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17100	https://www.nationalelfservice.net/dentistry/oral-medicine-and-pathology/recurrent-aphthous-stomatitis-topical-interventions/	Recurrent aphthous stomatitis – topical interventions - National Elf Service Menu ElfHelp Social Media Video Podcast Webinar Training Contact News Search National Elf Service No bias. No misinformation. No spin. Just what you need! 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Three forms are recognised minor, major and herpetiform with a peak age of onset of 10-19 years of age although they can persist into adulthood. Their aetiology and pathogenesis is not fully understood and treatments are mainly symptomatic with the main objectives being, relieve of pain, promoting lesion healing and increasing the interval period. Topical medication, laser, cryotherapy and cautery are considered to be effective topical treatments The aim of this review was to compare the efficacy and safety of topical interventions used for recurrent aphthous stomatitis. Methods A review protocol was registered with PROSPERO. Searches were conducted in the Cochrane Central Register of Controlled Trials (CENTRAL), PubMed, Embase and the Web of Science. Randomised controlled trials (RCTs) of local interventions in patients with a confirmed diagnosis of RAS were considered. Main outcomes were clinical efficacy assessed as ulcer size reduction, symptom reduction and time to healing and safety. Two reviewers screened and selected studies and extracted data. Study quality was assessed using the Cochrane domain-based approach. Standard pairwise and network meta-analysis were conducted. Risk difference (RD)with 95% confidence intervals (CI) was used for dichotomous variables and mean difference (MD) with 95%CI for continuous variables Results 72 studies were included including 3 three-arm studies and 2 four-arm studies. Most studies involved patients with minor RAS. Patients ages ranged from 6.8 to 71 years. Healing effect (26 RCTs ,1306 patients) 14 RCTs (820 patients) were pooled in 7 pairwise comparisons with triamcinolone, laser, silver nitrate, and honey showing a significant preference compared to placebo. A network meta-analysis (NMA) of 26 RCTs) showed shorter healing times and better healing effects relative to placebo for honey, insulin liposome gel and laser with no significant differences for other comparison. Compared to placebo the interventions were ranked in order as Insulin-liposomal gel, honey, laser, penicillin and aloe. Size-reducing effect (37 RCTs, 3587 patients) Data from 23 RCTs (1807 patients) were pooled in 8 eight pairwise comparisons for size-reducing effect triamcinolone, probiotics, glycyrrhiza and amlexanox were significantly better than placebo. Data from a NMA of 37 RCTs suggested that amlexanox, glycyrrhiza and triamcinolone were more effective than placebo with no significant differences for other interventions. Symptom-reducing effect (46 RCTs, 4020 patients) 9 pairwise comparisons (32 RCTs 2940 patients) were pooled for symptom-reducing effect. Triamcinolone, laser, glycyrrhiza, amlexanox and aloe were considered superior to placebo. Data from a NMA of 46 RCTs suggested that Amlexanox, laser and triamcinolone have a significant advantage over placebo with no statistically significant differences for other interventions. Conclusions The authors concluded: – most of the local interventions did not show significant differences in the efficacy evaluation and safety evaluation. Based on the currently available evidence, we recommend the use of laser as a short-term intervention to promote healing and reduce pain during the exacerbation phase of RAS and probiotics as a long-term intervention to prolong the inter-episode period and reduce recurrence during the exacerbation and remission phases of RAS. We call for more large-scale RCTs based on trustworthy standards. Comments 10 years have now elapsed since a Cochrane review looked at systemic treatment for recurrent aphthous stomatitis(Dental Elf – 14 th Sep 2012) and since then this blog has looked at a number of other reviews of specific topical and systemic interventions (Dental Elf RAS blogs). This new review had a protocol published on the PROSPERO database and searched a good range of databases for RCTs of topical treatments for RAS. The reviewers identified 72 studies examining some 29 interventions , including placebo, allicin, aloe, amlexanox, benzydamine, berberine gelatin, chitosan chlorhexidine, clobetasol, cryotherapy, curcumin, dexamethasone, diosmectite, doxycycline, glycyrrhiza, honey, insulin-liposomal gel, laser, minocycline, N-acetylcysteine, penicillin, probiotics, prostaglandin E2, quercetin, silver nitrate, sucralfate, triamcinolone, triester glycerol oxide and zinc. Of the 72 studies 23 were assessed as being at high risk of bias and only 4 at low risk with the remainder being at unclear risk. A number of the interventions were only tested in single RCTs and there is variation in the format, dosage and delivery and duration of the interventions in the included studies. In addition, studies on the recurrence of RAS are lacking. So, while this review provides a helpful overview of topical interventions that have been used for RAS the findings should be interpreted cautiously because of limitations in the quality of the evidence. Links Primary Paper Liu H, Tan L, Fu G, Chen L, Tan H. Efficacy of Topical Intervention for Recurrent Aphthous Stomatitis: A Network Meta-Analysis. Medicina (Kaunas). 2022 Jun 7;58(6):771. doi: 10.3390/medicina58060771. PMID: 35744034; PMCID: PMC9227309. Review protocol on PROSPERO Other references Dental Elf – 14 th Sep 2012 Dental Elf RAS blogs Picture Credits CC BY-SA 3.0, Share on Facebook Tweet this on Twitter Share on LinkedIn Share on Google+ Mark as read Share this post: Share on Facebook Tweet this on Twitter Share on LinkedIn Share on Google+Share via email Create a personal elf note about this blog Tagged with: aloe vera, benzydamine hydrochloride, berberine, chlorhexidine, clobetasol, cryotherapy, curcumin, dexamethasone, honey, lasers, minocycline, oral medicine, oral ulceration, probiotics, recurrent aphthae, recurrent aphthous stomatitis, zinc Derek Richards Derek Richards is a specialist in dental public health, Director of the Centre for Evidence-Based Dentistry and Specialist Advisor to the Scottish Dental Clinical Effectiveness Programme (SDCEP) Development Team. A former editor of the Evidence-Based Dentistry Journal and chief blogger for the Dental Elf website until December 2023. Derek has been involved with a wide range of evidence-based initiatives both nationally and internationally since 1994. Derek retired from the NHS in 2019 remaining as a part-time senior lecturer at Dundee Dental School until the end of 2023. More posts - Website Follow me here – Twitter LinkedIn Logging In... Profile cancel Sign in with TwitterSign in with Facebook or Comment Name Email Not published Website FOLLOW the Dental Elf Find us on Facebook Follow us on Twitter Follow us on Google+ Read our RSS feed Find us on LinkedIn Are You A Researcher? We can help you share your findings too — find out how. Recent Posts Does breastfeeding increase Early Childhood Caries? How do antiresorptive drugs affect the success of dental implants? Does antibiotic prophylaxis reduce endocarditis risk after dental procedures? Cracked Teeth Syndrome: Impact of Various Treatments on Survival Rates Does motivational Interviewing improve oral health? 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17101	https://www.dir.ca.gov/title8/5220d.html	California Code of Regulations, Title 8, Section 5220. Ethylene Oxide, Appendix D Skip to Main Content This information is provided free of charge by the Department of Industrial Relations from its web site at www.dir.ca.gov. These regulations are for the convenience of the user and no representation or warranty is made that the information is current or accurate. See full disclaimer at Subchapter 7. General Industry Safety Orders Group 16. Control of Hazardous Substances Article 110. Regulated Carcinogens Return to index New query §5220. Ethylene Oxide, Appendix D Sampling and Analytical Method for Ethylene Oxide This appendix contains details for the method which has been tested at the OSHA Analytical Laboratory in Salt Lake City. Inclusion of this method in the appendix does not mean that this method is the only one which will be satisfactory. Employers who note problems with sample breakthrough using the OSHA or other charcoal methods should try larger charcoal tubes. Tubes of larger capacity are available. In addition, lower flow rates and shorter sampling times should be beneficial in minimizing breakthrough problems. ETHYLENE OXIDE METHOD NO. : 30 Matrix: Air. Target Concentration: 1.0 ppm (1.8 mg/m 3). Procedure: Samples are collected on two charcoal tubes in series and desorbed with 1% CS 2 in benzene. The samples are derivatized with HBr and treated with sodium carbonate. Analysis is done by gas chromatography with an electron capture detector. Recommended Air Volume and Sampling Rate: 1 liter and 0.05 Lpm. Detection Limit of the Overall Procedure: 13.3 ppb (0.024 mg/m 3) (Based on 1.0 liter air sample). Reliable Quantitation Limit: 52.2 ppb (0.094 mg/m 3) (Based on 1.0 liter air sample). Standard Error of Estimate: 6.59%. Special Requirements: Samples must be analyzed within 15 days of sampling date. Status of Method: The sampling and analytical method has been subjected to the established evaluation procedures of the Organic Method Evaluations Branch. General Discussion. 1.1 Background. 1.1.1 History of Procedure. In studies to develop a method for the analysis of EtO at very low concentrations, it was found that the reaction of EtO with HBr (hydrobromic acid) gave a derivative, 2-bromoethanol, readily detectable by an ECD (electron capture detector) due to the presence of the bromine. Of solvents tested for their response on the ECD and their ability to desorb EtO from the charcoal, benzene was the only solvent tested that gave a suitable response on the ECD and a high desorption. The desorption efficiency was improved by using 1% CS 2 (carbon disulfide) with the benzene. 1.1.2 Physical Properties. See Section 5220, Appendix B. 1.2 Limit Defining Parameters. 1.2.1 Detection Limit of the Analytical Procedure. The detection limit of the analytical procedure is 12.0 picograms of ethylene oxide per injection. This is the amount of analyte which will give a peak whose height is five times the height of the baseline noise. 1.2.2 Detection Limit of the Overall Procedure. The detection limit of the overall procedure is 24.0 nanograms of ethylene oxide per sample. This is the amount of analyte spiked on the sampling device which allows recovery of an amount of analyte equivalent to the detection limit of the analytical procedure. 1.2.3 Reliable Quantitation Limit. The reliable quantitation limit is 94.0 nanograms of ethylene oxide per sample. This is the smallest amount of analyte which can be quantiated within the requirements of 75% recovery and 95% confidence limits. It must be recognized that the reliable quantitation limit and detection limit are based upon optimization of the instrument for the smallest possible amount of analyte. When the target concentration of an analyte is exceptionally higher than these limits, they may not be attainable at the routine operating parameters. The limits reported on analysis reports must be based on the operating parameters used during the analysis of the samples. 1.2.4 Sensitivity. The sensitivity of the analytical procedure over a concentration range representing 0.5 to 2 times the target concentration based on the recommended air volume is 34105 area units per µg/mL. The sensitivity is determined by the slope of the calibration curve. The sensitivity will vary somewhat with the particular instrument used in the analysis. 1.2.5 Recovery. The recovery of analyte from the collection medium must be 75% or greater. The average recovery from spiked samples over the range of 0.5 to 2 times the target concentration is 88.0%. At lower concentrations the recovery appears to be non-linear. 1.2.6 Precision (Analytical Method Only). The pooled coefficient of variation obtained from replicate determinations of analytical standards at 0.5X, 1X and 2X the target concentration is 0.036. 1.2.7 Precision (Overall Procedure). The overall procedure must provide results at the target concentration that are 25% or better at the 95% confidence level. The precision at the 95% confidence level for the 15-day storage test is plus or minus 12.9%. This includes an additional plus or minus 5% for sampling error. 1.3 Advantages. 1.3.1 The sampling procedure is convenient. 1.3.2 The analytical procedure is very sensitive and reproducible. 1.3.3 Reanalysis of samples is possible. 1.3.4 Samples are stable for at least 15 days at room temperature. 1.3.5 Interferences are reduced by the long GC retention time of the derivative. 1.4 Disadvantages. 1.4.1 Two tubes in series must be used because of possible breakthrough and migration. 1.4.2 The precision of the sampling rate may be limited by the reproducibility of the pressure drop across the tubes. The pumps are usually calibrated for one tube only. 1.4.3 The use of benzene as the desorption solvent increases the hazards of analysis because of the potential carcinogenic effects of benzene. 1.4.4 After repeated injections there can be a buildup of residue formed on the electron capture detector which decreases sensitivity. 1.4.5 Recovery from the charcoal tubes appears to be nonlinear at low concentrations. Sampling Procedure. 2.1 Apparatus. 2.1.1 A calibrated personal sampling pump whose flow can be determined within plus or minus 5% of the recommended flow. 2.1.2 Charcoal tubes: glass tube with both ends flame sealed, 7 cm long with a 6-mm O.D. and a 4-mm I.D., containing 2 sections of coconut shell charcoal separated by a 3-mm portion of urethane foam. The adsorbing section contains 100 mg of charcoal, the backup section 50 mg. A 3-mm portion of urethane foam is placed between the outlet end of the tube and the backup section. A plug of silylated glass wool is placed in front of the adsorbing section. 2.2 Reagents. 2.2.1 None required. 2.3 Sampling Technique. 2.3.1 Immediately before sampling, break the ends of the charcoal tubes. All tubes must be from the same lot. 2.3.2 Connect two tubes in series to the sampling pump with a short section of flexible tubing. A minimum amount of tubing is used to connect the two sampling tubes together. The tube closer to the pump is used as a backup. This tube should be identified as the backup tube. 2.3.3 The tubes should be placed in a vertical position during sampling to minimize channeling. 2.3.4 Air being sampled should not pass through any hose or tubing before entering the charcoal tubes. 2.3.5 Seal the charcoal tubes with plastic caps immediately after sampling. Also, seal each sample with OSHA seals lengthwise. 2.3.6 With each batch of samples, submit at least one blank tube from the same lot used for samples. This tube should be subjected to exactly the same handling as the samples (break, seal, transport) except that no air is drawn through it. 2.3.7 Transport the samples (and corresponding paperwork) to the lab for analysis. 2.3.8 If bulk samples are submitted for analysis, they should be transported in glass containers with Teflon-lined caps. These samples must be mailed separately from the container used for the charcoal tubes. 2.4 Breakthrough. 2.4.1 The breakthrough (5% breakthrough) volume for a 3.0 mg/m 3 ethylene oxide sample stream at approximately 85% relative humidity, 22° C and 633 mm is 2.6 liters sampled at 0.05 liters per minute. This is equivalent to 7.8 µg of ethylene oxide. Upon saturation of the tube it appeared that the water may be displacing ethylene oxide during sampling. 2.5 Desorption Efficiency. 2.5.1 The desorption efficiency, from liquid injection onto charcoal tubes, averaged 88.0% from 0.5 to 2.0 x the target concentration for a 1.0-liter air sample. At lower ranges it appears that the desorption efficiency is non-linear. 2.5.2 The desorption efficiency may vary from one laboratory to another and also from one lot of charcoal to another. Thus, it is necessary to determine the desorption efficiency for a particular lot of charcoal. 2.6 Recommended Air Volume and Sampling Rate. 2.6.1 The recommended air volume is 1.0 liter. 2.6.2 The recommended maximum sampling rate is 0.05 Lpm. 2.7 Interferences. 2.7.1 Ethylene glycol and Freon 12 at target concentration levels did not interfere with the collection of ethylene oxide. 2.7.2 Suspected interferences should be listed on the sample data sheets. 2.7.3 The relative humidity may affect the sampling procedure. 2.8 Safety Precautions. 2.8.1 Attach the sampling equipment to the employee so that it does not interfere with work performance. 2.8.2 Wear safety glasses when breaking the ends of the sampling tubes. 2.8.3 If possible, place the sampling tubes in a holder so the sharp end is not exposed while sampling. Analytical Method. 3.1 Apparatus. 3.1.1 Gas chromatograph equipped with a linearized electron capture detector. 3.1.2 GC column capable of separating the derivative of ethylene oxide (2-bromoethanol) from any interferences and the 1% CS 2 in benzene solvent. The column used for validation studies was: 10 ft x 1/8-inch stainless steel 20% SP-2100, 0.1% Carbowax 1500 on 100/120 Supelcoport. 3.1.3 An electronic integrator or some other suitable method of measuring peak areas. 3.1.4 Two milliliter vials with Teflon-lined caps. 3.1.5 Gas tight syringe--500 L or other convenient sizes for preparing standards. 3.1.6 Microliter syringes--10 L or other convenient size for diluting standards and 1 L for sample injections. 3.1.7 Pipets for dispensing the 1% CS 2 in benzene solvent. The Glenco 1 mL dispenser is adequate and convenient. 3.1.8 Volumetric flasks--5 mL and other convenient sizes for preparing standards. 3.1.9 Disposable Pasteur Pipets. 3.2 Reagents. 3.2.1 Benzene, reagent grade. 3.2.2 Carbon disulfide, reagent grade. 3.2.3 Ethylene oxide, 99.7% pure. 3.2.4 Hydrobromic acid, 48% reagent grade. 3.2.5 Sodium carbonate, anhydrous, reagent grade. 3.2.6 Desorbing reagent, 99% benzene/1%CS 2. 3.3 Sample Preparation. 3.3.1 The front and back sections of each sample are transferred to separate 2-mL vials. 3.3.2 Each sample is desorbed with 1.0 mL of desorbing reagent. 3.3.3 The vials are sealed immediately and allowed to desorb for one hour with occasional shaking. 3.3.4 Desorbing reagent is drawn off the charcoal with a disposable pipet and put into clean 2-mL vials. 3.3.5 One drop of HBr is added to each vial. Vials are resealed and HBr is mixed well with the desorbing reagent. 3.3.6 About 0.15 gram of sodium carbonate is carefully added to each vial. Vials are again resealed and mixed well. 3.4 Standard Preparation. 3.4.1 Standards are prepared by injecting the pure ethylene oxide gas into the desorbing reagent. 3.4.2 A range of standards are prepared to make a calibration curve. A concentration of 1.0 µL of ethylene oxide gas per 1 mL desorbing reagent is equivalent to 1.0 ppm air concentration (all gas volumes at 25°C and 760 mm) for the recommended 1-liter air sample. This amount is uncorrected for desorption efficiency. 3.4.3 One drop of HBr per mL of standard is added and mixed well. 3.4.4 About 0.15 grams of sodium carbonate is carefully added for each drop of HBr (a small reaction will occur). 3.5 Analysis. 3.5.1 GC Conditions. Nitrogen flow rate--10mL/min. Injector Temperature--250°C Detector Temperature--300°C Column Temperature--100°C Injection size--0.8 µL Elution time--3.9 minutes 3.5.2 Peak areas are measured by an integrator or other suitable means. 3.5.3 The integrator results are in area units and a calibration curve is set up with concentration vs. area units. 3.6 Interferences. 3.6.1 Any compound having the same retention time of 2-bromoethanol is a potential interference. Possible interferences should be listed on the sample data sheets. 3.6.2 GC parameters may be changed to circumvent interferences. 3.6.3 There are usually trace contaminants in benzene. These contaminants, however, posed no problem of interference. 3.6.4 Retention time data on a single column is not considered proof of chemical identity. Samples over the 1.0 ppm target level should be confirmed by GC/Mass Spec or other suitable means. 3.7 Calculations. 3.7.1 The concentration in g/mL for a sample is determined by comparing the area of a particular sample to the calibration curve, which has been prepared from analytical standards. 3.7.2 The amount of analyte in each sample is corrected for desorption efficiency by use of a desorption curve. 3.7.3 Analytical results (A) from the two tubes that compose a particular air sample are added together. 3.7.4 The concentration for a sample s calculated by the following equation: EtO, mg/m 3 = (A x B) / C where: A = µg/mL B = desorption volume in milliliters C = air volume in liters. 3.7.5 To convert mg/m 3 to parts per million (ppm) the following relationship is used: EtO, ppm = (mg/m 3 x 24.45) / 44.05 where: mg/m 3 = results from 3.7.4 24.45 = molar volume at 25°C and 760 mm Hg 44.05 = molecular weight of EtO. 3.8 Safety Precautions 3.8.1 Ethylene oxide and benzene are potential carcinogens and care must be exercised when working with these compounds. 3.8.2 All work done with the solvents (preparation of standards, desorption of samples, etc.) should be done in a hood. 3.8.3 Avoid any skin contact with all of the solvents. 3.8.4 Wear safety glasses at all times. 3.8.5 Avoid skin contact with HBr because it is highly toxic and a strong irritant to eyes and skin. NOTE: Authority cited: Sections 142.3, 9020, 9030 and 9040, Labor Code. Reference: Sections 142.3, 9004(d), 9009, 9020, 9030 and 9040, Labor Code. HISTORY Editorial correction moving Note and Histories 1-7 to precede Appendix A of section 5220 (Register 99, No. 28). For prior history of Appendices, see section 5220. Go Back to Article 110 Table of Contents
17102	https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_7?srsltid=AfmBOoreO9wseWJqt5DBfUKAsvQj6dbw05YjhMp0i1SV-o6JxP4hqRZ3	Art of Problem Solving 2007 AMC 12B Problems/Problem 7 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2007 AMC 12B Problems/Problem 7 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2007 AMC 12B Problems/Problem 7 Problem All sides of the convexpentagon are of equal length, and . What is the degree measure of ? Solution Since and are right angles, and equals , and is a square. Since the length of and are also 5, triangle is equilateral. Angle is therefore See Also 2007 AMC 12B (Problems • Answer Key • Resources) Preceded by Problem 6Followed by Problem 8 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Introductory Geometry Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
17103	https://www.youtube.com/watch?v=trXItdsn2TA	[Proof] de Morgan's Distributive Law \| A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) math et al 18900 subscribers 383 likes Description 33812 views Posted: 15 Apr 2017 Proving de Morgan's distributive law A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). Thanks for watching!! ❤️ Tip Jar 👉🏻👈🏻 ☕️ ♫ Eric Skiff - Chibi Ninja 16 comments Transcript: Hey guys, in this video I'm going to go over a proof problem. It says prove de Morgan's distributive law A intersect B Union C equal to A intersect B union A intersect C valid for any sets A, B, C. This is a proof that you could see early on in a real analysis class because all you need are basic set theory properties or definitions to be able to prove that this is true. One property that we're going to use is that if you have some set A that's the subset of B and B is a subset of A, so they are subsets of each other, then this therefore means that A is equal to B. Sometimes the entries or elements of the set A can be an improper subset of B so B can have all the elements of A plus other elements but if you have that all the elements of A are a subset of B and all the elements of B are a subset of A, then that must mean that they have the same elements or entries and so the sets are equal to each other. We're going to use this property to say that if we have some entries or some entry in the left-hand side, then it must therefore be a subset of the right-hand side and we're going to prove the same thing with the second part that says if you have some entry on the right-hand side, it therefore is a subset of the left-hand side to prove that this distributive law is true. Let's go ahead and get started. I'm going to say let some X belong to the left-hand side first. So let X belong to A intersect B union C and we're just going to write out the properties of what this means until we get it to look something like the right-hand side. That means that X belongs to A and X belongs to B or C. This means that X belongs to A and X belongs to B or X belongs to C. This means that X belongs to A and X belongs to B or X belongs to A and X belongs to C. So this is equivalent to saying that X belongs to A intersect B or X belongs to A intersect C, which is the same saying that A belongs to A intersect B union A intersect C and if we look up A intersect B union A intersect C is the right-hand side. And since X is some arbitrary value on the left-hand side, we can say that we have proved that the left-hand side has to be a subset of the right-hand side. So we can say thus A intersect B union C, which is the left-hand side, is a subset of A intersect B union A intersect C, which is the right-hand side. And I'm going to put a one and circle it because this is the first part and we're going to combine a first and second part together at the end to finish our proof. For the second part, I'm going to do the same thing but start on the right-hand side and then show that it must be a subset of the left-hand side. I'm going to let there be some Y that belongs to the right-hand side. So let Y belong to A intersect B union A intersect C. This is equivalent to saying that Y belongs to A and B or Y belongs to A and C. There's an A in either case, so we can say that this is equal to saying that Y belongs to A and Y belongs to B or Y belongs to C. This is equal to saying that Y belongs to A and Y belongs to B union C, which is equal to saying that Y belongs to A intersect B union C. This is our left-hand side and we started with our right-hand side. Thus we've proved that the right-hand side, A intersect B union A intersect C, is a subset of A intersect B union C, which is the left-hand side and I'm going to label that as part two. And in the final step, we're just going to say by combining parts one and two, they are subsets of each other and so they are equal to each other. So combining one and two, we have that A intersects B union C is equal to A intersect B union A intersect C and we are done with this proof. chiptune music
17104	https://www.ncbi.nlm.nih.gov/books/NBK594278/	Lensmaker’s Equation - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Lensmaker’s Equation Jillian Gallegos; Thomas J. Stokkermans. Author Information and Affiliations Authors Jillian Gallegos 1; Thomas J. Stokkermans 2. Affiliations 1 Creighton University School of Medicine 2 University Hospitals of Cleveland Last Update: July 8, 2023. Go to: Definition/Introduction The lensmaker equation is a mathematical computation to determine the focal length of a lens in air. The focal length of a lens is defined as the distance from the center of the lens to an optical point to which all light rays are refracted. The lensmaker equation quantifies the relationship between the focal length, refractive index, thickness, and curvature of a lens and is used to produce a lens with a desired focal point; this is the ideal lens for a given refraction. The lensmaker equation may be used to design simple lenses, double-convex or double-concave, meniscus, and plano-convex or plano-concave lenses. The lensmaker equation also has utility in determining the optical power of intraocular lenses placed after cataract surgery. Go to: Issues of Concern The lensmaker equation is given by: Equation 1. Lensmaker Equation 1/f = (n-1) ⋅ [(1/R 1) – (1/R 2) + {[(n-1)d]/(n⋅R 1⋅R 2)}], where f: Lens focal length n: Lens material refractive index R 1: Radius of curvature of the lens surface closest to the light source R 2: Radius of curvature of the lens surface furthest from the light source d:Lens thickness This is a fundamental equation in lens production, utilizing the refractive index, radii of curvature, and thickness of lenses to produce a lens with a targeted focal point based on the desired refraction.The reciprocal of the focal length of the lens, 1/f, is equivalent to the optical power measured in diopters. Sign Convention The sign of a lens signifies an important property of the surface of the radii of curvature of the lens. Sign convention dictates that a positive radius, R, occurs when the center of curvature is further along in the path of the light ray, and a negative radius, R, occurs when the rays have already passed by the center of the surface of curvature. Convex lenses, which converge light in a positive lens, have an R 1>0 and R 2<0. Conversely, concave lenses have R 1<0 and R 2>0. By convention, lenses that converge are assigned a positive sign, and lenses that diverge are assigned a negative sign. Thin Lens Approximation Thin lenses can be defined as lenses whose thickness, t, is significantly less than the radii of curvature of its surfaces.The thin lens approximation assumes that light rays are only refracted once when they pass through a lens, as would occur with a lens width of zero, rather than refracted twice, as occurs with lenses in real life. While most lenses have a nonnegligible thickness and are thus thick lenses, this assumption is helpful as it removes the phenomenon of “spherical aberration.”The use of the thin lens approximation allows the lensmaker equation to be simplified to: Equation 2. Thin Lens Approximation 1/f= (n-1) ⋅ [(1/R 1) – (1/R 2)] Thus, the simplified equation suggests that light travels in a straight line through the center of a thin lens because the light is refracted equally upon exiting the lens as it does when it enters. One limitation of this simplification is that by assuming a thin lens model for the cornea and lens, the true physiological estimate of the power of the cornea and lens is not accurately described. Also, spherical aberrations of the cornea, which lead to astigmatism, are not accounted for in the simplified lensmaker formula. Lens in a Medium Other Than Air A lens placed inside a medium with a refractive index that does not approach 1, as air does, will have a different refractive power that can be calculated using a modified lensmaker equation. In these cases, the index of refraction (n 1) and the index of refraction of the medium (n 0) replace the (n-1)component of the lensmaker equation and the thin lens approximation with [(n 1-n 0)/n 0]. In essence, increasing the index of refraction of the medium relative to the index of refraction of the lens will reduce the[(n 1-n 0)/n 0] portion of the lensmaker equation and thus reduce the refracting power of the lens. Convex Lens Model Using the thin lens model, it can be assumed that rays of light travel in a straight line through the center of the thin lens. Other rays travel parallel to the optical axis as collimated rays to a focal point on the opposite end of the lens. Furthermore, other rays travel as collimated rays parallel to the optical axis on the same side of the lens. Ultimately, convex lenses refract all collimated rays to converge at the focal point on the opposite side of the lens relative to the light source. Concave Lens Model For concave lenses, rays of light once again pass straight through the center of the lens using the thin lens assumption. Other rays begin as collimated rays, travel parallel to the optical axis, and are refracted so that their imaginary continuation passes through the focal point on the opposite side of the lens. The same occurs for rays on the same side of the lens. This occurs because concave lenses diverge light rays so that the imaginary continuations of the light rays converge at the focal point on the same side of the lens as the light source. This convergence of imaginary continuations is manifested as virtual images only seen when looking through the lenses. Magnification and Aniseikonia Altering the optical power and, thus, the focal point of a lens with the lensmaker formula can impact the magnification of resultant images when looking through the lens. Differences in spectacle magnification between the right and left lens can be used to calculate a phenomenon called optical aniseikonia. Aniseikonia is characterized by a difference in the perceived shape or size of the image and is a cause of amblyopia. Optical aniseikonia is caused by anisometropia, and retinal aniseikonia is caused by disturbances to the retina.Iseikonic spectacle lenses are used to manage aniseikonia by altering the vertex distance, base curve, center thickness, and index of refraction of spectacle lens material to provide differing magnification of one lens compared to the other. The magnification produced by iseikonic lenses is given by: Equation 3. Magnification M t = M s x M p, where _M_ t: magnification M s: lens shape factor determined by 1/{[1-(t⋅D 1)]/n} M p: lens power factor determined by 1/[1-(h⋅D v)], and n: refractive index D 1: base curve t: lens thickness D v: dioptrical power of lens h: vertex distance in meters Thus, magnification is impacted by altering any one of these lens variables in the creation of a lens: Increased refractive index (n) leads to decreased magnification and increased minification. Increased, or steepening, base curve (D 1) causes increased magnification and decreased minification. Increased thickness of the lens (t) increases magnification and decreases minification. Increased lens power (D v) leads to increased magnification for plus lenses and minification for minus lenses. Increased vertex distance (h) causes increased magnification in plus lenses and minification in minus lenses. Optical aniseikonia, caused by anisometropia, has a fixed lens power (D v); the base curve, center of the thickness, refractive index, and vertex distance of the iseikonic lens have an important effect. These lenses are incredibly useful in treating aniseikonia if the eyes differ by ≤3 diopters. At this maximum difference, the iseikonic lenses will have a visibly different base curve and thickness. Limitations When calculating lens power using the lensmaker equation, using a constant equivalent refractive index for the human crystalline lens will result in slight errors dependent on age. The human lens has a gradient refractive index that increases when moving from the outer edges toward the center, ranging from a maximum of 1.42 at the core to a minimum of 1.37 at the surface.Thus, the common usage of a fixed refractive index for the human lens leads to the overestimation of power in old lenses and the underestimation of power in young lenses.This can subsequently lead to errors in calculations of the lens thickness and radii of curvature. Go to: Clinical Significance Spectacle Lenses The lensmaker formula quantifies an important optical principle that lens manufacturers employ to produce different types of spectacle lenses,including double-convex or double-concave lenses, plano-convex or plano-concave lenses, and the newer meniscus lenses. Individuals with myopia, or near-sightedness, have a spherical equivalent of fewer than 0 diopters and thus require correction with minus power, produced with concave lenses. Hyperopia, or farsightedness, is the opposite and requires correction with positive power produced by convex lenses. Optics manufacturing has roots in the early 17 th century with the invention of telescopes. The invention of grinding machines has allowed large-scale manufacturing of spectacle lenses with well-defined properties.Further advances in computerized design have led to the current-day manufacturing of lenses using UV optics, high-power laser devices, and other technological advances. Recent developments have explored varifocal lenses with tunable, liquid-based microlenses in which the focal length is changed by changing the shape of the lens, as well as lenses that change focal length using liquid crystal-embedded dielectric metasurfaces.These new lenses broaden applicability by maximizing the functionality of multiple optical elements rather than focusing on a single element. Despite these increasingly complex advances, the fundamentals of the lensmaker formula are essential to define the properties of a lens, achieve the targeted focal point and optical power, and permit lens manufacture using this advanced technology. Aspheric lenses are a new approach to myopia. Aspheric lenses have a flattened curve, making the lens thinner, lighter, and more flattering in overall appearance. Contact Lenses Several different types of contact lenses are available, and each uses properties of the lensmaker equation to correct refractive errors without using physical spectacles. Contact lenses differ in compositional material: soft or hydrogel contact lenses made of hydroxyethyl methacrylate (HEMA), rigid gas-permeable contact lenses made of silicon and cellulose acetate butyrate, and rigid non-gas permeable contact lenses made of polymethyl methacrylate (PMMA). Contacts also differ by design; they can be single-cut or lenticular-cut. Single-cut lenses have a front surface with a single continuous curve and a back surface with a base curve and peripheral curve. Lenticular lenses have a front surface with a central optical and peripheral carrier portion thinner and flatter in the radius. The back surface also has a peripheral and base curve. Modifying material, curvature, and other properties of these contacts allow for a better fit and optical outcome. Intraocular Lenses The lensmaker equation has great utility for determining lens power by taking the inverse of the focal length of the lens. Calculating the power of the intraocular lens (IOL) has become of greater importance in recent years as advancements in cataract and refractive surgery have allowed for reduced surgically induced astigmatism, more accurate measurements of the eye, and the expectation of accurate refractive outcomes by patients. As a result, accurate calculations of IOL power are ever more important. Cataract surgeons commonly employ adapted thin-lens formulas for calculating the IOL power by assuming that the cornea is a thin optical lens with an index of refraction near 1.3375, depending on the generation of the formula employed. Advancing models for aspheric and multifocal IOLs may allow a patient to become spectacle free after cataract surgery if the optical power of the implanted IOL is correctly calculated.Since the IOL is placed inside the fluid of the eye with an index of refraction of about 1.34, the lensmaker equation can be applied to determine the refractive power of IOLs of different refractive indices and front and back surface radii. Using the principles of the lensmaker formula and thin lens formula, several intraocular telescopic implants have been developed which improve acuity by using high magnification to create a magnified retinal image in patients with age-related macular degeneration. This reduces the need for hand-held magnification and improves overall visual acuity. Keratoprostheses and Corneal Inlays Similar to IOLs, corneal implants and other prosthetic devices have been proposed for use in the cornea to prevent corneal blindness in the setting of corneal disease.Various materials and polymers have been developed to serve as keratoprostheses to replace a diseased cornea with an artificial cornea, manipulating properties to produce optimal vision.Similarly, corneal inlays, or the placement of a synthetic lens in the corneal stroma to correct presbyopia, may also utilize the principles of the lensmaker formula to develop an ideal patient-specific implant. Meniscus Lenses Using an applied lensmaker equation is an effective technique for developing more complex optical lenses, such as three-dimensional (3-D) plasmonic meniscus lenses for surface plasmon polaritons focusing at optical frequencies. These lenses priorly were only made using their ideal two-dimensional (2-D) configurations, but the successful application of the lensmaker equation suggests potential applications of producing optic forces for use as optical tweezers. Go to: Nursing, Allied Health, and Interprofessional Team Interventions The entire healthcare team plays a vital role in measuring for, fitting,and developing spectacles, contact lenses, and other ophthalmic optical devices with the ideal corrective refraction. Fitting spectacles properly centered over the eyes for ideal vision is an important task performed by opticians, nurses, and other interprofessional team members. Understanding optical power and the principles of positive and negative lenses ensures the avoidance of simple mistakes in spectacle fit and prescription. The lensmaker formula measures optical power in diopters, which opticians and optical technologists can use to understand patient-specific vision needs,including the adjustment of additive lenses should the patient need additional power for activities like reading. Ophthalmologists, optometrists, and opticians utilize the lensmaker formula to solve issues of aniseikonia when the difference between eyes is more than 1 diopter. Both contact lenses and iseikonic spectacles can be used to treat aniseikonia, and opticians should be trained in the manufacturing and optics of these special lenses to avoid poor lens adaptation, diplopia, headaches, and amblyopia. For all patients requiring refractive care, the collaboration of cataract surgeons, refractive surgeons, contact lens fitters, refractionists, and opticians should aim to provide the best optical outcome for patients. Clinical judgment and physical examination skills are necessary to determine when spectacle lenses or contacts will provide the ideal refraction for a patient, or further intervention, such as cataract surgery, may be required. After interventions such as cataract or refractive surgery, the surgeons should continue to work with opticians and other healthcare team members to ensure that any necessary postoperative refraction is provided to the patient for the best optical outcomes. Patient preference should be considered when medically possible, including a preference for contact lenses versus spectacles, a desire for further refractive surgeries, and overall expectations of the optical outcome. Go to: Review Questions Access free multiple choice questions on this topic. Comment on this article. Figure Lensmaker Formula with Converging Lens Contributed by Jillian Gallegos Figure Lensmaker Formula with Diverging Lens Contributed by Jillian Gallegos Figure Thin Lens Approximation Contributed by Jillian Gallegos Figure Lensmaker Equation and Thin Lens Assumption in medium other than air Contributed by Jillian Gallegos Go to: References 1. Tenney SR, Moshirfar M, Ronquillo Y. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Oct 31, 2022. Concave And Convex Lenses. [PubMed: 36512657] 2. Flynn RA, Fleet EF, Beadie G, Shirk JS. Achromatic GRIN singlet lens design. Opt Express. 2013 Feb 25;21(4):4970-8. [PubMed: 23482029] 3. Riley JA, Healy N, Pacheco-Peña V. Plasmonic meniscus lenses. Sci Rep. 2022 Jan 18;12(1):894. [PMC free article: PMC8766487] [PubMed: 35042917] 4. Olsen T. Calculation of intraocular lens power: a review. Acta Ophthalmol Scand. 2007 Aug;85(5):472-85. [PubMed: 17403024] 5. Stokkermans TJ, Day SH. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Aug 8, 2023. Aniseikonia. [PubMed: 36256755] 6. South J, Gao T, Collins A, Turuwhenua J, Robertson K, Black J. Aniseikonia and anisometropia: implications for suppression and amblyopia. Clin Exp Optom. 2019 Nov;102(6):556-565. [PubMed: 30791133] 7. Borja D, Manns F, Ho A, Ziebarth N, Rosen AM, Jain R, Amelinckx A, Arrieta E, Augusteyn RC, Parel JM. Optical power of the isolated human crystalline lens. Invest Ophthalmol Vis Sci. 2008 Jun;49(6):2541-8. [PMC free article: PMC2785024] [PubMed: 18316704] 8. Gallegos J, Stokkermans TJ. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): May 20, 2023. Refractive Index. [PubMed: 37276310] 9. Gurnani B, Kaur K. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Jun 11, 2023. Contact Lenses. [PubMed: 35593861] 10. Ji S, Ponting M, Lepkowicz RS, Rosenberg A, Flynn R, Beadie G, Baer E. A bio-inspired polymeric gradient refractive index (GRIN) human eye lens. Opt Express. 2012 Nov 19;20(24):26746-54. [PubMed: 23187529] 11. Atchison DA, Thibos LN. Optical models of the human eye. Clin Exp Optom. 2016 Mar;99(2):99-106. [PubMed: 26969304] 12. Mutti DO, Zadnik K, Adams AJ. The equivalent refractive index of the crystalline lens in childhood. Vision Res. 1995 Jun;35(11):1565-73. [PubMed: 7667914] 13. Banerjee S, Horton J. Lenses and Spectacles to Prevent Myopia Worsening in Children [Internet]. Canadian Agency for Drugs and Technologies in Health; Ottawa (ON): Apr, 2021. [PubMed: 34255449] 14. Gerhard C. On the History, Presence, and Future of Optics Manufacturing. Micromachines (Basel). 2021 Jun 09;12(6) [PMC free article: PMC8226999] [PubMed: 34207746] 15. Reichelt S, Zappe H. Design of spherically corrected, achromatic variable-focus liquid lenses. Opt Express. 2007 Oct 17;15(21):14146-54. [PubMed: 19550687] 16. Song X, Zhang H, Li D, Jia D, Liu T. Electrowetting lens with large aperture and focal length tunability. Sci Rep. 2020 Oct 01;10(1):16318. [PMC free article: PMC7530674] [PubMed: 33004850] 17. Bosch M, Shcherbakov MR, Won K, Lee HS, Kim Y, Shvets G. Electrically Actuated Varifocal Lens Based on Liquid-Crystal-Embedded Dielectric Metasurfaces. Nano Lett. 2021 May 12;21(9):3849-3856. [PubMed: 33900774] 18. Jalie M. Modern spectacle lens design. Clin Exp Optom. 2020 Jan;103(1):3-10. [PubMed: 31222837] 19. Ghaffari R, Abdi P, Moghaddasi A, Heidarzadeh S, Ghahvhechian H, Kasiri M. Ray Tracing versus Thin-Lens Formulas for IOL Power Calculation Using Swept-Source Optical Coherence Tomography Biometry. J Ophthalmic Vis Res. 2022 Apr-Jun;17(2):176-185. [PMC free article: PMC9185206] [PubMed: 35765642] 20. Chang YC, Mesquita GM, Williams S, Gregori G, Cabot F, Ho A, Ruggeri M, Yoo SH, Parel JM, Manns F. In vivo measurement of the human crystalline lens equivalent refractive index using extended-depth OCT. Biomed Opt Express. 2019 Feb 01;10(2):411-422. [PMC free article: PMC6377882] [PubMed: 30800489] 21. Dunbar HMP, Dhawahir-Scala FE. A Discussion of Commercially Available Intra-ocular Telescopic Implants for Patients with Age-Related Macular Degeneration. Ophthalmol Ther. 2018 Jun;7(1):33-48. [PMC free article: PMC5997589] [PubMed: 29700786] 22. Sharma S, Donthineni PR, Iyer G, Chodosh J, de la Paz MF, Maskati Q, Srinivasan B, Agarwal S, Basu S, Shanbhag SS. Keratoprosthesis in dry eye disease. Indian J Ophthalmol. 2023 Apr;71(4):1154-1166. [PMC free article: PMC10276669] [PubMed: 37026247] 23. Fu L, Hollick EJ. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Apr 20, 2023. Artificial Cornea Transplantation. [PubMed: 33760451] 24. Sweeney DF, Vannas A, Hughes TC, Evans MD, McLean KM, Xie RZ, Pravin VK, Prakasam RK., Vision CRC Inlay Team. Synthetic corneal inlays. Clin Exp Optom. 2008 Jan;91(1):56-66. [PubMed: 18045251] Disclosure:Jillian Gallegos declares no relevant financial relationships with ineligible companies. Disclosure:Thomas Stokkermans declares no relevant financial relationships with ineligible companies. Definition/Introduction Issues of Concern Clinical Significance Nursing, Allied Health, and Interprofessional Team Interventions Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK594278 PMID: 37603669 Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page Definition/Introduction Issues of Concern Clinical Significance Nursing, Allied Health, and Interprofessional Team Interventions Review Questions References Related information PMCPubMed Central citations PubMedLinks to PubMed Similar articles in PubMed The Spherical Equivalent.[StatPearls. 2025]The Spherical Equivalent.Enaholo ES, Musa MJ, Zeppieri M. StatPearls. 2025 Jan Toward individually tunable compound eyes with transparent graphene electrode.[Bioinspir Biomim. 2017]Toward individually tunable compound eyes with transparent graphene electrode.Shahini A, Jin H, Zhou Z, Zhao Y, Chen PY, Hua J, Cheng MM. Bioinspir Biomim. 2017 Jun 8; 12(4):046002. Epub 2017 Jun 8. Focal length, EFL, and the eye.[Appl Opt. 2023]Focal length, EFL, and the eye.Simpson MJ. Appl Opt. 2023 Mar 1; 62(7):1853-1857. Review Observations on the optical effects of a cataract.[J Cataract Refract Surg. 1999]Review Observations on the optical effects of a cataract.Campbell C. J Cataract Refract Surg. 1999 Jul; 25(7):995-1003. Review Physics of optical tweezers.[Methods Cell Biol. 2007]Review Physics of optical tweezers.Nieminen TA, Knöner G, Heckenberg NR, Rubinsztein-Dunlop H. Methods Cell Biol. 2007; 82:207-36. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Lensmaker’s Equation - StatPearlsLensmaker’s Equation - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. Cite this Page Close Gallegos J, Stokkermans TJ. Lensmaker’s Equation. [Updated 2023 Jul 8]. 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17105	https://library.fiveable.me/key-terms/ap-calc/extreme-value-theorem	printables ♾️ap calculus ab/bc review key term - Extreme Value Theorem Citation: MLA Definition The Extreme Value Theorem states that if a function is continuous on a closed interval, then it must have both a maximum value and a minimum value on that interval. Related terms Critical Points: These are points on a function where the derivative is either zero or undefined. Absolute Extrema: These are the highest and lowest values of a function over its entire domain. Closed Interval Method: This method involves evaluating the function at its critical points and endpoints to find local extrema. "Extreme Value Theorem" also found in: Subjects (2) Calculus I Multivariable Calculus Guided Practice Practice AP Calculus AB/BC questions
17106	https://s28543.pcdn.co/wp-content/uploads/sites/39/2021/03/Necessary-vs.-Sufficient-worksheet.pdf	Necessary vs. Suﬃcient: Under the Right Conditions - Reasoning Series \| Academy 4 Social Change Necessary vs. Suﬃcient: Worksheet Matching Match each statement with the combination of necessity and suﬃciency that best suits the condition described. Your Turn Create your own examples of conditional statements that satisfy the four combinations of necessity and suﬃciency outlined in the previous exercise. 1. Necessary but not suﬃcient 2. Suﬃcient but not necessary 3. Both necessary and suﬃcient 4. Neither necessary nor suﬃcient 1. Scoring a touchdown as a condition for getting six points in a football game. 2. First obtaining a learner’s permit as a condition for getting a driver’s license. 3. Putting the key into the ignition as a condition for starting a car. 4. For positive whole numbers, being divisible by two as a condition for being even. a. Necessary but not suﬃcient b. Suﬃcient but not necessary c. Both necessary and suﬃcient d. Neither necessary nor suﬃcient Necessary vs. Suﬃcient: Under the Right Conditions - Reasoning Series \| Academy 4 Social Change Necessary vs. Suﬃcient: Worksheet Answers Matching Match each statement with the combination of necessity and suﬃciency that best suits the condition described. Answers: 1. b, 2. a, 3. d, 4. c 1. Scoring a touchdown as a condition for getting six points in a football game. 2. First obtaining a learner’s permit as a condition for getting a driver’s license. 3. Putting the key into the ignition as a condition for starting a car. 4. For positive whole numbers, being divisible by two as a condition for being even. a. Necessary but not suﬃcient b. Suﬃcient but not necessary c. Both necessary and suﬃcient d. Neither necessary nor suﬃcient
17107	https://www.engineeringtoolbox.com/flow-liquid-water-tank-d_1753.html	Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! Liquid Flow from Containers - Emptying Time Calculate liquid velocity, volume flow and draining time when emptying a container. Base Apertures The liquid outlet velocity when draining a tank or a container can be calculated v = Cv (2 g H)1/2 (1a) v = outlet velocity (m/s) Cv = velocity coefficient (water 0.97) g = acceleration of gravity (9.81 m/s2) H = height (m) The liquid volume flow can be calculated V = Cd A (2 g H)1/2 (1b) where V = volume flow (m3/s) A = area of aperture - flow outlet (m2) Cd = discharge coefficient where Cd = Cc Cv where Cc = contraction coefficient (sharp edge aperture 0.62, well rounded aperture 0.97) A = area aperture (m2) Example - Volume Flow when draining a Container The height from the surface to the outlet aperture in a water filled container is 3 m. The aperture is sharp edged with diameter 0.1 m. The discharge coefficient can be calculated as Cd = 0.62 0.97 = 0.6 The area of the aperture can be calculated as A = π ((0.1 m) / 2)2 = 0.008 m2 The volume flow through the aperture can be calculated as V = 0.6 (0.008 m2) (2 (9.81 m/s2) (3 m))1/2 = 0.037 m3/s For height 1.5 m the volume flow is 0.026 m3/s. For height 0.5 m the volume flow is 0.015 m3/s. Draining Tank Calculator This calculator is based on eq. (1b) and can be used to estimate the volume flow and time used to drain a container or tank through an aperture. The calculator divides the container in "slices" and makes an iterative average calculation for each slice. The accuracy of the calculation can be increased by increasing the number of slices. results in table below! Note! - the flow is reduced and the time is increased with reduced height. Make a Shortcut to this Calculator on Your Home Screen? Small Lateral Apertures Outlet velocity can be expressed as v = Cv (2 g H)1/2 (2a) Distance s can be expressed as s = 2 (H h)1/2 (2b) Volume flow can be expressed as V = Cd A (2 g H)1/2 (2c) Reaction force can be expressed as F = ρ V v (2d) where ρ = density (kg/m3) (water 1000 kg/m3) Large Lateral Apertures Volume flow can be expressed as V = 2/3 Cd b (2 g)1/2 (H23/2 - H13/2) (3a) where b = width of aperture (m) Excess Pressure in Container Outlet velocity can be expressed as v = Cv (2 (g H + p / ρ))1/2 (4a) where p = excess pressure in container or tank (N/m2, Pa) Volume flow can be expressed as V = Cd A (2 (g H + p / ρ))1/2 (4b) Unit Converter . Cookie Settings \| \| \| --- \| \| \| \| \| \| \| --- \| \| \| \|
17108	https://www.ncbi.nlm.nih.gov/books/NBK534227/	An official website of the United States government The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Burn Fluid Resuscitation Patrycja Popowicz; Abby Regan; David T. Hotwagner. Author Information and Affiliations Authors Patrycja Popowicz1; Abby Regan2; David T. Hotwagner3. Affiliations 1 East Carolina University Health 2 William Carey University 3 St Lucie Medical Center Last Update: April 6, 2025. Continuing Education Activity Burn fluid resuscitation is a critical component of managing burn injuries, which affect approximately 500,000 individuals annually in the United States. Among civilians, these injuries commonly result from house fires, motor vehicle accidents, and workplace incidents. In combat settings, burns account for 5% to 10% of all casualties. The severity of burns is determined by several factors, including the intensity of the thermal energy, duration of exposure, and the affected body surface area. These factors help determine whether a patient requires treatment at a specialized burn center. Fluid resuscitation is required for burns exceeding 20% total body surface area in adults and 10% to 15% in children. Lactated Ringer solution is the preferred crystalloid for initial resuscitation, while pediatric patients may require dextrose supplementation. Over-resuscitation, or "fluid creep," can cause complications such as pulmonary edema and compartment syndrome. Individualized fluid management remains essential to improving patient outcomes. This activity underscores the importance of fluid resuscitation in burn management and highlights the importance of interprofessional collaboration among healthcare providers in optimizing patient outcomes. Objectives: Identify the signs and symptoms of burn severity and assess the need for fluid resuscitation based on total body surface area, burn depth, and patient-specific factors. Implement evidence-based burn resuscitation formulas, including the Parkland and modified Brooke, to guide fluid replacement therapy. Select the appropriate fluid type, such as Lactated Ringer, and determine the role of adjunct therapies, such as albumin, based on the patient’s condition. Collaborate with the interdisciplinary healthcare team to ensure timely, coordinated interventions and provide comprehensive care for burn patients. Access free multiple choice questions on this topic. Introduction In the United States, approximately 500,000 people seek care for burn injuries each year. Among civilians, these injuries most commonly result from house fires, motor vehicle crashes, and work-related accidents. Burn injuries account for 5% to 10% of combat casualties. The World Health Organization (WHO) and the Centers for Disease Control and Prevention (CDC) identify burns as one of the most common causes of home injuries in individuals aged 19 or younger. Several practice guidelines have been released to emphasize the importance of optimal care and management of burn injuries. Organizations such as the International Society for Burn Injuries (ISBI) and the American Burn Association (ABA) have published guidelines and updates to address the needs of both resource-limited and resource-abundant regions. Burn injuries can result from various sources, including thermal elements, grease, friction, electricity, and chemicals. The severity of a burn depends on factors such as the affected area, duration of contact, and the patient’s preexisting health conditions. Accurate classification is essential for the proper assessment and treatment of burn patients. Burns are classified by depth—superficial burns affect only the epidermis; superficial partial-thickness burns extend into the papillary dermis; deep partial-thickness burns reach the reticular dermis; and full-thickness burns damage both skin layers and the underlying subcutaneous tissue. Total body surface area (TBSA) is used to measure the percentage of burned skin. Only partial and full-thickness burns are considered when determining the need for fluid resuscitation. Fluid resuscitation is required for burns exceeding 20% TBSA in adults and 10% to 15% in children, although specific cutoffs may vary by institution. Various resuscitation formulas provide guidelines for initiating fluid replacement to prevent complications and improve patient outcomes. Anatomy and Physiology Fluid loss is arguably the most significant issue for patients who sustain burn injuries, therefore making volume replacement crucial. Burn management is typically divided into 3 phases—early resuscitative, wound management, and rehabilitative or reconstructive. This activity primarily focuses on the early resuscitative phase following initial stabilization. While fluid management guidelines have been in place since the early 1960s, protocols for the first 24 hours remain controversial. In the early resuscitative phase, the primary concern is hypovolemia due to increased capillary permeability. Thermal injury triggers the release of inflammatory markers both systemically and locally at the site of injury. These markers induce vasodilation and increase capillary permeability throughout the body, leading to a significant fluid shift out of the intravascular compartment. The inflammatory response to thermal burns is far more intense than that observed in trauma or sepsis patients. Contributing factors include histamines, leukotrienes, and a damaged glycocalyx. Burn shock is a complex condition involving distributive, cardiogenic, and hypovolemic shock. This is characterized by decreased cardiac output, increased systemic vascular resistance, and depleted intravascular volume, which exacerbates the shock state. Therefore, replacing fluid in the intravascular compartment is crucial to preserve tissue perfusion and maintain the function of vital organs. Indications A subset of patients with burn injuries also sustain additional traumatic injuries. The initial provider must not become so focused on the visible burn injury that they overlook other potentially life-threatening injuries that may be less apparent. Airway assessment, hemorrhage control, and adherence to advanced trauma life support guidelines are critical for proper management. As with any trauma patient, fluid management should be based on weight and burn size and initiated immediately after the primary airway evaluation. However, administering a fluid bolus in burn patients without evidence of hypovolemia is unnecessary and can have several negative consequences, including worsening edema formation. Therefore, fluid bolus administration should be avoided in these cases. An intact gastrointestinal tract can serve as a conduit for fluid resuscitation. Notably, many burn patients with a significantly involved TBSA may not tolerate oral resuscitation. Therefore, oral fluid resuscitation is recommended only for patients with less than 30% TBSA. Burns covering more than 20% TBSA in adults and over 10% to 15% TBSA in children require formal fluid resuscitation. Commonly used methods for calculating TBSA include the Lund and Browder chart for both children and adults, the Rule of Nines for adults, and the Palmar method. Please see StatPearls' companion resource, "Rule of Nines," for more information. Generally, the Rule of Nines estimates the TBSA affected by burns by assigning specific percentages to different body regions. The entire head is estimated to account for 9% of the TBSA (4.5% for the anterior and 4.5% for the posterior). The trunk is estimated at 36% (18% allocated to the anterior and 18% to the posterior components). The anterior aspect of the trunk can be further divided into the chest (9%) and abdomen (9%). The upper extremities contribute a total of 18%, with each arm accounting for 9%. Each arm can be divided into anterior and posterior sections (4.5% each). The lower extremities are estimated at 36%, with 18% for each leg. Each leg can be divided into anterior and posterior sections (9% each). The groin is estimated at 1%. The Lund and Browder chart is considered the most accurate method, offering precise estimations of TBSA by dividing body areas into smaller components, thus improving accuracy. For minor burns, the Palmar method can be used, utilizing the patient's palm and fingers as a template to estimate the affected TBSA with an accuracy of approximately 1%. Technique or Treatment Crystalloid fluids are the preferred choice for burn resuscitation due to their isotonic composition, which closely resembles plasma and helps achieve better outcomes. Hypotonic fluids are avoided as they can exacerbate edema. Lactated Ringer solution is recommended for initial resuscitation across all age groups. However, infants are more prone to hypoglycemia due to limited glycogen stores, necessitating the addition of dextrose. Pediatric burn resuscitation formulas may include maintenance fluids with or without dextrose, depending on the child's age. The ABA recently published clinical guidelines addressing the role of colloids in burn resuscitation. Albumin may be administered throughout the resuscitation process or as a rescue intervention to reduce overall fluid requirements. These guidelines recommend considering the administration of albumin in situations where resuscitative efforts are ineffective and crystalloid needs are increasing. These recommendations are stronger for patients with larger TBSA and weaker for those with smaller TBSA. However, no definitive recommendations exist for centers using albumin in resuscitation protocols on whether it should be administered before or after 12 hours. The guidelines also discuss additional recommendations for using plasma, vitamin C, and other resuscitative adjuncts. Various resuscitation formulas are used for burn management, with the Parkland and modified Brooke formulas being the most common in adults. Developed in 1968 by Dr Charles Baxter, the Parkland formula is widely recognized for fluid replacement in burn injuries. The formula recommends administering 4 mL of lactated Ringer solution per kilogram of body weight per percentage of TBSA burned (4 mL/kg/%TBSA) in adults and 3 mL/kg/%TBSA in pediatric patients, with half given in the first 8 hours and the remainder over the next 16 hours. Discrepancies persist in patients where body surface area (BSA) calculations are unreliable, such as in pediatric patients and those with obesity. When first introduced, the Parkland formula was unique for recommending higher fluid volumes compared to its predecessors. The modified Brooke formula, similar to the Parkland formula, recommends 2 mL/kg/%TBSA for adults. Please see StatPearls' companion resource, "Parkland Formula," for more information. One of the most significant differences in pediatric burn resuscitation is the variation in BSA and skin thickness. Children have a greater BSA-to-body mass ratio, rendering them more susceptible to hypothermia and increasing their fluid requirements for burns of any size. Children have different body proportions, such as larger heads and smaller legs, which must be considered when calculating their TBSA. Although several widely used formulas have been adopted in clinical practice, variations exist based on whether TBSA or BSA is the primary determinant. The formulas that use TBSA include the Parkland formula, which administers 4 mL/kg/%TBSA, and the modified Brooke formula, which uses 3 mL/kg/%TBSA. The Cincinnati formula retains TBSA while also incorporating BSA, recommending 4 mL/kg/%TBSA of lactated Ringer solution plus 1500 mL/m² of total BSA. In contrast, the Galveston formula uses only BSA for fluid resuscitation, recommending 5000 mL/m² of BSA burned plus 2000 mL/m² of total BSA of lactated Ringer solution. Both the Cincinnati and Galveston formulas may include colloid in their resuscitation protocols. A retrospective review by Stevens et al in 2023 compared pediatric resuscitative formulas incorporating BSA to the Parkland formula. No significant difference in fluid volume administered was observed between the Parkland and Cincinnati groups across all weight categories. The Galveston formula underpredicted fluid delivery. However, patients with higher body mass received greater fluid volumes compared to patients with obesity. In adults, fluid resuscitation goals generally include urine output (UOP) of 30 to 50 mL/h or 0.5 to 1.0 mL/kg/h with an indwelling catheter, a base deficit of less than 2, systolic blood pressure above 90 mm Hg, palpable peripheral pulses, and no altered mental status. Although studies indicate these variables are reliable predictors of fluid resuscitation, many physicians rely solely on UOP. In 1991, Dries and Waxman found that vital signs and UOP did not significantly change after volume repletion, while measurements from pulmonary artery catheterization were much more significant. This suggested that cardiac output was the most sensitive measure for guiding fluid therapy. However, this approach requires the placement of a pulmonary artery catheter, leading many burn units to be reluctant to adopt this method for guidance. Other proposed methods for goal-directed therapy include transpulmonary thermodilution and arterial pressure wave analysis. For children with a body weight of less than 30 kg, the recommended UOP goal is 1 mL/kg/h, while for those with a body weight of more than 30 kg, the target is 0.5 mL/kg/h. Similar to adults, relying solely on UOP as a measure of efficacy is controversial and often misleading. Sheridan et al suggest that in infants, resuscitation goals should be determined by sensorium, physical examination, pulse, and systolic blood pressure, in addition to UOP. Additional parameters for therapy end points in the pediatric population include lactate levels, invasive transpulmonary thermodilution, and central venous pressures. These factors highlight areas that require further investigation for both children and adults. An exception to the previously stated UOP goal applies to patients with rhabdomyolysis and/or acute renal failure, where mortality can reach as high as 70% in severe burns. In these cases, fluids should be administered at a rate that achieves a UOP of 1 mL/kg/h. However, notably, "more is not better," as the risk of fluid overload can be as life-threatening as the burn injury itself. Neither mannitol nor sodium bicarbonate is more effective in treating acute renal failure than fluid loading alone. Complications Patients often arrive at burn centers after receiving excessive hydration during transport, a result of inexperienced first responders or clinicians who may overestimate the burn size. Additional factors have been identified that predispose burn patients to increased fluid requirements, including inhalation injury, delays in resuscitation, polytrauma, or high-voltage electrical injury. The phenomenon referred to as "fluid creep" poses a challenge in burn management, as over-resuscitation can lead to detrimental outcomes, such as pulmonary and cerebral edema, or compartment syndrome of the extremities or abdomen. Abdominal compartment syndrome can develop in patients with intra-abdominal pressure exceeding 20 mm Hg and signs of end-organ dysfunction. This condition may arise in patients receiving excessive fluid resuscitation. A common symptom of abdominal compartment syndrome is oliguria, which can lead to excessive fluid administration if clinical data are misinterpreted, particularly if other related signs, such as elevated inspiratory pressures and decreased tidal volumes, are not considered. Extremity compartment syndrome can occur in both burned and unburned limbs due to over-resuscitation, leading to increased osteofascial compartment pressures exceeding 30 mm Hg, along with signs of impairment of circulation. Clinical Significance Extensive research on the pathophysiology and outcomes of burn patients has established that prompt fluid resuscitation is critical for survival in these patients. The implementation of more efficient fluid replacement protocols has led to a reduction in mortality within the first 48 hours. Suboptimal resuscitation can increase burn depth and prolong the shock period, ultimately raising the risk of mortality. On the other hand, over-resuscitation can result in additional complications, including compartment syndrome affecting the abdomen, extremities, and orbits, as well as worsening acute respiratory distress syndrome. Therefore, while resuscitation formulas provide guidelines for initial fluid replacement, ongoing resuscitation must be tailored to the individual patient's needs and current clinical condition. Enhancing Healthcare Team Outcomes The practice guidelines for fluid management in burn patients are designed for healthcare professionals involved in providing acute care in this area. Effective management of burn patients requires clear and open communication among emergency medical services, nursing staff, and physicians. Proper fluid resuscitation is a critical component of treatment, as there is a limited timeframe in which to administer the correct volume of fluids; both excessive and insufficient fluids can lead to severe consequences. Generally, intensivists, plastic surgeons, or general surgeons lead the management efforts. However, caring for burn patients requires a multidisciplinary approach to achieve the best outcomes, with valuable contributions from nursing staff, ancillary teams, nutritionists, and physical and occupational therapists. Nursing staff must collaborate with the clinical team to monitor fluids, UOP, vital signs, and breath sounds, ensuring adequate hydration while preventing fluid overload. Any changes in patient status should be promptly documented and communicated to the appropriate interprofessional healthcare team members to facilitate timely interventions. Treating burn patients is complex, as they are at risk for complications such as sepsis, cardiac dysfunction, neurogenic issues, and fluid imbalances. Clinical success requires a highly skilled and attentive team. Additionally, a cross-sectional, descriptive, analytic study identified communication barriers faced by burn patients, including the fast-paced nature of the intensive care unit and difficulty expressing symptoms due to their medical condition. Therefore, providing a safe and supportive environment is essential to promoting effective patient communication and ensuring their needs are met. Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. References 1. : White CE, Renz EM. Advances in surgical care: management of severe burn injury. Crit Care Med. 2008 Jul;36(7 Suppl):S318-24. [PubMed: 18594259] 2. : Korzeniowski T, Mertowska P, Mertowski S, Podgajna M, Grywalska E, Strużyna J, Torres K. 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R, Kordis P, Patel S, Grover V, Kuchyn I, Bielka K, Aidoni Z, Grosomanidis V, Kotzampassi K, Stavrou G, Fyntanidou B, Patsatzakis S, Skourtis C, Lee SD, Williams K, Weltes ID, Berhane S, Arrowsmith C, Peters C, Robert S, Caldas J, Panerai RB, Robinson TG, Camara L, Ferreira G, Borg-Seng-Shu E, De Lima Oliveira M, Mian NC, Santos L, Nogueira R, Zeferino SP, Jacobsen Teixeira M, Galas F, Hajjar LA, Killeen P, McPhail M, Bernal W, Maggs J, Wendon J, Hughes T, Taniguchi LU, Siqueira EM, Vieira Jr JM, Azevedo LC, Ahmad AN, Abu-Habsa M, Bahl R, Helme E, Hadfield S, Loveridge R, Shak J, Senver C, Howard-Griffin R, Wacharasint P, Fuengfoo P, Sukcharoen N, Rangsin R, Sbiti-Rohr D, Schuetz P, Na H, Song S, Lee S, Jeong E, Lee K, Cooper M, Milinis K, Williams G, McCarron E, Simants S, Patanwala I, Welters ID, Zoumpelouli E, Volakli EA, Chrysohoidou V, Georgiou S, Charisopoulou K, Kotzapanagiotou E, Panagiotidou V, Manavidou K, Stathi Z, Sdougka M, Salahuddin N, AlGhamdi B, Marashly Q, Zaza K, 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T, Fister M, Knafelj R, Muraray Govind P, Brahmananda Reddy N, Pratheema R, Arul ED, Devachandran J, Velasco MB, Dalcomune DM, Knafelj R, Fister M, Chin-Yee N, D’Egidio G, Thavorn K, Heyland D, Kyeremanteng K, Murchison AG, Swalwell K, Mandeville J, Stott D, Guerreiro I, Devine H, MacTavish P, McPeake J, Quasim T, Kinsella J, Daniel M, Goossens C, Marques MB, Derde S, Vander Perre S, Dufour T, Thiessen SE, Güiza F, Janssens T, Hermans G, Vanhorebeek I, De Bock K, Van den Berghe G, Langouche L, Devine H, MacTavish P, Quasim T, Kinsella J, Daniel M, McPeake J, Miles B, Madden S, Devine H, Weiler M, Marques P, Rodrigues C, Boeira M, Brenner K, Leães C, Machado A, Townsend R, Andrade J, MacTavish P, McPeake J, Devine H, Kinsella J, Daniel M, Kishore R, Fenlon C, Quasim T, Fiks T, Ruijter A, Te Raa M, Spronk P, Chiew YS, Docherty P, Dickson J, Moltchanova E, Scarrot C, Pretty C, Shaw GM, Chase JG, Hall T, Ngu WC, Jack JM, Morgan P, Avard B, Pavli A, Gee X, Bor C, Akin Korhan E, Demirag K, Uyar M, Shirazy M, Fayed A, Gupta S, Kaushal A, Dewan S, Varma A, Ghosh E, Yang L, Eshelman L, Lord B, Carlson E, Helme E, Broderick R, Hadfield S, Loveridge R, Ramos J, Forte D, Yang F, Hou P, Dudziak J, Feeney J, Wilkinson K, Bauchmuller K, Shuker K, Faulds M, Raithatha A, Bryden D, England L, Bolton N, Tridente A, Bauchmuller K, Shuker K, Tridente A, Faulds M, Matheson A, Gaynor J, Bryden D, S South Yorkshire Hospitals Research Collaboration. Ramos J, Peroni B, Daglius-Dias R, Miranda L, Cohen C, Carvalho C, Velasco I, Forte D, Kelly JM, Neill A, Rubenfeld G, Masson N, Min A, Boezeman E, Hofhuis J, Hovingh A, De Vries R, Spronk P, Cabral-Campello G, Aragão I, Cardoso T, Van Mol M, Nijkamp M, Kompanje E, Ostrowski P, Omar A, Kiss K, Köves B, Csernus V, Molnár Z, Hoydonckx Y, Vanwing S, Stessel B, Van Assche A, Jamaer L, Dubois J, Medo V, Galvez R, Miranda JP, Stone C, Wigmore T, Arunan Y, Wheeler A, Bauchmuller K, Bryden D, Wong Y, Poi C, Gu C, Molmy P, Van Grunderbeeck N, Nigeon O, Lemyze M, Thevenin D, Mallat J, Ramos J, Correa M, Carvalho RT, Forte D, Fernandez A, McBride C, Koonthalloor E, Walsh C, Webber A, Ashe M, Smith K, Jeanrenaud P, Marudi A, Baroni S, Ragusa F, Bertellini E, Volakli EA, Chochliourou E, Dimitriadou M, Violaki A, Mantzafleri P, Samkinidou E, Vrani O, Arbouti A, Varsami T, Sdougka M, Bollen JA, Van Smaalen TC, De Jongh WC, Ten Hoopen MM, Ysebaert D, Van Heurn LW, Van Mook WN, Sim K, Fuller A, Roze des Ordons A, Couillard P, Doig C, Van Keer RV, Deschepper RD, Francke AF, Huyghens LH, Bilsen JB, Nyamaizi B, Dalrymple C, Molokhia A, Dobru A, Marrinan E, Ankuli A, Molokhia A, McPeake J, Struthers R, Crawford R, Devine H, Mactavish P, Quasim T, Morelli P, Degiovanangelo M, Lemos F, MArtinez V, Verga F, Cabrera J, Burghi G, Rutten A, Van Ieperen S, De Geer S, Van Vugt M, Der Kinderen E, Giannini A, Miccinesi G, Marchesi T, Prandi E. 36th International Symposium on Intensive Care and Emergency Medicine : Brussels, Belgium. 15-18 March 2016. Crit Care. 2016 Apr 20;20(Suppl 2):94. [PMC free article: PMC5493079] [PubMed: 27885969] 30. : Coca SG, Bauling P, Schifftner T, Howard CS, Teitelbaum I, Parikh CR. Contribution of acute kidney injury toward morbidity and mortality in burns: a contemporary analysis. Am J Kidney Dis. 2007 Apr;49(4):517-23. [PubMed: 17386319] 31. : Brown CV, Rhee P, Chan L, Evans K, Demetriades D, Velmahos GC. Preventing renal failure in patients with rhabdomyolysis: do bicarbonate and mannitol make a difference? J Trauma. 2004 Jun;56(6):1191-6. [PubMed: 15211124] 32. : Homsi E, Barreiro MF, Orlando JM, Higa EM. Prophylaxis of acute renal failure in patients with rhabdomyolysis. Ren Fail. 1997 Mar;19(2):283-8. [PubMed: 9101605] 33. : Saffle JI. The phenomenon of "fluid creep" in acute burn resuscitation. J Burn Care Res. 2007 May-Jun;28(3):382-95. [PubMed: 17438489] 34. : Bacomo FK, Chung KK. A primer on burn resuscitation. J Emerg Trauma Shock. 2011 Jan;4(1):109-13. 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[PubMed: 27600122] 41. : Norouzinia R, Aghabarari M, Shiri M, Karimi M, Samami E. Communication Barriers Perceived by Nurses and Patients. Glob J Health Sci. 2015 Sep 28;8(6):65-74. [PMC free article: PMC4954910] [PubMed: 26755475] : Disclosure: Patrycja Popowicz declares no relevant financial relationships with ineligible companies. : Disclosure: Abby Regan declares no relevant financial relationships with ineligible companies. : Disclosure: David Hotwagner declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK534227PMID: 30480960 Share Views PubReader Print View Cite this Page Popowicz P, Regan A, Hotwagner DT. Burn Fluid Resuscitation. [Updated 2025 Apr 6]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. In this Page Continuing Education Activity Introduction Anatomy and Physiology Indications Technique or Treatment Complications Clinical Significance Enhancing Healthcare Team Outcomes Review Questions References Related information PMC PubMed Central citations PubMed Links to PubMed Similar articles in PubMed EMS Burn Rule of Tens.[StatPearls. 2025] EMS Burn Rule of Tens. Oboli VN, Waseem M. StatPearls. 2025 Jan Efficacy of hydrosurgical excision combined with skin grafting in the treatment of deep partial-thickness and full-thickness burns: A two-year retrospective study.[Burns. 2023] Efficacy of hydrosurgical excision combined with skin grafting in the treatment of deep partial-thickness and full-thickness burns: A two-year retrospective study. Cao YL, Liu ZC, Chen XL. Burns. 2023 Aug; 49(5):1087-1095. Epub 2022 Jul 29. Resuscitation of severely burned military casualties: fluid begets more fluid.[J Trauma. 2009] Resuscitation of severely burned military casualties: fluid begets more fluid. Chung KK, Wolf SE, Cancio LC, Alvarado R, Jones JA, McCorcle J, King BT, Barillo DJ, Renz EM, Blackbourne LH. J Trauma. 2009 Aug; 67(2):231-7; discussion 237. Review The Immune and Regenerative Response to Burn Injury.[Cells. 2022] Review The Immune and Regenerative Response to Burn Injury. Burgess M, Valdera F, Varon D, Kankuri E, Nuutila K. Cells. 2022 Sep 29; 11(19). Epub 2022 Sep 29. Review E-cigarette burn injuries: Comprehensive review and management guidelines proposal.[Burns. 2019] Review E-cigarette burn injuries: Comprehensive review and management guidelines proposal. Jones CD, Ho W, Gunn E, Widdowson D, Bahia H. Burns. 2019 Jun; 45(4):763-771. Epub 2018 Nov 12. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Burn Fluid Resuscitation - StatPearls Burn Fluid Resuscitation - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers
17109	https://math.stackexchange.com/questions/2126055/foot-of-perpendicular-in-3-dimensions	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Foot of perpendicular in 3 dimensions Ask Question Asked Modified 5 years, 5 months ago Viewed 13k times -1 $\begingroup$ Let us suppose that $Q$ is a foot of perpendicular from a point $P (2,4,3) $on the line joining the points $A(1,2,4)$ and $B(3,4,5)$; then what are the coordinates of $Q$? My try: Had this been a two dimensional problem, I would have found the equation of line and proceeded further to get the answer. But in 3D neither do I know to find the equation of a line nor do I know how to proceed further. Kindly Help. 3d Share asked Feb 2, 2017 at 17:14 ZlatanZlatan 66922 gold badges66 silver badges1818 bronze badges $\endgroup$ 4 $\begingroup$ Are you familiar with vector calculus, or at least linear algebra? $\endgroup$ Triatticus – Triatticus 2017-02-02 17:17:27 +00:00 Commented Feb 2, 2017 at 17:17 1 $\begingroup$ I am. I just need to brush up the concepts, so if you could give me a clue how to proceed? $\endgroup$ Zlatan – Zlatan 2017-02-02 17:20:08 +00:00 Commented Feb 2, 2017 at 17:20 $\begingroup$ You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points. $\endgroup$ Triatticus – Triatticus 2017-02-02 17:22:10 +00:00 Commented Feb 2, 2017 at 17:22 1 $\begingroup$ Could you start with the solution please? $\endgroup$ Zlatan – Zlatan 2017-02-02 17:28:49 +00:00 Commented Feb 2, 2017 at 17:28 Add a comment \| 1 Answer 1 Reset to default 1 $\begingroup$ We know $A(1,2,4)$ and $B(3,4,5)$, so $\vec{AB}=\langle2,2,1\rangle$. A vector equation for $\overleftrightarrow{AB}$ is $\vec{r}=\langle1,2,4\rangle+t\langle2,2,1\rangle$ for $t\in\mathbb{R}$. Point $Q$ must lie on $\overleftrightarrow{AB}$, so the position vector for $Q$ is $\vec{Q}=\langle1,2,4\rangle+q\langle2,2,1\rangle=\langle1+2q,2+2q,4+q\rangle$ for some $q\in\mathbb{R}$. We need to find $q$ in order to find $Q$. We know $Q$ is the foot of the perpendicular from $P$ to $Q$. In other words, $\vec{PQ}\perp\vec{AB}$. We find $\vec{PQ}=\langle2q-1,2q-2,q+1\rangle$. Since $\vec{PQ}\perp\vec{AB}$, then $\vec{PQ}\cdot\vec{AB}=0$. \begin{align} \vec{PQ}\cdot\vec{AB}&=0\ \langle2q-1,2q-2,q+1\rangle\cdot\langle2,2,1\rangle&=0\ 4q-2+4q-4+q+1&=0\ 9q&=5\ q&=\frac{5}{9} \end{align} Using $q=5/9$, we find that $\vec{Q}=\langle1+2(5/9),2+2(5/9),4+(5/9)\rangle=\langle19/9,28/9,41/9\rangle$. Thus we have $Q\left(\frac{19}{9},\frac{28}{9},\frac{41}{9}\right)$. Share edited Feb 2, 2017 at 18:42 answered Feb 2, 2017 at 18:23 Tim ThayerTim Thayer 1,37377 silver badges1111 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. 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17110	https://www.cdc.gov/contraception/hcp/usmec/fertility-awareness-based-methods.html	Appendix F: Classifications for Fertility Awareness-Based Methods \| Contraception \| CDC Skip directly to site contentSkip directly to searchSkip directly to On This Page An official website of the United States government Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Contraception Explore Topics Search Search Clear Search For Everyone Birth Control Methods View all Health Care Providers CDC Contraceptive Guidance for Health Care Providers U.S. MEC, 2024 U.S. SPR, 2024 U.S. MEC and U.S. SPR Provider Tools View all Related Topics: Reproductive Health\|HEAR HER Campaign\|Maternal Mortality Prevention\|PRAMS View All search close search clear search ContraceptionMenu clear search For Everyone Birth Control Methods View AllHome Health Care Providers CDC Contraceptive Guidance for Health Care Providers U.S. MEC, 2024 U.S. SPR, 2024 U.S. MEC and U.S. SPR Provider Tools View All Related Topics Reproductive Health HEAR HER Campaign Maternal Mortality Prevention PRAMS View All Contraception CDC Contraceptive Guidance for Health Care ProvidersU.S. MEC, 2024U.S. SPR, 2024U.S. MEC and U.S. SPR Provider ToolsView All Health Care Providers November 19, 2024 Appendix F: Classifications for Fertility Awareness-Based Methods Summary for U.S. MEC, 2024 \| Page 11 of 17 \| All pages At a glance This page includes recommendations for health care providers for the use of fertility awareness-based methods for persons who have certain characteristics or medical conditions. This information comes from the 2024 U.S. Medical Eligibility Criteria for Contraceptive Use (U.S. MEC). Overview Fertility awareness–based (FAB) methods involve identifying the fertile days of the menstrual cycle, whether by observing fertility signs, such as cervical secretions and basal body temperature or by monitoring cycle days, and might include use of Food and Drug Administration–cleared contraceptive software applications (Box F1) (Table F1). FAB methods can be used in combination with abstinence or barrier methods during the fertile time. If barrier methods are used, see Classifications for Barrier Methods (Appendix E). No medical conditions worsen because of FAB methods. In general, FAB methods can be used without concern for health effects in persons who choose them. However, multiple conditions make their use more complex. The existence of these conditions suggests that use of these methods should be delayed until the condition is corrected or resolved, or persons using FAB methods need special counseling; and a provider with particular training in use of these methods is generally necessary to ensure correct use. FAB methods do not protect against sexually transmitted infections (STIs), including HIV infection, and patients using FAB methods should be counseled that consistent and correct use of external (male) latex condoms reduces the risk for STIs, including HIV infection.Use of internal (female) condoms can provide protection from transmission of STIs, although data are limited. Patients also should be counseled that pre-exposure prophylaxis, when taken as prescribed, is highly effective for preventing HIV infection. Box F1. Definitions for terms associated with fertility awareness–based methods Symptoms-based methods: FAB methods based on observation of fertility signs (e.g., cervical secretions or basal body temperature) such as the cervical mucus method, the symptothermal method, and the TwoDay method. Calendar-based methods: FAB methods based on calendar calculations such as the calendar rhythm method and the standard days method. Accept: No medical reason exists to deny the particular FAB method to a patient in this circumstance. Caution: The method normally is provided in a routine setting but with extra preparation and precautions. For FAB methods, this usually means that special counseling might be needed to ensure correct use of the method by a patient in this circumstance. Delay: Use of this method should be delayed until the condition is evaluated or corrected. Alternative temporary methods of contraception should be offered. Abbreviation: FAB = fertility awareness–based. Table F1. Fertility awareness–based methods, including symptoms-based and calendar-based methods TABLE F1 CONDITIONS Personal characteristics and reproductive history Reproductive tract infections and disorders Other Table F1. Fertility awareness–based methods, including symptoms-based and calendar-based methods\| Condition \| Category \| Clarification/Evidence/Comment \| --- \| Symptoms-based method \| Calendar-based method \| \| Personal Characteristics and Reproductive History \| \| Pregnancy \| NA \| NA \| Clarification: FAB methods are not relevant during pregnancy. \| \| Life stage \| Comment: Menstrual irregularities are common in postmenarche and perimenopause and might complicate the use of FAB methods. \| \| a. Postmenarche \| Caution \| Caution \| \| b. Perimenopause \| Caution \| Caution \| \| Breastfeeding \| Comment: Use of FAB methods when breastfeeding might be less effective than when not breastfeeding. \| \| a. <6 weeks postpartum \| Delay \| Delay \| Comment: Persons who are primarily breastfeeding and are amenorrheic are unlikely to have sufficient ovarian function to produce detectable fertility signs and hormonal changes during the first 6 months postpartum. However, the likelihood of resumption of fertility increases with time postpartum and with substitution of breast milk by other foods. \| \| b. ≥6 weeks postpartum \| Caution \| Delay \| \| c. After menses begin \| Caution \| Caution \| Clarification: Once fertility signs are noted, particularly cervical secretions, then symptoms-based methods can be used. First postpartum menstrual cycles while breastfeeding vary significantly in length. Return to regularity takes several cycles. When there have been at least three postpartum menses and cycles are regular again, a calendar-based method can be used. When there have been at least four postpartum menses and the most recent cycle lasted 26–32 days, the standard days method can be used. Before that time, a barrier method should be offered if the patient plans to use a FAB method later. \| \| Postpartum (nonbreastfeeding) \| \| a. <4 weeks \| Delay \| Delay \| Clarification: Nonbreastfeeding persons are not likely to have detectable fertility signs or hormonal changes before 4 weeks postpartum. Although the risk for pregnancy is low, ovulation before first menses is common; therefore, a method appropriate for the postpartum period should be offered. \| \| b. ≥4 weeks \| Accept \| Delay \| Clarification: Nonbreastfeeding persons are likely to have sufficient ovarian function to produce detectable fertility signs, hormonal changes, or both at this time; likelihood increases rapidly with time postpartum. Calendar-based methods can be used as soon as three postpartum menses have been completed. Methods appropriate for the postpartum period should be offered before that time. \| \| Postabortion(spontaneous or induced) \| Caution \| Delay \| Clarification: After abortion, it is possible for ovarian function to produce detectable fertility signs, hormonal changes, or both; likelihood increases with time postabortion. Calendar-based methods can be used following at least one postabortion menses (e.g., persons who before this pregnancy primarily had cycles of 26–32 days can then use the standard days method). Methods appropriate for the postabortion period should be offered before that time. \| Back to Top Table F1. Fertility awareness–based methods, including symptoms-based and calendar-based methods\| Condition \| Category \| Clarification/Evidence/Comment \| --- \| Symptoms-based method \| Calendar-based method \| \| Reproductive Tract Infections and Disorders \| \| Irregular vaginal bleeding \| Delay \| Delay \| Clarification: Presence of this condition makes FAB methods unreliable. Therefore, barrier methods should be recommended until the bleeding pattern is compatible with proper method use. The condition should be evaluated and treated as necessary. \| \| Vaginal discharge \| Delay \| Accept \| Clarification: Because vaginal discharge makes recognition of cervical secretions difficult, the condition should be evaluated and treated if needed before providing methods based on cervical secretions. \| Back to Top Table F1. Fertility awareness–based methods, including symptoms-based and calendar-based methods\| Condition \| Category \| Clarification/Evidence/Comment \| --- \| Symptoms-based method \| Calendar-based method \| \| Other \| \| Use of drugs that affect cycle regularity, hormones, or fertility signs \| Caution /Delay \| Caution/Delay \| Clarification: Use of certain mood-altering drugs (e.g., lithium, tricyclic antidepressants, and antianxiety therapies), as well as certain antibiotics and anti-inflammatory drugs, might alter cycle regularity or affect fertility signs. The condition should be carefully evaluated and a barrier method offered until the degree of effect has been determined or the drug is no longer being used. \| \| Diseases that elevate body temperature \| \| a. Chronic diseases \| Caution \| Accept \| Clarification: Elevated temperatures might make basal body temperature difficult to interpret but have no effect on cervical secretions. Thus, use of a method that relies on temperature should be delayed until the acute febrile disease abates. Temperature-based methods are not appropriate for persons with chronically elevated temperatures. In addition, certain chronic diseases interfere with cycle regularity, making calendar-based methods difficult to interpret. \| \| b. Acute diseases \| Delay \| Accept \| Abbreviations: FAB = fertility awareness–based; NA = not applicable. Back to Top References Workowski KA, Bachmann LH, Chan PA, et al. Sexually transmitted infections treatment guidelines, 2021. MMWR Recomm Rep 2021;70(No. RR-4):1–187.PMID:34292926 CDC. US Public Health Service preexposure prophylaxis for the prevention of HIV infection in the United States—2021 update: a clinical practice guideline. Atlanta, GA: US Department of Health and Human Services, CDC; 2021. Back to Top On This Page Overview Box F1. Definitions for terms associated with fertility awareness–based methods Table F1. Fertility awareness–based methods, including symptoms-based and calendar-based methods References Related Pages U.S. MEC, 2024 Combined Hormonal Contraceptives Barrier Methods Lactational Amenorrhea Method Coitus Interruptus (Withdrawal) View All Contraception Lactational Amenorrhea Method Table of Contents \| Summary for U.S. MEC, 2024 Table of Contents Introduction Methods Keeping Guidance Up to Date How to Use This Document Summary of Changes from 2016 Intrauterine Devices Progestin-Only Contraceptives Combined Hormonal Contraceptives Barrier Methods Fertility Awareness-Based Methods Lactational Amenorrhea Method Coitus Interruptus (Withdrawal) Permanent Contraception Emergency Contraception Summary of Classifications for Hormonal Contraceptive Methods and Intrauterine Devices Acknowledgments, Contributors, and Participants Show More November 18, 2024 SourcesPrintShare FacebookLinkedInTwitterSyndicate Content Source: National Center for Chronic Disease Prevention and Health Promotion (NCCDPHP); Division of Reproductive Health Related Pages U.S. MEC, 2024 Combined Hormonal Contraceptives Barrier Methods Lactational Amenorrhea Method Coitus Interruptus (Withdrawal) View All Contraception Back to Top Contraception There are safe and highly effective methods of contraception available to prevent or reduce the chances of unintended pregnancy. CDC provides clinical guidance to assist health care providers in counseling women, men, and couples about contraceptive method choice. They also offer clinical guidance to reduce medical barriers to contraception access and use. View All For Everyone Birth Control Methods Health Care Providers CDC Contraceptive Guidance for Health Care Providers U.S. MEC, 2024 U.S. SPR, 2024 View All Sign up for Email Updates Contact Us Contact Us Call 800-232-4636 Contact CDC About CDC About CDC Pressroom Organization Budget & Funding Careers & Jobs Policies Accessibility External Links Privacy Web Policies FOIA OIG No Fear Act Nondiscrimination Vulnerability Disclosure Policy Languages Languages Español Language Assistance Español 繁體中文 Tiếng Việt 한국어 Tagalog Русский العربية Kreyòl Ayisyen Français Polski Português Italiano Deutsch 日本語 فارسی English Archive CDC Archive Public Health Publications Contact Us Contact Us Call 800-232-4636 Contact CDC About CDC Pressroom Organization Budget & Funding Careers & Jobs About CDC Policies Accessibility External Links Privacy Web Policies FOIA OIG No Fear Act Nondiscrimination Vulnerability Disclosure Policy Languages Languages Español Language Assistance Español 繁體中文 Tiếng Việt 한국어 Tagalog Русский العربية Kreyòl Ayisyen Français Polski Português Italiano Deutsch 日本語 فارسی English Archive CDC Archive Public Health Publications HHS.govUSA.gov
17111	https://www.teacherspayteachers.com/Product/Motion-and-Stability-Forces-and-Interactions-5-Unit-Bundle-MS-PS2-Editable-11959176	Motion and Stability - Forces and Interactions 5 Unit Bundle: MS-PS2 Editable Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 2nd grade ELA 3rd grade ELA 4th grade ELA 5th grade ELA 6th grade ELA 7th grade ELA 8th grade ELA High school ELA Elementary ELA Reading Writing Phonics Vocabulary Grammar Spelling Poetry ELA test prep Middle school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep High school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep Math Math by grade PreK math Kindergarten math 1st grade math 2nd grade math 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math High school math Elementary math Basic operations Numbers Geometry Measurement Mental math Place value Arithmetic Fractions Decimals Math test prep Middle school math Algebra Basic operations Decimals Fractions Geometry Math test prep High school math Algebra Algebra 2 Geometry Math test prep Statistics Precalculus Calculus Science Science by grade PreK science Kindergarten science 1st grade science 2nd grade science 3rd grade science 4th grade science 5th grade science 6th grade science 7th grade science 8th grade science High school science By topic Astronomy Biology Chemistry Earth sciences Physics Physical science Social studies Social studies by grade PreK social studies Kindergarten social studies 1st grade social studies 2nd grade social studies 3rd grade social studies 4th grade social studies 5th grade social studies 6th grade social studies 7th grade social studies 8th grade social studies High school social studies Social studies by topic Ancient history Economics European history Government Geography Native Americans Middle ages Psychology U.S. History World history Languages Languages American sign language Arabic Chinese French German Italian Japanese Latin Portuguese Spanish Arts Arts Art history Graphic arts Visual arts Other (arts) Performing arts Dance Drama Instrumental music Music Music composition Vocal music Special education Speech therapy Social emotional Social emotional Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Motion and Stability - Forces and Interactions 5 Unit Bundle: MS-PS2 Editable Rated 5 out of 5, based on 1 reviews 5.0(1 rating) $53.95 $37.77 Add to cart Wish List DescriptionReviews 1Q&AStandards 5 More from Mr Shark Tooth Thumbnail 1 Thumbnail 2 Thumbnail 3 Thumbnail 4 View Preview Share Description Teacher guides, answer keys, labs, and modeling - this bundle has it all!This mega-bundle covers all five Motions and Stability - Forces and Interactions MS-PS2 Standards (Newton's Third Law, Sum of Forces, Electromagnetism, Mass and Gravity, and the Existence of Fields). Take care of planning yourscience lessons with this 5 unit bundle (+ editable versions in Google Slides and PowerPoint). It addresses MS-PS2-1 through MS-PS2-5! With 60+ engaging activities, you will save time and money while watching your students learn! What's Included: ⭐️ 5 Complete Units ⭐️ Teacher Guides ⭐️ Answer Keys ⭐️ Unit Plan/Pacing Guide for Each Unit ⭐️ Cloze Notes ⭐️ Assessments ⭐️ Labs ⭐️ Modeling Activities ⭐️ Worksheets Why Your Students Will Love It: They get to experience fun labs that will make meaningful connections to the content! They will spend less time copying notes and more time interacting with the content! Why You Will Love It: You can simply read the teaching order, follow the teacher guide, obtain any necessary materials, and print the materials! You are ready to go! You will see deep connections being made by students as they proceed through the unit. You will have more time to focus on your class instead of preparing materials. 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This no-prep five-day resource fully addresses how electricity and magnetism are related. It includes everything you need to teach Science Standard MS-PS2-3. It includes preformatted PDF and editable versions. It Google Apps, PDF Preview Gravity Activities - MS-PS2-4 Gravitational Force Relationship Unit + Labs Looking for fun activities for your gravity and mass unit (MS-PS2-4)? This five-day resource fully addresses how the Moon and planets move through space. It includes preformatted PDF and editable versions. It will save you time by filling your lesson plans with ten exciting activities and letting yo Google Apps, PDF Preview Newton's Laws of Motion Unit: MS-PS2-1 Activities + Colliding Objects Lab Are you looking for fun activities and labs to teach Newton's Laws of Motion? This seven-day unit fully addresses the motion of two colliding objects, forces, and the 3rd law.It's everything you need to teach Science Standard MS-PS2-1. It includes preformatted PDF versions and editable versions. It Google Apps, PDF Preview Static Electricity and Magnetism - Evidence of Fields Activities - MS-PS2-5 Unit Are you looking for fun activities for your static electricity and magnetism unit (MS-PS2-5)? This five-day resource fully addresses how scientists develop evidence of fields from magnets and electrical sources. It includes preformatted PDF and editable versions. It will save you time by filling you Google Apps, PDF Preview Sum of Forces Unit - MS-PS2-2 Newton's Laws of Motion Labs and Activities Are you looking for fun activities to teach your sum of forces unit (Science Standard MS-PS2-2)? This ten-day resource fully addresses Newton's Laws of Motion, balanced and unbalanced forces, and the connection between mass and gravity. It includes preformatted PDF versions and editable versions. It Google Apps, PDF Preview Save even more with bundles Motion and Stability Forces and Interactions MEGA BUNDLE - MS-PS2 Units + Review Teacher guides, answer keys, labs, and modeling - this bundle has it all! This mega-bundle covers all five Motions and Stability - Forces and Interactions MS-PS2 Standards (Newton's Third Law, Sum of Forces, Electromagnetism, Mass and Gravity, and the Existence of Fields).Take care of planning your $55.23 Price $55.23$78.90 Original Price $78.90 Save $23.67 10 8th Grade Integrated Model Curriculum MEGA BUNDLE: Units and Review - FULL YEAR Teacher guides, activities answer keys, labs, and modeling - this bundle has it all! This mega-bundle covers all 22 8th grade integrated model standards. From Newton's laws of motion to the solar system, this full-year resource will have you prepared to teach!Take care of planning your entire 8th gr $180.77 Price $180.77$277.64 Original Price $277.64 Save $96.87 36 Physical Science Curriculum MEGA BUNDLE: 18 Units + 14 Escape Room Review Games Teacher guides, answer keys, labs, and modeling - this bundle has it all! This mega-bundle covers all 19 Physical Science standards. From atoms to forces and motion, this full-year resource will have you prepared to teach and engage your students with inquiry strategies!Take care of planning your cu $166.20 Price $166.20$255.68 Original Price $255.68 Save $89.48 32 1 2 3 4 Reviews 5.0 Rated 5 out of 5, based on 1 reviews 1 rating This product (1) All products (1,101) All verified TPT purchases All ratings 5 stars All grades 8th grade Sort by: Most recent Most relevant Most recent Highest rating Lowest rating Rated 5 out of 5 May 13, 2025 My students loved using this resource. They had fun with these activities! Amazing! Kailyn T. 234 reviews Grades taught:8th Questions & Answers Please log into post a question. Be the first to ask Mr Shark Tooth a question about this product. Standards Log in to see state-specific standards (only available in the US). NGSS MS-PS2-4 Construct and present arguments using evidence to support the claim that gravitational interactions are attractive and depend on the masses of interacting objects. Examples of evidence for arguments could include data generated from simulations or digital tools; and charts displaying mass, strength of interaction, distance from the Sun, and orbital periods of objects within the solar system. Assessment does not include Newton’s Law of Gravitation or Kepler’s Laws. NGSS MS-PS2-3 Ask questions about data to determine the factors that affect the strength of electric and magnetic forces. Examples of devices that use electric and magnetic forces could include electromagnets, electric motors, or generators. Examples of data could include the effect of the number of turns of wire on the strength of an electromagnet, or the effect of increasing the number or strength of magnets on the speed of an electric motor. Assessment about questions that require quantitative answers is limited to proportional reasoning and algebraic thinking. NGSS MS-PS2-2 Plan an investigation to provide evidence that the change in an object’s motion depends on the sum of the forces on the object and the mass of the object. Emphasis is on balanced (Newton’s First Law) and unbalanced forces in a system, qualitative comparisons of forces, mass and changes in motion (Newton’s Second Law), frame of reference, and specification of units. Assessment is limited to forces and changes in motion in one-dimension in an inertial reference frame, and to change in one variable at a time. Assessment does not include the use of trigonometry. Show more Meet the Teacher-Author ### Mr Shark Tooth Follow I am a curriculum designer with 12 years of middle school experience and a passion for helping students comprehend the world around them, making connections with everyday life, and (most importantly) asking questions. My goal is to design resources that help students do these three things. Greenville, South Carolina, United States 4.75 Store rating after 1.1k reviews 1.5k Followers My favorite moment in a class is when students debate an answer by citing facts and then having a class discussion. 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17112	https://www.uptodate.com/contents/overview-of-the-diagnosis-and-staging-of-head-and-neck-cancer	Overview of the diagnosis and staging of head and neck cancer - UpToDate Subscribe Sign in English Deutsch Español 日本語 Português Why UpToDate? Product Editorial Subscription Options How can UpToDate help you?Select the option that best describes you Medical Professional Resident, Fellow, or Student Hospital or Institution Group Practice Subscribe Overview of the diagnosis and staging of head and neck cancer View Topic Share Font Size Small Normal Large Bookmark Rate Feedback Tools Formulary drug information for this topic No drug references linked in this topic. Find in topic Formulary Print Share Feedback Font Size Small Normal Large Outline SUMMARY AND RECOMMENDATIONS INTRODUCTION ANATOMY CLINICAL PRESENTATION When to suspect Clinical features by tumor site DIAGNOSTIC EVALUATION Overview of evaluation Physical exam Imaging studies - Choice of imaging study - CT scan - PET/CT - MRI Establishing a diagnosis with biopsy PATHOLOGY Histology Future molecular methods STAGING Evaluation of neck nodes TNM staging system SOCIETY GUIDELINE LINKS INFORMATION FOR PATIENTS SUMMARY AND RECOMMENDATIONS REFERENCES GRAPHICS Algorithms - Evaluation of adults with suspected head and neck cancer Tables - Clinical characteristics of HPV associated HNSCC - Oral cavity cancer TNM, 8th edition - Nasopharyngeal cancer TNM, 8th edition - Oropharyngeal (p16 negative) CA TNM clinical staging, 8th ed - Oropharyngeal (p16 negative) CA TNM pathologic staging, 8th ed - HPV-related oropharyngeal carcinoma TNM clinical staging, 8th ed - HPV related oropharyngeal CA TNM pathologic staging, 8th edition - Hypopharyngeal cancer TNM clinical staging, 8th edition - Hypopharyngeal cancer TNM pathologic staging, 8th edition - Larynx cancer TNM, 8th edition - Nasal cavity and paranasal sinuses TNM, 8th edition - Major salivary gland tumors TNM clinical staging, 8th edition - Major salivary gland tumors TNM pathologic staging, 8th edition Figures - Anatomy head and neck - Anatomy of the salivary glands and ducts - Lymph node levels of the neck Pictures - Tonsil cancer Diagnostic Images - CT carcinoma tongue RELATED TOPICS Bell's palsy: Pathogenesis, clinical features, and diagnosis in adults Cancer of the nasal vestibule Diagnostic approach to and treatment of thyroid nodules in adults Differentiated thyroid cancer: Clinicopathologic staging Epidemiology and risk factors for head and neck cancer Epidemiology, etiology, and diagnosis of nasopharyngeal carcinoma Epidemiology, staging, and clinical presentation of human papillomavirus associated head and neck cancer Head and neck sarcomas Head and neck squamous cell carcinogenesis: Molecular and genetic alterations Head and neck squamous cell carcinoma of unknown primary Locoregional mucosal melanoma: Epidemiology, clinical diagnosis, and treatment Medullary thyroid cancer: Clinical manifestations, diagnosis, and staging Melanoma: Clinical features and diagnosis Overview of treatment for head and neck cancer Papillary thyroid cancer: Clinical features and prognosis Paranasal sinus cancer Pathology of head and neck neoplasms Patient education: Laryngeal cancer (The Basics) Patient education: Mouth sores (The Basics) Patient education: Throat cancer (The Basics) Patient education: Tongue cancer (The Basics) Salivary gland tumors: Epidemiology, diagnosis, evaluation, and staging Second primary malignancies in patients with head and neck cancers Society guideline links: Head and neck cancer Treatment of early (stage I and II) head and neck cancer: The hypopharynx Treatment of early (stage I and II) head and neck cancer: The larynx Treatment of early (stage I and II) head and neck cancer: The oropharynx Treatment of early and locoregionally advanced nasopharyngeal carcinoma Treatment of human papillomavirus associated oropharyngeal cancer Treatment of stage I and II (early) head and neck cancer: The oral cavity Tumors of the nasal cavity Overview of the diagnosis and staging of head and neck cancer Authors:Colin S Poon, MD, PhD, FRCPCKerstin M Stenson, MD, FACSSection Editors:Bruce E Brockstein, MDMarvin P Fried, MD, FACSDeputy Editor:Melinda Yushak, MD, MPH Contributor Disclosures All topics are updated as new evidence becomes available and our peer review process is complete. Literature review current through:Aug 2025. This topic last updated:Jun 28, 2024. Please read the Disclaimerat the end of this page. INTRODUCTION— Head and neck cancers can arise in the oral cavity, pharynx, larynx, nasal cavity, paranasal sinuses, and salivary glands. The most common histology is squamous cell carcinoma. Other common types include nasopharyngeal carcinoma and mucosal melanoma. This topic will review the initial evaluation, diagnosis, and staging of head and neck cancer. Epidemiology, risk factors, pathology, and treatment of head and neck cancer are discussed separately: ●(See "Epidemiology and risk factors for head and neck cancer".) ●(See "Pathology of head and neck neoplasms".) ●(See "Overview of treatment for head and neck cancer".) Specific details regarding the diagnosis of a selection of specific subtypes of head and neck cancers are discussed separately: ●(See "Epidemiology, staging, and clinical presentation of human papillomavirus associated head and neck cancer".) ●(See "Locoregional mucosal melanoma: Epidemiology, clinical diagnosis, and treatment".) ●(See "Epidemiology, etiology, and diagnosis of nasopharyngeal carcinoma".) ●(See "Head and neck sarcomas".) The diagnosis, staging and treatment of thyroid cancer is also discussed separately: ●(See "Diagnostic approach to and treatment of thyroid nodules in adults".) ●(See "Differentiated thyroid cancer: Clinicopathologic staging".) ●(See "Medullary thyroid cancer: Clinical manifestations, diagnosis, and staging".) ●(See "Papillary thyroid cancer: Clinical features and prognosis".) ANATOMY— Head and neck cancers can arise from several anatomic areas. These include: ●Oral cavity – The oral cavity includes the mucosa of the lips (not the external, dry lip), the buccal mucosa, the anterior tongue, the floor of the mouth, the hard palate, and the upper and lower gingiva. The anterior border of the oral cavity is defined by the portion of the lip that contacts the opposed lip (wet mucosa). The posterior border is defined by the circumvallate papillae of the tongue, the anterior tonsillar pillars (palatoglossus muscles), and the posterior margin of the hard palate. The hard palate defines the superior boundary of the oral cavity. Inferiorly, the oral cavity is defined by the mylohyoid muscles. The lateral boundary of the oral cavity is defined by the buccomasseteric region (buccal mucosa of the cheeks) and the retromolar trigone (which is located behind the mandibular third molar). ●Pharynx – The pharynx is divided into the nasopharynx, oropharynx, and hypopharynx (figure 1A). •Nasopharynx – The nasopharynx forms the continuation of the nasal cavity. The boundary between the nasal cavity and nasopharynx is defined by the posterior choanae of the nasal cavity. The nasopharynx is defined superiorly by the basisphenoid and basiocciput (clivus) and inferiorly by the hard and soft palate. The prevertebral muscle and anterior margin of the cervical spine at C1 and C2 levels form the posterior margin of the nasopharynx. The posterolateral boundary of the nasopharynx includes several important structures. The lateral wall of the nasopharynx is elevated by the torus tubarius, a cartilaginous structure that constitutes the opening of the Eustachian tube. The Eustachian tube allows communication between the middle ear and the nasopharynx through a defect (sinus of Morgagni) of the pharyngobasilar fascia, which lines the nasopharynx. The sinus of Morgagni may allow nasopharyngeal cancer to gain access into the skull base. The fossa of Rosenmüller (lateral nasopharyngeal recess), a common site of nasopharyngeal cancer, is located posterior to the torus tubarius. The nasopharynx also includes the adenoids (nasopharyngeal tonsils), located in the midline roof of the nasopharynx. •Oropharynx – The soft palate defines the boundary between the nasopharynx and oropharynx. The oropharynx is separated anteriorly from the oral cavity by the circumvallate papillae and the anterior tonsillar pillars. The palatine tonsils, posterior tonsillar pillars, tongue base (posterior one-third of the tongue), valleculae, soft palate and the posterior pharyngeal wall are structures of the oropharynx. Inferiorly, the oropharynx is defined by the hyoid and the pharyngoepiglottic folds. •Hypopharynx – The hypopharynx includes the pyriform sinuses, the posterior surface of the larynx (postcricoid area), and the inferior, posterior, and lateral pharyngeal walls. ●Larynx – The larynx is divided into three anatomic regions: the supraglottic region, the glottic larynx (true vocal cords and mucosa of the anterior and posterior commissures), and the subglottic larynx, which extends to the inferior border of the cricoid cartilage. ●Nasal cavity and paranasal sinuses – These include the nasal cavity and the paranasal sinuses (maxillary, ethmoid, sphenoid, and frontal). ●Salivary glands – These include the major (parotid, submandibular, and sublingual) and minor salivary glands (figure 1B). Minor salivary glands are found within the submucosa throughout the oral cavity, palate, paranasal sinuses, pharynx, larynx, trachea and bronchi, but are most concentrated in the buccal, labial, palatal, and lingual regions. CLINICAL PRESENTATION— The clinical presentation of head and neck cancer varies widely depending upon the primary site and type of malignancy. There is overlap in some symptoms regardless of where the tumor originated from because similar anatomic structures will be involved based on the close proximity of structures in the head and neck region. When to suspect—Head and neck cancer may be suspected in patients who present with one or more of the following symptoms if they are otherwise unexplained: ●Otalgia – Cranial nerves 5, 7, 9, and 10 contribute afferents to the external and middle ear. The presence of referred otalgia should prompt evaluation for a head and neck malignancy. ●Neck mass – A neck mass in an adult is abnormal and should prompt evaluation. Patients can present with a neck mass due to lymph node involvement. The lymph nodes are often firm and fixed or with reduced mobility. There may be ulceration of the overlying skin. ●Hoarseness or voice change. ●Nasal congestion or epistaxis. ●Odynophagia or dysphagia. ●Hemoptysis or blood in saliva. ●Mouth or skin ulcerations. ●Unilateral tonsil enlargement. ●Palpable lesions in the salivary glands, especially when they are asymptomatic. ●Solitary masses in the thyroid or a change in a pre-existing goiter. (See "Diagnostic approach to and treatment of thyroid nodules in adults".) Clinical features by tumor site—Tumors presenting in specific anatomic locations (eg, oral cavity, pharynx, larynx, or sinuses) will have a distinct set of presenting symptoms. Examples of these include: ●Oral cavity tumors – Patients may present with mouth pain or nonhealing mouth ulcers, loosening of teeth, ill-fitting dentures, dysphagia, odynophagia, weight loss, bleeding, or referred otalgia. •Tongue cancer may grow as an infiltrative and/or exophytic lesion. The presenting symptom is often pain, with or without dysarthria. Dysarthria implies deep muscle invasion of advanced tumor stage. There may be a history of longstanding leukoplakia or erythroplakia. Up to two-thirds of patients with primary tongue lesions have cervical lymph node involvement. •Lip cancer usually presents as an exophytic or ulcerative lesion of the lower lip, occasionally associated with bleeding or pain. Some patients complain of numbness of the skin of the chin due to involvement of the mental nerve. ●Nasopharynx – Patients may present with hearing loss (associated with serous otitis media), tinnitus, nasal obstruction, and pain. The growth of the tumor into adjacent anatomical structures can lead to muscle involvement and impaired function of cranial nerves 2 to 6. Nasopharyngeal carcinoma begins in the nasopharynx. However, the most frequent presenting complaint for this malignancy is a neck mass due to regional lymph node metastasis, which occurs in nearly 90 percent of patients. (See "Epidemiology, etiology, and diagnosis of nasopharyngeal carcinoma", section on 'Clinical presentation'.) ●Oropharyngeal tumors – Presenting complaints can include dysphagia, pain (odynophagia, otalgia), obstructive sleep apnea or snoring, bleeding, or a neck mass (picture 1). Unlike patients with human papillomavirus (HPV)-negative tumors who often present with odynophagia and otalgia, patients with HPV-positive oropharyngeal cancers often present with a neck mass (often cystic) as their only complaint (table 1). These cystic neck masses are often mistaken for branchial cleft cyst carcinomas. In reality, branchial cleft cyst carcinoma is exceptionally rare and its diagnosis should be one of exclusion rather than presumption . (See "Epidemiology, staging, and clinical presentation of human papillomavirus associated head and neck cancer".) ●Hypopharyngeal tumors– Patients with these tumors often remain asymptomatic for a longer period and are therefore more likely to be seen in the later stages of the disease. Dysphagia, odynophagia, otalgia, weight loss, hemoptysis, dyspnea, and neck mass are common presenting symptoms. ●Laryngeal cancer – The symptoms associated with cancer of the larynx depend upon location. Persistent hoarseness may be the initial complaint in glottic cancers; later symptoms may include dysphagia, referred otalgia, chronic cough, hemoptysis, and stridor. Supraglottic cancers are often discovered later and may present with airway obstruction or palpable metastatic lymph nodes. Primary subglottic tumors are rare. Affected patients typically present with stridor or complaints of dyspnea on exertion. ●Nasal cavity – Patients with nasal cavity tumors often have symptoms of locally advanced disease at presentation including nasal obstruction and epistaxis. As the disease progresses the tumor can involve adjacent structures and cause facial swelling/pain, proptosis, cranial nerve dysfunction, seizure, and lymph node enlargement. (See "Tumors of the nasal cavity".) ●Sinus tumors – Common presenting symptoms of sinus tumors include epistaxis and unilateral nasal obstruction. Facial and/or head pain may be seen in later stages, due to pressure or tumor infiltration into nerves or periosteum. (See "Paranasal sinus cancer", section on 'Clinical presentation'.) ●Salivary gland tumors – Patients with salivary gland tumors will often have swelling or a mass form at the tumor location. Symptoms of more advanced disease depend on which gland is involved. (See "Salivary gland tumors: Epidemiology, diagnosis, evaluation, and staging", section on 'Clinical presentation'.) Gradual facial weakness or twitching can be a sign of nerve involvement from a parotid gland malignancy. This must be distinguished from Bell's palsy, which is more sudden in onset. (See "Bell's palsy: Pathogenesis, clinical features, and diagnosis in adults".) DIAGNOSTIC EVALUATION— The initial evaluation of a patient with a suspected head and neck cancer involves a physical exam, imaging studies, and tissue diagnosis (algorithm 1). Overview of evaluation—Patients suspected of having head and neck cancer based on the initial history and physical examination should: ●Be referred to a specialist (eg, otolaryngologist or ear, nose, and throat specialist) for a targeted head and neck examination in the office or under anesthesia. ●Undergo a neck computed tomography (CT) or magnetic resonance imaging (MRI) evaluation with intravenous contrast. A full description of imaging studies for complete staging and alternatives such as positron emission tomography (PET)/CT are discussed separately. (See 'Imaging studies' below.) ●Undergo biopsy (typically a fine-needle aspiration, although some may require core or open biopsy). Physical exam—The initial assessment of the primary tumor is based upon a thorough physical exam. We perform the following: ●Examination of the nasal cavity and oral cavity with visual examination and/or palpation of mucous membranes, the floor of the mouth, the anterior two-thirds of the tongue, tonsillar fossae and tongue base, palate, buccal and gingival mucosa, and posterior pharyngeal wall. ●A flexible laryngoscopy is used to examine the mucosa in the nasopharynx, oropharynx, hypopharynx, and larynx. Indirect mirror examination is an alternative but can only visualize the tongue base and laryngeal structures. Aside from mucosal irregularities, other abnormalities that should be specifically searched for are impairment of vocal cord mobility, pooling of secretions, anatomic asymmetries, and bleeding. ●Examination of nodal drainage areas and the parotid glands (figure 2). ●External auditory canal examination and anterior rhinoscopy. Imaging studies Choice of imaging study—Imaging is recommended for all patients with newly-suspected or diagnosed head and neck cancer to help determine the extent of both locoregional and metastatic disease. There is debate about the best imaging modality to use. We use the following approach: ●CT of the neck with contrast and CT of the chest with contrast are our preferred imaging modality for most patients. ●MRI of the neck with contrast is an acceptable alternative to CT of the neck, especially for patients with oral primaries or nasopharyngeal cancer. ●PET/CT is also an acceptable alternative and our preferred modality for patients with bulky disease or risk factors for a second primary (eg, history of heavy alcohol or tobacco use), and in patients with head and neck carcinoma of unknown primary [2,3]. However, insurance companies may not approve PET imaging without a cancer diagnosis. If the diagnosis of head and neck cancer is established, CT and/or MRI of the neck with contrast is often required for radiation or surgical planning as it has superior anatomic imaging. The preferred imaging modality is evolving, and some institutions may order a PET/CT as the initial imaging for most patients with established new head and neck cancer. Ideally, imaging should take place prior to biopsy, which may distort anatomy and create a false-positive finding on PET/CT scanning. Each imaging modality has advantages and disadvantages over other modalities. As examples: ●Compared with MRI, CT provides greater spatial resolution, can be performed with faster acquisition times (thereby virtually eliminating motion artifact), and is better at evaluating bone destruction. ●MRI provides superior soft tissue contrast. It can provide more accurate definition of tumors of the tongue and is more sensitive for superficial tumors. MRI is also better than CT for discriminating tumor from mucus and in detecting bone marrow invasion. For this reason, MRI can be useful for evaluation of cartilage invasion, particularly for non-ossified cartilage that can pose difficulty for CT. ●PET/CT is useful for finding occult distant metastases, unknown primary lesions, and synchronous second primary tumors. Imaging studies assess the degree of local invasion, involvement of regional lymph nodes, and presence of distant metastases or second primary malignancies. Common sites of metastatic disease are the lungs, liver, and bone, while the most common sites of second primary malignancies are the head and neck, followed by the lungs and esophagus for squamous cell carcinoma. Distant metastases at initial diagnosis are usually asymptomatic. Screening tests such as chest radiograph, serum alkaline phosphatase, and liver function tests are insensitive to the presence of distant metastases [4-6]. The reported incidence of metastatic disease for squamous cell carcinoma is between 2 and 26 percent and varies based on locoregional control, nodal involvement (number and presence of extracapsular extension), primary site (particularly hypopharynx), histologic grade, and T stage [7-10]. CT scan—Almost all patients presenting with a head and neck cancer will receive a CT during the evaluation period. CT can identify tumors of the head and neck based upon either anatomic distortion or specific tumor enhancement (image 1). In general, tumors enhance more than normal head and neck structures except for mucosa, extraocular muscles, and blood vessels . Slice thickness of 3 mm is generally optimal, while slice thickness greater than 5 mm does not offer sufficient spatial resolution. Images should be reconstructed and viewed in both soft tissue and bone windows. CT technology that reduces metallic artifact is also being incorporated into routine clinical practice . ●Primary site – For cancers of the oral cavity, contrast-enhanced CT can help determine the extent of tumor infiltration into deep tongue musculature and whether or not the mandible is involved. The "puffed cheek" technique improves evaluation of lesions of the oral cavity. This technique requires patients to self-insufflate their oral cavity with air by puffing out their cheek . For other head and neck cancers, CT is particularly useful in upstaging cancers that have deeper local invasion or infiltration into adjacent structures that is difficult to detect on physical examination. In one review of 81 patients with head and neck cancer, CT resulted in a change in assigned clinical stage in 54 percent of cases . This was most likely to occur with hypopharyngeal tumors and least likely with glottic laryngeal tumors (90 and 16 percent, respectively). CT can provide information on invasion of the pre-epiglottic space, laryngeal cartilage, paraglottic space and subglottic extension, and can evaluate retropharyngeal, parapharyngeal, upper mediastinal, and paratracheal nodes. In addition, bone and cartilage invasion, a criterion for stage T4 disease, can be more readily detected. Improved technologies such as dual energy and multispectral CT improves upon the accuracy in assessing cartilage invasion compared with conventional CT [15-17]. ●Regional nodes – Imaging is complementary to the clinical examination for the staging of the neck lymph nodes. CT evaluation of regional lymph nodes primarily relies upon size criteria and lymph node appearance to differentiate involved from uninvolved lymph nodes. CT is also highly sensitive for detection of extracapsular spread of tumor. However, CT is unable to detect extracapsular spread confined within the radiologically defined margin of nodes; borderline-sized, non-necrotic nodes; and microscopic nodal metastasis. Pathologic lymphadenopathy is usually defined radiologically as a node greater than 10 to 11 mm in minimal axial diameter or one that contains central necrosis [18-25]. In a meta-analysis of 647 neck dissections that used nodal size and/or evidence of necrosis to define lymph node involvement, CT was superior to physical examination in terms of sensitivity (83 versus 74 percent), specificity (83 versus 81 percent), accuracy (83 versus 77 percent), and detection of pathologic cervical adenopathy (91 versus 75 percent) . Sensitivity is limited by its inability to detect microscopic nodal metastasis. One study of almost 1000 patients found 67 percent of lymph nodes containing tumor cells were under 10 mm in size and therefore below the size cutoff used to define radiologically detectable lymph node involvement . ●Metastatic sites – A chest CT scan identifies metastases in 4 to 19 percent of patients with newly diagnosed head and neck cancer [6,28-33]. Although chest CT detects distant metastases more frequently than chest radiograph, it fails to detect the 2 to 5 percent of patients who will have distant metastases outside the chest [28,34,35]. There is increased awareness of patient exposure to radiation dose during medical imaging. Although the overall risk of radiation-induced malignancy is small, it is non-negligible when population-based screening is considered. Previous literature estimates that approximately 1.5 to 2 percent of all cancers in the United States may be attributable to radiation from CT studies . The risk estimation of CT radiation to individual patients is a difficult subject, as evolving CT technology continues to help reduce radiation dose and the risk of induced cancer decreases with patient age. The potential risk versus the anticipated benefits of CT scans must be taken into account . PET/CT—PET/CT is often obtained after other imaging modalities when there is an indeterminate finding or need for further evaluation. PET/CT is useful for finding occult distant metastases, unknown primary lesions, and synchronous second primary tumors. It may also alter radiation fields and doses for patients who are not undergoing neck dissection. False negatives of PET may be seen in lymph nodes less than 5 mm, necrotic or cystic lymph nodes, tumors of low metabolic activity, or tumors located at pharyngeal lymphoid tissues with high background physiologic activity. In addition, false positives are common and findings that would change treatment decisions should be confirmed with a biopsy. ●Primary lesions – PET appears to be at least as sensitive and specific as CT and MRI in detecting primary head and neck tumors [39-42]. In one series of 30 patients with newly diagnosed head and neck cancer, PET/CT (98 percent) was better than CT (70 percent) and MRI alone (80 percent) in identifying primary tumor invasion of specific anatomical structures . Findings on PET/CT imaging altered management in 7 of 30 patients (23 percent). In patients presenting with cervical nodal metastases of unknown origin, the sensitivity of PET for detection of primary tumors is approximately 97 percent . ●Synchronous, second primary lesions – Patients with a history of heavy alcohol or tobacco use are at increased risk of a synchronous, second primary malignancy, which would impact treatment decisions. We typically obtain a PET/CT in these patients to evaluate for a synchronous primary. Panendoscopy is an alternative, but is less commonly performed because PET/CT provides more detail and accuracy and is less invasive. This is discussed in more detail separately. (See "Second primary malignancies in patients with head and neck cancers".) Both modalities have benefits in this high-risk patient population [44,45]. Panendoscopy may find synchronous primaries that are too small to be identified with PET/CT, while PET/CT may identify lower aerodigestive tract tumors not seen with panendoscopy. In a retrospective study of 190 patients with unknown primary sites, panendoscopy with directed biopsies and tonsillectomy complemented PET/CT studies especially in those with a negative PET/CT scan . Other studies have found that PET/CT is superior to panendoscopy in finding synchronous tumors in patients with human papillomavirus (HPV)-positive oropharyngeal cancer . Also, synchronous primaries found with panendoscopy correlated highly with PET avid lesions, which supports performing panendoscopy on an individual basis . ●Metastatic disease – In many institutions, PET and integrated PET/CT have greatly replaced other tests for detection of distant metastases [35,39,49-52]. PET/CT is superior to both CT and MRI for detecting regional nodal metastases, as well as distant metastases [53-59]. In a multicenter prospective study, the addition of PET/CT imaging to CT and/or MRI improved the Tumor, Node, Metastasis (TNM) staging of primary cancer and altered the management in 14 percent of patients . ●Clinically N0 neck – In patients with a clinically node-negative (N0) neck, the role of PET/CT is evolving in helping to determine the changes in the surgical plan. In one multicenter study of 212 patients with a clinically N0 neck that used neck pathology as the gold standard, the negative predictive value of PET/CT in patients with T2 to T4 cancers was 87 percent . Moreover, the neck dissection surgical treatment plan was modified in 22 percent of patients based upon the PET/CT findings. Despite high sensitivity with PET/CT, false-positive findings are common, underscoring the need to undertake histologic confirmation of any sites of abnormal uptake. In a series of 349 patients with head and neck tumors who underwent preoperative head/neck CT or MRI and whole body PET/CT, there were 14 second primary tumors (4 percent) and 26 patients with distant metastases (7.4 percent) detected during staging or within 15 months of surgery . PET/CT correctly identified all but one (a patient with CT-demonstrable metastases in the lungs and gluteus muscle). However, there were 23 false-positive PET/CT results. Overall, the sensitivity and specificity were 98 and 93 percent, respectively. MRI—MRI is often complementary to other imaging modalities. We obtain MRI in the following situations: ●Evaluation of perineural spread, skull base invasion, and intracranial extension, especially for nasopharyngeal cancers ●Tumors of the tongue ●Evaluation of cartilage invasion, particularly for non-ossified cartilage that can pose difficulty for CT ●Evaluation of bone marrow involvement MRI may also provide additional benefits compared with CT in the evaluation of the base of tongue and parotid glands. MRI scan is the imaging modality recommended by the National Comprehensive Cancer Network guidelines to evaluate skull base erosion for nasopharyngeal cancer . (See "Epidemiology, etiology, and diagnosis of nasopharyngeal carcinoma", section on 'Initial diagnostic evaluation'.) The most important imaging sequences for head and neck imaging include noncontrast-enhanced T1-weighted images, contrast-enhanced T1-weighted images with fat suppression, and fat-suppressed fluid-sensitive sequences, such as T2-weighted images with fat suppression or short-tau inversion recovery images. Optional diffusion weighted imaging provides additional characterization of lesion cellularity and may improve visualization of some primary tumors and nodal metastasis. Images in axial and coronal plane are the most useful. For general purpose, slice thickness should be no more than 5 mm. Some applications, such as evaluation of skull base and perineural spread, may require thinner slice thickness, typically around 3 mm. Establishing a diagnosis with biopsy—A diagnosis of head and neck cancer is commonly made by either a fine needle aspiration (FNA) of a suspected neck lymph node or a biopsy of the primary lesion [43,64-70]. ●Suspected metastatic disease – If a patient has a suspected site of metastatic disease (eg, lung metastases), this site should be biopsied to confirm the diagnosis and accurately stage the disease. ●No metastases, abnormal neck node present – If an abnormal neck node is present, an FNA can both establish the diagnosis and provide information about nodal staging. As such, it is often used to make an initial tissue diagnosis of a head and neck cancer. •When a patient presents with a neck mass (metastatic cervical lymph node) without an obvious primary mucosal/upper aerodigestive tract site, this can confirm the diagnosis. Additional evaluation is needed to locate the primary tumor. •When the primary is already known or suspected, tissue confirmation of regional spread of disease is important as this changes the management. There is no need to biopsy the primary site if there is already an adequate tissue diagnosis from either regional or metastatic disease. An FNA has high sensitivity, specificity, and diagnostic accuracy that ranges from 89 to 98 percent [64-66]. Nondiagnostic aspirations occur in 5 to 16 percent of cases, most commonly in cystic neck masses, as is common in the presentation of patients with HPV-associated oropharyngeal cancers. If an initial FNA is negative from a suspicious neck node, repeat FNA may be considered before doing an excisional biopsy. (See "Head and neck squamous cell carcinoma of unknown primary".) ●No metastases or abnormal neck nodes present – For patients who do not have a suspicious lymph node to biopsy, tissue sampling of the primary site is indicated. For easily accessible lesions this can be done in the office setting. ●When to performexam under anesthesia – Due to improved radiologic and in-office biopsy techniques, an examination under anesthesia is most often performed only to obtain a tissue diagnosis, for surgical (eg, robotic) planning, and to search for carcinoma of unknown primary [71,72]. This examination is particularly useful for patients with laryngeal and hypopharyngeal malignancies. PATHOLOGY Histology—Multiple different histologies can arise in the head and neck region. A diagnosis requires tissue confirmation and sometimes additional testing on the biopsy specimen to determine additional therapy options (eg, molecular testing for melanoma or human papillomavirus (HPV) testing for squamous cell carcinoma). Squamous cell carcinoma is the most common histology in the head and neck area. Less common histologies include melanoma, nasopharyngeal carcinoma, sarcoma, adenocarcinoma, adenoid cystic carcinoma, and mucoepidermoid carcinomas. Examples of histologies in the head and neck region include: ●Squamous cell carcinoma – Squamous cell carcinomas account for 90 to 95 percent of the lesions in the oral cavity and larynx. They can be categorized as well differentiated (greater than 75 percent keratinization), moderately differentiated (25 to 75 percent keratinization), and poorly differentiated (less than 25 percent keratinization) tumors. Verrucous carcinoma is a less common variant of squamous cell carcinoma. (See "Pathology of head and neck neoplasms", section on 'Squamous cell carcinoma'.) Squamous cell carcinoma of the head and neck often develops through a series of changes from premalignant entities due to carcinogen exposure. Leukoplakia, erythroplakia, and dysplasia can be precursor lesions. (See "Pathology of head and neck neoplasms", section on 'Squamous cell carcinoma precursors'.) •HPV testing – HPV infection is a causative agent for oropharyngeal squamous cell carcinoma, and HPV tumor status is incorporated into the staging system of these tumors. We agree with the approach to HPV evaluation proposed by the College of American Pathologists and endorsed by the American Society of Clinical Oncology [73,74]. HPV tumor status should be determined for all cases of newly diagnosed oropharyngeal squamous cell carcinoma. HPV tumor status is not appropriate for the routine evaluation of nonsquamous carcinoma of the oropharynx or nonoropharyngeal squamous cell carcinoma of the head and neck. It may be useful in select cases of oropharyngeal cancer with uncertain histology. The preferred method for determining HPV tumor status is surrogate marker p16 immunohistochemistry [73,74]. Overexpression of this surrogate marker is strongly associated with transcriptionally active high-risk HPV. The threshold for positivity is at least 70 percent nuclear and cytoplasmic expression with at least moderate to strong intensity. However, there can be false-positive and false-negative results; HPV in situ hybridization or polymerase chain reaction can be used to clarify HPV status when the clinical scenario and p16 results are discordant. Additional information about HPV-associated head and neck cancer is discussed separately. (See "Epidemiology, staging, and clinical presentation of human papillomavirus associated head and neck cancer".) ●Melanoma – Melanoma can originate in the mucosal surfaces. Once a diagnosis of melanoma is confirmed, additional testing should look for targetable mutations such as a KIT mutation. (See "Melanoma: Clinical features and diagnosis", section on 'Diagnosis confirmation'.) ●Nasopharyngeal carcinoma – Nasopharyngeal carcinoma has three separate histopathologic types. These are discussed elsewhere. (See "Epidemiology, etiology, and diagnosis of nasopharyngeal carcinoma", section on 'Histology'.) ●Head and neck sarcoma – Several types of sarcoma can occur in the head and neck including osteosarcoma, chondrosarcoma, rhabdomyosarcoma, and angiosarcoma. (See "Head and neck sarcomas", section on 'Histologic grade and tumor size'.) ●Salivary gland tumors – The most common types of salivary gland tumors include adenoid cystic carcinoma and mucoepidermoid carcinoma. These are discussed elsewhere. (See "Pathology of head and neck neoplasms", section on 'Malignant salivary gland tumors'.) Future molecular methods—Analyzing differences in gene expression patterns across individual patients with a certain type of cancer may reveal molecular differences that permit refinements in their classification, prognostication, and treatment selection. However, this is still an evolving area and further research is needed before this becomes because incorporated into patient care. (See "Head and neck squamous cell carcinogenesis: Molecular and genetic alterations".) As examples: ●Molecular profiling and deoxyribonucleic acid (DNA) genotyping has been shown to predict radiosensitivity and radiotoxicity in patients with head and neck cancer . ●COMT and MATE1 genotyping were found to predict cisplatin-induced ototoxicity in 206 head and neck cancer patients . ●Circulating tumor DNA (ctDNA) is being used to determine driver mutations, which could subsequently direct targeted molecular therapy. The role of ctDNA is also being investigated to detect disease recurrence as part of posttreatment surveillance, which are discussed separately [77-79]. (See "Treatment of human papillomavirus associated oropharyngeal cancer", section on 'Surveillance'.) STAGING— The staging of head and neck cancer is based on a combination of physical exam, imaging, and pathologic findings. There are specific staging classification systems for tumors arising from different anatomic regions of the head and neck. Evaluation of neck nodes—Evaluation of the neck nodes is an essential part of staging patients with head and neck cancer. Depending on clinical factors along with imaging findings some patients may have an elective neck dissection while others may have a sentinel lymph node biopsy (SLNB) to help increase the accuracy of staging. ●Imaging-directed FNA – Fine needle aspiration (FNA) is often helpful in diagnosing malignant lymph nodes in the neck. Several studies have compared ultrasound plus FNA biopsy versus neck CT. In some studies the procedures were comparable while others noted better results with FNA [43,67-69]. In one report of 86 patients with clinically node-negative (N0) necks, ultrasound with FNA detected malignancy in five who did not fulfill radiologic criteria for malignancy by CT scan . Conversely, several enlarged nodes on CT scan were negative by cytology (ie, they potentially represented a false-positive finding). ●Role of SLNB – SLNB is a promising strategy for increasing the accuracy of overall staging of head and neck cancer [80-83]. Further details on the use of SLNB in patients with early-stage oral cavity tumors (including technique, approach, and subsequent management) are discussed separately. (See "Treatment of stage I and II (early) head and neck cancer: The oral cavity", section on 'Sentinel lymph node biopsy'.) When performed at centers with clinical expertise in this approach, SLNB at or before surgery is a reliable and reproducible method for staging the clinically and radiologically N0 neck in patients with early-stage oral cavity cancer. SLNB can also be used to determine the best treatment for contralateral N0 neck in patients with midline malignancies and node-positive, ipsilateral disease. SLNB is technically feasible, reliable (high sensitivity if the three highest intensity nodes are sampled), oncologically safe, and associated with less morbidity than elective neck dissection. This technique is becoming more widely used in the United States and has been integrated into the National Comprehensive Cancer Network treatment guidelines [84,85]. ●Elective neck dissection – Elective neck dissection provides pathologic staging and is often performed in patients with resectable oral cavity tumors. (See "Treatment of stage I and II (early) head and neck cancer: The oral cavity", section on 'Elective neck dissection'.) TNM staging system—The eighth edition of the American Joint Committee on Cancer and the Union for International Cancer Control Tumor, Node, Metastasis (TNM) staging system is used to classify cancers of the head and neck . The T classifications indicate the extent of the primary tumor and are site specific; there is considerable overlap in the cervical N classifications. TNM staging varies depending upon the primary tumor site: ●Oral cavity (table 2) (see "Treatment of stage I and II (early) head and neck cancer: The oral cavity", section on 'Staging') ●Nasopharynx (table 3) (see "Treatment of early and locoregionally advanced nasopharyngeal carcinoma", section on 'Staging') ●Oropharynx (see "Treatment of early (stage I and II) head and neck cancer: The oropharynx", section on 'Staging') •HPV-negative tumors, clinical staging (table 4) •HPV-negative tumors, pathologic staging (table 5) •HPV-positive tumors, clinical staging (table 6) •HPV-positive tumors, pathologic staging (table 7) ●Hypopharynx (see "Treatment of early (stage I and II) head and neck cancer: The hypopharynx", section on 'Anatomy and staging') •Clinical staging (table 8) •Pathologic staging (table 9) ●Larynx (table 10) (see "Treatment of early (stage I and II) head and neck cancer: The larynx", section on 'Staging and anatomy') ●Nasal cavity and paranasal sinuses (table 11) (see "Cancer of the nasal vestibule", section on 'Staging' and "Paranasal sinus cancer", section on 'Diagnosis and staging' and "Tumors of the nasal cavity") ●Salivary glands (see "Salivary gland tumors: Epidemiology, diagnosis, evaluation, and staging", section on 'Staging') •Clinical staging (table 12) •Pathologic staging (table 13) SOCIETY GUIDELINE LINKS— Links to society and government-sponsored guidelines from selected countries and regions around the world are provided separately. (See "Society guideline links: Head and neck cancer".) INFORMATION FOR PATIENTS— UpToDate offers two types of patient education materials, "The Basics" and "Beyond the Basics." The Basics patient education pieces are written in plain language, at the 5 th to 6 th grade reading level, and they answer the four or five key questions a patient might have about a given condition. These articles are best for patients who want a general overview and who prefer short, easy-to-read materials. Beyond the Basics patient education pieces are longer, more sophisticated, and more detailed. These articles are written at the 10 th to 12 th grade reading level and are best for patients who want in-depth information and are comfortable with some medical jargon. Here are the patient education articles that are relevant to this topic. We encourage you to print or e-mail these topics to your patients. (You can also locate patient education articles on a variety of subjects by searching on "patient info" and the keyword(s) of interest.) ●Basics topics (see "Patient education: Mouth sores (The Basics)" and "Patient education: Tongue cancer (The Basics)" and "Patient education: Laryngeal cancer (The Basics)" and "Patient education: Throat cancer (The Basics)") SUMMARY AND RECOMMENDATIONS ●Sites of disease – Head and neck cancer encompasses a variety of histologies, mostly squamous cell carcinomas, which can arise from a variety of sites, such as the oral cavity, pharynx, larynx, nasal cavity and paranasal sinuses, or salivary glands (figure 1A-B). (See 'Anatomy' above.) ●Clinical presentation – For patients with head and neck cancer, initial symptoms and subsequent staging depend upon the site of the primary tumor. Head and neck cancer may be suspected in patients who present with one or more of the following symptoms if they are otherwise unexplained: otalgia, neck mass, voice changes, nasal congestion or epistaxis, odynophagia or dysphagia, hemoptysis or blood in the saliva, mouth or skin ulcers, unilateral tonsil enlargement, palpable lesions of the salivary glands, solitary masses in the thyroid or a change in a pre-existing goiter. Tumors presenting in specific anatomic locations (eg, oral cavity, pharynx, larynx, or sinuses) will have a distinct set of presenting symptoms related to their location. (See 'Clinical presentation' above.) ●Diagnostic evaluation – Proper staging of the primary tumor, regional lymph nodes, and distant metastases is required to develop an optimal treatment plan (algorithm 1). (See 'Diagnostic evaluation' above.) •Physical exam – The initial assessment of the primary tumor includes a combination of inspection, palpation, and flexible laryngoscopy. (See 'Physical exam' above.) •Imaging – Imaging is recommended for all patients with newly suspected or diagnosed head and neck cancer to help determine the extent of both locoregional and metastatic disease. There is debate about the best imaging modality to use. We use the following approach (see 'Imaging studies' above): -CT of the neck with contrast and CT of the chest with contrast are our preferred imaging modality for most patients. -MRI of the neck with contrast is an acceptable alternative to CT of the neck, especially for patients with oral primaries or nasopharyngeal cancer. -Positron emission tomography (PET)/CT is also an acceptable alternative and our preferred modality for patients with bulky disease or risk factors for a second primary (eg, patients with a history of heavy alcohol and tobacco use), and in patients with neck carcinoma of unknown primary. If the diagnosis of head and neck cancer is established, CT and/or MRI of the neck with contrast is often required for radiation or surgical planning. •Biopsy – If a patient has a suspected site of metastatic disease (eg, lung metastases), this site should be biopsied to confirm the diagnosis and accurately stage the disease. For patients without suspected metastases who present with a neck mass (metastatic cervical lymph node) without an obvious primary mucosal/upper aerodigestive tract site, a fine needle aspiration biopsy is frequently used to make an initial tissue diagnosis of a head and neck cancer or to help with staging in patients with a known primary. If no enlarged nodes are present, the primary site should be biopsied. (See 'Establishing a diagnosis with biopsy' above.) ●Pathology – Multiple histologies can arise from the head and neck region. Squamous cell carcinomas account for 90 to 95 percent of the lesions in the oral cavity and larynx. (See 'Pathology' above.) 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Ong TK, Kerawala CJ, Martin IC, Stafford FW. The role of thorax imaging in staging head and neck squamous cell carcinoma. J Craniomaxillofac Surg 1999; 27:339. 33. Keski-Säntti HT, Markkola AT, Mäkitie AA, et al. CT of the chest and abdomen in patients with newly diagnosed head and neck squamous cell carcinoma. Head Neck 2005; 27:909. 34. Brouwer J, de Bree R, Hoekstra OS, et al. Screening for distant metastases in patients with head and neck cancer: is chest computed tomography sufficient? Laryngoscope 2005; 115:1813. 35. Brouwer J, Senft A, de Bree R, et al. Screening for distant metastases in patients with head and neck cancer: is there a role for (18)FDG-PET? Oral Oncol 2006; 42:275. 36. Brenner DJ, Hall EJ. Computed tomography--an increasing source of radiation exposure. N Engl J Med 2007; 357:2277. 37. Zinsser D, Marcus R, Othman AE, et al. Dose Reduction and Dose Management in Computed Tomography - State of the Art. Rofo 2018; 190:531. 38. Albert JM. Radiation risk from CT: implications for cancer screening. AJR Am J Roentgenol 2013; 201:W81. 39. Zanation AM, Sutton DK, Couch ME, et al. Use, accuracy, and implications for patient management of [18F]-2-fluorodeoxyglucose-positron emission/computerized tomography for head and neck tumors. Laryngoscope 2005; 115:1186. 40. Rudmik L, Lau HY, Matthews TW, et al. Clinical utility of PET/CT in the evaluation of head and neck squamous cell carcinoma with an unknown primary: a prospective clinical trial. Head Neck 2011; 33:935. 41. Zhu L, Wang N. 18F-fluorodeoxyglucose positron emission tomography-computed tomography as a diagnostic tool in patients with cervical nodal metastases of unknown primary site: a meta-analysis. Surg Oncol 2013; 22:190. 42. Ha PK, Hdeib A, Goldenberg D, et al. The role of positron emission tomography and computed tomography fusion in the management of early-stage and advanced-stage primary head and neck squamous cell carcinoma. Arch Otolaryngol Head Neck Surg 2006; 132:12. 43. Wong WL, Hussain K, Chevretton E, et al. Validation and clinical application of computer-combined computed tomography and positron emission tomography with 2-[18F]fluoro-2-deoxy-D-glucose head and neck images. Am J Surg 1996; 172:628. 44. Rodriguez-Bruno K, Ali MJ, Wang SJ. Role of panendoscopy to identify synchronous second primary malignancies in patients with oral cavity and oropharyngeal squamous cell carcinoma. Head Neck 2011; 33:949. 45. Haerle SK, Strobel K, Hany TF, et al. (18)F-FDG-PET/CT versus panendoscopy for the detection of synchronous second primary tumors in patients with head and neck squamous cell carcinoma. Head Neck 2010; 32:319. 46. Sokoya M, Chowdhury F, Kadakia S, Ducic Y. Combination of panendoscopy and positron emission tomography/computed tomography increases detection of unknown primary head and neck carcinoma. Laryngoscope 2018; 128:2573. 47. Muller RG, Weidenbecher M, Ludlow D. PET/CT versus triple endoscopy in initial workup of HPV+ oropharyngeal squamous cell carcinoma. Head Neck 2022; 44:1164. 48. Sawaf T, Quereshy HA, Cabrera CI, et al. Incidence of synchronous malignancies found during triple endoscopy in head and neck cancer. Am J Otolaryngol 2022; 43:103349. 49. Schmid DT, Stoeckli SJ, Bandhauer F, et al. Impact of positron emission tomography on the initial staging and therapy in locoregional advanced squamous cell carcinoma of the head and neck. Laryngoscope 2003; 113:888. 50. Kim SY, Roh JL, Yeo NK, et al. Combined 18F-fluorodeoxyglucose-positron emission tomography and computed tomography as a primary screening method for detecting second primary cancers and distant metastases in patients with head and neck cancer. Ann Oncol 2007; 18:1698. 51. Teknos TN, Rosenthal EL, Lee D, et al. Positron emission tomography in the evaluation of stage III and IV head and neck cancer. Head Neck 2001; 23:1056. 52. Basu D, Siegel BA, McDonald DJ, Nussenbaum B. Detection of occult bone metastases from head and neck squamous cell carcinoma: impact of positron emission tomography computed tomography with fluorodeoxyglucose F 18. Arch Otolaryngol Head Neck Surg 2007; 133:801. 53. Johnson JT, Branstetter BF 4th. PET/CT in head and neck oncology: State-of-the-art 2013. Laryngoscope 2014; 124:913. 54. Escott EJ. Role of positron emission tomography/computed tomography (PET/CT) in head and neck cancer. Radiol Clin North Am 2013; 51:881. 55. Xu G, Li J, Zuo X, Li C. Comparison of whole body positron emission tomography (PET)/PET-computed tomography and conventional anatomic imaging for detecting distant malignancies in patients with head and neck cancer: a meta-analysis. Laryngoscope 2012; 122:1974. 56. Lonneux M, Hamoir M, Reychler H, et al. Positron emission tomography with [18F]fluorodeoxyglucose improves staging and patient management in patients with head and neck squamous cell carcinoma: a multicenter prospective study. J Clin Oncol 2010; 28:1190. 57. Schöder H, Yeung HW, Gonen M, et al. Head and neck cancer: clinical usefulness and accuracy of PET/CT image fusion. Radiology 2004; 231:65. 58. Schwartz DL, Ford E, Rajendran J, et al. FDG-PET/CT imaging for preradiotherapy staging of head-and-neck squamous cell carcinoma. Int J Radiat Oncol Biol Phys 2005; 61:129. 59. Fleming AJ Jr, Smith SP Jr, Paul CM, et al. Impact of [18F]-2-fluorodeoxyglucose-positron emission tomography/computed tomography on previously untreated head and neck cancer patients. Laryngoscope 2007; 117:1173. 60. Lowe VJ, Duan F, Subramaniam RM, et al. Multicenter Trial of [18F]fluorodeoxyglucose Positron Emission Tomography/Computed Tomography Staging of Head and Neck Cancer and Negative Predictive Value and Surgical Impact in the N0 Neck: Results From ACRIN 6685. J Clin Oncol 2019; 37:1704. 61. Sakata K, Hareyama M, Tamakawa M, et al. Prognostic factors of nasopharynx tumors investigated by MR imaging and the value of MR imaging in the newly published TNM staging. Int J Radiat Oncol Biol Phys 1999; 43:273. 62. Rasch C, Keus R, Pameijer FA, et al. The potential impact of CT-MRI matching on tumor volume delineation in advanced head and neck cancer. Int J Radiat Oncol Biol Phys 1997; 39:841. 63. Pfister DG, Spencer S, Adelstein D, et al. Head and Neck Cancers, Version 2.2020, NCCN Clinical Practice Guidelines in Oncology. J Natl Compr Canc Netw 2020; 18:873. 64. Feldman PS, Kaplan MJ, Johns ME, Cantrell RW. Fine-needle aspiration in squamous cell carcinoma of the head and neck. Arch Otolaryngol 1983; 109:735. 65. Tandon S, Shahab R, Benton JI, et al. Fine-needle aspiration cytology in a regional head and neck cancer center: comparison with a systematic review and meta-analysis. Head Neck 2008; 30:1246. 66. Pisharodi LR. False-negative diagnosis in fine-needle aspirations of squamous-cell carcinoma of head and neck. Diagn Cytopathol 1997; 17:70. 67. Righi PD, Kopecky KK, Caldemeyer KS, et al. Comparison of ultrasound-fine needle aspiration and computed tomography in patients undergoing elective neck dissection. Head Neck 1997; 19:604. 68. Takes RP, Righi P, Meeuwis CA, et al. The value of ultrasound with ultrasound-guided fine-needle aspiration biopsy compared to computed tomography in the detection of regional metastases in the clinically negative neck. Int J Radiat Oncol Biol Phys 1998; 40:1027. 69. Dammann F, Horger M, Mueller-Berg M, et al. Rational diagnosis of squamous cell carcinoma of the head and neck region: comparative evaluation of CT, MRI, and 18FDG PET. AJR Am J Roentgenol 2005; 184:1326. 70. Atula TS, Varpula MJ, Kurki TJ, et al. Assessment of cervical lymph node status in head and neck cancer patients: palpation, computed tomography and low field magnetic resonance imaging compared with ultrasound-guided fine-needle aspiration cytology. Eur J Radiol 1997; 25:152. 71. Noor A, Stepan L, Kao SS, et al. Reviewing indications for panendoscopy in the investigation of head and neck squamous cell carcinoma. J Laryngol Otol 2018; 132:901. 72. Golusinski P, Di Maio P, Pehlivan B, et al. Evidence for the approach to the diagnostic evaluation of squamous cell carcinoma occult primary tumors of the head and neck. Oral Oncol 2019; 88:145. 73. Fakhry C, Lacchetti C, Rooper LM, et al. Human Papillomavirus Testing in Head and Neck Carcinomas: ASCO Clinical Practice Guideline Endorsement of the College of American Pathologists Guideline. J Clin Oncol 2018; 36:3152. 74. Lewis JS Jr, Beadle B, Bishop JA, et al. Human Papillomavirus Testing in Head and Neck Carcinomas: Guideline From the College of American Pathologists. Arch Pathol Lab Med 2018; 142:559. 75. Drobin K, Marczyk M, Halle M, et al. Molecular Profiling for Predictors of Radiosensitivity in Patients with Breast or Head-and-Neck Cancer. Cancers (Basel) 2020; 12. 76. Teft WA, Winquist E, Nichols AC, et al. Predictors of cisplatin-induced ototoxicity and survival in chemoradiation treated head and neck cancer patients. Oral Oncol 2019; 89:72. 77. Egyud M, Sridhar P, Devaiah A, et al. Plasma circulating tumor DNA as a potential tool for disease monitoring in head and neck cancer. Head Neck 2019; 41:1351. 78. Payne K, Spruce R, Beggs A, et al. Circulating tumor DNA as a biomarker and liquid biopsy in head and neck squamous cell carcinoma. Head Neck 2018; 40:1598. 79. Chera BS, Kumar S, Shen C, et al. Plasma Circulating Tumor HPV DNA for the Surveillance of Cancer Recurrence in HPV-Associated Oropharyngeal Cancer. J Clin Oncol 2020; 38:1050. 80. Alkureishi LW, Ross GL, Shoaib T, et al. Sentinel node biopsy in head and neck squamous cell cancer: 5-year follow-up of a European multicenter trial. Ann Surg Oncol 2010; 17:2459. 81. Barzan L, Sulfaro S, Alberti F, et al. An extended use of the sentinel node in head and neck squamous cell carcinoma: results of a prospective study of 100 patients. Acta Otorhinolaryngol Ital 2004; 24:145. 82. Stoeckli SJ, Broglie MA. Sentinel node biopsy for early oral carcinoma. Curr Opin Otolaryngol Head Neck Surg 2012; 20:103. 83. Thompson CF, St John MA, Lawson G, et al. Diagnostic value of sentinel lymph node biopsy in head and neck cancer: a meta-analysis. Eur Arch Otorhinolaryngol 2013; 270:2115. 84. Monroe MM, Lai SY. Sentinel lymph node biopsy for oral cancer: supporting evidence and recent novel developments. Curr Oncol Rep 2014; 16:385. 85. National Comprehensive Cancer Network Guidelines for Head and Neck Cancer (Accessed on March 17, 2021). 86. Part II Head and Neck. In: AJCC Cancer Staging Manual, 8th ed, Amid MB (Ed), Springer, New York 2017. p.53, corrected at 4th printing, 2018. Disclaimer: This generalized information is a limited summary of diagnosis, treatment, and/or medication information. 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Topic 3393 Version 49.0 Topic Feedback Algorithms Evaluation of adults with suspected head and neck cancer Evaluation of adults with suspected head and neck cancer Tables Clinical characteristics of HPV associated versus non-HPV associated squamous cell carcinoma of the head and neck (HNSCC)Cancer of the mucosal lip and oral cavity TNM staging AJCC UICC 8th editionNasopharyngeal cancer TNM staging AJCC UICC 8th editionOropharyngeal (p16 negative) cancer TNM clinical staging AJCC UICC 8th editionOropharyngeal (p16 negative) cancer TNM pathologic staging AJCC UICC 8th editionHPV-related oropharyngeal carcinoma TNM clinical staging AJCC UICC 8th editionHPV related oropharyngeal carcinoma TNM pathologic staging AJCC UICC 8th editionHypopharyngeal cancer TNM clinical staging AJCC UICC 8th editionHypopharyngeal cancer TNM pathologic staging AJCC UICC 8th editionCancer of the larynx TNM staging AJCC UICC 8th editionCancer of the nasal cavity and paranasal sinuses TNM staging AJCC UICC 8th editionMajor salivary gland tumors TNM clinical staging AJCC UICC 8th editionMajor salivary gland tumors TNM pathologic staging AJCC UICC 8th edition Clinical characteristics of HPV associated versus non-HPV associated squamous cell carcinoma of the head and neck (HNSCC)Cancer of the mucosal lip and oral cavity TNM staging AJCC UICC 8th editionNasopharyngeal cancer TNM staging AJCC UICC 8th editionOropharyngeal (p16 negative) cancer TNM clinical staging AJCC UICC 8th editionOropharyngeal (p16 negative) cancer TNM pathologic staging AJCC UICC 8th editionHPV-related oropharyngeal carcinoma TNM clinical staging AJCC UICC 8th editionHPV related oropharyngeal carcinoma TNM pathologic staging AJCC UICC 8th editionHypopharyngeal cancer TNM clinical staging AJCC UICC 8th editionHypopharyngeal cancer TNM pathologic staging AJCC UICC 8th editionCancer of the larynx TNM staging AJCC UICC 8th editionCancer of the nasal cavity and paranasal sinuses TNM staging AJCC UICC 8th editionMajor salivary gland tumors TNM clinical staging AJCC UICC 8th editionMajor salivary gland tumors TNM pathologic staging AJCC UICC 8th edition Figures Anatomy of the head and neckAnatomy of the salivary glands and ductsLymph node levels of the neck Anatomy of the head and neckAnatomy of the salivary glands and ductsLymph node levels of the neck Pictures Tonsil cancer Tonsil cancer Diagnostic Images Carcinoma of the tongue by CT scan Carcinoma of the tongue by CT scan Company About Us Editorial Policy Testimonials Wolters Kluwer Careers Support Contact Us Help & Training Citing Our Content News & Events What's New Clinical Podcasts Press Announcements In the News Events Resources UpToDate Sign-in CME/CE/CPD Mobile Apps Webinars EHR Integration Health Industry Podcasts Follow Us Sign up today to receive the latest news and updates from UpToDate. 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17113	https://www.theoremoftheday.org/GeometryAndTrigonometry/DescartesCircle/TotDDescartesCircle.pdf	THEOREM OF THE DAY The Descartes Circle Theorem If four circles forming a Descartes conﬁguration have respective curva-tures b1, b2, b3 and b4, then b2 1 + b2 2 + b2 3 + b2 4 = 1 2 b1 + b2 + b3 + b4 2. For pairs of lips to kiss maybe Involves no trigonometry. ’Tis not so when four circles kiss Each one the other three. To bring this oﬀthe four must be As three in one or one in three. If one in three, beyond a doubt Each gets three kisses from without. If three in one, then is that one Thrice kissed internally. Four circles to the kissing come. The smaller are the benter. The bend is just the inverse of The distance from the centre. Though their intrigue left Euclid dumb There’s now no need for rule of thumb. Since zero bend’s a dead straight line And concave bends have minus sign, The sum of the squares of all four bends Is half the square of their sum. To spy out spherical aﬀairs An oscular surveyor Might ﬁnd the task laborious, The sphere is much the gayer, And now besides the pair of pairs A ﬁfth sphere in the kissing shares. Yet, signs and zero as before, For each to kiss the other four The square of the sum of all ﬁve bends Is thrice the sum of their squares. “The Kiss Precise” Frederick Soddy Nature, 137, 1936. A Descartes conﬁguration of circles consists of four mutually tangent circles in the plane. Descartes was con-cerned with conﬁgurations such as that at A in the above display, with bend, or curvature, deﬁned as the reciprocal of radius. For example, the circles at A might have radii 1/4, 1/12, 1/13 and 1/61. Then (b1 + b2 + b3 + b4)2/2 = (4 + 12 + 13 + 61)2/2 = 8100/2 = 4050 and 16 + 144 + 169 + 3721 = 4050 also, conforming to Descartes’ theorem. If we adopt the convention that negative curvature corresponds to the interior of a circle being outside it, then conﬁguration B is equally valid. For example the circles at B might have radii −1/7 (the outside circle) and 1/12, 1/17 and 1/24 (the inside circles). Then (−7 + 12 + 17 + 24)2/2 = 1058 = 49 + 144 + 289 + 576 as required. If horizontal lines are taken as circles of inﬁnite radius—“zero bend’s a dead straight line”—then the conﬁgurations at C and D are again valid. But conﬁguration E is not valid. As the third stanza of Soddy’s poem celebrates, conﬁgurations of ﬁve spheres in R3 can also be made to work, the fraction in the theorem becoming 1/3; with the fraction 1/n, we can go to n dimensions; and adding curvature of space into the equation, even non-Euclidean Descartes Circle Theorems obtain! Although he did not prove it completely, this result appears originally in a letter sent by Ren´ e Descartes to Princess Elisabeth of Bohemia in 1643. There have been many rediscoveries with complete proofs, Jakob Steiner’s, in 1826, being perhaps the ﬁrst. The Rn version is due to Thorold Gosset in 1937, with n = 3 being given by Robert Lachlan in 1886. Web link: arxiv.org/abs/math/0101066. Additions by Gosset and Fred Lunnon to Soddy’s poem, which is reproduced courtesy of Nature Publishing Group, can be found here: pballew.blogspot.com/2014/10/the-kiss-precise-soddys-circle-theorem.html. Further reading: Introduction to Circle Packing: The Theory of Discrete Analytic Functions, by Kenneth Stephenson, CUP, 2005. Created by Robin Whitty for www.theoremoftheday.org
