. 10 Beyond the limiting law, analytical expressions for single-ion activities or related analytical expressions are only approximately available for dilute solutions (below 1 mol l–1) in the theoretical framework of statistical thermodynamics of electrolytes (Wiechert et al., 1978; Ebeling and Scherwinski, 1983). At ion concentrations such as 1.2 mol kg–1 typically encountered in the ocean (Feistel and Marion, 2007), convenient simplifying theoretical models such as hard spherical ions immersed in an unstructured background solvent become increasingly inappropriate. The real microscopic interactions of 15 electrolytes are not precisely known and can only approximately be represented mathematically. A workable practical approach to this problem may be the use of so-called Pitzer equations. They approximate single-ion activities as series expansions with respect to the ion concentrations and adjust the unknown empirical coefficients to other, measurable properties (Nesbitt, 1980; Marion and Grant, 1994; Prausnitz et al., 1999; Marion and Kargel, 2008; Marion et al., 2011). Pitzer equations for seawater ions successfully describe colligative properties while other thermodynamic properties such as 20 sound speed may not yet be represented as accurate as by TEOS-10 (Feistel and Marion, 2007; Feistel, 2008a; Sharp et al., 2015). Based on the Reference Composition model of TEOS-10 (Millero et al., 2008), studies are underway aiming at a definition of seawater pH in terms of Pitzer equations as functions of SI-traceable measurands (Waters and Millero, 2013; Dickson et al., 2016; Turner et al., 2016; Camoes et al., 2016). 1.2 Relative Humidity 25 In TEOS-10, equations for two different definitions of relative humidity (RH) are provided, for the “WMO definition” and the “CCT definition” (IOC et al., 2010; Feistel et al., 2010a,b; IOC et al., 2010). These choices made for both names turned out to be misleading, unfortunately. The so-called “WMO definition” was taken over from recent textbooks on atmospheric sciences (Gill, 1982; Rogers and Yau, 1989; Pruppacher and Klett, 1997; Jacobson, 2005; Pierrehumbert, 2010), however, the definition reported in those books incorrectly retains an obsolete definition of the OMI (1951) which was superseded by 30 the actual WMO definition already in 1954. What is called the “CCT definition” in TEOS-10, on the other hand, is in fact the current WMO definition (WMO, 2008), while the Consultative Committee on Thermometry (CCT) of the BIPM has not recommended yet any RH definition by now. Subsequently, a closer inspection showed that this regrettable naming Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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22 confusion in TEOS-10 reveals only the “tip of the iceberg” (Lovell-Smith et al., 2016). Several mutually inconsistent definitions of relative humidity prevail in different branches of meteorology, climatology, physical chemistry, food engineering or air conditioning, all of them typically denoted by just “RH” without additional specification. The global ocean surface presents a vast open window to the very heart of the dynamics governing the climate system. 5 Properties and processes may be observed there by research vessels, automatic buoys and remotely sensing satellites over long times with exceptional spatial coverage. On the other hand, clouds, wind, waves, corrosive and depositing sea salt, floating icebergs, plastic garbage or oil spills, commercial ship traffic and fishery, sensor fouling by biological processes, political, economic and military interests of riparian states, and large distances and high costs for device maintenance may pose severe difficulties for systematic scientific investigations of the air-sea interface at the open ocean. While TEOS-10 10 cannot overcome such practical obstacles, it has been developed with the intention to provide a consistent thermodynamic framework for studies of the transfer of energy and water between ocean and atmosphere as key players of the terrestrial climate. Water in the troposphere has a mean residence time of 8-10 days until it precipitates as rain or snow. About 80 % to 90 % of 15 that water is replaced by evaporation from the oceans (Reid and Valdés, 2011). Evaporation is driven by the difference between the chemical potentials of water in humid air and in seawater or ice. Per mass of water, these chemical potentials are available from TEOS-10; for humid air from the Helmholtz function (15) by the relation (IAPWS G8-10, 2010), 𝜇W AV = 𝑓AV + 𝜌 𝜕𝑓AV 𝜕𝜌−𝐴 𝜕𝑓AV 𝜕𝐴, (16) for seawater (and in the limit SA = 0, also for pure liquid water) from the Gibbs function (14) by the relation (IAPWS R13-20 08, 2008), 𝜇W SW = 𝑔SW −𝑆A 𝜕𝑔SW 𝜕𝑆A , (17) and for ice from its Gibbs function, gIh, by the relation (IAPWS R10-06, 2009), 𝜇W Ih = 𝑔Ih. (18) Chemical potentials per mole of water, µ, may similarly be expressed by so-called fugacities, f, (symbol not to be confused 25 here with the Helmholtz functions, f, considered earlier), according to the relation (Guggenheim, 1949; Prausnitz et al., 1999) 𝜇= 𝜇id + 𝑅𝑇𝑙𝑛 𝑓 𝑥𝑝. (19) Here, x is the mole fraction of water in the substance considered, µid is the ideal-gas chemical potential of water vapour with the same number of moles, and R is the molar gas constant. It is obvious from eq. (19) that the fugacity, f, plays the role for a 30 real gas which the partial pressure, xp, plays for an ideal gas (Lewis, 1901). Regardless of the wind conditions, evaporation from the ocean ceases when the fugacity of water in seawater equals the fugacity of water in humid air. This fact makes hot Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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23 humid tropics so unpleasant to humans because sweating provides no cooling there. A statement similar to that for the fugacities may not be made for partial pressures, however, as neither a fictitious “partial pressure of water in seawater” is a well-defined quantity, nor does the partial pressure of water in humid air state any thermodynamic criterion for phase equilibria. For these physical reasons, TEOS-10 suggests a replacement of the usual definition of relative humidity in terms of partial pressures by a new definition in terms of fugacities (IOC et al., 2010; Feistel et al. 2010b, 2015, 2016; Feistel, 5 2012; IAPWS G11-15, 2015; Feistel and Lovell-Smith, 2017). In fact, the small real-gas effects of the atmosphere are of similar magnitude as the expected changes of relative humidity associated with greenhouse effects (Hellmuth et al., 2018). 6 Summary TEOS-10 is the present international standard for thermodynamic properties of water, ice, seawater and humid air, recommended for geosciences by leading international organisations such as UNESCO/IOC and IUGG. TEOS-10 10 incorporates an extended manifold of different experimental data for those substances, collected from publications over many decades, in an unprecedentedly compact, consistent, comprehensive and accurate way, based on an axiomatic approach that offers various advantages over otherwise used, relatively arbitrary collections of empirical property equations with often unclear mutual consistency. However, certain important problems have remained to be addressed even after the introduction of TEOS-10. In particular, the climatological key parameters ocean salinity, seawater pH and atmospheric relative humidity 15 still pose metrological challenges that are currently investigated by the standing SCOR/IAPSO/IAPWS Joint Committee on the Properties of Seawater, JCS, in cooperation with other international organisations. References Abbot, C.G., and Fowle, F.E. Jr.: Income and Outgo of Heat from the Earth, and the Dependence of Its Temperature Thereon, Annals of the Astrophysical Observatory (Smithsonian Institution, Washington DC) 2, 159-176, 1908. 