17114	http://sites.science.oregonstate.edu/~walshke/COURSES/ph201/lecture/Vector%20Worksheet%202.pdf	Vector Worksheet Much of the physical world can be described in terms of numbers. Examples of this are the mass of an object, its temperature and its volume. These are called scalar quantities. But some quantities also need a direction to fully describe it. What if you wanted directions to the nearest physics lab? The answer ”Go three miles” is not very helpful. More useful would be ”Go three miles south.” A quantity having both size (magnitude) and direction is called a vector quantity. Vectors are not just about position. Such quantities as velocity, acceleration, force and momentum all have both a size and a direction. For example, what we know as ”speed” is just the magnitude of the velocity vector. To fully describe velocity, we must have both the speed and the direction in which we are going. We represent vectors geometrically with an arrow. This arrow will have a length that represents magnitude and point in a direction. This worksheet will walk you through some basic vector operations. In your textbooks, you will see vectors denoted in boldface (v), but when writing a vector, we denote it by writing an arrow above the letter (⃗ v). We can also, and will do here, use this notation: ⃗ vAB to denote a vector that has its tail at a point A and the tip of its arrow at another point B. In this case we say ”The vector from A to B.” Vector Facts There are many diﬀerent types of vector quantities, but they all have similar properties. To explore these properties, let’s look at one type of vector quantity, position vectors. Suppose you are trying to get from Valsetz Dining Hall to the Natural Sciences Building.On the map below, you begin at Valsetz (point A) and you walk to Monmouth Ave., due east, about 400 feet (point B). The vector ⃗ vAB describes this displacement by starting where you started and ending at the place you will turn. It points in the direction you will go, and has a length (magnitude) that represents the 400 feet you will walk. When you reach point B, you turn and walk down Monmouth Avenue about 1500 feet, and end up at point C, in front of the Natural Sciences Building where you will have your favorite class, physics. Vector ⃗ vBC represents this displacement in the same way as ⃗ vAB, but with the new direction and magnitude. If you were a bird and you wanted to ﬂy from Valsetz to NSS, you would probably ﬂy along vector ⃗ vAC, which describes your ﬂight path, stating point, ending point, direction, and magnitude in much the same way. 1 Answers to the questions can be written on a seperate sheet of paper. A B C As you can see, vectors represent some very important information and, once you get used to using them, will be instrumental in helping you to visualize and solve most physics problems. Vector Addition and Subtraction Back to the map above. You start out at point, A, and move to point B. Then you turn and go to point C. If you were the bird, you could have accomplished the same net direction by ﬂying directly from A to C. The resulting displacement vector ⃗ vAC is the sum of ⃗ vAB and ⃗ vBC and is written ⃗ vAC = ⃗ vAB + ⃗ vBC. This is illustrated in Figure 2. Figure two is useful when we are presented with vectors that are placed with the tip of one 2 vector at the tail of another. This placement is common when we are discussing position, but more often we want to add vectors that look more like this: What do we do then? Well, we can move vectors, as long as we don’t change their magnitude or direction so that the tip of one is at the tail of the other, like this: Now we can draw the vector that is ⃗ VAB + ⃗ VCD: 3 We can also subtract vectors. This is not very diﬀerent from vector addition. Really, if we have two vectors, ⃗ vA and ⃗ vB, and we want to know ⃗ vA −⃗ vB, all we do is add to ⃗ vA a vector that is in the direction exactly opposite of ⃗ vB. Thus, ⃗ vA −⃗ vB = ⃗ vA + (−⃗ vB). Figure 3 illustrates this point. In subtracting vectors, like adding them, if we are presented with vectors that are not placed ”tip to tail”, we can move them so they are, and then perform our calculations. Vector Components Drawing pictures of vectors is great to get a sense of what we have been learning conceptu-ally, but to obtain accurate results, we have to use some math, speciﬁcally trigonometry and geometry. Going back to our map above, we can place a coordinate system over the map, with the origin at Valsetz, and see that ⃗ vAB represents motion along the x-axis, and ⃗ vBC represents displacement along the y-axis. The resultant vector ⃗ vC has components in both the x and y directions. Thus, if ⃗ vx and ⃗ vy are the x and y components of ⃗ v, then ⃗ v = ⃗ vx + ⃗ vy. ⃗ vx and ⃗ vy are called component vectors. The x-component vector is the projection of ⃗ v along the x-axis, and the y-component vector is the projection of ⃗ v along the y-axis. To visualize a projection, imagine a ﬂashlight on the vector pointing from top to bottom will leave a shadow, or projection, on the x-axis. Figure 4 will be of use to shed some light on this idea. 4 Here’s how you determine the components of a vector, ⃗ v: 1. The absolute value \|vx\| of the x-component of ⃗ v is the magnitude of ⃗ vx. 2. The sign of \|vx\| is positive if ⃗ vx points in the positive x direction, and negative if ⃗ vx points in the negative x direction. 3. The y-component ⃗ vy is determined in the same way. Let’s practice drawing some vectors. One the axes below, start at the origin and draw the given component vectors and the resultant vector. Each line is one unit. (a) \|vx\| = 2, \|vy\| = 3 (b) \|vx\| = −4, \|vy\| = 4 (c) \|vx\| = −3, \|vy\| = −5 Now , lets talk about how one ﬁnds the exact numerical values for the magnitudes of vectors. Consider the vector in ﬁgure 5. Lets place our ⃗ Ax component vector (see ﬁgure 5) along the x-axis. Lets also place our ⃗ Ay component vector so that it runs from the x-axis to the tip of ⃗ A. Why did we do this? We did this because this placement of the component vectors makes the computations for ﬁnding the length of them (and the length of ⃗ A) much easier. Notice also that ⃗ A has an angle of θ above the x-axis. Trigonometry gives us the lengths of our component vectors, and since our component vectors form a right triangle with ⃗ A as its hypotenuse, the Pythagorean Theorem can be used to ﬁnd the length of ⃗ A. An example may help. 5 Example: Find the length of the x and y components of vector ⃗ R if ⃗ R is at an angle of 30◦above the x-axis and has a magnitude of 25 cm. Solution: \|Rx\| = R cos θ = 25 cos(30) = 21.65 cm \|Ry\| = R sin θ = 25 sin(30) = 12.5 cm Is this solution correct? We can check it using the Pythagorean Theorem: ⃗ R = q \|R2 x + R2 y\|, so √ 21.652 + 12.52 = 25 cm. Check! Going the other way, we can also determine the length and angle of a vector from its x and y components. As in the check of the example above, we can use Pythagorean Theorem to ﬁnd the length of a vector. To ﬁnd the angle, we can use the fact that the tangent of angle θ is the ratio of the opposite side to the adjacent side (SOHCAHTOA, remember?), so θ = tan−1 \|Ry\| \|Rx\| . Using components, we can now add (and subtract) vectors exactly. Thus, if ⃗ R = ⃗ A + ⃗ B, then \|Rx\| = \|Ax + Bx\| and \|Ry\| = \|Ay + By\|. See ﬁgure 6. Here’s an example: Example: Find ⃗ R = ⃗ A + ⃗ B, given that ⃗ A has magnitude 4 and is at an angle of 30◦above the positive x-axis and ⃗ B has magnitude 2 and is at an angle of 55◦above the positive x-axis. Solution: First ﬁnd the x and y components of ⃗ A and ⃗ B. \|Ax\| = A cos θA = 4 cos(30) = 3.46 \|Ay\| = A sin θA = 4 sin(30) = 2 \|Bx\| = B cos θB = 2 cos(55) = 1.15 \|By\| = B sin θB = 2 sin(55) = 1.64 6 Now add them to ﬁnd the x and y components of ⃗ R. \|Rx\| = \|Ax + Bx\| = 4.61 \|Ry\| = \|Ay + By\| = 3.64 Use the Pythagorean Theorem to ﬁnd the length of ⃗ R. \|R\| = q \|R2 x + R2 y\| = p (4.61)2 + (3.64)2 = 5.87 Now use tan to ﬁnd the measure of ⃗ R above the x-axis. tan θ = \|Ry\| \|Rx\| = 3.64 4.61 = 0.790. tan−1(0.790) = 38.3◦. Thus, ⃗ R has a magnitude 5.87 and is at an angle of 38.3◦above the positive x-axis. So far, we have only been measuring our angle θ from the positive x-axis. However, using the laws of Trigonometry, we can easily measure θ from any axis we choose. This is simply a matter of keeping track of the component vectors and where they are in relation to the angle of measurement. Lets say, for example, that you are on a jet plane that takes oﬀ from PDX. Its velocity is 550 miles per hour due west. There is a wind blowing with a velocity of 150 miles per hour from the south. A) Use vector addition to diagram the two vectors and calculate the resultant vector. B) What is the direction of the jet’s velocity vector measured west of north? The ﬁrst step in solving any physics problem is to draw a diagram including all of the relevant information. So lets do that: This diagram gives us a clear picture of the information we have been given in the problem and the information we are looking for. To solve part (A), we use what we know about adding vector components. \|V \| = q \|V 2 x \| + \|V 2 y \| = p (−550)2 + 1502 = 570m/h To solve part (B), we need to be careful. Notice that our angle is being measured from the 7 y-axis toward the negative x-axis. This means that our x component is the side opposite θ and the y component is adjacent to θ. Thus, our calculation for the angle measured west of north looks like this: tan θ = Vx Vy Thus, θ = tan−1 Vx = 74.74◦ Vy So we can conclude that the jet is moving at 570 m/h at an angle of 74.74◦ west of north. Let’s tie all of these ideas up in a nice, fun physics problem. You may be familiar with Newton’s Second Law of Motion, Σ⃗ F = m⃗ a. In English, this equation says: The sum, or net, (Σ) of the forces (⃗ F) acting upon an object equals (=) the mass (m) of the object multiplied by the object’s acceleration (⃗ a). Even if you are familiar with this famous equation, did you notice the arrows above F and a before? If so, you may have wondered what they mean. They are vectors! This means that they have both a magnitude and a direction and can be expressed in terms of their components. These arrows are a ﬂashing red light that alerts you to the fact that these quantities imply two equations; one in the x direction and one in the y direction. We’ll work through a problem together: A pack of four Arctic wolves are exerting four diﬀerent forces upon the carcass of a 500 kg dead polar bear. Wolf one is pulling on the bear with a force of 20 Newtons (N) due north. Wolf two is pulling on the bear with a force of 30 N, 30 degrees north of west. Wolf three is pulling on the bear with a force of 25 N due west, and wolf four is pulling on the bear with a force of 35 N, 10 degrees west of south. Determine the net force acting upon the polar bear and the direction of that force measured north of west. Then compute the magnitude of the acceleration of the polar bear. Wow. This much information can be scary at ﬁrst but we are just going to use the techniques we have learned above and work our way through. Take a deep breath. You will be ﬁne. Your ﬁrst step should always be to draw a diagram of the situation. Do so here: (Hint: Draw this as though you are looking down on it from directly above with the bear at the origin of your coordinate system, like a two-dimensional tug-o’-war. 8 We are trying to ﬁnd the length (magnitude) of the total force vector and its direction. To ﬁnd the length, we will need to ﬁnd the x and y components of each of the forces being exerted on the bear. Lets take it a vector at a time. Wolf One \|Fx\|= \|Fy\|= Wolf Two \|Fx\|= \|Fy\|= Wolf Three \|Fx\|= \|Fy\|= Wolf Four \|Fx\|= \|Fy\|= Now add up all of the x components: (Careful of minus signs) Add up all of the y components: (Careful of minus signs) Now use Pythagorean Theorem to ﬁnd the magnitude of the total force vector: 9 What direction is the total force in? Now ﬁnd the magnitude of the acceleration by using the equation Σ\|F\| = m\|a\| and what you know about solving for stuﬀ. A word on notation: So far we have been expressing vectors and their components as ⃗ v = ⃗ vx + ⃗ vy. This notation is acceptable but you may run into something that looks like this: ⃗ v = ⟨vx, vy⟩. This notation is more formal, but accomplishes the same thing. In this new notation, the wolf problem above would look like this: Σ⃗ F = ⟨ΣFx, ΣFy⟩ = ⟨F1x + F2x + F3x + F4x, F1y + F2y + F3y + F4y⟩ In place of all of the Fix and Fiy ’s, you would put the numbers you found in the problem and calculate the same way. 10 Now that you have the idea, here are some practice problems to cement the concepts in your mind. Practice Problems (a) (b) (c) 1. Determine the x and y components of each of the force vectors above. 11 2. Are the following quantities vectors or scalars? Explain. (a) The cost of a theater ticket. (b) The current in a river. (c) The initial ﬂight path from Houston to Dallas. (d) The population of the world. 3. Of the following quantities, which are vectors? If it is, explain why. If it isn’t, explain why. (a) The speed of a car. (b) The velocity of a car. (c) How long it takes to do a physics assignment. (d) The magnitude of the force required to push a pumpkin oﬀan abandoned overpass. (e) The acceleration of said pumpkin. 12 4. Write each combination of vectors as a single vector. See ﬁgure below. (a) − → PQ + − → QR (b) − → RP + − → PS (c) − → QS −− → PS (d) − → RS + − → SP + − → PQ 5. If a child pulls a sled through the snow with a force of 50 N exerted at an angle of 38◦above the horizontal, ﬁnd the horizontal and vertical components of the force. 13 6. A plane ﬂies with a velocity of 52 m/s east through a 12 m/s cross wind blowing the plane south. Find the magnitude and direction (relative to due north) of the resultant velocity at which it travels. 7. Draw the vector ⃗ v and ﬁnd the magnitude and direction, counter clockwise from the +x-axis. (a) vx=2.5, vy=1 (b) vx=-0.5, vy=1 (c) vx=-1.5, vy=-1 14 8. Construct c=a+b by drawing and calculating the direction and magnitude of c. The direction should be measured from the +x-axis. (a) (b) 9. Construct c=a-b by drawing and calculating the direction and magnitude of c. The direction should be measured from the +x-axis. (a) (b) 15 10. A racing car is accelerating at 20 m/s2 70◦N of W. Find the acceleration of the car in the north direction and in the west direction. 11. Your uncle Mike’s boat can travel 4.0 m/s in still water. One sunny afternoon, you and Unk Mike decide to go ﬁshing. While waiting for a bite, you begin thinking, ”If this river is ﬂowing at 5.5 m/s southward, and we are heading eastward, directly across the river, what are the direction and magnitude of our total velocity?” Answer your own question. 16 12. The ﬁgure above shows three ropes tied together in a knot. One of your friends pulls on a rope with a force of 3 Newtons and another pulls on a second rope with a force of 5 Newtons. How hard and in what direction must you pull on the third rope to keep the knot from moving? (Hint: If you don’t want the knot to move, the net force must be zero.) 17 References: Knight, Randall D., Physics: A Strategic Approach, Volume one, Pearson Education, 2004. Stewart, James, Calculus: Concepts and Contexts, Third Edition, Thompson Brooks/Cole, 2005. The Physics Hypertextbook, Vector Addition and Subtraction, Worksheet created by Laura K. Waight 18
17115	https://www.learnalberta.ca/content/sep20u/html/java/vector_addition_numerical/applethelp/lesson_1.html	Lesson 1 - Vector Addition: Numerical Vector Addition: Numerical calculates the magnitude and direction of the resultant given the magnitudes and directions of an arbitrary number of vectors to be added. Prerequisites Students should understand the vector properties of magnitude and direction and be familiar with adding vectors graphically by the Tip-to-Tail method. They should also have a working knowledge of basic trigonometry. Learning Outcomes Students will learn to calculate the magnitude and direction of the sum of two vectors (resultant) given the magnitudes and directions of the two vectors to be added. They will learn to use the Law of Cosines and the Law of Sines for this purpose. They will also learn how to calculate the sum of two vectors using components. Instructions Students should understand the applet functions that are described in Help and ShowMe. The applet should be open. The step-by-step instructions on this page are to be done in the applet. You may need to toggle back and forth between instructions and applet if your screen space is limited. Contents Describing the Sum of Two Vectors Method 1 - Calculating the Resultant using the Law of Cosines and Sines Method 2 - Calculating the Resultant using Vector Components The Law of Cosines The Law of Sines Components 1. Describing the Sum of Two Vectors Imagine you are re-entering the atmosphere in the space shuttle. In order to land the shuttle safely, the pilot must make an exact approach to the runway. In order to do this, the pilot needs to know the shuttle's exact velocity relative to the ground at all times. On re-entry, the shuttle is traveling at 130 m/s at 140° relative to the air when it enters the jet stream (high altitude, global air circulation) which moves at 100 m/s at 60° relative to the ground. What is the shuttle's velocity relative to the ground after it enters the jet stream? You may be tempted to say 100 m/s + 130 m/s = 230 m/s. This however, would be incorrect. Velocity is a vector quantity and as such the two velocities must be added as vectors. The applet will be used to illustrate how two vectors are correctly added. The applet will be used to determine the shuttle's velocity relative to the ground at re-entry. \| \| \| 1. If the applet window is not empty, clear it by clicking "Reset" (). 2. Enter the first vector, 1, with a magnitude of 100 at an angle of 60° in the Polar (positive) mode (). 3. Enter the second vector, 2, with a magnitude of 130 at an angle of 140° in the Polar (positive) mode () and drag both vectors to the centre of the screen. 4. After entering the two vectors, click the radio button labeled "Show Resultant" to display the resultant (sum of the vectors). If you use the Polar (positive) mode () for the resultant, you should see a display similar to Figure 1. Figure 1 According to the applet, the shuttle's velocity relative to the ground at re-entry would be 177.24 m/s at 106.25° in the polar positive mode. \| Circle the correct answer below. The graphical construction of the resultant in Figure 1 illustrates the Tip-to-Tail Method of vector addition Parallelogram Method of vector addition The resultant from Exercise 1 may be displayed in three modes. Click the mode control button () for the resultant, , three times and cycle to the "Cartesian" mode (). In this mode, the resultant's x- and y- components (rx,ry) are displayed, as shown in Figure 2. \| \| \| \| --- \| Button \| Mode \| Example \| \| \| Magnitude and polar positive direction (r,θ) \| \| \| \| Magnitude and navigational direction (r,θ E of N ) \| \| \| \| Cartesian Components (rx,ry) \| \| Using the mode control button, describe the shuttle's velocity relative to the ground using all three modes. The magnitude and polar (positive) direction of the resultant () is (________ , ________ deg) The magnitude and navigational direction of the resultant () is (________ , ________ E of N ) The Cartesian components of the resultant () are (________ , ________) How is the resultant being calculated? It can be found using two different methods. Method 1 uses the Law of Cosines to find the resultant magnitude and the Law of Sines for the resultant direction. Method 2 uses vector components to derive both the resultant magnitude and direction. Each method is presented below. 2. Method 1 - Calculating the Resultant using the Law of Cosines and Sines \| \| \| To calculate the magnitude and direction of the resultant as displayed in Figure 1, we need a diagram showing all relevant quantities as in Figure 2 below. Figure 2 We want to calculate the magnitude r of the resultant and the angle α. The direction angle θ of the resultant in the Polar (positive) specification is then θ = α + 60°. The Law of Cosines is used to calculate the magnitude (r) and the Law of Sines is used to calculate the angle (α). For a description of these laws, see the appendix. \| \| According to Figure 2, the Law of Cosines can be used to calculate the magnitude (r) of the resultant vector: (Note: the angle opposite to vector is equal to 60° + 40° = 100°.) \| \| The Law of Sines can then be used to calculate the direction (θ) of the resultant vector. To apply the Law of Sines, pair the angle (α) with the opposite side of magnitude (v2) and the 100° angle with the opposite side of magnitude (r). Therefore, the resultant vector has a magnitude of 177.24 at an angle of 106.25° in the polar (positive) direction: \| Using the Law of Cosine and Sines, calculate the resultant (sum) of the following two vectors. Add the vectors on the applet in order to view the correct Tip-to-Tail vector diagram and verify the resultant. 1 = 150, 50° polar (positive) 2 = 200, 150° polar (positive) \| \| \| \| --- \| Tip-to-Tail vector diagram: \| Resultant magnitude (r): (Use the Law of Cosine.) \| Resultant direction (θ): (Use the Law of Sines and polar (positive) specification.) \| Using the Law of Cosine and Sines, calculate the resultant (sum) of the following two vectors. Add the vectors on the applet in order to view the correct Tip-to-Tail vector diagram and verify the resultant. 1 = 100, 150° polar (positive) 2 = 75, 250° polar (positive) \| \| \| \| --- \| Tip-to-Tail vector diagram: \| Resultant magnitude (r): (Use the Law of Cosine.) \| Resultant direction (θ): (Use the Law of Sines and polar (positive) specification.) \| 3. Method 2 - Calculating the Resultant using Vector Components \| \| \| We will now calculate the sum of the same two vectors 1 and 2 as in the preceding section, but this time using components. For the following calculations, you will need to know the (scalar) components of a vector. For an overview of vector Components, see the appendix. It is particularly easy if the vectors are already given in terms of their (x, y) components, (vx, vy)1 and (vx, vy)2. However, we will assume the vectors are given in terms of magnitude and direction [(v1, θ1) and (v2, θ2)]. Angles are measured in the polar (positive) specification (or navigation N of E). See Figure 3. Figure 3 The magnitude and direction of 1 and 2 are: v1 = 100, θ1 = 60° v2 = 130, θ2 = 140° \| \| The components of each vector are calculated using the appropriate trigonometric functions: \| \| \| \| --- \| Vector Diagram \| x-component \| y-component \| \| \| \| \| \| \| \| \| To add the two vectors is to add the respective components. If the components of the resultant are denoted (rx,ry), we get: \| \| \| rx = v1x + v2xrx = (+50.00) + (-99.59) rx = -49.59 ry = v1y + v2y ry = (+86.60) + (+83.56) ry = +170.16 \| \| \| You could specify the resultant in terms of the components and stop the calculation at this point. However, if the magnitude and direction angle of the resultant are required, these can be calculated from the components as follows. \| \| \| Again, it is very helpful to make a diagram to illustrate the angles and other relevant quantities involved. See Figure 4 below. Figure 4 \| \| The triangle formed by the vector and the two components is a right-angle triangle, with the vector as the hypotenuse. Therefore, the Pythagorean theorem can be used to calculate the resultant magnitude: \| \| From Figure 4 we can see that the direction can be calculated using the definition of the tangent of an angle. This implies for the direction angle of the value 180 - 73.75 = 106.24° Therefore, the resultant vector is 177.24 at 106.25° in the polar (positive) direction. You can verify these values on the applet by selecting the magnitude and polar positive direction for the resultant: \| In summary, if two vectors 1 and 2 are given in terms of magnitude and direction, a resultant can be calculated by doing the following: Use a vector diagram and trigonometric functions to convert the vectors to component form. Add the components (xtotal = x1 + x2) and (ytotal = y1 + y2). Remember to include positive or negative directions. Draw the resultant vector using the xtotal component and the ytotal component. Remember to include positive or negative directions. Calculate the resultant magnitude using the Pythagorean theorem (c2 = a2 + b2). Calculate the direction using the appropriate trigonometric function (tangent function). This may be a bit more work than to calculate the resultant using the Law of Cosines and the Law of Sines as was done in the preceding section. However, if the two vectors 1 and 2 are already given in component form and if one wants the resultant in component form as well, as will often be the case, the calculation is simpler. Using the component method, calculate the resultant (sum) of the following two vectors. Show all required calculations and diagrams below and identify the direction using the polar (positive) specification. Add the vectors on the applet in order to verify the resultant magnitude and direction. 1 = 175, 70° polar (positive) 2 = 200, 200° polar (positive) \| \| \| --- \| \| a. Vector diagram and calculation of the components for 1. \| b. Vector diagram and calculation of the components for 2. \| c. Addition of the components and drawing of the resultant vector. d. Calculation of the resultant magnitude using the Pythagorean theorem. e. Calculation of the resultant direction using the tangent function. Express the direction in terms of the polar (positive) specification. Using the component method, calculate the resultant (sum) of the following two vectors. Show all required calculations and diagrams, and identify the direction using the polar (positive) specification. Add the vectors on the applet in order to verify the resultant magnitude and direction. 1 = 185, 45° polar (positive) 2 = 95, 320° polar (positive) \| \| \| --- \| \| a. Vector diagram and calculation of the components for 1. \| b. Vector diagram and calculation of the components for 2. \| c. Addition of the components and drawing of the resultant vector. d. Calculation of the resultant magnitude using the Pythagorean theorem. e. Calculation of the resultant direction using the tangent function. Express the direction in terms of the polar (positive) specification. Using the component method, calculate the resultant (sum) of the following two vectors. Show all required calculations and diagrams below and identify the direction using the polar (positive) specification. Add the vectors on the applet in order to verify the resultant magnitude and direction. 1 = (+135, -120) - components 2 = (-200, -45) - components a. Addition of the components and drawing of the resultant vector: b. Calculation of the resultant magnitude using the Pythagorean theorem: c. Calculation of the resultant direction using the tangent function: (express the direction in terms of the Polar (positive) specification) Appendix 4. The Law of Cosines \| \| \| The Law of Cosines is a general equation relating three sides and one angle in a triangle. There are no restrictions on the triangle's shape. Three elements determine a triangle. If any three of the four elements in the law-of-cosines equation are given, the equation allows you to calculate the fourth one. Figure A1 illustrates a general triangle. The three sides are labeled a, b, c, and the three angles are labeled α, β, γ. Figure A1 \| \| \| There are three law-of-cosines equations, depending on which angle is included: c2 = a2 + b2 - 2ab cos γ (A1) a2 = b2 + c2 - 2bc cos α (A2) b2 = c2 + a2 - 2ca cos β (A3) Note that the Pythagorean theorem is a special case of these equations if one of the angles is equal to 90°. For example, if γ = 90°, then cos γ = 0 and Equation (A1) reduces to the Pythagorean theorem: c2 = a2 + b2 (A4) Also note the minus sign in front of the cosine term in these equations. This has the following effect. Let's consider Equation (1). If γ < 90° , the cosine is positive. With the minus sign in front of the cosine term, Equation (A1) gives a value for c that is less than the value given by the Pythagorean theorem (4). If γ > 90°, the cosine is negative. Combined with the minus sign in front of the cosine term, the term now makes a positive contribution to the right-hand side of Equation (1) that yields a value of c that is greater than the one given by the Pythagorean theorem. 5. The Law of Sines \| \| \| The Law of Sines is a set of equations true for any triangle. It states that the ratio "sine of an angle divided by the length of the opposite side" is the same for any pair of angle and opposite side. Figure A2 illustrates a general triangle. The three sides are labeled a, b, c, and the three angles are labeled α, β, γ. Figure A2 \| The law-of-sines equations are: (A5) A triangle is determined by three of its elements. Given two sides and an angle opposite to one of the sides, the Law of Sines lets you to determine the angle opposite to the other side. 6. Components \| \| \| Vectors can be described in terms of their scalar components. A vector in two dimensions has two scalar components, one along the x-axis and one along the y-axis. For a vector , these components are denoted ax and ay, respectively. Figure A3 illustrates the components for a vector that is in the first quadrant. Figure A3 The scalar components of a vector are the vector's projections onto the x and y axes. In Figure A3, they are shown in green and yellow, respectively. They are called scalar components because they are numbers. The scalar components are equal to the x and y coordinates of the tip of the vector if the tail end of the vector is at the origin of the coordinate system, as it is here. \| The vector in Figure A3 has a magnitude of 8 and an angle θ with the positive x-axis equal to 30°. Its scalar components have the values: ax = 6.93, ay = 4.00 (A6) For vectors in the first quadrant, both components are positive, but for vectors in one of the other three quadrants, one or both components are negative. For example, with a vector in the second quadrant, the x-component is negative while the y-component is still positive. The definition of the sine and cosine imply that: ax = a cos θ, ay = a sin θ (A7) Substituting a = 8.00 and θ = 30.0° into these equations gives the values listed in Equations (A6) and illustrated in Figure A3. Note that Equations (A7) are correct even if the vector is in the second, third, or fourth quadrant. No signs need to be changed. Any sign changes are automatically taken care of by the signs of the cosine and sine functions for values of θ in any of these other quadrants. Physics 20-30 v1.0 ©2004 Alberta Learning (www.learnalberta.ca) Last Updated: June 16, 2004
17116	https://www.meaningfulmathco.com/mm_math_terms/acute-angle/	Acute Angle - Meaningful Math Skip to main content Join the Movement to Make Math Meaningful! Log In Account Get Started Navigation Menu Meaningful Math Resources Videos Workshops Articles Math Dictionary Podcasts View All Math Concepts STUDENT-FRIENDLY DEFINITION Acute Angle An angle that measures greater than 0° and less than 90° is an acute angle. Understanding Acute Angles In Mathematics An angle is formed when two rays share a common endpoint, called the vertex. The amount of turn between these two rays is measured in degrees (°). An acute angle has a measure between 0° and 90°, meaning it represents a small turn compared to other types of angles. Why Understanding Acute Angles Is Important Acute angles appear in many mathematical and real-world contexts. Understanding acute angles builds the groundwork for reasoning about shapes, symmetry, measurement, and transformations in geometry: Comparing and classifying angles: Recognizing whether an angle is acute, right, or obtuse strengthens students’ ability to analyze shapes. Triangle properties: Any triangle that contains only acute angles is called an acute triangle, which connects angle understanding to the geometry of triangles. Estimation and reasoning: Learning to compare angles visually before measuring them helps students develop flexible thinking about size, rotation, and space. Identifying And Measuring Acute Angles Using A Right Angle As A BenchMark For Acute Angles Students first learn to recognize acute angles by comparing them to known references. Using benchmarks, like right angles, can help students with this. Since a right angle measures exactly 90°, any angle that has less turn than a right angle is acute. This comparison builds a strong conceptual foundation, helping students develop their spatial sense and estimation skills. However, recognizing an acute angle visually isn’t always enough – measurement ensures accuracy. To determine if an angle is acute, a protractor can be used. If the measurement is between 0° and 90°, it is an acute angle. Teaching Strategies For Acute Angles Hands-On Exploration of Acute Angles Before introducing formal definitions or measurements, students benefit from building an intuitive sense of what acute angles look like and how they compare to other types of angles. Early experiences should focus on observing, classifying, and reasoning about angles they can see and manipulate. To support this, offer students a collection of angles in different forms. These could be drawn on cards, cut from paper, or created using manipulatives like straws or craft sticks. Include a mix of acute, right, and obtuse angles to encourage comparison. Invite students to sort the angles into groups and explain their thinking. Ask guiding questions such as, “What makes this angle belong in that group?” or “How do you know this angle is smaller than a right angle?” This type of activity is important because instead of relying on memorized definitions, students develop a relational understanding of acute angle as they compare to right and obtuse angles. Visual Models for Understanding Acute Angles As students progress, visual representations help them refine their reasoning. A right angle remains an essential benchmark, but students begin to engage with acute angles in shapes and diagrams rather than just physical objects. A simple and effective classroom strategy is to have students create a right angle template by folding a square piece of paper in half diagonally or edge-to-edge. This folded corner becomes a tool students can use to test other angles. When they place an angle inside their template: If the angle fits inside the right angle, it is acute. If it extends beyond the right angle, it is obtuse. This develops proportional reasoning, helping students estimate acute angle measures before they work with more formal tools and measurements. Abstract Reasoning With Acute Angles Once students can recognize and compare acute angles, they transition to measuring and reasoning about acute angles numerically. This is where precision becomes important, and estimation skills begin to connect with formal measurement. To support this transition, begin with estimation. Present students with a variety of angles and ask them to predict whether an angle is closer to 0° or 90°, using a right angle as their familiar benchmark. This encourages proportional thinking and strengthens their intuitive sense of angle size. Once students have made their estimates, introduce the protractor. Guide them in measuring each angle and then comparing the measured result to their original estimate. Encourage discussion around how close their predictions were, what clues helped them estimate, and how their sense of angle size is developing. Extend the conversation by having students compare measured angles in different orientations to see if an angle’s size is independent of its direction (an important concept in abstract reasoning). Common Misconceptions About Acute Angles Misconception: rotating an angle changes its measure Students often assume that if an angle is flipped, turned, or drawn in a different orientation, its measure changes. This is because they associate the angle’s appearance with its size rather than understanding that an angle is a measure of turn. Encourage students to physically rotate their papers and re-measure their angles with a protractor to confirm that the degree measure remains the same. 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17117	https://www.youtube.com/watch?v=TBgCZjJ1Soc	Probability Theory: Sample Space and Events - Part 1 Maths and Stats 23500 subscribers 492 likes Description 69589 views Posted: 29 Mar 2017 This short video introduces two important concepts in Probability, that of a sample space (outcome space) and that of an event. 3 comments Transcript: hi guys this is Jonathan Lambert with the mathematics development and support center at the National College of Ireland and this short video is going to I suppose it's going to be an introduction to probability Theory a very foundational introduction uh but more importantly it's going to concentrate on two important things uh the sample space associated with an experiment and also what we mean by an event when we put an event occurring or the likelihood of an event happening when we perform an experiment uh I haven't I haven't indicated at this particular stage what we mean by an experiment and I suppose an experiment could be anything from tossing a coin rolling a particular dice randomly selecting a card from a deck of cards an experiment could also possibly be the determination of maybe the sex of a newborn child uh it could be it could be anything at all okay but one thing about this particular experiment is that we don't know with any certainty and what the outcome of the experiment is going to be so there's some sort of uncertainty or we it's it's not predictable what the certainty in advance of the experiment of the outcome of the experiment is going to be and but one thing hopefully that we will be able to do and we probably won't be able to rationalize this in all cases yeah but for every experiment we do we can actually we should be able to list uh the set of all possible outcomes and I suppose last set of all possible outcomes we call uh the sample space so let's consider some examples here okay so some example experiments okay some example experiments and let's say the first experiments that we're going to consider okay let's say one okay the experiment is okay let's say where it's it's it's simply uh tossing a coin yeah okay so it's tossing tossing a single a single coin now we know when we toss a single client that there's one of two possible things that could happen okay and you could get a head or you could get a tail okay so this list of all possible outcomes when you perform an experiment is what's known as the as the sample space okay we usually symbolize the sample space by an S so for example in relation to the experimental tossing a kind the sample space space symbolized by an S is equal to the set of all possible outcomes so in Decline tossing experiment it's equal to a head symbolized by a h or a tail symbolized by a t so that's the sample space for a single clientos okay let's consider another experiment okay this time this experiment is in relation to Rolling rolling a single die okay so rolling a single dice or die okay and we know that when we roll a die let's say a six-sided die that the possible things that could happen with respect to the value that's face up on the die okay that's shown face up is the numbers one two three four five or six cut a core in that particular type of experiments so in this case here the sample space of Interest okay the sample space of Interest symbolized by S is simply the face values of the of the die one two three four five and six uh another experiment uh might be and maybe randomly selecting a card from a deck of cards okay so let's have a look at the sample space there okay so in this case here we're selecting selecting a single card a single card from from a deck of cards okay now the typical deck of cards that's that's let's say a basic deck of cards that was used in most in most card games has 52 cards in it okay and the sample space okay if I wanted to list the sample space out okay s there's 52 cards so the sample space is going to have 52 entries here okay and the cards themselves are broken into two colors there's red cards and there's black cards okay so we have red cards and black cards when we consider the red cards so half the 52 cards are red that gives us 26 cards red cards and half are black so there's 26 black cards within the red cards there's two suits just hearts and there's diamonds so tortain of the red cards are hearts and torting her diamonds similarly tortion of the black cards are spades and torten are clubs okay and also in relation to each individual card there's a there's a face value associated with the cards and those face values range from the number two three four all the way through to 10 followed by a jacket a queen and king and an ace so if I wanted to list out the sample space associated with the single selection of a card I'd probably have the two of hearts The Tree Of Hearts the four of Hearts the five of Hearts the six of Hearts the seven of Hearts the eight of Hearts the nine of hearts ten of Hearts the Jack of Hearts Queen of Hearts the king of hearts and the Ace of Hearts yet that would be the list of all the cards with respect to hearts when I look at diamonds we have the two of diamonds the tree of diamonds and I'm not gonna do all this I'm just gonna say dot dot dot maybe up to the ten diamonds followed by the jack of diamonds the Queen of Diamonds the King of Diamonds the ace of diamonds when we look at the Spades we could have the two of spades the tree of Spades all the way true to the Ten of Spades followed by the Jack of Spades all the way through to the Ace of Spades and finally we would have we would have the two of clubs the tree of clubs all the way true to the Ten of clubs the Jack of clubs all the way through to the Ace of clubs and that will be the sample space okay of all possible Atomic events that could occur when we randomly select a single card from a deck of cards okay another experiment might be I suppose an experiment where we don't know the outcome of it okay okay but we we don't know what might actually happen in the experiment so it's it's unknown there's some non-predictability associated with some uncertainty okay it might be an experiment where we've to I suppose determine the term determination okay of the sex of a of an a newborn child okay of a newborn of a newborn child in which case Okay the sample space sample space will be symbolized by S is equal to let's say G for a girl and B for a boy okay there's some basic sample spaces that we have it's just simply a list of all the possible outcomes associated with a particular experiment okay let's consider another one let's say consider something a little bit more complicated okay let's call this sample space five okay and sample space five is where we toss let's say we toss clients so when we toss two coins okay maybe we'll just keep track of the clients we have we've labeled the force kind the force kind and the second kind the second kind okay uh so what we'd like to know is what are all the possible pairs of observations that could occur there's one way we could figure this out okay so we're interested in the sample space and the sound space is going to be all the possible pairings okay and where we have the fourth and second kind so let's just think about it from a cross product perspective okay let's say the first here second here okay second kind the first coin could be a head or a tail the second could be a head or a tail which gives us ordered pairs HH t Okay t h t t so the sample space when we flip two clients would be the ordered pairs the force client is ahead the second is ahead the force client is ahead or the second has or the second kinds of tail okay the force climbs a tail or the second is a head and finally the force kinds of tail and the second kind is a tail okay so this is the experiment in relation to tossing two coins and let's consider something a little bit more complicated yeah let's say when we roll to to dice okay what does the sample space look like here well the sample space is going to be once again all the ordered pairs all the values of the force die paired off with all the values of the second die once again we could do that to a table where we have the force die the second Dias outcomes for slide one two three four five six the second I one two three four five six so the ordered pairs would be one one one two one three one four one five one six we might have two one two two two three two four to five to six and I'm actually going to exhaust this out I'm gonna do this two three one let's say four one five one six one we have three two three sorry four two five two six two We have tree tree let's say we have here we have four tree here we have five tree six
17118	https://test-english.com/explanation/a1-vocabulary/food-and-meals-a1-english-vocabulary/	Food and meals – A1 English Vocabulary Explanations / A1 Vocabulary Explanations / Food and meals – A1 English Vocabulary Exercises Explanation Downloads Food and Meals In this A1 Elementary Vocabulary lesson, you will learn essential vocabulary to talk about Food and Meals. Check the pictures and read the texts and definitions below. ### Meals, breakfast and snacks #### Meals People usually have three main 1 meals every day. They eat 2 breakfast in the morning before they go to school or work. Then they have 3 lunch around midday, usually from 12 pm to 2 pm. In the evening, it’s common to have 4 dinner with family, which is often between 6 pm and 8 pm. If people get hungry between lunch and dinner, they might have a small meal or something to eat. In the UK, this small meal can be called 5 tea and often includes a cup of tea with food. In the US, they usually call this a 5 snack. The word snack is also used in the UK to describe a small amount of food that you eat between meals. #### Breakfast People often eat 6 cereal with milk for breakfast. Some people like to eat 7 bread with 8 cheese or 9 ham. Fried 10 eggs are also a very common breakfast option. People who like sweet food often eat 10 toast with 11 butter and 12 jam. Strawberry jam is very popular for breakfast. And people who like to eat healthy food often have 14 yogurt and 15 fruit. #### Snacks For a snack, many people like 16 chips (or crisps) because they are crunchy and salty. People often eat chips while they watch TV on the sofa. Kids love 17 cookies with milk after school. If you are very hungry, you can take some bread and make a 18 sandwich. Ham and cheese sandwiches are very popular. Sometimes, at a party, you can have a piece of 19 cake. And for a quick sweet snack, some people eat a 20 candy bar. ### Food and Drink #### Food 1 Rice is white and it’s a very popular food in Asian countries. 2 Pasta comes in shapes like spaghetti and macaroni. 3 Pizza is round with cheese, tomato, and other toppings on bread. 4 Meat comes from animals, like cows and chickens. 5 Fish live in water, and people can cook and eat them. 6 Vegetables are plants like carrots and broccoli that are good for your health. 7 Soup is hot and liquid, and you eat it with a spoon. 8 Salad is a mix of cold vegetables, sometimes with dressing. 9 You eat dessert at the end of your lunch or dinner. Common desserts are fruit or cake. In the UK, people often use the word pudding to mean ‘dessert’. #### Drink 10 People drink water when they are thirsty. Water is the healthiest drink. 11 Milk is white, comes from cows, and is rich in calcium. 12 Juice is a sweet drink made from fruits like oranges and apples. 13 Tea is a hot drink, and in the UK, people often drink it in the afternoon. 14 Coffee is a hot, dark drink that contains caffeine. 15 Beer is the most common drink in British pubs. 16 Wine is made from grapes and is an alcoholic drink. ### Listen & Practice: Vocabulary with Pronunciation Here’s a video to help you review the vocabulary from this lesson and improve your pronunciation. Flashcards After completing the exercises in this lesson on Food and Meals, you can use the unit’s Vocabulary Flashcards to revise and help you memorize the terms. Related tests: Daily routines – A1 English Vocabulary Food and cooking – B1 English Vocabulary
17119	https://math.libretexts.org/Courses/Cosumnes_River_College/Math_373%3A_Trigonometry_for_Calculus/07%3A_Trigonometric_Equations/7.02%3A_The_Remaining_Inverse_Trigonometric_Functions	7.2: The Remaining Inverse Trigonometric Functions - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode Chapter 7: Trigonometric Equations Math 373: Trigonometry for Calculus { "7.2.01:_Resources_and_Key_Concepts" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.2.02:_Homework" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "7.01:The_Fundamental_Inverse_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_The_Remaining_Inverse_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Solving_Trigonometric_Equations-_Algebraic_Techniques" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Solving_Trigonometric_Equations_Using_Identities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Triangles_and_Circles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Coordinate_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Right_Triangle_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Radian_Measure_and_the_Circular_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Graphs_of_the_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Analytic_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Trigonometric_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Non-Right_Triangle_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Polar_Coordinates_and_Complex_Numbers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_The_Language_of_Mathematics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sun, 20 Jul 2025 22:21:14 GMT 7.2: The Remaining Inverse Trigonometric Functions 158843 158843 Roy Simpson { } Anonymous Anonymous 2 false false [ "article:topic", "arccosecant", "stage:final", "license:ccbysa", "showtoc:no", "author@Roy Simpson", "authorname:roysimpson", "licenseversion:12", "arcsecant", "arccotangent" ] [ "article:topic", "arccosecant", "stage:final", "license:ccbysa", "showtoc:no", "author@Roy Simpson", "authorname:roysimpson", "licenseversion:12", "arcsecant", "arccotangent" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Cosumnes River College 4. Math 373: Trigonometry for Calculus 5. Chapter 7: Trigonometric Equations 6. 7.2: The Remaining Inverse Trigonometric Functions Expand/collapse global location 7.2: The Remaining Inverse Trigonometric Functions Last updated Jul 20, 2025 Save as PDF 7.1.2: Homework 7.2.1: Resources and Key Concepts Page ID 158843 Roy Simpson Cosumnes River College ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Note to the Instructor- Where's the "Stewart Calculus" Definition? 2. Learning Objectives 3. The Inverse Cotangent Function 1. Definition: Inverse Cotangent Function 1. Theorem: Properties of the Inverse Cotangent Function 2. Example 7.2.1) 3. Checkpoint 7.2.1) The Inverse Secant and Cosecant Functions Definition: The Inverse Secant Function Definition: The Inverse Cosecant Function Theorem: Properties of the Inverse Secant and Inverse Cosecant Functions Example 7.2.2 Checkpoint 7.2.2 Calculators and the Non-Fundamental Inverse Trigonometric Functions Example 7.2.3 Checkpoint 7.2.3 Example 7.2.4 Note to the Instructor- Where's the "Stewart Calculus" Definition? For those who don't know, the Stewart Calculus textbook and, subsequently, Cosumnes River College's OER Calculus textbook use a different range definition for the secant and cosecant functions than introduced in this section.The redefinition that occurs in Calculus is to make some developments in Calculus II work more smoothly; however, I chose to stick with the conventional definitions in this text so students going to other campuses (or for those who are using the online homework) don't have to worry about things. Learning Objectives Find the exact value of a non-fundamental inverse trigonometric function. Use technology to approximate the value of a non-fundamental inverse trigonometric function. Evaluate compositions involving trigonometric functions and non-fundamental inverse trigonometric functions. Simplify expressions involving trigonometric functions and non-fundamental inverse trigonometric functions. Hawk A.I. Section-Specific Tutor Please note that, to access the Hawk A.I. Tutor, you will need a (free) OpenAI account. In the previous section, we completely ignored the inverses of the secant, cosecant, and cotangent. This is not because they are unimportant - in fact, the inverses of the secant and cosecant are essential parts of Integral Calculus (also known as Calculus II). We have delayed their introduction because it's best to master the inverses of the fundamental trigonometric functions before jumping into the remaining inverse trigonometric functions. The Inverse Cotangent Function Like the rest of the trigonometric functions, the cotangent is not naturally one-to-one. Therefore, we restrict the domain of g⁡(x)=cot⁡(x) to its principal cycle on (0,π) to obtain a one-to-one segment of the cotangent. The vertical asymptotes, x=0 and x=π, of the graph of g⁡(x)=cot⁡(x) become the horizontal asymptotes, y=0 and y=π, of the graph of g−1⁡(x)=arccot⁡(x) (which, recall, is a reflection of the graph of y=cot⁡(x) about the line y=x). Figure 7.2.1 The definition of the arccotangent follows the same pattern as the definitions of our other inverse trigonometric functions. Definition: Inverse Cotangent Function The function f⁡(x)=cot−1⁡(x) is defined as follows:cot−1⁡(x)=θ if and only if cot⁡(θ)=x and 0<θ<π. Again, I want to be clear:Very technically, the function cot−1⁡(x) returns an angle and arccot⁡(x) returns a real number. This is true for all inverse trigonometric functions. Therefore, it is valid for cot−1⁡(x) to return an angle in degree measure or radian measure; however, it is technically not valid for arccot⁡(x) to return a number in degree measure (as this is an angle). Since radians is a unitless measure, a radian is just a real number, so arccot⁡(x) can return a number in radian measure. The moral of the story is that you are always safe to work in radian measure when dealing with inverse trigonometric functions, but working in degree measure is a little risky. Theorem: Properties of the Inverse Cotangent Function Properties of f⁡(x)=cot−1⁡(x) θ=cot−1⁡(x) if and only if cot⁡(θ)=x where 0<θ<π Domain: (−∞,∞) Range: (0,π) As x→−∞, cot−1⁡(x)→π−, and as x→∞, cot−1⁡(x)→0+ cot⁡(cot−1⁡(x))=x for all x∈R cot−1⁡(cot⁡(x))=x provided 0<x<π Example 7.2.1 Find the exact values of the following. arccot⁡(−3) cot⁡(arccot⁡(−5)) Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid.cos⁡(arccot⁡(2⁢x)) Solutions 1. 1. arccot⁡(−3)=θ means cot⁡(θ)=−3, where θ∈(0,π). Quadrant:Since the cotangent is negative, we further restrict the angle to θ∈(π 2,π), or Quadrant II. Reference Angle: The reference angle for the cotangent that gives cot⁡(θ^)=3 is θ^=π 6. Given the quadrant annd reference angle, we get arccot⁡(−3)=5⁢π 6. 2. It is helpful to remember that trig⁡(trig−1⁡(x))=x for all of the trigonometric functions as long as x is in the domain of trig−1 (which is the range of trig). Therefore, cot⁡(arccot⁡(−5))=cot⁡(cot−1⁡(−5))=−5 because −5 is definitely in the domain of the inverse cotangent. Another way to think about this is to let θ=arccot⁡(−5). Then θ belongs to the interval (0,π) and cot⁡(θ)=−5. Hence, cot⁡(arccot⁡(−5))=cot⁡(θ)=−5. To get started, we let θ=arccot⁡(2⁢x) so that cot⁡(θ)=2⁢x where 0<θ<π. We then sketch two reference triangles, one in QI and the other in QII. Figure 7.2.2 In terms of θ,cos⁡(arccot⁡(2⁢x))=cos⁡(θ).Therefore, we can use our reference triangles to evaluate cos⁡(θ). Looking at the reference triangle in QI, cos⁡(θ)=2⁢x 4⁢x 2+1. Likewise, the reference triangle from the QII sketch yields cos⁡(θ)=2⁢x 4⁢x 2+1. Hence, it appears that cos⁡(arccot⁡(2⁢x))=cos⁡(θ)=2⁢x 4⁢x 2+1.Being good mathematicians, we take a moment to reflect and make sure our logic is solid. What might give us pause is the fact that our second sketch, where θ∈QII, leads to cosine being negative. Still, our result,cos⁡(θ)=2⁢x 4⁢x 2+1, looks to be positive; however, looks can be deceiving! The expression, 2⁢x, in the second sketch,is, in fact, negative. The negativity of it is "hidden" within the variable x. That is, if the x-value of the point on the upper-left vertex of that reference triangle is −4, then 2⁢x=−4⟹x=−2. Hence, when θ∈QII,cos⁡(θ)=2⁢x 4⁢x 2+1 is negative (which is what we want). We now concern ourselves with the domain of the function. We know that the domain of the arccotangent is all real numbers. Hence, the domain of arccot⁡(2⁢x) is also all real numbers. Moreover, the range of the arccotangent is (0,π). Since the outputs of arccot⁡(2⁢x) are the inputs of cos⁡(arccot⁡(2⁢x))and the cosine can take any value as an input, the domain remains unchanged. That is, the domain of y=cos⁡(arccot⁡(2⁢x)) is (−∞,∞). Checkpoint 7.2.1 Find the exact value of the expression.sin⁡(cot−1⁡(−5 2)) Answer 2 29 The Inverse Secant and Cosecant Functions The last two functions to invert are secant and cosecant. The principal cycle for each is highlighted in the figure below. Figure 7.2.3 It is clear from the graph of secant that we cannot find one single, continuous piece of its graph which covers its entire range of (−∞,−1]∪[1,∞), and which restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus, in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely [1,∞), and another piece to cover the bottom, namely (−∞,−1].To make the definitions of the inverse secant and cosecant functions align with our previous work, we restrict their domains to coincide with the restrictions on cosine and sine, respectively. It is critical to note that there are two ways to restrict the domains of the secant and cosecant functions. The method shown in this text aligns with all other texts in Trigonometry and Precalculus; however, there is a _much_ better choice of domain restriction for these functions that makes Calculus (specifically, Integral Calculus) work much more smoothly. Despite this, we give the traditional definition here because it aligns with all of our previous work. When defining the arccosine, we restricted the cosine to the interval [0,π]; however, such a restriction needs to be modified for the secant because the secant is undefined at x=π 2. Thus,we restrict the domain of the secant to [0,π 2)∪(π 2,π] to help us define the arcsecant. Figure 7.2.4 This gives us the following definition. Definition: The Inverse Secant Function The function f⁡(x)=sec−1⁡(x) is defined as follows:sec−1⁡(x)=θ if and only if sec⁡(θ)=x and θ∈[0,π 2)∪(π 2,π]. Likewise, we restricted the sine function to the interval [−π 2,π 2] when defining the arcsine, but such a restriction will not work for the arccosecant because the cosecant is undefined at x=0. Therefore, we restrict g⁡(x)=csc⁡(x) to [−π 2,0)∪(0,π 2] to define the arccosecant. Figure 7.2.5 Note that, for both arcsecant and arccosecant, the domain is (−∞,−1]∪[1,∞). We can rewrite this using set-builder notation as {x∣\|x\|≥1}. This is often done in Calculus textbooks, so we include it here for completeness. Definition: The Inverse Cosecant Function The function f⁡(x)=csc−1⁡(x) is defined as follows:csc−1⁡(x)=θ if and only if csc⁡(θ)=x and θ∈[−π 2,0)∪(0,π 2]. Using these definitions, we get the following properties of the arcsecant and arccosecant functions. Theorem: Properties of the Inverse Secant and Inverse Cosecant Functions Properties of f⁡(x)=arcsec⁡(x) sec−1⁡(x)=θ if and only if sec⁡(θ)=x where θ∈[0,π 2)∪(π 2,π] Domain: {x:\|x\|≥1}=(−∞,−1]∪[1,∞) Range: [0,π 2)∪(π 2,π] As x→−∞, arcsec⁡(x)→π 2+, and as x→∞, arcsec⁡(x)→π 2− sec⁡(arcsec⁡(x))=x provided \|x\|≥1 arcsec⁡(sec⁡(x))=x provided 0≤x<π 2 or π 2<x≤π Properties of f⁡(x)=arccsc⁡(x) csc−1⁡(x)=θ if and only if csc⁡(θ)=x where θ∈[−π 2,0)∪(0,π 2] Domain: {x:\|x\|≥1}=(−∞,−1]∪[1,∞) Range: [−π 2,0)∪(0,π 2] As x→−∞, arccsc⁡(x)→0−, and as x→∞, arccsc⁡(x)→0+ csc⁡(arccsc⁡(x))=x provided \|x\|≥1 arccsc⁡(csc⁡(x))=x provided −π 2≤x<0 or 0<x≤π 2 Additionally, arccosecant is an odd function. Example 7.2.2 Find the exact values of the following. sec−1⁡(2) arccsc⁡(−2) arcsec⁡(sec⁡(8⁢π 7)) cot⁡(arccsc⁡(−3)) Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. tan⁡(arcsec⁡(x)) cos⁡(arccsc⁡(4⁢x)) Solutions 1. 1. Let θ=sec−1⁡(2). Then sec⁡(θ)=2, where θ∈[0,π 2)∪(π 2,π]. Of these two intervals, the only one where secant is positive is [0,π 2). Thus, θ∈QI. Moreover, sec⁡(θ^)=2 implies cos⁡(θ^)=1 2. Hence, θ^=π 3. Therefore,sec−1⁡(2)=θ=π 3. 2. We now start to streamline our arguments.arccsc⁡(−2)=θ(inverse trigonometric functions return angles,)⟹csc⁡(θ)=−2,where θ∈−π 2,0)∪(0,π 2⟹θ∈[−π 2,0)(cosecant is negative)and θ^=π 6⟹arccsc⁡(−2)=−π 6(reference angle of π 6 in the interval[−π 2,0)) 3. Our choice of non-standard angle for this example is purposeful. It's to force you to think about the process instead of focusing solely on computing.arcsec⁡(sec⁡(8⁢π 7))=θ,where θ∈0,π 2)∪(π 2,π⟹sec⁡(θ)=sec⁡(8⁢π 7)Therefore, the reference angle for θ matches the reference angle for 8⁢π 7.That is,θ^=π 7.We now need to determine in which quadrant θ belongs. Since 8⁢π 7∈QIII, sec⁡(8⁢π 7) is negative. Moreover, since sec⁡(θ)=sec⁡(8⁢π 7), sec⁡(θ) must be negative. Hence, θ∈(π 2,π]. Thus, θ=6⁢π 7. This means that arcsec⁡(sec⁡(8⁢π 7))=θ=6⁢π 7. 4. As usual, we begin by addressing the fact that the inverse trigonometric function is an angle.arccsc⁡(−3)=θ,where θ∈−π 2,0)∪(0,π 2⟹csc⁡(θ)=−3 However, we are interested in cot⁡(arccsc⁡(−3))=cot⁡(θ).Since the cosecant is negative, we know that θ∈[−π 2,0).Let sketch a reference triangle in QIV to help us determine cot⁡(θ). Figure 7.2.6 From the reference triangle, we see that cot⁡(arccsc⁡(−3))=cot⁡(θ)=−2⁢2. As per usual, let θ=arcsec⁡(x) so that sec⁡(θ)=x, where we restrict θ to the interval [0,π 2)∪(π 2,π]. We then sketch two reference triangles, one for θ∈[0,π 2) and the other for θ∈(π 2,π]. Figure 7.2.7 We now use these reference triangles to help us determine the value of tan⁡(arcsec⁡(x))=tan⁡(θ). In this case, we run into a very interesting problem. In Quadrant I, tan⁡(θ)=x 2−1; however, in Quadrant II, tan⁡(θ)=−x 2−1. Pausing for a moment, let's think about this before going further. Does this make sense? We know tangent is supposed to be positive in Quadrant I.Since sec⁡(θ)=x, we know either x≥1 or x≤−1. Luckily, x 2−1 is definitely positive for these values of x (unless x=±1, but that would mean tan⁡(θ)=0⟹θ=0 which is not in Quadrant I). So the statement, tan⁡(θ)=x 2−1 when θ∈(0,π 2) makes sense. In Quadrant II, the tangent is supposed to be negative. It's easy to see that −x 2−1 is negative whenever θ∈(π 2,π). Thus, the statement,tan⁡(θ)=−x 2−1 when θ∈(π 2,π) also makes sense. Therefore,tan⁡(θ)={x 2−1,if 0≤θ<π 2−x 2−1,if π 2<θ≤π Now we need to determine what these conditions on θ mean for x. Since x=sec⁡(θ), when 0≤θ<π 2, x≥1, and when π 2<θ≤π, x≤−1. Since we encountered no further restrictions on θ, the equivalence below holds for all x in (−∞,−1]∪[1,∞).tan⁡(arcsec⁡(x))={x 2−1,if x≥1−x 2−1,if x≤−1 2. While I always use the graphing tactic, you do not have to think geometrically to work with inverse trigonometric functions. You can rely solely on identities. For example, to simplify cos⁡(arccsc⁡(4⁢x)), we start by letting θ=arccsc⁡(4⁢x). Then csc⁡(θ)=4⁢x for θ∈[−π 2,0)∪(0,π 2]. We now set about finding an expression for cos⁡(arccsc⁡(4⁢x))=cos⁡(θ). Since cos⁡(θ) is defined for all θ, we do not encounter any additional restrictions on θ. From csc⁡(θ)=4⁢x, we get sin⁡(θ)=1 4⁢x, so to find cos⁡(θ), we can make use of the Pythagorean Identity, cos 2⁡(θ)+sin 2⁡(θ)=1. Substituting sin⁡(θ)=1 4⁢x gives cos 2⁡(θ)+(1 4⁢x)2=1.Solving, we get cos⁡(θ)=±16⁢x 2−1 16⁢x 2=±16⁢x 2−1 4⁢\|x\|.Since θ∈[−π 2,0)∪(0,π 2], we know cos⁡(θ)≥0. Therefore, we choose cos⁡(θ)=16−x 2 4⁢\|x\|.(The absolute values here are necessary, since x could be negative.) To find the values for which this equivalence is valid, we look back at our original substitution, θ=arccsc⁡(4⁢x). Since the domain of arccsc⁡(x) requires its argument x to satisfy \|x\|≥1, the domain of arccsc⁡(4⁢x) requires \|4⁢x\|≥1. This implies x≤−1 4 or x≥1 4. Since we had no additional restrictions on θ, the equivalence cos⁡(arccsc⁡(4⁢x))=16⁢x 2−1 4⁢\|x\|holds for all x in (−∞,−1 4]∪[1 4,∞). Checkpoint 7.2.2 Rewrite the function as one that does not involve trigonometric functions and state the domain.f⁡(x)=sin⁡(sec−1⁡(2⁢x))(Your answer might be a piecewise function or it might involve absolute values.) Answer There are two ways to write this function: f⁡(x)=4⁢x 2−1 2⁢\|x\|,or f⁡(x)={4⁢x 2−1 2⁢x,if x≥1 2−4⁢x 2−1 2⁢x,if x<−1 2 The domain of this function is all x∈(−∞,−1 2]∪[1 2,∞). Calculators and the Non-Fundamental Inverse Trigonometric Functions In the sections to come, we will need to approximate the values of the inverse trigonometric functions.We have already discussed how to use technology to approximate values of the arcsine, arccosine, and arctangent functions; however, if we are asked for an approximation of the arccotangent, arcsecant, or arccosecant, how is this to be done when our calculators do not have these functions? Example 7.2.3 Use a calculator to approximate the following values to four decimal places. cot−1⁡(2) arcsec⁡(5) arccot⁡(−2) csc−1⁡(−3 2) Find the domain and range of f⁡(x)=π 2−sec−1⁡(x 5). Solutions 1. 1. cot−1⁡(2)=θ means cot⁡(θ)=2, where θ∈(0,π). Quadrant:Since the cotangent is positive, we know θ∈(0,π 2). Reference Angle:2 is not a special ratio for the cotangent, so we will need technology to help us find the reference angle; however, our calculators do not have an inverse cotangent button. What do we do? The answer is quite simple - we use the Reciprocal Identities. Since cot⁡(θ)=2, tan⁡(θ)=1 2. To compute the reference angle, we evaluate θ^=tan−1⁡(\|1 2\|)=tan−1⁡(1 2)≈0.4636 Now that we have a reference angle (≈0.4636) and a quadrant (QII), we get cot−1⁡(2)≈0.4636. 2. arcsec⁡(5)=θ means sec⁡(θ)=5, where θ∈[0,π 2)∪(π 2,π]. Quadrant:Since the secant is positive, we further restrict the angle to θ∈(0,π 2) (Quadrant I). Reference Angle:Again, since our technology does not have an inverse secant button, we use the Reciprocal Identities.sec⁡(θ)=5⟹cos⁡(θ)=1 5.Now we use a calculator to compute θ^=cos−1⁡(\|1 5\|)=cos−1⁡(1 5)≈1.3694. Using our reference angle (≈1.3694) and quadrant (QI), we get arcsec⁡(5)≈1.3694. 3. arccot⁡(−2)=θ means cot⁡(θ)=−2, where θ∈(0,π). Quadrant: Since cot⁡(θ)<0, we know π 2<θ<π. That is, θ∈QII. Reference Angle:Once again, we do not have an appropriate inverse trigonometric function on the calculator, so we use the Reciprocal Identities.cot⁡(θ)=−2⟹tan⁡(θ)=−1 2.Thus, we compute the reference angle as follows:θ^=tan−1⁡(\|−1 2\|)=tan−1⁡(1 2)≈0.4636. We have a reference angle (≈0.4636) and a quadrant (QII). To find the actual angle, we subtract our reference angle from π to get arccot⁡(−2)=θ=π−θ^=π−arctan⁡(1 2)≈2.6779. 4. csc−1⁡(−3 2)=θ means csc⁡(θ)=−3 2, where θ∈[−π 2,0)∪(0,π 2]. Quadrant:Since csc⁡(θ)<0, we further restrict the angle to the interval θ∈[−π 2,0) (Quadrant IV). Reference Angle:We do not have an appropriate inverse trigonometric button on our calculator for the cosecant. Therefore, we must use the Reciprocal Identities.csc⁡(θ)=−3 2⟹sin⁡(θ)=−2 3.Hence,θ^=sin−1⁡(\|−2 3\|)=sin−1⁡(2 3)≈0.7297. Now that we have a reference angle (≈0.7297) and a quadrant (QIV), we get csc−1⁡(−3 2)≈−0.7297. The domain of the arcsecant is normally (−∞,−1]∪[1,∞). That is, if we were given sec−1⁡(r), we know that allowable values of r are r∈(−∞,−1]∪[1,∞). In this case, however, the argument of the arcsecant is x 5. Hence, x 5∈(−∞,−1]∪[1,∞). Rewriting this in interval notation will be helpful.x 5≤−1 or x 5≥1⟹x≤−5 or x≥5 Thus, the domain of f is (−∞,−5]∪[5,∞). We know the range of the arcsecant is normally [0,π 2)∪(π 2,π]. Given the function f⁡(x)=π 2−sec−1⁡(x 5), the only transformation that affects the range is the addition of π 2 to the output values. Thus, the range of f is [π 2,π)∪(π,3⁢π 2]. Checkpoint 7.2.3 Use a calculator to approximate arcsec⁡(−7 3). Answer ≈2.0137 The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. The following example presents one such scenario. Example 7.2.4 The roof on the house below has a "6/12 pitch." This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Figure 7.2.8 Solution If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. We find the angle of inclination, labeled θ below, satisfies tan⁡(θ)=6 12=1 2. Since θ is an acute angle, we can use the arctangent function and we find θ=arctan⁡(1 2)radians≈26.56∘. Figure 7.2.9 This page titled 7.2: The Remaining Inverse Trigonometric Functions is shared under a CC BY-SA 12 license and was authored, remixed, and/or curated by Roy Simpson. Back to top 7.1.2: Homework 7.2.1: Resources and Key Concepts Was this article helpful? Yes No Recommended articles 10.2: The Remaining Inverse Trigonometric FunctionsThis section introduces the inverse trigonometric functions for cotangent, secant, and cosecant. It covers their definitions, properties, and domains,... 10.2: The Remaining Inverse Trigonometric FunctionsThis section introduces the inverse trigonometric functions for cotangent, secant, and cosecant. It covers their definitions, properties, and domains,... 5.3: Inverse Trigonometric FunctionsWe have briefly mentioned the inverse trigonometric functions before, but we will now define those inverse functions and determine their graphs. 7.3: Solving Trigonometric Equations - Algebraic TechniquesThis section focuses on solving trigonometric equations using algebraic techniques. It covers methods for solving basic trigonometric equations, findi... 7.4: Solving Trigonometric Equations Using IdentitiesThis section focuses on solving trigonometric equations using identities. It discusses strategies for equations involving different trigonometric func... Article typeSection or PageAuthorRoy SimpsonLicenseCC BY-SALicense Version1.2Show Page TOCnoStageFinal Tags arccosecant arccotangent arcsecant author@Roy Simpson © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? 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17120	https://www.skillsyouneed.com/num/percent-change.html	NUMERACY SKILLS Percentage Change \| Increase and Decrease Search SkillsYouNeed: Numeracy Skills: A - Z List of Numeracy Skills How Good Are Your Numeracy Skills? Numeracy Quiz Money Management and Financial Skills Real-World Maths Numbers \| An Introduction Special Numbers and Mathematical Concepts Systems of Measurement Common Mathematical Symbols and Terminology Apps to Help with Maths Arithmetic: Addition + Subtraction - Multiplication × Division ÷ Positive and Negative Numbers Ordering Mathematical Operations - BODMAS Mental Arithmetic – Basic Mental Maths Hacks Essentials of Numeracy: Fractions Decimals Ratio and Proportion Percentages % Percentage Calculators Percentage Change \| Increase and Decrease Calculating with Time Estimation, Approximation and Rounding Geometry: Introduction to Geometry: Points, Lines and Planes Introduction to Cartesian Coordinate Systems Polar, Cylindrical and Spherical Coordinates Angles Properties of Polygons Simple Transformations of 2-Dimensional Shapes Circles and Curved Shapes Perimeter and Circumference Calculating Area Three-Dimensional Shapes Net Diagrams of 3D Shapes Calculating Volume Area, Surface Area and Volume Reference Sheet Data Analysis: Graphs and Charts Averages (Mean, Median & Mode) Simple Statistical Analysis Statistical Analysis: Types of Data Understanding Correlations Understanding Statistical Distributions Significance and Confidence Intervals Developing and Testing Hypotheses Multivariate Analysis More Advanced Mathematical Concepts: Introduction to Algebra Simultaneous and Quadratic Equations Introduction to Trigonometry Introduction to Probability Set Theory Subscribe to our FREE newsletter and start improving your life in just 5 minutes a day. Subscribe You'll get our 5 free 'One Minute Life Skills' and our weekly newsletter. We'll never share your email address and you can unsubscribe at any time. Percentage Change \| Increase and Decrease Being able to calculate percentage change is an important skill used in everyday life to measure growth, decline, or variation over time. For example, businesses use it to track changes in turnover, helping them identify whether they are expanding or facing challenges. Shoppers can calculate percentage discounts during sales to see exactly how much they are saving. In education, percentage change is frequently used to measure improvements or declines in student performance. By understanding and learning how to calculate percentage change, you can make more informed decisions in many aspects of life - see our various examples below. Percentage Change Calculator Use this calculator to work out the percentage change of two numbers More: Percentage Calculators For an explanation and everyday examples of using percentages generally see our page Percentages: An Introduction. For more general percentage calculations see our page Percentage Calculators. To calculate the percentage increase: First: work out the difference (increase) between the two numbers you are comparing. Increase = New Number - Original Number Then: divide the increase by the original number and multiply the answer by 100. % increase = Increase ÷ Original Number × 100. If your answer is a negative number, then this is a percentage decrease. To calculate percentage decrease: First: work out the difference (decrease) between the two numbers you are comparing. Decrease = Original Number - New Number Then: divide the decrease by the original number and multiply the answer by 100. % Decrease = Decrease ÷ Original Number × 100 If your answer is a negative number, then this is a percentage increase. If you wish to calculate the percentage increase or decrease of several numbers then we recommend using the first formula. Positive values indicate a percentage increase whereas negative values indicate percentage decrease. Further Reading from Skills You Need The Skills You Need Guide to Numeracy This four-part guide takes you through the basics of numeracy from arithmetic to algebra, with stops in between at fractions, decimals, geometry and statistics. Whether you want to brush up on your basics, or help your children with their learning, this is the book for you. Examples - Percentage Increase and Decrease In January Dylan worked a total of 35 hours, in February he worked 45.5 hours – by what percentage did Dylan’s working hours increase in February? To tackle this problem first we calculate the difference in hours between the new and old numbers. 45.5 - 35 hours = 10.5 hours. We can see that Dylan worked 10.5 hours more in February than he did in January – this is his increase. To work out the increase as a percentage it is now necessary to divide the increase by the original (January) number: 10.5 ÷ 35 = 0.3 (See our division page for instruction and examples of division.) Finally, to get the percentage we multiply the answer by 100. This simply means moving the decimal place two columns to the right. 0.3 × 100 = 30 Dylan therefore worked 30% more hours in February than he did in January. In March Dylan worked 35 hours again – the same as he did in January (or 100% of his January hours). What is the percentage difference between Dylan’s February hours (45.5) and his March hours (35)? First calculate the decrease in hours, that is: 45.5 - 35 = 10.5 Then divide the decrease by the original number (February hours) so: 10.5 ÷ 45.5 = 0.23 (to two decimal places). Finally multiply 0.23 by 100 to give 23%. Dylan’s hours were 23% lower in March than in February. You may have thought that because there was a 30% increase between Dylan’s January hours (35) and February (45.5) hours, that there would also be a 30% decrease between his February and March hours. As you can see, this assumption is incorrect. The reason is because our original number is different in each case (35 in the first example and 45.5 in the second). This highlights how important it is to make sure you are calculating the percentage from the correct starting point. Sometimes it is easier to show percentage decrease as a negative number – to do this follow the formula above to calculate percentage increase – your answer will be a negative number if there was a decrease. In Dylan’s case the increase in hours between February and March is -10.5 (negative because it is a decrease). Therefore -10.5 ÷ 45.5 = -0.23. -0.23 × 100 = -23%. Dylan's hours could be displayed in a data table as: \| \| \| \| --- \| Month \| Hours Worked \| Percentage Change \| \| January \| 35 \| \| \| February \| 45.5 \| 30% \| \| March \| 35 \| -23% \| Calculating Values Based on Percentage Change Sometimes it is useful to be able to calculate actual values based on the percentage increase or decrease. It is common to see examples of when this could be useful in the media. You may see headlines like: UK rainfall was 23% above average this summer. Unemployment figures show a 2% decline. Bankers’ bonuses slashed by 45%. These headlines give an idea of a trend – where something is increasing or decreasing, but often no actual data. Without data, percentage change figures can be misleading. Ceredigion, a county in West Wales, has a very low violent crime rate. Police reports for Ceredigion in 2011 showed a 100% increase in violent crime. This is a startling number, especially for those living in or thinking about moving to Ceredigion. However, when the underlying data is examined it shows that in 2010 one violent crime was reported in Ceredigion. So an increase of 100% in 2011 meant that two violent crimes were reported. When faced with the actual figures, perception of the amount of violent crime in Ceredigion changes significantly. In order to work out how much something has increased or decreased in real terms we need some actual data. Take the example of “UK rainfall this summer was 23% above average” – we can tell immediately that the UK experienced almost a quarter (25%) more rainfall than average over the summer. However, without knowing either what the average rainfall is or how much rain fell over the period in question we cannot work out how much rain actually fell. Calculating the actual rainfall for the period if the average rainfall is known. If we know the average rainfall is 250mm, we can work out the rainfall for the period by calculating 250 + 23%. First work out 1% of 250, 250 ÷ 100 = 2.5. Then multiply the answer by 23, because there was a 23% increase in rainfall. 2.5 × 23 = 57.5. Total rainfall for the period in question was therefore 250 + 57.5 = 307.5mm. Calculating the average rainfall if the actual amount is known. If the news report states the new measurement and a percentage increase, “UK rainfall was 23% above average... 320mm of rain fell…”. In this example we know the total rainfall was 320mm. We also know that this is 23% above the average. In other words, 320mm equates to 123% (or 1.23 times) of the average rainfall. To calculate the average we divide the total (320) by 1.23. 320 ÷ 1.23 = 260.1626. Rounded to one decimal place, the average rainfall is 260.2mm. The difference between the average and the actual rainfall can now be calculated: 320 - 260.2 = 59.8mm. We can conclude that 59.8mm is 23% of the average rainfall amount (260.2mm), and that in real terms, 59.8mm more rain fell than average. 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17122	https://adac.ee/index.php/ma/article/download/134/83	©2023 Ada Academica Eur. J. Math. Anal. 3 (2023) 7 doi: 10.28924/ada/ma.3.7 A Note on the Stability of Functional Equations via a Celebrated Direct Method Dongwen Zhang 1, John Michael Rassias 2, Qi Liu 3, Yongjin Li 4,∗ 1 School of Mathematics (Zhuhai), Sun Yat-sen University, Zhuhai 519082, P.R. China zhangdw25@mail2.sysu.edu.cn 2 National and Kapodistrian University of Athens, Department of Mathematics and Informatics, Attikis 15342, Greece jrassias@primedu.uoa.gr 3 School of Mathematics and Physics, Anqing Normal University, Anqing 246133, P.R. China liuq325@mail2.sysu.edu.cn 4 Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, P.R. China stslyj@mail.sysu.edu.cn ∗ Correspondence: stslyj@mail.sysu.edu.cn Abstract. More than ten years after Justyna Sikorska attempted to solve the Heyers-Ulam sta-bility of a single variable equation by using direct method. In this paper, we will improve the results of Justyna Sikorska by using a more efficient approach. Relations between the generalized functional equation, the dependence of their different parameters and several properties are also further ex-plored. To achieve the problem, we try to develop some new techniques to overcome the fundamental difficulties caused by the different properties of the function and the presence of several variables in the equation. Furthermore, we continue to construct and study a couple of functional equations by making a new direct method. Introduction The core idea of the Hyers-Ulam stability for functional equations has been dated back to a well-known problem concerning about group homomorphisms solved by S.M. Ulam and D.H. Hyers (see [1–3]). In the last decades, a great number of papers treating the stability problem about functional equations has already been achieved and a great deal of important problems about this field has been studied ( [4–7]). It follows that the most efficient methods have been stated in many papers ( [10, 18–24, 27]) such as the direct approach, the shadowing approach, and invariant mean approach and so on. In particular, the direct method is always the main studying tool on the investigation of functional equations of different types. Received: 15 Sep. 2022. Key words and phrases. Stability; Several functional equations; Approximations; Odd function; Even function. 1Eur. J. Math. Anal. 10.28924/ada/ma.3.7 2 The stability problems for an appropriate simple variable functional equations have earlier been investigated by direct method. The direct method is familiar with many readers to derive the solutions of the equations. The author in have made full use of quite a general way to solve the Hyers-Ulam stability problems on the functional equations under which many excellent outcomes have been achieved without reduplicating the similar procedure in the whole process of computation. However, her results can only be used to derive the solutions of the equation where the mediate function is odd. This is exactly our contribution to the paper. In fact, a straightforward observation is that the inequality ‖f (x) − uf (e(x)) − v f (−e(x)) ‖ 6 δ(x) can be solved if the function h is even. Next, the present studying approach calls us to investigate the following functional inequality, by using a direct method, under which the result can not be covered by earlier works ‖f (x + y + z) + f (x) + f (y ) + f (z) − f (x + y ) − f (z + y ) − f (x + z)‖ 6 K (‖x‖r + ‖y ‖r + ‖z‖r ) . (1.1) In fact, the functional inequality (1 .1) comes from some equivalent characterizations of Hilbert space in . The investigator described several properties of an inner product space and applies these results to solve many interesting functional inequalities such as: Zarantone’s inequality, Hayashi’s inequality and so on. However, the more far reaching work can be done M. Fréchet in under which he ascertained that the corresponding equation is a necessary prerequisite condition when complex or real normed completed spaces become Hilbert spaces. Investigator in studied the stability of Fréchet functional equation from which a characterization of inner product spaces had been achieved by using a stationary point theorem in Banach spaces. Compared with the before studying approaches, we further explored solutions of the equation (1 .1) in this literature. Of course, to the best of our knowledge, it has also already been solved by under which a direct method was to derive solutions of the equation (1 .1) and to look for some improvement approximations. However, in this literature we make a new direct method to achieve the solution of equation (1 .1) must be close to the approximate solution, approximately satisfying the corresponding equation. Besides this, we will consider that the functions on the functional equation of different types have been defined in a more general domain. For instance, the papers [11, 12] have defined an additive ρ-functional inequalities in nonArchimedean normed spaces and Banach spaces. However, this phenomenon can not attract enough attention to the study of functional equations in the more general and complex nonlinear structure of F-spaces (see the definition in [13, 14]). But, the nonlinear structure of space has always stood in a very important position of leadership in functional analysis. Based on the above analysis, it is of great significance that the functional inequality is considered in β-homogeneous F-space. Eur. J. Math. Anal. 10.28924/ada/ma.3.7 3 In section 2, the counterpart of Theorem 2.1 from where the mediate function is odd will be considered. In the subsequent part, a new direct method for solving (1 .1) in F-space will be described and some new extended results of Theorem 2.1 from will be presented. With it, two new different applications of the results will be described in the final part. 2. A simple variable of abstract equation In Theorem 2.1 from , Sikorska solved the equation (2 .1) where the mediate function e is odd and the related parameters u, v are restricted on the real field. For simplicity in notation, we provide traditionally our first result with the studying mapping defined in Banach space. By making use of small conjectures the more general form of the results will be provided in β-homogeneous F -space in section 3. Therefore, our first result is simply considered in Banach space. Theorem 2.1 Let (X, +) be a group, and (Y, ‖ · ‖ ) be a Banach space, and assume the mapping f : X → Y satisfying the inequality ‖f (x) − uf (e(x)) − v f (−e(x)) ‖ 6 δ(x), x ∈ X, (2.1) where u, v ∈ (−∞ , +∞), and the mappings e : X → X, δ : X → [0 , ∞) satisfy that e is even ( i.e., e(−x) = e(x) for every x ∈ X). Let the infinite progression ∑∞ n=0 [\|un\| δ (en(x)) + \|vn\|δ (−en(x))] with u0 := 1 , un := [u(u + v )n−1] ,v0 := 0 , vn := [v (u + v )n−1] , n ∈ N (em states the m-th composition of function e ), be assumed convergence for every x ∈ X. Then there has a unique even mapping g : X → Y satisfying g(x) = ung(en(x)) + vng(−en(x)) , and ‖f (x) − g(x)‖ 6 ∞ ∑ i=0 [\|ui \| δ (ei (x)) + \|vi \| δ (−ei (x))] , x ∈ X and n ∈ N. (2.2) Proof. We will prove that ‖f (x) − unf (en(x)) − vnf (−en(x)) ‖ 6 γn(x), x ∈ X, (2.3) where γn(x) := n−1 ∑ i=0 [\|ui \| δ (ei (x)) + \|vi \| δ (−ei (x))] , x ∈ X, n ∈ N. First of all, consider with every m, n ∈ N and it is easy to observe that un+1 = uu n + uv n, vn+1 = v v n + v u n, and un+m = umun + vmun, vn+m = umvn + vmvn. (2.4) Eur. J. Math. Anal. 10.28924/ada/ma.3.7 4 From the definition of sequences (un) and (vn) we also have uv n = v u n, v mun = umvn. First, (2.1) gives (2.3) with setting n = 1 , and by mathematical induction, later we suppose that (2 .3) establishes for some n ∈ N. We prove that in the case for n + 1 by virtue of (2 .1) ‖f (x) − un+1 f (en+1 (x)) − vn+1 f (−en+1 (x)) ‖ 6 ‖f (x) − unf (en(x)) − vnf (−en(x)) ‖ \|un\| ∥∥f (en(x)) − uf (en+1 (x)) − v f (−en+1 (x))∥ ∥ \|vn\| ∥∥f (−en(x)) − uf (en+1 (x)) − v f (−en+1 (x))∥ ∥ 6 n−1 ∑ i=0 [\|ui \| δ (ei (x)) + \|vi \| δ (−ei (x))] + \|un\| δ (en(x)) + \|vn\| δ (−en(x)) = n ∑ i=0 [\|ui \| δ (ei (x)) + \|vi \| δ (−ei (x))] . Since the series ∑∞ i=0 [\|ui \| δ (ei (x)) + \|vi \| δ (−ei (x))] is convergent for every x ∈ X, combined with (2.3) and by virtue of the completeness of Y , the mapping can be well defined as in the following: g(x) := lim n→∞ [unf (en(x)) + vnf (−en(x))] , x ∈ X, (2.5) and we prove the following properties of the function g.An easy computation is to prove that ug (e(x)) + v g (−e(x)) = u lim n→∞ [unf (en+1 (x)) + vnf (−en+1 (x))] + v lim n→∞ [unf (en+1 (x)) + vnf (−en+1 (x))] = lim n→∞ [(uu n + v u n) f (en+1 (x)) + ( uv n + v v n) f (−en+1 (x))] = g(x). Furthermore, we will prove the more general property of gg(x) = ung (en(x)) + vng (−en(x)) , f or all x ∈ X and n ∈ N. (2.6) By induction, we assume that the equation is true for all natural number k with k ≤ n for some n ∈ N. Let us calculate with k = n + 1 g(x) = ung (en(x)) + vng (−en(x)) = un(ug (en+1 (x)) + v g (−en+1 (x))) + vn(ug (en+1 (x)) + v g (−en+1 (x)))= un+1 g (en+1 (x)) + vn+1 g (−en+1 (x))). In particular, we also have that the function g is even. An easy computation is to state that g(−x) = ung (en(x)) + vng (−en(x)) = g(x), f or ev er y x ∈ X and n ∈ N.Eur. J. Math. Anal. 10.28924/ada/ma.3.7 5 In order to achieve the uniqueness of g, suppose further that ¯g : X → Y is the another mapping such that (2 .2) and (2 .6) hold. Then ‖g(x) − ¯g(x)‖ 6 2 ∞ ∑ i=0 [\|ui \| δ (ei (x)) + \|vi \| δ (−ei (x))] , x ∈ X. Moreover, we have g(x) − ¯g(x) = un [g (en(x)) − ¯g (en(x))] + vn [g (−en(x)) − ¯g (−en(x))] , x ∈ X, and on account of (2 .4) and (2 .6) we can rewrite ‖g(x) − ¯g(x)‖ 6 \|un + vn\| ‖ g (en(x)) − ¯g (en(x)) ‖ 6 2\|un + vn\| ∞ ∑ i=0 [\|ui \| δ (ei+n(x)) + \|vi \| δ (−ei+n(x))] = 2 ∞ ∑ i=0 [( \|ui (un + vn)\|) δ (ei+n(x)) + \|vi (un + vn)\|δ (−ei+n(x))]= 2 ∞ ∑ i=0 [\|ui+n\| δ (ei+n(x)) + \|vi+n\| δ (−ei+n(x))] = 2 ∞ ∑ j=n [∣ ∣uj ∣∣ δ (ej (x)) + ∣∣vj ∣∣ δ (−ej (x))] for every x ∈ X and n ∈ N, where it states that g = ¯ g as n → ∞ . This proves the theorem. The purpose of stating and proving this results is of particular interest and give out a solution of a simple variable functional equation (2 .1) at least. In section 3, we will extend the results of Theorem 2.1 form to a more general setting. In particular, the related parameters u, v can be extended to complex numbers. According to the above analysis, we give out a corollary of Theorem 2.1 (still quite general). First of all, we must state that the absolute of an element x ∈ X can be given out in the real field considering that the function h is even for the meaningful of the results, for example h(x) = a\|x\|.As a matter of fact, we can also present the absolute of x = ( x1, x 2, · · · , x n) ∈ Rn by \|x\| =(\|x1\|, \|x2\|, · · · , \|xn\|). Thus, it worth stating the results. In particular, we can present the following results in the Euclidean space if the more general setting can not be judged. Corollary 2.1 Assume that (X, +) is a real or complex normed linear space and set (Y, ‖ · ‖ ) is a Banach space. Suppose further that the mapping f : X → Y fulfils the inequality ∥∥∥∥f (x) − a + 1 2a2 f (a\|x\|) + a − 12a2 f (−a\|x\|) ∥∥∥∥ 6 δ(x), x ∈ X, (2.7) where a ∈ R with a > 1 and mappings e : X → X, δ : X → [0 , ∞) make that e is an even function ( i.e., e(−x) = e(x) for every x ∈ X). The infinite progression ∑∞ i=0 1 ai δ (ai \|x\|) is convergence for Eur. J. Math. Anal. 10.28924/ada/ma.3.7 6 every x ∈ X. Then there has a unique mapping g : X → Y fulfilling the following equations for all x ∈ Xg(x) = a + 1 2a2 g(a\|x\|) − a − 12a2 g(−a\|x\|), and ‖f (x) − g(x)‖ 6 ∆( x) + Λ( x), where ∆( x) := 12 ∑∞ i=0 1 ai [δ (ai \|x\|) + δ (−ai \|x\|)] , Λ( x) := 12 ∑∞ i=0 1 a2i [δ (ai \|x\|) − δ (−ai \|x\|)] , x ∈ X. Furthermore, g can be obtained in the following limiting equality g(x) := lim n→∞ ( an + 1 2a2n f (an\|x\|) − an − 12a2n f (−an\|x\|) ) , x ∈ X. Proof. By using the results of Theorem 2.1, u := 1+ a 2a2 , v := 1−a 2a2 and together e(x) := a\|x\|, for all x ∈ X, a computation is to prove that un := 1 + a 2a2n , vn := 1 − a 2a2n , n ∈ N. For the convergent series ∑∞ i=0 1 ai δ (ai \|x\|) with x ∈ X, therefore ∑∞ i=0 1 a2i δ (ai \|x\|) is convergence. Applying Theorem 2.1, there has a unique limiting function g : X → Y fulfilling ‖f (x) − g(x)‖ 6 ∞ ∑ i=0 [∣ ∣∣∣ 1 + ai 2a2i ∣∣∣∣ δ (ai \|x\|) + ∣∣∣∣ 1 − ai 2a2i ∣∣∣∣ δ (−ai \|x\|)] = ∞ ∑ i=0 1 a2i [δ (ai \|x\|) − δ (−ai \|x\|)] + ∞ ∑ i=0 1 ai [δ (ai \|x\|) + δ (−ai \|x\|)] = Λ( x) + ∆( x). Function g has been dated back to derived from (2 .5) . We complete the proof. Remark 2.2 If u = 1 , v = 0 and e(x) = \|x\|, the above results may be trivial and meaningless. In the above results, suppose that a ∈ (−∞ , ∞) which is not equal to −1, 0, 1. Exchanging a with −a, this transformation may not be different from primary inequality (2 .7) . This is a basic fact leaving to the reader to check it. Assume that the convergent series ∑∞ i=0 1 \|a\|i δ (\|a\|i \|x\|) establishes for every x ∈ X. In fact, the assertions with \|a\| exchanging for a has also been achieved by a similar way. Corollary 2.2 Assume that (X, +) is a group and set (Y, ‖ · ‖ ) is a Banach space. Suppose further that the mapping f : X → Y fulfils the inequality ∥∥∥∥f (x) − a + 1 2a2 f (a\|x\|) + a − 12a2 f (−a\|x\|) ∥∥∥∥ 6 δ, x ∈ X, Eur. J. Math. Anal. 10.28924/ada/ma.3.7 7 where a ∈ (−∞ , ∞) with \|a\| > 1 and δ > 0 is constant. Then there is a unique limiting even function g : X → Y fulfilling ‖f (x) − g(x)‖ 6 \|a\|δ \|a\| − 1 . Proof. Since the function δ is a positive constant, thus ∆( x) = \|a\|\|a\|− 1 δ and Λ( x) = 0 for every x ∈ X. The mapping g has been stated in the following shape: g(x) := lim n→∞ ( \|a\|n + 1 2a2n f (\|a\|n\|x\|) − \|a\|n − 12a2n f (−\| a\|n\|x\|) ) , x ∈ X. This proves the proof. Remark 2.3 The above corollaries 2.1 and 2.2 will still establish in β-homogeneous F -space with a ∈ (−∞ , ∞) and \|a\| > 1. If we exchange a for 1 a in the equation f (x) − a + 1 2a2 f (a\|x\|) + a − 12a2 f (−a\|x\|) from (2 .7) , the second group of results will also be obtained with a is a positive constant stated in the following results. Corollary 2.3 Assume that (X, +) is a group divisible by a with a ∈ (−∞ , ∞) and \|a\| > 1 and set (Y, ‖ · ‖ ) is a Banach space. Suppose further that the mapping f : X → Y fulfils the inequality ∥∥∥∥f (x) − a2 + a 2 f ( 1 a \|x\| ) − a2 − a 2 f ( − 1 a \|x\| )∥ ∥∥∥ 6 δ(x), x ∈ X, with δ : X → [0 , ∞) is such that the convergent series ∑∞ i=0 a2i δ ( 1 ai \|x\|) holds for every x ∈ X.Then there has a unique even limiting mapping g : X → Y fulfilling for every x ∈ X, g(x) = a2 + a 2 g ( 1 a \|x\| ) a2 − a 2 g ( − 1 a \|x\| ) , and ‖f (x) − g(x)‖ 6 ˜∆( x) + ˜Λ( x). Furthermore, the mapping g can be stated in the following shape: g(x) := lim n→∞ [ a2n + an 2 f ( 1 an \|x\| ) a2n − an 2 f ( − 1 an \|x\| )] , x ∈ X. Proof. Applying for Theorem 2.1 for u := a2+a 2 , v := a2−a 2 and e(x) := 1 a \|x\|, for all x ∈ X, an easy computation is to show that un := a2n + a2n−1 2 , vn := a2n − a2n−1 2 , n ∈ N. According to the convergent series ∑∞ i=0 a2i δ ( 1 ai \|x\|) for all x ∈ X, hence the series ∑∞ i=0 a2i−1δ ( 1 ai \|x\|) is convergence, and there has a unique even limiting mapping g : X → YEur. J. Math. Anal. 10.28924/ada/ma.3.7 8 fulfilling ‖f (x) − g(x)‖ 6 ∞ ∑ i=0 [∣ ∣∣∣ a2i + ai 2 ∣∣∣∣ δ ( 1 ai \|x\| ) + ∣∣∣∣ a2i − ai 2 ∣∣∣∣ δ ( − 1 ai \|x\| )] = ∞ ∑ i=0 a2i 2 [ δ ( 1 ai \|x\| ) δ ( − 1 ai \|x\| )] + ∞ ∑ i=0 ai 2 [ δ ( 1 ai \|x\| ) − δ ( − 1 ai \|x\| )] = ˜∆( x) + ˜Λ( x). The definition of g is derived from (2 .5) . We complete the proof. Remark 2.4 Corollary 2.3 can be used to investigate the function from which it could be split into even and odd parts. There is a good point of the approach achieved here where the functions split into two two parts of odd and even functions can give more concise approximations than the before approximations in Theorem 2.1. The above results is the counterpart of the corresponding results of Sikorska’s paper. However, it is not copied word by word. It is the counterpart of even function. 3. The stability of functional equations in F-space An F -space is called β-homogeneous if it satisfies ‖tx ‖ = \|t\|β ‖x‖ for every x ∈ X, t ∈ C. In this section of the first two theorems, β1, β 2 are to be 0 < β 1 ≤ 1 and 0 < β 2 ≤ 1. Furthermore, we suppose X is β1-homogeneous F-space and Y is β2-homogeneous F-space. Before applying Theorem 2.1 we would like to make an answer that all roads lead to Rome. Therefore another approach to prove the following functional inequality has been stated in the following. In fact, there is also a similar solution about functional equation being stated in . Theorem 3.1 Assume the mapping f : X → Y fulfilling for some K ≥ 0 and r < β2 β1 ‖f (x + y + z) + f (x) + f (y ) + f (z) − f (x + y ) − f (z + y ) − f (x + z)‖ 6 K (‖x‖r + ‖y ‖r + ‖z‖r ) (3.1) for x, y , z ∈ X. Then there has a unique limiting mapping ψ1 : X → Y such that ‖f (x) − ψ1(x)‖ 6 (2 + 2 r β 1 + 3 · 2β2 )K (2 β1r − 22β2 )(2 β1r − 2β2 ) ‖x‖r for x ∈ X. Moreover, ψ1 satisfying the above inequality is also satisfying the following equation ψ1(x + y + z) + ψ1(x) + ψ1(z) + ψ1(y ) = ψ1(x + y ) + ψ1(z + y ) + ψ1(x + z) (3.2) for all x, y , z ∈ X. Proof. From (x, x, x ) in place of (x, y , z ) in (3 .1) we have ‖f (3 x) + 3 f (x) − 3f (2 x)‖ 6 3K (‖x‖r ) .Eur. J. Math. Anal. 10.28924/ada/ma.3.7 9 Hence ‖2f (3 x) + 6 f (x) − 6f (2 x)‖ 6 3 · 2β2 K (‖x‖r ) . Substitute (x, x, 2x) in place of (x, y , z ) in (3 .1) , yielding that ‖f (4 x) + 2 f (x) − 2f (3 x)‖ 6 (2 + 2 r β 1 )K (‖x‖r ) . And combining the above two inequalities, we get ‖f (4 x) + 8 f (x) − 6f (2 x)‖ 6 (2 + 2 r β 1 + 3 · 2β2 )K (‖x‖r ) . (3.3) Let us define g(x) = f (2 x) − 4f (x) for all x ∈ X. Hence ‖g(2 x)/2 − g(x)‖ 6 (2 + 2 r β 1 + 3 · 2β2 )K (‖x‖r ) /2β2 (3.4) for all x ∈ X. Therefore ‖g(2 nx)/2n − g(2 mx)/2m‖ 6 n−1 ∑ j=m (2 + 2 r β 1 + 3 · 2β2 )K 2jβ 1r 2β2 2jβ 2 (‖x‖r ) (3.5) for m, n ∈ N with n > m and all x ∈ X. Since the sequence {g(2 nx)/2n} is a Cauchy sequence in Y for all x ∈ X and Y is complete, the mapping can be well defined as: φ(x) = lim n→∞ g(2 nx)/2n for all x ∈ X. In particular, letting m = 0 and setting n → ∞ in (3 .4) , we have ‖φ(x) − g(x)‖ 6 (2+2 r β 1 +3 ·2β2 )K 2β1r−2β2 (‖x‖r ) . (3.6) Now, we prove the mapping φ is additive and is unique. From (x, y , y + x) in (3 .1) yields that ‖f (2 x + 2 y ) + f (x) + f (y ) − f (x + 2 y ) − f (2 x + y )‖ 6 K (‖x‖r + ‖y ‖r + ‖x + y ‖r ) . From (x, x, y ) in equation (3 .1) yields that ‖f (2 x + y ) + 2 f (x) + f (y ) − f (2 x) − 2f (x + y )‖ 6 K (2 ‖x‖r + ‖y ‖r ) . From (x, y , y ) in (3 .1) we have ‖f (x + 2 y ) + f (x) + 2 f (y ) − f (2 y ) − 2f (x + y )‖ 6 K (‖x‖r + 2 ‖y ‖r ) .Eur. J. Math. Anal. 10.28924/ada/ma.3.7 10 Combining the above three inequalities, we have that for x, y , z ∈ X ‖φ(x + y ) − φ(x) − φ(y )‖ = lim n→∞ 12β2n ‖f (2 n+1 x + 2 n+1 y ) + 4 f (2 nx) + 4 f (2 ny ) − f (2 n+1 x) − f (2 n+1 y ) − 4f (2 nx + 2 ny )‖ 6 lim n→∞ 12β2n ∥∥f (2 n+1 x + 2 n+1 y ) + f (2 nx) + f (2 ny ) − f (2 n+1 x + 2 ny ) − f (2 n+1 y + 2 nx)∥∥ lim n→∞ 12β2n ∥∥f (2 n+1 x + 2 ny ) + 2 f (2 nx) + f (2 ny ) − f (2 n+1 x) − 2f (2 nx + 2 ny )∥∥ lim n→∞ 12β2n ∥∥f (2 nx + 2 n+1 y ) + f (2 nx) + 2 f (2 ny ) − f (2 n+1 y ) − 2f (2 nx + 2 ny )∥∥ 6 lim n→∞ 2β1r n 2β2n K (4 ‖x‖r + 4 ‖y ‖r + ‖x + y ‖r ) . So we have φ(x + y ) = φ(x) + φ(y ) for all x, y ∈ X.Next, the uniqueness of the mapping φ will be proved. Let u(x) be another additive mapping such that for some K2 ≥ 0 and r < β2 β1 , ‖g(x) − u(x)‖ 6 K2‖x‖r2 . Hence ‖φ(x) − u(x)‖ =‖φ(nx ) − u(nx )‖/n β2 6‖φ(nx ) − g(nx )‖/n β2 + ‖g(nx ) − u(nx )‖/n β2 6 (2 + 2 r β 1 + 3 · 2β2 )K 2β1r − 2β2 ‖x‖r nr β 1−β2 + K2‖x‖r2 nr2β1−β2 for all x ∈ X. Therefore φ(x) = u(x) for all x ∈ X. by the condition r < β2 β1 . So there has a unique additive limiting mapping φ fulfilling ‖(f (x) − 12 φ(x)) − (f (2 x) − 12 φ(2 x)) /4‖ ≤ (2 + 2 r β 1 + 3 · 2β2 )K 2β1r − 2β2 ‖x‖r /22β2 . Hence ‖(f (x) − 12 φ(x)) − (f (2 nx) − 12 φ(2 nx)) /4n‖ ≤ n−1 ∑ j=0 2β1r j 22β2j (2 + 2 r β 1 + 3 · 2β2 )K 22β2 (2 β1r − 2β2 ) ‖x‖r . Then the mapping can be well defined as ψ(x) = lim n→∞ (f (2 nx) − 12 φ(2 nx)) /4n for all x ∈ X, by the completeness of the space Y . Thus ‖ψ(x) − f (x) + φ(x)/2‖ ≤ (2 + 2 r β 1 + 3 · 2β2 )K (2 β1r − 2β2 )(2 β1r − 22β2 ) ‖x‖r . Let u(x) be another limiting mapping which has the same property to the function ψ(x) such that, ‖u(x) − φ(x)‖ 6 ‖u(x) − (f (2 nx) − 12 φ(2 nx)) /4n‖ + ‖(f (2 nx) − 12 φ(2 nx)) /4n − φ(x)‖ 6 2 ∞ ∑ j=n 2β1r j 22β2j (2 + 2 r β 1 + 3 · 2β2 )K 22β2 (2 β1r − 2β2 ) ‖x‖rEur. J. Math. Anal. 10.28924/ada/ma.3.7 11 which shows that the approximation function φ(x) is unique. Finally, it remains to prove that φ(x) satisfies (3 .2) and we obtain 14n ‖f (2 nx + 2 ny + 2 nz) + f (2 nx) + f (2 ny ) + f (2 nz)− f (2 nx + 2 ny ) − f (2 nz + 2 ny ) − f (2 nx + 2 nz)‖ 6 2β1n 4n K (‖x‖r + ‖y ‖r + ‖z‖r ) . (3.7) Letting n → ∞ , and we get our assertion by using the additivity of φ(x). In another direction, we will describe the similar stability results of the above Theorem 3.1. Theorem 3.2 Let r > β2 β1 and assume that f : X → Y is a mapping satisfying the equation (3 .1) .Then there has a unique limiting mapping ψ1 : X → Y satisfying ‖f (x) − ψ1(x)‖ 6 (2 + 2 r β 1 + 3 · 2β2 )K (2 β1r − 22β2 )(2 β1r − 2β2 ) ‖x‖r for all x ∈ X. Moreover, ψ1 solves also the following equation ψ1(x + y + z) + ψ1(x) + ψ1(z) + ψ1(y ) = ψ1(x + y ) + ψ1(z + y ) + ψ1(x + z) (3.8) for all x, y , z ∈ X. Proof. According to the equation (3 .3) , we obtain ‖g(x) − 2g( x 2 )‖ 6 (2 + 2 r β 1 + 3 · 2β2 )K (‖x‖r ) /2β1r . Therefore ‖2ng( x 2n ) − 2mg( x 2m )‖ 6 n−1 ∑ j=m (2 + 2 r β 1 + 3 · 2β2 )K 2jβ 2 2jβ 1r 2β1r (‖x‖r ) for m, n ∈ N with n > m and x ∈ X. Since the sequence {2ng( x 2n )} is a Cauchy sequence in Y for all x ∈ X and Y is complete, the mapping can be well defined as: φ(x) = lim n→∞ 2ng( x 2n ) for all x ∈ X. Using a similar manner, we can complete the rest part. If f (x) is odd, then (x, y , −x − y ) in (3.2) can give a precise condition to ascertain the additive property of the function f (x) (See ). Obviously, the additive property is stronger than the property of the equation (3 .2) , but vice versa is not true. In contrast with the subadditive property, we can not get obvious strong or weak property temporarily. By using another approach to solve the Theorem 3.1, according to (3 .6) , we have ‖f (2 nx)/22n − f (x) − n−1 ∑ j=0 φ(2 j x)/22( j+1) ‖ ≤ n−1 ∑ j=0 2β1r j 22β2j (2 + 2 r β 1 + 3 · 2β2 )K 22β2 (2 β1r − 2β2 ) ‖x‖r . Then the mapping can be well defined as ψ(x) = lim n→∞ f (2 nx)/22nEur. J. Math. Anal. 10.28924/ada/ma.3.7 12 for all x ∈ X, by the completeness of the space Y . Thus ‖ψ(x) − f (x) − φ(x)/2‖ ≤ (2 + 2 r β 1 + 3 · 2β2 )K (2 β1r − 2β2 )(2 β1r − 22β2 ) ‖x‖r . In a similar way, we can use two steps to prove that the mapping ψ(x) is unique. The first step we show that the mapping satisfies the property: ψ(kx ) = k2ψ(x) for all k ∈ N, x ∈ X. We prove this by mathematical induction, for a fixed element x ∈ X. We will prove that the property is true for k = 2 . From (x, −x, x ) in equation (3 .1) , we can get that ‖3f (x) + f (−x) − f (2 x)‖ 6 3K (‖x‖r ) for all x ∈ X. Thus ‖f (−x) − f (x) − φ(x)‖ ≤ ‖ f (2 x) − 4f (x) − φ(x)‖ + ‖3f (x) + f (−x) − f (2 x)‖ 6 ( (2 + 2 r β 1 + 3 · 2β2 )K 2β1r − 2β2 + 3 K) ( ‖x‖r ) for all x ∈ X. Using the similar above argumentation together the above inequality and equation (3 .1) , yields ψ(−x) = ψ(x) + lim n→∞ φ(x)2n and ψ(x + y + z) + ψ(x) + ψ(z) + ψ(y ) = ψ(x + y ) + ψ(z + y ) + ψ(x + z) (3.9) for all x, y , z ∈ X. From (x, −x, x ) in equation (3 .7) , we achieve ψ(2 x) = 3 ψ(x) + ψ(−x) = 4 ψ(x). Fixed x ∈ X, we prove this by induction. We have already proved that the property is true for n = 2 . Supposing that ψ(nx ) = n2ψ(x) for all natural n ≤ 2k, with k ≥ 1, let us calculate ψ((2 k + 1) x). From (kx, kx, x ) in (3.7), we know ψ((2 k + 1) x) = ψ(2 kx ) + 2 ψ(( k + 1) x) − 2ψ(kx ) − ψ(x)= (4 k2 + 2( k + 1) 2 − 2k2 − 1) ψ(x)= (2 k + 1) 2ψ(x). Now, we show the mapping ψ satisfies the property ψ(kx ) = k2ψ(x) for all k ∈ N, x ∈ X. The second step, we claim that the mapping φ is unique. Let u(x) be another limiting mapping such that for some K2 ≥ 0 and r < β2 β1 , ‖u(x) − f (x) − φ(x)/2‖ 6 K2‖x‖r2Eur. J. Math. Anal. 10.28924/ada/ma.3.7 13 which satisfies the property u(kx ) = k2u(x) for all k ∈ N and x ∈ X. Therefore ‖ψ(x) − u(x)‖ =‖ψ1(kx ) − u(kx )‖/k 2β2 6‖u(xk ) − f (kx ) − φ(kx )/2‖/k 2β2 + \|ψ(xk ) − f (kx ) − φ(kx )/2‖/k 2β2 6 (2 + 2 r β 1 + 3 · 2β2 )K (2 β1r − 22β2 )(2 β1r − 2β2 ) ‖x‖r kr β 1−2β2 + \|K2‖x‖r2 kr2β1−2β2 . Hence φ(x) = u(x) for all x ∈ X. This shows that ψ is unique. Let ψ1(x) = ψ(x) − φ(x)/2. This completes the uniqueness of ψ1(x). We have ‖ψ1(x) − f (x)‖ 6 (2 + 2 r β 1 + 3 · 2β2 )K (2 β1r − 22β2 )(2 β1r − 2β2 ) ‖x‖r for all x ∈ X and also the equation (3 .2) holds by using the additive property of φ and equation (3 .7) . We complete the proof. We may also assume that lim n→∞ φ(x)2n = lim n→∞ g(2 n x)/2n 2n = 0 .Otherwise, this limit may not be convergence to zero. Conversely, we may add some similar small additional assumptions to guarantee the convergence in Theorem 3.2. 4. The stability of functional equations in Banach space In this section, we will prove the counterpart of the results of Theorem 2.1 from to more general case. We generalize the results of Sikorska in 2010. In particular, the related parameters u, v can be extended to complex numbers by using a more efficient approach. Beyond that, we state that the first results in section 2 are presented and combined the first results in . Our contribution to the parameters u, v are complex numbers. The results is stated in this section in more detail. Theorem 4.1 Suppose that (X, +) is a group, and (Y, ‖ · ‖ ) is a Banach space, and let the mapping f : X → Y satisfy the inequality ‖f (x) − uf (e(x)) − v f (−e(x)) ‖ 6 δ(x), x ∈ X, where u, v ∈ C (C denotes the complex field.), and e : X → X, δ : X → [0 , ∞) are arbitrary given functions. (1): If e is a even function ( i.e., e(−x) = e(x) for x ∈ X) and the convergent series ∞ ∑ n=0 [\|un\| δ (en(x)) + \|vn\|δ (−en(x))] with u0 := 1 , un := [u(u + v )n−1] , n ∈ N,v0 := 0 , vn := [v (u + v )n−1] , n ∈ NEur. J. Math. Anal. 10.28924/ada/ma.3.7 14 (and where en states the n-th composition of the function e ), establishes for every x ∈ X. Then there has a unique even limiting function g : X → Y fulfilling g(x) = ung(en(x)) + vng(−en(x)) , x ∈ X and n ∈ N, (4.1) and ‖f (x) − g(x)‖ 6 ∞ ∑ i=0 [\|ui \| δ (ei (x)) + \|vi \| δ (−ei (x))] , x ∈ X. (4.2) (2): If e is odd ( i·e., e(−x) = −e(x) for all x ∈ X) and the convergent series ∞ ∑ n=0 [\|un\| δ (en(x)) + \|vn\| δ (−en(x))] . with u0 := 1 , un := 12 [( u + v )n + ( u − v )n] , n ∈ N,v0 := 0 , vn := 12 [( u + v )n − (u − v )n] , n ∈ N establishes for all x ∈ X. Then there has a unique limiting mapping g : X → Y fulfilling (3 .10) and (3 .11) . Proof. We only need to prove the uniqueness of the approximation function. (1): Let us suppose that ˜g : X → Y is another approximating mapping. So let’s first prove the inequality together with the equation (2 .5) and g(−x) = g(x) ‖f (em(x)) − um(unf (en+m(x)) + vnf (−em+n(x))) − vm(unf (en+m(x)) + vnf (−em+n(x)))‖ =‖f (em(x)) − un+mf (en+m(x)) − vn+mf (−en+m(x)) ‖ 6 n+m−1 ∑ j=m [∣ ∣uj ∣∣ δ (ej (x)) + ∣∣vj ∣∣ δ (−ej (x))] , and letting n → ∞ we have for any m ∈ N ‖f (em(x)) − umg (em(x)) − vmg (−em(x)) ‖ 6 ∞ ∑ j=m [∣ ∣uj ∣∣ δ (ej (x)) + ∣∣vj ∣∣ δ (−ej (x))] , and we can rewrite ‖g(x) − ˜g(x)‖ 6‖f (km(x)) − umg (em(x)) − vmg (−em(x)) ‖ ‖f (km(x)) − um ˜g (em(x)) − vm ˜g (em(x)) ‖ 62 ∞ ∑ j=m [∣ ∣uj ∣∣ δ (ej (x)) + ∣∣vj ∣∣ δ (−ej (x))] for any x ∈ X and m ∈ N, which yields g = ˜g in X as m → ∞ .(2): Combined with the results of Theorem 2.1 from where e is odd, we only need to prove the uniqueness of the approximation function. Let us suppose that ˜g : X → Y is another approximating Eur. J. Math. Anal. 10.28924/ada/ma.3.7 15 mapping. So let’s first prove the inequality together with the equation the results in Theorem 2.1 in ‖f (em(x)) − um(unf (en+m(x)) + vnf (−em+n(x))) − vm(unf (−en+m(x)) + vnf (em+n(x)))‖ =‖f (em(x)) − un+mf (en+m(x)) − vn+mf (−en+m(x)) ‖ 6 n+m−1 ∑ j=m [∣ ∣uj ∣∣ δ (ej (x)) + ∣∣vj ∣∣ δ (−ej (x))] , and letting n → ∞ we have for any m ∈ N ‖f (em(x)) − umg (em(x)) − vmg (−em(x)) ‖ 6 ∞ ∑ j=m [∣ ∣uj ∣∣ δ (ej (x)) + ∣∣vj ∣∣ δ (−ej (x))] , and we can rewrite ‖g(x) − ˜g(x)‖ 6‖f (km(x)) − umg (em(x)) − vmg (−em(x)) ‖ ‖f (km(x)) − um ˜g (em(x)) − vm ˜g (em(x)) ‖ 62 ∞ ∑ j=m [∣ ∣uj ∣∣ δ (ej (x)) + ∣∣vj ∣∣ δ (−ej (x))] for any x ∈ X and m ∈ N, which yields g = ˜g in X as m → ∞ . This completes the proof. For the Euler-Lagrange equation, we provide another method to solve it in contrast with . Theorem 4.2 Suppose that (X, +) is a group, and (Y, ‖ · ‖ ) is a Banach space and let the mapping f : X → Y satisfy the inequality for all x, y , z ∈ X and some ε > 0 ‖f (x + y + z) + f (x − y + z) + f (x + y − z) + f (x − y − z) − 4f (x) − 4f (y ) − 4f (z)‖ 6 ε. (4.3) Then there has a unique limiting function g : X → Y such that g(x) = 29 g(3 x) − 19 g(−3x), x ∈ X and ‖f (x) − g(x)‖ 6 3ε 8 x ∈ X. In particular, if X is Abelian, then g is a solution of the equation in the following f (x + y + z) + f (x − y + z) + f (x + y − z) + f (x − y − z) = 4 f (x) + 4 f (y ) + 4 f (y ), (4.4) for all x, y ∈ X. Proof. From (x, x, −x) in (4 .3) , we obtain ‖6f (x) + 3 f (−x) − f (3 x)‖ 6 ε, x ∈ X. Replacing x by −x in the above inequality we obtain ‖6f (−x) + 3 f (x) − f (−3x)‖ 6 ε, x ∈ X. Eur. J. Math. Anal. 10.28924/ada/ma.3.7 16 Consequently, combining the above two inequalities yield that ‖9f (x) + f (−3x) − 2f (3 x)‖ 6 3ε, x ∈ X. By using the results second part of Theorem 4.1, a computation is to prove that un := 3n + 1 2 · 9n , vn := 1 − 3n 2 · 9n , n ∈ N. and the convergent series can be described as ∞ ∑ n=0 [\|un\| δ (en(x)) + \|vn\|δ (−en(x))] = 3ε 8 . And we show that if X is commutative, by using (x, y , z ) = (3 nx, 3ny , 3nz), then ‖un[f (3 n(x + y + z)) + f (3 n(x − y + z)) + f (3 n(x + y − z)) + f (3 n(x − y − z)) − 4f (3 nx) − 4f (3 ny ) − 4f (3 ny )] + vn[f (3 n(x + y + z)) + f (3 n(x − y + z)) + f (3 n(x + y − z)) + f (3 n(x − y − z)) − 4f (3 nx) − 4f (3 ny ) − 4f (3 ny )] ‖ 6 ε 9n which we achieve our result (3 .13) by letting n → ∞ . Theorem 4.3 Suppose that X is a group, and (Y, ‖ · ‖ ) is a Banach space and let the mapping f : X → Y satisfy the inequality for all x, y ∈ X and some ε > 0 ‖f (x + y ) + f (x − y ) − 2f (x) − f (y ) − f (−y )‖ 6 ε, x, y ∈ X. (4.5) Then there has a unique limiting function g : X → Y fulfilling g(x) = 38 g(2 x) − 18 g(−2x), x ∈ X and ‖f (x) − g(x)‖ 6 2ε 3 x ∈ X. In particular, if X is commutative, then g also fulfils g(x + y ) + g(x − y ) = 2 g(x) + g(y ) + g(−y ), x, y ∈ X. Proof. Substituting in the sequel (x, x ) in (4 .5) , we obtain ‖f (2 x) + f (0) − 3f (x) − f (−x)‖ 6 ε, x ∈ X. (4.6) Replacing x by −x in (4 .6) we have ‖f (−2x) + f (0) − 3f (−x) − f (x)‖ 6 ε, x ∈ X. (4.7) Consequently, (4 .6) and (4 .7) yield that ‖8f (x) + f (−2x) − 3f (2 x)‖ 6 4ε, x ∈ X. Eur. J. Math. Anal. 10.28924/ada/ma.3.7 17 By using the results of Theorem 3.3, a computation is to prove that un := 2n + 1 2 · 4n , vn := 1 − 2n 2 · 4n , n ∈ N. and the convergent series ∞ ∑ n=0 [\|un\| δ (en(x)) + \|vn\|δ (−en(x))] = 2ε 3 . And we show that if X is commutative, by using (x, y ) = (2 nx, 2ny ), then ‖un[f (2 nx + 2 ny ) + f (2 nx − 2ny ) − 2f (2 nx) − f (2 ny ) − f (−2ny )] + vn[f (2 nx + 2 ny ) + f (2 nx − 2ny ) − 2f (2 nx) − f (2 ny ) − f (−2ny )] ‖ 6 ε 4n which we achieve our result (∗) by letting n → ∞ . If we can not set f (0) = 0 , then the approximate constat is 56 ε. Acknowledgments The authors express their gratitude to the anonymous reviewers and editor for their careful reading the manuscript and for many valuable remarks and suggestions. Conflict of Interest The author(s) declare(s) that there is no conflict of interest regarding this manuscript. Data Availability Data sharing not applicable to this article as no datasets were generated or analysed during the current study. Funding Statement This work was supported by the National Natural Science Foundation of China (11971493) and (12071491). References S.M. Ulam, Problems in modern mathematics, Science Editions John Wiley and Sons. Inc., New York (1964). S.M. Ulam, A collection of mathematical problems, Interscience Tracts in Pure and Applied Mathematics. no. 8 Interscience Publishers, New York-London (1960). D.H. Hyers, On the stability of the linear functional equation, Proc. Nat. Acad. Sci. U.S.A. 27 (1941) 222-224. . D.H. 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17123	https://mathcenter.oxford.emory.edu/site/math108/exponents/	Powers, Exponents, and their Dependencies About Statistics Number Theory Java Data Structures Cornerstones Calculus Powers, Exponents, and their Dependencies In the context of both braids and permutations, we have seen notations that mirror that of traditional powers of real numbers. In this section, we remind ourselves of what we know about these (multiplicative) powers by exploring their dependencies on the associativity, identities, inverses and commutativity. Doing so justifies the traditional exponent rules for real numbers - but also explains why braid concatenation and permutation composition can be treated so similarly (since these closed operations also are associative, have identities and inverses, and are -- in certain circumstances -- commutative). In all three contexts, we use exponents to more efficiently denote a repeated application of some operation. With braids, we were concatenating a braid to itself some number of times. With permutations, we were composing a permutation with itself over and over. With the more traditional (read "familiar") use of exponents that students have likely already seen, we are multiplying some value by itself several times. There are practical reasons why we developed this notation in the first place. One need only look to some of the values that naturally pop up in the sciences or elsewhere for motivation: the speed of light is roughly 300,000,000 meters per second the atomic mass of an element is given in terms of grams per 602,200,000,000,000,000,000,000 atoms of that element. the number of bits contained in a terabyte of data is 8,796,093,022,208 As was the case with "powers" of braids and permutations, working with the values shown above can be tedious if they are left in these long forms -- especially if the calculations are done by hand. However, with exponents we can of course write each of these in a much better (shorter) form... Note, in the first two examples the long strings of zeros... Recall, we can add zeros to the end of any number by multiplying by some number of 10's. So, for example, we could turn 3 into 300,000,000 by multiplying by ten exactly eight times. Similarly, we could multiply 6.022 by ten exactly twenty three times to arrive at the number used in the second example. Of course, there is nothing inherently special about the value 10, as the third example demonstrates. Here we have a number made solely from a product of two's -- forty three of them, to be exact. Given what we have seen with braids and permutations (and likely what you have seen in previous math courses) it should be no surprise that we adopt the following shorthand: We denote by x n the product resulting from multiplying n values of x together: x n=x⋅x⋅x⋯x⏟n times The following will undoubtedly be familiar to students, but as a matter of verbiage, x n is read as "x raised to the n t h power". We refer to n as the exponent on x, while x itself is called the base. While the choice of the word base probably makes sense (i.e., the number "on the bottom"), one might wonder from where the term exponent came. The first recorded use of this word appears in the book Arithemetica Integra written by Michael Stifel in 1544. In Latin, the prefix ex- means "out (of)", while the word pono is the verb "place". So exponent corresponds directly to the little number "placed outside of" of the base. As a natural extension of this, the act of raising a value to some power is called exponentiation. We call x 2 the square of x, as it equals the area of a square with side length x. Similarly, we call x 3 the cube of x as x 3 equals the volume of a cube with that same side length. The act of finding squares and cubes actually significantly predates the calculation of expressions involving more general exponents. As evidence of this, consider the ancient Babylonian tablet below (known as "Plimpton 322", and dating back to 1800 BC). The depressions made into this clay tablet can be deciphered in a systematic way to numbers, where for any given row the numbers in the second and third columns are part of a Pythagorean triplet (i.e., integers a, b, and c, such that a 2+b 2=c 2). An example demonstrating this for one row of the tablet is highlighted in red and blue below. The term power, which formally represents the resulting value of an exponentiation, connects to these older ideas of square and cube. Specifically, the relationship between different measurable magnitudes (e.g., length, area, volume) in segments, squares, cubes, shows up in the writing of Euclid, an ancient greek mathematician (somewhere around 300 BC). In his writings, he uses the equivalent of the phrase "in power" when he writes (in Greek, of course) that "magnitudes are commensurable in power when their squares are commensurable". Henry Billingsley, the first English translator of Euclid in 1570 cemented this verbiage in his translation of Euclid's Second book, where he writes "The power of a line is the square of the same line." By the way, two values are "commensurable" when their quotient is a rational value (i.e., equivalent to a quotient of two integers). The values of 2 and 1 are provably not commensurable, while their squares 2 and 1 certainly are. But we are getting ahead of ourselves. Wait until we have talked about square roots and why some are irrational, and then come back here and read this again. At the moment, let us assume that n≥2 is an integer, although we will quickly see that more general values of n can be considered. Using exponents, we can write all of the numbers in our earlier examples much more succinctly: the speed of light is roughly 3×10 8 meters per second the atomic mass of an element is given in terms of grams per 6.022×10 23 atoms of that element. the number of bits contained in a terabyte of data is 2 43 Of course, it would be silly to develop this shorthand for writing such products if we needed to go back to the original form every time we needed to use them in a calculation. Just like we saw with braids and permutations, there are some very useful (and very similar) rules that govern how we can combine powers in the context of repeated multiplication. One can combine powers in several ways, of course. We can add two powers together, subtract them, consider their product, or quotient -- we could even raise one power to another. Considering sums and differences of powers at this point in time would likely lead us into a discussion of polynomials. However, we choose to motivate polynomials in a slightly different way in the future -- so for now, let us just consider only products, quotients, and powers... Products and Powers of Powers Consider the following product: x 3⋅x 5=(x⋅x⋅x)⋅(x⋅x⋅x⋅x⋅x)=x⋅x⋅x⋅x⋅x⋅x⋅x⋅x=x 8 Was there really a need to expand the powers above to find the overall product? Absolutely not! We know x 3 contributes three factors of x to the overall product, and x 5 contributes five more, so x 3⋅x 5 must be the product of 3+5=8 factors of x (that is to say, x 8). This suggests the following rule: x m⋅x n=x m+n when m and n are positive integers. There is some potential ambiguity in the above argument though, often neglected by most textbooks. When we write something like x⋅x⋅x⋅x⋅x, in what order do those multiplications occur? Most will recall the so-called "order of operations" from grade school that insists when any parentheses are absent, a string of multiplications (and/or divisions) should be evaluated "left-to-right". As an example to clarify what is meant by this "left-to-right" multiplication, consider 2⋅3⋅4⋅5. First, we multiply 2 and 3 to get 6. Then we multiply that by 4 to get 24. Finally, we multiply that result by 5 to get 120. In other words, "left-to-right" multiplication means we are implicitly grouping things in a "leaning-left" way, as shown below: 2⋅3⋅4⋅5=(((2⋅3)⋅4)⋅5) This means that x m is really the following: x m=x⋅x⋅x⋯x⏟m times=((((x⋅x)⋅x)⋯)⋅x)⏟m occurrences of x So to accept what was done to x 3⋅x 5 above as a general argument for dealing with x m⋅x n, we would need to accept the following: x m⋅x n=(x⋅x⋅x⋯x)⏟m times⋅(x⋅x⋅x⋯x)⏟n times=((((x⋅x)⋅x)⋯)⋅x)⏟m occurrences of x⋅((((x⋅x)⋅x)⋯)⋅x)⏟n occurrences of x=(((((((x⋅x)⋅x)⋯)⋅x)⋅x)⋅x)⋯)⋅x⏟m+n occurrences of x=x⋅x⋅x⋯x⏟m+n times=x m+n But how do we get from the second step above to the third? The key is associativity! Remember, if an operation (like multiplication) is associative, it means (x y)z=x(y z). This in turn allows us to regroup larger products (i.e., products of more than 3 things) and evaluate them in many different ways, without changing the final result. As an example, notice how we can use the associative property as shown below to re-express (((a b)c)d)e as (a b)((c d)e): (((a b)c)d)e=((a b)c)(d e)=(a b)(c(d e))=(a b)((c d)e) In each of the steps above, we apply the associative property (x y)z=x(y z). The reader will benefit from asking what plays the role of x, y, and z in each step. In this light, consider the following argument (that uses only the associative property in each step) for re-expressing x 4⋅x 4 as x 5⋅x 3: x 4⋅x 4=(((x⋅x)⋅x)⋅x)⋅(((x⋅x)⋅x)⋅x)=(((x⋅x)⋅x)⋅x)⋅((x⋅(x⋅x))⋅x)=(((x⋅x)⋅x)⋅x)⋅(x⋅((x⋅x)⋅x))=((((x⋅x)⋅x)⋅x)⋅x)⋅((x⋅x)⋅x)=x 5⋅x 3 This is an incremental step in the right direction, as we have essentially moved one x from the power on the right to the power on the left. However, we can apply the exact same strategy again to move another x from the right to the left. (Can you figure out how?) Consequently, we can rewrite x 5⋅x 3 as x 6⋅x 2, and then (upon doing it again) as x 7⋅x, until finally we exhaust all of the x factors on the right, leaving only (((((((x⋅x)⋅x)⋅x)⋅x)⋅x)⋅x)⋅x)=x 8 This same strategy can be applied more generally, to turn x m⋅x n into x m+1⋅x n−1, and then that into x m+2⋅x n−2, and so on -- until we finally get to x m+n form. In this way, any time we have an associative "multiplication", whether that means normal real number multiplication, or braid concatenation, or permutation composition -- the same argument can be applied. As such, in all these situations one adds exponents when multiplying powers (presuming the exponents are integers): x m⋅x n=x m+n Taking this a bit further, consider a power that is in turn raised to some other power. As long as the "multiplication" in question is associative, the above rule holds for any integers m and n. This then allows us to also argue: (x m)n=x m⋅x m⋅x m⋯⋅x m⏟n times=(((x m⋅x m)⋅x m)⋯)⋅x m⏟n occurrences of x m=((x m+m⋅x m)⋯)⋅x m⏟product of n−1 powers=(x m+m+m+⋯)⋅x m⏟product of n−2 powers=x m+m+⋯+m⏞sum of n terms=x m n which suggests that raising a power to a power can be dealt with by multiplying the exponents: (x m)n=x m n Negative and Zero Exponents The above rules for products of powers, and powers of powers, both hold when the exponents involved are positive integers -- but what about exponents that are zero or negative integers? First, recall that for any real number x, we have x⋅1=1⋅x=x This means that the value 1 plays the role of the multiplicative identity. We have denoted identities for other operations (e.g., braid concatenation, permutation composition) by I in the past -- partly because I is the first letter of "identity" and partly because it looks a bit like a 1 too. Additionally, every non-zero real number x has a multiplicative inverse where x⋅x−1=x−1⋅x=1 As examples, 5−1=0.2 and 0.1−1=10. A similar result holds for braids, where 1 is replaced by the identity braid. The same can be said of permutations when 1 is replaced by the permutation that leaves the order untouched. Consider what this means for x 3, as we appeal to associativity and the properties of the identity and inverses: x 3⋅(x−1)3=(x⋅x⋅x)⋅(x−1⋅x−1⋅x−1)=(x⋅x)⋅(x⋅x−1)⋅(x−1⋅x−1)after using associativity=(x⋅x)⋅1⋅(x−1⋅x−1)taking advantage of inverses=(x⋅x)⋅(x−1⋅x−1)using the property of the identity=x⋅(x⋅x−1)⋅x−1 then we repeat the process, using associativity=x⋅1⋅x−1 again, as products of inverses yields the identity=x⋅x−1 again remembering multiplying by the inverse leaves things unchanged=1 we finish by taking advantage of inverses one last time Similar calculations reveal (x−1)3⋅x 3=1 More generally, for any integer n≥2 we have (x−1)n⋅x n=x n⋅(x−1)n=1 As we have seen, something similar holds for braids, for permutations, and -- of course -- for real numbers. Anytime we have an operation that is closed on some set, with associativity, an identity, and guaranteed inverses, it must be the case that (x−1)n is the inverse of x n for any integer n>1. Given this, we abbreviate (x−1)n by x−n, as this then allows us to extend our previous rules of (x m)n=x m n and x m⋅x n=x m+n to work with many more pairs of integer exponents, m and n. As a quick example, we can easily show x 5⋅x−3=x 2: x 5⋅x−3=x 2+3⋅x−3=(x 2⋅x 3)⋅x−3=x 2⋅(x 3⋅x−3)=x 2⋅(x 3⋅(x−1)3)=x 2⋅1=x 2 However, we need to make two more definitions so that these two rules work with ALL integer exponents m and n: Consider x 2⋅x−2. The properties of associativity, the identity, and inverses tell us the above must equal the multiplicative identity, 1. However, the rule x m x n=x m+n tells us this must be x 0. We said before to assume x n meant the product resulting from multiplying n values of x together -- but what we mean by "multiplying 0 values of x together" is not exactly clear. Thus, our current definition of x n can't really be used to tell us what x 0 is. However, to make everything consistent, we define: x 0=1 for all real numbers x≠0† In a like manner, when considering what x 1 should mean, we note that "multiplying 1 value of x together" is also somewhat suspicous (although possibly less so). Perhaps we could tweak the language we use to avoid this little hickup. That said, note if x m x n=x m+n is to work for all integers, then x 1=x 2+(−1)=x 2⋅x−1=(x⋅x)⋅x−1=x⋅(x⋅x−1)=x⋅1=x, encouraging us to agree that x 1=x must be true for all x. †Note: we leave 0 0 undefined here, as the topics we address are meant not only to expose students to the greater universe of mathematics, but also to prepare one for calculus -- where 0 0 is considered an undefinable "indeterminant form". However, in other contexts this is not always the best course of action. As Ronald L. Graham, Donald Knuth, and Oren Patashnik argue in their classic text Concrete Mathematics: "Some textbooks leave the quantity 0 0 undefined, because the functions x 0 and 0 x have different limiting values when x decreases to 0. But this is a mistake. We must define x 0=1, for all x, if the binomial theorem is to be valid when x=0,y=0, and/or x=−y. The theorem is too important to be arbitrarily restricted! By constrast, the function 0 x is quite unimportant." -- Graham, Knuth, & Patashnik Recipricals, Division, and the Consequences of Commutatitivy We define the division of one real number by another as the product of the first and the multiplicative inverse of the second, writing the result (called the quotient) either using the ÷ operator, or as a fraction, as seen below: x÷y=x y=x⋅y−1 It should hopefully be obvious that if two real values agree, then their products with some other number should also agree. That is to say, for any real values a,b, and c, if a=b, then a c=b c. But then, for any non-zero real value x we have: x−n⋅x n=1 which, if we consider c=x−n implies(x−n⋅x n)⋅x−n=1⋅x−n 1⋅x−n=1 x n x−n=1 x n As the quotient of 1 and any non-zero real value is called the reciprical of that value, we see that negative exponents are connected to recipricals of powers: x−n=1 x n Even more immediate, note that if all the rules above hold, we can also show the following for integers m and n (and non-zero x): x m x n=x m⋅x−n=x m+(−n)=x m−n which confirms (for integer exponents anyways) that we subtract exponents when dividing powers: x m x n=x m−n Having considered products and quotients of powers, we now turn our attention to the reverse -- powers of products and quotients. Recall how real number multiplication "distributes over addition and subtraction": (a+b)c=a c+b c and(a−b)c=a c−b c In a similar manner, we can show that because multiplication of real values is commutative (i.e., a b=b a for all real values a,b), (integer) exponents will "distribute" over products and quotients. That is to say: (a⋅b)c=a c⋅b c and(a÷b)c=a c÷b c To see how we know the first statement holds (i.e., (a⋅b)c=a c⋅b c), consider a representative case where the exponent involved is 3. That is, we will show (x y)3=x 3 y 3. Notice how the property of commutivity (along with associativity) allows us to do this by gradually move all the y values to the right, where they can be collected into single power -- and how this leaves all the x values on the left, where they too can be collected together into a single power: (x y)3=(x y)(x y)(x y)=x y x(y x)y=x y x(x y)y=x(y x)x y y=x(x y)x y y=x x(y x)y y=x x(x y)y y=x x x y y y=x 3 y 3 Interestingly, one can deduce the sequence of applications of the commutative property (like the one shown above) to accomplish any similar re-ordering by considering the permutation that would take the starting order of factors to the desired order of factors, and then considering a braid that accomplishes the same permutation of its threads to express that permutation as a series of transpositions. To see this, consider the below diagrams -- especially the one on the right which keeps crossings from happening simultaneously -- which suggest we can turn (x y)3=x y x y x y into x x x y y y=x 3 y 3 by first transposing the 4th and 5th factors (orange and magenta), and then the 2nd and 3rd (green and blue), and then the 3rd and 4th (green and magenta). Notice how this matches the applications of commutativity given in the chain of equality seen above. Demonstrating the second result (i.e., (a÷b)c=a c÷b c) through a representative example can be done similarly -- although we'll express the quotients as fractions instead of using the "÷" symbol: (x y)3=(x y−1)3=x 3(y−1)3(now we use(a b)n=a n b n, which of course requires commutativity)=x 3 y−3=x 3(y 3)−1=x 3 y 3 The upshot of this is that certainly for all integers n and real values x and y (where y≠0) -- but also when n is an integer and x and y are some elements from some set of things which under some operation (denoted in a "multiplicative way" here) is closed, associative, with an identity, inverses, and (fully) commutative -- the following are both true: (x y)n=x n y n and(x y)n=x n y n That said, in other contexts (especially when commutativity is not present) we may not be able to rely on this being true -- recall, such was the case with both braids and permutations! Simplified Forms While there will be exceptions, we often try to simplify expressions involving powers of real numbers to their shortest written forms. Notice in particular the rules x m x n=x m+n, x m/x n=x m−n, and (x m)n=x m n offer immediate reductions in how much pencil lead is required to write an expression. Consequently, one should typically seek to manipulate expressions as needed to create opportunities to apply these specific rules when simplifying. Also -- as a useful matter of convention when dealing with powers of real numbers -- when simplifying expressions using the above rules, one should avoid the presence of negative exponents in the final form. One advantage of such a convention is that it helps reduce the number of "equivalent forms" a completely simplified expression may take. Below an example of how an expression might be simplified is provided. Note, this is not the only way to simplify the expression given -- any legal application of the rules developed in all the discussion above that seeks to write the expression as briefly as possible should get you to the same final quotient. So feel free to follow a different path! 🙂 (x 2 y−5)−3 z−2 w−7 x 6 z−2 x 2 y 4=(x 2 y−5)−3 z−2 w−7 x 6 x 2 z−2 y 4 we used commutativity to get the bottom two powers of x near one another=(x 2 y−5)−3 z−2 w−7 x 8 z−2 y 4=x−6 y 15 z−2 w−7 x 8 z−2 y 4 we used commutativity to distribute the−3 exponent=x−6 y 15 z−2 w−7(x 8 z−2 y 4)−1=x−6 y 15 z−2 w−7 x−8 z 2 y−4 we used commutativity to restore the original order of powers after taking the inverse=x−6 x−8 y 15 y−4 z−2 z 2 w−7 we used commutativity several times to get powers with the same base adjacent=x−14 y 11 z 0 w−7=y 11 x 14 w 7 The reader will benefit by identifying each rule being applied to generate next steps in the calculation above. As a special note, be very careful about noticing when commutativity is being applied (see the several provided comments in the calculations above), recalling that can't always make similar calculations in other contexts (e.g., braids and permutations).
17124	https://math.stackexchange.com/questions/1894512/derivative-of-ex1	calculus - Derivative of $e^{x+1}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Derivative of e x+1 e x+1 Ask Question Asked 9 years, 1 month ago Modified9 years, 1 month ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I used an online derivative calculator to check my answer to this problem, I'm not sure what I did wrong. To me it seems like e x+1+1=(x+1)e x e x+1+1=(x+1)e x The online calculator gives e x+1 e x+1 as the derivative. What is the correct answer and why? Thanks. calculus derivatives exponential-function Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 16, 2016 at 23:16 gt6989b 55k 3 3 gold badges 40 40 silver badges 75 75 bronze badges asked Aug 16, 2016 at 23:03 vivi-dvivi-d 61 8 8 bronze badges 5 6 f(x)=e x+1=e×e x⟹f′(x)=e×e x=e x+1=f(x)f(x)=e x+1=e×e x⟹f′(x)=e×e x=e x+1=f(x).lulu –lulu 2016-08-16 23:05:54 +00:00 Commented Aug 16, 2016 at 23:05 1 How did you get your answer? Did you try to apply the chain rule? You might get better answers if you show your process so we can point out where things go wrong.Milo Brandt –Milo Brandt 2016-08-16 23:07:51 +00:00 Commented Aug 16, 2016 at 23:07 2 Perhaps you are thinking of the rule for differentiation of a polynomial being d d x x n=n⋅x n−1 d d x x n=n⋅x n−1. That rule only works for polynomial terms and does not work where the exponent is x x as opposed to the base.JMoravitz –JMoravitz 2016-08-16 23:08:10 +00:00 Commented Aug 16, 2016 at 23:08 You are confusing the derivative of x n x n is n x n−1 n x n−1 with the derivative of e x e x is e x e x. Use the chain rule. f(g(x))=f′(g(x))g′(x)f(g(x))=f′(g(x))g′(x). f(x)=e x f(x)=e x so f′(x)=e x f′(x)=e x. g(x)=x+1 g(x)=x+1 so g′(x)=1 g′(x)=1 so f′(g(x))g′(x)=e x+1∗1=e x+1 f′(g(x))g′(x)=e x+1∗1=e x+1.fleablood –fleablood 2016-08-16 23:09:40 +00:00 Commented Aug 16, 2016 at 23:09 ... or use lulu's slick argument. That is REALLY slick.fleablood –fleablood 2016-08-16 23:10:56 +00:00 Commented Aug 16, 2016 at 23:10 Add a comment\| 7 Answers 7 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The correct answer is: e x+1 e x+1 We use the chain rule which says: (f(g(x))′=f′(g(x))g′(x)(f(g(x))′=f′(g(x))g′(x) If we let, f(x)=e x f(x)=e x and g(x)=x+1 g(x)=x+1 then we have f(g(x))=e x+1 f(g(x))=e x+1. Now using the chain rule and the fact that d d x e x=e x d d x e x=e x we have: (e x+1)′=e x+1(x+1)′(e x+1)′=e x+1(x+1)′ But (x+1)′=1(x+1)′=1 so we have: (e x+1)′=e x+1(e x+1)′=e x+1 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 16, 2016 at 23:07 Ahmed S. AttaallaAhmed S. Attaalla 19.2k 8 8 gold badges 49 49 silver badges 78 78 bronze badges 2 You're welcome @vivi-d Ahmed S. Attaalla –Ahmed S. Attaalla 2016-08-16 23:24:44 +00:00 Commented Aug 16, 2016 at 23:24 @vivi-d Your confusion arises from muddling up polynomials of the form x n x n which have derivative of n x n−1 n x n−1 with exponentials such as a x a x which have derivative of a x⋅ln a a x⋅ln⁡a.Ian Miller –Ian Miller 2016-08-16 23:25:25 +00:00 Commented Aug 16, 2016 at 23:25 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Note that d d x[e x]=e x,d d x[e x]=e x, or equivalently, d d x[exp(x)]=exp(x).d d x[exp⁡(x)]=exp⁡(x). Then by the chain rule, with the choice f(x)=exp(x)f(x)=exp⁡(x) and g(x)=x+1 g(x)=x+1, we have d d x[f(g(x))]=f′(g(x))g′(x)=d f d g⋅d g d x=exp(x+1)⋅d d x[x+1]=exp(x+1),d d x[f(g(x))]=f′(g(x))g′(x)=d f d g⋅d g d x=exp⁡(x+1)⋅d d x[x+1]=exp⁡(x+1), and for a general differentiable function g g, we have d d x[exp(g(x))]=exp(g(x))g′(x)=g′(x)e g(x).d d x[exp⁡(g(x))]=exp⁡(g(x))g′(x)=g′(x)e g(x). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 16, 2016 at 23:08 heropupheropup 145k 15 15 gold badges 114 114 silver badges 201 201 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. For f(x)=e g(x)f(x)=e g(x) the derivative is f′(x)=g′(x)e g(x)f′(x)=g′(x)e g(x) by the Chain Rule of differentiation, provided g(x)g(x) is differentiable. In your case, if we let g(x)=x+1 g(x)=x+1 then g′(x)=1 g′(x)=1, leaving f(x)=e x+1 f(x)=e x+1 unchanged by differentiation. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 16, 2016 at 23:07 Edward EvansEdward Evans 4,789 1 1 gold badge 21 21 silver badges 47 47 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Let y=e x+1 y=e x+1 and let u=x+1 u=x+1. Then we can write, y=e u y=e u. By the chain rule, f′(x)=d y d x=d y d u⋅d u d x=d d u(e u)⋅d d x(x+1)=e u⋅(1)=e u=e x+1.f′(x)=d y d x=d y d u⋅d u d x=d d u(e u)⋅d d x(x+1)=e u⋅(1)=e u=e x+1. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 16, 2016 at 23:11 benguinbenguin 3,892 14 14 silver badges 19 19 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Notice e x+1=(e)e x e x+1=(e)e x. d d x(e)e x=e d d x e x=(e)e x=e x+1.d d x(e)e x=e d d x e x=(e)e x=e x+1. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 16, 2016 at 23:26 user141854 user141854 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Note: d x n d x=n x n−1 d x n d x=n x n−1. x x is the variable in the base, and n n is a constant in the exponent. But d e x d x≠x e x−1 d e x d x≠x e x−1. x x is a variable the exponent, while e e is a constant in the base. You derive based on "what the variable x x is doing". When x x is in the exponent it is a very different thing than when x x is in the base. d e x d x=e x d e x d x=e x. Taking a cute hint from lulu: d e x+1 d x=d(e∗e x)d x=e d(e x)d x=e∗e x=e x+1 d e x+1 d x=d(e∗e x)d x=e d(e x)d x=e∗e x=e x+1. Note: d(x n)d x=lim(x+h)n−x n h=lim x n+n∗x n−1∗h+a∗x n−2 h 2+.....+h n−x n h=lim n∗x n−1∗h+h 2(a x n−2+......+h n−2)h=lim n∗x n−1+h∗(a x n−2+.....)=n x n−1 d(x n)d x=lim(x+h)n−x n h=lim x n+n∗x n−1∗h+a∗x n−2 h 2+.....+h n−x n h=lim n∗x n−1∗h+h 2(a x n−2+......+h n−2)h=lim n∗x n−1+h∗(a x n−2+.....)=n x n−1. An entirely different reason for e x e x d(e x)d x=lim e x+h−e x h=lim e x e h−e x h=lim e x(e h−1)h=e x∗lim e h−1 h=e x∗1 d(e x)d x=lim e x+h−e x h=lim e x e h−e x h=lim e x(e h−1)h=e x∗lim e h−1 h=e x∗1. Completely different. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 16, 2016 at 23:27 fleabloodfleablood 132k 5 5 gold badges 52 52 silver badges 142 142 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. You can also use logarithmic differentiation First set your function in terms of x x and y y: y=e x+1 y=e x+1 Then take the natural logarithm of both sides. i.e l n(y)=l n(e x+1)l n(y)=l n(e x+1) l n(y)=x+1 l n(y)=x+1 Then use implicit differentiation to derive both sides of the equation 1 y 1 y d y d x d y d x=1=1 Move y y over to the right side by multiplying both sides by y y. So: d y d x d y d x=y=y Which means: d y d x d y d x=e x+1=e x+1 The derivative of e x+1 e x+1 is itself. Hope this can be helpful. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Aug 16, 2016 at 23:28 BensstatsBensstats 531 4 4 silver badges 18 18 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus derivatives exponential-function See similar questions with these tags. 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17125	https://math.stackexchange.com/questions/482580/linear-system-with-constrained-solutions	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Linear System with constrained solutions Ask Question Asked Modified 12 years ago Viewed 208 times 1 $\begingroup$ After a model my problem I found a rectangular linear system : $$Ax=b$$ I can easely solve it with a least square with QR/SVD... But the model include constrains for each solution $x_i$, the $\vec{x}$ have n elements. ($1\leqslant i\leqslant n$) For each $x_i$ I have a min and max: $$ x_i \geqslant min_i $$ $$ x_i \leqslant max_i $$ And exist some couple constraint with $(i,j)$ as: $$ (i,j) \in \left (\frac{\mathbb{Z}}{n\mathbb{Z}} \right )^2 \| i \neq j $$ $i$ and $j$ is on $\mathbb{Z}$ and is included on $[1,n]$ I have 2 others kinds of constraints $L^p_{ij}$ and $E^q_{ij}$ on "99.9%" of my cases : $$ L^p_{ij} = L^p_{ji} $$ $$ E^q_{ij} = E^q_{ji} $$ I have N ($1 \leqslant p \leqslant N$) contrains L and K ($1 \leqslant q \leqslant K$) constains E. $$ L^p_{ij} : x_i + x_j = 1 \space \mathbf{or} \space 0 $$ $$E^p_{ij} : x_i > 0 \Rightarrow x_j = 0$$ I can rewrite $E^q_{ij}$ like : $$ \begin{matrix} x_i+x_j=\max(x_i,x_j) \Rightarrow \ x_i + x_j = \frac{1}{2}\left(x_i + x_j + \left \| x_i - x_j \right \|\right) \Rightarrow \ x_i + x_j = \frac{1}{2}\left (x_i + x_j + \sqrt{ (x_i - x_j)^2 }\right ) \end{matrix} $$ I search a way to minimize $\\|Ax=b\\|$ with my $min_i$, $max_i$, $L^p_{ij}$, $E^q_{ij}$. All $x_i$ have one $min_i$ and $max_i$, but for $L^p_{ij}$ and $E^q_{ij}$ I have no control on number or "topology" constrains. As you see I have an $\mathbf{or}$ on $L^p_{ij}$ and the derivation of $E^q_{ij}$ is not continue so I can't simply use the Lagrange Multiplier. Any comment or solutions are welcome. Thanks linear-algebra optimization convex-optimization numerical-optimization Share asked Sep 3, 2013 at 0:46 chkonechkone 5555 bronze badges $\endgroup$ Add a comment \| 1 Answer 1 Reset to default 0 $\begingroup$ $Ax = b$ with bounds on $x$ is dealt with in linear programming. Your $L$ and $E$ constraints will take you into mixed integer linear programming. Share answered Sep 3, 2013 at 1:41 Robert IsraelRobert Israel 472k2828 gold badges376376 silver badges714714 bronze badges $\endgroup$ 5 $\begingroup$ I'm not a specialist of "Mixed Integer Linear Programming" but this method was not developped to constrain the solution as Interger ? In $\mathbb{Z}$. I forget to specify something, in major part of my cases $min_i = 0$ and $max_i = 1$ So my $x_i$ is not integer. Thanks $\endgroup$ chkone – chkone 2013-09-03 14:18:24 +00:00 Commented Sep 3, 2013 at 14:18 1 $\begingroup$ Your $x_i$ are not integer, but you'd have additional variables that are integer. That's why it's "mixed". For example, $L^p_{ij}$ would be handled with an additional binary variable $u_{ij}$ (i.e. its allowed values are $0$ and $1$), with constraint $x_i + x_j = u_{ij}$. For $E^p_{ij}$ you'd have binary variables $u_i$ and $u_j$, with constraints $max_i u_i \ge x_i$, $max_j u_j \ge x_j$, $u_i + u_j \le 1$. $\endgroup$ Robert Israel – Robert Israel 2013-09-03 15:12:37 +00:00 Commented Sep 3, 2013 at 15:12 $\begingroup$ I have no link for $L^p_{ij}$ and $E^q_{ij}$. For that I'm not sur to understand the second part for $E^p_{ij}$ : $$ max_i u_i \geqslant x_i ? $$ $$ max_j u_j \geqslant x_j ? $$ $$ u_i+u_j\leqslant 1 $$ $max_i u_i$ it is a multiplication ? Thanks $\endgroup$ chkone – chkone 2013-09-03 15:44:38 +00:00 Commented Sep 3, 2013 at 15:44 1 $\begingroup$ Yes, it's multiplication. The point is that $\max_i u_i \ge x_i$ means that when $x_i > 0$, $u_i$ can't be $0$ (so it must be $1$), while $u_i = 1$ is always allowed. $u_i + u_j \le 1$ means you can't have both $u_i = 1$ and $u_j = 1$. $\endgroup$ Robert Israel – Robert Israel 2013-09-04 04:04:22 +00:00 Commented Sep 4, 2013 at 4:04 $\begingroup$ I see ! Thanks you. I will find a C++ implementation for a MIP solver. $\endgroup$ chkone – chkone 2013-09-06 21:45:17 +00:00 Commented Sep 6, 2013 at 21:45 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra optimization convex-optimization numerical-optimization See similar questions with these tags. 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17126	https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-8-for-ccss/section/9.2/primary/lesson/properties-of-exponents-in-division-expressions-msm8-ccss/	Properties of Exponents in Division Expressions \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! Skip to content What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Start Practice 9.2 Properties of Exponents in Division Expressions FlexBooks 2.0> CK-12 Interactive Middle School Math 8> Properties of Exponents in Division Expressions Written by:Larame Spence \|Sean Regan Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Lesson Review Asked on Flexi Related Content Lesson Astronomical Numbers Performing calculations is an important responsibility for astronomers. Astronomers need to calculate distance, volumes, surface areas, speeds and much more for the largest objects in the universe. For example, the distance from the Earth to the International Space Station (ISS)is approximately 3 5 miles, and the distance from the Earth to Alpha Centauri is approximately 3 28 miles. How many times further is Alpha Centauri's distance than the ISS's distance? How do you find out how many times further something is? What operation must you apply to solve this question? Since you are trying to find out how many times further away Alpha Centauri is, you should divide the numbers. The distance to Alpha Centauri compared to the distance to the ISS would look like this: Distance to Alpha Centauri Distance to the ISS≈22,876,792,454,961 243 However, this number is so big that some calculators would not even be able to display it. How could you find a solution without using a calculator? Dividing With Exponents As done in the lessonProperties of Exponents in Multiplication Expressions, you will begin by writing the problem in factored form. Using a factored form will help you see the problem more clearly. Additionally, using the factored form is a great strategy for solving problems using exponent properties if you ever forget the properties. Example Simplify the expression 4 6 4 2. When written in factored form you get the following: 4⋅4⋅4⋅4⋅4⋅4 4⋅4 U se the same property to cancel factors in the numerator and denominator that you used to simplify fractions: INTERACTIVE Dividing with Exponents by Factoring Drag the sliders to observe what happens when you divide terms of the same base by factoring. Try It Your device seems to be offline. Please check your internet connection and try again. + Do you want to reset the PLIX? Yes No Progress 0 / 0 Discussion Questions Why does this work? Why wouldn’t this rule work if the bases were different from each other? Quotient of Powers Property The property which was derived in the previous activity is called the quotient of powers property. The Quotient of Powers Property states that when dividing two exponents with the same base, you can subtract the exponents and keep the base. This property will always work as long as the base is not zero because you cannot divide by zero. Use the interactive below to explore this property further. INTERACTIVE Quotient of Powers Drag the sliders to observe what happens when you divide terms with the same base. Try It Your device seems to be offline. Please check your internet connection and try again. + Do you want to reset the PLIX? Yes No Discussion Questions Returning to the original question of 22 7 22 4,how could you use the Quotient of Powers Property to solve the problem? Astronomers typically use exponent rules to solve problems with large numbers. Why do you think this is? Dividing With Terms To simplify a quotient of two terms, you will need to use the Quotient of Powers Property to simplify individual parts. Use the interactive below to explore this idea. INTERACTIVE Quotient of Powers with Terms Drag the sliders to observe what happens when you divide terms. Try It Your device seems to be offline. Please check your internet connection and try again. + Do you want to reset the PLIX? Yes No When terms have coefficients, simplify the coefficient like a regular fraction. Example Simplify the expression 4 r 3 s 5 6 r s 5. To solve this,you can either write the expression in expanded form or use the Quotient of Powers Property. Here is a look at both. Expanded Form Quotient of Powers 4 r 3 s 5 6 r s 5 4 r 3 s 5 6 r s 5=4⋅r⋅r⋅r⋅s⋅s⋅s⋅s⋅s 2⋅3⋅r⋅s⋅s⋅s⋅s⋅s⋅4 6=2 3=2⋅r⋅r 3⋅r 3 r 1=r 3−1=r 2=2 3 r 2⋅s 5 s 5=1=2 3 r 2 Summary Use exponent property to simplify exponential expressions: Product of Powers Property:a m×a n=a m+n Quotient of Powers Property:a m a n=a m−n The Quotient of Powers Property can be used to simplify fractions with variables by canceling pairs of variables in the numerator and denominator. For example: x 3 y 1 x 2 y 7=x y 6 Add Note CancelSave Discuss with Flexi NOTES / HIGHLIGHTS Please Sign In to create your own Highlights / Notes Add Note Edit Note Remove Highlight Asked by Students Here are the top questions that students are asking Flexi for this concept: Simplify the expression the quantity 'three-sevenths squared'. To simplify the expression (3 7)2, you just need to square both the numerator and the denominator: (3 7)2=3 2 7 2 9 49 How to divide polynomials with different exponents? For dividing a polynomial by a polynomial, we may follow the following steps: Step 1: Arrange the terms of the dividend and divisor in descending order of their degrees. Step 2: Divide the first term if the dividend by the first term of the divisor to obtain the first term of the quotient Step 3: Multiply the divisor by the first term of the quotient and subtract the result from the dividend to obtain the remainder. Step 4: Take down the remaining terms of the dividend next to the remainder so obtained. Treat this expression as the new dividend. Step 5: Repeat the above process till we obtain a remainder which is either 0 or a polynomial of degree less than that of the divisor. To learn about polynomial division, click here! How do you write a number in standard form? Standard form of a number is written as a number between 1 and 10, multiplied by a power of 10. For example, the number 1234 can be written in standard form as 1.234×10 3. What is the factorial of 200? The factorial of a number is the product of all positive integers less than or equal to that number. For a number n, it's denoted as n!. However, calculating the factorial of 200 directly would be quite large and is impractical to display in its complete form. In math notation we will write: 200!=1×2×3×…×198×199×200 The exact number is extremely large, much bigger than the number of atoms in the known universe! It has 375 digits: 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000000000000000000000 As a future student of higher-level mathematics, you will rarely, if ever, need to calculate such large factorials precisely. Instead, you'll learn about tools like the Stirling's approximation to handle large factorials. For all practical purposes, it's usually enough to know that 200! is a very large number indeed. What is the result when a number is raised to the zero power? Any non-zero number a that is raised to the power zero always results in 1. a 0=1 Note: We never take 0 to the power of 0 as it is an indeterminant number in mathematics. Overview Use exponent property to simplify exponential expressions: Product of Powers Property:a m×a n=a m+n Quotient of Powers Property:a m a n=a m−n The Quotient of Powers Property can be used to simplify fractions with variables by canceling pairs of variables in the numerator and denominator. For example: x 3 y 1 x 2 y 7=x y 6 Vocabulary distance operation factored form Exponent Property Quotient of Powers Property Term Coefficient fraction Expression expanded form Test Your Knowledge Question 1 Simplify: (3 x−2 y)3 a 37 x 3−8 y 3 b 27 x 3 4 y 3 c 9 x 3−8 y 3 d 27 x 3−8 y 3 Check It Simplify the expression: (3 x−2 y)3 First, apply the power of a quotient property. (3 x−2 y)3(3 x)3(−2 y)3 3 3 x 3(−2)3 y 3 Next, expand to simplify. 3 3 x 3(−2)3 y 3 3⋅3⋅3⋅x⋅x⋅x(−2)⋅(−2)⋅(−2)⋅y⋅y⋅y 27 x 3−8 y 3 Option D is the correct answer. FlexCard™ Question 2 Evaluate the expression: 4 5 4 3 a 16 b 1 16 c 32 d 1 32 Check It Evaluate the expression: 4 5 4 3 4 5⋅4−3 4 5−3 4 2 16 Option A is the correct answer. FlexCard™ Study Guide Go to Study Guide Asked by Students Ask your question Here are the top questions that students are asking Flexi for this concept: Simplify the expression the quantity 'three-sevenths squared'. To simplify the expression (3 7)2, you just need to square both the numerator and the denominator: (3 7)2=3 2 7 2 9 49 How to divide polynomials with different exponents? For dividing a polynomial by a polynomial, we may follow the following steps: Step 1: Arrange the terms of the dividend and divisor in descending order of their degrees. Step 2: Divide the first term if the dividend by the first term of the divisor to obtain the first term of the quotient Step 3: Multiply the divisor by the first term of the quotient and subtract the result from the dividend to obtain the remainder. Step 4: Take down the remaining terms of the dividend next to the remainder so obtained. Treat this expression as the new dividend. Step 5: Repeat the above process till we obtain a remainder which is either 0 or a polynomial of degree less than that of the divisor. To learn about polynomial division, click here! How do you write a number in standard form? Standard form of a number is written as a number between 1 and 10, multiplied by a power of 10. For example, the number 1234 can be written in standard form as 1.234×10 3. What is the factorial of 200? The factorial of a number is the product of all positive integers less than or equal to that number. For a number n, it's denoted as n!. However, calculating the factorial of 200 directly would be quite large and is impractical to display in its complete form. In math notation we will write: 200!=1×2×3×…×198×199×200 The exact number is extremely large, much bigger than the number of atoms in the known universe! It has 375 digits: 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000000000000000000000 As a future student of higher-level mathematics, you will rarely, if ever, need to calculate such large factorials precisely. Instead, you'll learn about tools like the Stirling's approximation to handle large factorials. For all practical purposes, it's usually enough to know that 200! is a very large number indeed. What is the result when a number is raised to the zero power? Any non-zero number a that is raised to the power zero always results in 1. a 0=1 Note: We never take 0 to the power of 0 as it is an indeterminant number in mathematics. Related Content Properties of Exponents Study Guide Quotient of Powers Property - Overview Power of a Quotient: A Sample Application Quotient of Powers Property - Example 1 Exponential Properties Involving Quotients Back to Properties of Exponents in Division Expressions Ask me anything! CK-12 Foundation is a non-profit organization that provides free educational materials and resources. FLEXIAPPS ABOUT Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials SUPPORT Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us BYCK-12 Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps CONNECT TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn v2.11.10.20250923073248-4b84c670be © CK-12 Foundation 2025 \| FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation. Terms of usePrivacyAttribution guide Curriculum Materials License Student Sign Up Are you a teacher? Sign up here Sign in with Google Having issues? Click here Sign in with Microsoft Sign in with Apple or Sign up using email By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Sign In Adaptive Practice I’m Ready to Practice! Get 10 correct to reach your goal Estimated time to complete: 9 min Start Practice I need help Save this section to your Library in order to add a Practice or Quiz to it. Title (Edit Title)47/ 100 Save Go Back This lesson has been added to your library. Got It No Results Found Your search did not match anything in . Got It Searching in: CK-12 Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for: Go to the Table of Contents Ok Are you sure you want to restart this practice? Restarting will reset your practice score and skill level.