20 Alberty, R.A.: Use of Legendre transforms in chemical thermodynamics, Pure and Applied Chemistry 73, 1349–1380, 2001. Arrhenius, S.: On the Influence of Carbonic Acid in the Air upon the Temperature of the Ground. London, Edinburgh, and Dublin, Philosophical Magazine and Journal of Science (fifth series) 41, 237–275, 1896. Bielska, K., Havey, D.K., Scace, G.E., Lisak, D., Harvey, A.H., and Hodges, J.T.: High-accuracy measurements of the vapor pressure of ice referenced to the triple point, Geophysical Research Letters 40, 6303–6307, 2013. 25 Bigg, P.H.: Density of water in S.I. units over the range 0-40 C. British Journal of Applied Physics 18, 521-537, 1967. Bjerrum, N.: On the activity-coefficient for ions, Meddelanden från Kungliga Vetenskapsakademiens Nobelinstitut 5, 1-21, 1919. German translation: Der Aktivitätskoeffizient der lonen, Zeitschrift für anorganische und allgemeine Chemie 109, 275–292, 1920. Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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37 Wagner, W., Riethmann, T., Feistel, R., and Harvey, A. H.: New Equations for the Sublimation Pressure and Melting Pressure of H2O Ice Ih, J. Phys. Chem. Ref. Data, 40, 043103, 2011. Wallace, W.J.: The Development of the Chlorinity/Salinity Concept in Oceanography, Elsevier, Amsterdam, London, New York, 1974. Waters, J.F., and Millero, F.J.: The free proton concentration scale for seawater pH, Marine Chemistry, 149, 8–22, 2013. 5 Wells, N. C.: The Atmosphere and Ocean, Wiley-Blackwell, Oxford, 2012. Wiechert, H., Krienke, H., Feistel, R., and Ebeling, W.: Interionische Wechselwirkung und Thermodynamik starker Elektrolyte, Zeitschrift für physikalische Chemie Leipzig, 259, 1057-1076, 1978. Wirth, H.E.: The Problem of the Density of Seawater, Journal of Marine Research, 3, 230-247, 1940. WMO: Guide to Meteorological Instruments and Methods of Observation, 7th edn, World Meteorological Organization, 10 Geneva, Switzerland, 2008. www.wmo.int/pages/prog/gcos/documents/gruanmanuals/CIMO/CIMO_Guide-7th_Edition-2008.pdf WMO: Report on the WMO-BIPM workshop on Measurement Challenges for Global Observation Systems for Climate Change Monitoring: Traceability, Stability and Uncertainty. 30 March - 1 April 2010, WMO Headquarters, Geneva, Switzerland, 2010. 15 Wolf, H.: Determination of water density: limitations at the uncertainty level of 1×10−6, Accred. Qual. Assur., 13, 587–591, 2008. Woosley, R.J., Huang, F., and Millero, F.J.: Estimating absolute salinity (SA) in the world’s oceans using density and composition, Deep-Sea Research I, 93, 14–20, 2014. Wright, D. G., Feistel, R., Reissmann, J.H., Miyagawa, K., Jackett, D.R., Wagner, W., Overhoff, U., Guder, C., Feistel, A., 20 and Marion, G.M.: Numerical implementation and oceanographic application of the thermodynamic potentials of water, vapour, ice, and seawater. Part 2: The library routines, Ocean Science, 6, 695 – 718, 2010a. Wright, D.G., Pawlowicz, R., McDougall, T.J., and Feistel, R.: Progress Report for the SCOR/IAPSO Working Group 127 on "Thermodynamics and Equation of State of Seawater", Canadian Ocean Science Newsletter 49, March 30, 2010, Canadian National Committee for SCOR, 2010b. 25 Wright, D.G., Pawlowicz, R., McDougall, T.J., Feistel, R., and Marion, G.M.: Absolute Salinity, “Density Salinity” and the Reference-Composition Salinity Scale: present and future use in the seawater standard TEOS-10, Ocean Sci., 7, 1–26, 2011. Appendix A: The Jacobi Method Consider N functions, y1, y2, ... yN, depending on N independent variables, x1, x2, ... xN. The matrix 𝑨= {𝑎𝑖𝑗} consisting of 30 their pairwise partial derivatives, 𝑎𝑖𝑗= 𝜕𝑦𝑖𝜕𝑥𝑗 ⁄ , as its elements is regarded as the Jacobian matrix of this set of functions. The determinant 𝐽= |𝑨| = det{𝑎𝑖𝑗} of this matrix is commonly referred to as the functional determinant, or the Jacobian Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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38 (Bronstein and Semendjajew, 1979; Kaplan, 1984; Gradshteyn and Ryshik, 2000). It is instructive to write J in the form of a fraction, 𝐽= det { 𝜕𝑦𝑖 𝜕𝑥𝑗} = 𝜕(𝑦1,𝑦2,…𝑦𝑁) 𝜕(𝑥1,𝑥2,…𝑥𝑁). (A.1) For N = 1, the Jacobian agrees with the partial derivative, 𝐽= 𝜕(𝑦1) 𝜕(𝑥1) = 𝜕𝑦1 𝜕𝑥1 . (A.2) 5 Determinants of higher dimensions N can be computed by means of the so-called Laplace expansion with respect to a selected row or column of the matrix and the related remaining submatrices, the minors of dimension N – 1. For N = 2, the Jacobian reads 𝐽= 𝜕(𝑦1, 𝑦2) 𝜕(𝑥1, 𝑥2) = 𝜕𝑦1 𝜕𝑥1 × 𝜕(𝑦2) 𝜕(𝑥2) − 𝜕𝑦1 𝜕𝑥2 × 𝜕(𝑦2) 𝜕(𝑥1) = 𝜕𝑦1 𝜕𝑥1 𝜕𝑦2 𝜕𝑥2 − 𝜕𝑦1 𝜕𝑥2 𝜕𝑦2 𝜕𝑥1 (A.3) where the 2-dimensional Jacobian is expressed in terms of 1-dimensional Jacobians. 10 For N = 3, the Laplace expansion takes the form 𝐽= 𝜕(𝑦1, 𝑦2, 𝑦3) 𝜕(𝑥1, 𝑥2, 𝑥3) = 𝜕𝑦1 𝜕𝑥1 × 𝜕(𝑦2, 𝑦3) 𝜕(𝑥2, 𝑥3) − 𝜕𝑦1 𝜕𝑥2 × 𝜕(𝑦2, 𝑦3) 𝜕(𝑥1, 𝑥3) + 𝜕𝑦1 𝜕𝑥3 × 𝜕(𝑦2, 𝑦3) 𝜕(𝑥1, 𝑥2) (A.4) where the 3-dimensional Jacobian is expanded into products of 1- und 2-dimensional Jacobians. Eqs. (A.3) and (A.4) are the most frequently used relations required for deriving TEOS-10 functions with 2 or 3 independent variables. 15 Here are some useful manipulation rules for Jacobians. General properties of determinants imply that the exchange of any two variables switches the sign of the Jacobian, e.g., 𝜕(𝑦1,…,𝑦𝑖,…,𝑦𝑗,…𝑦𝑁) 𝜕(𝑥1,𝑥2,…𝑥𝑁) = − 𝜕(𝑦1,…,𝑦𝑗,…,𝑦𝑖,…𝑦𝑁) 𝜕(𝑥1,𝑥2,…𝑥𝑁) , (A.5) and similarly, 20 𝜕(𝑦1,…𝑦𝑁) 𝜕(𝑥1,…𝑥𝑖,…𝑥𝑗,…𝑥𝑁) = − 𝜕(𝑦1,…𝑦𝑁) 𝜕(𝑥1,…𝑥𝑗,…𝑥𝑖,…𝑥𝑁). (A.6) For N = 2, this means 𝜕(𝑦1, 𝑦2) 𝜕(𝑥1, 𝑥2) = − 𝜕(𝑦2, 𝑦1) 𝜕(𝑥1, 𝑥2) = − 𝜕(𝑦1, 𝑦2) 𝜕(𝑥2, 𝑥1) = 𝜕(𝑦2, 𝑦1) 𝜕(𝑥2, 𝑥1) (A.7) As special cases for N = 3, the reversal of variables inverts the sign, 𝜕(𝑦1, 𝑦2, 𝑦3) 𝜕(𝑥1, 𝑥2, 𝑥3) = − 𝜕(𝑦3, 𝑦2, 𝑦1) 𝜕(𝑥1, 𝑥2, 𝑥3) = − 𝜕(𝑦1, 𝑦2, 𝑦3) 𝜕(𝑥3, 𝑥2, 𝑥1) = 𝜕(𝑦3, 𝑦2, 𝑦1) 𝜕(𝑥3, 𝑥2, 𝑥1) (A.8) 25 while the rotation of variables preserves the sign, 𝜕(𝑦1, 𝑦2, 𝑦3) 𝜕(𝑥1, 𝑥2, 𝑥3) = 𝜕(𝑦3, 𝑦1, 𝑦2) 𝜕(𝑥1, 𝑥2, 𝑥3) = 𝜕(𝑦2, 𝑦3, 𝑦1) 𝜕(𝑥1, 𝑥2, 𝑥3) = 𝜕(𝑦1, 𝑦2, 𝑦3) 𝜕(𝑥3, 𝑥1, 𝑥2) = 𝜕(𝑦1, 𝑦2, 𝑦3) 𝜕(𝑥2, 𝑥3, 𝑥1) (A.9) If one (or more) of the functions is an identity, say, 𝑦𝑁(𝑥1, … 𝑥𝑁) ≡𝑥𝑁, the Jacobian reduces by one dimension (or more), due to the Laplace expansion, Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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39 𝜕(𝑦1, 𝑦2,…. 𝑦𝑁−1,𝑥𝑁) 𝜕(𝑥1, 𝑥2,…. 𝑥𝑁−1,𝑥𝑁) = 𝜕(𝑦1, 𝑦2,… 𝑦𝑁−1) 𝜕(𝑥1, 𝑥2,….𝑥𝑁−1). (A.10) In particular, the identical Jacobian equals unity, 𝜕(𝑥1, 𝑥2,…. 𝑥𝑁−1,𝑥𝑁) 𝜕(𝑥1, 𝑥2,…. 𝑥𝑁−1,𝑥𝑁) = 1. (A.11) The product rule for functional determinants of higher dimensions is a generalisation of the usual chain rule for partial derivatives. If a set of functions z depends on the variables y, which in turn depend on the variables x, the relation between 5 their Jacobians is (Bronstein and Semenjajew, 1979), 𝜕(𝑧1,𝑧2,…𝑧𝑁) 𝜕(𝑥1,𝑥2,…𝑥𝑁) = 𝜕(𝑧1,𝑧2,…𝑧𝑁) 𝜕(𝑦1,𝑦2,…𝑦𝑁) × 𝜕(𝑦1,𝑦2,…𝑦𝑁) 𝜕(𝑥1,𝑥2,…𝑥𝑁). (A.12) In particular, if the functions z are chosen identical to x, for the inverse functions there follows from (A.12), 𝜕(𝑦1,𝑦2,…𝑦𝑁) 𝜕(𝑥1,𝑥2,…𝑥𝑁) = [ 𝜕(𝑥1,𝑥2,…𝑥𝑁) 𝜕(𝑦1,𝑦2,…𝑦𝑁)] −1 (A.13) Developed by Shaw (1935), the Jacobi method is the mathematically most elegant way of transforming the various partial 10 derivatives of different thermodynamic potentials into one another, exploiting the convenient formal calculus of functional determinants (Margenau and Murphy, 1943; Landau and Lifschitz, 1966). Namely, if any thermodynamic derivative in two variables, (𝜕𝑢𝜕𝑥 ⁄ )𝑦, is considered, it can formally be written as a Jacobian, eq. (A.10) ( 𝜕𝑢 𝜕𝑥) 𝑦≡ 𝜕(𝑢,𝑦) 𝜕(𝑥,𝑦). (A.14) If this Jacobian is to be expressed in specific variables, say T and p, the transformation into these independent variables 15 follows from (A.12) as ( 𝜕𝑢 𝜕𝑥) 𝑦= 𝜕(𝑢,𝑦) 𝜕(𝑇,𝑝) 𝜕(𝑥,𝑦) 𝜕(𝑇,𝑝) . (A.15) If any of the variables u, x, y equals T or p, the numerator or denominator of (A.15) can be simplified by means of the rules (A.7) and (A.10). If not, we get the result for (A.15) from (A.3), ( 𝜕𝑢 𝜕𝑥) 𝑦= 𝜕(𝑢,𝑦) 𝜕(𝑇,𝑝) 𝜕(𝑥,𝑦) 𝜕(𝑇,𝑝) = (𝜕𝑢 𝜕𝑇)𝑝(𝜕𝑦 𝜕𝑝) 𝑇 −(𝜕𝑦 𝜕𝑇)𝑝(𝜕𝑢 𝜕𝑝) 𝑇 (𝜕𝑥 𝜕𝑇)𝑝(𝜕𝑦 𝜕𝑝) 𝑇 −(𝜕𝑦 𝜕𝑇)𝑝(𝜕𝑥 𝜕𝑝) 𝑇 . (A.16) 20 As an