17127	https://www.cuemath.com/questions/prove-that-sin-pi-x-sinx/	Prove that sin (π - x) = sin (x). Sine is one of the primary functions of trigonometry. To prove this, we will use trigonometric identity. Answer: Hence proved that sin (π - x) = sin (x) Let's prove. Explanation: Given that LHS = sin (π - x) By using trigonometric identity: sin (A - B) = sin A cos B - cos A sin B, we get sin (π - x) = sin (π) cos (x) - cos (π) sin (x) As we know that sin (π) = sin 180º = 0 and cos (π) = cos 180º = - 1 sin (π - x) = 0 × cos (x) - (- 1) × sin (x) 0 + sin (x) = RHS Hence Proved. Thus, sin (π - x) = sin (x). Math worksheets andvisual curriculum FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers MATH TEST Math Kangaroo AMC 8 MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 ABOUT US Our Mission Our Journey Our Team MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math Terms and ConditionsPrivacy Policy
17128	https://www.unitconverters.net/power/calorie-it-second-to-calorie-it-hour.htm	Convert Calorie (IT)/second to Calorie (IT)/hour Home / Power Conversion / Convert Calorie (IT)/second to Calorie (IT)/hour Convert Calorie (IT)/second to Calorie (IT)/hour Please provide values below to convert calorie (IT)/second [cal/s] to calorie (IT)/hour [cal/h], or vice versa. From:calorie (IT)/second To:calorie (IT)/hour Calorie (IT)/second to Calorie (IT)/hour Conversion Table \| Calorie (IT)/second [cal/s] \| Calorie (IT)/hour [cal/h] \| --- \| \| 0.01 cal/s \| 36 cal/h \| \| 0.1 cal/s \| 360 cal/h \| \| 1 cal/s \| 3600 cal/h \| \| 2 cal/s \| 7200 cal/h \| \| 3 cal/s \| 10800 cal/h \| \| 5 cal/s \| 18000 cal/h \| \| 10 cal/s \| 36000 cal/h \| \| 20 cal/s \| 72000 cal/h \| \| 50 cal/s \| 180000 cal/h \| \| 100 cal/s \| 360000 cal/h \| \| 1000 cal/s \| 3600000 cal/h \| How to Convert Calorie (IT)/second to Calorie (IT)/hour 1 cal/s = 3600 cal/h 1 cal/h = 0.0002777778 cal/s Example: convert 15 cal/s to cal/h: 15 cal/s = 15 × 3600 cal/h = 54000 cal/h Popular Power Unit Conversions hp to kw kw to hp hp to watts watts to hp BTU to Ton Ton to BTU Convert Calorie (IT)/second to Other Power Units Calorie (IT)/second to Watt Calorie (IT)/second to Exawatt Calorie (IT)/second to Petawatt Calorie (IT)/second to Terawatt Calorie (IT)/second to Gigawatt Calorie (IT)/second to Megawatt Calorie (IT)/second to Kilowatt Calorie (IT)/second to Hectowatt Calorie (IT)/second to Dekawatt Calorie (IT)/second to Deciwatt Calorie (IT)/second to Centiwatt Calorie (IT)/second to Milliwatt Calorie (IT)/second to Microwatt Calorie (IT)/second to Nanowatt Calorie (IT)/second to Picowatt Calorie (IT)/second to Femtowatt Calorie (IT)/second to Attowatt Calorie (IT)/second to Horsepower Calorie (IT)/second to Horsepower (550 Ftlbf/s) Calorie (IT)/second to Horsepower (metric) Calorie (IT)/second to Horsepower (boiler) Calorie (IT)/second to Horsepower (electric) Calorie (IT)/second to Horsepower (water) Calorie (IT)/second to Pferdestarke (ps) Calorie (IT)/second to Btu (IT)/hour Calorie (IT)/second to Btu (IT)/minute Calorie (IT)/second to Btu (IT)/second Calorie (IT)/second to Btu (th)/hour Calorie (IT)/second to Btu (th)/minute Calorie (IT)/second to Btu (th)/second Calorie (IT)/second to MBtu (IT)/hour Calorie (IT)/second to MBH Calorie (IT)/second to Ton (refrigeration) Calorie (IT)/second to Kilocalorie (IT)/hour Calorie (IT)/second to Kilocalorie (IT)/minute Calorie (IT)/second to Kilocalorie (IT)/second Calorie (IT)/second to Kilocalorie (th)/hour Calorie (IT)/second to Kilocalorie (th)/minute Calorie (IT)/second to Kilocalorie (th)/second Calorie (IT)/second to Calorie (IT)/minute Calorie (IT)/second to Calorie (th)/hour Calorie (IT)/second to Calorie (th)/minute Calorie (IT)/second to Calorie (th)/second Calorie (IT)/second to Foot Pound-force/hour Calorie (IT)/second to Foot Pound-force/minute Calorie (IT)/second to Foot Pound-force/second Calorie (IT)/second to Pound-foot/hour Calorie (IT)/second to Pound-foot/minute Calorie (IT)/second to Pound-foot/second Calorie (IT)/second to Erg/second Calorie (IT)/second to Kilovolt Ampere Calorie (IT)/second to Volt Ampere Calorie (IT)/second to Newton Meter/second Calorie (IT)/second to Joule/second Calorie (IT)/second to Exajoule/second Calorie (IT)/second to Petajoule/second Calorie (IT)/second to Terajoule/second Calorie (IT)/second to Gigajoule/second Calorie (IT)/second to Megajoule/second Calorie (IT)/second to Kilojoule/second Calorie (IT)/second to Hectojoule/second Calorie (IT)/second to Dekajoule/second Calorie (IT)/second to Decijoule/second Calorie (IT)/second to Centijoule/second Calorie (IT)/second to Millijoule/second Calorie (IT)/second to Microjoule/second Calorie (IT)/second to Nanojoule/second Calorie (IT)/second to Picojoule/second Calorie (IT)/second to Femtojoule/second Calorie (IT)/second to Attojoule/second Calorie (IT)/second to Joule/hour Calorie (IT)/second to Joule/minute Calorie (IT)/second to Kilojoule/hour Calorie (IT)/second to Kilojoule/minute All Converters Common Converters LengthWeight and MassVolumeTemperatureAreaPressureEnergyPowerForceTimeSpeedAngleFuel ConsumptionNumbersData StorageVolume - DryCurrencyCase Engineering ConvertersHeat ConvertersFluids ConvertersLight ConvertersElectricity ConvertersMagnetism ConvertersRadiology ConvertersCommon Unit Systems about us \| terms of use \| privacy policy \| sitemap © 2008 - 2025 unitconverters.net
17129	https://nrich.maths.org/problems/hidden-rectangles	Skip to main content Hidden rectangles Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard? Age 11 to 14 Challenge level Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving Being curious Being resourceful Being resilient Being collaborative Problem Image Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn along the lines on a chessboard? How many of the rectangles are squares? What happens on an n by n board? Getting Started What shape rectangles can you draw and in how many ways can these be fitted? Student Solutions A neat solution to the first part of the problem was received from Sana, Jenny, Chris and Rosion of Madras College, St. Andrews. I like the use of the lines rather than thinking about the rectangles in the same way as the squares (which is what I did). This idea can be generalised quite easily to an n x n chess board. Other correct solutions were received from Andrei of School 205 Bucharest, Mary of Birchwood Community High School and Chen of The Chinese High School, Singapore. Well done to all of you. There are 1296 different rectangles on the chess board. 204 of these rectangles are squares. First looking at the squares: Consider placing a square of size 1 x 1 along the left hand edge of the chessboard. This square can be in any one of 8 positions (as there are 8 by 8 squares on a chessboard). Similarly, the square can be placed in any one of eight positions along the top edge. So the total number of 1 x 1 squares = 8 x 8 = 64. A 2 x 2 square can occupy a 7 positions along the left hand edge and 7 positions along the top edge 7, giving 7 x 7 = 49 squares of size 2 x 2. Continuing in this way we get squares of size 3 x 3, 4 x 4, and so on. \| \| \| --- \| \| Size of Square \| Number of squares \| \| 1 x 1 \| 8 x 8 = 64 \| \| 2 x 2 \| 7 x 7 = 49 \| \| 3 x 3 \| 6 x 6 = 36 \| \| 4 x 4 \| 5 x 5 = 25 \| \| 5 x5 \| 4 x 4 = 16 \| \| 6 x 6 \| 3 x 3 = 9 \| \| 7 x 7 \| 2 x 2 = 4 \| \| 8 x 8 \| 1 x 1 = 1 \| So there are 204 squares. Then looking at rectangles: There are 9 vertical lines and 9 horizontal lines on the chess board. To form a rectangle you must choose 2 of the 9 vertical lines and 2 of the 9 horizontal lines. For the two horizontal lines: the first line can be chosen in 9 ways the second in eight ways. This would imply that you could tell the difference between lines 1 and 3 say and 3 and 1, which is not the case so you need to divide 9 x8 8 by 2, making 36. Similarly you can choose the two vertical lines in 36 ways. So the number of rectangles is given by 362 362 = 1296 The following approach utilises several of the other solutions sent in. A rectangle (or square) will have a height between 1 and 8 units and a width between 1 and 8 units. Tis can be represented by a table with each possible width represented by a column and each eight by a row. The entries in the table below then indicate the number of each size rectangle on the chessboard (using similar arguments to first part of the problem above). The final column of the table gives the total number of rectangles in each row. \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Height/Width \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| Total \| \| 1 \| 8 x 8 \| 7 x 8 \| 6 x 8 \| 5 x 8 \| 4 x 8 \| 3 x 8 \| 2 x 8 \| 1 x 8 \| 8 x(1+2+3+4+5+6+7+8}= 8 x 36 \| \| 2 \| 8 x 7 \| 7 x 7 \| 6 x 7 \| 5 x 7 \| 4 x 7 \| 5 x 7 \| 2 x 7 \| 1 x 7 \| 7 x(1+2+3+4+5+6+7+8}= 7 x 36 \| \| 3 \| 8 x 6 \| 7 x 6 \| 6 x 6 \| 5 x 6 \| 4 x 6 \| 3 x 6 \| 2 x 6 \| 1 x 6 \| 6 x(1+2+3+4+5+6+7+8}= 6 x 36 \| \| 4 \| 8 x 5 \| 7 x 5 \| 6 x 5 \| 5 x 5 \| 4 x 5 \| 3 x 5 \| 2 x 5 \| 1 x 5 \| 5 x(1+2+3+4+5+6+7+8} \| \| 5 \| 8 x 4 \| 7 x 4 \| 6 x 4 \| 4 x 4 \| 5 x 4 \| 3 x 4 \| 2 x 4 \| 1 x 4 \| 4 x(1+2+3+4+5+6+7+8} \| \| 6 \| 8 x 3 \| 7 x 3 \| 6 x 3 \| 4 x 3 \| 5 x 3 \| 3 x 3 \| 2 x 3 \| 1 x 3 \| 3 x(1+2+3+4+5+6+7+8} \| \| 7 \| 8 x 2 \| 7 x 2 \| 6 x 2 \| 4 x 2 \| 5 x 2 \| 3 x 2 \| 2 x 2 \| 1 x 2 \| 2 x(1+2+3+4+5+6+7+8} \| \| 8 \| 8 x 1 \| 7 x 1 \| 6 x 1 \| 4 x 1 \| 5 x 1 \| 3 x 1 \| 2 x 1 \| 1 x 1 \| 1 x(1+2+3+4+5+6+7+8} \| \| \| \| \| \| \| \| \| \| \| (1+2+3+4+5+6+7+8}x 36 = 362 \| Therefore the total number of rectangles in an 8 x 8 grid is (1 + 2 + 3 + ... + 8)2 The total number of rectangles in an n x n grid is (1 + 2 + 3 + ... + n)2 = (n2 (n+1)2 )/4. This uses the formula for the sum of the first n natural numbers, an arithmetic progression. For more details of this look at the proof sorter (link). The number of squares is: (12 + 22 + 32 + ... n2 ) = (n(n+1)(2n+1))/6 ( for a proof see Telescoping series )
17130	https://www.vedantu.com/physics/epsilon-naught-value	Physics Epsilon Naught (ε₀) Value, Units & Application in Physics Epsilon Naught (ε₀) Value, Units & Application in Physics Reviewed by: Gaurang Shah Download PDF NCERT Solutions NCERT Solutions for Class 12 NCERT Solutions for Class 11 NCERT Solutions for Class 10 NCERT Solutions for class 9 NCERT Solutions for class 8 NCERT Solutions for class 7 NCERT Solutions for class 6 NCERT Solutions for class 5 NCERT Solutions for class 4 NCERT Solutions for Class 3 NCERT Solutions for Class 2 NCERT Solutions for Class 1 CBSE CBSE class 3 CBSE class 4 CBSE class 5 CBSE class 6 CBSE class 7 CBSE class 8 CBSE class 9 CBSE class 10 CBSE class 11 CBSE class 12 NCERT CBSE Study Material CBSE Sample Papers CBSE Syllabus CBSE Previous Year Question Paper CBSE Important Questions Marking Scheme Textbook Solutions RD Sharma Solutions Lakhmir Singh Solutions HC Verma Solutions TS Grewal Solutions DK Goel Solutions NCERT Exemplar Solutions CBSE Notes CBSE Notes for class 12 CBSE Notes for class 11 CBSE Notes for class 10 CBSE Notes for class 9 CBSE Notes for class 8 CBSE Notes for class 7 CBSE Notes for class 6 How is epsilon naught used in Coulomb’s Law and electrostatics? Epsilon naught, commonly denoted as ε₀, is a fundamental constant in Physics. It represents the permittivity of free space and serves as the baseline for measuring how electric fields interact in vacuum versus other materials. Understanding ε₀ is essential for students studying electrostatics, electric fields, or related concepts. This constant directly features in major formulas like Coulomb’s Law, Gauss’s Law, and the equations for capacitance of capacitors. When studying electric fields, ε₀ helps determine the strength and transmission of electric forces in a vacuum. It provides a reference point for comparing the behavior of dielectrics and insulating materials. Mastery of this constant improves accuracy in Physics problem-solving and is vital for exams. What is Epsilon Naught (ε₀)? Epsilon naught, or ε₀, is the permittivity of free space. It quantifies how easily an electric field can develop in a vacuum. In formulas, ε₀ appears as a denominator—indicating the minimum resistance a vacuum offers to the formation and movement of electric field lines. The standard SI value of ε₀ is: ε₀ = 8.854 × 10-12 C2 N-1 m-2 This constant is known as the electric constant or the permittivity of free space. It underpins all electrostatic calculations in vacuum or air. Understanding Permittivity and Its Role Permittivity refers to the ability of a medium to permit electric fields. While materials have their own permittivity (ε), a vacuum’s permittivity is the lowest possible and is called ε₀. If we introduce a material into a vacuum, electric field lines pass through the material more easily if the permittivity is higher. For instance, the permittivity of a material is denoted by ε, and its relation to ε₀ is: ε = εr × ε₀ εr here represents the relative permittivity or dielectric constant of the material. Units and Dimensional Formula of Epsilon Naught Epsilon naught has standardized units for consistent calculations: \| Expression \| Value \| Unit \| Fundamental Quantities \| --- --- \| \| ε₀ \| 8.854187817 × 10-12 \| C2 N-1 m-2 \| A2 s4 kg-1 m-3 \| The dimensional formula for ε₀ is [M-1 L-3 T4 A2]. It is important to use this value exactly as given in exams and Physics problems since calculation errors often arise from unit mistakes. Epsilon Naught in Key Physics Formulas \| Formula \| Application \| Key Notes \| --- \| F = (1/4πε₀) × (q1q2/r2) \| Coulomb’s Law \| Force between charges in vacuum \| \| E = (1/4πε₀) × (q/r2) \| Electric Field (Point Charge) \| Vacuum or air medium \| \| C = ε₀ × A/d \| Capacitance (Parallel Plate) \| A: area, d: plate separation \| \| ΦE = Q/ε₀ \| Gauss’s Law \| Electric flux through closed surface \| Example Problem: Using Epsilon Naught in Calculation Let’s solve a basic numericals step-by-step using ε₀. Problem: Find the force between two charges of 1 μC each, separated by 1 m in vacuum. Step 1: Use Coulomb’s Law: F = (1/(4πε₀)) × (q1q2/r2) Step 2: Substitute values: q1 = q2 = 1 × 10-6 C, r = 1 m. Step 3: Calculate: 1/(4πε₀) = 9 × 109 N m2 C-2 F = 9 × 109 × (1 × 10-6)2 / 12 = 9 × 10-3 N. Final Answer: F = 9 milli-Newtons (mN) Common Mistakes and Best Practices with Epsilon Naught \| Mistake \| How to Avoid \| --- \| \| Incorrect units for ε₀ \| Always use C2 N-1 m-2 (SI) \| \| Using ε₀ in place of ε for materials \| For materials: Use ε = εr × ε₀ \| \| Confusing ε₀ (electric) and μ₀ (magnetic) \| ε₀ is for electric field calculations, μ₀ is for magnetic field \| \| Memorizing the wrong numeric value \| Use 8.854 × 10-12 as standard in exams \| How to Master Epsilon Naught for Physics Problems Memorize the units and numeric value of ε₀ accurately. Always check if your calculation is for vacuum (use ε₀) or a material (use ε = εr × ε₀). Apply ε₀ in all electrostatics problems involving forces, fields, capacitance, and Gauss’s law. Review solved problems and keep a formula table handy during practice. Check for similar constants (like μ₀) to avoid conceptual mix-ups. Practice and Next Steps Deepen understanding at Permittivity & Permeability. Enhance problem-solving skills with Electric Field Numericals and Practice on Capacitors. Explore related topics like Gauss’s Law, Electrostatics, and Dielectric Constant for a broader foundation. Quick Recap ε₀ is the permittivity of free space, key in all electrostatics formulas. Value: 8.854 × 10-12 C2 N-1 m-2. Use the correct units and context (vacuum or material) for accurate Physics problem-solving. Refer to Vedantu resources for more practice and concept clarity. Keep practicing related questions and always verify the value and units of ε₀ while solving problems. This approach ensures accuracy and builds confidence in Physics. FAQs on Epsilon Naught (ε₀) Value, Units & Application in Physics What is the value of epsilon naught (ε₀) in SI units? Epsilon naught (ε₀), the permittivity of free space, has the SI value of 8.854 × 10-12 C2 N-1 m-2. This constant is fundamental in Physics for calculations in electrostatics and appears in key formulas such as Coulomb’s Law and Gauss’s Law. What does epsilon naught (ε₀) represent in Physics? Epsilon naught (ε₀) is the permittivity of free space (vacuum). It measures how much the vacuum opposes the formation of an electric field. Its value is essential for understanding and calculating electric force, electric fields, and capacitance in vacuum or air. Why is epsilon naught important in electrostatics? Epsilon naught (ε₀) determines how strongly electric charges interact in vacuum. It is crucial in: Coulomb’s Law: Calculating force between point charges Gauss’s Law: Determining electric flux and field distribution Capacitance: Computing the capacity of capacitors in vacuum/air What is the unit of epsilon naught (ε₀)? The SI unit of epsilon naught (ε₀) is C2 N-1 m-2. It may also be expressed in base units as A2 s4 kg-1 m-3. Always use these units when solving problems in the SI system. How is epsilon naught used in Coulomb’s Law? In Coulomb’s Law, epsilon naught (ε₀) appears in the denominator as 1/(4πε₀). The formula for the force between two point charges is: F = (1/4πε₀) × (q₁q₂/r²), where F is force, q₁ and q₂ are charges, and r is the distance between them. Can epsilon naught be used for materials other than vacuum? Epsilon naught (ε₀) specifically represents the permittivity of free space. For other materials, use: ε = εr × ε₀, where εr is the relative permittivity or dielectric constant of the material. What is the value of 1/(4πε₀)? 1/(4πε₀) is a constant used in electrostatics with the value 8.99 × 109 N m2 C-2. It simplifies the calculation of electric force between point charges in SI units. How can I memorize the value and units of epsilon naught (ε₀) for exams? To memorize ε₀ effectively: Remember the rounded value: 8.85 × 10-12 C2 N-1 m-2 Create mnemonics or visual aids linking ε₀ to vacuum and electrostatics Write and practice using ε₀ in different formulas until it becomes familiar What is the relation between epsilon naught (ε₀), the speed of light (c), and permeability (μ₀)? Epsilon naught (ε₀) is related to the speed of light (c) and the permeability of free space (μ₀) by the formula: ε₀ = 1 / (μ₀c²) This connects electric and magnetic constants in Physics. What are some common mistakes when using epsilon naught (ε₀)? Common mistakes include: Using the wrong units (always check for C2 N-1 m-2 in SI) Applying ε₀ in non-vacuum problems (use ε = εr × ε₀ for dielectrics) Confusing ε₀ (electrostatics) with μ₀ (magnetism) Forgetting to use accurate or rounded values as per exam guidelines Which key formulas in Physics use epsilon naught (ε₀)? Key formulas involving ε₀ include: Coulomb’s Law: F = (1/4πε₀)(q₁q₂/r²) Gauss’s Law: ΦE = Q/ε₀ Electric field due to point charge: E = (1/4πε₀)(q/r²) Capacitance of parallel plate capacitor: C = ε₀A/d Is epsilon naught (ε₀) included in the latest Physics syllabus for competitive exams? Yes, epsilon naught (ε₀) is a fundamental Physics constant included in the latest syllabus for exams such as JEE, NEET, and CBSE. It is essential for topics including electrostatics, capacitance, and electromagnetic theory. Always refer to the current syllabus and official notifications for updates. 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17131	https://www.hmhco.com/blog/strategies-for-teaching-math-to-english-language-learners-ells?srsltid=AfmBOorcr1umqv4frc9OKJVo9BXmym4S1yQ2Z4sMsZAE9qcyT3ndM9rA	English Language Learner (ELL) Strategies for Math \| HMH Curriculum Literacy Core Curriculum Into Literature, 6-12 Into Reading, K-6 See all Literacy Performance Suite Intervention English 3D, K-12 Read 180, 3-12 See all Reading Intervention Assessment Supplemental A Chance in the World SEL, 8-12 Amira Learning, PreK–8 Classcraft, K-8 JillE Literacy, K-3 Waggle, K-8 Writable, 3-12 See all Reading Supplemental Personalized Path Math Core Curriculum Arriba las Matematicas, K-8 Go Math!, K-6 Into Algebra 1, Geometry, Algebra 2, 8-12 Into Math, K-8 Math in Focus, K-8 See all Math Performance Suite Supplemental Classcraft, K-8 Waggle, K-8 See all Supplemental Math Personalized Path Intervention Math 180, 3-12 See all Math Intervention See all Assessment Science Core Curriculum Into Science, K-5 Into Science, 6-8 Science Dimensions, K-12 See all Science Readers ScienceSaurus, K-8 Social Studies Core Curriculum HMH Social Studies, 6-12 See all Social Studies Supplemental Writable Professional Development For Teachers Coachly Teacher's Corner Live Online Courses Program-Aligned Courses See all Professional Development For Leaders The Center for Model Schools More AI Tools on HMH Ed Assessment Early Learning English Language Development Homeschool Intervention Literacy Mathematics Professional Development Science School Improvement Social Studies Special Education Summer School See all Solutions Resources Browse Resources AI Tools Classroom Activities Customer Resource Hub Customer Success Stories Digital Samples Events Grants & Funding Family & Caregiver Resources International Product Updates Research Library Shaped, HMH Blog Webinars Customer Support Contact Sales Customer Service & Technical Support Portal Platform Login Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA Shop Support Platform Login From setup to support, HMH will ensure your back-to-school season runs smoothly. 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Core Curriculum Arriba las Matematicas, K-8 Go Math!, K-6 Into Algebra 1, Geometry, Algebra 2, 8-12 Into Math, K-8 Math in Focus, K-8 See all Math Performance Suite Supplemental Classcraft, K-8 Waggle, K-8 See all Supplemental Math Personalized Path Intervention Math 180, 3-12 See all Math Intervention See all Assessment HMH Personalized Path Discover a solution that provides K–8 students in Tiers 1, 2, and 3 with the adaptive practice and personalized intervention they need to excel. Optimizing the Math Classroom: 6 Best Practices Our compilation of math best practices highlights six ways to optimize classroom instruction and make math something all learners can enjoy. 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Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA HMH Careers Exploring a career at HMH? Do work that matters. Learn about our culture, benefits, and available job opportunities. Accessibility Explore HMH’s approach to designing affirming and accessible curriculum materials and learning tools for students and teachers. Shop Support 0 Log in Back to Shaped Math 7 Strategies for Teaching Math to English Language Learners Richard Blankman September 10, 2021 8 Min Read Mathematics is sometimes regarded as a “universal language,” as it is grounded in abstract concepts such as numbers and shapes. And there is some truth to that! After all, two people who speak different languages may still be able to solve the same math problem and get the same solution. However, language is nevertheless essential for teaching math, as mathematics is inextricably linked to the language in which it is taught. Proficiency in any language relies on two factors: language comprehension and language production. When it comes to teaching math to ELL students, we can look closely across lessons and consider, “What is the language load at this moment? Comprehension or production?” Then we can leverage targeted strategies to promote academic language development while also deepening mathematical understanding. Students learning English carry a different cognitive load when they are also learning mathematics. What follows are ELL strategies for math that lighten the students' cognitive load so that they can show mathematical understanding while you maintain rigor in the lesson. A note on language: we employ the common phrase English language learner, along with the acronym ELL, but we also recognize that this is imperfect nomenclature. Students who are learning English do not fit neatly into a single label. ELL Strategies for Math: Language Comprehension 1: Use Routines That Break Down Word Problems Word problems present an especially tough problem to students learning English. They’re hard enough for students who are further along in their English language development! Word problems demand that the reader parse plenty of non-mathematical vocabulary such as names, objects, jobs, and places. Mathematics instruction will always include word problems, as that is how to articulate complex mathematical situations. Strategies like Three Reads, Stronger and Clearer Each Time, and Compare and Connect help break down the context of a problem and guide students in focusing on one aspect of the language at a time. 2: Focus on Mathematical Vocabulary Learning mathematics is so much more than learning mathematical vocabulary. Spend time on the words. For multilingual learners, vocabulary can sometimes be an entryway into the math. Students may have existing notions about words such as product, times, and one, for example, and by discussing them, you are connecting about ideas on both language and mathematics. Your students will help you learn other languages as you help them learn math, a symbiotic relationship. Provide vocabulary instruction upfront both for mathematical vocabulary and non-mathematical vocabulary that appears in word problems students will confront in the lesson. Vocabulary instruction involves more than just reviewing definitions. It can include playing vocabulary games, reading math readers, and using tools such as word banks. 3:Connect with Familiar Contexts What would engage your students, regardless of language? Think about what sports they play, music they listen to, TV shows they watch, food they eat, or holidays they celebrate (to name a few). If you can draw on your students’ cultural knowledge, you can open paths to engage them even when language serves as a temporary barrier. Look for ways to modify contexts of word problems and possible bridges between math that they understand and the content you’re trying to teach. Position your students as mathematically competent regardless of how well they can articulate their reasoning in English. If you can, let them make up their own word problems! Then you learn more about what is familiar to them. ELL Strategies for Math: Language Production 4: Use Sentence Frames Sentence frames allow students to practice speaking but may help them feel less intimidated. They offer a scaffold to students who understand mathematical ideas but get caught up in English grammar. The scaffold extends to students whose first language is English but nevertheless struggle to articulate mathematical ideas. By employing sentence frames in your classroom, you also support learning vocabulary (see Strategy 2: Focus on Mathematical Vocabulary). It gives students clear contexts in which to practice the vocabulary, and the vocabulary itself can be part of the sentence frame, for example “This polygon (is/is not) a quadrilateral because _____.” Seek out sentence frames that are both lesson-specific and content-agnostic. HMH math professional learning includes a treasure trove of examples like the ones below: A _ (is/is not) a because __. I can conclude that _____. While a _ and a both have a __, they are different because _____. If _, then will __. A _ is a polygon because and __. 5: Create a Low-Stakes Conversation Space It is important that students can test out ideas aloud in a low-stakes environment before being asked to share with the whole class. Look for ways that students who don’t participate much yet can join the conversation. For example: Ask questions that students can answer using gestures or drawings. Provide students with sentence frames to help them express their ideas in English. Have students practice sharing their ideas in lower-stakes partner or small group discussions before being asked to share out with the whole class. There is no need to reduce the complexity of a math lesson even when you must think about the complexity of language. Incorporate routines and structures that equip students to participate in discussions and share their ideas. As you plan instruction, ask yourself questions like the examples below to help promote student participation along the way: Whose ideas will I record? How will students express their ideas? What visual models can students use to express their ideas? When will I provide sentence frames? When is there time in the lesson for students to think independently and discuss with a partner or small group? When possible, encourage students to think, write, and enter math conversations in any language available to them. It’s unreasonable to task you with picking up another language, but long gone are the days where you have to look everything word by word in the dictionary. Translate letters using Google Translate or, if your school has the option, look for learning software that’s available in multiple languages. 6: Use Tools, Visual Models, and Manipulatives It is important to remember that the barrier is language, not necessarily knowledge. These strategies aren't just good practice for multilingual learners. All learners who have working eyesight benefit from visual learning, and the same applies to physical learning with manipulatives. Math is unique in how often and how flexibly it can make use of visual and physical models. Yet when the learner is multilingual, models can go from best practice to critical mathematical scaffolds. When learning is digital, the line between visual and physical models becomes blurred. For many students, in fact, digital models are especially helpful. Some learners feel more comfortable using tools when they can do so independently, and potentially anonymously. When it comes to arithmetic in particular, students may draw calculations using algorithms and models that work but are new to you. Arithmetic falls into this gray area between procedural math and conceptual math, and as such is prone to regional differences. Just look at how division "looks" to a Venezuelan, U.S., and French student: In general, have students show you what they know. And lean into the differences! There is rich discourse to be had when comparing different ways to divide (or multiply or add or subtract) numbers, along with other possible cultural differences such as how to notate numbers or draw graphs. 7: Collaborate with Colleagues This strategy is possibly the biggest one of all and transcends simply helping with language production. It helps with comprehension too; in fact, it helps with practically everything! Only you know the roles within your community, school, or district. Are there translation services? School-community liaisons? Dedicated English language development teachers? Ask your colleagues what is available, and use it! Don’t forget about ELA instructors either. It doesn’t matter what grade you teach. If you are teaching in English, then it is impossible to fully separate math content from English content. We don’t have a “left brain” and a “right brain,” we just have brains. The skills that come from learning to read and write in a language improve one’s math ability, too, and in this way, the ELA educators in your school or district could become valuable contributors to your classroom. How to Support ELL Students in Math When supporting English language learners in math class, we often don’t know if the problem stems from a language need or a math need. Yet it is important to think constantly in terms of assets, not deficits. A class that collectively speaks multiple languages is a cultural gold mine! With Spanish language components, multilingual family letters, and ELL activity guides, HMH mathematics solutions are designed with the multilingual learner in mind. 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17132	https://brainly.com/question/39822831	[FREE] Which of the following shows the correct number of protons, neutrons, and electrons in a neutral cesium-134 - brainly.com 7 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +77,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +36,5k Ace exams faster, with practice that adapts to you Practice Worksheets +5k Guided help for every grade, topic or textbook Complete See more / Chemistry Expert-Verified Expert-Verified Which of the following shows the correct number of protons, neutrons, and electrons in a neutral cesium-134 atom? A. Protons: 55, Neutrons: 55, Electrons: 55 B. Protons: 55, Neutrons: 79, Electrons: 55 C. Protons: 55, Neutrons: 79, Electrons: 79 D. Protons: 79, Neutrons: 55, Electrons: 79 E. Protons: 134, Neutrons: 55, Electrons: 134 1 See answer Explain with Learning Companion NEW Asked by onmilevel6842 • 10/10/2023 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 1081413 people 1M 0.0 0 Upload your school material for a more relevant answer The correct number of protons, neutrons, and electrons in a neutral cesium-134 atom is 55 protons, 79 neutrons, and 55 electrons. Cesium's atomic number is 55, meaning it has 55 protons (and in a neutral state, 55 electrons), while the number of neutrons is found by subtracting the atomic number from the atomic mass. Explanation The correct option for number of protons, neutrons, and electrons in a neutral cesium-134 atom would be B. ProtonsNeutronsElectrons557955. To explain, in a neutral atom, the number of protons equals the number of electrons. Cesium has an atomic number of 55, meaning it has 55 protons and therefore, in a neutral state, also has 55 electrons. The number of neutrons can be found by subtracting the atomic number from the atomic mass, so in the case of cesium-134, 134 - 55 equals 79 neutrons. Learn more about electrons here: brainly.com/question/7328982 SPJ11 Answered by Priyanka2811 •11.6K answers•1.1M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 1081413 people 1M 0.0 0 Upload your school material for a more relevant answer In a neutral cesium-134 atom, there are 55 protons, 79 neutrons, and 55 electrons. The atomic number indicates the number of protons and electrons, while neutrons are found by subtracting the atomic number from the mass number. The correct multiple-choice answer is B. Explanation To find the correct number of protons, neutrons, and electrons in a neutral cesium-134 atom, we first look at the atomic number and the mass number. Protons: The atomic number of cesium (Cs) is 55. This means there are 55 protons in a cesium atom. Electrons: In a neutral atom, the number of electrons equals the number of protons. Therefore, a neutral cesium-134 atom also has 55 electrons. Neutrons: The mass number of cesium-134 is 134. Neutrons can be calculated by subtracting the atomic number from the mass number. Thus, the number of neutrons is: Neutrons=Mass Number−Atomic Number=134−55=79 Therefore, cesium-134 has 79 neutrons. To summarize: Protons: 55 Neutrons: 79 Electrons: 55. The correct option from the provided choices is B. Protons: 55, Neutrons: 79, Electrons: 55. Examples & Evidence For example, to find the number of neutrons in oxygen-16, you subtract the atomic number of oxygen (8) from the mass number (16), which gives you 8 neutrons. Similarly, cesium-134 follows this same method of calculation. The atomic number and mass number are standard concepts in chemistry that help identify the structure of atoms, supported by the periodic table and atomic theory. Thanks 0 0.0 (0 votes) Advertisement onmilevel6842 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. 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Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. Community Answer 8 For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval mc015-1. Jpg? mc015-2. Jpg mc015-3. Jpg mc015-4. Jpg mc015-5. Jpg. Community Answer 4.8 268 How did Mendeleev come up with the first periodic table of the elements? (1 point) A He determined the mass of atoms of each element. B He estimated the number of electrons in atoms of each element. C He arranged the elements by different properties to find a pattern. D He organized the elements by their atomic number. New questions in Chemistry State what happens to the rate constant. This overall reaction between hydrogen and nitrogen monoxide takes place by a two-step mechanism. The first step is much slower than the second step. Suggest a possible two-step mechanism for the overall reaction. step 1: step 2: Ionic compounds are compounds that A. consist only of metallic elements. B. have atoms that transfer electrons. C. exist as individual ions. D. have atoms that share electrons. What is the term for a combination of two or more substances in any proportions in which the substances do not combine chemically to form a new substance? A. mixture B. compound C. molecule D. element What happens to the acidity of solutions A and B after dry ice is added? \| Acidity Changes after Dry Ice Is Added \| \| Time (sec) \| \| pH of Solution A \| pH of Solution B \| \| 0 \| \| 11.0 \| 8.5 \| \| \| \| 10.8 \| 8.0 \| \| \| \| 10.2 \| 7.2 \| \| \| \| 9.6 \| 6.5 \| \| \| \| 9.0 \| 5.8 \| A. The pH of solution A decreases, and the pH of solution B increases. B. The pH of solution A increases, and the pH of solution B decreases. C. The pH of solution A increases, and the pH of solution B increases. D. The pH of solution A decreases, and the pH of solution B decreases. Which of the following is unique for any given element? 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17133	https://math.stackexchange.com/questions/1796312/what-is-the-degree-of-the-zero-polynomial-and-why-is-it-so	Skip to main content What is the degree of the zero polynomial and why is it so? [duplicate] Ask Question Asked Modified 3 years, 3 months ago Viewed 49k times This question shows research effort; it is useful and clear 21 Save this question. Show activity on this post. My teacher says- The degree of the zero polynomial is undefined. My book says- The degree of the zero polynomial is defined to be zero. Wikipedia says- The degree of the zero polynomial is −∞. I am totally confused and want to know which one is true or are all true? polynomials Share CC BY-SA 3.0 Follow this question to receive notifications edited May 23, 2016 at 7:25 psmears 79055 silver badges1111 bronze badges asked May 23, 2016 at 5:58 SohamSoham 10.2k88 gold badges4747 silver badges9696 bronze badges 7 13 Wanna be more confused? I think I have also seen it be defined as −1. Haha. – Em. Commented May 23, 2016 at 6:04 4 It's a matter of convention, and there are a few reasons to argue that the best convention is −∞. – Qiaochu Yuan Commented May 23, 2016 at 8:29 It happens all the time that a Wikipedia page only reflects the page author's thought. Just don't consider it to be the ultimate truth. On the other hand, I very much prefer to consider the degree of the zero polynomial either equal to −∞ or to leave it undefined. When it comes to definitions, none is right or wrong: definitions don't admit “proofs”. – egreg Commented May 23, 2016 at 9:36 5 Yet, I'd argue that the book is wrong, since defining deg(0)=0 thwarts one of the main uses of degree (for univariate polynomials over a field), namely to get the property that Euclidean division of any a by any nonzero element b is possible leaving a remainder with strictly lower degree than deg(b). – Marc van Leeuwen Commented May 23, 2016 at 10:17 1 I agree, the book is wrong. What book is that?! The definition used by your teacher and the definition used by Wikipedia are both useful and common. So you have to be careful what definition the author you read uses. – Jeppe Stig Nielsen Commented Mar 14, 2017 at 8:57 \| Show 2 more comments 3 Answers 3 Reset to default This answer is useful 12 Save this answer. Show activity on this post. Well, it depends. Mathematical practice shows that sometimes it is useful to define the degree of the zero polynomial to be zero, sometimes to define it to be −∞ and sometimes to leave is undefined. Which option one chooses depends on what one is trying to do. This is quite different with what happens with the degree of all other polynomials, which is always defined in the same way () But don't think that if for the slightiest of reasons we were to fnd it useful to change the definition to do something we wanted, we would. () Actually, that is not exactly true: we sometimes put degrees on polynomials which are different from the usual ones, but usually only on polynomials with more than one variable. Share CC BY-SA 3.0 Follow this answer to receive notifications edited May 23, 2016 at 6:03 answered May 23, 2016 at 6:01 Mariano Suárez-ÁlvarezMariano Suárez-Álvarez 140k1414 gold badges260260 silver badges395395 bronze badges 8 1 Can you give an example of where each kind is used? – Soham Commented May 23, 2016 at 6:03 14 Well, for example: the degree of the product of two nonzero polynomials is the sum of the degrees of the factors. If you want to extend this to include the possibility that the factors be zero, one needs to do something, and a natural option is to define the degree of 0 to be −∞. – Mariano Suárez-Álvarez Commented May 23, 2016 at 6:05 10 For some purposes it i useful to know that the degree of a polynomial f to be the least number i such that the ith derivative f(i+1) is equal to zero, and if this is going to work for the zero polynomial too, then you need to define its degree to be −1. &c. – Mariano Suárez-Álvarez Commented May 23, 2016 at 6:09 9 So it has yet to be found where deg0=0 is any good. – Vim Commented May 23, 2016 at 6:25 1 There are also contexts for which it makes sense to define the degree of the zero polynomial to be +∞. For example, for nonzero polynomials P and Q, it's true that if P divides Q then degP≤degQ; setting deg0=+∞ is the only way to extend this fact to the zero polynomial. (Another reason: degP equals the number of roots of a polynomial over C.) – Greg Martin Commented May 23, 2016 at 8:19 \| Show 3 more comments This answer is useful 12 Save this answer. Show activity on this post. I think −∞ make sense. Indeed, let P a polynomial of degree ≥1. Then, you have that deg(PQ)=deg(P)+deg(Q), for every polynomial Q. Now, if you define deg(0)=0, you'll get deg(0⋅P)=0+deg(P)>0, which is not compatible with the degree formula. The only way to give a sense to this formula is to define deg(0)=−∞. Same if you defined deg(0)=−1, the formula won't be compatible if deg(P)≥2. Share CC BY-SA 4.0 Follow this answer to receive notifications edited May 13, 2022 at 9:42 Dietrich Burde 141k88 gold badges9898 silver badges176176 bronze badges answered May 23, 2016 at 6:15 SurbSurb 57.3k1111 gold badges6868 silver badges119119 bronze badges 5 It makes sense if what you want is your first displayed formula to hold. If you are not interested in that, then it does not make a lot of sense for you migh be interested in some other property to hold which forces you to define deg0 differently. This is a matter of convention, and conventions are chosen with some purpose in mind. – Mariano Suárez-Álvarez Commented May 23, 2016 at 6:20 1 Indeed, only −∞ can act as a black hole. But anyway this's only one aspect of mathematical practices. – Vim Commented May 23, 2016 at 6:24 4 @Vim This isn't really enough to justify −∞: why not +∞ which also acts as a black hole? (Yes, I understand why, but this isn't explained in the answer). In fact, naively ∞ is a very natural suggestion since it generalizes the theorem "a polynomial of degree d has at most d roots". – Erick Wong Commented May 23, 2016 at 6:34 @ErickWong I'd say its because there are some theorems that say "if degP>degQ, then blah blah" but unfortunately I've forgotten most of the algebra I learned so I can't name any :( – Vim Commented May 23, 2016 at 6:48 @Vim You're probably thinking of "if degP>degQ then deg(P+Q)=degP". If deg0=+∞ then we could take P=0, and then we'd have degQ=+∞ for any Q. – Erick Wong Commented May 23, 2016 at 7:46 Add a comment \| This answer is useful 10 Save this answer. Show activity on this post. Defining it as −∞ makes the most sense. As mentioned in Surb's answer and comments, some properties of degrees are kept intact this way, e.g. deg(PQ)=degP+degQ If degP>degQ then deg(P+Q)=degP It also starts making more sense if you consider expressions that can take on negative powers as well. That is, instead of ∑nk=0akxk, consider ∑nk=−∞akxk. So you could have 3x2+2x and x+1+3x−2 and 2x−3−45x−5. Then degree is just "the supremum of all ks for which ak≠0. The degrees of these 3 expressions are 2, 1 and -3 respectively. Then it's easy to see that 0, which has no nonzero coefficients, has a degree of −∞. The same works if you consider expressions than can also have fractional degrees. Then the degree of, say, 3x−−√−x−3 is 1/2. Of course, this inspires the definition of a "dual" degree, which is the infimum instead of the supremum. Then the degree of 3x4+2x3+5x2 will be 2, and the degree of 0 will be ∞. Keeping the degree of 0 undefined is understandable (not everyone wants to deal with infinities). Defining it as −1 has merits (if you don't consider negative powers, 0 is one step down from nonzero constants). But there is absolutely no sense in defining the degree as 0. The 0 polynomial has as much similarity with constants, as constants have with linear polynomials. Share CC BY-SA 3.0 Follow this answer to receive notifications answered May 23, 2016 at 8:52 Meni RosenfeldMeni Rosenfeld 4,11322 gold badges1717 silver badges2626 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions polynomials See similar questions with these tags. 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17134	https://math.stackexchange.com/questions/839029/find-all-triples-of-positive-integers-a-b-and-c-satisfying-the-equations-a-b	elementary number theory - Find all triples of positive integers a,b and c satisfying the equations: $ (a,b,c)=10$ and$ [a,b,c]=100$ simultaneously. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find all triples of positive integers a,b and c satisfying the equations: (a,b,c)=10(a,b,c)=10 and[a,b,c]=100[a,b,c]=100 simultaneously. Ask Question Asked 11 years, 3 months ago Modified17 days ago Viewed 4k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. This method below worked for gcd and lcm of two integers a,b, but I'm having trouble using the same argument. Perhaps that is the reason, but then I'm stuck with how to move forward. (a,b,c)=10,(a,b,c)=10, Let a=10 a′,b=10 b′,c=10 c′a=10 a′,b=10 b′,c=10 c′. Then (a,b,c)=(10 a′,10 b′,10 c′)=10.(a,b,c)=(10 a′,10 b′,10 c′)=10. But, (10 a′,10 b′,10 c′)=10(a′,b′,c′)=10⇒(a′,b′,c′)=1(10 a′,10 b′,10 c′)=10(a′,b′,c′)=10⇒(a′,b′,c′)=1 By the product of the greatest common factor and least common multiple, [a,b,c]=a b c(a,b,c)⇒100=1000 a′b′c 10⇒1=a′b′c′[a,b,c]=a b c(a,b,c)⇒100=1000 a′b′c 10⇒1=a′b′c′ This leads to an erroneous conclusion. Without rigor I can see that solutions include (10,20,50),(20,20,50),(10,10,100),(10,100,100),(20,50,50),(20,50,100),(10,20,100),(10,50,100)(10,20,50),(20,20,50),(10,10,100),(10,100,100),(20,50,50),(20,50,100),(10,20,100),(10,50,100), but how do I show all solutions? elementary-number-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Dec 7, 2016 at 15:43 user371838 asked Jun 18, 2014 at 23:21 LalaloopsyLalaloopsy 2,063 2 2 gold badges 24 24 silver badges 43 43 bronze badges 2 You got them all.Ross Millikan –Ross Millikan 2014-06-19 00:19:37 +00:00 Commented Jun 19, 2014 at 0:19 Thanks. I was more looking for the reason that these 8 were the ones, so thanks to you and the rest for helping clarify my missing understanding.Lalaloopsy –Lalaloopsy 2014-06-19 00:24:29 +00:00 Commented Jun 19, 2014 at 0:24 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. The only mistake you've got is to assume that [a,b,c]=a b c(a,b,c)[a,b,c]=a b c(a,b,c), which is not true. You shurely inspired in the old formula for 2 numbers [a,b]=a b(a,b)[a,b]=a b(a,b). If you want to derive a formula for 3 numbers you should thing as follows, and I warn you by now that you should try this by yourself, the following is not so nice to read :/ - [a,b,c]=[a′d,b′d,c′d]=d[a′,b′,c′][a,b,c]=[a′d,b′d,c′d]=d[a′,b′,c′] where d=(a,b,c),(a′,b′,c′)=1 d=(a,b,c),(a′,b′,c′)=1, you want to know what is [a′,b′,c′][a′,b′,c′] and at this point you can see that it's not always a′b′c′a′b′c′. Set a′=∏p∣a′p α p a′=∏p∣a′p α p, b′=∏p∣b′p β p b′=∏p∣b′p β p, c′=∏p∣c′p γ p c′=∏p∣c′p γ p. Looking at the prime factorisation, [a′,b′,c′]=∏p∣a′b′c′p m a x(α p,β p,γ p)[a′,b′,c′]=∏p∣a′b′c′p m a x(α p,β p,γ p) and you know by the fact (a′,b′,c′)=1(a′,b′,c′)=1 that for each p p prime, one of α p,β p,γ p α p,β p,γ p is zero, so p m a x(α p,β p,γ p)=p α p p β p p γ p p m a x 2(α p,β p,γ p)p m a x 3(α p,β p,γ p)p m a x 3(α p,β p,γ p)p m a x(α p,β p,γ p)=p α p p β p p γ p p m a x 2(α p,β p,γ p)p m a x 3(α p,β p,γ p)p m a x 3(α p,β p,γ p), where m a x 2,m a x 3 m a x 2,m a x 3 stands for the second maximum and third maximum. Multiplying for each prime we get [a′,b′,c′]=a′b′c′(a′,b′)(b′,c′)(c′,a′)=a b c(a,b)(b,c)(c,a)[a′,b′,c′]=a′b′c′(a′,b′)(b′,c′)(c′,a′)=a b c(a,b)(b,c)(c,a) so [a,b,c]=(a,b,c)a b c(a,b)(b,c)(c,a)[a,b,c]=(a,b,c)a b c(a,b)(b,c)(c,a) You should anyways keep the job without using this formula, as [a,b,c]=100⇒[a′,b′,c′]=10[a,b,c]=100⇒[a′,b′,c′]=10 is small enough to brute-force all the cases: assume a≤b≤c a≤b≤c then for a=1,b=1 a=1,b=1 we get c=10 c=10, for a=1,b=2 a=1,b=2 we get c=10 c=10 or c=5 c=5, etc... Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 19, 2014 at 1:27 PenasRaulPenasRaul 1,234 7 7 silver badges 10 10 bronze badges 1 I stared at the formula from my example and knew that it was most likely wrong. Thank you for deriving the gcd triple! I realy appreciate it.Lalaloopsy –Lalaloopsy 2014-06-19 01:42:03 +00:00 Commented Jun 19, 2014 at 1:42 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. If a=p e 1 1⋯p e k k a=p 1 e 1⋯p k e k and b=p f 1 1⋯p f k k b=p 1 f 1⋯p k f k are prime factorizations, then gcd(a,b)=p min{e 1,f 1}1⋯p min{e k,f k}k gcd(a,b)=p 1 min{e 1,f 1}⋯p k min{e k,f k}. Similarly for [a,b][a,b] you look for the max of the exponents in the prime factorization. This holds also for three numbers a,b,a,b, and c c. So our numbers can only have prime factors 2 and 5 and we need minimum and maximum exponents to statisfy (a,b,c)=10=2 1⋅5 1(a,b,c)=10=2 1⋅5 1 and [a,b,c]=2 2⋅5 2[a,b,c]=2 2⋅5 2. This doesn't leave many options. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 18, 2014 at 23:54 John MachacekJohn Machacek 3,026 2 2 gold badges 15 15 silver badges 17 17 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Every one of a,b,c a,b,c is a multiple of 10 10 and divides into 100 100, so they can each only be 10,20,50,100 10,20,50,100 To have the GCD be exactly 10 10, we either need one to be 10 10 or to have one be 20 20 and another be 50 50. To have the LCM be exactly 100 100 we either need one to be 100 100 or one be 20 20 and another 50 50. We might as well assume a≤b≤c a≤b≤c and permute. This means we can have (10,10,100),(10,20,100),(10,50,100),(10,100,100),(10,20,50),(20,20,50),(20,50,50),(20,50,100)(10,10,100),(10,20,100),(10,50,100),(10,100,100),(10,20,50),(20,20,50),(20,50,50),(20,50,100) and all permutations of these. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 19, 2014 at 0:17 Ross MillikanRoss Millikan 384k 28 28 gold badges 264 264 silver badges 472 472 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Since GCD is 10 10 All numbers are divisible by 2 2 and 5 5, there is at least one number exactly divisible by 2 2 and at least one number exactly divisible by 5 5. Also no numbers are divisible by other primes. Since L C M L C M is 100 there is at least 1 number exactly divisible by 4 4 and at least 1 number exactly divisible by 25 25. No number is divisible by powers of 3 3 or higher. We look at possibilities of powers of 2 2 in prime factorization: 2,4,4 2,4,4 or 2,2,4 2,2,4 (6 6 ways counting permutations) and now we look at the powers of 5 5 5,25,25 5,25,25 or 5,5,25 5,5,25 (6 6 ways counting permutations). Thus there are 36 36 triples. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 11 at 8:27 answered Jun 18, 2014 at 23:28 AsinomásAsinomás 108k 25 25 gold badges 139 139 silver badges 290 290 bronze badges 15 I also need [a,b,c]=100[a,b,c]=100 and your solutions do not satisfy that.Lalaloopsy –Lalaloopsy 2014-06-18 23:30:47 +00:00 Commented Jun 18, 2014 at 23:30 It is fixed now Asinomás –Asinomás 2014-06-18 23:43:52 +00:00 Commented Jun 18, 2014 at 23:43 This is way off Eleven-Eleven –Eleven-Eleven 2014-06-18 23:45:35 +00:00 Commented Jun 18, 2014 at 23:45 Really, then how many triples are there?Asinomás –Asinomás 2014-06-18 23:47:42 +00:00 Commented Jun 18, 2014 at 23:47 1 @19021605 it does not, not all of those triples produce 6 distinct permutations.Asinomás –Asinomás 2025-09-09 18:39:18 +00:00 Commented Sep 9 at 18:39 \|Show 10 more comments You must log in to answer this question. 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17135	https://www.youtube.com/playlist?list=PLSQl0a2vh4HBvATVRLb7XhazoLvhAwTZX	Exponent expressions and equations \| Algebra I \| Khan Academy - YouTube Back Skip navigation Search Search with your voice Sign in Home HomeShorts ShortsSubscriptions SubscriptionsYou YouHistory History Play all Exponent expressions and equations \| Algebra I \| Khan Academy by Khan Academy • Playlist•48 videos•69,962 views Play all PLAY ALL Exponent expressions and equations \| Algebra I \| Khan Academy 48 videos 69,962 views Last updated on Jul 9, 2018 Save playlist Shuffle play Share Show more Khan Academy Khan Academy Subscribe Play all Exponent expressions and equations \| Algebra I \| Khan Academy by Khan Academy Playlist•48 videos•69,962 views Play all 1 2:36 2:36 Now playing Exponent properties 1 \| Exponent expressions and equations \| Algebra I \| Khan Academy Khan Academy Khan Academy • 378K views • 14 years ago • 2 7:32 7:32 Now playing Fractional exponent expressions 2 \| Exponent expressions 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17136	https://demonstrations.wolfram.com/BlendedQuadrilaterals/	Blended Quadrilaterals \| Wolfram Demonstrations Project 13,000+ Open Interactive Demonstrations From the makers of Wolfram Language and Mathematica Wolfram Demonstrations Project Topics Latest Random Authoring Notebook Blended Quadrilaterals p a i n t i n g 2 1 o f 3 0 s r s e l e c t<<<>>> z o o m1. s h i f t n e s w N E S W c l i c k s: 0 d e n s i t y1 6 5. t i l e 1 o f 3 ,n e w- [x] (n o e d i t) <<<>>> (a c t i o n) - [x] C o l o r s e l e c t i o n s R e d B l u e s w a p- [x] r a n d o m- [x] r o t- [x] r e s e t- [x] d e l e t e- [x] s n a p- [x] s h o w t i l e n u m b e r s- [x] p o i n t s- [x] a r e a s- [x] b o a r d- [x] g r i d- [x] Initializing live version Open Notebook in CloudCopy Manipulate to Clipboard Source Code Use this Demonstration to create a piece of art with color-blending tiles. Contributed by: Karl Scherer(2017) Open content licensed under CC BY-NC-SA Snapshots Details Introduction Let us start with painting 21. The small square and two small triangles are examples of shapes called tiles. Some of these tile types can be used to cover two-dimensional areas without gaps or overlaps. The blending of colors on these tiles gives mosaics with a fascinating effect that cannot be achieved with monochrome tiles (e.g. paintings 3, 22, 23 and 24), especially when adjacent sides match. Try to put lines of solid color next to lines of solid color (with the same hue), and place lines of blending colors next to blending lines of the same hue. Painting 3 shows a good example of this. Each tile is defined by four mouse clicks on the board (while "no edit" is selected), resulting in a square or triangle (any quadrilateral can be used), each with a blend of two colors. Certain tile sets force a special type of result. Here are just a few: Tile set 1: only tiles of the type in painting 21 (square and two triangles) plus their rotations (multiples of 90 degrees on the orthogonal grid, multiples of 60 degrees on the isometric grid) and their mirror images. Tile set 2: In addition to tile set 1, tiles that are monochrome (both colors are the same) are also allowed. Tile set 3: In addition to tile set 2 any quadrilaterals (and triangles) of any shape are also allowed. Tile set 4: In addition to tile set 3 gaps and overlaps are allowed. First Exercises Look at painting 21. It shows one large, light gray area, a small square and two triangles. The small square and the two triangles show a mix of red and blue hues. We get back to this later. The small square has four vertices colored black, red, blue and green (and marked A, B, C and D). Two sides of the square are of pure color, and the other two sides show a blend of red and blue. These four colors of the vertices A, B, C and D are NOT related to the two colors red and blue that the content of the square is made of! Now, start in an empty area and click the board at four positions in a counterclockwise direction, as shown in the small square. Congratulations! You have created another square with the same hues as the first square, showing blue at the top and red at the bottom, and a blend of red and blue in between. (The system does not check if you actually clicked inside the gray square playing area.) Do a few experiments by clicking four positions that do not result in a square. Explore! What happens if the fourth click is placed exactly in the same place as the third click (i.e. if D=C )? The first small triangle shows the result: the bottom is red, close to vertex B it is blue, and a blend of red and blue in between. Only one side shows a pure color; the other two show a blending of colors. Finally, the second triangle shows what happens in the case D=A : two sides show a pure color, and only one side shows a blending of two colors. Controls "painting": create a new painting or modify an existing one. Paintings 1 to 4 have a triangular playing area with side length 9. Paintings 5 to 8 have a triangular playing area with side length 10. Paintings 9 to 12 have a hexagonal playing area with side length 5. Paintings 13 to 16 have a hexagonal playing area with side length 7. Paintings 17 to 20 have a hexagonal playing area with side length 9. Paintings 21 to 30 have a square playing area with side length 16. "s": click to save your painting "r": click to restore painting "select": jump to the first/previous/next/last project. "zoom": this slider is used to zoom in and out. "shift": n/e/s/w: shift vertex; N/E/S/W: shift tile (by half a unit step at a time) "clicks": shows the number of vertices in the quadrilateral being created. If an incomplete tile (consisting of 1, 2 or 3 clicks) exists when changing to a different control, it will be deleted. "density": use this slider to set the density; a higher density means more lines, and hence a more completely painted quadrilateral. "tile": use this popup menu to select the existing tile to work on. "new": click this button to insert a copy of the current tile at the position of the current tile. At first, it is hidden because it is placed on top of the current tile. "(no edit)": use this popup menu to • select one of the four vertices (A, B, C, D) of the current tile. Then click the board anywhere to relocate the selected vertex. • or choose option "edit tile". If "edit tile" is chosen, click the board anywhere to relocate the current tile. "(action)": use this popup menu to choose from these options: • "mirror tile hor": mirror tile horizontally immediately (with the black vertex on the mirror axis). • "mirror tile ver": mirror tile vertically immediately (with the black vertex on the mirror axis). • "rotate 1°/5°/30°/60°/90°": rotate the current tile around vertex A by the selected degree. Click the toggle next to it to repeat the rotation. • "shift all N/E/S/W": shift the whole painting to the north/east/south/west. Click the toggle next to it to repeat the shift. • "swap all colors": swap the two colors of the current tile. • "color all randomly": select random color combinations for all tiles. Color selections: Use the popup menus to pick the two colors to blend. "swap": swap the two colors of the current tile. "random": the system chooses two random colors (out of 24) for the current tile. "rot": rotates the set of four vertices (A, B, C, D) associated to the current tile: A=> B => C => D => A. "reset": deletes all tiles except the current one. The last tile cannot be deleted. "delete": deletes the current tile. The last tile cannot be deleted. "snap": if active, clicking the board will find the lattice point closest to the grid. (using the 60° grid for painting 1 to 20, and the 90° grid for painting 21 and later). "show": use these checkboxes to display the following: • "tile numbers": the sequence number of each tile is displayed at the center of the tile. • "points": shows/hides the markers of the clicked positions during the creation of a new quadrilateral. • "areas": shows/hides the colors of the tiles, instead the outlines of the tiles are shown. Go to painting 24 and click "areas". What happens? • "board": a large background quadrilateral indicates an area you might want to fill in.shows/hides the playing area. • "grid": shows/hides the grid. Paintings 1 to 20 use the isometric grid; paintings 21 to 30 use the orthogonal grid. Tips The system can be slow when using a high density. Hence it is recommended that you use a low density (around 50) while you design your layout, and only as a last step (or for checking in between) move to a high density. Permanent Citation Cite this as Karl Scherer (2017), "Blended Quadrilaterals" Wolfram Demonstrations Project. demonstrations.wolfram.com/BlendedQuadrilaterals/ Related Demonstrations Color-Blended Circular Tiles Karl Scherer Chain Puzzle on Triangular Grid Karl Scherer Color Blades Karl Scherer Tiles with Finite Convex Spectra Karl Scherer, Erich Friedman, Ed Pegg Jr, Joe DeVincentis and Rainbow Art Karl Scherer Nowhere-Neat Tilings of the Plane Karl Scherer, Dave Walton Line Art Karl Scherer, Dennis Van Straten Irreptiles Karl Scherer, Michael Reid, Rodolfo Kurchan, Ernesto Amezcua, George Freeman, and George Sicherman MAX Puzzle Karl Scherer Spinning Action Sandra McLellan, Abby Brown-McLellan Sliding Puzzle with Periodic Boundary Conditions Frank Brechtefeld RGB Spheres Ed Pegg Jr Ragged Automaton Disk Michael Schreiber Pentagon Tilings Ed Pegg Jr Pentagonal Tiling Sándor Kabai Op Art Motifs on Quadrilaterals Izidor Hafner Named Colors Ed Pegg Jr Moiré Patterns and Commensurability in Rotated Graphene Bilayers Jessica Alfonsi Making 2D Art Using 3D Graphics Izidor Hafner Related Topics Polygons Art Color Browse all topics Wolfram LanguageMathematicaWolfram UCommunityMore» ©2025 Wolfram Demonstrations Project &Contributors \| Terms of Use \| Privacy Policy \| Give feedback AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA
17137	https://www.mathworksheets4kids.com/bundles-tens-ones.php	Bundles of Tens and Ones Worksheets [x] Login Child Login Main MenuMathLanguage ArtsScienceSocial StudiesInteractive WorksheetsBrowse By GradeBecome a Member Become a Member Math Interactive Worksheets Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Worksheets by Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Number Sense and Operations Number Recognition Counting Skip Counting Place Value Number Lines Addition Subtraction Multiplication Division Word Problems Comparing Numbers Ordering Numbers Odd and Even Prime and Composite Roman Numerals Ordinal Numbers Properties Patterns Rounding Estimation In and Out Boxes Number System Conversions More Number Sense Worksheets Measurement Size Comparison Time Calendar Money Measuring Length Weight Capacity Metric Unit Conversion Customary Unit Conversion Temperature More Measurement Worksheets Financial Literacy Writing Checks Profit and Loss Discount Sales Tax Simple Interest Compound Interest Statistics and Data Analysis Tally Marks Pictograph Line Plot Bar Graph Line Graph Pie Graph Scatter Plot Mean, Median, Mode, Range Mean Absolute Deviation Stem-and-leaf Plot Box-and-whisker Plot Factorial Permutation and Combination Probability Venn Diagram More Statistics Worksheets Geometry Positions Shapes - 2D Shapes - 3D Lines, Rays and Line Segments Points, Lines and Planes Angles Symmetry Transformation Area Perimeter Rectangle Triangle Circle Quadrilateral Polygon Ordered Pairs Midpoint Formula Distance Formula Slope Parallel, Perpendicular and Intersecting Lines Scale Factor Surface Area Volume Pythagorean Theorem More Geometry Worksheets Pre-Algebra Factoring GCF LCM Fractions Decimals Converting between Fractions and Decimals Integers Significant Figures Percents Convert between Fractions, Decimals, and Percents Ratio Proportions Direct and Inverse Variation Order of Operations Exponents Radicals Squaring Numbers Square Roots Scientific Notations Logarithms Speed, Distance, and Time Absolute Value More Pre-Algebra Worksheets Algebra Translating Algebraic Phrases Evaluating Algebraic Expressions Simplifying Algebraic Expressions Algebraic Identities Equations Quadratic Equations Systems of Equations Functions Polynomials Inequalities Sequence and Series Matrices Complex Numbers Vectors More Algebra Worksheets Trigonometry Calculus Math Workbooks English Language Arts Summer Review Packets Science Social Studies Holidays and Events Support Worksheets> Math> Number Sense> Place Value> Bundles of Tens and Ones Bundles of Tens and Ones Worksheets Pique the interest of your eager-to-learn kids with our printable bundles of tens and ones worksheets featuring adequate exercises in grouping and splitting numbers into tens and ones. Kids in kindergarten, grade 1, and grade 2 get acquainted with the fact that the two digits of a two-digit number represent amounts of tens and ones and a group of 10 tens makes a 100. Deliberate practice in place values being the motto of these pdfs, the exercises reiterate these simple facts, and help kids comprehend that the position of each digit in a number is significant. Our answer key feature accelerates evaluation. Begin your practice with our free bundles of tens and ones worksheets. Composing and Decomposing \| Tens and Ones The numbers from 11 to 19 are composed of a ten and few ones. Kids in kindergarten count the individual units and form the numeral in Part A and group objects into bundles of ten, and decompose them into tens and ones in Part B. Download the set Counting Bundles of Sticks \| No Regrouping Buff up the place value skills of your 1st grade kid with our printable bundles of tens and ones worksheets. Count each bundle made up of 10 sticks and the individual sticks separately and write the numeral. Download the set Bundling Shapes into Tens \| Regrouping Ones Push boundaries with our bundling shapes into tens worksheet pdfs that involve regrouping. Kids count and fuse the ones and check if they can make a ten, and trade the extra ones for tens and write the total. Download the set Grouping Real-life Objects into Tens and Ones Cracking these bundles of tens and ones worksheets is a true measure of your place value skills. Bundle real-life objects into groups of tens by circling them. Count the 10s and the leftover 1s, and write the number. Download the set Bundles of Tens and Ones (10 tens + ones) \| 3-digit Get 2nd grade kids out of their comfort zones as they understand that 100 refers to a bundle of 10 tens. Let them count the bundles of hundred, tens and the individual units, and form the 3-digit number. 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17138	https://math.stackexchange.com/questions/3326780/10-digit-numbers-with-constraints	Skip to main content 10-digit numbers with constraints Ask Question Asked Modified 6 years ago Viewed 332 times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. How many 10-digit numbers can be made by using the digits {5,6,7} (all of them) and with the additional constraints that no two consecutive digits must be the same and also that the first and last digits of the number must be the same? I am trying to find a solution by using combinatorics. I start from the 1st leftmost digit, which can have any value from {5,6,7} (3 possibilities). Then we move to the 2nd digit, which can have 2 values (since it can't be the same with the 1st) and so on, and for the last digit we only have 1 option. But this is not correct, because for the 9th digit we have the restriction that it must be different from the 8th and also different from the 10th, which, in turn, is equal to the 1st. I don't know how to express this. I therefore tried to find a recursive relation. I found that the general relation is a(n)=2∗a(n−1) if n odd and 2∗a(n−1)+6 if n is even. For n=4, we have 6 such numbers (4-digit numbers, but with the given restrictions). Then if we add one more digit to the right, we remove the rightmost (4th) digit, which had to be the same with the first, and now for the 3rd digit we have 2 options instead of 1 (we can also add the options that were rejected because they were neighboring with the 4th digit). So in total we now have 2x6=12 options. Therefore, a(4)=6 and a(5)=12. I don't understand, however, where this +6 (in the recursive relation) comes from! By the way, the correct answer is 510. Many thanks in anticipation. combinatorics Share CC BY-SA 4.0 Follow this question to receive notifications asked Aug 18, 2019 at 9:29 Marius StephantMarius Stephant 64533 silver badges1515 bronze badges Add a comment \| 4 Answers 4 Reset to default This answer is useful 3 Save this answer. Show activity on this post. I don't understand the recursion in the OP, but here is another recursive solution. For simplicity, let's suppose the number starts with 5; we will need to remember to multiply our result by 3 in the end, to account for 6 and 7. Let's say a number is "acceptable" if it starts with 5 and has no two consecutive digits the same, ignoring for now the restriction that it must end in 5. Define ai(n) to be the number of n-digit acceptable strings ending in i, for i=5,6,7. Then we have a5(1)=1, a6(1)=a7(1)=0, and a5(n+1)=a6(n)+a7(n)a6(n+1)=a5(n)+a7(n)a7(n+1)=a5(n)+a6(n) because of the requirement that no two consecutive digits are the same. These recursive relations allow us to calculate a5(n),a6(n) and a7(n) for n as large as we like. The answer to the original question is 3a5(10). Share CC BY-SA 4.0 Follow this answer to receive notifications answered Aug 18, 2019 at 14:33 awkwardawkward 15.6k11 gold badge2323 silver badges3333 bronze badges Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. We have a cycle of nine digits xi∈Z3 (1≤i≤9), and can forget about the tenth digit x10=x1. If no two consecutive digits may be the same we automatically use all three of the digits −1,0,1∈Z3. Put xi−xi−1=:yi cyclically. It is then required that yi∈{−1,1}. Otherwise the yi∈Z3 are arbitrary, except for the condition s:=∑i=19yi=0mod 3 . This s will automatically be odd. We can have s=9 or s=−9 in 1 way each. The value s=3 is realized by six yi=1 and three yi=−1. This can be done in (93)=84 ways. The same holds for s=−3. It follows that there are 2(1+84)=170 admissible choices for the yi. We still are free to choose x1, so that we obtain 510 admissible three digit numbers in all. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Aug 20, 2019 at 9:23 Christian BlatterChristian Blatter 232k1414 gold badges203203 silver badges489489 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Equivalently, we want to 3-color the labeled cycle with 9 vertices. If we know (e.g. see the wiki for chromatic polynomials) that the chromatic polynomial of Cn, the labeled cycle with n≥3 vertices, is χ(Cn,t)=(t−1)n+(−1)n(t−1) then the answer is easy: χ(C9,3)=(3−1)9+(−1)9(3−1)=510 Share CC BY-SA 4.0 Follow this answer to receive notifications edited Aug 20, 2019 at 9:43 answered Aug 20, 2019 at 7:57 Brian MoehringBrian Moehring 24.1k11 gold badge2828 silver badges5050 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. I have another solution. We call a number satisfying the condition of the problem "accepted". Let a(n) denote the number of accepted numbers with n digits. Imagine an accepted number with n−2 digits and put two empty places at the left side of the number with n−2 digits. You can easily make two different accepted numbers with n digits. To make this claim clear, assume 1,3,...,1 is an accepted number with n−2 digits, we can build two n-digit numbers as below: 1,2,1,3,...,1 1,3,1,3,...,1 Now, assume we have an accepted number with n−1 digits. We can make an accepted number with n digits from this number. For example, assume 1,3,...,1 is such number. we can write: 1,2,3,...,1 Just notice that we added 1 to the left side and the first 1 in1,3,...,1 changed into 2. Now, assume n is even. An accepted number can be made from 1,2,1,2,1,2,...,1,2 with n−2 digit, as below: 2,3,1,2,...,1,2 It is easy to understand that 1,2,1,2,1,2,...,1,2 doesn't belong to the set of accepted numbers with n−2 digits. If n is odd, the same goes for it, consider 1,2,1,2,...,1,2 with n−1 digits. It's possible to make an accepted number with n digits from this as below: 2,3,2,1,2,1,2,...,1,2 In each situation there are 6 accepted numbers (why?) with n digits which can't be made from accepted numbers with n−1 or n−2 digits. So, we will get a(n)=a(n−1)+2a(n−2)+6. To be done, one should check two statements below: A n-digit accepted number made from n−2-digit accepted numbers doesn't coincide with any n-digit accepted numbers made from n−1-digit accepted numbers and vice versa. Each n-digit accepted number can be obtained through either n−2-digit accepted numbers, n−1-digit accepted numbers or those six numbers. Note: What I wrote implies the formula you mentioned. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Aug 22, 2019 at 13:38 answered Aug 20, 2019 at 11:31 Reza RajaeiReza Rajaei 6,27111 gold badge1111 silver badges2626 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Linked 0 This question is based on basic counting. 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17139	https://math.jhu.edu/~brown/courses/s19/Lectures/Lecture5.pdf	February 11, 2019 LECTURE 5: THE RULES OF DIFFERENTIATION. 110.211 HONORS MULTIVARIABLE CALCULUS PROFESSOR RICHARD BROWN Synopsis. Here, we bring back the rules for diﬀerentiation (used to derive new functions constructed using various combinations of other functions) from Calculus I and use them in our new context. The basic frame for this discussion is, ”the rules are the same, but only precisely when they actually make sense.” What this means is the focus of this class. Also, we will look at higher derivatives and the notion of a function being diﬀerentiable more than once. This involves deﬁning the kth partial of a real-valued (and vector-valued) function and what it means for mixed partials to be equal. The diﬀerentiable class of a function is discussed, along with just what kind of object the kth derivative of a real-valued function on n variables is and how it encompasses its nk partials. The Rules of Diﬀerentiation. The nice thing about calculating derivatives in multivari-able calculus is that, in many ways, they follow the same rules as they did in single variable calculus, suitably generalized, of course. The Constant Multiple Rule. Multiplying a function f : X ⊂Rn →Rm by a real constant c ∈R aﬀects only the functions values, as in Calculus I functions. The new function (cf) : X ⊂Rn →Rm has a vector output, and a constant times a vector means simply multiplying each component by that constant. Indeed, (cf)(x) = c ·     f1(x) f2(x) . . . fm(x)    =     (cf1)(x) (cf2)(x) . . . (cfm)(x)    . The partial derivatives of each fi are single variable derivatives, where the Constant Multiple Rule held, so ∂(cfi) ∂xj (x) = lim h→0 (cfi)(x1, . . . , xj−1, xj + h, xj+1, . . . , xn) −(cf)(x) h = lim h→0 c (fi(x1, . . . , xj−1, xj + h, xj+1, . . . , xn) −f(x)) h = c · lim h→0 fi(x1, . . . , xj−1, xj + h, xj+1, . . . , xn) −f(x) h = c · ∂fi ∂xj (x) . The eﬀect is that the entire derivative matrix is multiplied by c, just as in matrix multipli-cation by a scalar, so that D(cf)(x) = c · Df(x). 1 2 110.211 HONORS MULTIVARIABLE CALCULUS PROFESSOR RICHARD BROWN The Sum/Diﬀerence Rule. The Sum Rule (and hence the Diﬀerence Rule when one of the functions is multiplied by −1), is also precisely the same as that for single variable calculus, except that the two functions in the sum must have the same domain and codomain. Only then will the two derivative matrices have the same dimensions, allowing us to actually sum the derivative matrices. So consider f, g : X ⊂Rn →Rm to m-vector-valued functions on X ⊂Rn, and let h = f + g. Then Dh(x) = D (f + g) (x) = Df(x) + Dg(x). The Product Rule. The Product Rule, in multivariable calculus, can be a bit trickier, given that the product of two vectors may or may not be a vector of the same size: It is for the cross product in R3). But the dot product of two vectors in Rn is a scalar. And sometimes, the output of a product of vectors may not even be a vector of any size (look up outer product, for example). The tricky part is to ensure that, if two vector-valued functions are diﬀerentiable, then so should be the product of those two functions, however, that is deﬁned. And further, to ﬁnd a rule to write the derivative of the product using the derivatives of the factors. The nice part of all this is that the Product Rule will always hold, as long as the parts and the product make sense. Indeed, let f : X ⊂Rn →Rm and g : X ⊂Rn →Rp be two vector-valued functions, possibly of diﬀerent sizes, but deﬁnitely deﬁned on the same domain (why is this necessary?) The deﬁne h(x) = f(x) · g(x). However that product is deﬁned, it does need to make sense. But when it does, it means that the m × 1-matrix of outputs f and the p × 1-matrix of outputs of g are multiplied together in a well-deﬁned way. But then the Product Rule is Dh(x) = D (f · g) (x) = Df(x) · g(x) + f(x) · Dg(x). Following the dimensions, at least, we would get that, if one could multiply an m-vector and a p-vector, then one can also multiple a m × n-matrix to a p-vector, and add to that the product of an m-vector with a p × n-matrix. Write this out to verify, but the genral idea is that an m × n matrix is just a collection of n m-vectors. Here are some examples: Example 5.1. Let p = 1. Then g(x) is a scalar function, and Dg(x) is a 1 × n-matrix. Then h = f · g makes sense, as the output is the product of an m-vector f(x) with a scalar g(x) at every input. We get h : X ⊂Rn →Rm, h(x) = f(x)g(x), and Dh(x) \| {z } m×n = Df(x) \| {z } m×n g(x) \|{z} scalar + f(x) \|{z} m×1 Dg(x) \| {z } 1×n . Here is a concrete example: Suppose that we wanted to create a function that was the product of f(x, y, z) = xy + y2z x4z , and g(x, y, z) = ln(yz). Then we can certainly deﬁne h(x, y, z) = f(x, y, z)g(x, y, z) = (xy + y2z) ln(yz) (x4z) ln(yz) , but we have to carefully choose our domain so that h makes sense. f is deﬁned on all of R3, but the largest domain of g is the set X = (x, y, z) ∈R3 yz > 0 . LECTURE 5: THE RULES OF DIFFERENTIATION. 3 Then h : X ⊂R3 →R2 is deﬁned as above. And on this open domain X, h will be diﬀerentiable (in fact, it is the product of diﬀerentiable functions). So we calculate the derivative in two ways: Directly, and via the Product Rule. Directly, Dh(x) = " ∂h1 ∂x (x) ∂h1 ∂y (x) ∂h1 ∂z (x) ∂h2 ∂x (x) ∂h2 ∂y (x) ∂h2 ∂z (x) # = " y ln(yz) (x + 2yz) ln(yz) + x + yz y2 ln(yz) + xy z + y2 4x3z x4z y x4 ln(yz) + x4 # . Via the Product Rule, we have Dh(x) = Df(x) · g(x) + f(x) · Dg(x) = " ∂f1 ∂x (x) ∂f1 ∂y (x) ∂f1 ∂z (x) ∂f2 ∂x (x) ∂f2 ∂y (x) ∂f2 ∂z (x) # · g(x) + f1(x) f2(x) ·    ∂g ∂x(x) ∂g ∂y(x) ∂g ∂z(x)    = y x + 2yz y2 4x3z 0 x4 ln(yz) + xy + y2z x4z · 0 1 y 1 z . I will leave it to the reader to see that these two matrices of functions are the same. Example 5.2. Now let p = m > 1, with the Dot Product on vectors. Note that, for ease of calculation here, we let n = 1. Then, for f, g : X ⊂R →Rm, the dot-product function is h : X ⊂R →R, where h(x) = f(x) · g(x). Note that the product function h here is scalar-valued, but still has n inputs. Then Dh(x) = Dx1h(x) · · · Dxnh(x) , with Dxih(x) = ∂ ∂xi h(x) = ∂ ∂xi " m X j=1 fj(x) · gj(x) # = n X j=1 ∂ ∂xi [fj(x)gj(x)] (Sum Rule) = m X j=1 ∂fj ∂xi (x)gj(x) + fj(x)∂gj ∂xi (x) (Calc I Product Rule) = m X j=1 ∂fj ∂xi (x)gj(x) + m X j=1 fj(x)∂gj ∂xi (x) (Sum Rule) = Df(x) · g(x) + f(x) · Dg(x), where each of the four pieces in this last sum of products is an m-vector. Notice that in the middle of this last calculation, we were simply multiplying together scalar-valued functions, so there was no · present. Now, as a special note of caution: Be careful with vector products. The two examples above are symmetric products, named because f(x) · g(x) = g(x) · f(x) 4 110.211 HONORS MULTIVARIABLE CALCULUS PROFESSOR RICHARD BROWN If the product is not symmetric, then the order of the factors matters. But the Product Rule will still work correctly. One jsut needs to pay attention to the order of the factors inside the product rule. For example, for a, b ∈R4, the cross product is called antisymmetric, since a × b = −b × a. Perhaps you already know this via a detailed calculation. However, we will see why this is true geometrically in a while. Hence for f, g : X ⊂R →R3, and h(x) = f(x) × g(x), we have Dh(x) = D (f × g) (x) = Df(x) × g(x) + f(x) × Dg(x). And lastly, on this note, the Quotient Rule also hold, but only where it makes sense. One thing to keep in mind for the Quotient Rule is that the denonimator function myust be scalar-valued for even a quotient of functions to make sense. (Why?) At that point, the square of the denominator function in the Quotient Rule will also make sense. A Note on Partial Derivatives. Given a diﬀerentiable real-valued function f : X ⊂R3 → R, say, we know that all of ∂f ∂x, ∂f ∂y , ∂f ∂z : X →R are all continuous in a neighborhood of x =   x y z  . In this case, where all partials of a function are continuous on a domain, we say that the function is a C1-function, or write f ∈C1. Deﬁnition 5.3. A second partial derivative of f with respect to a variable (x, say) is any one of ∂ ∂x ∂f ∂x : X →R, ∂ ∂x ∂f ∂y : X →R, ∂ ∂x ∂f ∂z : X →R. We also write ∂ ∂x ∂f ∂x = ∂2f ∂x2 = fxx ∂ ∂x ∂f ∂y = ∂2f ∂x∂y = fyx, ∂ ∂x ∂f ∂z = ∂2f ∂x∂z = fzx. Pay attention to the order of the variables, and hence the derivatives here. Indeed, the order of diﬀerentiation is written diﬀerently in the two notations, fractional and subscript-wise. Be careful here. Now If all 9 of these second partial derivative of f exist and are continuous on the domain X, then we say that f ∈C2. Generalizing, let f : X ⊂Rn →R. For i1, i2, . . . , ik ∈{1, 2, . . . , n}, the kth partial derivative of f with respect to xi1, . . . , xik, is ∂kf ∂xik · · · ∂xi1 (x) = ∂ ∂xik · · · ∂f ∂xi1 · · · = fxi1···xik. LECTURE 5: THE RULES OF DIFFERENTIATION. 5 Example 5.4. Let f : R3 →R be deﬁned by f(x, y, z) = z cos(2xy). It should be readily apparent, and can be rigorously shown, that polynomials in many variables are continu-ous everywhere, and diﬀerentiable everywhere. The same it true for the cosine function. And since f is a product of a polynomial and a composition of the cosine function and a polynomial, f ∈C1, and we can calculate fx(x, y, z) = −z sin(2xy)2y = −2yz sin(2xy), fy(x, y, z) = −z sin(2xy)2x = −2xz sin(2xy), and fz(x, y, z) = cos(2xy). But all three of these partial derivatives of f are also sums, products and compositions of diﬀerentiable functions, so that f ∈C2 also. Thus, we ﬁnd fxy(x, y, z) = ∂2f ∂y∂x = ∂ ∂y ∂f ∂x = −2z sin(2xy) −4xyz cos(2xy), and fyx(x, y, z) = ∂2f ∂x∂y = ∂ ∂x ∂f ∂y = −2z sin(2xy) −4xyz cos(2xy). Notice immediately that these two functions are the same. It turns out that this is true always when f ∈C2: Theorem 5.5. Suppose f : X ⊂Rn →R is C2 on an open X. Then, for any choice of i1, i2 ∈{1, 2, . . . , n}, ∂2f ∂xi1∂xi2 = ∂2f ∂xi2∂xi1 . The proof is constructive and in the book. We will not do it in class. Deﬁnition 5.6. A function f : X ⊂Rn →R is of class Ck, k ∈N if it has continuous partial derivative up to and including order k. A function g : X ⊂Rn →Rm is of class Ck if each component function gi : X →R is of class Ck. And ﬁnally, a function like the above is of class C∞if it is smooth. This means that it has continuous partial derivatives of all orders. Some notes: • This should be an obvious fact, but worth stating explicitly: If f ∈Ck, then f ∈Cℓ for all ℓ< k. • A continuous function is said to be of class C0. So here is a thought experiment: Suppose f : X ⊂Rn →R is of class C∞, so it is a smooth function. Then we know the following: (1) f has n ﬁrst partial derivatives, and (2) f has n2 second partials, since each of the n ﬁrst partials has n second partials. (3) Thus f has nk kth partials, for each k ∈N. Now Df(x) is a row matrix with n entries, each entry a real-valued function on n variables. Each of these entries is also diﬀerentiable. Plug in a value to evaluate the derivative of f at a point, and one gets a matrix of numbers. But without evaluation, Df(x) is a (row)-matrix of functions. Just for a moment, view this row matrix as a column matrix. Then, in a way, Df : X ⊂Rn →Rn. And then D (Df) = D2f(x) will be an n × n matrix of functions, with 6 110.211 HONORS MULTIVARIABLE CALCULUS PROFESSOR RICHARD BROWN each entry, ∂2f ∂xi∂xj a real-valued function on n variables. If we, for the moment think of D2f as a function on X, then what is its codomain? And, since f is smooth, the object D (D2f) = D3f exists! What kind of object is this? And in general, what kind of object is Dkf(x), for k ∈N? These objects will play a role in the multivariable Taylor expansion of a function like f, since Taylor series’ of functions exist in multivariable calculus and will (must) account for all of the derivatives of a function. Think about this....
17140	https://www.youtube.com/watch?v=y-VSilp0JA4	What Does It Mean To Be Necessary And Sufficient? Amour Learning 18800 subscribers 2 likes Description 136 views Posted: 14 Jun 2020 The transcript used in this video was heavily influenced by Dr. Oscar Levin's free open-access textbook: Discrete Mathematics: An Open Introduction. Please visit his website to get the full textbook for free: Prerequisites: (This will be updated soon!) Hi! My name is Kody Amour, and I make free math videos on YouTube. My goal is to provide free open-access online college math lecture series on YouTube using free open-access resources. I rely on generous donations and ad revenue - a special thanks to everyone who helped this channel grow. Please only give if you have the financial ability to do so. Support this Amour Learning channel directly on Patreon: Credit to all of the authors of all of the resources below that greatly helped influence this video. My newer videos are much more original than these videos. Free open-access online math textbooks: My favorite number theory textbooks: Number Theory: In Context and Interactive by Dr. Karl-Dieter Crisman: Elementary Number Theory: Primes, Congruences, and Secrets by Dr. William Stein: A Computational Introduction to Number Theory and Algebra by Dr. Victor Shoup: My favorite programming language: SageMath (which in my view is categorically more productive than Python, R, Mathematica, Matlab, Maple, Magma, etc - if you're using any of these and they aren't working out, consider switching to Sage): A free in-depth tutorial for SageMath: Learn LaTeX in 30 minutes: My favorite math/science YouTube channels that I absolutely love (in no particular order): Zach Star: Numberphile: 3Blue1Brown: MIT OpenCourseWare: Khan Academy: PBS Infinite Series: Lex Fridman: Transcript: Introduction [Music] in this video we will discuss necessary and sufficient conditions which is yet another way to frame implications now I want to give some advice to everyone here the the words necessary and sufficient are words that we don't really use that often in our day-to-day lives and so it's understandable that many students get confused between the two they'll use necessary when they mean sufficient and they'll use sufficient when they mean necessary this is totally common please take my advice right now you are gonna want to dedicate the extra time to memorize when to use necessary and when to use sufficient properly hey everyone real quick I just want to mention that this video is a part of a whole course that I made you can find a link to this entire course in the description below and make sure to click on that subscribe button so let's go through an example together where we use some of this phrasing let's Examples rephrase the implication if I dream then I am asleep and as many different ways as possible so let's before we get to the converse let's get to the original implication so I can say I am asleep if I dream I could say I dream only if I am asleep we could say in order to dream I must be asleep we could say and this is where we use the word necessary to dream it is necessary that I am asleep again to dream it is necessary that I am asleep notice how we put the necessary with the asleep portion of this implication it's really important you use necessary and sufficient properly the next one we're going to use sufficient we're gonna say to be asleep it is sufficient that I dream in this case sufficient is attached to the hypothesis the premise of the implication where as necessary is the conclusion so the conclusion is necessary that hypothesis the premise is sufficient and then last but not least technically this is equivalent but this is the contrapositive I am NOT dream dreaming unless I am asleep so that's another way we can phrase it the following are equivalent to the converse so this is where we pretty much just switch everything so I dream if I am asleep I am asleep only if I dream and this is where we can use necessary and sufficient it is necessary that I dream in order to be asleep and we could say it is sufficient that I be asleep in order to dream and then the contrapositive of this is if I don't dream then I'm not asleep so hopefully you agree with this example I include the necessary and sufficient versions though because those are common when discussing mathematics in fact let's agree once and for all what they mean exactly so P is necessary Conclusion for Q means Q implies P P is sufficient for Q means that P implies Q again sufficient means the premise necessary means you're the conclusion if P is necessary and sufficient for Q then P and Q are have this by conditional connective relative to each other to be honest I have trouble with these if I'm not very careful I find it helps to keep a standard example for a reference though thinking about the necessity and sufficiency of conditions can also help when writing proofs and justifying conclusions if you want to establish some math the medical fact it is helpful to think what other facts would be enough or be sufficient to prove your fact if you have an assumption think about what must also be necessary if that hypothesis is true anyways thank you so much I'll see in the next lecture
17141	https://suresolv.com/reasoning-and-logic-analysis/how-solve-sbi-po-level-reasoning-and-logic-analysis-problems-family	How to solve SBI PO level reasoning and logic analysis problems on family relations in a few simple steps 3 Jump to Navigation Brain teaser Maze puzzles Riddles Inventive puzzles Reasoning puzzles Paradoxes Mathematical puzzles Logic puzzles Number lock puzzles Missing number puzzles River crossing puzzles Ball weighing puzzles Matchstick puzzles Problem solving Real life problem solving Problem solving techniques Problem solving skills Problem solving tools Health Cough Sleep Critical thinking HOW TOs Innovation Innovation concepts Systematic innovation Inventive principles Innovations around us Explorations Urbanization and Environment Global warming Trees and Plants Mahua AI Inquiries Our life Basics Basic maths Reasoning and logic analysis Exams School math Efficient Math problem solving Exam guidelines SSC CGL Tier II SSC CGL SSC CHSL SSC CPO Bank PO UGC/CSIR Net Maths WBCS Main Variety Sudoku Sudoku beginner level 1 and 2 Medium Sudoku level 3 Hard Sudoku level 4 NYTimes Sudoku hard Expert Sudoku Books Reflections Innovative creations Unusual thoughts You are here Home » Reasoning and logic analysis » How to solve SBI PO level reasoning and logic analysis problems on family relations in a few simple steps 3 Submitted by Atanu Chaudhuri on Sat, 25/02/2017 - 15:17 SBI PO level Reasoning Family Relation problems can also be solved quickly by Collapsed Column Technique We have discussed logic analysis in our earlier sessions on Simple logic analysis, Complex logic analysisand efficient method of collapsed column logic analysis technique for solving SBI PO or higher level logic puzzles in our previous sessions on How to solve SBI PO level logic puzzles in a few simple steps 1 and How to solve SBI PO level logic puzzles in a few simple steps 2. In this session we will expand the scope of the powerful method further through solving a series of SBI PO level Reasoning problems on family relations with characteristic ease. Before going through this session, you are urged to go through the above-mentioned first and second session on efficient solution process for logic puzzles. Warning: The collapsed column logic analysis technique may seem to be easy and attractive for solving logic puzzles of any difficulty level efficiently in quick time. But these problems are inherently confusing and potentially a mine-field of errors. Unless the problem solver understands the nature and structural variation of this problem class, absorbs the efficient collapsed column technique with sufficient clarity and builds up confidence through the essential activity of solving many such problems of different structures, attempting such a tricky logic puzzle in actual test environment is not recommended. There is nothing wrong with the method, but large puzzles being inherently complex and confusing, chances of losing your way wasting critically valuable time will be high. We will take up a series of SBI PO level reasoning problems on family relations now for efficient solutions using Collapsed column logic analysis technique that simplifies the solution process and reduces the time to solve considerably. Assignment logic analysis using Collapsed column logic analysis technique can be applied to a number of Reasoning problem types In Assignment logic analysis problems there is a set of objects that are to be assigned (or mapped) to a second set of objects determined by a set of logic statements or conditions. The objects that are to be assigned belong to the to-be-assigned object set (TBA object set) and the objects to which the to-be-assigned objects are to be assigned belong to the to-be-assigned-to object set (or TBAT object set). In case of one-to-one unique assignment between two object sets, any of the two object sets are interchangeable with the other with respect to assignment as an assignment between two objects is a mutual two directional operation. In the simplest form of assignment logic analysis problem, one set of objects are to be assigned to a second set of objects according to a third set of given conditions. The conventional method to solve this form of problems is to represent the values of the two object sets, one as rows and the second as columns. This representation is the simplest form of 2 dimensional logic table. The cell values represent assignment between the members of the two sets. For example one set may be husbands and the second wives; problem will be to determine who is married to whom according to a set of conditions. The names of husbands may form row labels and names of wives column labels. Cross section of a row and column will represent marital status of the pair involved. Commonly we encounter logic analysis problems of assignment type in logic puzzles which frequently turn out to be quite confusing. We have already used the efficient technique of collapsed logic table columns in solving logic puzzles in our earlier sessions. Now we will apply the same technique to solve a different category of Reasoning problems, namely Reasoning Family Relation problems. In future sessions we will show how the same method can be applied for efficient solution of Floor type and Sitting arrangement type of Reasoning problems as well. Use of same concept and similar method in solving varieties of Reasoning problems eases the overall problem of solving a large portion of difficult reasoning problems considerably. Usually the subcategory of Family relation problems is referred to as Blood relation problems which is nothing but a subset of Family relation problems. Blood related family is always a subset of extended family in real life. Stages in Reasoning logic analysis problems One of the most important stages in solving reasoning logic analysis problems of any type is the first stage of Problem structure analysis and representation. Unless such a problem is represented in appropriate efficient form, attempt to solve the problem in any which way most possibly ends up in confusion. For example, in our earlier logic puzzle solution we have identified its nature as an assignment type logic puzzle problem and used the final collapsed column logic table of the form, In the second stage, we Analyze logic statements in conjunction with the problem state to determine the sequence of executing or processing the logic statements. Any logic problem will have a prime component in the form of a set of logic or condition statements. In second stage we determine the efficient sequence of executing the conditional statements. On an appropriate problem representation if we can apply an efficient sequence of execution of logic conditions, often the problem is solved much faster. In the third repetitive stage we actually carry out analysis and execution of one or more than one logic statement at each step. This is the last level and forms the core of elementary logic analysis. With proper execution of the individual logic statements coupled with proper sequencing or selection of logic statements for execution, the problem logic table can be simplified greatly at a single step. At this stage the clarity in domain knowledge plays a critical role. For example, in the process of solving a reasoning problem on family relations, you must be very clear about the commonly accepted form of relations between various members of a common family. Family relation logic problems The following is a snapshot of commonly accepted relations between various members of a family. This in fact is a network of relations with every node connected to every other node signifying a specific relation. Technically, a family is a fully connected graph with each node connected to all the other nodes. Though every node has a connected edge from every other node, we have not shown a number of connections to keep the picture uncluttered. An important point to note in family relation network is the nature of relationships between any two members in a family. Any such relation can be described in two different ways depending on which member is mentioned first in the relation statement. For example, for the married couple [c-a], the woman [a] is wife of the man [c] while the man [c] is the husband of the woman [a]. Thus "husband" is the attribute or role of [c] while "wife" is the role of [a] in this single relationship. Consequently the relationship is represented as a bi-directional arrow. Same holds good for any other relationship between two members of a family. Problem example 1: Reasoning Family Relation problem 1: What are the relationships between the members of the family? Problem description There are five persons, A, B, C, D, and E of a family. Conditional statements B is unmarried and also A's daughter. E is C's brother. C is the only husband of a married couple in this family. A is father of two sons and a daughter. He also has a daughter-in-law. Questions Question 1-1. Who is D's husband? E C A Data insufficient None of the above Question 1-2. Which of the following statements is definitely correct? D is sister-in-law of C C is married to B D is the daugher of A B is the daughter-in-law of A D is sister-in-law of E Question 1-3. Which of the following statements is definitely wrong? A is a father-in-law C's wife is D B has a sister-in-law D is daughter-in-law of A C is E's brother-in-law Question 1-4. Who is D's father-in-law? E C A Data insufficient None of the above Question 1-5. Who is the brother of D? C E A Data insufficient None of the above Problem example 1: Solution to the family relation reasoning problem 1: Problem analysis Note that there is only one object set of 5 family members. The set of relationships are not physical objects. Instead these are binary attributes involving a pair of objects, that is, two family members. This characteristic of family relationships is true for most types of relationships. Another specialty of a relationship is, there are always two roles involved in a relationship, one each for each member involved in the relationship. For example, in the case of a married couple, "wife" is the role or attribute of the woman while "husband" is the role of the man. But then what is to be assigned? The answer in this case is simple. Inevitably the only physical set of family members will form the set of objects to-be-assigned-to and will represent the row labels in the collapsed column logic table (effectively a member of same set is assigned to another member and the assignment is represented by a specific bi-directional relationship, but this might seem to be confusing). As soon as we are able to assign a specific relationship role to a family member with respect to a second family member, the role of the second member with respect to the first is also determined automatically (from the family relation concept network). Solution Strategy As usual we will first execute those conditional statements that directly define a relation between two members, and next we will process the statements bearing largest amount of information. Solution steps Step 1. According to our efficient strategy we will execute Statements 1 and 2 sequentially but together in the first step as both these statements specify definitive relationships between two members. Following is the result problem state in this step, Note that as soon as we encounter one statement assigning a role to a member such as "E is C's brother", we create two entries against row for E and also against the row for C. Note. Recording "C's brother" against row of E is straightforward and with complete certainty, but the opposite is not so. While recording the second role of this relationship against the row of C, you have to take care of the fact that C may be a sister or brother of E. Essentially then, it is safer to assume this relationship in more abstract form of gender-independent "SIBLING". If you mentally convert any mention of brother or sister as sibling, chances of error in recording the reverse relationship should be reduced to a great extent. Another example of such gender-independent family relationship is "SPOUSE" for wife-husband relation. In this problem we have kept two cells for each row label and the single assignment column for recording relation statements and analysis results. For larger problems we may need to use more number of cells per row-column intersection. In this first step, we have used both the cells for row B to keep things clear. We could have though combined the two conditions using one cell only. It always is a better practice to record one result in one cell (in actual exam, these cells will be just white space left between rows conveniently). Step 2. Out of the two statements left, we find the statement 4 to contain largest amount of information. In fact it clearly indicates that A is the head of the family having 2 sons and 1 daughter with 1 son married. Obviously the 5th member will be the daughter-in-law. Let us see the problem state after processing statement 4 in this second step. As expected, we just have recorded in this step that A has 2 sons one of whom is married. Note: At this stage itself you can identify C and E as the two sons of A and D as the daughter-in-law. Logic chain: A has one daughter B, one daughter-in-law and two sons, 1 of them married. As C and E are siblings they must be the two sons of A with unassigned D as the daughter-in-law. At this stage, only one information is missing—which one of C and E is married. Step 3. With this problem state when we process the last statement 3, "C is the only husband of a married couple in this family.", automatically C is identified as the only son of A who is married. As deduced already, E is C's brother and is the second son of A with D automatically identified as the wife of C and so daughter-in-law of A. This is what we call the third stage of logic analysis in the domain of family relationship concepts. Without clarity of family relationship concepts successful execution of this step may not be possible for a complex problem on family relationships. The final fully assigned logic table will be, Now we are ready to answer the questions and it should take only about a minute's time to answer the five questions. Answers Question 1-1. Who is D's husband? Answer 1-1. Option 2: C. Question 1-2. Which of the following statements is definitely correct? Answer 1-2. Option 5: D is sister-in-law of E. Note: You should be clear about the two-way relationship concept and the fully enumerated family relation network for answering this type of question with speed and accuracy. Question 1-3. Which of the following statements is definitely wrong? Answer 1-3. Option 5: C is E's brother-in-law Question 1-4. Who is D's father-in-law? Answer 1-4. Option 3: A. Question 1-5. Who is the brother of D? Answer 1-5. Option 4: Data insufficient. Note:About Option values "Data insufficient" and "None of the above": According to the problem nothing is mentioned about D, the daughter-in-law. She might have a brother who is not mentioned, or in other words, which information is not included (or negated) in the given Data. That's why the answer is Option 4: Data insufficient. On the other hand, if the Option 4 value were "B", the answer would have been "None of the above". In the given problem, as "Data insufficient" becomes a correct choice, the fifth option of "None of the above" violates this fourth option value, and hence is incorrect. This is a simple problem on family relationships only, though the framework used may very well cater to efficient solution of any complex problem on purely family relationships, however large the number of members be. Next we will explain the solution process of a similar problem on family relationships but with an additional complexity of Professions of family members. This forms a second object set along with the family members and we will use the professions as the to-be-assigned-to row labels for this problem. Solution process will again be through the same method using one collapsed column and five rows with compound row-column intersection cells. Problem example 2: Reasoning Family Relation problem 2: What are the professions and relationships between the members of the family? Problem description There are five persons, A, B, C, D, and E of a family. They are by profession, a businessman, a lawyer, a teacher, a farmer and a doctor but not necessarily in the same order. Conditional statements B is an unmarried teacher and also A's daughter. E is a lawyer and C's brother. C is the only husband of a married couple in this family. A is a farmer, father of two sons and a daughter. A's daughter-in-law is a doctor. Questions Question 2-1. Who is C's brother and what is his profession? B; Teacher D; Farmer A; Businessman E; Lawyer Data insufficient Question 2-2. Who is the businessman in the family? E C D B None of the above Question 2-3. Who is E's wife and what is her profession? D; Doctor B; Teacher C; Doctor Data insufficient None of the above Question 2-4. Which of the statement below is definitely wrong? A is head of the family D is a single child C is a businessman C is B's sister D's father-in-law is A Question 2-5. Who is the doctor in the family? C D E Data insufficient None of the above Problem example 2: Solution to the family relation reasoning problem 2: Problem analysis In addition to the special assignment requirement for binary attribute of family relation, we have a specific object set of family members to be assigned to the row labels corresponding to the five professions. The same single collapsed column representation will be used for recording the results of analysis and statement execution. Only, instead of using 2 cells for each row label, we will use 3 cells-just a little more extra space, to record the family members' identifications. Solution Strategy As usual we will first execute those conditional statements that directly define an assignment between a family member and a profession, and next we will process the statements bearing largest amount of information. Solution steps Step 1. According to our efficient strategy we will execute Statements 1, 2 and 4 sequentially but together in the first step as only these statements make specific assignment of a member to a profession. Mark that relationship takes secondary precedence here. Following is the result problem state in this step, Step 2. In the second step, we will process both the remaining two statements 3 and 5 to create full assignment of the logic table. Let us show the final logic table state first and then will explain the logic behind these last stage assignments. C being the only husband of a married couple and A having a daughter B with two sons, the definitive implication is, C is a son of A and is married, further implying the existence of a daughter-in-law of A. This is logic analysis in family relations domain. As E is C's brother, E is identified as the second son of A. The person left is D (B has already been identified as the daughter of A, the head of the family) and so D must be wife of C and daughter-in-law of A and hence the doctor by the last statement. The remaining unassigned profession of businessman goes to C because from the problem description it follows that every member has a separate profession. As before, after forming the final logic table, answering the five questions should take about only a minute more. Answers Question 2-1. Who is C's brother and what is his profession? Answer 2-1. Option 4: E; Lawyer. Question 2-2. Who is the businessman in the family? Answer 2-2. Option 2: C. Question 2-3. Who is E's wife and what is her profession? Answer 2-3. Option 5: None of the above. Note:As per the problem description, C is the only married son of A. So the other son E is unmarried and thus the question of his wife does not arise. All the first four choice values are thus incorrect and this implies, the fifth choice value "None of the above" is the correct choice. Question 2-4. Which of the statement below is definitely wrong? Answer 2-4. Option 4: C is B's sister. Note: You need to be alert while answering this type of question. Consider the choice 2: "D is a single child". You can't say for sure that the statement is incorrect as no further information about D, the daughter-in-law, is supplied in the problem description, she might very well be a single child.. Question 2-5. Who is the doctor in the family? Answer 2-5. Option 2: D. These problems are solvable by single collapsed column logic tables as these involve one-to-one unique assignments. However complex the logic conditions are, if the assignments are unique one-to-one, with judicious sequencing of statement execution and suitable third stage logic analysis, such a problem is always solvable with ease, assurance and efficiency. Now we increase the complexity by destroying the one-to-one assignment condition. In such an assignment problem, number of objects in a to-be-assigned set is more than the number of objects in a to-be-assigned-to set creating the complexity of one-to-many assignment condition. For example if we increase the number of family members to 7 keeping the professions numbering still 5, there is bound to be at least one profession corresponding to more than one family member (for 1 profession having more than one family member, the number of members having this profession will be 3 in this case if rest 4 professions belong to 4 unique family member). Problem example 3: Reasoning Family Relation problem 3: What are the professions and relationships between the members of the family? Problem description There are seven persons, A, B, C, D, E, F and G in a family. Their individual professions are one of businessman, lawyer, teacher, farmer and doctor (not necessarily in the same order). At most two family members may have the same profession and each profession will have at least one family member. Conditional statements B is an unmarried teacher and also A's daughter. G's sister is a teacher. A is a farmer, father of two sons and a daughter. A's daughter-in-law, an only child, is a doctor. F's sister and her husband both are farmers. E is a lawyer and C's brother. Questions Question 3-1. Who is C's sister? F G B Data insufficient None of the above Question 3-2. What is G's profession? Businessman Doctor Teacher Farmer None of the above Question 3-3. Who is F's husband? A C E Data insufficient None of the above Question 3-4. Which of the statements below is definitely wrong? A is head of the family C's profession is businessman C is F's sister F is G's sister D's father-in-law is A Question 3-5. Which of the statements below is definitely correct? F's son is a lawyer E's wife is a doctor D's mother is a teacher G is a mother E is D's sister-in-law Problem example 3: Solution to the family relation reasoning problem 3: Problem analysis Out of the two object sets Family members and Professions, the second having less number of objects than the first, it will be used as the to-be-assigned-to object set to which the larger number of family members will be assigned. The relationship analysis will be carried out simultaneously with profession assignment analysis. With the condition: "At most two family members may have the same profession and each profession will have at least one family member", the number of collapsed columns will be two to accommodate two family members having the same profession. Number of professions with more than one family member will also be 2 with the rest 3 professions having 1 family member. The five professions will form five compound rows. To record assigned family member name and the relation analysis results, we will use three cells for each row-column intersection pair. The logic table will then have two columns and five compound-rows each row comprising of three cells. Solution Strategy As usual we will first execute those conditional statements that directly define an assignment between a family member and a profession, and next we will process the statements bearing largest amount of information. Solution steps Step 1. According to our efficient strategy we will execute Statements 1, 3 and 6 sequentially but together in the first step as only these statements make specific assignment of a family member to a profession. Mark that relationship takes secondary precedence here. Following is the result problem state in this step, As decided, the five professions form the five rows and the logic table has two columns to record at most two family members having the same profession. Each row-column intersection has enough space to record three analysis results. For larger number of family members this space would have been more. In the first step we have recorded just the results of processing the statements. Breakthroughs will come later based on this data. Observethat we couldn't record the reverse role of "E is lawyer and C's brother" as we don't have C's profession known at this stage. Notice that we have jotted down the names of seven family members at the top left of the logic table to keep track of the members who have already been assigned to a profession and who are left to be assigned. The assigned member names are colored red (in actual exam we will cross-out the assigned names). We will shortly see how this action of keeping track of who are left to be assigned helps analysis in a significant way. Step 2. Out of the three statements left we will execute the statement 4 and statement 5 together as these two contain more information than the statement 2. The result is shown below. Explanation will follow. The content of statement 4 is duly noted against profession doctor without analysis or drawing any conclusion. But on analyzing statement 5 we could conclude that F's sister is the second farmer and also the wife of A. We could not decide who is this F's sister but could be sure that it would be one of C, D or G. This subset of possible members has been deduced from the reasoning, "A, B and E have already been assigned professions with only C, D, F and G remaining. So, wife of A who is F's sister must be one of C, D or G." At this penultimate stage a large portion of the logic table is yet to be assigned when only one condition statement is left to be processed. Naturally, you have to do extensive elementary logic analysis. To keep it simple is a challenge. Let's see how best we can meet the challenge. Step 3. The last statement 2 will be processed at this step. Final assignments are shown below. Explanations will follow. G's sister is a teacher.Question is, who is G's sister? Resolving the identity of G's sister We will take the simplest approach of analysing the possibilities of two-way sibling relationship between G and his or her sister. Prior to this stage we had "wife of A as F's sister". Possibilities of identity of wife of A were C, D or G. Now with the last statement we have, "G's sister is a teacher". So we have two sisters—"F's sister and G's sister". If F and G are not sisters to each other, C and D must then be the two sisters of F and G. So F and G cannot be any of the two sons of A because B is the only sister to the sons. If F and G were not sisters to each other then, A won't have any son—violation of a basic condition. Thus only possible sister to sister relationship that does not violate any of the existing relationships is "F and G are sisters to each other." We call this technique of analysing the implications of a two way relationshop as Two way relationship analysis. Resolving profession and relationships of C and D We already know that E is C's brother and daughter-in-law of A is an only child and a doctor. Then C must be the second son with E as the first son, and daughter-in-law must be D. As every profession must belong to at least one family member, the remaining yet-to-be-assigned-to profession of business goes to C. With the given information, we have been able to discover much of the logic table cell values, but still, who among C or E is the husband of D, the daughter-in-law of A remains unknown. Note: We have broken up the yet to be resolved questions into two neatly separated questions and resolved the two questions by not-so-simple but easy to understand elementary logic analysis based on family relationships. Answering the five questions should not take much time now. Answers Question 3-1. Who is C's sister? Answer 3-1. Option 3: B. Question 3-2. What is G's profession? Answer 3-2. Option 4: Farmer. Question 3-3. Who is F's husband? Answer 3-3. Option 4: Data insufficient. F is G's sister, but there has been no mention about F's husband. Question 3-4. Which of the statements below is definitely wrong? Answer 3-4. Option 3: C is F's sister. Question 3-5. Which of the statements below is definitely correct? Answer 3-5. Option 4: G is a mother. Option 2 - E's wife is a doctor can't be the definitely correct answer, as who among C and E is the husband of D couldn't be resolved. Recommendation The collapsed column logic analysis technique is a structured, systematic, clear, and efficient framework for analyzing logic puzzles and other types of logic analysis problems of any complexity with ease, speed and confidence without being confused. But one must be thoroughly conversant with variations of this class of problems by solving the puzzles, family relation or other types of problems using this efficient method. Without solving a sufficient number of such logic assignment problems during timed practice sessions one may not gain enough confidence and ability to solve a tricky logic puzzle or logic analysis question in an important competitive test. End note Solving reasoning puzzles does not need knowledge on any subject—it is just identifying useful patterns by analysis of the problem and using appropriate methods. It improves problem solving skill, because patterns and methods lie at the heart of any problem solving. Other resources for learning how to discover useful patterns and solve logic analysis problems Einstein's puzzle or Einstein's riddle The puzzle popularly known as Einstein's puzzle or Einstein's riddle is a six object set assignment logic analysis problem. Going through the problem and its efficient solution using collapsed column logic analysis technique in the session Method based solution of Einstein's logic analysis puzzle whose fish should be a good learning experience. Playing Sudoku As a powerful method of enhancing useful pattern identification and logic analysis skill, play Sudoku in a controlled manner. But beware, this great learning game, popularly called Rubik's Cube of 21st Century, is addictive. To learn how to play Sudoku, you may refer to our Sudoku pagesstarting from the very beginning and proceeding to hard level games. Reading list on SBI PO and Other Bank PO level Reasoning puzzles Tutorials How to solve SBI PO level logic puzzles in a few simple steps 1 How to solve SBI PO level logic puzzles in a few simple steps 2 How to solve SBI PO level family relation problems in a few simple steps 3 How to solve SBI PO level floor stay Reasoning Puzzle in a few confident steps 4 How to solve high level circular seating reasoning puzzles for SBI PO in confident steps 5 How to solve high level hard two row seating reasoning puzzles for SBI PO in confident steps 6 How to solve high level circular seating arrangement reasoning puzzles for SBI PO quickly 7 How to solve high level nine position circular seating easoning puzzles for SBI PO quickly 8 _How to solve high level box positioning reasoning puzzle for SBI PO quickly 9_ Solved reasoning puzzles SBI PO type SBI PO type high level floor stay reasoning puzzle solved in a few confident steps 1 SBI PO type high level reasoning puzzle solved in a few confident steps 2 SBI PO type high level reasoning puzzle solved in a few confident steps 3 SBI PO type high level circular seating reasoning puzzle solved in confident steps 4 SBI PO type high level hard reasoning puzzle solved in confident steps 5 SBI PO type high level one to many valued group based reasoning puzzle solved in confident steps 6 SBI PO type high level hard two in one circular seating reasoning puzzle solved in confident steps 7 SBI PO type hard facing away circular seating reasoning puzzle solved in confident steps 8 SBI PO type high level four dimensional reasoning puzzle solved in confident steps 9 SBI PO type hard two row seating reasoning puzzle solved in confident steps 10 SBI PO type high level floor stay reasoning puzzle solved in confident steps 11 Solved reasoning puzzles Bank PO type Bank PO type two row hybrid reasoning puzzle solved in confident steps 1 Bank PO type four variable basic assignment reasoning puzzle solved in a few steps 2 Bank PO type basic floor based reasoning puzzle solved in a few steps 3 Bank PO type high level floor stay reasoning puzzle solved in quick steps 4 Search form About us Terms
17142	https://www.mometrix.com/academy/slope-intercept-and-point-slope-forms/	Skip to content Slope-Intercept and Point-Slope Forms Point-Slope Form and Slope-Intercept Form (Video & Practice Questions) On this page Hi, and welcome to this review of linear equation forms! Specifically, we’ll be talking about slope-intercept and point-slope forms. We will also review the terminology and the process of graphing linear equations in these forms. Let’s get started! A linear equation can be expressed in many different ways, but no matter which form you use, it just represents a straight line. Standard Form The standard form of a linear equation is written as: (Ax +By=C), where (A), (B), and (C) are constants, and (x) and (y) represent variables. This form of the equation is very useful for some purposes in math. For example, a line can be quickly graphed when it is in this form by finding the (x)– and (y)-intercepts. There are also methods of solving systems of equations that require each equation in the system to be written in this form. Slope-Intercept Form Rearranging the standard form equation into slope-intercept form, (y=mx + b), reveals other key features of the line, namely, the slope and the (y)-intercept. The slope of a line describes the slant, or steepness, and the (y)-intercept is the point on the graph where the line crosses the (y)-axis. Notationally, slope is represented by (m) and the (y)-intercept is represented by (b). Because the (y)-intercept is an actual point on the coordinate plane, it is represented as an ordered pair, ((0,b)). Converting Between Forms Sometimes you will be asked to rearrange an equation from one form to another. Here’s an example: Remember, the standard form of a linear equation is (Ax +By=C), so this is currently in standard form. If we want to change it to slope-intercept form, we are going to need to rearrange it so that (y) is by itself our the left side. Our first step is to subtract (2x) from both sides. (3y=12-2x) is what we have now. Then, we’re going to divide everything by 3, which gives us (y=4- \frac{2}{3}x). This is almost right, but if we look again at slope-intercept form we see that we need the (x)-term to be in front. Thankfully we can think of this as (4+(- \frac{2}{3}x)) and use the commutative property of addition to swap their places. This gives us: The resulting equation is more informative about the line than the original equation in standard form. The coefficient of (x), (-\frac{2}{3}), is the slope. A negative slope tells us that the line slants downward, from left to right. The (y)-intercept of 4 tells us that the line crosses the (y)-axis at the point ((0,4)). Slope Example Problems Now that we have seen how to convert a standard form equation into slope-intercept form, let’s practice recognizing the key features of slope and the (y)-intercept with a few examples. For these examples, we want to name the slope, describe the slant of the line, and name the (y)-intercept as an ordered pair. Example #1 (y=2x+3) The slope of this equation is (m=2), the coefficient of the (x)-variable. A positive slope indicates that the line slants upward from left to right. The (y)-intercept is (b=3), which indicates that the line crosses the (y)-axis at the point, ((0,3)). Example #2 Let’s try another one: (y=\frac{3}{5}x-\frac{2}{3}) The slope of this equation is (m=\frac{3}{5}). Because the slope is positive, the line slants upward from left to right. The (y)-intercept is (b=-\frac{2}{3}), which indicates that the line crosses the (y)-axis at the point, ((0,-\frac{2}{3})). Example #3 Let’s try one more: (y= -5x-2) The slope is (m=-5). Negative slope means that the line slants downward from left to right; the line crosses the (y)-axis at the point ((0,-2)). Graphing from the Y-Intercept As you can imagine, knowing where the line crosses the (y)-axis and the slope of the line will make the line very easy to graph. Now, let’s take a look at how slope provides you with instructions to graph from the (y)-intercept. Any value of slope can be looked at as a fraction, where the numerator indicates where to move along the (y)-axis, and the denominator indicates where to move along the (x)-axis. (\text{Slope}=\frac{\text{Vertical Change}}{\text{Horizontal Change}}) Movement along the (y)-axis is typically referred to as the “rise.” A positive rise value would instruct a move up the (y)-axis, while a negative rise would indicate a move down the (y)-axis. Likewise, a positive run value would mean a shift to the right, and a negative run would mean a shift to the left. Rise and Run Examples Here are a few examples to practice identifying the rise and run indicated by a given slope: Example #1 (m=5) This slope is not written as a fraction, but any whole number can be rewritten as a fraction over 1: (m=\frac{5}{1}) The numerator, 5, is the rise. Positive value means UP, 5. The denominator is the run. Positive value means RIGHT, 1. Example #2 Let’s try another one. (m=\frac{-2}{3}) When you have a negative slope, you can consider either the numerator or the denominator to be negative, not both! For this example, let’s consider the numerator, the rise, to be the negative value. Negative value means DOWN, 2. The denominator, 3, is the run. Positive value means RIGHT, 3. Example #3 Here’s one last example: (m= – 3) First, we need to rewrite the whole number as a fraction: (m=-\frac{3}{1}). Let the numerator be the negative value. Rise is -3, DOWN, 3. Run is 1, RIGHT, 1. Graphing Examples Now onto some graphing. People sometimes find it helpful to use the notation of slope-intercept form to get started. “Begin” at (b), and “Move” according to (m). Example #1 Let’s graph the linear equation in slope-intercept form: (y= \frac{2}{3}x-2) Step 1: Begin at (b) Plot the (y)-intercept, ((0,-2)), 1st point. Step 2: Move by (m=\frac{2}{3}) Rise equals UP, 2. Run equals RIGHT, 3. Plot the 2nd point at ((3,0)). Step 3: Repeat. Rise equals UP, 2 Run equals RIGHT, 3 Plot the 3rd point at ((6,2)). Step 4: Draw a straight line through the three points. Got the hang of it? Let’s try one more: Example #2 (y=-2x-3) Step 1: Begin at (b), Plot the (y)-intercept, ((0,-3)), 1st point. Step 2: Move by (m=-\frac{2}{1}). (Let the rise be negative!) Rise equals DOWN, 2 Run equals RIGHT, 1 Plot the 2nd point at ((1, -5)). Step 3: Repeat. Rise equals DOWN, 2 Run equals RIGHT, 1 Plot the 3rd point at ((2, -7)). Step 4: Draw a straight line through the three points. Note the slant downward, from left to right, due to the negative slope in this equation. Point-Slope Form Now that we have had some review of the key features of linear equations, we have the tools to explore the point-slope form. This form is of special use if we know one point that is on the line, and the slope. The general form of this arrangement is (y-y_{1}=m(x-x_{1})), where (m=\text{slope}) and ((x_{1},y_{1})) is another point that is known on the line. Using this template, let’s practice identifying the slope and the point from the following examples of point-slope form: Example #1 (y-5=3(x-2)) This is a straightforward example. First, identify the slope as the coefficient outside the parentheses, (m=3). When naming the point on the line, note that in the general form the (x)-coordinate is being subtracted from (x), and the (y)-coordinate is being subtracted from (y), so the ordered pair of the point will be ((2,5)). Example #2 Here’s another one: (y+3=-\frac{1}{2}(x-4)) The slope in this equation is (m=-\frac{1}{2}). How is this equation different from the general form (y-y_{1}=m(x-x_{1}))? You may have noticed that the (y)-value of the point, 3, is being added. To identify the point that is on this line, the equation must look like the general form, which subtracts the coordinates of the point. Therefore, the point can be seen more clearly if the equation is written as: (y-(-3)=-\frac{1}{2}(x-4))(Subtracting a negative value is the same as addition.) Now, we can see that this line travels through the point, ((4,-3)) and has a slope of (m=-\frac{1}{2}). Example #3 Let’s look at one more example: (y+12= -3(x+5)) By now, you can quickly see that the slope of this equation is (m=-3). This equation also does not “match” the general form but it can be rewritten as follows: (y-(-12)=-3(x-(-5))). This adjustment reveals that the point that is on the line, ((-5,-12)). Once you feel comfortable with identifying the slope and the point from this form, you can graph the line as we did before. (y-5=\frac{1}{2}(x-2)) Step 1: Identify the slope, (m=\frac{1}{2}). Step 2: Identify the point on the line, ((2,5)). Some students find it helpful to “switch the sign” of the given formula to determine the coordinates of the point! Step 3: Plot the point, ((2,5)), as the first point. Step 4: Move by (m=\frac{1}{2}). Rise equals UP, 1 Run equals RIGHT, 2 Plot the 2nd point at ((4,6)). Step 5: Repeat. Rise equals UP, 1 Run equals RIGHT, 2 Plot the 3rd point at ((6,7)). Step 6: Draw a straight line through the three points. Alright, we’ve covered a lot of ground in this video regarding the different ways linear equations can be written. While the structure of the equations looks different, they all represent a line. The use of each depends on what you are given or what you are asked to do. That’s all for this review! Thanks for watching, and happy studying! Frequently Asked Questions Q What is slope-intercept form? A Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. Q How do you graph slope-intercept form? A Graph slope-intercept form by first plotting the y-intercept, then using the slope to find a second point and plotting that point, and finally drawing a line through the two points.Ex. Graph y = 2x – 3Plot the y-intercept: (0, -3)Find a second point using the slope and plot: (1, -1)Draw a line through the 2 points Q What is b in slope-intercept form? A In slope-intercept form, b stands for the y-intercept of the line. Q What is point-slope form? A Point-slope form is a linear equation in the form:(y-y_1=m(x-x_1)) Q How do you graph point-slope form? A To graph point-slope form (y – y1 = m(x – x1)), first plot the point (x1,y1). Then, use the slope (m) to find a second point on the line. Plot that point. Finally, draw a straight line through the two points. Ex. Graph y – 7 = -4(x – 1) First, plot the point (x1,y1), which in this case is (1, 7). Then, use the slope to find a second point. The slope is -4, so move right 1 and down 4. The new point is (2, 3). Plot this point. Finally, draw a line through these two points. Slope-Intercept and Point-Slope Practice Questions Question #1: Which of the following equations is in slope-intercept form? (y=8(x+12)) (3x+7y=19) (y=4x-3) (y-7=4(x-17)) Answer: The correct answer is (y=4x-3). Equations in slope-intercept are in the form (y=mx+b), where m is the slope and b is the y-intercept. Question #2: Which of the following equations is in point-slope form? (7x+2=19y) (y=3x+5) (2x-9y=21) (y-11=2(x+14)) Answer: The correct answer is (y-11=2(x+14)). Equations in point-slope form have the form (y-y_1=m(x-x_1)), where ((x_1,y_1)) is a point on the line and m is the slope of the line. Question #3: Which of the following equations is in standard form? (19x+3y=27) (y-4=7(x+3)) (y=2x+14) (y-11=2(x-11)) Answer: The correct answer is (19x+3y=27). Equations in standard form have the form (Ax+By=C). Question #4: What is the equation (3x+4y=12) in slope-intercept form? (y-12=3(x+4)) (y-3=4(x+12)) (y=-\frac{3}{4}x+3) (y=-3x+8) Answer: The correct answer is (y=-\frac{3}{4}x+3). Equations in slope-intercept form have the form (y=mx+b). To get (3x+4y=12) in that form, manipulate the equation to isolate y. (3x+4y=12) Subtract 3x from both sides. (4y=12-3x) Divide both sides by 4. (y=3-\frac{3}{4}x) Rearrange the right side so it is in the proper form. (y=-\frac{3}{4}x+3) Question #5: What is the point-slope equation (y-7=13(x+3)) in slope-intercept form? (3x-7y=13) (7x+3y=13) (y=7x-13) (y=13x+46) Answer: The correct answer is (y=13x+46). Equations in slope-intercept form have the form (y=mx+b). To get (y-7=13(x+3)) in that form, manipulate the equation to isolate y. (y-7=13(x+3)) Distribute 13 to ((x+3)). Add 7 to both sides. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding non missing value percentage of each group across DataFrame columns Ask Question Asked 4 years, 11 months ago Modified4 years, 11 months ago Viewed 605 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I use the following code whenever I wish to calculate the % of non missing values in a column (say a) when groupby an id column: python df.groupby('id')['a'].apply(lambda x: x.notnull().sum()/len(x)100) If I want to calculate for multiple columns I have to loop through. Is there a better way to avoid looping over the columns? Thanks python pandas Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked Oct 12, 2020 at 19:10 oceanbeach96oceanbeach96 634 10 10 silver badges 23 23 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Lets try python df.groupby('id')['a','b'].apply(lambda x: x.notnull().sum()/len(x)100) Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 12, 2020 at 20:58 wwndewwnde 26.7k 6 6 gold badges 21 21 silver badges 38 38 bronze badges 2 Comments Add a comment oceanbeach96 oceanbeach96Over a year ago Ahh, so instead of looping, just put in a list of columns 2020-10-12T21:05:45.49Z+00:00 0 Reply Copy link wwnde wwndeOver a year ago Yap, that should be able to do. Using theloc accessor, you can pull a range of columns and parse them if needed. If happy, please accept and or upvote so that other can use your question and my answer confidently. 2020-10-12T21:08:00.197Z+00:00 0 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. 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17144	https://cs.uwaterloo.ca/journals/JIS/VOL17/Carroll/carroll2.pdf	23 11 Article 14.4.8 Journal of Integer Sequences, Vol. 17 (2014), 2 3 6 1 47 Cyclic Products of Purely Periodic Irrationals C. R. Carroll Department of Mathematics Texas A&M University Kingsville, TX 78363 USA c-carroll@tamuk.edu Abstract Let (a0, · · · , ak−1) be a sequence of positive integers and m a positive integer. We prove that “almost every” real quadratic unit ǫ of norm (−1)k admits at least m distinct factorizations into a product of purely periodic irrationals of the form ǫ = [a0; a1, . . . , ak−1, x, y] × [a1; a2, . . . , x, y, a0] × · · · × [y; a0, . . . , ak−1, x]. Periods exhibited in this expression are not assumed minimal. The analogous assertion holds for real quadratic units ǫ > 1 with prime trace and m = 1. The proofs are based on the fact that an integral polynomial map of the form f(x, y) = axy + by + cx + d, gcd(a, bc) = 1, a > 1, b, c > 0, assumes almost every positive integral value and almost every prime value when evaluated over the positive integers. 1 Introduction To a sequence of positive integers ν = (a0, a1, a2, . . . , ak−1), k ≥1, we associate the real quadratic unit ǫ > 1 obtained by taking the following product of purely periodic quadratic irrationals ǫ = k−1 Y i=0 [ai; ai+1, . . . , ak−1, a0, ai−1]. (1) Deviating from the standard convention we allow periods exhibited in this expression to be multiples of the minimal period. This convention is retained throughout. An induction 1 shows that ǫ > 1 is a quadratic unit . Alternatively, this fact follows easily from the matrix approach to the continued fraction algorithm (see van der Poorten and §4); from this point of view (1) corresponds to the matrix relation a0 1 1 0 a1 1 1 0 · · · ak−1 1 1 0 α 1 = ǫ α 1 (2) where α = [a0; a1, . . . , ak−1]. Note from (2) that ǫ has norm (−1)k. Every real quadratic unit ǫ > 1 has a trivial factorization of the form given in (1). Let N denote the trace of ǫ. For ǫ with norm −1 we have ǫ = [N;]; otherwise when the norm is +1 we have ǫ = [1; N −2] × [N −2; 1]. We call a product of the form given in (1) a cyclic factorization of the corresponding unit ǫ. Such factorizations are of interest from a variety of diﬀerent points of view. Distinct factorizations of ǫ may be taken to represent distinct conjugacy classes of hyperbolic matrices in GL(2, Z) with dominant eigenvalue ǫ. These classes, in the study of hyperbolic automor-phisms of the torus, correspond to topological conjugacy classes of homeomorphisms, the invariants ǫ and ν having natural topological interpretations . By a result due to Latimer and MacDuﬀee there is a further identiﬁcation with ideal classes of the order Z[ǫ]. If ǫ0 is the fundamental unit of Q[ √ d] (d square-free), then, as explained by Yamamoto , the ﬁeld class number h(d) can be identiﬁed with the number of cyclic factorizations of ǫ0 satisfying the special condition that the factors have discriminant d. With the aid of a computer the complete family of cyclic factorizations of a real quadratic unit ǫ > 1, for ǫ not too large, can be determined. Looking at such families one observes in general a profusion of cyclic factorizations, the associated sequences ν displaying a great deal of randomness. A natural question is the extent to which these sequences ν may vary over the family of factorizations determined by a unit ǫ. We call a sequence of integers whose terms are known with the exception of d integers in ﬁxed positions a d-pattern. We consider the extent to which units ǫ admit cyclic factorizations instantiating a given d-pattern. If the length of the pattern is odd, it may be assumed the corresponding units have norm −1, a condition noted above to be necessary. For a 1-pattern ν(x) = (a0, . . . , ak−1, x), k ≥2, it is not hard to see that a signiﬁcant proportion of real quadratic units ǫ > 1 of norm (−1)k+1 have no factorization instantiating ν(x). This is a consequence of the fact that the existence of a corresponding cyclic factor-ization would induce a constraint on the congruence class, relative to certain moduli, of the trace of ǫ. The same constraint arises also for certain d-patterns, d > 1, for instance for the patterns (a, x, a −1, a + 1, y), a > 1. However, if a d-pattern has at least two adjacent free variables, a corresponding cyclic factorization almost always exits. We prove Theorem 1. Let ν = (a0, a1, · · · , ak−1) be a sequence of positive integers and m be an arbitrary positive integer. Almost every real quadratic unit ǫ > 1 of norm (−1)k admits a 2 factorization over Q[ǫ] of the form ǫ = [a0; a1, . . . , ak−1, x, y] × [a1; a2, . . . , x, y, a0] × · · · × [y; a0, . . . , ak−1, x]. (3) The integers x, y may be selected in at least m diﬀerent ways. Furthermore, if ν ̸= (1) almost every unit whose trace belongs to a ﬁxed integer translate of the primes has at least one such factorization. The argument given extends to certain 2-patterns whose free variables are not adjacent. Moreover, Theorem 1 immediately generalizes to ﬁnite families of sequences. Fix a ﬁnite family of sequences F with lengths of a single parity. For example we may take F to consist of all sequences of odd (resp., even) length whose length and terms satisfy ﬁxed upper bounds. Then almost every real quadratic unit ǫ > 1 of norm −1 (resp., +1) has a cyclic factorization of the form given in (3) for each ν ∈F. In the contrary direction, we see in §2 that given a ﬁxed sequence ν of length k there exist inﬁnitely many real quadratic units ǫ > 1 of norm (−1)k that have no corresponding cyclic factorization. If Dickson’s conjecture is correct, then given a ﬁnite family F as above, these units may be taken to have no cyclic factorization corresponding to any sequence ν ∈F. The sense in which units are to be considered generic can be made precise as follows. Let ǫ(i,+), i ≥3, (resp., ǫ(i,−), i ≥1) denote the real quadratic unit of magnitude greater than one with trace i and norm +1 (resp., −1). The magnitude of such a unit is well approximated by its trace (since ǫ(i,±) −i < 1). Let T+ (resp., T−) be the set of traces i of units of the form ǫ(i,+) (resp., ǫ(i,−)) that satisfy a given statement. The statement will be said to hold of almost every unit ǫk of norm +1 (resp., −1) or, also, of a generic unit of this type, if and only if the corresponding subset of T+ (resp., of T−) is of asymptotic density one in the integers. The same approach can be employed to deﬁne genericity relative to any inﬁnite subset of units ǫ(i,+) (resp., ǫ(i,−)); in particular, the trace i may be taken to be prime. It is not hard to show that almost every real quadratic unit ǫ(i,+) (resp., ǫ(i,−)) is a fundamental unit . The proof of Theorem 1 rests on the following result on the representation of integers by polynomials. Let f(x, y) = axy + by + cx + d be an integral polynomial such that a > 1, b, c ∈Z+, and gcd(a, bc) = 1. Let P denote the set of primes. Theorem 2. Let Rf = {f(x, y) : x, y ∈Z+}. Fix m ∈Z+. Then: (i) Rf ∩Z+ is of asymptotic density one within the set of positive integers, each element of the set being represented by at least m distinct pairs of positive integers (x, y); (ii) Rf ∩P is of relative asymptotic density one in P. Assertion (i) follows quickly from the fact that, given a ﬁxed modulus a, almost every integer has at least one divisor in each residue class r mod a, gcd(r, a) = 1 . S.K. Stein obtains a more general factorization result from which he derives (i) with m = 1 and d = 0. The proof of (ii) relies on Dirichlet’s and de la Vall´ ee Poussin’s theorems, together with a modern variant of Euler’s product identity for arithmetic progressions. The relative 3 density of the set of primes that is represented by a general two-variable integral quadratic polynomial over the integers was investigated in well-known work of Iwaniec . Their results imply that the above polynomials f(x, y) assume a set of prime values of positive relative density when evaluated over the integers. We should note that in contrast with the restricted class of polynomials considered here, which assume a set of integer values of asymptotic density one, the usual number-theoretic focus is on polynomials that assume a sparse set of integer values. It is possible to extend the argument given for (ii) to a more general class of integral polynomials of the form f(x, y) = (rx + ty)(ux + my) + bx + cy + d, subject to various restrictions on the parameters, but this level of generality is not needed for our purposes. In §2-3 a proof of Theorem 2 is given. The derivation of Theorem 1 is given in §4. 2 The representation of positive integers Let f(x, y) = a xy + by + cx + d be an integral polynomial whose coeﬃcients satisfy a > 1, b, c > 0, gcd(a, bc) = 1. (4) We consider the positive integral values taken by f(x, y) over the positive integers. This set may be identiﬁed with f(Z+ ×Z+) up to a possible ﬁnite number of non-positive values that may arise when d < 0. Given N ∈Z+ we say f(x, y) represents N if there exist integers x, y ∈Z+ such that f(x, y) = N; it should be emphasized that the representation we are considering is over the positive integers, an assumption necessitated by our later arguments (see §4). We will say f(x, y) represents a set B ⊂Z+ if it represents each element of B in the above sense. Fix an inﬁnite set of positive integers B and let A ⊂B. Then the asymptotic density (or natural density) of A relative to B is deﬁned to be the limit dB(A) = lim k→∞ \|A ∩{1, 2, . . . , k}\| \|B ∩{1, 2, . . . , k}\| provided it exists. If dB(A) = 1 we will say that almost every element of B lies in A or, alternatively, that A is of full density in B. On occasion the reference to B is omitted, in which case it is to be assumed B = Z+ . In this section we show that any polynomial f(x, y) satisfying the conditions stated in (4) represents almost every positive integer, or dZ+(f(Z+ × Z+) ∩Z+) = 1. We recall some basic facts. Let B be an inﬁnite set of integers and let C be the collection of all subsets of B having well-deﬁned asymptotic density relative to B. It is well-known that C is closed under ﬁnite disjoint unions and under complementation but is not closed under intersection and hence does not form an algebra. A useful property of dB is that it is additive over disjoint sets in C. Below we list several additional useful properties [8, pp. 79–80]. Lemma 3. Let B be an inﬁnite subset of positive integers and A ⊂B. 4 (i) If A is ﬁnite then dB(A) = 0. (ii) Assume A has well-deﬁned density, given by dB(A) = d. If Ac is the complement of A in B then dB(Ac) = 1 −d. (iii) Let Ai, 1 ≤i ≤k, be a collection of subsets of B for which dB(Ai) = 1. Then the asymptotic density of Tk i=1 Ai exists and is given by dB(Tk i=1 Ai) = 1. (iv) Assume A1 ⊂A2 are nested subsets of B with well-deﬁned asymptotic densities. Then dB(A1) ≤dB(A2). (v) Let A be any subset of positive integers. Fix k1 ∈Z+ and k2 ∈Z. Then dZ+(A) = d if and only if dZ+(k1 A + k2 ∩Z+) = d/k1. (vi) If A1, A2 ⊂B satisfy dB(A1) = 1 and dB(A2) = d then the density of A1 ∩A2 relative to B is well-deﬁned and given by dB(A1 ∩A2) = d. We consider now the positive integers N represented by f(x, y). A standard approach to ﬁnding solutions of an equation of the form a xy + by + cx + d = N is to rewrite it as aN −(ad −bc) = (ax + b)(ay + c). (5) From this equation we see that N is represented by f(x, y) exactly when aN −(ad −bc) is the product of two integers greater than a which belong to suitable residue classes modulo a. Of course, such a factorization need not exist; in particular aN −(ad −bc) may be prime. By Dirichlet’s theorem Theorem 4 (Dirichlet). Every arithmetic progression of the form a x+b, with a, b non-zero integers satisfying a > 0 and gcd(a, b) = 1, contains inﬁnitely many primes. In fact, the sum of the reciprocals of the primes generated by such a progression diverges. Given that the ﬁxed parameters of (5) satisfy gcd(a, ad −bc) = 1, by Dirichlet’s theorem the integer aN −(ad −bc) is prime for inﬁnitely many positive integral values of N; hence these values cannot be represented by f(x, y). On the other hand we will see now that f(x, y) represents almost every positive integer. This is a consequence of the fact that the divisors of an integer are in general well-distributed over residue classes. As stated by Erd˝ os , Proposition 5. Let a ∈Z+, a > 1. Let F(a) denote the set of positive integers that have at least one divisor congruent to k modulo a for every integer k relatively prime to a. Then F(a) is of full asymptotic density in Z+. By Lemma 3 (vi) this observation extends to any subset S ⊂Z+ of positive asymptotic density; that is, dS(S ∩F(a)) = 1. With this fact in hand it is easy to see that for almost every positive integer N there exist non-negative integers x, y satisfying (5). Let T be the aﬃne transformation given by T(N) = a N −(ad −bc). The set T(Z+) may include a ﬁnite number of non-positive integers which can be ignored since for any 5 solution x, y > 0 of (5) the right-hand side of the equation must be positive. Accordingly let B = T(Z+)∩Z+. Consider the subset B1 = B ∩F(a). Since by Lemma 3 (v) dZ+(B) = 1/a, we have dZ+(B1) = 1/a as well. An element m = T(N ′) of B1 must satisfy the congruence m ≡bc (mod a). Given that gcd(b, a) = 1, it follows m has a factorization of the form m = m1 m2 with m1 ≡b (mod a) and hence with m2 ≡c (mod a). Therefore (5) has a solution in non-negative integers x, y for N = N ′ and indeed for any N ∈T −1(B1). Consider now the asymptotic density of T −1(B1). To simplify notation put l = ad −bc. Let Dk = T −1(B1) ∩{1, 2, . . . , k}. Then T(Dk) = B1 ∩{1, 2, . . . , ak −l} must be a set of the same cardinality, say nk. Hence we may express the asymptotic density of B1 as dZ+(B1) = lim k→∞ nk ak + l = 1 a lim k→∞ nk k . Recalling that dZ+(B1) = 1/a this shows limk→∞ nk k = 1. Therefore T −1(B1) is of full density. We now know that for almost every positive integer N there exist non-negative integers x, y satisfying (5). It remains to be shown that there exist multiple positive solutions. Let Z(a; s) denote the set of all positive integers m that are congruent to s modulo a. Proposition 6. Let k be an arbitrary positive integer and let a, b, c ∈Z+ such that a > 1 and gcd(a, bc) = 1. Then almost every integer m ∈Z(a; bc) admits at least k distinct factorizations of the form m = (xa + b)(ya + c), where x, y ∈Z+. Proof. Let s, k ∈Z+ and q be a prime such that s > 2k + c a and q > b + s a. Since gcd(a, b) = 1 the integers b + i a, 0 < i ≤s, are relatively prime to qa. Note that they represent s distinct reduced residue classes modulo qa. By the remark following Proposition 5, Z(a; bc) contains a subset Z1(a; bc) of full relative density whose elements have divisors in each of these residue classes. Fix m′ ∈Z1(a; bc). Then, given any residue class b+i0 a mod qa, 0 < i0 ≤s, we may write m′ as a product of positive integers m′ = m1m2 where m1 ≡b + i0 a (mod qa). Since m1m2 ≡bm2 ≡b c (mod a) it follows that m2 ≡c (mod a). Put c = c− c a . We may write m′ = m1 m2 = (b + i0 a + i1 qa) (c + j0 a) for non-negative integers i1 and j0. Notice that m1 = b + xa with x = i0 + i1q > 0. When i0 varies the factor m1 determines at least s distinct residue classes modulo qa. We have j0 ≤ c a for at most c a + 1 values. In the remaining cases m2 is of the form c + ya with y > 0. Thus we have products m′ = (b + xa)(c + ya) with x, y positive and with at least 2 k choices for the ﬁrst factor. These factorizations are distinct unless b = c, in which case there are still at least k distinct factorizations. We now are in a position to conclude 6 Corollary 7. Fix k ∈Z+. Let f(x, y) = axy + by + cx + d be an integral polynomial such that a, b, c > 0, a > 1, and gcd(a, bc) = 1. Then there exists a set of positive integers N of full asymptotic density such that f(x, y) = N for at least k distinct pairs of positive integers (x, y). An alternative proof of Corollary 7 can be given starting from the following reformulation of a xy + by + cx + d = N, a ̸= 0: N = (ax + b)y + cx + d. The idea, roughly, is as follows. Since gcd(a, b) = 1 Dirichlet’s theorem yields an inﬁnite sequence of distinct primes pi = a xi + b. Each value of the index i in turn determines a corresponding arithmetic progression p(xi, y) = pi y + (c xi + d) in the variable y. By a result due to C. A. Rogers [3, p. 242] the asymptotic density of the positive integers that are realized by at least one of these progressions is bounded below by the asymptotic density of the set of positive integers that are divisible by at least one of the primes pi. This density is known to be 1 provided the sum P∞ j=1 1 pi diverges , a fact which in the case at hand is a consequence of Dirichlet’s theorem. Consequently T(x, y) represents almost every positive integer over Z+ × Z+. A trick similar to that used in the proof of Proposition 6 is needed to obtain multiple representations. 3 The representation of primes Corollary 7 gives no information about the representation of sets of integers of asymptotic density zero. In this section we consider the representation of primes. We show Proposition 8. Let f(x, y) = a xy +by +cx+d be an integral polynomial whose coeﬃcients satisfy a > 1, b, c > 0, and gcd(a, bc) = 1. Then f(x, y) represents (over the positive integers) a set of primes p of full relative density. Remark 9. Since the integer d is arbitrary, f(x, y) also represents a subset of full relative density of any ﬁxed integer translate of the primes. The proof of Proposition 8 is based on Dirichlet’s theorem and the following additional classical results. De la Vall´ ee Poussin established that the primes generated by an arithmetic progression are equidistributed among the possible residue classes. Let P be the set of primes and Pa,b the set of primes generated by the arithmetic progression a x+b, a > 0, gcd(a, b) = 1. We have Theorem 10 (de la Vall´ ee Poussin). lim k→∞ \|Pa,b ∩{1, 2, . . . , k}\| \|P ∩{1, 2, . . . , k}\| = 1 φ(a) , 7 where φ denotes Euler’s totient function. We also rely on the following variant of Euler’s famous product identity for primes gen-erated by an arithmetic progression : Theorem 11. Let a, b be integers such that a > 0 and gcd(a, b) = 1. Then Y q∈Pa,b 1 −1 q = 0. To establish Proposition 8, we consider the families of primes generated by arithmetic progressions obtained by evaluating the polynomials f(x, y) = (a x+b) y+cx+d at selected values of x. To ﬁx ideas let us consider an arbitrary pair of arithmetic progressions q1 y +r1, q2 y +r2, with q1, q2 distinct odd primes and r1, r2 relatively prime to q1, q2, respectively. We wish to determine the relative asymptotic density of the set of primes A generated by these progressions, A = Pq1,r1 ∪Pq2,r2. This is most conveniently accomplished by computing the relative asymptotic density of the complementary set of primes P −A. Notice that only ﬁnitely many primes qc ∈P −A can belong to one of the four residue classes 0 (mod qi), ri (mod qi), i = 1, 2. The remaining primes qc are distributed across (q1 −2)(q2 −2) possible pairs of residue classes. Let k1 (mod q1), k2 (mod q2) be such a pair, with the residues ki taken to be reduced modulo qi, i = 1, 2. By the Chinese remainder theorem the condition that the prime qc belong to this pair of classes is equivalent to a condition of the form qc ≡s3 (mod q1q2), (6) with s3 a ﬁxed integer such that 1 ≤s3 ≤q1q2−1. Since k1, k2 ̸= 0, we have gcd(s3, q1 q2) = 1. Applying Theorem 10 we see that the set of all primes satisfying (6) has relative density dP(Pq1q2,s3) = lim k→∞ \|Pq1q2,s3 ∩{1, 2, . . . , k}\| \|P ∩{1, 2, . . . , k}\| = 1 φ(q1 q2) = 1 (q1 −1)(q2 −1). Since this computation does not depend on which of the (q1 −2)(q2 −2) pairs of residue classes under consideration is selected, the total number of primes in P −A is asymptotic to (q1 −2) (q2 −2) 1 (q1 −1)(q2 −1) = 1 − 1 q1 −1 1 − 1 q2 −1 The argument generalizes to any number k of progressions. Thus we have Lemma 12. Let qj, 1 ≤j ≤k, be k distinct primes and rj corresponding integers such that gcd(qj, rj) = 1. Let A = Sk j=1 Pqj,rj. Then the asymptotic density of A within the set of primes is given by 1 − k Y j=1 1 − 1 qj −1 8 This fact can be extended to the case of an inﬁnite union S∞ j=1 Pqj,rj by applying Lemma 3 (iv) to the nested sequence of sets Ai = i [ j=1 Pqj,rj, i ∈Z+. We obtain Lemma 13. Let qj be an inﬁnite sequence of distinct primes and rj a corresponding sequence of integers satisfying gcd(qj, rj) = 1. Then S∞ j=1 Pqj,rj is a set of full density within the set of primes provided ∞ Y j=1 1 − 1 qj −1 = 0. The proof of Proposition 8 is now straightforward. Assume the polynomial f(x, y) = axy + by + cx + d = (ax + b)y + cx + d satisﬁes the hypotheses of Proposition 8. By Dirichlet’s theorem the values of ax + b are prime for an increasing sequence of positive integers xi. We write pi = a xi + b, ri = cxi + d. Consider the arithmetic progression in y given by f(xi, y) = pi y + ri. If gcd (pi, ri) = 1, Dirichlet’s theorem can be applied a second time. Lemma 14. There exists an integer i0 such that for i ≥i0 gcd(pi, ri) = gcd(axi + b, c xi + d) = 1. Proof. Since a ̸= 1 and gcd(a, c) = 1, it follows a ̸= c. Case (i): a > c. For xi suﬃciently large necessarily 0 < cxi + d < axi + b. Since pi = axi + b is prime gcd(pi, cxi + d) = 1. Case (ii): a < c. Since gcd(a, c) = 1 we have c = k a + c′ for positive integers k and c′ with 0 < c′ < a. Thus we may write f(xi, y) = (axi + b)y + (k a + c′)xi + d = (axi + b)(y + k) + c′xi + d′ where d′ = d −k b. Again, for xi suﬃciently large 0 < c′xi + d′ < axi + b. Hence gcd(axi + b, cxi + d) = gcd(axi + b, (cxi + d) −k(axi + b)) = gcd(axi + b, c′xi + d′) = 1. Hence we may apply Dirichlet’s theorem to the progressions piy + ri for suﬃciently large values of the index i, say, i ≥i0; accordingly, f(x, y) represents the primes Ppi,ri for i ≥i0. Consider S∞ i=i0 Ppi,ri. Theorem 11 yields ∞ Y i=i0 1 − 1 pi −1 ≤ ∞ Y i=i0 1 −1 pi = 0. It follows by Lemma 13 that the set of primes S∞ i=i0 Ppi,ri is of full density in the primes. Therefore f(x, y) represents, over the positive integers, a set of primes of full relative density, completing the proof of Proposition 8. 9 4 Periods prescribed up to two adjacent integers There exists a well-known correspondence, mentioned already in §1, between certain ma-trix products and continued fractions. In particular, the following two statements may be regarded as equivalent: (i) α = [a0; a1, . . . , ak−1] with k a ﬁxed multiple of the minimal period). (ii) For some unit ǫ ∈Q[α], ǫ > 1, a0 1 1 0 a1 1 1 0 · · · ak−1 1 1 0 α 1 = ǫ α 1 . (7) From the second equation we obtain a corresponcing factorization of ǫ, as follows. Let αm, m ≥0 denote the purely periodic irrational obtained by cyclically permuting the period of the expansion of α so that am occurs as the initial element, i.e., αm = [am; am+1, · · · , ak−1, a0, · · · , am−1]. Notice that αi+k = αi. A computation yields am 1 1 0 −1 αm 1 = 1 αm+1 αm+1 1 . (8) The k matrices occurring in (7) may be eliminated by repeatedly multiplying each side of the equation by the inverse of the left-most matrix and applying the equality of Equation (8) to the right-hand side of the equation. After the k-th multiplication we are left with α 1 = ǫ α0αk−1 · · · α1 α 1 . Hence ǫ = α0 α1 . . . αk−1. (9) We call a matrix of the type arising in Equation (7)—that is, either a single matrix of the form a 1 1 0 , a ∈Z+, or a ﬁnite product of such matrices—a CF-matrix of depth k. Such matrices have a simple characterization [12, Proposition 3]: namely, they are the non-negative elements A = s1 s2 s3 s4 of GL(2, Z) for which s1 ≥max (s2, s3). By an elementary induction if s1 ̸= 1 the inequality is strict, s1 > max (s2, s3). One may easily verify that an arbitrary CF-matrix A of depth k in addition satisﬁes the following properties: Det(A) = (−1)k; s1, s2, s3 > 0; gcd(s1, s2) = gcd(s1, s3) = 1. Given a sequence of positive integers ν = (a0, a1, · · · , ak−1), let Mν(x, y) = a0 1 1 0 · · · ak−1 1 1 0 x 1 1 0 y 1 1 0 (10) 10 and let ǫν(x, y) > 1 denote the dominant eigenvalue of Mν(x, y). Writing b1 b2 b3 b4 = Y 0≤i≤k−1 ai 1 1 0 , we obtain the following expression for the trace of Mν(x, y) (which by the Cayley-Hamilton theorem also gives the trace of its eigenvalues) Tν(x, y) = b1xy + b2y + b3x + b1 + b4. Thus if N ∈Z+ is represented by the polynomial Tν(x, y)—that is, N = Tν(x0, y0) for some choice of positive integers x0, y0—then ǫν(x0, y0) is the dominant eigenvalue of the matrix Mν(x0, y0) and by (9) it follows that ǫν(x0, y0) admits the cyclic factorization [a0; a1, . . . , ak−1, x0, y0] × [a1; a2, . . . , x0, y0, a0] × · · · × [y0; a0, . . . , ak−1, x0]. By the characterization of CF-matrices provided above the polynomial Tν(x, y) satisﬁes the hypotheses of Theorem 2 provided ν ̸= (1). When ν = (1), Tν(x, y) = (x + 1)(y + 1). In the former case, applying Theorem 2, and in the latter, noting that composite integers form a set of integers of asymptotic density one, we may conclude that Tν(x, y) represents almost every positive integer N in at least m diﬀerent ways. Turning to the representation of primes, assuming ν ̸= (1), by Theorem 2 almost every prime is represented as well. This yields Theorem 1. 5 Acknowledgement The author is grateful to an anonymous referee for helpful comments and corrections. References R. Adler, C. Tresser, and P. A. Worfolk. Topological conjugacy of linear endomorphisms of the 2-torus, Trans. Amer. Math. Soc. 349 (1997), 1633–1652. P. Erd˝ os. On the distribution of divisors of integers in the residue classes (mod d), Bul. Soc. Math. Gr´ ece (N.S.) 6 (1965), fasc.1, 27–36. H. Halberstam and K. F. Roth. Sequences, Springer, New York, 1983. G. H. Hardy and E. M. Wright. An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008. H. Iwaniec. Primes represented by quadratic polynomials in two variables, Acta Arith. 24 (1974), 435–459. 11 A. Languasco and A. Zaccagnini. A note on Mertens’ formula for arithmetic progres-sions, J. Number Theory 127 (2007), 37–46. C. Latimer and C. C. MacDuﬀee. A correspondence between classes of ideals and classes of matrices, Ann. of Math. 34 (1933), 313–316. W. Narkiewicz. Number Theory, World Scientiﬁc, Singapore, 1983. P. A. B. Pleasants. The representation of primes by quadratic and cubic polynomials, Acta Arith. 12 (1966), 131–163. V. G. Sprindz˘ uk. The distribution of the fundamental units of real quadratic ﬁelds, Acta Arith. 25 (1973/74), 405–409. S. K. Stein. The density of the product of arithmetic progression, Fibonacci Quart. 11 (1973), 145–152. A. J. van der Poorten. Fractions of the period of the continued fraction expansion of quadratic integers, Bull. Austral. Math. Soc. 44 (1991), 155–169. Y. Yamamoto. Real quadratic number ﬁelds with large fundamental units, Osaka J. Math. 8 (1971), 261–270. 2010 Mathematics Subject Classiﬁcation: Primary 11A55; Secondary 11N32, 37D20. Keywords: simple continued fraction, purely periodic irrational, prime represented by quadratic polynomial, linear automorphism of the torus, Anosov automorphism. Received May 3 2013; revised version received March 21 2014. Published in Journal of Integer Sequences, March 22 2014. Return to Journal of Integer Sequences home page. 12
17145	https://math.stackexchange.com/questions/4629409/find-the-equation-of-a-system-of-coaxial-circles-of-which-the-points-pm-k-0	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Find the equation of a system of coaxial circles of which the points $(\pm k,0)$ are the limiting points. Ask Question Asked Modified 1 year, 8 months ago Viewed 285 times 0 $\begingroup$ Find the equation of a system of coaxial circles of which the points $(\pm k,0)$ are the limiting points. The definition of some terms relevant to the question in 2D geometry is as follows: The coaxial circles is the set of all circles such that any two of the circles have the same radical axis. The radical axis of two circles is the locus of the points which have the same powers with respect to them.(The power of a point with respect to a circle is the length of the tangents drawn from the point to the circle) My solution goes like this: The equation of the radical axis which is $y-0=0(x-k)$ and thus, $y=0.$ Now, we know that the limiting points and centres in a coaxial system of circles are collinear. Given, $(\pm k,0)$ are the limiting points, we can say, the centre of any circle in the coaxial system lies on the $x-$ axis. Thus, the equation of the point circles corresponding to limiting points $(\pm k,0)$ are $(x\pm k)^2+y^2=0.$ Thus, the equation of the system of coaxial circles is $$(x-k)^2+y^2+\lambda y=0.$$ Is the above solution correct? If not, where is it going wrong? solution-verification analytic-geometry Share edited Feb 7, 2023 at 13:22 ArthurArthur asked Jan 31, 2023 at 7:44 ArthurArthur 2,64811 gold badge99 silver badges2929 bronze badges $\endgroup$ 3 1 $\begingroup$ @AnneBauval I have added the definitions...By slope I mean the slope that it synonymous in 2-D geometry. $\endgroup$ Arthur – Arthur 2023-01-31 14:13:54 +00:00 Commented Jan 31, 2023 at 14:13 $\begingroup$ Thank you! I also learned the definition of limiting points. Does this page of MathWorld answer your question? $\endgroup$ Anne Bauval – Anne Bauval 2023-01-31 14:46:17 +00:00 Commented Jan 31, 2023 at 14:46 $\begingroup$ @AnneBauval I have edited my post. Mind taking a look at it?... $\endgroup$ Arthur – Arthur 2023-02-07 13:09:22 +00:00 Commented Feb 7, 2023 at 13:09 Add a comment \| 2 Answers 2 Reset to default 1 $\begingroup$ Since $(\pm k, 0)$ are limiting points of the system, these two point circles $(x\pm k)^2+y^2=0\tag{}$ belong to the system. Label their LHS as $S_1(x,y)$ and $S_2(x,y)$ then the equation of the system is $S_1+\lambda(S_1-S_2)=0$. $$x^2+y^2+(2k+4\lambda k)x+k^2=0$$ Since $k\neq 0$, you might want to replace $2k+4\lambda k$ with $\lambda$ (because it's just a parameter). $$\boxed{x^2+y^2+\lambda x+k^2=0}$$ Visual graph for demonstration. Share edited Jan 16, 2024 at 10:52 answered Jan 16, 2024 at 10:43 Nothing specialNothing special 3,88011 gold badge88 silver badges2828 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ Your equation $(x-k)^2+y^2+\lambda y=0$ is not correct: the centers of these circles are not on the $x$-axis. Following the equation of a system of coaxial circles with limiting points $(±k,0)$ is: $$x^2+y^2+2\lambda x+k^2=0$$ or equivalently $$(x+\lambda)^2+y^2+k^2-\lambda^2=0.$$ Share answered Feb 7, 2023 at 13:39 Anne BauvalAnne Bauval 50.6k1212 gold badges3535 silver badges7979 bronze badges $\endgroup$ 2 $\begingroup$ But aren't circles in a system of coaxial circles have there centers collinear or equivalently upon a straight line? $\endgroup$ Arthur – Arthur 2023-02-07 13:45:57 +00:00 Commented Feb 7, 2023 at 13:45 $\begingroup$ Yes, this is why your equation was not correct: your centers were $(k,-\lambda/2),$ and $(-k,0)$ is not on this line. As you said yourself, since the limit points are on the $x$-axis, all centers have to be on this axis (not on the vertical line $x=k$). With MathWord's equation, the centers are $(-\lambda,0)$ and belong to the $x$-axis. $\endgroup$ Anne Bauval – Anne Bauval 2023-02-07 13:53:19 +00:00 Commented Feb 7, 2023 at 13:53 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions solution-verification analytic-geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 0 Find equations of two circles drawn through the origin which cut another given circle orthogonally and touch a given line 0 Finding the equation of a coaxial circle with its diameter falls on the radical line 0 Limiting points subtend right angle at the centre 1 If tangents are drawn from two points which are equidistant from given point, then find the locus Find the locus of centres of circles 0 Find the locus of a point which moves so that the tangents from it to a circle are at right angles. 0 Find the common tangents for the given circles. 1 What is the locus of points equidistant from two circles? 1 Locus of centre of circle which bisects two smaller circles Hot Network Questions How can the problem of a warlock with two spell slots be solved? 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17146	https://www.youtube.com/watch?v=ncORPosDrjI	The Water Cycle \| The Dr. Binocs Show \| Learn Videos For Kids Peekaboo Kidz 6760000 subscribers 57968 likes Description 7873742 views Posted: 11 Mar 2015 Learn about the water cycle with Dr. Binocs. Hey kids! Ever wondered how it rains? Where does the water vapour disappear? Don't worry, Dr. Binocs is here with all the answers. For more fun filled facts and learn videos stay tuned to The Dr. Binocs Show. Share on Facebook - Tweet about this - Catch More Of Dr.Binocs - To Watch Popular Nursery Rhymes Go To - To Watch Alphabet Rhymes Go To - To Watch Compilations Go To - Watch Popular Nursery Rhymes with Lyrics - Subscribe : Voice-Over Artist: Joseph D'Souza Script Writer: Sreejoni Nag Background Score: Agnel Roman Sound Engineer: Mayur Bakshi Animation: Qanka Animation Studio Creative Team (Rajshri): Kavya Krishnaswamy, Alisha Baghel, Sreejoni Nag Producer: Rajjat A. Barjatya Copyrights and Publishing: Rajshri Entertainment Private Limited All rights reserved. SUBSCRIBE to Peekaboo Kidz: Catch Dr.Binocs At - To Watch More Popular Nursery Rhymes Go To - To Watch Alphabet Rhymes Go To - To Watch Compilations Go To - Catch More Lyricals At - Like our Facebook page: Transcript: “Ahhh.. It's raining” Hey! Have you ever wondered where the rain comes from?” Or how the clouds are formed? Well, this is what the water cycle is all about. Come let's explore! Zoooom In When the sun heats up the rivers and oceans, water becomes water vapour and it rises up in the air. This process is called evaporation. It is the first step of the water cycle. You too can see water vapour at home! Just tell your mommy to heat some water. And as the water gets heated, you'll be able to see the water vapour rising up in the air. When the water vapour reaches up in the sky, it turns into tiny droplets of water. These water droplets along with various gases and dust particles, come together to form clouds. This is known as condensation. Now, hold a cold lid over the vessel in which you heated water. When you open the lid after sometime, you'll be able to see water droplets on the lid. That's exactly what condensation is! When the cloud becomes too heavy and it cannot hold any more water inside, it bursts open to give out rain, hail or snow. This is known as precipitation. As it rains, water gets collected in oceans, lakes and rivers. It even seeps through the soil and becomes ground water. Thus water cycle is a continuous process of evaporation, condensation and precipitation. Did you know that even plants sweat? That's called transpiration. That's why it rains more in places with more trees, like hill stations and forests. Sometimes snow directly turns into water vapour without melting into water. That's called sublimation. This happens a lot in cold countries. Oh, I need to run now. It's raining again. So this is me zooming out, tune in next time for more fun facts.