example, we compute the sound speed, c, eq. (7), in terms of derivatives of the Gibbs function g(T, p), and of the Helmholtz function, f(T, ρ). For the Gibbs function we get from (A.16) 𝑐2 = ( 𝜕𝑝 𝜕𝜌) 𝜂 = 𝜕(𝑝,𝜂) 𝜕(𝜌,𝜂) = 𝜕(𝑝,𝜂) 𝜕(𝑇,𝑝) 𝜕(𝜌,𝜂) 𝜕(𝑇,𝑝) = −(𝜕𝜂 𝜕𝑇) 𝑝 (𝜕𝜌 𝜕𝑇)𝑝(𝜕𝜂 𝜕𝑝) 𝑇 −(𝜕𝜂 𝜕𝑇)𝑝(𝜕𝜌 𝜕𝑝) 𝑇 . (A.17) Substituting in (A.17) the variables ρ and η, respectively, by means of 𝑔𝑝= 𝑣= 1 𝜚 ⁄ and 𝑔𝑇= −𝜂, where partial derivatives of g are now conveniently written as subscripts, the final formula for the sound speed expressed in terms of the Gibbs 25 function reads 𝑐= 𝑔𝑝√ 𝑔𝑇𝑇 𝑔𝑇𝑝 2 −𝑔𝑇𝑇𝑔𝑝𝑝. (A.18) Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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40 Alternatively, for the Helmholtz function we get from (A.16) the sound speed formula, replacing in (A.17) the independent variable p by ρ, 𝑐2 = ( 𝜕𝑝 𝜕𝜌) 𝜂 = 𝜕(𝑝,𝜂) 𝜕(𝜌,𝜂) = 𝜕(𝑝,𝜂) 𝜕(𝑇,𝜌) 𝜕(𝜌,𝜂) 𝜕(𝑇,𝜌) = (𝜕𝑝 𝜕𝑇)𝜌(𝜕𝜂 𝜕𝜌) 𝑇 −(𝜕𝜂 𝜕𝑇)𝜌(𝜕𝑝 𝜕𝜌) 𝑇 −(𝜕𝜂 𝜕𝑇)𝜌 . (A.19) Substituting in (A.19) the variables p and η, respectively, by means of 𝜚2𝑓 𝜚= 𝑝 and 𝑓 𝑇= −𝜂, the final formula for the sound speed expressed in terms of the Helmholtz function reads 5 𝑐= √ −𝜚2𝑓 𝑇𝜚 2 𝑓𝑇𝑇 + 2𝜚𝑓 𝜚+ 𝜚2𝑓 𝜚𝜚 . (A.20) For more independent variables, such as the additional mass fractions of sea salt or dry air, the method applies correspondingly. If a derivative (𝜕𝑢𝜕𝑥 ⁄ )𝑦,𝑧 is to be calculated, we apply the rule (A.10), ( 𝜕𝑢 𝜕𝑥) 𝑦,𝑧≡ 𝜕(𝑢,𝑦,𝑧) 𝜕(𝑥,𝑦,𝑧) . (A.21) If this derivative of an arbitrary function u(x, y, z) needs to be expressed in, say, the particular variables S, T, p of the Gibbs 10 function of seawater, eq. (A.21) needs to be transformed to those variables by the thermodynamic “chain rule” ( 𝜕𝑢 𝜕𝑥) 𝑦,𝑧≡ 𝜕(𝑢,𝑦,𝑧) 𝜕(𝑥,𝑦,𝑧) = 𝜕(𝑢,𝑦,𝑧) 𝜕(𝑆,𝑇,𝑝) 𝜕(𝑥,𝑦,𝑧) 𝜕(𝑆,𝑇,𝑝) ⁄ . (A.22) Then, the Jacobians in the numerator and denominator are evaluated by the rules (A.4), (A.8), (A.9) and (A.10), until (A.22) is eventually expressed by the Gibbs function g and its derivatives with respect to S, T and p. Appendix B: Meetings of SCOR/IAPSO WG127 15 WG127 was formed of a small group of specialists from countries all over the world. After the WG had been established in 2005, the WG Chair, Trevor McDougall, was its first, natural member who decided to invite Rainer Feistel as a second member. Together possible further candidates were then commonly discussed and contacted as the next potential member. So the democratic invitation process went on up to the intended maximum size of the group. Rather than face to face, many of the members knew each other only from publications before. Despite this, the work of WG127 was very constructive and 20 intense from the first day on. Year by year, the membership varied slightly, bringing new expertise into the discussions and decisions. Until the adoption of TEOS-10 in 2009, the participants of the four WG127 meetings may briefly be reported here as an appreciation and gratitude for their substantial contributions and very successful cooperation during those years. It was very sad, however, that two extremely productive and creative friends and colleagues were lost quite untimely: Dan Wright died in 2010, and David Jackett in 2012. In memory, TEOS-10 will always be associated with their names. 25 2006 Initial Meeting at Warnemünde, Germany, 2 to 5 May, see Fig. B.1 2007 Meeting at Reggio/Calabria, Italy, 7 to 10 May, see Fig. B.2 2008 Meeting at Goetz near Berlin, Germany, 3 to 9 September, see Fig. B.3 2009 Final Meeting at Arnhem, The Netherlands, 2 to 5 September: 30 Participants (alphabetically, no photo available): Rainer Feistel (Germany), Brian A. King (UK), Norge Larson (USA), Trevor J. McDougall (Australia), Richard Pawlowicz (Canada), Daniel G. Wright (Canada). Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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41 Figure 1: Temperature-pressure diagram of water (from IAPWS R14-08, 2011, permitted). The curves indicate phase transitions between stable gaseous, liquid and solid states; pmelt indicates the melting line of ambient hexagonal ice Ih, psubl its sublimation line, and ps the saturation line, or boiling line, between liquid and gas. Several additional ice phases are not indicated. Within this diagram, geophysical conditions for ice, liquid water and water vapour cover only small regions (not shown) in some vicinity of the 5 triple point. Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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42 Figure 2: The hexagonal elementary crystal of ice Ih (Penny, 1948; Schulson, 1999) consists of 27 oxygen (O) atoms (spheres) and 28 hydrogen (H) bonds between them (bars). Of the four H-atoms adjacent to each O-atom, two are placed closer than the other pair, thus retaining the structure of individual H2O molecules within the crystal lattice. Despite the crystal’s anisotropy, 5 thermodynamic ice properties are isotropic to within measurement uncertainty. Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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43 Figure 3: Example for differences between directly measured densities and values calculated from TEOS-10 with measured Practical Salinities, SP. Samples taken in 2008, 2011 and 2013 from eastern Atlantic surface waters between 33 °N and 18 °S show systematic negative density anomalies up to 13 ppm (diagram courtesy Stefan Weinreben, IOW, priv. comm.). 5 Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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44 Figure 4: Brine salinity computed from TEOS-10 as a function of temperature at atmospheric pressure, compared with measured results for brine pockets of Antarctic sea ice. Symbol “F”: data of Fischer (2009), “G”: data of Gleitz et al. (1995). As these 5 observational data were not included in the construction of the 2008 Gibbs function, note that the curve is a theoretical prediction derived from the thermodynamic potentials of seawater and ice. Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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45 Figure 5: Isobaric evaporation enthalpy of water, indicated by “0 g/kg”, and of seawater with salinities 120 g kg–1 at 101325 Pa and 40 g kg–1 at 5 MPa and 10 MPa, as functions of the temperature, computed from TEOS-10. Here, the latent heat of seawater is derived from the total heat capacity of a 2-phase seawater-air composite, reduced by the two separate single-phase heat capacities 5 involved. The salinity corrections to the latent heats of pure water are very small. At high pressure the validity of the Gibbs function of seawater is restricted to maxima of 40 g kg–1 and 40 °C, but at atmospheric pressure it is valid up to 120 g kg–1 and 80 °C. The lower temperature bounds shown are the particular freezing points of water or seawater (Feistel et al., 2010b). Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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46 Figure 6: Phase diagrams of composite air-water-ice systems (“clouds”), depending on the entropy and dry-air mass fraction of a given parcel, computed from TEOS-10 (Feistel et al., 2010a). Natural clouds with air fractions of 98-99 % at pressures between 5 20 000 and 101 325 Pa are located in the down-right corner of the diagrams. Metastable states, such as subcooled liquid droplets (WMO, 2008), are not considered here. Upper panel: In the triangle-shaped, 2-phase regions “Wet Air”, humid air is in equilibrium with liquid water. The tips of these wet-air wedges are triple points, located along the “Triple Line” depending on the pressure (or altitude) as indicated. Above the wedge, between the “Dewpoint” curve and “Triple Line”, humid air is the only stable phase. Between “Dewpoint” and “Ice 10 Formation”, liquid water is in equilibrium with humid air. Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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47 Lower panel: In the triangle-shaped, 3-phase regions “WIA” (wet ice air), humid air is in equilibrium with both liquid water and ice at the same time. The tips of theses WIA wedges are triple points, located along the “Triple Line” depending on the pressure (or altitude) as indicated. Above the wedge, the cloud contains only liquid water (wet air), if at all, while below the wedge, ice is the only condensed stable phase (ice air) in equilibrium with humid air. When adiabatically ascending, entropy and dry-air fraction of an air parcel remain unchanged, even when passing the transition to 5 cloud formation (the isentropic condensation level), and so does the parcel’s representative point in the diagrams. With decreasing pressure, the “wet-air” and “wet-ice-air” regions will pan to the left, and states located initially between the “Dewpoint” and “Triple Line” get into the wet-air region first, then become wet ice air, and finally ice air. Points located above the “Triple Line” condensate to ice directly without passing any intermediate liquid phase. Using such diagrams, the parcel’s phase changes during convection may be predicted from its conservative properties at the ground. 10 Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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48 Figure 7: Hierarchical structure of TEOS-10, based on four fundamental thermodynamic potentials defined in IAPWS documents for ice Ih, fluid water, dissolved sea salt, and dry air, including air-water interaction properties. Derived from those fundamental equations and their empirical coefficients by numerical and analytical mathematical methods, without introducing any additional 5 empirical functions or coefficients, any thermodynamic properties can be computed for the pure substances (such as water and ice), their mixtures (such as seawater and humid air) and multi-phase composites (such as sea ice, ocean-atmosphere equilibria, and clouds). Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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49 front of the Baltic Sea Research Institute, IOW. 5 Figure B.1: Participants from left to right: Chen-Tung Arthur Chen (Taiwan), Frank J. Millero (USA), Brian A. King (UK), Rainer Feistel (Germany), Daniel G. Wright (Canada), Trevor J. McDougall (Australia), Giles M. Marion (USA). Photo taken in Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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50 Figure B.2: Participants from left to right: Giles M. Marion (USA), Rainer Feistel (Germany), Trevor J. McDougall (Australia), Brian A. King (UK), Chen-Tung Arthur Chen (Taiwan), David Jackett (Australia), Daniel G. Wright (Canada), Petra Spitzer (Germany), Frank J. Millero (USA). Photo taken at the shore of the Strait of Messina, with Sicily Island in the back. 5 Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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51 Figure B.3: Participants from left to right: Frank J. Millero (USA), Petra Spitzer (Germany), Nigel Higgs (UK), Trevor J. McDougall (Australia), Chen-Tung Arthur Chen (Taiwan), Giles M. Marion (USA), David Jackett (Australia), Rainer Feistel (Germany), Steffen Seitz (Germany), Brian A. King (UK), Daniel G. Wright (Canada). 5 Ocean Sci. Discuss., Manuscript under review for journal Ocean Sci.
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8393 | https://theclassicjournal.uga.edu/index.php/2024/04/18/permutations-of-n-with-k-cycles-and-left-to-right-maxima/ | The Classic Journal
a journal of undergraduate writing and research, from WIP at UGA
Permutations of [n] with k Cycles and Left-to-Right Maxima
by Blake Voyles, Mathematics
Background
What is a permutation?
A permutation is an ordered arrangement of elements. More specifically, a permutation of n is a unique arrangement of the set {1, 2, 3, …, (n − 1), n}. For example, the permutations of 3 are the following:
{1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1}
The total amount of permutations of [n] is always n!. Similarly, permutations of [n − 1] are always n!, and the same is true for sets starting at 2 ({2, 3, 4, …, n}).
What is a cycle?
A cycle is a subset of permutations with all elements in the subset trading positions. A permutation cycle is cyclic meaning that the last element of the cycle goes to the first element of the cycle. Cycle notation is a way to write a permutation that breaks down the elements into their cycles. There are also three ways to write permutations in cycle notation: two line, one line, and cycle notation. For the purposes of this paper, only one line and cycle notation will be used. For example, the cycle notations of {2, 1, 3} (a specific permutation of 3) are the following:
One Line: (2, 1, 3) Cycle Notation: (12)(3)
The elements 1,2 trade positions in the permutation, and the 3 is in its own cycle because it stays in the same position.
What are left-to-right maxima in a permutation?
Given an element in a permutation of numbers (1 to n), it is a left-to-right maxima when the given element is greater than all the other elements to the left of the given element. For example, given the permutation (1,3,2,5,4) of 5. There are exactly 3 left- to-right maxima, those being 1,3, and 5. The first element of a permutation will always be a left-to-right maxima.
What does it mean for a permutation of [n] to have exactly k cycles?
Every permutation of [n] can be written as a sequence of cycles. The cycle of a permutation is a subset of the permutation where the elements are moved to the position of the next element in the cycle. A permutation cycle is cyclic meaning that the last element of the cycle goes to the first element of the cycle. A permutation of [n] with k cycles is a permutation that has exactly k of the subset cycles, including cycles that are a single element (which are said to be mapped to themselves).
Claim
Let f : [n] → [n] be a permutation of [n]. The number of permutations of [n] with k left-to-right maxima is equal to the number of permutations of [n] with exactly k cycles (including fixed points). To prove this claim, we will show that the amount of both sets of permutations follows the recursive formula for the unsigned Stirling numbers of the first kind and that the two sets start with equivalent base cases.
For example, let n = 3 and k = 2:
Permutations with 2 cycles: (1)(32), (13)(2), (12)(3)
Permutations with 2 left-to-right maxima: {1,3,2},{2,1,3},{2,3,1}
Unsigned Stirling Numbers of the First Kind
It is well known that the unsigned Stirling numbers of the first kind are the number of permutations of [n] with k cycles, but we will still break down how the recursive formula is found from the cycles. The recursive formula for the unsigned Stirling numbers of the first kind is only made up of two terms and they are as follows (formula 1.0):
Recursion for Permutations of [n] with Exactly k Cycles
As mentioned earlier the recursive formula for the unsigned Stirling numbers of the first kind gives you the number of per- mutations of [n] with exactly k cycles. To start we need to establish the trivial cases or the base cases. The true base case is that any permutation of 0 with 0 cycles is 1. The rest is assuming the initial condition that n > 0. Any permutation of [n] into 0 cycles is 0, because there is no permutation with n > 1 with no cycles. The third trivial case is that the number of permutations of [n] with n cycles is only 1 because this would be the permutation where every cycle is of length 1. This gives n cycles of length one, in other words, every element is a fixed point. The final trivial case is the number of permutations of [n] with 1 cycle is equivalent to (n − 1)!, and this is because a permutation of n with 1 cycle is equivalent to fixing the first value and permuting the (n − 1) elements. The total amount of possibilities for a permutation of length n is n!, so with the first element fixed the rest of the permutation is equal to (n − 1).