17147	https://www.cuemath.com/calculus/extreme-value-theorem/	Extreme Value Theorem The extreme value theorem is an important theorem in calculus that is used to find the maximum and minimum values of a continuous real-valued function in a closed interval. This theorem is used to prove Rolle's theorem in calculus. The extreme value theorem is specific as compared to the boundedness theorem which gives the bounds of the continuous function on a closed interval. In this article, we will discuss the concept of extreme value theorem, its statement, and its proof. We will also learn how to use the theorem with the help of a few solved examples for a better understanding of the concept. \| \| \| --- \| \| 1. \| What is Extreme Value of Theorem? \| \| 2. \| Extreme Value Theorem Statement \| \| 3. \| Extreme Value Theorem Proof \| \| 4. \| How to Use Extreme Value Theorem? \| \| 5. \| FAQs on Extreme Value Theorem \| What is Extreme Value of Theorem? The extreme value theorem helps in proving the existence of the maximum and minimum values of a real-valued continuous function over a closed interval. Once the existence of maximum and minimum values is proved, we might be asked to determine those values using the derivative of the function and finding the critical points. Rolle's theorem and Mean value theorem are the consequences of the extreme value theorem. Let us understand the meaning of extreme value below as we proceed to state the theorem and prove it. Extreme Value Meaning Extreme values of a function f(x) are the values y = f(x) which a function attains for a specific input x such that no other value of f(x) in the range is greater or less than these values. We have two types of extreme values: maximum and minimum. The maximum value of a function is a value such that no other value of the function can be greater than this and the minimum value of a function is a value such that no other value of the function is less than this value. Extreme Value Theorem Statement The extreme value theorem states that 'If a real-valued function f is continuous on a closed interval [a, b] (with a < b), then there exist two real numbers c and d in [a, b] such that f(c) is the minimum and f(d) is the maximum value of f(x). Mathematically, we can write the formula for the extreme value theorem as, f(c) ≤ f(x) ≤ f(d), ∀ x ∈ [a, b]. The extreme value theorem can also be stated as 'If a real-valued function f is continuous on [a, b], then f attains its maximum and minimum of [a, b]. Extreme Value Theorem Proof Now that we have understood the extreme value theorem and its statement, let us now prove it using the contradiction method and the boundedness theorem. We will prove that f attains its maximum on the closed interval [a, b]. The proof that f attains its minimum on [a, b] can be proved on similar lines. By hypothesis, f is continuous on [a, b], so f is bounded on [a, b] such that there exist m, M such that we have m ≤ f(x) ≤ M using the Boundedness Theorem. Here, suppose M is the least upper bound of f. Now, if there exists c in [a, b] such that f(c) = M, then this implies f attains maximum on [a, b]. We have proved the required result. Now, assume there is no such c in [a, b], then we have f(x) < M for all x in [a, b]. Define a function h(x) = 1 / [M - f(x)] on [a, b]. Now, we know that h(x) > 0 because f(x) < M for all x in [a, b] and h is also continuous on [a, b]. So, using the boundedness theorem, we have h(x) is bounded on [a, b]. This implies there exists K > 0 such that h(x) ≤ K, for all x in [a, b]. ⇒ 1 / [M - f(x)] ≤ K ⇒ M - f(x) ≥ 1/K Adding f(x) - 1/K on both sides, we have ⇒ M - 1/K ≥ f(x) ⇒ f(x) ≤ M - 1/K This contradicts the fact that M is the least upper bound of f(x). Hence, our assumption that there exists no such c in [a, b] such that f(c) = M is wrong. Therefore, f attains its maximum on [a, b]. We can prove that f attains its minimum on [a, b] on similar lines. How to Use Extreme Value Theorem? Now that we have proved the extreme value theorem, let us learn how to use it with the help of an example. Consider function f(x) = x3 - 27x + 2. Find the maximum and minimum values of f(x) on [0, 4] using the extreme value theorem. Solution: Since f(x) = x3 - 27x + 2 is differentiable, therefore it is continuous. Since [0, 4] is closed and bounded, therefore we can apply the extreme value theorem. Differentiate f(x) = x3 - 27x + 2. f'(x) = 3x2 - 27 Setting f'(x) = 0, we have 3x2 - 27 = 0 ⇒ 3x2 = 27 ⇒ x2 = 27/3 = 9 ⇒ x = -3, 3 So, x = -3, 3 are the critical points. Now, we find the value of f(x) at critical points and the endpoints of the interval. f(-3) = (-3)3 - 27(-3) + 2 = -27 + 81 + 2 = 56 f(3) = (3)3 - 27(3) + 2 = 27 - 81 + 2 = -52 f(0) = (0)3 - 27(0) + 2 = 2 f(4) = (4)3 - 27(4) + 2 = -42 So the minimum value of f(x) on [0, 4] is -52 and its maximum value on [0, 4] is 56. Important Notes on Extreme Value Theorem ☛ Related Articles: Extreme Value Theorem Examples Example 1: Find the maximum and minimum values of f(x) = x4 - 3x3 - 1 on [-2, 2]. Solution: Since f(x) is differentiable, so it is continuous on [-2, 2]. So, we can apply extreme value theorem. Now, differentiate f(x). f'(x) = 4x3 - 9x2 = x2(4x - 9) Setting f'(x) = 0, we have x2(4x - 9) = 0 ⇒ x = 0, x = 9/4 So, 0 and 9/4 are the critical points. Since 9/4 does lie in the interval [-2, 2], therefore we will only consider one critical point x = 0. Now, we will find the value of f(x) at x = 0, -2 and 2. f(0) = 04 - 3(0)3 - 1 = -1 f(-2) = (-2)4 - 3(-2)3 - 1 = 16 + 24 - 1 = 39 f(2) = (2)4 - 3(2)3 - 1 = 16 - 24 - 1 = -9 So, the maximum value of f(x) = x4 - 3x3 - 1 on [-2, 2] is 39 at x = -2 and minimum value is -9 at x = 2. Answer: Maximum value = 39; Minimum value = -9 Example 2: Determine the maximum and minimum values of f(x) = sin x + cos x on [0, 2π] using the extreme value theorem. Solution: f(x) = sin x + cos x on [0, 2π] is continuous. So we can apply extreme value theorem and find the derivative of f(x). f'(x) = cos x - sin x Setting f'(x) = 0, we have cos x - sin x = 0 ⇒ cos x = sin x ⇒ x = π/4, 5π/4 which lie in [0, 2π] So, we will find the value of f(x) at x = π/4, 5π/4, 0 and 2π. f(π/4) = sin (π/4) + cos (π/4) = 1/√2 + 1/√2 = √2 f(5π/4) = sin (5π/4) + cos (5π/4) = -1/√2 - 1/√2 = -√2 f(0) = sin 0 + cos 0 = 0 + 1 = 1 f(2π) = sin 2π + cos 2π = 0 + 1 = 1 So, the maximum value of f(x) = sin x + cos x on [0, 2π] is √2 at x = π/4 and minimum value is -√2 at x = 5π/4. Answer: Maximum value = √2; Minimum value = -√2 go to slidego to slide Book a Free Trial Class Extreme Value Theorem Questions go to slidego to slide FAQs on Extreme Value Theorem What is Extreme Value Theorem in Math? The extreme value theorem is an important theorem in calculus that is used to find the maximum and minimum values of a continuous real-valued function in a closed interval. What is Extreme Value Theorem Formula? Mathematically, we can write the formula for the extreme value theorem as, f(c) ≤ f(x) ≤ f(d), ∀ x ∈ [a, b], where f is a continuous function on closed interval [a, b] and c, d lie in [a, b]. What is the Extreme Value Theorem Statement? The extreme value theorem states that 'If a real-valued function f is continuous on a closed interval [a, b] (with a < b), then there exist two real numbers c and d in [a, b] such that f(c) is the minimum and f(d) is the maximum value of f(x). How to Use the Extreme Value Theorem? The extreme value theorem is used in proving the existence of the maximum and minimum values of a real-valued continuous function over a closed interval. Once the existence of maximum and minimum values is proved, we might be asked to determine those values using the derivative of the function and finding the critical points. We find the value of the function at critical points and the endpoints of the interval to the maximum and minimum values. When Does Extreme Value Theorem Not Apply? The extreme value theorem cannot be applied if the function is not continuous on the closed and bounded interval [a, b]. What is the Condition of Extreme Value Theorem? The necessary condition of the extreme value theorem is that the function should be continuous on the closed and bounded interval [a, b]. How to Prove Extreme Value Theorem? The extreme value theorem can be proved using the contradiction and boundedness theorem. We can prove the existence of the maximum value and similarly for the minimum value of the function.
17148	https://jamanetwork.com/journals/jamaotolaryngology/fullarticle/562017	THROMBOSIS OF THE LATERAL SINUS: A SURVEY OF CURRENT OPINION AND RECORDS \| JAMA Otolaryngology–Head & Neck Surgery \| JAMA Network [Skip to Navigation] Our website uses cookies to enhance your experience. By continuing to use our site, or clicking "Continue," you are agreeing to our Cookie Policy\|Continue Navigation HomeIssuesMultimediaFor Authors JAMA+ AIWomen's HealthLearn More JOURNALS JAMAJAMA Network OpenJAMA CardiologyJAMA DermatologyJAMA Health ForumJAMA Internal MedicineJAMA NeurologyJAMA OncologyJAMA OphthalmologyJAMA Otolaryngology–Head & Neck SurgeryJAMA PediatricsJAMA PsychiatryJAMA SurgeryArchives of Neurology & Psychiatry (1919-1959) JAMA NETWORK CMESUBSCRIBEJOBSINSTITUTIONS / LIBRARIESREPRINTS & PERMISSIONS Search Individual Sign In Sign inCreate an Account Access through your institution Sign In Search All Search All JAMA JAMA Network Open JAMA Cardiology JAMA Dermatology JAMA Forum Archive JAMA Health Forum JAMA Internal Medicine JAMA Neurology JAMA Oncology JAMA Ophthalmology JAMA Otolaryngology–Head & Neck Surgery JAMA Pediatrics JAMA Psychiatry JAMA Surgery Archives of Neurology & Psychiatry Type text and use arrow keys to navigate suggestions HomeIssuesMultimediaFor Authors Home JAMA Otolaryngology–Head & Neck Surgery Vol. 28, No. 6 Sections PDF Share Close LinkedInXWhatsAppFacebookThreadsBlueskyWeChatCopy URLEmail Article THROMBOSIS OF THE LATERAL SINUS: A SURVEY OF CURRENT OPINION AND RECORDS WILLIAM H. EVANS, M.D. Author Affiliations YOUNGSTOWN, OHIO Cite This### Citation EVANS WH. THROMBOSIS OF THE LATERAL SINUS: A SURVEY OF CURRENT OPINION AND RECORDS. Arch Otolaryngol. 1938;28(6):959–986. doi:10.1001/archotol.1938.00650040972007 Manage citations: Select Format Download citation Copy citation Close Permissions View Metrics Arch Otolaryngol Published Online: December 1938 1938;28;(6):959-986. doi:10.1001/archotol.1938.00650040972007 related icon Related Articlesmultimedia icon Media Close Trending Audio Highlights: September 26, 2025JAMA This Week in JAMA September 26, 2025 Addressing Food Insecurity in Children with Chronic ConditionsJAMA Network Open Invited Commentary September 2025 Error in BylineJAMA Network Open Correction September 2025 Close Abstract While thrombosis of the lateral sinus is the intracranial complication most frequently encountered after operations for mastoiditis, the number of cases observed by the average otologist is so small that he feels obliged to draw on the larger collective experience of his confréres in formulating criteria for the management of this serious condition. When he consults the literature, he finds many excellent reports of cases and the expression of a wide diversity of opinion regarding the efficacy of the surgical procedures used to combat thrombosis of the lateral sinus. It is difficult, if not impossible, to draw definite conclusions, for there are records which seem to substantiate each contrasting opinion. Since the individual series of cases reported are, as a rule, small, it occurred to me that to obtain the recent records from various parts of the country on the incidence of thrombosis of the lateral sinus following operations for Get Access View Full TextDownload PDF Button To Top Select Your Interests Customize your JAMA Network experience by selecting one or more topics from the list below. 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17149	https://www.ncbi.nlm.nih.gov/books/NBK559320/	Follicular Adenoma - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Follicular Adenoma Joshua J. Norris; Fabiola Farci. Author Information and Affiliations Authors Joshua J. Norris 1; Fabiola Farci 2. Affiliations 1 Government Medical College Aurangabad 2 Pisa University Last Update: April 23, 2023. Go to: Continuing Education Activity Follicular adenomas are a common benign neoplasm encountered in clinical practice. They usually present as a solitary thyroid nodule, but other presentations can occur in varied patients. It needs to be distinguished from follicular carcinoma, which is malignant and has a poor prognosis if undiagnosed. This activity reviews the evaluation and treatment of follicular adenomas, identifies features that distinguish it from its malignant counterpart, and highlights the role of the interprofessional healthcare team in evaluating and treating patients with this condition. Objectives: Describe the pathophysiology of follicular adenomas. Outline the typical clinical presentation of a follicular adenoma. Review special clinical features of follicular neoplasms and management considerations for patients with the same. Outline the importance of collaboration and coordination among the interprofessional team dealing with follicular neoplasms especially when it comes to differentiating between a benign or malignant follicular neoplasm to improve patient outcomes with the right line of management. Access free multiple choice questions on this topic. Go to: Introduction Andreas Vesalius (1514 -1564) first formally described the thyroid in his anatomy manuscript,De humani corporis Fabrica Libri septem. Since then, we have come a long way in understanding the anatomy, physiology, and workings of this endocrine gland.The thyroid gland is a midline structure in the neck. Derived from the primitive pharynx and neural crest cells, embryologically, the tissue mass divides, forming the isthmus and the two lateral lobes of the thyroid. Bu the age of 2, the gland is already half the adult size. Developmental anomalies could result in aberrant masses of ectopic thyroid tissue, cysts, and sinuses.Apart from that, the thyroid gland is no stranger to neoplastic lesions. Follicular lesions of the thyroid include many subvariants from the benign follicular adenoma right up to malignant follicular carcinomas, follicular variants of other malignant thyroid lesions. Follicular adenomas are one subset of benign neoplasms that can occur in the thyroid gland or ectopic thyroid tissue. They typically present as a solitary thyroid nodule or in association with nodular hyperplasia or thyroiditis. Thyroid nodules are palpable in 4 to 7 % of individuals, but the prevalence of nodules detected incidentally by ultrasound shows a higher prevalence of 19 to 67 percent. The majority of thyroid nodules are asymptomatic. Similar to worldwide incidence, 60 to 70% of the US population present with thyroid nodules. Most of these are benign, although 5% exhibit malignant features.Although the distinguishing line between the adenoma itself and its malignant counterpart is tricky, this is for all practical purposes a benign neoplasm. Go to: Etiology Although follicular adenomas are mostly sporadic, multiple other etiological factors have been identified. Other causes implicated in the development of a follicular adenoma include: Iodine deficiency: This is a known risk factor seen in individuals consuming non-iodized sea salt regularly. Genetic alterations: PTEN hamartoma tumor syndrome (PHTS) includes several syndromes like Cowden syndrome and BRRS. Multiple follicular adenomas may occur along with other clinical manifestations. PTHS is due to mutations causing inactivation of the PTEN tumor suppressor gene. Other genetic mutations in BRAF,NRAS, RET, and KRAS can also cause unexplained follicular adenomas. Follicular adenomas are also part of various syndromes like Familial adenomatous polyposis(FAP), Carney Complex syndrome. Genetic rearrangement of the PAX8-PPAR gene causes loss of follicular growth inhibition, thus facilitating the development of follicular neoplasms. Prior I-131 radiation exposure also increases the risk of malignant lesions of the thyroid. Go to: Epidemiology Follicular adenomas usually present as a solitary thyroid nodule. They are benign neoplasms in nearly 2 to 4.3 % of the population.A study done on 300 consecutive autopsies revealed a 3% incidence of follicular adenomas compared to a 2.7% incidence for follicular carcinoma; small nodules, measuring less than 15 mm in diameter,have a prevalence of 5 to 10% (based on autopsy studies). The data vary between different geographic areas, Finland having the highest reported incidence of occult microcarcinomas. Thyroid adenomas are typically more common in females.Data in the US, more or less, reflect the same incidence for development. During the past 30 years, the incidence of thyroid cancer has raised from 4.3 cases per 100,000 in 1980 to 12.9 cases per 100,000 in 2008, and there is a strong debate in whether the increase in incidence should be considered an artifact of improved detection. The availability of ultrasound imaging, the technological improvement in medical equipment, the increased awareness in the general practice and among patients, the screening campaigns in higher-risk populationare all factors that contributed to an increase in incidence rates. The incidence rates for thyroid nodules are also increasing in the pediatric population; increased detection is surely a great component but can only explain a part of the cases.Birth cohort analyses showed the correlation with environmental exposures (such as diagnostic radiography, cosmic radiation, and other exposures).Studies on individuals exposed to I-131 radiation after Chernobyl's nuclear accident revealed an excess odds ratio per Grey of 2.22 equal for both sexes, with the odds ratio increasing with higher doses.Children aged less than two years were found to have the highest risks.Population studiesindicate that when a specific etiology (i.e. radiation) is involved, the incidence differs from the normal population. The peak in incidence seen in the last 30 years cannot be fully explained by the increased detection, suggesting that there might be a true increase, but the specific causing agent has not been identified yet. Go to: Pathophysiology Iodine deficiency implicated in many studies for follicular lesions. Endemic goiter is a predisposing factor for follicular carcinoma with a higher rate of follicular carcinomas seen in low iodine areas.Studies show that 4 to 13% of follicular adenomas exhibit rearrangement of the PAX8-PPAR. PAX8 helps follicular cell differentiation by encoding a nuclear protein product necessary for transcription of thyroid-specific factors. Martelli ML determined mutations in PAX8 that led to dysfunction and loss of optimal follicular growth inhibition.A mutation causing inactivation of the PTEN tumor suppressor gene causes PTHS. Likewise, mutations in BRAF,NRAS, RET, and KRAS can explain aberrant neoplasms, including the thyroid parenchyma. Functioning follicular adenomas arise as a result of a monoclonal increase of thyroid follicular cells with a predominance of activating mutations in the gene for the TSH receptor and less frequently in the adenylate cyclase-stimulating G alpha protein gene that results in increased thyroid hormone secretion independent of TSH. A monoclonal follicular cell expansion with activating mutations on the TSH receptor or in the adenylate cyclase-G protein mostly explains functioning follicular adenomas.20% of nonfunctioning follicular adenomas that have oncogene mutations can later progress to follicular carcinoma. Go to: Histopathology Follicular adenoma is grossly described as a solitary, encapsulated nodule; the size can be extremely variable, ranging from a few millimeters to 10-15 cm. The color vary from tan to light brown with solid and fleshy appearance. It can resemble multinodular goiter due to secondary changes in hemorrhage and cystic degeneration. Histological examination of the nodule reveals a follicular architecture present in the entire or nearly entire lesion.The nodule can be described as either microfollicular or macrofollicular growth pattern, and the thyrocytes have a normal cytological appearance. Follicular adenoma is usually a solitary encapsulated lesion and does not have any features suggestive of vascular or/and invasion of the adjoining capsule or to the neighboring thyroid tissue. Degenerative changes within the nodule may be seen in cases of prior FNA (fine needle aspiration performed for cytological diagnosis), which could result in degenerative changes.The only way to determine whether a follicular neoplasm is a carcinoma is after histopathological examination. The two key characteristics that make the follicular lesion malignant is the evidence of capsular invasion and angioinvasion. There is much debate about the extent to which tumor cells need to invade the capsule to determine it's malignant potential.However, in a follicular adenoma, this is not the case. If the neoplasm only enters into the capsule without passing through the entirety of the capsule, then it is still an adenoma. When tumor cells invade a large vessel having an endothelial lining and attached to the wall, that is taken as a reliable sign of malignancy.Many studies suggest a minimum of 10 tissue blocks with tumor capsule and thyroid tissue; this method has a better success rate in ruling out follicular carcinoma before definitely labeling the neoplasm as a follicular adenoma. Go to: Toxicokinetics Exposure to radiation has caused an increased incidence of follicular adenomas, as documented by studies carried out in Chernobyl radiation affected areas as well as in the areas surrounding Fukushima nuclear plant. This supports the fact that radiation is a likely etiology. Go to: History and Physical Most patients with a follicular adenoma present with solitary thyroid nodule in an otherwise normal thyroid gland, however, it may occur in association with thyroiditis or nodular hyperplasia. Although most solitary nodules are asymptomatic and euthyroid, there is a <1% chance of hyperthyroidism. So while taking history and performing a physical of a patient with neck swelling, it is imperative to keep in mind the features of hyperthyroidism or hypothyroidism that could accompany the presentation. Thyroid medication history could help determine the etiology of many of these nodules. Family history of autoimmune disease (Hashimoto disease, Grave disease) thyroid carcinoma or familial syndromes ( like Gardners) all are valid points to consider. Commonly patients present with and visually describe a slowly growing mass in the neck, pressure sensation over the neck. Pain seldom accompanies a thyroid nodule unless spontaneous hemorrhage or cystic degeneration has occurred within the nodule. Some patients may even complain of the cosmetic appearance of the mass. Pressure symptoms could cause dyspnea due to tracheal compression, increased coughing, voice hoarseness, and choking spells due to recurrent laryngeal nerve irritation and dysphagia secondary to esophagus compression. On physical examination, a diligent head to toe examination is imperative to assess if any signs are suggestive of a clinically non-euthyroid state. Patients with a follicular adenoma present with a thyroid nodule that's palpable on examination or identified on an imaging study. Nodules less than 1 cm are usually challenging to palpate unless located anteriorly on the gland. Careful consideration should be taken to assess the size, location, shape, borders, and consistency of the nodule. There may be more than one nodule that is palpated. However, even after neck palpation, nearly half of these nodules go undetected and can only be picked up on ultrasonography. Go to: Evaluation Thyroid nodules are evaluated according to the American Thyroid Association Guidelines.The four main parameters to assess in a thyroid nodule evaluation are detailed patient history, physical examination, baseline serum TSH assay followed by an ultrasound. The ultrasound examination provides important informations regarding size and feature of the nodule so it determines the following steps of evaluation.In general, follicular adenomas are benign but could have characteristics akin to follicular carcinoma; also an important differential diagnosis is with the non-invasive follicular thyroid neoplasm with papillary-like nuclear features or NIFT. 5% of microfollicular adenomas, when subjected to histopathological examination, are reported as follicular cancers. Follicular carcinoma has a worse prognosis and hence should be carefully ruled out after thorough history, examination and investigations. Nodules with a size >1 cm should be completely evaluated since they have the potential to be, or evolve to, cancer. Occasionally nodules <1 cm require evaluation is symptomatic or associated with lymphadenopathy.Preliminary laboratory assessment, including a baseline thyroid function test, helps determine the status of the gland (euthyroid, or hyper/hypothyroidism). When indicated, aspiration by a fine-needle (FNA) to obtain a cytological sample can aid in narrowing down to a diagnosis. Sonographic features of both follicular adenoma and follicular carcinoma are similar and careful examination can help distinguish the two entities. An extensive lesion size, hypoechogenicity, mixed or solid echotexture, absence of sonographic halo, micro/rim calcifications favor the malignant counterpart.The lack of internal flow, or a predominantly peripheral flow are associated with a reduced probability of thyroid follicular malignancy.A neck US should be performed during the thyroid evaluation. It is a time-efficient manner to evaluate all patients with a thyroid nodule, nodular goiter, or in cases of a thyroid nodule incidentally detected in other ways. The presence of a solid hypoechoic component in nodules with irregular margins, microcalcifications, taller than wide, rim calcifications are strongly suspicious and should be evaluated with FNA if the size is at least 1 cm. Nodules that are isoechoic or hyperechoic solid or partially cystic without the suspicious feature of irregular margins, microcalcifications, taller than wide, rim calcifications, should be evaluated with FNA only if bigger than 1,5 cm. Purely cystic nodules are considered not at risk and do not need evaluation with cytology. In thyroid nodules >1 cm (in any diameter) it is recommended to measure serum thyrotropin (TSH) during the initial evaluation.If the TSH is low a I thyroid scan should be performed, if TSH is normal or elevated the radionuclide scan is not recommended.Radionuclide scanning can be evaluated together with US findings, FNA reports and clinical information.Simple follicular adenomas show less uptake on radioiodine scintigraphy comparing with functioning follicular adenomas, which increase absorption and concentrate radioiodine within the nodule; the internal control is the neighboring tissue of the gland showing a suppressed uptake. Serum Tg levels should not be measured in the initial workup of a thyroid nodule because has a low specificity rate in the general evaluation but should be monitored in selected cases (i.e. after total thyroidectomy). Laboratory screening for calcitonin is neither recommended nor inappropriate: serum calcitonin can be evaluated for screening purposes because it may detect incidentally C-cell hyperplasia or medullary thyroid cancer at an earlier stage, improving overall survival. CT scan and MRI have a limited role in the initial evaluation f solitary thyroid nodule.Indications for these imaging techniques include suspected tracheal involvement, either by invasion or compression, extension into the mediastinum, or recurrent disease. FNA is performed to provide a cytological examination of the nodule, and it remains the mainstay for assessing these lesions, although it may not provide a diagnosis in all cases.The success rate of FNA is improved when the procedure is done under ultrasound guidance. The cytopathological findings are reported using one of the three major international nomenclatures: the British Thyroid Association and Royal College of Pathologists (Thy), the Italian Consensus (TIR) and the Bethesda System for Reporting Thyroid Cytopathology. Thyroid nodule fine-needle aspiration (FNA) washout may be examined for calcitonin in the preoperative evaluation in patients with elevated serum calcitonin (20–100 pg/mL). In patients with follicular adenoma the cytological sample has abundant follicular epithelial cells organized in sheets with crowding and overlapping of cells, microfollicle formation with scant or absent colloid. Some patients have follicular cells with abnormal architecture with atypia that is more significant than usually seen with benign lesions but not sufficient enough to call it a neoplasm. According to the Bethesda classification system, these patients are characterized as having "atypia of undetermined significance" or "follicular lesion of undetermined significance".Molecular assessment of indeterminate cytology can be helpful. A final diagnosis of follicular adenoma is made only after ruling out capsular and vessel invasion by standard nodulectomy or thyroidectomy with subsequent histological examination. Frozen section should not be requested in clinically benign nodules but is a common practice in cases of suspect papillary thyroid carcinoma hence its evaluation intraoperatively can guide in performing lymphadenectomy. Go to: Treatment / Management Usually, surgical management is preferred when it comes to follicular lesions. Medical Management If the patient is not clinically euthyroid, medical therapy done to achieve a clinically euthyroid state. Those with a low serum TSH level could signify a "toxic adenoma." In such lesions, additional, free T3, T4 assay must be done. As mentioned above, an iodine-123 thyroid scan may help determine the functionality of the nodule. These patients must be appropriately treated with medication. The patient may undergo observation or levothyroxine suppression therapy as an initial treatment modality. Levothyroxine is administered for six months to determine if the nodule decreases in size. If the nodule falls in size with levothyroxine treatment, this medication is stopped with a follow-up examination of the nodule in three to six months. The growth of nodules during levothyroxine therapy is a strong indication for surgery. Almost all cases require surgical management. Surgical Management In the case of follicular neoplasms determined by FNA, the risk of malignancy is less than 1% in a hyperfunctioning nodule, with a higher 20% risk if the nodule is hypo functioning. If the FNA result shows a follicular neoplasm, surgical management in the form of a thyroid lobectomy with isthmusectomy is the norm. If the ultrasound shows suspicious features, the surgeon should keep in mind the likelihood of malignancy. After due consideration of additional risk factors like family history, the presence of other comorbidities, previous history of neck/head radiation, a decision of performing a total thyroidectomy may be taken. Patients having a solitary toxic nodule, after determination of the functionality of the same, could undergo therapeutic iodine-131 therapy. Surgically, a unilateral thyroid lobectomy is adequate. If the histopathological examination (HPE) examination confirms the Folllciular neoplasm as an adenoma, no further intervention is required. Although beyond the purview of the topic, If the lesion fits HPE features of malignancy, then: A definitive total thyroidectomy is performed. Prophylactic neck node dissection is not recommended as <10% show nodal spread. The tumor must be staged according to the TNM staging system by the American Joint Committee on Cancer. About six weeks of postoperative, some studies recommend treatment with I-131 to ablate residual thyroid tissue and reduce the risk of recurrence. Adequate postoperative surveillance must be done to rule out recurrence/ metastasis. Serum thyroglobulin levels, routine ultrasonography is recommended modalities for monitoring the outcome of the disease. Metastatic lesions must be jointly dealt with the surgeon and the oncologist for better outcomes. Main Advantages of Surgical Therapy Relief from pressure-compressive symptoms like dyspnea, dysphagia, hoarseness, etc Removal of the lesion helps alleviate the patient's anxiety Resolves the issues of thyrotoxicosis in a toxic follicular adenoma Avoids unnecessary radiation exposure to the healthy part of the thyroid Go to: Differential Diagnosis Follicular adenomas typically present as a solitary nodule; thus, all other likely causes of a solitary thyroid nodule should be considered. Follicular carcinoma Other benign neoplasms Hurtle cell adenoma Non-invasive follicular tumor with papillary like nuclear features (NIFTP) Other carcinomas Papillary thyroid carcinoma Medullary carcinoma Thyroiditis:rarely can present as a nodule Go to: Prognosis Follicular adenomas are benign neoplasms. They are slow-growing and can progress to a size that may cause compressive symptoms such as dyspnea, dysphagia, hoarseness. After due evaluation of symptoms, the patient should ideally be reassured that these do not signify a malignant process but rather that the mass is causing compression on another structure. 20% of nonfunctioning follicular adenomas have oncogene mutations that may progress to develop into a follicular carcinoma.N-RAS and K-RAS have been implicated in this transformation.If a follicular neoplasm is proven to be a carcinoma, then prompt oncological management is necessary. Patients with biopsy-confirmed follicular adenoma ideally do not need new therapy. Thyroxine supplementation is not recommended to suppress the gland unless hypothyroidism occurs after gland lobectomy.The risk of malignancy is more in the Chernobyl post-radiation studies. Go to: Complications Pressure symptoms are not usually complications but are a known effect once the nodule reaches a size where it can compress on adjacent structures. A dull dragging sensation owing to the mass and size of the lesion is almost always the complaint of the patient Dyspnea, due to compression of the trachea Dysphagia, due to oesophageal compression Hoarseness of voice if the lesions press on the recurrent laryngeal nerve Pain over the nodule: this is more concerning to the patient. It could mean that spontaneous hemorrhage or cystic degeneration has occurred within the nodule. A toxic nodule could, in the worst-case scenario, precipitate thyroid storm requiring additional patient care. Surgical complications: Bleeding Infection Scarring: hypertrophic or keloid formation Thyroid crises Temporary hypocalcemia or permanent if the parathyroids entirely removed Recurrent laryngeal nerve transection could lead to a loss of voice and, in worse cases, even respiratory failure necessitating invasive ventilation Go to: Postoperative and Rehabilitation Care When surgical intervention is done due to compressive symptoms, the following points should be taken care of in the postoperative period: Routine wound care Watch out for complications listed above Assess for parathyroid insufficiency and treat accordingly Continuity clinic until biopsy-proven benign follicular neoplasm Go to: Consultations The management of follicular adenoma is a multidisciplinary approach, that's why liaison between the following departments is essential for a good outcome: General or endocrine surgery Endocrinology Radiology Nuclear Medicine Pathology Go to: Deterrence and Patient Education Patients who identify a growing mass over the neck should get evaluated by a licensed provider. Although many thyroid neoplasms exist, follicular neoplasms are relatively common. FNA cytology cannot rule out if the follicular neoplasm is malignant as tissue architecture for invasion can only be identified in a tissue biopsy specimen. Complications occurring in a follicular neoplasm should not be identified as a cause for malignant transformation. A proven benign follicular tumor rarely, if not ever, transforms into a malignant lesion. Go to: Pearls and Other Issues Follicular adenomas are common and usually present as solitary thyroid nodules. Fine needle aspiration cytology cannot differentiate between a follicular adenoma or carcinoma. A definitive diagnosis is by a tissue biopsy specimen. A lobectomy and isthmusectomy are adequate for such lesions. Rarely, if other comorbidities are present, a total thyroidectomy may be necessary. An interdisciplinary team approach between the surgeon, the pathologist, the radiologist must perform to optimize patient outcomes. Go to: Enhancing Healthcare Team Outcomes Thyroid nodules remain a challenge when it comes to diagnosis, evaluation, and clinical management. Usually, these neck swellings develop gradually as a solitary thyroid nodule, more often at the edge of the thyroid gland. These involve the isthmus or may be present over the lobes. Neck examination helps determine the location of these nodules that often are palpated or visualized over the neck. Follicular adenomas are usually the most common neoplasms that present as a solitary thyroid nodule. Adequate history and factors like age, gender, diet history( especially that of inadequate iodine supplementation), medical history, comorbidities, and radiation exposure must be considered. While often these cases usually present to the endocrine or general surgeon, to ensure best patient outcomes, an interdisciplinary team approach must be performed. The careful gaze of a pathologist helps determine whether the neoplasm is benign or malignant. The expertise of a radiologist helps identify sonographic features of the nodule/s or if any suspicious features coexist. This enables the primary care provider to understand the disease process better and formulate a definitive plan for management. Before surgery, routine clearances should be obtained depending on the comorbidities of the patient. Collaboration, teamwork, and careful decision making are paramount for a good outcome. Nurses assess the patient regularly and could identify surgical complications when they occur. Complications like bleeding, surgical site infection, thyroid crises, hypocalcemia may occur postoperatively. Scarring causing the patient distress, be it a hypertrophic scar or keloid formation, could be referred to as a plastic surgeon for further cosmetic management. In an ideal scenario, the patient should be followed up until the biopsy returns with a report confirming a benign follicular adenoma. If a malignancy is determined, aggressive management is required. The oncologist, the oncologic surgeon, must be involved at this stage to better the prognosis of the patient. Integrated interprofessional care provided to the patient, combined with an evidence-based approach, helps achieve the best results. Go to: Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Go to: References 1. Laios K, Lagiou E, Konofaou V, Piagkou M, Karamanou M. From thyroid cartilage to thyroid gland. Folia Morphol (Warsz). 2019;78(1):171-173. [PubMed: 30009365] 2. Hoyes AD, Kershaw DR. Anatomy and development of the thyroid gland. Ear Nose Throat J. 1985 Jul;64(7):318-33. [PubMed: 3837720] 3. Suster S. Thyroid tumors with a follicular growth pattern: problems in differential diagnosis. Arch Pathol Lab Med. 2006 Jul;130(7):984-8. [PubMed: 16831055] 4. Welker MJ, Orlov D. Thyroid nodules. 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Zablotska LB, Nadyrov EA, Polyanskaya ON, McConnell RJ, O'Kane P, Lubin J, Hatch M, Little MP, Brenner AV, Veyalkin IV, Yauseyenka VV, Bouville A, Drozdovitch VV, Minenko VF, Demidchik YE, Mabuchi K, Rozhko AV. Risk of thyroid follicular adenoma among children and adolescents in Belarus exposed to iodine-131 after the Chornobyl accident. Am J Epidemiol. 2015 Nov 01;182(9):781-90. [PMC free article: PMC4751233] [PubMed: 26443421] 10. Zablotska LB, Ron E, Rozhko AV, Hatch M, Polyanskaya ON, Brenner AV, Lubin J, Romanov GN, McConnell RJ, O'Kane P, Evseenko VV, Drozdovitch VV, Luckyanov N, Minenko VF, Bouville A, Masyakin VB. Thyroid cancer risk in Belarus among children and adolescents exposed to radioiodine after the Chornobyl accident. Br J Cancer. 2011 Jan 04;104(1):181-7. [PMC free article: PMC3039791] [PubMed: 21102590] 11. Zablotska LB, Bogdanova TI, Ron E, Epstein OV, Robbins J, Likhtarev IA, Hatch M, Markov VV, Bouville AC, Olijnyk VA, McConnell RJ, Shpak VM, Brenner A, Terekhova GN, Greenebaum E, Tereshchenko VP, Fink DJ, Brill AB, Zamotayeva GA, Masnyk IJ, Howe GR, Tronko MD. A cohort study of thyroid cancer and other thyroid diseases after the Chornobyl accident: dose-response analysis of thyroid follicular adenomas detected during first screening in Ukraine (1998-2000). Am J Epidemiol. 2008 Feb 01;167(3):305-12. [PubMed: 17989057] 12. Cameselle-Teijeiro J, Fachal C, Cabezas-Agrícola JM, Alfonsín-Barreiro N, Abdulkader I, Vega-Gliemmo A, Hermo JA. Thyroid Pathology Findings in Cowden Syndrome: A Clue for the Diagnosis of the PTEN Hamartoma Tumor Syndrome. Am J Clin Pathol. 2015 Aug;144(2):322-8. [PubMed: 26185318] 13. Carney JA, Lyssikatos C, Seethala RR, Lakatos P, Perez-Atayde A, Lahner H, Stratakis CA. The Spectrum of Thyroid Gland Pathology in Carney Complex: The Importance of Follicular Carcinoma. Am J Surg Pathol. 2018 May;42(5):587-594. [PMC free article: PMC5925749] [PubMed: 29635258] 14. Bisi H, Fernandes VS, de Camargo RY, Koch L, Abdo AH, de Brito T. The prevalence of unsuspected thyroid pathology in 300 sequential autopsies, with special reference to the incidental carcinoma. Cancer. 1989 Nov 01;64(9):1888-93. [PubMed: 2676140] 15. Silverberg SG, Vidone RA. Adenoma and carcinoma of the thyroid. Cancer. 1966 Aug;19(8):1053-62. [PubMed: 5912322] 16. Arem R, Padayatty SJ, Saliby AH, Sherman SI. Thyroid microcarcinoma: prevalence, prognosis, and management. Endocr Pract. 1999 May-Jun;5(3):148-56. [PubMed: 15251688] 17. Harach HR, Franssila KO, Wasenius VM. Occult papillary carcinoma of the thyroid. A "normal" finding in Finland. A systematic autopsy study. Cancer. 1985 Aug 01;56(3):531-8. [PubMed: 2408737] 18. Mazzaferri EL. Management of a solitary thyroid nodule. N Engl J Med. 1993 Feb 25;328(8):553-9. [PubMed: 8426623] 19. Li N, Du XL, Reitzel LR, Xu L, Sturgis EM. Impact of enhanced detection on the increase in thyroid cancer incidence in the United States: review of incidence trends by socioeconomic status within the surveillance, epidemiology, and end results registry, 1980-2008. Thyroid. 2013 Jan;23(1):103-10. [PMC free article: PMC3539256] [PubMed: 23043274] 20. Morris LG, Myssiorek D. Improved detection does not fully explain the rising incidence of well-differentiated thyroid cancer: a population-based analysis. Am J Surg. 2010 Oct;200(4):454-61. [PMC free article: PMC2943969] [PubMed: 20561605] 21. Jacob P, Kaiser JC, Ulanovsky A. Ultrasonography survey and thyroid cancer in the Fukushima Prefecture. Radiat Environ Biophys. 2014 May;53(2):391-401. [PMC free article: PMC3996282] [PubMed: 24398917] 22. Tsuda T, Tokinobu A, Yamamoto E, Suzuki E. Thyroid Cancer Detection by Ultrasound Among Residents Ages 18 Years and Younger in Fukushima, Japan: 2011 to 2014. Epidemiology. 2016 May;27(3):316-22. [PMC free article: PMC4820668] [PubMed: 26441345] 23. Boice JD, Blettner M, Auvinen A. Epidemiologic studies of pilots and aircrew. Health Phys. 2000 Nov;79(5):576-84. [PubMed: 11045533] 24. Ron E, Lubin JH, Shore RE, Mabuchi K, Modan B, Pottern LM, Schneider AB, Tucker MA, Boice JD. Thyroid cancer after exposure to external radiation: a pooled analysis of seven studies. Radiat Res. 1995 Mar;141(3):259-77. [PubMed: 7871153] 25. Zhu C, Zheng T, Kilfoy BA, Han X, Ma S, Ba Y, Bai Y, Wang R, Zhu Y, Zhang Y. A birth cohort analysis of the incidence of papillary thyroid cancer in the United States, 1973-2004. Thyroid. 2009 Oct;19(10):1061-6. [PMC free article: PMC2833179] [PubMed: 19732011] 26. Marques AR, Espadinha C, Catarino AL, Moniz S, Pereira T, Sobrinho LG, Leite V. Expression of PAX8-PPAR gamma 1 rearrangements in both follicular thyroid carcinomas and adenomas. J Clin Endocrinol Metab. 2002 Aug;87(8):3947-52. [PubMed: 12161538] 27. Baloch ZW, Livolsi VA. Follicular-patterned lesions of the thyroid: the bane of the pathologist. Am J Clin Pathol. 2002 Jan;117(1):143-50. [PubMed: 11789719] 28. Wong R, Farrell SG, Grossmann M. Thyroid nodules: diagnosis and management. Med J Aust. 2018 Jul 16;209(2):92-98. [PubMed: 29996756] 29. D'Avanzo A, Treseler P, Ituarte PH, Wong M, Streja L, Greenspan FS, Siperstein AE, Duh QY, Clark OH. Follicular thyroid carcinoma: histology and prognosis. Cancer. 2004 Mar 15;100(6):1123-9. [PubMed: 15022277] 30. Yamashina M. Follicular neoplasms of the thyroid. Total circumferential evaluation of the fibrous capsule. Am J Surg Pathol. 1992 Apr;16(4):392-400. [PubMed: 1373582] 31. Haugen BR, Sawka AM, Alexander EK, Bible KC, Caturegli P, Doherty GM, Mandel SJ, Morris JC, Nassar A, Pacini F, Schlumberger M, Schuff K, Sherman SI, Somerset H, Sosa JA, Steward DL, Wartofsky L, Williams MD. American Thyroid Association Guidelines on the Management of Thyroid Nodules and Differentiated Thyroid Cancer Task Force Review and Recommendation on the Proposed Renaming of Encapsulated Follicular Variant Papillary Thyroid Carcinoma Without Invasion to Noninvasive Follicular Thyroid Neoplasm with Papillary-Like Nuclear Features. Thyroid. 2017 Apr;27(4):481-483. [PubMed: 28114862] 32. Haugen BR, Alexander EK, Bible KC, Doherty GM, Mandel SJ, Nikiforov YE, Pacini F, Randolph GW, Sawka AM, Schlumberger M, Schuff KG, Sherman SI, Sosa JA, Steward DL, Tuttle RM, Wartofsky L. 2015 American Thyroid Association Management Guidelines for Adult Patients with Thyroid Nodules and Differentiated Thyroid Cancer: The American Thyroid Association Guidelines Task Force on Thyroid Nodules and Differentiated Thyroid Cancer. Thyroid. 2016 Jan;26(1):1-133. [PMC free article: PMC4739132] [PubMed: 26462967] 33. Nikiforov YE, Seethala RR, Tallini G, Baloch ZW, Basolo F, Thompson LD, Barletta JA, Wenig BM, Al Ghuzlan A, Kakudo K, Giordano TJ, Alves VA, Khanafshar E, Asa SL, El-Naggar AK, Gooding WE, Hodak SP, Lloyd RV, Maytal G, Mete O, Nikiforova MN, Nosé V, Papotti M, Poller DN, Sadow PM, Tischler AS, Tuttle RM, Wall KB, LiVolsi VA, Randolph GW, Ghossein RA. Nomenclature Revision for Encapsulated Follicular Variant of Papillary Thyroid Carcinoma: A Paradigm Shift to Reduce Overtreatment of Indolent Tumors. JAMA Oncol. 2016 Aug 01;2(8):1023-9. [PMC free article: PMC5539411] [PubMed: 27078145] 34. Katsakhyan L, Song S, Lepe M, Shojaei H, Montone KT, LiVolsi VA, Baloch ZW. Practice Paradigms Before and After Introduction of the Diagnosis-Noninvasive Follicular Thyroid Neoplasm with Papillary-Like Nuclear Features (NIFTP): an Institutional Experience. Endocr Pathol. 2020 Jun;31(2):174-181. [PubMed: 32146581] 35. Mazzaferri EL. Thyroid cancer in thyroid nodules: finding a needle in the haystack. Am J Med. 1992 Oct;93(4):359-62. [PubMed: 1415298] 36. Seo HS, Lee DH, Park SH, Min HS, Na DG. Thyroid follicular neoplasms: can sonography distinguish between adenomas and carcinomas? J Clin Ultrasound. 2009 Nov-Dec;37(9):493-500. [PubMed: 19746451] 37. Zhang JZ, Hu B. Sonographic features of thyroid follicular carcinoma in comparison with thyroid follicular adenoma. J Ultrasound Med. 2014 Feb;33(2):221-7. [PubMed: 24449724] 38. Iared W, Shigueoka DC, Cristófoli JC, Andriolo R, Atallah AN, Ajzen SA, Valente O. Use of color Doppler ultrasonography for the prediction of malignancy in follicular thyroid neoplasms: systematic review and meta-analysis. J Ultrasound Med. 2010 Mar;29(3):419-25. [PubMed: 20194937] 39. Wu CW, Dionigi G, Lee KW, Hsiao PJ, Paul Shin MC, Tsai KB, Chiang FY. Calcifications in thyroid nodules identified on preoperative computed tomography: patterns and clinical significance. Surgery. 2012 Mar;151(3):464-70. [PubMed: 21911238] 40. Collins J, Rossi ED, Chandra A, Ali SZ. Terminology and nomenclature schemes for reporting thyroid cytopathology: An overview. Semin Diagn Pathol. 2015 Jul;32(4):258-63. [PubMed: 25680862] 41. Challeton C, Bounacer A, Du Villard JA, Caillou B, De Vathaire F, Monier R, Schlumberger M, Suárez HG. Pattern of ras and gsp oncogene mutations in radiation-associated human thyroid tumors. Oncogene. 1995 Aug 03;11(3):601-3. [PubMed: 7630645] 42. McHenry CR, Slusarczyk SJ. Hypothyroidisim following hemithyroidectomy: incidence, risk factors, and management. Surgery. 2000 Dec;128(6):994-8. [PubMed: 11114634] 43. Gopalakrishna Iyer N, Shaha AR. Complications of thyroid surgery: prevention and management. Minerva Chir. 2010 Feb;65(1):71-82. [PubMed: 20212419] Disclosure:Joshua Norris declares no relevant financial relationships with ineligible companies. Disclosure:Fabiola Farci declares no relevant financial relationships with ineligible companies. 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Thyroid Nodule Sphericity Metrics Discriminate Benign and Malignant Follicular and Oncocytic Neoplasms.[Thyroid. 2025]Thyroid Nodule Sphericity Metrics Discriminate Benign and Malignant Follicular and Oncocytic Neoplasms.Bell C, White SL, Tylee T, Dighe M, Greca A, Goldner W, Mayson S, Haugen BR, Pozdeyev N. Thyroid. 2025 Mar; 35(3):291-297. Epub 2025 Feb 25. Andreas Vesalius and his De humani corporis Fabrica libri septem.[Vesalius. 2014]Andreas Vesalius and his De humani corporis Fabrica libri septem.Steele L. Vesalius. 2014 Summer; 20(1):5-10. Thyroid Adenoma.[StatPearls. 2025]Thyroid Adenoma.Mulita F, Anjum F. StatPearls. 2025 Jan Review Overview of the 2022 WHO Classification of Thyroid Neoplasms.[Endocr Pathol. 2022]Review Overview of the 2022 WHO Classification of Thyroid Neoplasms.Baloch ZW, Asa SL, Barletta JA, Ghossein RA, Juhlin CC, Jung CK, LiVolsi VA, Papotti MG, Sobrinho-Simões M, Tallini G, et al. Endocr Pathol. 2022 Mar; 33(1):27-63. Epub 2022 Mar 14. 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17150	https://www.khanacademy.org/test-prep/mcat/physical-processes/stoichiometry/v/empirical-formula-from-mass-composition-edited	Empirical formula from mass composition edited (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Florida B.E.S.T. Math: Pre-K - 8th grade Math: Get ready courses Math: High school & college Math: Multiple grades Science Test prep Computing Reading & language arts Economics Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Skip to lesson content MCAT Course: MCAT>Unit 9 Lesson 25: Stoichiometry Stoichiometry questions Stoichiometry article Stoichiometry and empirical formulae Empirical formula from mass composition edited Molecular and empirical formulas The mole and Avogadro's number Stoichiometry example problem 1 Stoichiometry Limiting reactant example problem 1 edited Specific gravity Test prep> MCAT> Foundation 4: Physical processes> Stoichiometry © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Empirical formula from mass composition edited Google Classroom Microsoft Teams About About this video Transcript Visit us ( for health and medicine content or ( for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video.Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Majid Khorchid 6 years ago Posted 6 years ago. Direct link to Majid Khorchid's post “For anyone who was confus...” more For anyone who was confused when Sal got HgCl2, this is how he did it. When Sal got HgCl2, I did dimension analysis to keep things simple and to the point. For example, I did the following down below. 73.0g Hg / 1 1 mol Hg / 200.59g Hg = 0.364 moles of Hg 27.0g Cl / 1 1 mol Cl / 35.453g Cl = 0.761 moles of Cl (See how efficiently grams are crossed out and left with moles?) The next step is to DIVIDE by the LOWEST of the moles to get an approximate close whole number value. Which in this case, this is what I did below. Hg: 0.364 / 0.364 = 1 Cl: 0.761 / 0.364 = 2 Therefore working from left to right, you have HgCl2. Answer Button navigates to signup page •Comment Button navigates to signup page (16 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript [Voiceover] What I want to do in this video is start with mass composition and then see if we can figure out the empirical formula of the molecule that we're dealing with based on the mass composition. So let's say that we have a bag, and we're able to measure that this bag is 73%, it's 73% mercury, and it is the remainder of the bag, 27% chlorine. Now based just on this, can we figure out the likely empirical formula for the molecule that we have in that bag? And I encourage you to pause the video and try to see if you can figure it out on your own. Well, one way to think about it, well let's just assume a number. This is just all the information we have, but let's just assume we have 100 grams of it. We could assume 1,000 grams or 10,000 grams or 57 grams, but I'll pick 100 grams because it will make the numbers, it will make the numbers easy to work with in our head. So let's just assume, let me make it clear that I'm assuming this, I'm going to assume that I have 100 grams of this molecule that is 73% mercury and 27% chlorine. And if I assume that, that means that the 73% that is mercury is going to be 73 grams, and the 27% that is chlorine is going to be 27 grams of chlorine. I'm going to make it clear this is mercury, and this is chlorine. Now I just need to think about, well how many moles of mercury is 73 grams? And how many moles of chlorine is 27 grams? And to do that, I'll just look up this periodic table right here. I have the atomic weight, which is of course the weighted average of the atomic masses as kind of found in nature. And the atomic weight here for mercury is 200.59. So that means, so let me write this right over here, so one mole of mercury is, we could say, is 200.59 grams. And similarly, we could look up the atomic weight for chlorine. Chlorine, right over here, 35.453, and so we could say, one mole of chlorine, and once again this is a weighted average of all of the isotopes of chlorine as found in nature, and I guess we'll just go with that number. So one mole of chlorine is going to be 35.453, 35.453 grams. So given this information right over here, how many moles of mercury is this, roughly, and how many moles of chlorine is this, roughly? And I say roughly because getting an empirical formula from measurements of mass composition, it's going to be necessarily a messy affair. It's not going to come out completely, the numbers aren't going to work out completely exact, so that's why I said roughly. So how many moles is this? Well this is going to be, this is going to be, 73 over 200.59 of a mole. If a mole is 200.59 and we have 73, this is the fraction of a mole that we have. Moles of mercury. And remember, moles are just a number, Avogadro's number of something. But let's just figure out what it is. So, so if we take 73 divided by 200.59 we get .36, I'll just say 0.364, and once again, this is, so approximately 0.364. That's how many moles of mercury that we have, and we could do the same thing for chlorine. This is going to be 27 over 35.453 moles of chlorine, which is approximately equal to 0.762. Now, moles of chlorine. So, what's going to be the ratio of mercury to chlorine? Or I guess I could say, since chlorine, there's more of it, chlorine to mercury, remember, this is just a number. When I say 0.762 moles, this is just 0.762 times Avogadro's number of chlorine atoms. This is 0.364 times Avogadro's number of mercury atoms. And so this, we can literally think of this as the ratio. This is a certain number of moles, this is another number of moles. Well, what's the ratio? Let's say, what's the ratio of chlorine to mercury? Well, you can eyeball it. It looks like it's roughly 2:1, and because of that, you could say, well this is likely to be, so likely for every mercury, you have two chlorines. So, based on these measurements right over here, it's very likely that you have mercury two chloride. And the reason why it's called mercury two chloride is because, well, I won't go into too much detail right over here, but chlorine is highly electronegative. It's an oxidizing agent. It likes to take other people's electrons, or hog other people's electrons. In this case it's hogging, since there's, each of the chlorine likes to hog at least, likes to hog one electron, so in this case, two chlorines are going to hog two electrons, so it's hogging two electrons from the mercury, when you lose electrons, or when your electrons are being hogged, you're being oxidized. So the oxidation state on mercury right over here is two. Two of its electrons are being hogged, one by each of the two chlorines. So this is mercury two chloride, where the two is the oxidation state of the mercury. But this is what we likely have, the ratio, we have two chlorines for every one mercury, roughly. 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17151	https://www.ipcms.fr/uploads/2021/06/chap6.pdf	Chapter 6 The Lagrangian Method Copyright 2007 by David Morin, morin@physics.harvard.edu (draft version) In this chapter, we’re going to learn about a whole new way of looking at things. Consider the system of a mass on the end of a spring. We can analyze this, of course, by using F = ma to write down m¨ x = −kx. The solutions to this equation are sinusoidal functions, as we well know. We can, however, ﬁgure things out by using another method which doesn’t explicitly use F = ma. In many (in fact, probably most) physical situations, this new method is far superior to using F = ma. You will soon discover this for yourself when you tackle the problems and exercises for this chapter. We will present our new method by ﬁrst stating its rules (without any justiﬁcation) and showing that they somehow end up magically giving the correct answer. We will then give the method proper justiﬁcation. 6.1 The Euler-Lagrange equations Here is the procedure. Consider the following seemingly silly combination of the kinetic and potential energies (T and V , respectively), L ≡T −V. (6.1) This is called the Lagrangian. Yes, there is a minus sign in the deﬁnition (a plus sign would simply give the total energy). In the problem of a mass on the end of a spring, T = m ˙ x2/2 and V = kx2/2, so we have L = 1 2m ˙ x2 −1 2kx2. (6.2) Now write d dt µ∂L ∂˙ x ¶ = ∂L ∂x . (6.3) Don’t worry, we’ll show you in Section 6.2 where this comes from. This equation is called the Euler-Lagrange (E-L) equation. For the problem at hand, we have ∂L/∂˙ x = m ˙ x and ∂L/∂x = −kx (see Appendix B for the deﬁnition of a partial derivative), so eq. (6.3) gives m¨ x = −kx, (6.4) which is exactly the result obtained by using F = ma. An equation such as eq. (6.4), which is derived from the Euler-Lagrange equation, is called an equation of motion.1 If the 1The term “equation of motion” is a little ambiguous. It is understood to refer to the second-order diﬀerential equation satisﬁed by x, and not the actual equation for x as a function of t, namely x(t) = A cos(ωt + φ) in this problem, which is obtained by integrating the equation of motion twice. VI-1 VI-2 CHAPTER 6. THE LAGRANGIAN METHOD problem involves more than one coordinate, as most problems do, we just have to apply eq. (6.3) to each coordinate. We will obtain as many equations as there are coordinates. Each equation may very well involve many of the coordinates (see the example below, where both equations involve both x and θ). At this point, you may be thinking, “That was a nice little trick, but we just got lucky in the spring problem. The procedure won’t work in a more general situation.” Well, let’s see. How about if we consider the more general problem of a particle moving in an arbitrary potential V (x) (we’ll stick to one dimension for now). The Lagrangian is then L = 1 2m ˙ x2 −V (x), (6.5) and the Euler-Lagrange equation, eq. (6.3), gives m¨ x = −dV dx . (6.6) But −dV/dx is the force on the particle. So we see that eqs. (6.1) and (6.3) together say exactly the same thing that F = ma says, when using a Cartesian coordinate in one dimension (but this result is in fact quite general, as we’ll see in Section 6.4). Note that shifting the potential by a given constant has no eﬀect on the equation of motion, because eq. (6.3) involves only the derivative of V . This is equivalent to saying that only diﬀerences in energy are relevant, and not the actual values, as we well know. In a three-dimensional setup written in terms of Cartesian coordinates, the potential takes the form V (x, y, z), so the Lagrangian is L = 1 2m( ˙ x2 + ˙ y2 + ˙ z2) −V (x, y, z). (6.7) It then immediately follows that the three Euler-Lagrange equations (obtained by applying eq. (6.3) to x, y, and z) may be combined into the vector statement, m¨ x = −∇V. (6.8) But −∇V = F, so we again arrive at Newton’s second law, F = ma, now in three dimensions. Let’s now do one more example to convince you that there’s really something nontrivial going on here. Example (Spring pendulum): Consider a pendulum made of a spring with a mass m on the end (see Fig. 6.1). The spring is arranged to lie in a straight line (which we can arrange θ l+x m Figure 6.1 by, say, wrapping the spring around a rigid massless rod). The equilibrium length of the spring is ℓ. Let the spring have length ℓ+ x(t), and let its angle with the vertical be θ(t). Assuming that the motion takes place in a vertical plane, ﬁnd the equations of motion for x and θ. Solution: The kinetic energy may be broken up into the radial and tangential parts, so we have T = 1 2m ³ ˙ x2 + (ℓ+ x)2 ˙ θ2´ . (6.9) The potential energy comes from both gravity and the spring, so we have V (x, θ) = −mg(ℓ+ x) cos θ + 1 2kx2. (6.10) The Lagrangian is therefore L ≡T −V = 1 2m ³ ˙ x2 + (ℓ+ x)2 ˙ θ2´ + mg(ℓ+ x) cos θ −1 2kx2. (6.11) 6.1. THE EULER-LAGRANGE EQUATIONS VI-3 There are two variables here, x and θ. As mentioned above, the nice thing about the La-grangian method is that we can just use eq. (6.3) twice, once with x and once with θ. So the two Euler-Lagrange equations are d dt ³∂L ∂˙ x ´ = ∂L ∂x = ⇒ m¨ x = m(ℓ+ x) ˙ θ2 + mg cos θ −kx, (6.12) and d dt ³∂L ∂˙ θ ´ = ∂L ∂θ = ⇒ d dt ¡ m(ℓ+ x)2 ˙ θ¢ = −mg(ℓ+ x) sin θ = ⇒ m(ℓ+ x)2¨ θ + 2m(ℓ+ x) ˙ x ˙ θ = −mg(ℓ+ x) sin θ. = ⇒ m(ℓ+ x)¨ θ + 2m ˙ x ˙ θ = −mg sin θ. (6.13) Eq. (6.12) is simply the radial F = ma equation, complete with the centripetal acceleration, −(ℓ+ x) ˙ θ2. And the ﬁrst line of eq. (6.13) is the statement that the torque equals the rate of change of the angular momentum (this is one of the subjects of Chapter 8). Alternatively, if you want to work in a rotating reference frame, then eq. (6.12) is the radial F = ma equation, complete with the centrifugal force, m(ℓ+ x) ˙ θ2. And the third line of eq. (6.13) is the tangential F = ma equation, complete with the Coriolis force, −2m ˙ x ˙ θ. But never mind about this now. We’ll deal with rotating frames in Chapter 10.2 Remark: After writing down the E-L equations, it is always best to double-check them by trying to identify them as F = ma and/or τ = dL/dt equations (once we learn about that). Sometimes, however, this identiﬁcation isn’t obvious. And for the times when everything is clear (that is, when you look at the E-L equations and say, “Oh, of course!”), it is usually clear only after you’ve derived the equations. In general, the safest method for solving a problem is to use the Lagrangian method and then double-check things with F = ma and/or τ = dL/dt if you can. ♣ At this point it seems to be personal preference, and all academic, whether you use the Lagrangian method or the F = ma method. The two methods produce the same equations. However, in problems involving more than one variable, it usually turns out to be much easier to write down T and V , as opposed to writing down all the forces. This is because T and V are nice and simple scalars. The forces, on the other hand, are vectors, and it is easy to get confused if they point in various directions. The Lagrangian method has the advantage that once you’ve written down L ≡T −V , you don’t have to think anymore. All you have to do is blindly take some derivatives.3 When jumping from high in a tree, Just write down del L by del z. Take del L by z dot, Then t-dot what you’ve got, And equate the results (but quickly!) But ease of computation aside, is there any fundamental diﬀerence between the two meth-ods? Is there any deep reasoning behind eq. (6.3)? Indeed, there is. . . 2Throughout this chapter, I’ll occasionally point out torques, angular momenta, centrifugal forces, and other such things when they pop up in equations of motion, even though we haven’t covered them yet. I ﬁgure it can’t hurt to bring your attention to them. But rest assured, a familiarity with these topics is by no means necessary for an understanding of what we’ll be doing in this chapter, so just ignore the references if you want. One of the great things about the Lagrangian method is that even if you’ve never heard of the terms “torque,” “centrifugal,” “Coriolis,” or even “F = ma” itself, you can still get the correct equations by simply writing down the kinetic and potential energies, and then taking a few derivatives. 3Well, you eventually have to solve the resulting equations of motion, but you have to do that with the F = ma method, too. VI-4 CHAPTER 6. THE LAGRANGIAN METHOD 6.2 The principle of stationary action Consider the quantity, S ≡ Z t2 t1 L(x, ˙ x, t) dt. (6.14) S is called the action. It is a quantity with the dimensions of (Energy)×(Time). S depends on L, and L in turn depends on the function x(t) via eq. (6.1).4 Given any function x(t), we can produce the quantity S. We’ll just deal with one coordinate, x, for now. Integrals like the one in eq. (6.14) are called functionals, and S is sometimes denoted by S[x(t)]. It depends on the entire function x(t), and not on just one input number, as a regular function f(t) does. S can be thought of as a function of an inﬁnite number of values, namely all the x(t) for t ranging from t1 to t2. If you don’t like inﬁnities, you can imagine breaking up the time interval into, say, a million pieces, and then replacing the integral by a discrete sum. Let’s now pose the following question: Consider a function x(t), for t1 ≤t ≤t2, which has its endpoints ﬁxed (that is, x(t1) = x1 and x(t2) = x2, where x1 and x2 are given), but is otherwise arbitrary. What function x(t) yields a stationary value of S? A stationary value is a local minimum, maximum, or saddle point.5 For example, consider a ball dropped from rest, and consider the function y(t) for 0 ≤ t ≤1. Assume that we somehow know that y(0) = 0 and y(1) = −g/2. 6 A number of possibilities for y(t) are shown in Fig. 6.2, and each of these can (in theory) be plugged into y -g/2 t 1 Figure 6.2 eqs. (6.1) and (6.14) to generate S. Which one yields a stationary value of S? The following theorem gives us the answer. Theorem 6.1 If the function x0(t) yields a stationary value (that is, a local minimum, maximum, or saddle point) of S, then d dt µ ∂L ∂˙ x0 ¶ = ∂L ∂x0 . (6.15) It is understood that we are considering the class of functions whose endpoints are ﬁxed. That is, x(t1) = x1 and x(t2) = x2. Proof: We will use the fact that if a certain function x0(t) yields a stationary value of S, then any other function very close to x0(t) (with the same endpoint values) yields essentially the same S, up to ﬁrst order in any deviations. This is actually the deﬁnition of a stationary value. The analogy with regular functions is that if f(b) is a stationary value of f, then f(b + ϵ) diﬀers from f(b) only at second order in the small quantity ϵ. This is true because f ′(b) = 0, so there is no ﬁrst-order term in the Taylor series expansion around b. Assume that the function x0(t) yields a stationary value of S, and consider the function xa(t) ≡x0(t) + aβ(t), (6.16) where a is a number, and where β(t) satisﬁes β(t1) = β(t2) = 0 (to keep the endpoints of the function ﬁxed), but is otherwise arbitrary. When producing the action S[xa(t)] in (6.14), the t is integrated out, so S is just a number. It depends on a, in addition to t1 and 4In some situations, the kinetic and potential energies in L ≡T −V may explicitly depend on time, so we have included the “t” in eq. (6.14). 5A saddle point is a point where there are no ﬁrst-order changes in S, and where some of the second-order changes are positive and some are negative (like the middle of a saddle, of course). 6This follows from y = −gt2/2, but pretend that we don’t know this formula. 6.2. THE PRINCIPLE OF STATIONARY ACTION VI-5 t2. Our requirement is that there be no change in S at ﬁrst order in a. How does S depend on a? Using the chain rule, we have ∂ ∂aS[xa(t)] = ∂ ∂a Z t2 t1 L dt = Z t2 t1 ∂L ∂a dt = Z t2 t1 µ ∂L ∂xa ∂xa ∂a + ∂L ∂˙ xa ∂˙ xa ∂a ¶ dt. (6.17) In other words, a inﬂuences S through its eﬀect on x, and also through its eﬀect on ˙ x. From eq. (6.16), we have ∂xa ∂a = β, and ∂˙ xa ∂a = ˙ β, (6.18) so eq. (6.17) becomes7 ∂ ∂aS[xa(t)] = Z t2 t1 µ ∂L ∂xa β + ∂L ∂˙ xa ˙ β ¶ dt. (6.19) Now comes the one sneaky part of the proof. We will integrate the second term by parts (you will see this trick many times in your physics career). Using Z ∂L ∂˙ xa ˙ β dt = ∂L ∂˙ xa β − Z µ d dt ∂L ∂˙ xa ¶ β dt, (6.20) eq. (6.19) becomes ∂ ∂aS[xa(t)] = Z t2 t1 µ ∂L ∂xa −d dt ∂L ∂˙ xa ¶ β dt + ∂L ∂˙ xa β ¯ ¯ ¯ ¯ t2 t1 . (6.21) But β(t1) = β(t2) = 0, so the last term (the “boundary term”) vanishes. We now use the fact that (∂/∂a)S[xa(t)] must be zero for any function β(t), because we are assuming that x0(t) yields a stationary value. The only way this can be true is if the quantity in parentheses above (evaluated at a = 0) is identically equal to zero, that is, d dt µ ∂L ∂˙ x0 ¶ = ∂L ∂x0 . (6.22) The E-L equation, eq. (6.3), therefore doesn’t just come out of the blue. It is a con-sequence of requiring that the action be at a stationary value. We may therefore replace F = ma by the following principle. • The Principle of Stationary Action: The path of a particle is the one that yields a stationary value of the action. This principle (also known as Hamilton’s principle) is equivalent to F = ma because the above theorem shows that if (and only if, as you can show by working backwards) we have a stationary value of S, then the E-L equations hold. And the E-L equations are equivalent to F = ma (as we showed for Cartesian coordinates in Section 6.1, and as we’ll prove for any coordinate system in Section 6.4). Therefore, “stationary action” is equivalent to F = ma. 7Note that nowhere do we assume that xa and ˙ xa are independent variables. The partial derivatives in eq. (6.18) are very much related, in that one is the derivative of the other. The use of the chain rule in eq. (6.17) is still perfectly valid. VI-6 CHAPTER 6. THE LAGRANGIAN METHOD If we have a multidimensional setup where the Lagrangian is a function of the variables x1(t), x2(t), . . ., then the above principle of stationary action is still all we need. With more than one variable, we can now vary the path by varying each coordinate (or combinations thereof). The variation of each coordinate produces an E-L equation which, as we saw in the Cartesian case, is equivalent to an F = ma equation. Given a classical mechanics problem, we can solve it with F = ma, or we can solve it with the E-L equations, which are a consequence of the principle of stationary action (often called the principle of “least action” or “minimal action,” but see the fourth remark below). Either method will get the job done. But as mentioned at the end of Section 6.1, it is often easier to use the latter, because it avoids the use of force which can get confusing if you have forces pointing in all sorts of complicated directions. It just stood there and did nothing, of course, A harmless and still wooden horse. But the minimal action Was just a distraction; The plan involved no use of force. Let’s now return to the example of a ball dropped from rest, mentioned above. The Lagrangian is L = T −V = m ˙ y2/2−mgy, so eq. (6.22) gives ¨ y = −g, which is simply the F = ma equation (divided through by m), as expected. The solution is y(t) = −gt2/2+v0t+y0, as we well know. But the initial conditions tell us that v0 = y0 = 0, so our solution is y(t) = −gt2/2. You are encouraged to verify explicitly that this y(t) yields an action that is stationary with respect to variations of the form, say, y(t) = −gt2/2 + ϵt(t −1), which also satisﬁes the endpoint conditions (this is the task of Exercise 6.30). There is, of course, an inﬁnite number of other ways to vary y(t), but this speciﬁc result should help convince you of the general result of Theorem 6.1. Note that the stationarity implied by the Euler-Lagrange equation, eq. (6.22), is a local statement. It gives information only about nearby paths. It says nothing about the global nature of how the action depends on all possible paths. If we ﬁnd that a solution to eq. (6.22) happens to produce a local minimum (as opposed to a maximum or a saddle), there is no reason to conclude that it is a global minimum, although in many cases it turns out to be (see Exercise 6.32, for the case of a thrown ball). Remarks: 1. Theorem 6.1 is based on the assumption that the ending time, t2, of the motion is given. But how do we know this ﬁnal time? Well, we don’t. In the example of a ball thrown upward, the total time to rise and fall back to your hand can be anything, depending on the ball’s initial speed. This initial speed will show up as an integration constant when solving the E-L equations. The motion must end sometime, and the principle of stationary action says that for whatever time this happens to be, the physical path has a stationary action. 2. Theorem 6.1 shows that we can explain the E-L equations by the principle of stationary action. This, however, simply shifts the burden of proof. We are now left with the task of justifying why we should want the action to have a stationary value. The good news is that there is a very solid reason for this. The bad news is that the reason involves quantum mechanics, so we won’t be able to discuss it properly here. Suﬃce it to say that a particle actually takes all possible paths in going from one place to another, and each path is associated with the complex number eiS/¯ h (where ¯ h = 1.05 · 10−34 Js is Planck’s constant). These complex numbers have absolute value 1 and are called “phases.” It turns out that the phases from all possible paths must be added up to give the “amplitude” of going from one point to another. The absolute value of the amplitude must then be squared to obtain the probability.8 8This is one of those remarks that is completely useless, because it is incomprehensible to those who 6.2. THE PRINCIPLE OF STATIONARY ACTION VI-7 The basic point, then, is that at a non-stationary value of S, the phases from diﬀerent paths diﬀer (greatly, because ¯ h is so small compared with the typical size of the action for a macroscopic particle) from one another, which eﬀectively leads to the addition of many random vectors in the complex plane. These end up canceling each other, yielding a sum of essentially zero. There is therefore no contribution to the overall amplitude from non-stationary values of S. Hence, we do not observe the paths associated with these S’s. At a stationary value of S, however, all the phases take on essentially the same value, thereby adding constructively instead of destructively. There is therefore a nonzero probability for the particle to take a path that yields a stationary value of S. So this is the path we observe. 3. But again, the preceding remark simply shifts the burden of proof one step further. We must now justify why these phases eiS/¯ h should exist, and why the Lagrangian that appears in S should equal T −V . But here’s where we’re going to stop. 4. The principle of stationary action is sometimes referred to as the principle of “least” action, but this is misleading. True, it is often the case that the stationary value turns out to be a minimum value, but it need not be, as we can see in the following example. Consider a harmonic oscillator which has a Lagrangian equal to L = 1 2m ˙ x2 −1 2kx2. (6.23) Let x0(t) be a function that yields a stationary value of the action. Then we know that x0(t) satisﬁes the E-L equation, m¨ x0 = −kx0. Consider a slight variation on this path, x0(t)+ξ(t), where ξ(t) satisﬁes ξ(t1) = ξ(t2) = 0. With this new function, the action becomes Sξ = Z t2 t1 ³m 2 ³ ˙ x2 0 + 2 ˙ x0 ˙ ξ + ˙ ξ2´ −k 2 ³ x2 0 + 2x0ξ + ξ2´´ dt. (6.24) The two cross-terms add up to zero, because after integrating the ˙ x0 ˙ ξ term by parts, their sum is m ˙ x0ξ ¯ ¯ ¯ t2 t1 − Z t2 t1 (m¨ x0 + kx0)ξ dt. (6.25) The ﬁrst term is zero, due to the boundary conditions on ξ(t). The second term is zero, due to the E-L equation. We’ve basically just reproduced the proof of Theorem 6.1 for the special case of the harmonic oscillator here. The terms in eq. (6.24) involving only x0 give the stationary value of the action (call it S0). To determine whether S0 is a minimum, maximum, or saddle point, we must look at the diﬀerence, ∆S ≡Sξ −S0 = 1 2 Z t2 t1 (m ˙ ξ2 −kξ2) dt. (6.26) It is always possible to ﬁnd a function ξ that makes ∆S positive. Simply choose ξ to be small, but make it wiggle very fast, so that ˙ ξ is large. Therefore, it is never the case that S0 is a maximum. Note that this reasoning works for any potential, not just a harmonic oscillator, as long as it is a function of position only (that is, it contains no derivatives, as we always assume). You might be tempted to use the same line of reasoning to say that it is also always possible to ﬁnd a function ξ that makes ∆S negative, by making ξ large and ˙ ξ small. If this were true, then we could put everything together and conclude that all stationary points are saddle points, for a harmonic oscillator. However, it is not always possible to make ξ large enough and ˙ ξ small enough so that ∆S is negative, due to the boundary conditions ξ(t1) = ξ(t2) = 0. If ξ changes from zero to a large value and then back to zero, then ˙ ξ may also have to be large, if the time interval is short enough. Problem 6.6 deals quantitatively with this issue. haven’t seen the topic before, and trivial to those who have. My apologies. But this and the following remarks are by no means necessary for an understanding of the material in this chapter. If you’re interested in reading more about these quantum mechanical issues, you should take a look at Richard Feynman’s book (Feynman, 1998). Feynman was, after all, the one who thought of this idea. VI-8 CHAPTER 6. THE LAGRANGIAN METHOD For now, let’s just recognize that in some cases S0 is a minimum, in some cases it is a saddle point, and it is never a maximum. “Least action” is therefore a misnomer. 5. It is sometimes said that nature has a “purpose,” in that it seeks to take the path that produces the minimum action. In view of the second remark above, this is incorrect. In fact, nature does exactly the opposite. It takes every path, treating them all on equal footing. We end up seeing only the path with a stationary action, due to the way the quantum mechanical phases add. It would be a harsh requirement, indeed, to demand that nature make a “global” decision (that is, to compare paths that are separated by large distances), and to choose the one with the smallest action. Instead, we see that everything takes place on a “local” scale. Nearby phases simply add, and everything works out automatically. When an archer shoots an arrow through the air, the aim is made possible by all the other arrows taking all the other nearby paths, each with essentially the same action. Likewise, when you walk down the street with a certain destination in mind, you’re not alone. . . When walking, I know that my aim Is caused by the ghosts with my name. And although I can’t see Where they walk next to me, I know they’re all there, just the same. 6. Consider a function, f(x), of one variable (for ease of terminology). Let f(b) be a local minimum of f. There are two basic properties of this minimum. The ﬁrst is that f(b) is smaller than all nearby values. The second is that the slope of f is zero at b. From the above remarks, we see that (as far as the action S is concerned) the ﬁrst property is completely irrelevant, and the second one is the whole point. In other words, saddle points (and maxima, although we showed above that these never exist for S) are just as good as minima, as far as the constructive addition of the eiS/¯ h phases is concerned. 7. Given that classical mechanics is an approximate theory, while quantum mechanics is the (more) correct one, it is quite silly to justify the principle of stationary action by demon-strating its equivalence with F = ma, as we did above. We should be doing it the other way around. However, because our intuition is based on F = ma, it’s easier to start with F = ma as the given fact, rather than calling upon the latent quantum-mechanical intuition hidden deep within all of us. Maybe someday. . . At any rate, in more advanced theories dealing with fundamental issues concerning the tiny building blocks of matter (where actions are of the same order of magnitude as ¯ h), the approximate F = ma theory is invalid, and you have to use the Lagrangian method. 