When trying to form all the permutations of [n] with k cycles, there are two options. The new element can be added as a new cycle, or it can be added into an existing cycle. To show how the permutations of [n] with k cycles follows formula 1.0, we will go through the reasoning for the first and then the second term. The first term in 1.0 counts all the permutations where the new term is added as a single cycle. If we are trying to get permutations of [n] with k cycles, we can add a cycle of length 1 to all the permutations of (n – 1) with (k – 1) cycles. The additional cycle of length 1 would be the nth term as a fixed point. The new permutations are now all permutations of [n] with (k – 1) + 1 cycles, or better put permutations of [n] with k cycles. For example, let’s say that we want a permutation of 4 with 2 cycles. Consider the following permutation of 3: (2, 3, 1). This is a permutation of (4 − 1) with exactly (2 − 1) cycles. To get the target permutation of 4 with 2 cycles, it is easy to add a single cycle (4) to the end. The new permutation is now (2, 3, 1, 4), which is a permutation of 4 with exactly 2 cycles those being (1, 3, 2)(4).
This gives us the first part formula 1.0, which would be
The second term in 1.0 counts all the permutations where the new term is added to an existing cycle. To do this and still end with a permutation of n with k cycles, we have to start with a permutation of (n – 1) with k cycles. This is because when adding the new term anywhere we are increasing the permuta- tion from one of (n – 1) to a permutation of n. The reason we are starting with k cycles this time is because we aren’t adding the nth element as a new cycle, so the total number of cycles before the nth is added stays the same. The new term can be added anywhere within the permutation, and there are (n – 1) options for where to place the new term. For example, let’s say that we want another permutation of 4 with 2 cycles. Consider the following permutation of 3: (1, 3, 2). In this permutation, there are already two cycles, (1), (2, 3). We can add the 4 after the 1, 3, or 2. That is a total of (4 − 1) options, and say we add the 4 after the 3. The permutation is now (1, 3, 4, 2), with the two cycles being (1)(2, 4, 3).
This gives us the second term of formula 1.0 which would be
Because the nth term can only be added as a new cycle of length 1 or into an existing cycle, these two terms account for all the permutations of [n] with k cycles. This means that the recursive formula for the number of permutations of [n] with k cycles is the following (1.0):
Recursion for permutations of [n] with k left-to-right maxima
Again, to start we need to establish the trivial cases or the base cases. The true base case is that any permutation of 0 with 0 left-to-right maxima is 1. The rest is assuming the initial condition that n > 0. The first trivial case is the amount of permutations of [n] with 0 left-to-right maxima is also 0. This is because the first element of every permutation is a left-to-right maxima. The second trivial case is the number of permuta- tions of [n] with n left-to-right maxima is only 1, and this is because the only possible way for every element to be a left-to- right maxima is when the permutation of n is just the elements in ascending order. If all elements are not in ascending order, then there will be at least one element that is not a left-to-right maxima. The last trivial case is the amount permutation of n with 1 left-to-right maxima is equivalent to (n − 1)!. Since the first element is always a left-to-right maxima, this is the case when the largest element is also the first element in the permutation. Because every element is less than the largest, the only left-to-right maxima would be the first element. The amount of permutations that this applies to is equivalent to (n − 1)! because the first term is fixed (having to be the largest element) and this leaves (n − 1) elements with the freedom to be placed in any order. This is essentially a permutation of (n − 1) which is exactly (n − 1)!.
When trying to form all the permutations of [n] with k left-to- right maxima, there are two categories of permutations. Those two categories are permutations of [n] with k left-to-right max- ima that start with a 1 or don’t start with a 1, and these two categories are represented by the first and second term in the recursive formula mentioned previously.
The first term counts all the permutations of [n] with k left- to-right maxima that start with a 1. Because of the rule that the first term is always a left-to-right maxima, the 1 being in the first term is significant. Other than permutations of just 1, the first element being 1 is the only way that 1 can be a left-to-right maxima. If we were trying to get permutations of [n] with k left- to-right maxima, we could start with a permutation of (n − 1). Instead of this permutation being the usual permutation containing the elements 1 to (n− 1), let’s say that the permutation contains elements from 2 to n. There are still a total of (n − 1) elements so all the properties hold for this newly shifted set of permutations. The first term counts all these shifted permuta- tions of 2 to (n – 1) with (k – 1) left-to-right maxima. To get the permutations of [n] with k left-to-right maxima we take the shifted permutations of (n – 1) with (k – 1) left-to-right maxima, and add a 1 to the beginning of the permutation. When this is done, the permutation is now the complete permutation of n, and because the first element is always a left-to-right maxima, adding the 1 increases the count of left-to-right-maxima from (k − 1) to k. This operation gives results in a permutation of n with k left-to-right maxima, which is exactly what we needed. For example, let’s say we want a permutation of 4 with 2 left-to- right maxima. Consider the following permutation of 3: (3, 1, 2). We then would add 1 to all the elements, resulting in the per- mutation (4, 2, 3). Then to complete the transformation add a 1 to the beginning of the permutation, giving us the permutation (1, 4, 2, 3). This is now a permutation of 4 with 2 left-to-right maxima, which was derived from a permutation of (4 − 1) with (2 − 1) left-to-right maxima.
This gives us the first part of the recursive formula, which would be
The second term counts all the permutations where the 1 is not the first element. When 1 is not the first element is always true that it is not a left-to-right maxima within a permutation. We will start by doing the same thing we did when explaining the first term. Instead of the normal permutation of (n – 1), we will use the permutation starting at 2 going to n. Again, this new set of permutations has all the same properties as the nor- mal set, but with every term shifted up 1. In order to add the 1, and still end with a permutation of n with k left-to-right max- ima we have to start with the shifted permutation of (n−1), but with k left-to-right maxima before adding the 1. This is exactly the amount of permutations of (n – 1) with k left-to-right max- ima. We now need to add the one into the permutation, with the only restriction being that it can’t go in front of the first el- ement. That leaves us with exactly (n− 1) possible positions to place the 1. This second term is the amount of possible positions (n− 1) multiplied by the amount of permutations of (n – 1 with k left-to-right maxima. For example, let’s say that we want a permutation of 4 with exactly 2 left-to-right maxima. Consider the following permutation of 3: (2, 3, 1). This permutation al- ready has 2 left-to-right maxima. We then again add 1 to all the elements of the permutation, resulting in the new permutation being (3, 4, 2). This permutation still only has 2 left-to-right- maxima. We can add the 1 after the 3, 4, or 2, and doing so will not increase the number of left-to-right maxima. Let’s add the 1 after the 3, which gives us the final permutation (3, 1, 4, 2). This is a permutation of 4 with 2 left-to-right maxima that was derived by a permutation of 3 with 2 left-to-right maxima. This gives us the following second term for the recursive formula for permutations of [n] with k left-to-right maxima:
Because all permutations fit into one of the two categories of starting with a 1 or not starting with a 1, the number of permu- tations of [n] with k left-to-right maxima can be counted using only these two terms. This means that the recursive formula for the number of permutations of [n] with k left-to-right maxima is the following (1.0):
Conclusion
We have justified the recursive formula for both the amount of permutations of [n] with k cycles and the amount of permutations of [n] with k left-to-right maxima. The two have the exact same recursive formula being (1.0),
This doesn’t necessarily mean that the two are equal. For the two to be equivalent, they must also have the same base cases. Both have the following trivial and base cases:
The number of permutations of [n] with exactly k cycles and the number of permutations of [n] with k left-to-right maxima have identical recursive formulas and have the exact same base cases, therefore the two are equivalent. They both start out the same because they have equivalent base cases. Any n and k when plugged into the recursive formula will undergo the same operations and will end with the same base cases, so when deter- mining the amounts for any given n and k for both the number of permutations of [n] with exactly k cycles and the number of permutations of [n] with k left-to-right maxima the resulting amounts are the same.
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Basics on the topic Side and Angle Conditions for a Triangle
Side and Angle Conditions for a Triangle
Triangles are among the simplest and most studied shapes in geometry, used in everything from architectural design to trigonometry classes. A triangle is a three-sided polygon, and its properties are fundamental to understanding many aspects of geometry. Constructing triangles may seem easy at first, but there are important factors to take into account.
Triangles are shapes with three straight sides and three angles. The sum of the interior angles in a triangle always equals 180 degrees, and the length of any side must be less than the sum of the other two sides (triangle inequality theorem).
Side and Angle Conditions for a Triangle – Explanation
The properties of triangles vary depending on the lengths of their sides and the measures of their angles. Here are the basic rules and conditions:
| Rule Name | Description |
--- |
| Triangle Inequality Theorem | The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. |
| Sum of Angles | The sum of the interior angles of a triangle is always 180 degrees. |
Side and Angle Conditions for a Triangle – Examples
Let's explore these concepts with an example:
Can a triangle have sides with lengths of 7 cm, 10 cm, and 18 cm?
To determine if these can form a triangle, apply the triangle inequality theorem. Add the lengths of any two sides and compare it to the third:
7 cm + 10 cm = 17 cm, which is less than 18 cm.
Therefore, these sides do not meet the triangle inequality theorem, and they cannot form a triangle.
Is it possible to form a triangle with sides measuring 8 cm, 15 cm, and 22 cm?
To check if these sides can form a triangle, apply the triangle inequality theorem:
8 cm + 15 cm = 23 cm, which is greater than 22 cm,
15 cm + 22 cm = 37 cm, which is greater than 8 cm,
8 cm + 22 cm = 30 cm, which is greater than 15 cm.
All conditions satisfy the triangle inequality theorem, so these sides can form a triangle.
A triangle has two angles measuring 65 degrees and 45 degrees. What is the measure of the third angle?
To find the measure of the third angle, use the angle sum theorem, which states that the sum of the interior angles of a triangle is always 180 degrees. Add the known angles and subtract from 180 degrees:
$65^\circ + 45^\circ = 110^\circ$,
$180^\circ - 110^\circ = 70^\circ$.
Therefore, the third angle measures $70^\circ$.
If a triangle has one angle measuring $90^\circ$ and another measuring $30^\circ$, what is the measure of the third angle?
Using the angle sum theorem:
The sum of the angles in any triangle is $180^\circ$,
With a right angle ($90^\circ$) and another angle of $30^\circ$, add these angles: $90^\circ + 30^\circ = 120^\circ$,
Subtract this sum from $180^\circ$ to find the third angle: $180^\circ - 120^\circ = 60^\circ$.
Thus, the third angle measures $60^\circ$.
Side and Angle Conditions for a Triangle – Practice
Practice what you have learned on your own!
A triangle has angles measuring $50^\circ$ and $60^\circ$. What is the measure of the third angle?
Using the angle sum theorem, calculate the third angle: $180^\circ - (50^\circ + 60^\circ) = 70^\circ$.
If two angles of a triangle are $35^\circ$ and $95^\circ$, what is the measure of the remaining angle?
The sum of the angles in a triangle is $180^\circ$. Thus, the remaining angle is $180^\circ - (35^\circ + 95^\circ) = 50^\circ$.
Determine the measure of the smallest angle in a triangle if the other two angles are $75^\circ$ and $55^\circ$.
Calculate the smallest angle: $180^\circ - (75^\circ + 55^\circ) = 50^\circ$.
Can a triangle have sides measuring $2$ cm, $3$ cm, and $5$ cm?
No, because $2 \text{ cm} + 3 \text{ cm} < 5 \text{ cm}$, which violates the triangle inequality theorem.
Is it possible to form a triangle with side lengths of $7$ cm, $8$ cm, and $15$ cm?
No, $7 \text{ cm} + 8 \text{ cm} = 15 \text{ cm}$, which means these sides lie on a straight line, not forming a triangle.
Verify if side lengths $6$ cm, $7$ cm, and $12$ cm can construct a triangle.
No, since $6 \text{ cm} + 7 \text{ cm} = 13 \text{ cm}$, which is not greater than $12 \text{ cm}$, they cannot form a triangle.
If a triangle has one angle measuring $120^\circ$, and the second angle is half the third angle, find the measures of the other two angles.
Let the third angle be $2x$ and the second angle be $x$. Solve $120^\circ + x + 2x = 180^\circ$, giving $x = 20^\circ$. So, the angles are $20^\circ$ and $40^\circ$.
A triangle with sides $a$, $b$, and $c$ has $a = 3b$ and area $50$ square units. If $b = 5$ cm and $c = 10$ cm, find the area of the triangle.
Since $b = 5$ cm, $a = 3 \times 5 = 15$ cm. Apply Heron's formula with these sides to confirm if the area is $50$ square units. The formula will not hold, indicating incorrect side lengths.
Can a triangle with an angle of $90^\circ$ have the remaining angles measuring $50^\circ$ and $60^\circ$?
No, because $90^\circ + 50^\circ + 60^\circ = 200^\circ$, which exceeds $180^\circ$, violating the angle sum theorem.
Side and Angle Conditions for a Triangle – Summary
Key Learnings from this Text:
Triangles are three-sided figures with interior angles that add up to 180 degrees.
The triangle inequality theorem is crucial for understanding whether a set of three lengths can form a triangle.
Understanding these basic properties helps in solving more complex problems involving triangles and their properties in both academic studies and real-life applications.
Side and Angle Conditions for a Triangle – Frequently Asked Questions
What is the triangle inequality theorem?
The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This ensures that the sides can meet to form a triangle.
Why must the sum of the interior angles of any triangle equal 180 degrees?
The sum of the interior angles in a triangle always equals 180 degrees due to the geometric properties of flat, two-dimensional spaces where triangles exist. This is a fundamental rule in Euclidean geometry.
Can a triangle have a 90-degree angle and still adhere to the angle sum theorem?
Yes, a triangle can have a 90-degree angle, known as a right triangle, and still adhere to the angle sum theorem. The other two angles must sum to 90 degrees, maintaining the total of 180 degrees.
What happens if the sum of two sides equals the third side?
If the sum of two sides equals the third side, the figure cannot form a triangle and is instead considered a straight line. This situation does not meet the conditions of the triangle inequality theorem.
How can the angle sum theorem be applied to determine the measure of an unknown angle in a triangle?
To find an unknown angle in a triangle, add the measures of the known angles and subtract from 180 degrees. The result is the measure of the unknown angle.
Is it possible for a triangle to have all angles of equal measure?
Yes, a triangle with all angles of equal measure is called an equilateral triangle, and each angle is 60 degrees, satisfying the angle sum theorem.
Can the triangle inequality theorem be used to determine if three lengths can form a triangle without actually drawing it?
Yes, by applying the triangle inequality theorem (checking if the sum of any two sides is greater than the third), you can determine if three lengths can form a triangle without needing to draw it.
What are some real-world applications of understanding triangle side and angle conditions?
Understanding triangle side and angle conditions is crucial in fields such as architecture, engineering, and construction, where the structural integrity often relies on the properties of triangles.
Can a triangle have one angle greater than 90 degrees?
Yes, a triangle can have one angle greater than 90 degrees, known as an obtuse triangle, as long as the sum of all three angles remains 180 degrees.
How does knowing the side and angle conditions help in constructing triangles?
Knowing these conditions helps ensure that the sides and angles chosen can form a valid triangle, which is essential in geometric constructions and various design tasks where specific triangle types are needed.
Transcript Side and Angle Conditions for a Triangle
Pharaoh Ahmose enjoys a luxurious lifestyle, but there's something missing. He has heard stories of the kites which thrill the emperors in China. He wants a kite to call his own. So, Ahmose summons his engineers, and challenges them to build a triangular kite that will soar like a falcon. The engineers aren't sure they can actually form a triangle from the materials they have. They'll need to experiment to understand Side and Angle Conditions for a Triangle. The engineers are given three rods to form the supporting elements for the kite. The rods are 150 cm 70 cm and 30 cm long. We'll label the rods 'a', 'b,' and 'c,' respectively. Let's leave 'a,' the largest segment, here. Now let's try and assemble a triangle by placing the remaining segments on its ends. Can we complete a triangle? No! We can change the angles of 'b' and 'c' as much as we like, but no matter how much we try, they won't reach each other to complete a triangle. Together, they are just too short. Notice that 'b' and 'c' sum to be 100 cm which is less than 150 cm, the length of 'a'. What can we do to make this a complete triangle? Let's replace the 30 cm rod with one of 100 cm. Look! That makes a complete triangle! The two shorter sides of 70 cm and 100 cm sum to be 170 cm. That's greater than the longest side of 150 cm. In general, the two other sides must sum to be more than the longest side. Given three sides this is one of the conditions we must meet in order to form a triangle. The engineers want to try out different side lengths given two sides. Let's remove the 100 cm side and experiment. Given any two sides, what are the lower and upper limits for the third side? Let's start by finding the smallest measurement that this side, 'c,' can be. Let's rotate 'a' so that its angle is really small. Can 'c' be 80 cm? No! Because that would just give us a straight line and not a triangle. But could 'c' be 81 cm? Or even 85 cm? Yes! In fact, 'c' can be anything more than 80 cm. Because 80 is the difference of the two given sides, 150 and 70. Anything greater than that difference should be fine, right? Oh, but wait! What's the biggest that 'c' can be? If we swing side 'a' so that this angle is really big, then the side 'c' gets longer. Can 'c' be 220 cm? No! That just makes another straight line and not a triangle. So 'c' has to be less than 220 which is the SUM of the two given sides, 150 and 70. Anything less than the sum of the other two sides works. Now we can put these limits together. Given any two sides we can find the range of values for the third side. Any third side must be greater than the difference and less than the sum of the two given sides. In this case, 'c' must be greater than 80 and less than 220. But a triangle consists of more than just three sides. It also has three angles! How can we determine a triangle by just looking at two angles? Let's experiment again. First, let's construct a horizontal line segment 'AB'. Now, let's place our protractor at point A and mark a measure of 50 degrees. We'll use this mark to construct a ray, that is a line with a start point but no end point. Now from point 'B' let's measure a 150 degree angle. We'll construct a ray here as well. Do these two angles give us a triangle? No, clearly the rays will never intersect! Notice that the sum of the two angles is 200 degrees! But what if we made angle 'B' smaller? Notice that as the sum of the angles gets smaller we get closer to a triangle. In fact, any pair of angles in a triangle must sum to less than 180 degrees in order to form a triangle. Now the engineers understand the limits on the sides and angles they can use to build the triangular kite! Let's review side and angle conditions for forming a triangle as they complete their construction. Given three sides, the longest side must be less than the sum of the other two sides. Given any two sides, the third side must be greater than their difference, and less than their sum. Finally, two angle measurements determine many triangles provided that the sum of any two angles is less than 180 degrees. With all this knowledge the engineers construct a beautiful kite that soars like a falcon! But Ahmose, you can't just watch a kite, you've got to fly it.
Side and Angle Conditions for a Triangle exercise
Identify the measurements of the three sides of a triangle.
To form a triangle the sum of the two smaller sides must be greater than the longest side.
If we have a side of $150$ cm and the sum of the other two sides is less than that, they will not meet. For example, $70 + 30 = 100$
as $100 < 150$
There are $3$ correct answers here.
Find a possible measurement for a third side of a triangle, given two sides.
The third side has to ensure it creates a triangle.
In this example, one side is $9$ cm and the other side is $5$ cm. In order to form a triangle the third side needs to be $> 4$ as $4 + 5 = 9$ and that would form a line, not a triangle.
There are $2$ correct answers.
The correct answer is $18$ cm and $20$ cm.
Determine the side and angle conditions to form a triangle.
The sum of the angles in a triangle is $180^\circ$. Anything more and the angles will not meet.
For example, $120 + 70 > 180$, therefore will not form a triangle.
The sides must meet, so the longest side should be smaller than the sum of the other two sides.
For example, sides of $10$ cm, $6$ cm and $3$ cm will not meet as $6 + 3 < 10$.
There are $3$ correct answers.
The $3$ correct side and angle conditions are above.
The $2$ that are not correct are:
angles in a triangle have a sum of $180^\circ$ and not $360^\circ$
there can be a right angle in a triangle, but not in all triangles.
Create a triangular flowerbed.
Grandpa will need two sides which will make a fairly large triangle. As you can see above, if the two sides are small, Grandpa will not be able to get many flowers in there.
The sum of the two sides will have to be longer than $28$ ft or they will not form a triangle.
For example, with sides of $2$ ft, $3$ ft and $7$ ft we cannot form a triangle.
Sides $18$ ft and $21$ ft will add to $28$ ft to give a good-sized triangular flowerbed.
The others do not work as:
they either will not meet - Sides $14$ ft and $12$ ft
they form a straight line - Sides $14$ ft and $14$ ft
or a very small triangle - Sides $10$ ft and $19$ ft
Identify the angle pair that would be impossible for a triangle.
The sum of the angles in a triangle is $180^\circ$.
For example $40 + 35 + 105 = 180^\circ$
We are looking for all $3$ angles in a triangle to $= 180^\circ$. Therefore, $2$ angles that have a sum greater than $180^\circ$ cannot form a triangle.
For example, if we have angles of $100^\circ$ and $110^\circ$ then that cannot possibly form a triangle as $100 + 110 > 180$.
The correct answer is angles $132^\circ$ and $55^\circ$.
This will not work because
angles in a triangle $= 180^\circ$
here we have $132 + 55 = 187 > 180$
the lines will not meet
the other pairs of angles are $< 180^\circ$
Limits of sides to create a triangle.
For the lower limit it has to be greater than a value which will create a straight line.
For example, if one side is $9$ cm and the other side is $8$ cm then we could form a line if we added a side of $1$ cm.
$9 - 8 = 1$.
Therefore, we need the third side to be greater than $1$ cm.
For the upper limit it has to be greater than a value which will create a straight line when the two sides are added.
For example, if one side is $9$ cm and the other side is $8$ cm then we could form a line if we added $9 + 8 = 17$.
Therefore, we need the third side to be less than $17$ cm in order to form a triangle.
Constructing Triangles
Side and Angle Conditions for a Triangle
Determining a Unique Triangle Given Two Sides and a Non-included Angle.
The Angle Sum Theorem
Interior and Exterior Angles of a Triangle |
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a-d a a +d arithmetic progression | a-3d a-d a+d a+3d arithmetic progression | a-2d a-d a a+d a+2d
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a-d a a +d arithmetic progression | a-3d a-d a+d a+3d arithmetic progression | a-2d a-d a a+d a+2d
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a-d a a +d arithmetic progression | a-3d a-d a+d a+3d arithmetic progression | a-2d a-d a a+d a+2d , a-d, a , a + d three consecutive terms in an arithmetic progression, a-3d, a -d , a+d, a+3d four consecutive terms in an arithmetic progression, a-2d, a-d, a, a+d, a+2d five consecutive terms in an arithmetic progression, ------------------------------------------------------------------------------------------------------------------------ Why three consecutive terms in an arithmetic progression a -d, a , a +d , Why four consecutive terms in an arithmetic progression a-3d, a-d, a+d, a+3d, Why five consecutive terms in an arithmetic progression a-2d, a-d, a , a+d, a+2d, -------------------------------------------------------------------------------------------------------------------------- (i) If the sum of three terms in Arithmetic Progression be given, assume the numbers as a - d, a and a + d. Here common difference is d. (ii) If the sum of four terms in Arithmetic Progression be given, assume the numbers as a - 3d, a - d, a + d and a + 3d. (iii) If the sum of five terms in Arithmetic Progression be given, assume the numbers as a - 2d, a - d, a, a + d and a + 2d. Here common difference is 2d. (iv) If the sum of six terms in Arithmetic Progression be given, assume the numbers as a - 5d, a - 3d, a - d, a + d, a + 3d and a + 5d. Here common difference is 2d. -------------------------------------------------------------------------------------------------------- What are the 5 terms in AP, How do you assume 4 terms on an AP, What is the nth formula, what is A and D in AP, What are the first 4 terms of an AP, What is an in AP class 10? Can (a-3d), (a-d), (a+d), (a+3d) be taken as four consecutive terms in AP? If we want to choose 4 consecutive number then why do we choose (a - 3d), (a - d), (a + d), (a + 3d), why we have not chosen (a-2d), (a-d) ,(a+d), (a+2d). -------------------------------------------------------------------------------------------------------- class 10 maths chapter 5, class 10 maths ex 5.1, class 10 maths ex 5.2, class 10 maths arithmetic progression, class 10 maths ap, class 10 maths arithmetic progression full chapter ---------------------------------------------------------------------------------------------------------- #vipraminds#class10maths#arithmeticprogression…...more
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a-d a a +d arithmetic progression | a-3d a-d a+d a+3d arithmetic progression | a-2d a-d a a+d a+2d
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8396 | https://www.youtube.com/watch?v=UsMom3yNsX8 | Alternating series test | Series | AP Calculus BC | Khan Academy
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When a series alternates (plus, minus, plus, minus,...) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0.
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let's now expose ourselves to another test of convergence and that's the alternating series test and i'll explain the alternating series test and i'll apply it to an actual series while i do it to make the the the explanation of the alternating series test a little bit more concrete so let's say that i have some series some infinite series let's say it goes from n equals k to infinity of a sub n and let's say i can write it as or i can rewrite a sub n so let's say a sub n i can write so a sub n is equal to negative 1 to the n times b sub n or a sub n is equal to negative 1 to the n plus 1 times b sub n where where b sub n is greater than or equal to 0 for all the n's we care about so for all of these integer n's greater than or equal to k so if all of these things if all of these things are true and we know two more things and we know number one the limit as n approaches infinity of b sub n is equal to zero and number two b sub n is a decreasing sequence decreasing decreasing sequence then that lets us know that the original infinite series the original infinite series is going to converge so this might seem a little bit abstract right now let's actually you show let's let's use this with an actual series to make it a little bit more a little bit more concrete so let's say that i had the series let's say i had the series from n equals 1 to infinity of negative 1 to the n over n and we could write it out just to make this series a little bit more concrete when n is equal to 1 this is going to be negative 1 to the 1 power actually let's just make this a little bit let's make this a little bit more interesting let's make this negative 1 to the n plus 1. so when n is equal to 1 this is going to be negative 1 squared over 1 which is going to be 1 and then when n is 2 it's going to be negative 1 to the third power which is going to be negative one-half so it's minus one half plus one third minus one fourth plus minus and it just keeps going on and on and on forever now can we rewrite this a sub n like this well sure the negative one to the n plus 1 is actually explicitly called out we can rewrite we can rewrite our a sub n so let me do that so negative so a sub n which is equal to negative 1 to the n plus 1 over n this is clearly the same thing as negative 1 to the n plus 1 times 1 over n which is which we can then say this thing right over here could be our b sub n so this right over here is our b sub n and we can verify that our b sub n is going to be greater than or equal to zero for all the ends we care about so our b sub n is equal to one over n now clearly this is going to be greater than or equal to zero for any for any positive n now what's the limit as b sub n approach what's the limit of b sub n as n approaches infinity the limit of let me just write 1 over n 1 over n as n approaches infinity is going to be equal to 0 so we satisfy the first constraint and then this is clearly a decreasing sequence as n as n increases the denominators are going to increase and with a larger denominator you're going to have a lower value so we can also say 1 over n is a decreasing decreasing sequence for the ends that we care about so this is this satisfies this is satisfied as well and so based on that this thing this thing is always this thing right over here is always greater than or equal to zero the limit as one over n or is our b sub n as n approaches infinity is going to be zero it's a decreasing sequence therefore we can say that our original series actually converges so n equals 1 to infinity of negative 1 to the n plus 1 over n and that's kind of interesting because we've already seen that if all of these were positive if all of these terms were positive we just have the harmonic series and that one didn't converge but this one did putting these negatives here do the trick and actually we can prove this one over here converges using other techniques and maybe if we have time actually in particular the limit comparison test i'll just throw that out there in case you are curious so this is a pretty powerful tool it looks a little bit about like that divergence test but remember the divergence test is really is only useful if you want to show something diverges if the limit of if the limit of your terms do not approach zero then you say okay that thing is going to diverge this thing is useful because you can actually prove convergence now once again if something does not pass the alternating series test that does not necessarily mean that it diverges it just means that you couldn't use the alternating series test to prove that it converges |
8397 | https://virtuallabs.merlot.org/vl_chemistry.html | Chemistry
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8398 | https://math.stackexchange.com/questions/549467/how-many-ways-is-there-to-do-a-round-trip-if-at-least-one-of-the-roads-taken-on | combinatorics - How many ways is there to do a round trip if at least one of the roads taken on the return trip is different? - Mathematics Stack Exchange
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How many ways is there to do a round trip if at least one of the roads taken on the return trip is different?
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I'm stuck on part c) of this question. The answer key gives 182. I already know there are 14 ways to make the trip from city A and city B and vice versa. It appears 182 came from 14×13 14×13 but I don't get where 13 came from? If neither R 8 R 8 or R 9 R 9 were used on the trip there, one of the roads in the middle would need to be removed from the return trip so that would become 2×4 2×4 or 3×3 3×3.
By the way, what's the policy for this site regarding typing out questions instead of putting images for them? Normally I would try but if a person can't view images then they wouldn't be able to see the graph and wouldn't be able to help anyways.
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asked Nov 2, 2013 at 21:47
CeleritasCeleritas
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Hint: Round trip possibilities: #Trip forward ×× #trip backward. Thus 14 14-1 (1 road is different, mainly the original 1). Images are great if they are clear. As long as you show your work, or at an attempt at minimum!Don Larynx –Don Larynx 2013-11-02 21:53:13 +00:00 Commented Nov 2, 2013 at 21:53
@DonLarynx but I disagree; it shouldn't be 14-1 because if you take off R5 fore example, then there isn't 13 routes but 10.Celeritas –Celeritas 2013-11-02 21:58:45 +00:00 Commented Nov 2, 2013 at 21:58
R 5 R 5 is one path, what are you on mate?Don Larynx –Don Larynx 2013-11-02 22:00:42 +00:00 Commented Nov 2, 2013 at 22:00
3 Crack cocaine..Celeritas –Celeritas 2013-11-02 22:09:07 +00:00 Commented Nov 2, 2013 at 22:09
1 If there are 14 14 ways from A A to C C, to count the round trips not using exactly the same route backwards from C C to A A, do it by first choosing one of the 14 14 ways from A A to C C, and then choosing one of the 13 13 ways which differ from the way you already took from A A to C C, and going back from C C to A A by that route. There are then 14∗13 14∗13 ways as the answer says. [There is no need to get into specifics about the form of the routes involved, each route is reversible.]coffeemath –coffeemath 2013-11-03 01:35:03 +00:00 Commented Nov 3, 2013 at 1:35
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To plan a round trip from A A to C C and back, where the return route is not the reverse of the first part from A A to C C, we proceed as follows. Since there are 14 14 ways to go from A A to C C we must choose one of those ways. Once that has been done, exactly one of the 14 14 routes has been ruled out for reversal to get back to A A from C C, leaving us 13 13 choices for the return trip. This gives 14⋅13=182 14⋅13=182 possible round trips.
Note that we do not have to consider the specific types of these trips in terms of how they go through B B (or avoid B B). One of the trips for the first leg of the journey from A A to C C being chosen, it is the only one excluded on the way back.
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answered Nov 3, 2013 at 11:32
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The second part of the question mentions "to calculate the number of different round trips" won't the answer for that part be 1413 = 182, I say this because the question doesn't mention to consider repetition of paths.AnkitSablok –AnkitSablok 2014-07-04 18:57:05 +00:00 Commented Jul 4, 2014 at 18:57
@AnkitSablok No: Different round trips is 1414, that is, one of 14 ways A to C followed by one of 14 ways from C to A. By saying "different round trip" means the overall trips are different. In part (c) the extra requirement that the return trip differs from being the exact reverse of the initial trip thus rules only one of them out.coffeemath –coffeemath 2014-07-04 23:23:31 +00:00 Commented Jul 4, 2014 at 23:23
You are correct in saying that, but the language of the question seemed a little ambiguous to me :)AnkitSablok –AnkitSablok 2014-07-05 00:56:45 +00:00 Commented Jul 5, 2014 at 0:56
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8399 | https://www.vedantu.com/question-answer/how-do-you-solve-the-inequality-32x7-class-11-maths-cbse-5ff315b792cefc2e0d600e9d | How do you solve the inequality \[3-2x7\]?
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How do you solve the inequality 3−2 x>7?
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Hint: To solve this linear inequality in one variable, we have to take the variable terms to one side of the inequality, and the constant terms to the other side. Inequalities do not provide a fixed value as the solution, it gives a range. All the values in this range hold the inequality. To solve the inequality we should know some of the properties of the inequality as follows, given that a>b. We can state the following from this.
a+k>b+k,k∈Real numbers
a k>b k,k∈Positive real numbers
a k<b k,k∈Negative real numbers
Complete step-by-step solution:
We are given the inequality, 3−2 x>7.
Subtracting 7 from both sides of the above inequality, we get
⇒3−2 x−7>7−7
⇒−2 x−4>0
Adding 2 x to both sides of the above inequality, we get
⇒−2 x−4+2 x>0+2 x⇒−4>2 x
By multiplying or dividing an inequality by a positive quantity, the inequality sign does not change. Dividing both sides of the above inequality by 2, we get
⇒−4 2>2 x 2
Canceling out 2 as a common factor of numerator and denominator for the RHS of inequality, we get
⇒−2>x
Hence the solution of the given inequality is x<−2.
Note: We can also use a different property to solve the given inequality, as follows
3−2 x>7
Subtracting 3 from both sides of the inequality, we get
⇒3−2 x−3>7−3⇒−2 x>4
By multiplying or dividing an inequality by a negative quantity, the inequality sign changes. Dividing both sides of the above inequality by −2, we get
⇒−2 x−2<4−2⇒x<−2
We are getting the same range from both of the methods.
We can check whether the solution is correct or not by substituting any value in the range we got. Let’s substitute x=−3 in the inequality, the LHS of the inequality becomes
⇒3−2(−3)=9
And the RHS is 7. From this as, L H S>R H S the solution is correct.
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