8. When dealing with a system in which a non-conservative force such as friction is present, the Lagrangian method loses much of its appeal. The reason for this is that non-conservative forces don’t have a potential energy associated with them, so there isn’t a speciﬁc V (x) that you can write down in the Lagrangian. Although friction forces can in fact be incorporated in the Lagrangian method, you have to include them in the E-L equations essentially by hand. We won’t deal with non-conservative forces in this chapter. ♣ 6.3 Forces of constraint A nice thing about the Lagrangian method is that we are free to impose any given constraints at the beginning of the problem, thereby immediately reducing the number of variables. This is always done (perhaps without thinking) whenever a particle is constrained to move on a wire or surface, etc. Often we are concerned not with the exact nature of the forces doing the constraining, but only with the resulting motion, given that the constraints hold. By imposing the constraints at the outset, we can ﬁnd the motion, but we can’t say anything about the constraining forces. If we want to determine the constraining forces, we must take a diﬀerent approach. The main idea of the strategy, as we will show below, is that we must not impose the constraints 6.3. FORCES OF CONSTRAINT VI-9 too soon. This leaves us with a larger number of variables to deal with, so the calculations are more cumbersome. But the beneﬁt is that we are able to ﬁnd the constraining forces. Consider the setup of a particle sliding oﬀa ﬁxed frictionless hemisphere of radius R (see Fig. 6.3). Let’s say that we are concerned only with ﬁnding the equation of motion θ R Figure 6.3 for θ, and not the constraining force. Then we can write everything in terms of θ, because we know that the radial distance r is constrained to be R. The kinetic energy is mR2 ˙ θ2/2, and the potential energy (relative to the bottom of the hemisphere) is mgR cos θ, so the Lagrangian is L = 1 2mR2 ˙ θ2 −mgR cos θ, (6.27) and the equation of motion, via eq. (6.3), is ¨ θ = (g/R) sin θ, (6.28) which is equivalent to the tangential F = ma statement. Now let’s say that we want to ﬁnd the constraining normal force that the hemisphere applies to the particle. To do this, let’s solve the problem in a diﬀerent way and write things in terms of both r and θ. Also (and here’s the critical step), let’s be really picky and say that r isn’t exactly constrained to be R, because in the real world the particle actually sinks into the hemisphere a little bit. This may seem a bit silly, but it’s really the whole point. The particle pushes and sinks inward a tiny distance until the hemisphere gets squashed enough to push back with the appropriate force to keep the particle from sinking in any more (just consider the hemisphere to be made of lots of little springs with very large spring constants). The particle is therefore subject to a (very steep) potential arising from the hemisphere’s force. The constraining potential, V (r), looks something like the plot in Fig. 6.4. The true V(r) r R Figure 6.4 Lagrangian for the system is thus L = 1 2m( ˙ r2 + r2 ˙ θ2) −mgr cos θ −V (r). (6.29) (The ˙ r2 term in the kinetic energy will turn out to be insigniﬁcant.) The equations of motion obtained from varying θ and r are therefore mr2¨ θ + 2mr ˙ r ˙ θ = mgr sin θ, m¨ r = mr ˙ θ2 −mg cos θ −V ′(r). (6.30) Having written down the equations of motion, we will now apply the constraint condition that r = R. This condition implies ˙ r = ¨ r = 0. (Of course, r isn’t really equal to R, but any diﬀerences are inconsequential from this point onward.) The ﬁrst of eqs. (6.30) then reproduces eq. (6.28), while the second yields −dV dr ¯ ¯ ¯ ¯ r=R = mg cos θ −mR ˙ θ2. (6.31) But FN ≡−dV/dr is the constraint force applied in the r direction, which is precisely the force we are looking for. The normal force of constraint is therefore FN(θ, ˙ θ) = mg cos θ −mR ˙ θ2. (6.32) This is equivalent to the radial F = ma equation, mg cos θ−FN = mR ˙ θ2 (which is certainly a quicker way to ﬁnd the normal force in the present problem). Note that this result is valid only if FN(θ, ˙ θ) > 0. If the normal force becomes zero, then this means that the particle has left the sphere, in which case r no longer equals R. VI-10 CHAPTER 6. THE LAGRANGIAN METHOD Remarks: 1. What if we instead had (unwisely) chosen Cartesian coordinates, x and y, instead of polar coordinates, r and θ? Since the distance from the particle to the surface of the hemisphere is η ≡ p x2 + y2 −R, we obtain a true Lagrangian equal to L = 1 2m( ˙ x2 + ˙ y2) −mgy −V (η). (6.33) The equations of motion are (using the chain rule) m¨ x = −dV dη ∂η ∂x , and m¨ y = −mg −dV dη ∂η ∂y . (6.34) We now apply the constraint condition η = 0. Since −dV/dη equals the constraint force F, you can show that the equations we end up with (namely, the two E-L equations and the constraint equation) are m¨ x = F x R , m¨ y = −mg + F y R , and p x2 + y2 −R = 0. (6.35) These three equations are suﬃcient to determine the three unknowns ¨ x, ¨ y, and F as functions of the quantities x, ˙ x, y, and ˙ y. See Exercise 6.37, which should convince you that polar coordinates are the way to go. In general, the strategy is to take two time derivatives of the constraint equation and then eliminate the second derivatives of the coordinates by using the E-L equations (this process was trivial in the polar-coordinate case). 2. You can see from eq. (6.35) that the E-L equations end up taking the form, d dt µ ∂L ∂˙ qi ¶ = ∂L ∂qi + F ∂η ∂qi , (6.36) for each coordinate qi. The quantity η is what appears in the constraint equation, η = 0. In our hemisphere problem, we had η = r −R in polar coordinates, and η = p x2 + y2 −R in Cartesian coordinates. The E-L equations, combined with the η = 0 condition, give us exactly the number of equations (N + 1 of them, where N is the number of coordinates) needed to determine all of the N + 1 unknowns (all the ¨ qi, and F), in terms of the qi and ˙ qi. Writing down the equations in eq. (6.36) is basically the method of Lagrange multipliers, where the Lagrange multiplier turns out to be the force. But if you’re not familiar with this method, no need to worry; you can derive everything from scratch using the above technique involving the steep potential. If you do happen to be familiar with it, then there might in fact be a need to worry about how you apply it, as the following remark explains. 3. When trying to determine the forces of constraint, you can just start with eq. (6.36), without bothering to write down V (η). But you must be careful to make sure that η does indeed represent the distance the particle is from where it should be. In polar coordinates, if someone gives you the constraint condition as 7(r −R) = 0, and if you use the left-hand side of this as the η in eq. (6.36), then you will get the wrong constraint force; it will be too small by a factor of 7. Likewise, in Cartesian coordinates, writing the constraint as y − √ R2 −x2 = 0 will give you the wrong force. The best way to avoid this problem is, of course, to pick one of your variables as the distance the particle is from where it should be (up to an additive constant, as in the case of r −R = 0). ♣ 6.4 Change of coordinates When L is written in terms of Cartesian coordinates x, y, z, we showed in Section 6.1 that the Euler-Lagrange equations are equivalent to Newton’s F = ma equations; see eq. (6.8). But what about the case where we use polar, spherical, or some other coordinates? The equivalence of the E-L equations and F = ma isn’t so obvious. As far as trusting the E-L 6.4. CHANGE OF COORDINATES VI-11 equations for such coordinates goes, you can achieve peace of mind in two ways. You can accept the principle of stationary action as something so beautiful and profound that it simply has to work for any choice of coordinates. Or, you can take the more mundane road and show through a change of coordinates that if the E-L equations hold for one set of coordinates (and we know that they do hold for at least one set, namely Cartesian coordinates), then they also hold for any other coordinates (of a certain form, described below). In this section, we will demonstrate the validity of the E-L equations through the explicit change of coordinates.9 Consider the set of coordinates, xi : (x1, x2, . . . , xN). (6.37) For example, if N = 6, then x1, x2, x3 could be the Cartesian x, y, z coordinates of one particle, and x4, x5, x6 could be the r, θ, φ polar coordinates of a second particle, and so on. Assume that the E-L equations hold for these variables, that is, d dt µ ∂L ∂˙ xi ¶ = ∂L ∂xi (1 ≤i ≤N). (6.38) Consider a new set of variables that are functions of the xi and t, qi = qi(x1, x2, . . . , xN; t). (6.39) We will restrict ourselves to the case where the qi do not depend on the ˙ xi. (This is quite reasonable. If the coordinates depended on the velocities, then we wouldn’t be able to label points in space with deﬁnite coordinates. We’d have to worry about how the particles were behaving when they were at the points. These would be strange coordinates indeed.) We can, in theory, invert eq. (6.39) and express the xi as functions of the qi and t, xi = xi(q1, q2, . . . , qN; t). (6.40) Claim 6.2 If eq. (6.38) is true for the xi coordinates, and if the xi and qi are related by eq. (6.40), then eq. (6.38) is also true for the qi coordinates. That is, d dt µ ∂L ∂˙ qm ¶ = ∂L ∂qm (1 ≤m ≤N). (6.41) Proof: We have ∂L ∂˙ qm = N X i=1 ∂L ∂˙ xi ∂˙ xi ∂˙ qm . (6.42) (Note that if the xi depended on the ˙ qi, then we would have to include the additional term, P(∂L/∂xi)(∂xi/∂˙ qm). But we have excluded such dependence.) Let’s rewrite the ∂˙ xi/∂˙ qm term. From eq. (6.40), we have ˙ xi = N X m=1 ∂xi ∂qm ˙ qm + ∂xi ∂t . (6.43) Therefore, ∂˙ xi ∂˙ qm = ∂xi ∂qm . (6.44) 9This calculation is straightforward but a bit messy, so you may want to skip this section and just settle for the “beautiful and profound” reasoning. VI-12 CHAPTER 6. THE LAGRANGIAN METHOD Substituting this into eq. (6.42) and taking the time derivative of both sides gives d dt µ ∂L ∂˙ qm ¶ = N X i=1 d dt µ ∂L ∂˙ xi ¶ ∂xi ∂qm + N X i=1 ∂L ∂˙ xi d dt µ ∂xi ∂qm ¶ . (6.45) In the second term here, it is legal to switch the order of the total derivative, d/dt, and the partial derivative, ∂/∂qm. Remark: In case you have your doubts, let’s prove that this switching is legal. d dt µ ∂xi ∂qm ¶ = N X k=1 ∂ ∂qk µ ∂xi ∂qm ¶ ˙ qk + ∂ ∂t µ ∂xi ∂qm ¶ = ∂ ∂qm Ã N X k=1 ∂xi ∂qk ˙ qk + ∂xi ∂t ! = ∂˙ xi ∂qm . ♣ (6.46) In the ﬁrst term on the right-hand side of eq. (6.45), we can use the given information in eq. (6.38) and rewrite the (d/dt)(∂L/∂˙ xi) term. We then obtain d dt µ ∂L ∂˙ qm ¶ = N X i=1 ∂L ∂xi ∂xi ∂qm + N X i=1 ∂L ∂˙ xi ∂˙ xi ∂qm = ∂L ∂qm , (6.47) as we wanted to show. We have therefore demonstrated that if the Euler-Lagrange equations are true for one set of coordinates, xi (and they are true for Cartesian coordinates), then they are also true for any other set of coordinates, qi, satisfying eq. (6.39). If you’re inclined to look at the principle of stationary action with distrust, thinking that it might be a coordinate-dependent statement, this proof should put you at ease. The Euler-Lagrange equations are valid in any coordinates. Note that the above proof did not in any way use the precise form of the Lagrangian. If L were equal to T +V , or 8T +πV 2/T, or any other arbitrary function, our result would still be true: If eq. (6.38) is true for one set of coordinates, then it is also true for any other set of coordinates qi satisfying eq. (6.39). The point is that the only L for which the hypothesis is true at all (that is, for which eq. (6.38) holds) is L ≡T −V (or any constant multiple of this). Remark: On one hand, it is quite amazing how little we assumed in proving the above claim. Any new coordinates of the very general form in eq. (6.39) satisfy the E-L equations, as long as the original coordinates do. If the E-L equations had, say, a factor of 5 on the right-hand side of eq. (6.38), then they would not hold in arbitrary coordinates. To see this, just follow the proof through with the factor of 5. On the other hand, the claim is quite believable, if you make an analogy with a function instead of a functional. Consider the function f(z) = z2. This has a minimum at z = 0, consistent with the fact that d f/dz = 0 at z = 0. But let’s now write f in terms of the variable y deﬁned by, say, z = y4. Then f(y) = y8, and f has a minimum at y = 0, consistent with the fact that d f/dy = 0 at y = 0. So f ′ = 0 holds in both coordinates at the corresponding points y = z = 0. This is the (simpliﬁed) analog of the E-L equations holding in both coordinates. In both cases, the derivative equation describes where the stationary value occurs. 6.5. CONSERVATION LAWS VI-13 This change-of-variables result may be stated in a more geometrical (and friendly) way. If you plot a function and then stretch the horizontal axis in an arbitrary manner (which is what happens when you change coordinates), then a stationary value (that is, one where the slope is zero) will still be a stationary value after the stretching.10 A picture (or even just the thought of one) is worth a dozen equations, apparently. As an example of an equation that does not hold for all coordinates, consider the preceding example, but with f ′ = 1 instead of f ′ = 0. In terms of z, f ′ = 1 when z = 1/2. And in terms of y, f ′ = 1 when y = (1/8)1/7. But the points z = 1/2 and y = (1/8)1/7 are not the same point. In other words, f ′ = 1 is not a coordinate-independent statement. Most equations are coordinate dependent. The special thing about f ′ = 0 is that a stationary point is a stationary point no matter how you look at it. ♣ 6.5 Conservation Laws 6.5.1 Cyclic coordinates Consider the case where the Lagrangian does not depend on a certain coordinate qk. Then d dt µ ∂L ∂˙ qk ¶ = ∂L ∂qk = 0 = ⇒ ∂L ∂˙ qk = C, (6.48) where C is a constant, that is, independent of time. In this case, we say that qk is a cyclic coordinate, and that ∂L/∂˙ qk is a conserved quantity (meaning that it doesn’t change with time). If Cartesian coordinates are used, then ∂L/∂˙ xk is simply the momentum, m ˙ xk, because ˙ xk appears only in the the kinetic energy’s m ˙ x2 k/2 term (we exclude cases where the potential depends on ˙ xk). We therefore call ∂L/∂˙ qk the generalized momentum corresponding to the coordinate qk. And in cases where ∂L/∂˙ qk does not change with time, we call it a conserved momentum. Note that a generalized momentum need not have the units of linear momentum, as the angular-momentum examples below show. Example 1: Linear momentum Consider a ball thrown through the air. In the full three dimensions, the Lagrangian is L = 1 2m( ˙ x2 + ˙ y2 + ˙ z2) −mgz. (6.49) There is no x or y dependence here, so both ∂L/∂˙ x = m ˙ x and ∂L/∂˙ y = m ˙ y are constant, as we well know. The fancy way of saying this is that conservation of px ≡m ˙ x arises from spatial translation invariance in the x direction. The fact that the Lagrangian doesn’t depend on x means that it doesn’t matter if you throw the ball in one spot, or in another spot a mile down the road. The setup is independent of the x value. This independence leads to conservation of px. Example 2: Angular and linear momentum in cylindrical coordinates Consider a potential that depends only on the distance to the z axis. In cylindrical coordi-nates, the Lagrangian is L = 1 2m( ˙ r2 + r2 ˙ θ2 + ˙ z2) −V (r). (6.50) 10There is, however, one exception. A stationary point in one coordinate system might be located at a kink in another coordinate system, so that f′ is not deﬁned there. For example, if we had instead deﬁned y by z = y1/4, then f(y) = y1/2, which has an undeﬁned slope at y = 0. Basically, we’ve stretched (or shrunk) the horizontal axis by a factor of inﬁnity at the origin, and this is a process that can change a zero slope into an undeﬁned one. But let’s not worry about this. VI-14 CHAPTER 6. THE LAGRANGIAN METHOD There is no z dependence here, so ∂L/∂˙ z = m ˙ z is constant. Also, there is no θ dependence, so ∂L/∂˙ θ = mr2 ˙ θ is constant. Since r ˙ θ is the speed in the tangential direction around the z axis, we see that our conserved quantity, mr(r ˙ θ), is the angular momentum (discussed in Chapters 7-9) around the z axis. In the same manner as in the preceding example, conservation of angular momentum around the z axis arises from rotation invariance around the z axis. Example 3: Angular momentum in spherical coordinates In spherical coordinates, consider a potential that depends only on r and θ. Our convention for spherical coordinates is that θ is the angle down from the north pole, and φ is the angle around the equator. The Lagrangian is L = 1 2m( ˙ r2 + r2 ˙ θ2 + r2 sin2 θ ˙ φ2) −V (r, θ). (6.51) There is no φ dependence here, so ∂L/∂˙ φ = mr2 sin2 θ ˙ φ is constant. Since r sin θ is the distance from the z axis, and since r sin θ ˙ φ is the speed in the tangential direction around the z axis, we see that our conserved quantity, m(r sin θ)(r sin θ ˙ φ), is the angular momentum around the z axis. 6.5.2 Energy conservation We will now derive another conservation law, namely conservation of energy. The conserva-tion of momentum or angular momentum above arose when the Lagrangian was independent of x, y, z, θ, or φ. Conservation of energy arises when the Lagrangian is independent of time. This conservation law is diﬀerent from those in the above momenta examples, because t is not a coordinate which the stationary-action principle can be applied to. You can imagine varying the coordinates x, θ, etc., which are functions of t. But it makes no sense to vary t. Therefore, we’re going to have to prove this conservation law in a diﬀerent way. Consider the quantity, E = Ã N X i=1 ∂L ∂˙ qi ˙ qi ! −L. (6.52) E turns out (usually) to be the energy. We’ll show this below. The motivation for this expression for E comes from the theory of Legendre transforms, but we won’t get into that here. We’ll just accept the deﬁnition in eq. (6.52) and prove a very useful fact about it. Claim 6.3 If L has no explicit time dependence (that is, if ∂L/∂t = 0), then E is conserved (that is, dE/dt = 0), assuming that the motion obeys the E-L equations (which it does). Note that there is one partial derivative and one total derivative in this statement. Proof: L is a function of the qi, the ˙ qi, and possibly t. Making copious use of the chain rule, we have dE dt = d dt Ã N X i=1 ∂L ∂˙ qi ˙ qi ! −dL dt = N X i=1 µµ d dt ∂L ∂˙ qi ¶ ˙ qi + ∂L ∂˙ qi ¨ qi ¶ − Ã N X i=1 µ ∂L ∂qi ˙ qi + ∂L ∂˙ qi ¨ qi ¶ + ∂L ∂t ! . (6.53) There are ﬁve terms here. The second cancels with the fourth. And the ﬁrst (after using the E-L equation, eq. (6.3), to rewrite it) cancels with the third. We therefore arrive at the simple result, dE dt = −∂L ∂t . (6.54) 6.6. NOETHER’S THEOREM VI-15 In the event that ∂L/∂t = 0 (that is, there are no t’s sitting on the paper when you write down L), which is usually the case in the situations we consider (because we generally won’t deal with potentials that depend on time), we have dE/dt = 0. Not too many things are constant with respect to time, and the quantity E has units of energy, so it’s a good bet that it’s the energy. Let’s show this in Cartesian coordinates (however, see the remark below). The Lagrangian is L = 1 2m( ˙ x2 + ˙ y2 + ˙ z2) −V (x, y, z), (6.55) so eq. (6.52) gives E = 1 2m( ˙ x2 + ˙ y2 + ˙ z2) + V (x, y, z), (6.56) which is the total energy. The eﬀect of the operations in eq. (6.52) in most cases is just to switch the sign in front of the potential. Of course, taking the kinetic energy T and subtracting the potential energy V to obtain L, and then using eq. (6.52) to produce E = T + V , seems like a rather convoluted way of arriving at T + V . But the point of all this is that we used the E-L equations to prove that E is conserved. Although we know very well from the F = ma methods in Chapter 5 that the sum T + V is conserved, it’s not fair to assume that it is conserved in our new Lagrangian formalism. We have to show that this follows from the E-L equations. As with the translation and rotation invariance we observed in the examples in Sec-tion 6.5.1, we see that energy conservation arises from time translation invariance. If the Lagrangian has no explicit t dependence, then the setup looks the same today as it did yesterday. This fact leads to conservation of energy. Remark: The quantity E in eq. (6.52) gives the energy of the system only if the entire system is represented by the Lagrangian. That is, the Lagrangian must represent a closed system with no external forces. If the system is not closed, then Claim 6.3 (or more generally, eq. (6.54)) is still perfectly valid for the E deﬁned in eq. (6.52), but this E may simply not be the energy of the system. Problem 6.8 is a good example of such a situation. Another example is the following. Imagine a long rod in the horizontal x-y plane. The rod points in the x direction, and a bead is free to slide frictionlessly along it. At t = 0, an external machine is arranged to accelerate the rod in the negative y direction (that is, transverse to itself) with acceleration −g. So ˙ y = −gt. There is no internal potential energy in this system, so the Lagrangian is just the kinetic energy, L = m ˙ x2/2 + m(gt)2/2. Eq. (6.52) therefore gives E = m ˙ x2/2 −m(gt)2/2, which isn’t the energy. But eq. (6.54) is still true, because dE dt = −∂L ∂t ⇐ ⇒ m ˙ x¨ x −mg2t = −mg2t ⇐ ⇒ ¨ x = 0, (6.57) which is correct. However, this setup is exactly the same as projectile motion in the x-y plane, where y is now the vertical axis, provided that we eliminate the rod and consider gravity instead of the machine to be causing the acceleration in the y direction. But if we are thinking in terms of gravity, then the normal thing to do is to say that the particle moves under the inﬂuence of the potential V (y) = mgy. The Lagrangian for this closed system (bead plus earth) is L = m( ˙ x2 + ˙ y2)/2 −mgy, and so eq. (6.52) gives E = m( ˙ x2 + ˙ y2)/2 + mgy, which is indeed the energy of the particle. But having said all this, most of the systems we’ll deal with are closed, so you can usually ignore this remark and assume that the E in eq. (6.52) gives the energy. ♣ 6.6 Noether’s Theorem We now present one of the most beautiful and useful theorems in physics. It deals with two fundamental concepts, namely symmetry and conserved quantities. The theorem (due VI-16 CHAPTER 6. THE LAGRANGIAN METHOD to Emmy Noether) may be stated as follows. Theorem 6.4 (Noether’s Theorem) For each symmetry of the Lagrangian, there is a conserved quantity. By “symmetry,” we mean that if the coordinates are changed by some small quantities, then the Lagrangian has no ﬁrst-order change in these quantities. By “conserved quantity,” we mean a quantity that does not change with time. The result in Section 6.5.1 for cyclic coordinates is a special case of this theorem. Proof: Let the Lagrangian be invariant, to ﬁrst order in the small number ϵ, under the change of coordinates, qi − →qi + ϵKi(q). (6.58) Each Ki(q) may be a function of all the qi, which we collectively denote by the shorthand, q. Remark: As an example of what these Ki’s might look like, consider the Lagrangian, L = (m/2)(5 ˙ x2 −2 ˙ x ˙ y + 2 ˙ y2) + C(2x −y). We’ve just pulled this out of a hat, although it happens to be the type of L that arises in Atwood’s machine problems; see Problem 6.9 and Exercise 6.40. This L is invariant under the transformation x →x + ϵ and y →y + 2ϵ, because the derivative terms are unaﬀected, and the diﬀerence 2x −y is unchanged. (It’s actually invariant to all orders in ϵ, and not just ﬁrst order. But this isn’t necessary for the theorem to hold.) Therefore, Kx = 1 and Ky = 2. In the problems we’ll be doing, the Ki’s can generally be determined by simply looking at the potential term. Of course, someone else might come along with Kx = 3 and Ky = 6, which is also a symmetry. And indeed, any factor can be taken out of ϵ and put into the Ki’s without changing the quantity ϵKi(q) in eq. (6.58). Any such modiﬁcation will just bring an overall constant factor (and hence not change the property of being conserved) into the conserved quantity in eq. (6.61) below. It is therefore irrelevant. ♣ The fact that the Lagrangian does not change at ﬁrst order in ϵ means that 0 = dL dϵ = X i µ ∂L ∂qi ∂qi ∂ϵ + ∂L ∂˙ qi ∂˙ qi ∂ϵ ¶ = X i µ ∂L ∂qi Ki + ∂L ∂˙ qi ˙ Ki ¶ . (6.59) Using the E-L equation, eq. (6.3), we can rewrite this as 0 = X i µ d dt µ ∂L ∂˙ qi ¶ Ki + ∂L ∂˙ qi ˙ Ki ¶ = d dt ÃX i ∂L ∂˙ qi Ki ! . (6.60) Therefore, the quantity P(q, ˙ q) ≡ X i ∂L ∂˙ qi Ki(q) (6.61) does not change with time. It is given the generic name of conserved momentum. But it need not have the units of linear momentum. As Noether most keenly observed (And for which much acclaim is deserved), 6.6. NOETHER’S THEOREM VI-17 It’s easy to see That for each symmetry, A quantity must be conserved. Example 1: Consider the Lagrangian in the above remark, L = (m/2)(5 ˙ x2 −2 ˙ x ˙ y + 2 ˙ y2) + C(2x −y). We saw that Kx = 1 and Ky = 2. The conserved momentum is therefore P(x, y, ˙ x, ˙ y) = ∂L ∂˙ x Kx + ∂L ∂˙ y Ky = m(5 ˙ x −˙ y)(1) + m(−˙ x + 2 ˙ y)(2) = m(3 ˙ x + 3 ˙ y). (6.62) The overall factor of 3m isn’t important. Example 2: Consider a thrown ball. We have L = (m/2)( ˙ x2 + ˙ y2 + ˙ z2) −mgz. This is invariant under translations in x, that is, x →x + ϵ; and also under translations in y, that is, y →y + ϵ. (Both x and y are cyclic coordinates.) We need invariance only to ﬁrst order in ϵ for Noether’s theorem to hold, but this L is invariant to all orders. We therefore have two symmetries in our Lagrangian. The ﬁrst has Kx = 1, Ky = 0, and Kz = 0. The second has Kx = 0, Ky = 1, and Kz = 0. Of course, the nonzero Ki’s here can be chosen to be any constants, but we may as well pick them to be 1. The two conserved momenta are P1(x, y, z, ˙ x, ˙ y, ˙ z) = ∂L ∂˙ x Kx + ∂L ∂˙ y Ky + ∂L ∂˙ z Kz = m ˙ x, P2(x, y, z, ˙ x, ˙ y, ˙ z) = ∂L ∂˙ x Kx + ∂L ∂˙ y Ky + ∂L ∂˙ z Kz = m ˙ y. (6.63) These are simply the x and y components of the linear momentum, as we saw in Example 1 in Section 6.5.1. Note that any combination of these momenta, say 3P1 + 8P2, is also conserved. (In other words, x →x +3ϵ, y →y + 8ϵ, z →z is a symmetry of the Lagrangian.) But the above P1 and P2 are the simplest conserved momenta to choose as a “basis” for the inﬁnite number of conserved momenta (which is how many you have, if there are two or more independent continuous symmetries). Example 3: Consider a mass on a spring, with relaxed length zero, in the x-y plane. The Lagrangian, L = (m/2)( ˙ x2+ ˙ y2)−(k/2)(x2+y2), is invariant under the change of coordinates, x →x + ϵy and y →y −ϵx, to ﬁrst order in ϵ (as you can check). So we have Kx = y and Ky = −x. The conserved momentum is therefore P(x, y, ˙ x, ˙ y) = ∂L ∂˙ x Kx + ∂L ∂˙ y Ky = m( ˙ xy −˙ yx). (6.64) This is the (negative of the) z component of the angular momentum. The angular momentum is conserved here because the potential, V (x, y) ∝x2 +y2 = r2, depends only on the distance from the origin. We’ll discuss such potentials in Chapter 7. In contrast with the ﬁrst two examples above, the x →x + ϵy, y →y −ϵx transformation isn’t so obvious here. How did we get this? Well, unfortunately there doesn’t seem to be any fail-proof method of determining the Ki’s in general, so sometimes you just have to guess around, as was the case here. But in many problems, the Ki’s are simple constants which are easy to see. Remarks: 1. As we saw above, in some cases the Ki’s are functions of the coordinates, and in some cases they are not. VI-18 CHAPTER 6. THE LAGRANGIAN METHOD 2. The cyclic-coordinate result in eq. (6.48) is a special case of Noether’s theorem, for the following reason. If L doesn’t depend on a certain coordinate qk, then qk − →qk + ϵ is certainly a symmetry. Hence Kk = 1 (with all the other Ki’s equal to zero), and eq. (6.61) reduces to eq. (6.48). 3. We use the word “symmetry” to describe the situation where the transformation in eq. (6.58) produces no ﬁrst-order change in the Lagrangian. This is an appropriate choice of word, because the Lagrangian describes the system, and if the system essentially doesn’t change when the coordinates are changed, then we say that the system is symmetric. For example, if we have a setup that doesn’t depend on θ, then we say that the setup is symmetric under rotations. Rotate the system however you want, and it looks the same. The two most common applications of Noether’s theorem are the conservation of angular momentum, which arises from symmetry under rotations; and conservation of linear momentum, which arises from symmetry under translations. 4. In simple systems, as in Example 2 above, it is clear why the resulting P is conserved. But in more complicated systems, as in Example 1 above, the resulting P might not have an obvious interpretation. But at least you know that it is conserved, and this will invariably help in understanding a setup. 5. Although conserved quantities are extremely useful in studying a physical situation, it should be stressed that there is no more information contained in them than there is in the E-L equations. Conserved quantities are simply the result of integrating the E-L equations. For example, if you write down the E-L equations for Example 1 above, and then add the “x” equation (which is 5m¨ x −m¨ y = 2C) to twice the “y” equation (which is −m¨ x + 2m¨ y = −C), then you arrive at 3m(¨ x + ¨ y) = 0. In other words, 3m( ˙ x + ˙ y) is constant, as we found from Noether’s theorem. Of course, you might have to do some guesswork to ﬁnd the proper combination of the E-L equations that gives a zero on the right-hand side. But you’d have to do some guesswork anyway, to ﬁnd the symmetry for Noether’s theorem. At any rate, a conserved quantity is useful because it is an integrated form of the E-L equations. It puts you one step closer to solving the problem, compared with where you would be if you started with the second-order E-L equations. 6. Does every system have a conserved momentum? Certainly not. The one-dimensional prob-lem of a falling ball (m¨ z = −mg) doesn’t have one. And if you write down an arbitrary potential in 3-D, odds are that there won’t be one. In a sense, things have to contrive nicely for there to be a conserved momentum. In some problems, you can just look at the physical system and see what the symmetry is, but in others (for example, in the Atwood’s-machine problems for this chapter), the symmetry is not at all obvious. 7. By “conserved quantity,” we mean a quantity that depends on (at most) the coordinates and their ﬁrst derivatives (that is, not on their second derivatives). If we don’t make this restriction, then it is trivial to construct quantities that are independent of time. For example, in Example 1 above, the “x” E-L equation (which is 5m¨ x−m¨ y = 2C) tells us that 5m¨ x−m¨ y has its time derivative equal to zero. Note that an equivalent way of excluding these trivial cases is to say that the value of a conserved quantity depends on the initial conditions (that is, the velocities and positions). The quantity 5m¨ x−m¨ y doesn’t satisfy this criterion, because its value is always constrained to be 2C. ♣ 6.7 Small oscillations In many physical systems, a particle undergoes small oscillations around an equilibrium point. In Section 5.2, we showed that the frequency of these small oscillations is ω = r V ′′(x0) m , (6.65) 6.7. SMALL OSCILLATIONS VI-19 where V (x) is the potential energy, and x0 is the equilibrium point. However, this result holds only for one-dimensional motion (we’ll see below why this is true). In more compli-cated systems, such as the one described below, it is necessary to use another procedure to obtain the frequency ω. This procedure is a fail-proof one, applicable in all situations. It is, however, a bit more involved than simply writing down eq. (6.65). So in one-dimensional problems, eq. (6.65) is still what you want to use. We’ll demonstrate our fail-proof method through the following problem. Problem: A mass m is free to slide on a frictionless table and is connected by a string, which passes through a hole in the table, to a mass M which hangs below (see Fig. 6.5). M m l-r r θ Figure 6.5 Assume that M moves in a vertical line only, and assume that the string always remains taut. (a) Find the equations of motion for the variables r and θ shown in the ﬁgure. (b) Under what condition does m undergo circular motion? (c) What is the frequency of small oscillations (in the variable r) about this circular motion? Solution: (a) Let the string have length ℓ(this length won’t matter). Then the Lagrangian (we’ll call it “L” here, to save “L” for the angular momentum, which arises below) is L = 1 2M ˙ r2 + 1 2m( ˙ r2 + r2 ˙ θ2) + Mg(ℓ−r). (6.66) For the purposes of the potential energy, we’ve taken the table to be at height zero, but any other value could be chosen. The E-L equations of motion obtained from varying θ and r are d dt(mr2 ˙ θ) = 0, (M + m)¨ r = mr ˙ θ2 −Mg. (6.67) The ﬁrst equation says that angular momentum is conserved. The second equation says that the Mg gravitational force accounts for the acceleration of the two masses along the direction of the string, plus the centripetal acceleration of m. (b) The ﬁrst of eqs. (6.67) says that mr2 ˙ θ = L, where L is some constant (the angular momentum) which depends on the initial conditions. Plugging ˙ θ = L/mr2 into the second of eqs. (6.67) gives (M + m)¨ r = L2 mr3 −Mg. (6.68) Circular motion occurs when ˙ r = ¨ r = 0. Therefore, the radius of the circular orbit is given by r3 0 = L2 Mmg . (6.69) Since L = mr2 ˙ θ, eq. (6.69) is equivalent to mr0 ˙ θ2 = Mg, (6.70) which can also be obtained by letting ¨ r = 0 in the second of eqs. (6.67). In other words, the gravitational force on M exactly accounts for the centripetal acceleration of m if the motion is circular. Given r0, eq. (6.70) determines what ˙ θ must be in order to have circular motion, and vice versa. (c) To ﬁnd the frequency of small oscillations about the circular motion, we need to look at what happens to r if we perturb it slightly from its equilibrium value of r0. Our fail-proof procedure is the following. VI-20 CHAPTER 6. THE LAGRANGIAN METHOD Let r(t) ≡r0 + δ(t), where δ(t) is very small (more precisely, δ(t) ≪r0), and expand eq. (6.68) to ﬁrst order in δ(t). Using 1 r3 ≡ 1 (r0 + δ)3 ≈ 1 r3 0 + 3r2 0δ = 1 r3 0(1 + 3δ/r0) ≈1 r3 0 ³ 1 −3δ r0 ´ , (6.71) we obtain (M + m)¨ δ ≈L2 mr3 0 ³ 1 −3δ r0 ´ −Mg. (6.72) The terms not involving δ on the right-hand side cancel, by the deﬁnition of r0 given in eq. (6.69). This cancellation always occurs in such a problem at this stage, due to the deﬁnition of the equilibrium point. We are therefore left with ¨ δ + µ 3L2 (M + m)mr4 0 ¶ δ ≈0. (6.73) This is a good old simple-harmonic-oscillator equation in the variable δ. Therefore, the frequency of small oscillations about a circle of radius r0 is ω ≈ r 3L2 (M + m)mr4 0 = r 3M M + m r g r0 , (6.74) where we have used eq. (6.69) to eliminate L in the second expression. To sum up, the above frequency is the frequency of small oscillations in the variable r. In other words, if you have nearly circular motion, and if you plot r as a function of time (and ignore what θ is doing), then you will get a nice sinusoidal graph whose frequency is given by eq. (6.74). Note that this frequency need not have anything to do with the other relevant frequency in this problem, namely the frequency of the circular motion, which is p M/m p g/r0, from eq. (6.70). Remarks: Let’s look at some limits. For a given r0, if m ≫M, then ω ≈ p 3Mg/mr0 ≈0. This makes sense, because everything is moving very slowly. This frequency equals √ 3 times the frequency of circular motion, namely p Mg/mr0, which isn’t at all obvious. For a given r0, if m ≪M, then ω ≈ p 3g/r0, which isn’t so obvious, either. The frequency of small oscillations is equal to the frequency of circular motion if M = 2m, which, once again, isn’t obvious. This condition is independent of r0. ♣ The above procedure for ﬁnding the frequency of small oscillations can be summed up in three steps: (1) Find the equations of motion, (2) Find the equilibrium point, and (3) Let x(t) ≡x0 + δ(t), where x0 is the equilibrium point of the relevant variable, and expand one of the equations of motion (or a combination of them) to ﬁrst order in δ, to obtain a simple-harmonic-oscillator equation for δ. If the equilibrium point happens to be at x = 0 (which is often the case), then everything is greatly simpliﬁed. There is no need to introduce δ, and the expansion in the third step above simply entails ignoring powers of x that are higher than ﬁrst order. Remark: If you just use the potential energy for the above problem (which is Mgr, up to a constant) in eq. (6.65), then you will obtain a frequency of zero, which is incorrect. You can use eq. (6.65) to ﬁnd the frequency, if you instead use the “eﬀective potential” for this problem, namely L2/(2mr2) + Mgr, and if you use the total mass, M + m, as the mass in eq. (6.65), as you can check. The reason why this works will become clear in Chapter 7 when we introduce the eﬀective potential. In many problems, however, it isn’t obvious what the appropriate modiﬁed potential is that should be used, or what mass goes in eq. (6.65). So it’s generally much safer to take a deep breath and go through an expansion similar to the one in part (c) of the example above. ♣ 6.8. OTHER APPLICATIONS VI-21 The one-dimensional result in eq. (6.65) is, of course, a special case of our above expansion procedure. We can repeat the derivation of Section 5.2 in the present language. In one dimension, the E-L equation of motion is m¨ x = −V ′(x). Let x0 be the equilibrium point, so V ′(x0) = 0. And let x(t) ≡x0 + δ(t). Expanding m¨ x = −V ′(x) to ﬁrst order in δ, we have m¨ δ = −V ′(x0) −V ′′(x0)δ, plus higher-order terms. Since V ′(x0) = 0, we have m¨ δ ≈−V ′′(x0)δ, as desired. 6.8 Other applications The formalism developed in Section 6.2 works for any function L(x, ˙ x, t). If our goal is to ﬁnd the stationary points of S ≡ R L, then eq. (6.15) holds, no matter what L is. There is no need for L to be equal to T −V , or indeed, to have anything to do with physics. And t need not have anything to do with time. All that is required is that the quantity x depend on the parameter t, and that L depend only on x, ˙ x, and t (and not, for example, on ¨ x; see Exercise 6.34). The formalism is very general and powerful, as the following example demonstrates. Example (Minimal surface of revolution): A surface of revolution has two given rings as its boundary; see Fig. 6.6. What should the shape of the surface be so that it has the Figure 6.6 minimum possible area? We will present three solutions. A fourth is left for Problem 6.22. First solution: Let the surface be generated by rotating the curve y = y(x) around the x axis. The boundary conditions are y(a1) = c1 and y(a2) = c2; see Fig. 6.7. Slicing the surface y(x) a a 1 2 c2 c1 Figure 6.7 up into vertical rings, we see that the area is given by A = Z a2 a1 2πy p 1 + y′2 dx. (6.75) Our goal is to ﬁnd the function y(x) that minimizes this integral. We therefore have exactly the same situation as in Section 6.2, except that x is now the parameter (instead of t), and y in now the function (instead of x). Our “Lagrangian” is thus L ∝y p 1 + y′2. To minimize the integral A, we “simply” have to write down the E-L equation, d dx µ ∂L ∂y′ ¶ = ∂L ∂y , (6.76) and calculate the derivatives. This calculation however, gets a bit tedious, so I’ve relegated it to Lemma 6.5 at the end of this section. For now we’ll just use the result in eq. (6.86) which gives (with f(y) = y here) 1 + y′2 = By2. (6.77) At this point we can cleverly guess (motivated by the fact that 1 + sinh2 z = cosh2 z) that the solution is y(x) = 1 b cosh b(x + d), (6.78) where b = √ B, and d is a constant of integration. Or, we can separate variables to obtain (again with b = √ B) dx = dy p (by)2 −1 , (6.79) and then use the fact that the integral of 1/ √ z2 −1 is cosh−1 z, to obtain the same result. The answer to our problem, therefore, is that y(x) takes the form of eq. (6.78), with b and d determined by the boundary conditions, c1 = 1 b cosh b(a1 + d), and c2 = 1 b cosh b(a2 + d). (6.80) VI-22 CHAPTER 6. THE LAGRANGIAN METHOD In the symmetrical case where c1 = c2, we know that the minimum occurs in the middle, so we may choose d = 0 and a1 = −a2. Solutions for b and d exist only for certain ranges of the a’s and c’s. Basically, if a2 −a1 is too large, then there is no solution. In this case, the minimal “surface” turns out to be the two given circles, attached by a line (which isn’t a nice two-dimensional surface). If you perform an experiment with soap bubbles (which want to minimize their area), and if you pull the rings too far apart, then the surface will break and disappear as it tries to form the two circles. Problem 6.23 deals with this issue. Second solution: Consider the curve that we rotate around the x axis to be described now by the function x(y). That is, let x be a function of y. The area is then given by A = Z c2 c1 2πy p 1 + x′2 dy, (6.81) where x′ ≡dx/dy. Note that the function x(y) may be double-valued, so it may not really be a function. But it looks like a function locally, and all of our formalism deals with local variations. Our “Lagrangian” is now L ∝y √ 1 + x′2, and the E-L equation is d dy ³ ∂L ∂x′ ´ = ∂L ∂x = ⇒ d dy µ yx′ √ 1 + x′2 ¶ = 0. (6.82) The nice thing about this solution is the “0” on the right-hand side, which arises from the fact that L doesn’t depend on x (that is, x is a cyclic coordinate). Therefore, yx′/ √ 1 + x′2 is constant. If we deﬁne this constant to be 1/b, then we can solve for x′ and then separate variables to obtain dx = dy p (by)2 −1 , (6.83) in agreement with eq. (6.79). The solution proceeds as above. Third solution: The “Lagrangian” in the ﬁrst solution above, L ∝y p 1 + y′2, is indepen-dent of x. Therefore, in analogy with conservation of energy (which arises from a Lagrangian that is independent of t), the quantity E ≡y′ ∂L ∂y′ −L = y′2y p 1 + y′2 −y p 1 + y′2 = −y p 1 + y′2 (6.84) is constant (that is, independent of x). This statement is equivalent to eq. (6.77), and the solution proceeds as above. As demonstrated by the brevity of the second and third solutions here, it is highly advantageous to make use of conserved quantities whenever you can. Let us now prove the following lemma, which we invoked in the ﬁrst solution above. This lemma is very useful, because it is common to encounter problems where the quantity to be extremized depends on the arclength, p 1 + y′2, and takes the form of R f(y) p 1 + y′2 dx. We will give two proofs. The ﬁrst proof uses the Euler-Lagrange equation. The calculation gets a bit messy, so it’s a good idea to work through it once and for all and then just invoke the result whenever needed. This derivation isn’t something you’d want to repeat too often. The second proof makes use of a conserved quantity. And in contrast with the ﬁrst proof, this method is exceedingly clean and simple. It actually is something you’d want to repeat quite often. But we’ll still do it once and for all. 6.8. OTHER APPLICATIONS VI-23 Lemma 6.5 Let f(y) be a given function of y. Then the function y(x) that extremizes the integral, Z x2 x1 f(y) p 1 + y′2 dx, (6.85) satisﬁes the diﬀerential equation, 1 + y′2 = Bf(y)2, (6.86) where B is a constant of integration.11 First Proof: Our goal is to ﬁnd the function y(x) that extremizes the integral in eq. (6.85). We therefore have exactly the same situation as in Section 6.2, except with x in place of t, and y in place of x. Our “Lagrangian” is thus L = f(y) p 1 + y′2, and the Euler-Lagrange equation is d dx µ ∂L ∂y′ ¶ = ∂L ∂y = ⇒ d dx Ã f · y′ · 1 p 1 + y′2 ! = f ′p 1 + y′2, (6.87) where f ′ ≡d f/dy. We must now perform some straightforward (albeit tedious) diﬀerentia-tions. Using the product rule on the three factors on the left-hand side, and making copious use of the chain rule, we obtain f ′y′2 p 1 + y′2 + fy′′ p 1 + y′2 − fy′2y′′ (1 + y′2)3/2 = f ′p 1 + y′2. (6.88) Multiplying through by (1 + y′2)3/2 and simplifying gives fy′′ = f ′(1 + y′2). (6.89) We have completed the ﬁrst step of the proof, namely producing the Euler-Lagrange diﬀer-ential equation. We must now integrate it. Eq. (6.89) happens to be integrable for arbitrary functions f(y). If we multiply through by y′ and rearrange, we obtain y′y′′ 1 + y′2 = f ′y′ f . (6.90) Taking the dx integral of both sides gives (1/2) ln(1+y′2) = ln(f)+C, where C is a constant of integration. Exponentiation then gives (with B ≡e2C) 1 + y′2 = Bf(y)2, (6.91) as we wanted to show. In an actual problem, we would solve this equation for y′, and then separate variables and integrate. But we would need to be given a speciﬁc function f(y) to be able to do this. Second Proof: Our “Lagrangian,” L = f(y) p 1 + y′2, is independent of x. Therefore, in analogy with the conserved energy given in eq. (6.52), the quantity E ≡y′ ∂L ∂y′ −L = −f(y) p 1 + y′2 (6.92) 11The constant B and also one other constant of integration (arising when eq. (6.86) is integrated to obtain y) are determined by the boundary conditions on y(x); see, for example, eq. (6.80). This situation, where the two constants are determined by the values of the function at two points, is slightly diﬀerent from the situation in the physics problems we’ve done where the two constants are determined by the value (that is, the initial position) and the slope (that is, the speed) at just one point. But either way, two given facts must be used. VI-24 CHAPTER 6. THE LAGRANGIAN METHOD is independent of x. Call it 1/ √ B. Then we have easily reproduced eq. (6.91). For practice, you can also prove this lemma by considering x to be a function of y, as we did in the second solution in the minimal-surface example above. 6.9. PROBLEMS VI-25 6.9 Problems Section 6.1: The Euler-Lagrange equations 6.1. Moving plane A block of mass m is held motionless on a frictionless plane of mass M and angle of inclination θ (see Fig. 6.8). The plane rests on a frictionless horizontal surface. The θ m M Figure 6.8 block is released. What is the horizontal acceleration of the plane? (This problem already showed up as Problem 3.8. If you haven’t already done so, try solving it using F = ma. You will then have a greater appreciation for the Lagrangian method.) 6.2. Two falling sticks Two massless sticks of length 2r, each with a mass m ﬁxed at its middle, are hinged at an end. One stands on top of the other, as shown in Fig. 6.9. The bottom end m m r r ε Figure 6.9 of the lower stick is hinged at the ground. They are held such that the lower stick is vertical, and the upper one is tilted at a small angle ϵ with respect to the vertical. They are then released. At this instant, what are the angular accelerations of the two sticks? Work in the approximation where ϵ is very small. 6.3. Pendulum with an oscillating support A pendulum consists of a mass m and a massless stick of length ℓ. The pendulum support oscillates horizontally with a position given by x(t) = A cos(ωt); see Fig. 6.10. l m Figure 6.10 What is the general solution for the angle of the pendulum as a function of time? 6.4. Two masses, one swinging Two equal masses m, connected by a massless string, hang over two pulleys (of negli-gible size), as shown in Fig. 6.11. The left one moves in a vertical line, but the right m r m θ Figure 6.11 one is free to swing back and forth in the plane of the masses and pulleys. Find the equations of motion for r and θ, as shown. Assume that the left mass starts at rest, and the right mass undergoes small oscillations with angular amplitude ϵ (with ϵ ≪1). What is the initial average acceleration (averaged over a few periods) of the left mass? In which direction does it move? 6.5. Inverted pendulum A pendulum consists of a mass m at the end of a massless stick of length ℓ. The other end of the stick is made to oscillate vertically with a position given by y(t) = A cos(ωt), where A ≪ℓ. See Fig. 6.12). It turns out that if ω is large enough, and if the pendulum l m Figure 6.12 is initially nearly upside-down, then surprisingly it will not fall over as time goes by. Instead, it will (sort of) oscillate back and forth around the vertical position. If you want to do the experiment yourself, see the 28th demonstration of the entertaining collection in (Ehrlich, 1994). Find the equation of motion for the angle of the pendulum (measured relative to its upside-down position). Explain why the pendulum doesn’t fall over, and ﬁnd the frequency of the back and forth motion. Section 6.2: The principle of stationary action 6.6. Minimum or saddle (a) In eq. (6.26), let t1 = 0 and t2 = T, for convenience. And let ξ(t) be an easy-to-deal-with “triangular” function, of the form ξ(t) = ½ ϵt/T, 0 ≤t ≤T/2, ϵ(1 −t/T), T/2 ≤t ≤T. (6.93) VI-26 CHAPTER 6. THE LAGRANGIAN METHOD Under what condition is the harmonic-oscillator ∆S in eq. (6.26) negative? (b) Answer the same question, but now with ξ(t) = ϵ sin(πt/T). Section 6.3: Forces of constraint 6.7. Normal force from a plane A mass m slides down a frictionless plane that is inclined at an angle θ. Show, using the method in Section 6.3, that the normal force from the plane is the familiar mg cos θ. Section 6.5: Conservation Laws 6.8. Bead on a stick A stick is pivoted at the origin and is arranged to swing around in a horizontal plane at constant angular speed ω. A bead of mass m slides frictionlessly along the stick. Let r be the radial position of the bead. Find the conserved quantity E given in eq. (6.52). Explain why this quantity is not the energy of the bead. Section 6.6: Noether’s Theorem 6.9. Atwood’s machine Consider the Atwood’s machine shown in Fig. 6.13. The masses are 4m, 3m, and x y 4m 3m m Figure 6.13 m. Let x and y be the heights of the left and right masses, relative to their initial positions. Find the conserved momentum. Section 6.7: Small oscillations 6.10. Hoop and pulley A mass M is attached to a massless hoop (of radius R) which lies in a vertical plane. The hoop is free to rotate about its ﬁxed center. M is tied to a string which winds part way around the hoop, then rises vertically up and over a massless pulley. A mass m hangs on the other end of the string (see Fig. 6.14). Find the equation of motion for R M m Figure 6.14 the angle of rotation of the hoop. What is the frequency of small oscillations? Assume that m moves only vertically, and assume M > m. 6.11. Bead on a rotating hoop A bead is free to slide along a frictionless hoop of radius R. The hoop rotates with constant angular speed ω around a vertical diameter (see Fig. 6.15). Find the equation R ω θ Figure 6.15 of motion for the angle θ shown. What are the equilibrium positions? What is the frequency of small oscillations about the stable equilibrium? There is one value of ω that is rather special; what is it, and why is it special? 6.12. Another bead on a rotating hoop A bead is free to slide along a frictionless hoop of radius r. The plane of the hoop is horizontal, and the center of the hoop travels in a horizontal circle of radius R, with constant angular speed ω, about a given point (see Fig. 6.16). Find the equation of R ω (top view) r θ Figure 6.16 motion for the angle θ shown. Also, ﬁnd the frequency of small oscillations about the equilibrium point. 6.13. Mass on a wheel A mass m is ﬁxed to a given point on the rim of a wheel of radius R that rolls without slipping on the ground. The wheel is massless, except for a mass M located at its center. Find the equation of motion for the angle through which the wheel rolls. For the case where the wheel undergoes small oscillations, ﬁnd the frequency. 6.9. PROBLEMS VI-27 6.14. Pendulum with a free support A mass M is free to slide along a frictionless rail. A pendulum of length ℓand mass m hangs from M (see Fig. 6.17). Find the equations of motion. For small oscillations, M l m Figure 6.17 ﬁnd the normal modes and their frequencies. 6.15. Pendulum support on an inclined plane A mass M is free to slide down a frictionless plane inclined at an angle β. A pendulum of length ℓand mass m hangs from M; see Fig. 6.18 (assume that M extends a M m l β Figure 6.18 short distance beyond the side of the plane, so the pendulum can hang down). Find the equations of motion. For small oscillations, ﬁnd the normal modes and their frequencies. 6.16. Tilting plane A mass M is ﬁxed at the right-angled vertex where a massless rod of length ℓis attached to a very long massless rod (see Fig. 6.19). A mass m is free to move θ M m l x Figure 6.19 frictionlessly along the long rod (assume that it can pass through M). The rod of length ℓis hinged at a support, and the whole system is free to rotate, in the plane of the rods, about the support. Let θ be the angle of rotation of the system, and let x be the distance between m and M. Find the equations of motion. Find the normal modes when θ and x are both very small. 6.17. Rotating curve The curve y(x) = b(x/a)λ is rotated around the y axis with constant frequency ω (see Fig. 6.20). A bead moves frictionlessly along the curve. Find the frequency of small x a y b ω Figure 6.20 oscillations about the equilibrium point. Under what conditions do oscillations exist? (This problem gets a little messy.) 6.18. Motion in a cone A particle slides on the inside surface of a frictionless cone. The cone is ﬁxed with its tip on the ground and its axis vertical. The half-angle at the tip is α (see Fig. 6.21). r0 α Figure 6.21 Let r be the distance from the particle to the axis, and let θ be the angle around the cone. Find the equations of motion. If the particle moves in a circle of radius r0, what is the frequency, ω, of this motion? If the particle is then perturbed slightly from this circular motion, what is the frequency, Ω, of the oscillations about the radius r0? Under what conditions does Ω= ω? 6.19. Double pendulum Consider a double pendulum made of two masses, m1 and m2, and two rods of lengths ℓ1 and ℓ2 (see Fig. 6.22). Find the equations of motion. l l m m 1 1 2 2 Figure 6.22 For small oscillations, ﬁnd the normal modes and their frequencies for the special case ℓ1 = ℓ2 (and consider the cases m1 = m2, m1 ≫m2, and m1 ≪m2). Do the same for the special case m1 = m2 (and consider the cases ℓ1 = ℓ2, ℓ1 ≫ℓ2, and ℓ1 ≪ℓ2). Section 6.8: Other applications 6.20. Shortest distance in a plane In the spirit of Section 6.8, show that the shortest path between two points in a plane is a straight line. VI-28 CHAPTER 6. THE LAGRANGIAN METHOD 6.21. Index of refraction Assume that the speed of light in a given slab of material is proportional to the height above the base of the slab.12 Show that light moves in circular arcs in this material; see Fig. 6.23. You may assume that light takes the path of least time between two Figure 6.23 points (Fermat’s principle of least time). 6.22. Minimal surface Derive the shape of the minimal surface discussed in Section 6.8, by demanding that a cross-sectional “ring” (that is, the region between the planes x = x1 and x = x2) is in equilibrium; see Fig. 6.24. Hint: The tension must be constant throughout the x1 x2 Figure 6.24 surface (assuming that we’re ignoring gravity, which we are). 6.23. Existence of a minimal surface Consider the minimal surface from Section 6.8, and look at the special case where the two rings have the same radius r (see Fig. 6.25). Let 2ℓbe the distance between the r r 2l Figure 6.25 rings. What is the largest value of ℓ/r for which a minimal surface exists? You will need to solve something numerically here. 6.24. The brachistochrone A bead is released from rest at the origin and slides down a frictionless wire that connects the origin to a given point, as shown in Fig. 6.26. You wish to shape the wire x y Figure 6.26 so that the bead reaches the endpoint in the shortest possible time. Let the desired curve be described by the function y(x), with downward taken to be positive. Show that y(x) satisﬁes 1 + y′2 = B y . (6.94) where B is a constant. Then show that x and y may be written as x = a(θ −sin θ), y = a(1 −cos θ). (6.95) This is the parametrization of a cycloid, which is the path taken by a point on the rim of a rolling wheel. 12If you want to make the equivalent statement in terms of the material’s “index of refraction,” commonly denoted by n, then you can say: As a function of the height y, the index n is given by n(y) = y0/y, where y0 is some length that is larger than the height of the slab. This is equivalent to the original statement because the speed of light in a material equals c/n. 6.10. EXERCISES VI-29 6.10 Exercises Section 6.1: The Euler-Lagrange equations 6.25. Spring on a T A rigid T consists of a long rod glued perpendicularly to another rod of length ℓthat is pivoted at the origin. The T rotates around in a horizontal plane with constant frequency ω. A mass m is free to slide along the long rod and is connected to the intersection of the rods by a spring with spring constant k and relaxed length zero (see Fig. 6.27). Find r(t), where r is the position of the mass along the long rod. There is m l k ω (top view) Figure 6.27 a special value of ω; what is it, and why is it special? 6.26. Spring on a T, with gravity Consider the setup in the previous exercise, but now let the T swing around in a vertical plane with constant frequency ω. Find r(t). There is a special value of ω; what is it, and why is it special? (You may assume ω < p k/m.) 6.27. Coﬀee cup and mass A coﬀee cup of mass M is connected to a mass m by a string. The coﬀee cup hangs over a frictionless pulley of negligible size, and the mass m is initially held with the string horizontal, as shown in Fig. 6.28. The mass m is then released. Find the equations m r M Figure 6.28 of motion for r (the length of string between m and the pulley) and θ (the angle that the string to m makes with the horizontal). Assume that m somehow doesn’t run into the string holding the cup up. The coﬀee cup will initially fall, but it turns out that it will reach a lowest point and then rise back up. Write a program (see Section 1.4) that numerically determines the ratio of the r at this lowest point to the r at the start, for a given value of m/M. (To check your program, a value of m/M = 1/10 yields a ratio of about 0.208.) 6.28. Three falling sticks Three massless sticks of length 2r, each with a mass m ﬁxed at its middle, are hinged at their ends, as shown in Fig. 6.29. The bottom end of the lower stick is hinged at m m ε m r r Figure 6.29 the ground. They are held such that the lower two sticks are vertical, and the upper one is tilted at a small angle ϵ with respect to the vertical. They are then released. At this instant, what are the angular accelerations of the three sticks? Work in the approximation where ϵ is very small. (You may want to look at Problem 6.2 ﬁrst.) 6.29. Cycloidal pendulum The standard pendulum frequency of p g/ℓholds only for small oscillations. The frequency becomes smaller as the amplitude grows. It turns out that if you want to build a pendulum whose frequency is independent of the amplitude, you should hang it from the cusp of a cycloid of a certain size, as shown in Fig. 6.30. As the string α Figure 6.30 wraps partially around the cycloid, the eﬀect is to decrease the length of string in the air, which in turn increases the frequency back up to a constant value. In more detail: A cycloid is the path taken by a point on the rim of a rolling wheel. The upside-down cycloid in Fig. 6.30 can be parameterized by (x, y) = R(θ −sin θ, −1 + cos θ), where θ = 0 corresponds to the cusp. Consider a pendulum of length 4R hanging from the cusp, and let α be the angle the string makes with the vertical, as shown. (a) In terms of α, ﬁnd the value of the parameter θ associated with the point where the string leaves the cycloid. VI-30 CHAPTER 6. THE LAGRANGIAN METHOD (b) In terms of α, ﬁnd the length of string touching the cycloid. (c) In terms of α, ﬁnd the Lagrangian. (d) Show that the quantity sin α undergoes simple harmonic motion with frequency p g/4R, independent of the amplitude. (e) In place of parts (c) and (d), solve the problem again by using F = ma. This actually gives a much quicker solution. Section 6.2: The principle of stationary action 6.30. Dropped ball Consider the action, from t = 0 to t = 1, of a ball dropped from rest. From the E-L equation (or from F = ma), we know that y(t) = −gt2/2 yields a stationary value of the action. Show explicitly that the particular function y(t) = −gt2/2 + ϵt(t −1) yields an action that has no ﬁrst-order dependence on ϵ. 6.31. Explicit minimization For a ball thrown upward, guess a solution for y of the form y(t) = a2t2 + a1t + a0. Assuming that y(0) = y(T) = 0, this quickly becomes y(t) = a2(t2 −Tt). Calculate the action between t = 0 and t = T, and show that it is minimized when a2 = −g/2. 6.32. Always a minimum For a ball thrown up in the air, show that the stationary value of the action is always a global minimum. 6.33. Second-order change Let xa(t) ≡x0(t)+aβ(t). Eq. (6.19) gives the ﬁrst derivative of the action with respect to a. Show that the second derivative is d2 da2 S[xa(t)] = Z t2 t1 µ∂2L ∂x2 β2 + 2 ∂2L ∂x∂˙ xβ ˙ β + ∂2L ∂˙ x2 ˙ β2 ¶ dt. (6.96) 6.34. ¨ x dependence Assume that there is ¨ x dependence (in addition to x, ˙ x,t dependence) in the Lagrangian in Theorem 6.1. There will then be the additional term (∂L/∂¨ xa)¨ β in eq. (6.19). It is tempting to integrate this term by parts twice, and then arrive at a modiﬁed form of eq. (6.22): ∂L ∂x0 −d dt µ ∂L ∂˙ x0 ¶ + d2 dt2 µ ∂L ∂¨ x0 ¶ = 0. (6.97) Is this a valid result? If not, where is the error in the reasoning? Section 6.3: Forces of constraint 6.35. Constraint on a circle A bead of mass m slides with speed v around a horizontal hoop of radius R. What force does the hoop apply to the bead? (Ignore gravity.) 6.36. Atwood’s machine Consider the standard Atwood’s machine in Fig. 6.31, with masses m1 and m2. Find m1 m2 Figure 6.31 the tension in the string. 6.10. EXERCISES VI-31 6.37. Cartesian coordinates In eq. (6.35), take two time derivatives of the p x2 + y2 −R = 0 equation to obtain R2(x¨ x + y¨ y) + (x ˙ y −y ˙ x)2 = 0, (6.98) and then combine this with the other two equations to solve for F in terms of x, y, ˙ x, ˙ y. Convert the result to polar coordinates (with θ measured from the vertical) and show that it agrees with eq. (6.32). 6.38. Constraint on a curve Let the horizontal plane be the x-y plane. A bead of mass m slides with speed v along a curve described by the function y = f(x). What force does the curve apply to the bead? (Ignore gravity.) Section 6.5: Conservation Laws 6.39. Bead on stick, using F = ma After doing Problem 6.8, show again that the quantity E is conserved, but now use F = ma. Do this in two ways: (a) Use the ﬁrst of eqs. (3.51). Hint: multiply through by ˙ r. (b) Use the second of eqs. (3.51) to calculate the work done on the bead, and use the work-energy theorem. Section 6.6: Noether’s Theorem 6.40. Atwood’s machine Consider the Atwood’s machine shown in Fig. 6.32. The masses are 4m, 5m, and 3m. x y 4m 5m 3m Figure 6.32 Let x and y be the heights of the right two masses, relative to their initial positions. Use Noether’s theorem to ﬁnd the conserved momentum. (The solution to Problem 6.9 gives some other methods, too.) Section 6.7: Small oscillations 6.41. Spring and a wheel The top of a wheel of mass M and radius R is connected to a spring (at its equilibrium length) with spring constant k, as shown in Fig. 6.33. Assume that all the mass of M k Figure 6.33 the wheel is at its center. If the wheel rolls without slipping, what is the frequency of (small) oscillations? 6.42. Spring on a spoke A spring with spring constant k and relaxed length zero lies along a spoke of a massless wheel of radius R. One end of the spring is attached to the center, and the other end is attached to a mass m that is free to slide along the spoke. When the system is in its equilibrium position with the spring hanging vertically, how far (in terms of R) should the mass hang down (you are free to adjust k) so that for small oscillations, the frequency of the spring oscillations equals the frequency of the rocking motion of the wheel? Assume that the wheel rolls without slipping. 6.43. Oscillating hoop Two equal masses are glued to a massless hoop of radius R that is free to rotate about its center in a vertical plane. The angle between the masses is 2θ, as shown in Fig. 6.34. θ θ R Figure 6.34 Find the frequency of small oscillations. VI-32 CHAPTER 6. THE LAGRANGIAN METHOD 6.44. Oscillating hoop with a pendulum A massless hoop of radius R is free to rotate about its center in a vertical plane. A mass m is attached at one point, and a pendulum of length √ 2R (and also of mass m) is attached at another point 90◦away, as shown in Fig. 6.35. Let θ be the angle of R m m R 2 45 45 Figure 6.35 the hoop relative to the position shown, and let α be the angle of the pendulum with respect to the vertical. Find the normal modes for small oscillations. 6.45. Mass sliding on a rim A mass m is free to slide frictionlessly along the rim of a wheel of radius R that rolls without slipping on the ground. The wheel is massless, except for a mass M located at its center. Find the normal modes for small oscillations. 6.46. Mass sliding on a rim, with a spring Consider the setup in the previous exercise, but now let the mass m be attached to a spring with spring constant k and relaxed length zero, the other end of which is attached to a point on the rim. Assume that the spring is constrained to run along the rim, and assume that the mass can pass freely over the point where the spring is attached to the rim. To keep things from getting too messy here, you can set M = m. (a) Find the frequencies of the normal modes for small oscillations. Check the g = 0 limit, and (if you’ve done the previous exercise) the k = 0 limit. (b) For the special case where g/R = k/m, show that the frequencies can be written as p k/m( √ 5 ± 1)/2. This numerical factor is the golden ratio (and its inverse). Describe what the normal modes look like. 6.47. Vertically rotating hoop A bead is free to slide along a frictionless hoop of radius r. The plane of the hoop is vertical, and the center of the hoop travels in a vertical circle of radius R, with constant angular speed ω, about a given point (see Fig. 6.36). Find the equation of r R ω (side view) θ Figure 6.36 motion for the angle θ shown. For large ω (which implies small θ), ﬁnd the amplitude of the “particular” solution with frequency ω. What happens if r = R? 6.11. SOLUTIONS VI-33 6.11 Solutions 6.1. Moving plane Let x1 be the horizontal coordinate of the plane (with positive x1 to the left), and let x2 be the horizontal coordinate of the block (with positive x2 to the right); see Fig. 6.37. The θ m M x1 x2 Figure 6.37 relative horizontal distance between the plane and the block is x1 + x2, so the height fallen by the block is (x1 + x2) tan θ. The Lagrangian is therefore L = 1 2M ˙ x2 1 + 1 2m ³ ˙ x2 2 + ( ˙ x1 + ˙ x2)2 tan2 θ ´ + mg(x1 + x2) tan θ. (6.99) The equations of motion obtained from varying x1 and x2 are M ¨ x1 + m(¨ x1 + ¨ x2) tan2 θ = mg tan θ, m¨ x2 + m(¨ x1 + ¨ x2) tan2 θ = mg tan θ. (6.100) Note that the diﬀerence of these two equations immediately yields conservation of momentum, M ¨ x1 −m¨ x2 = 0 = ⇒(d/dt)(M ˙ x1 −m ˙ x2) = 0. Eqs. (6.100) are two linear equations in the two unknowns, ¨ x1 and ¨ x2, so we can solve for ¨ x1. After a little simpliﬁcation, we arrive at ¨ x1 = mg sin θ cos θ M + m sin2 θ . (6.101) For some limiting cases, see the remarks in the solution to Problem 3.8. 6.2. Two falling sticks Let θ1(t) and θ2(t) be deﬁned as in Fig. 6.38. Then the position of the bottom mass in θ θ 1 2 m m Figure 6.38 Cartesian coordinates is (r sin θ1, r cos θ1), and the position of the top mass is (2r sin θ1 − r sin θ2, 2r cos θ1 + r cos θ2). So the potential energy of the system is V (θ1, θ2) = mgr(3 cos θ1 + cos θ2). (6.102) The kinetic energy is somewhat more complicated. The kinetic energy of the bottom mass is simply mr2 ˙ θ2 1/2. Taking the derivative of the top mass’s position given above, we ﬁnd that the kinetic energy of the top mass is 1 2mr2³ (2 cos θ1 ˙ θ1 −cos θ2 ˙ θ2)2 + (−2 sin θ1 ˙ θ1 −sin θ2 ˙ θ2)2´ . (6.103) We can simplify this, using the small-angle approximations. The terms involving sin θ are fourth order in the small θ’s, so we can neglect them. Also, we can approximate cos θ by 1, because this entails dropping only terms of at least fourth order. So the top mass’s kinetic energy becomes (1/2)mr2(2 ˙ θ1 −˙ θ2)2. In retrospect, it would have been easier to obtain the kinetic energies of the masses by ﬁrst applying the small-angle approximations to the positions, and then taking the derivatives to obtain the velocities. This strategy shows that both masses move essentially horizontally (initially). You will probably want to use this strategy when solving Exercise 6.28. Using the small-angle approximation cos θ ≈1 −θ2/2 to rewrite the potential energy in eq. (6.102), we have L ≈1 2mr2³ 5 ˙ θ2 1 −4 ˙ θ1 ˙ θ2 + ˙ θ2 2 ´ −mgr ³ 4 −3 2θ2 1 −1 2θ2 2 ´ . (6.104) The equations of motion obtained from varying θ1 and θ2 are, respectively, 5¨ θ1 −2¨ θ2 = 3g r θ1 −2¨ θ1 + ¨ θ2 = g r θ2. (6.105) At the instant the sticks are released, we have θ1 = 0 and θ2 = ϵ. Solving eq. (6.105) for ¨ θ1 and ¨ θ2 gives ¨ θ1 = 2gϵ r , and ¨ θ2 = 5gϵ r . (6.106) VI-34 CHAPTER 6. THE LAGRANGIAN METHOD 6.3. Pendulum with an oscillating support Let θ be deﬁned as in Fig. 6.39. With x(t) = A cos(ωt), the position of the mass m is given θ l m x(t) Figure 6.39 by (X, Y )m = (x + ℓsin θ, −ℓcos θ). (6.107) Taking the derivative to obtain the velocity, we ﬁnd that the square of the speed is V 2 m = ˙ X2 + ˙ Y 2 = ℓ2 ˙ θ2 + ˙ x2 + 2ℓ˙ x ˙ θ cos θ, (6.108) which also follows from applying the law of cosines to the horizontal ˙ x and tangential ℓ˙ θ parts of the velocity vector. The Lagrangian is therefore L = 1 2m(ℓ2 ˙ θ2 + ˙ x2 + 2ℓ˙ x ˙ θ cos θ) + mgℓcos θ. (6.109) The equation of motion for θ is d dt(mℓ2 ˙ θ + mℓ˙ x cos θ) = −mℓ˙ x ˙ θ sin θ −mgℓsin θ = ⇒ ℓ¨ θ + ¨ x cos θ = −g sin θ. (6.110) Plugging in the explicit form of x(t), we have ℓ¨ θ −Aω2 cos(ωt) cos θ + g sin θ = 0. (6.111) In retrospect, this makes sense. Someone in the reference frame of the support, which has horizontal acceleration ¨ x = −Aω2 cos(ωt), may as well be living in a world where the acceler-ation from gravity has a component g downward and a component Aω2 cos(ωt) to the right. Eq. (6.125) is just the F = ma equation in the tangential direction in this accelerating world. A small-angle approximation in eq. (6.111) gives ¨ θ + ω2 0θ = aω2 cos(ωt), (6.112) where ω0 ≡ p g/ℓand a ≡A/ℓ. This equation is simply the equation for a driven oscillator, which we solved in Chapter 4. The solution is θ(t) = aω2 ω2 0 −ω2 cos(ωt) + C cos(ω0t + φ), (6.113) where C and φ are determined by the initial conditions. If ω happens to equal ω0, then it appears that the amplitude goes to inﬁnity. However, as soon as the amplitude becomes large, our small-angle approximation breaks down, and eqs. (6.112) and (6.113) are no longer valid. 6.4. Two masses, one swinging The Lagrangian is L = 1 2m ˙ r2 + 1 2m( ˙ r2 + r2 ˙ θ2) −mgr + mgr cos θ. (6.114) The last two terms are the (negatives of the) potentials of each mass, relative to where they would be if the right mass were located at the right pulley. The equations of motion obtained from varying r and θ are 2¨ r = r ˙ θ2 −g(1 −cos θ), d dt(r2 ˙ θ) = −gr sin θ. (6.115) The ﬁrst equation deals with the forces and accelerations along the direction of the string. The second equation equates the torque from gravity with the change in angular momentum 6.11. SOLUTIONS VI-35 of the right mass. If we do a (coarse) small-angle approximation and keep only terms up to ﬁrst order in θ, we ﬁnd that at t = 0 (using the initial condition, ˙ r = 0), eq. (6.115) becomes ¨ r = 0, ¨ θ + g r θ = 0. (6.116) These equations say that the left mass stays still, and the right mass behaves just like a pendulum. If we want to ﬁnd the leading term in the initial acceleration of the left mass (that is, the leading term in ¨ r), we need to be a little less coarse in our approximation. So let’s keep terms in eq. (6.115) up to second order in θ. We then have at t = 0 (using the initial condition, ˙ r = 0) 2¨ r = r ˙ θ2 −1 2gθ2, ¨ θ + g r θ = 0. (6.117) The second equation still says that the right mass undergoes harmonic motion. We are told that the amplitude is ϵ, so we have θ(t) = ϵ cos(ωt + φ), (6.118) where ω = p g/r. Plugging this into the ﬁrst equation gives 2¨ r = ϵ2g ³ sin2(ωt + φ) −1 2 cos2(ωt + φ) ´ . (6.119) If we average this over a few periods, both sin2 α and cos2 α average to 1/2, so we ﬁnd ¨ ravg = ϵ2g 8 . (6.120) This is a small second-order eﬀect. It is positive, so the left mass slowly begins to climb. 6.5. Inverted pendulum Let θ be deﬁned as in Fig. 6.40. With y(t) = A cos(ωt), the position of the mass m is given θ l m y(t) Figure 6.40 by (X, Y ) = (ℓsin θ, y + ℓcos θ). (6.121) Taking the derivative to obtain the velocity, we ﬁnd that the square of the speed is V 2 = ˙ X2 + ˙ Y 2 = ℓ2 ˙ θ2 + ˙ y2 −2ℓ˙ y ˙ θ sin θ, (6.122) which also follows from applying the law of cosines to the vertical ˙ y and tangential ℓ˙ θ parts of the velocity vector. The Lagrangian is therefore L = 1 2m(ℓ2 ˙ θ2 + ˙ y2 −2ℓ˙ y ˙ θ sin θ) −mg(y + ℓcos θ). (6.123) The equation of motion for θ is d dt ³∂L ∂˙ θ ´ = ∂L ∂θ = ⇒ ℓ¨ θ −¨ y sin θ = g sin θ. (6.124) Plugging in the explicit form of y(t), we have ℓ¨ θ + sin θ ³ Aω2 cos(ωt) −g ´ = 0. (6.125) In retrospect, this makes sense. Someone in the reference frame of the support, which has vertical acceleration ¨ y = −Aω2 cos(ωt), may as well be living in a world where the acceleration VI-36 CHAPTER 6. THE LAGRANGIAN METHOD 0.2 0.4 0.6 0.8 1 1.2 1.4 t 0.1 0.05 0.05 0.1 theta 0.2 0.4 0.6 0.8 1 1.2 1.4 t 0.25 0.5 0.75 1 1.25 1.5 1.75 theta θ θ t t Figure 6.41 2 4 6 8 t 0.1 0.05 0.05 0.1 theta 0.02 0.04 0.06 0.08 0.1 t 0.098 0.0985 0.099 0.0995 theta θ t θ t Figure 6.42 from gravity is g −Aω2 cos(ωt) downward. Eq. (6.125) is just the F = ma equation in the tangential direction in this accelerating world. Assuming θ is small, we may set sin θ ≈θ, which gives ¨ θ + θ ³ aω2 cos(ωt) −ω2 0 ´ = 0, (6.126) where ω0 ≡ p g/ℓ, and a ≡A/ℓ. Eq. (6.126) cannot be solved exactly, but we can still get a good idea of how θ depends on time. We can do this both numerically and (approximately) analytically. Fig. 6.41 shows how θ depends on time for parameters with values ℓ= 1 m, A = 0.1 m, and g = 10 m/s2. So a = 0.1, and ω2 0 = 10 s−2. We produced these plots numerically using eq. (6.126), with the initial conditions of θ(0) = 0.1 and ˙ θ(0) = 0. In the ﬁrst plot, ω = 10 s−1. And in the second plot, ω = 100 s−1. The stick falls over in ﬁrst case, but it undergoes oscillatory motion in the second case. Apparently, if ω is large enough the stick won’t fall over. Let’s now explain this phenomenon analytically. At ﬁrst glance, it’s rather surprising that the stick stays up. It seems like the average (over a few periods of the ω oscillations) of the tangential acceleration in eq. (6.126), namely −θ(aω2 cos(ωt) −ω2 0), equals the positive quantity θω2 0, because the cos(ωt) term averages to zero (or so it appears). So you might think that there is a net force making θ increase, causing the stick fall over. The fallacy in this reasoning is that the average of the −aω2θ cos(ωt) term is not zero, because θ undergoes tiny oscillations with frequency ω, as seen in the second plot in Fig. 6.42. Both of these plots have a = 0.005, ω2 0 = 10 s−2, and ω = 1000 s−1 (we’ll work with small a and large ω from now on; more on this below). The second plot is a zoomed-in version of the ﬁrst one near t = 0. The important point here is that the tiny oscillations in θ shown in the second plot are correlated with cos(ωt). It turns out that the θ value at the t where cos(ωt) = 1 is larger than the θ value at the t where cos(ωt) = −1. So there is a net negative contribution to the −aω2θ cos(ωt) part of the acceleration. And it may indeed be large enough to keep the pendulum up, as we will now show. 6.11. SOLUTIONS VI-37 To get a handle on the −aω2θ cos(ωt) term, let’s work in the approximation where ω is large and a ≡A/ℓis small. More precisely, we will assume a ≪1 and aω2 ≫ω2 0, for reasons we will explain below. Look at one of the little oscillations in the second plot in Fig. 6.42. These oscillations have frequency ω, because they are due to the support moving up and down. When the support moves up, θ increases; and when the support moves down, θ decreases. Since the average position of the pendulum doesn’t change much over one of these small periods, we can look for an approximate solution to eq. (6.126) of the form θ(t) ≈C + b cos(ωt), (6.127) where b ≪C. C will change over time, but on the scale of 1/ω it is essentially constant, if a ≡A/ℓis small enough. Plugging this guess for θ into eq. (6.126), and using a ≪1 and aω2 ≫ω2 0, we ﬁnd −bω2 cos(ωt) + Caω2 cos(ωt) = 0, to leading order.13 So we must have b = aC. Our approximate solution for θ is therefore θ ≈C¡ 1 + a cos(ωt)¢ . (6.128) Let’s now determine how C gradually changes with time. From eq. (6.126), the average acceleration of θ, over a period T = 2π/ω, is ¨ θ = −θ ³ aω2 cos(ωt) −ω2 0 ´ ≈ −C ³ 1 + a cos(ωt) ´³ aω2 cos(ωt) −ω2 0 ´ = −C ³ a2ω2cos2(ωt) −ω2 0 ´ = −C µ a2ω2 2 −ω2 0 ¶ ≡ −CΩ2, (6.129) where Ω= r a2ω2 2 −g ℓ. (6.130) But if we take two derivatives of eq. (6.127), we see that ¨ θ simply equals ¨ C. Equating this value of ¨ θ with the one in eq. (6.129) gives ¨ C(t) + Ω2C(t) ≈0. (6.131) This equation describes nice simple-harmonic motion. Therefore, C oscillates sinusoidally with the frequency Ωgiven in eq. (6.130). This is the overall back and forth motion seen in the ﬁrst plot in Fig. 6.42. Note that we must have aω > √ 2ω0 if this frequency is to be real so that the pendulum stays up. Since we have assumed a ≪1, we see that a2ω2 > 2ω2 0 implies aω2 ≫ω2 0, which is consistent with our initial assumption above. If aω ≫ω0, then eq. (6.130) gives Ω≈aω/ √ 2. This is the case if we change the setup and just have the pendulum lie ﬂat on a horizontal table where the acceleration from gravity is zero. In this limit where g is irrelevant, dimensional analysis implies that the frequency of the C oscillations must be a multiple of ω, because ω is the only quantity in the problem with units of frequency. It just so happens that the multiple is a/ √ 2. 13The reasons for the a ≪1 and aω2 ≫ω2 0 qualiﬁcations are the following. If aω2 ≫ω2 0, then the aω2 cos(ωt) term dominates the ω2 0 term in eq. (6.126). The one exception to this is when cos(ωt) ≈0, but this occurs for a negligibly small amount of time if aω2 ≫ω2 0. If a ≪1, then we can legally ignore the ¨ C term when eq. (6.127) is substituted into eq. (6.126). This is true because we will ﬁnd below in eq. (6.129) that our assumptions lead to ¨ C being roughly proportional to Ca2ω2. Since the other terms in eq. (6.126) are proportional to Caω2, we need a ≪1 in order for the ¨ C term to be negligible. In short, a ≪1 is the condition under which C varies slowly on the time scale of 1/ω. VI-38 CHAPTER 6. THE LAGRANGIAN METHOD As a double check that we haven’t messed up somewhere, the Ωvalue resulting from the parameters in Fig. 6.42 (namely a = 0.005, ω2 0 = 10 s−2, and ω = 1000 s−1) is Ω= p 25/2 −10 = 1.58 s−1. This corresponds to a period of 2π/Ω≈3.97 s. And indeed, from the ﬁrst plot in the ﬁgure, the period looks to be about 4 s (or maybe a hair less). For more on the inverted pendulum, see (Butikov, 2001). 6.6. Minimum or saddle (a) For the given ξ(t), the integrand in eq. (6.26) is symmetric around the midpoint, so we obtain ∆S = Z T/2 0 µ m ³ ϵ T ´2 −k ³ϵt T ´2¶ dt = mϵ2 2T −kϵ2T 24 . (6.132) This is negative if T > p 12m/k ≡2 √ 3/ω. Since the period of the oscillation is τ ≡2π/ω, we see that T must be greater than ( √ 3/π)τ in order for ∆S to be negative, assuming that we are using the given triangular function for ξ. (b) With ξ(t) = ϵ sin(πt/T), the integrand in eq. (6.26) becomes ∆S = 1 2 Z T 0 Ã m µ ϵπ T cos(πt/T) ¶2 −k ³ ϵ sin(πt/T) ´2! dt. = mϵ2π2 4T −kϵ2T 4 , (6.133) where we have used the fact that the average value of sin2 θ and cos2 θ over half of a period is 1/2 (or you can just do the integrals). This result for ∆S is negative if T > π p m/k ≡π/ω = τ/2, where τ is the period. Remark: It turns out that the ξ(t) ∝sin(πt/T) function gives the best chance of making ∆S negative. You can show this by invoking a theorem from Fourier analysis that says that any function satisfying ξ(0) = ξ(T) = 0 can be written as the sum ξ(t) = P∞ 1 cn sin(nπt/T), where the cn are numerical coeﬃcients. When this sum is plugged into eq. (6.26), you can show that all the cross terms (terms involving two diﬀerent values of n) integrate to zero. Using the fact that the average value of sin2 θ and cos2 θ is 1/2, the rest of the integral yields ∆S = 1 4 ∞ X 1 c2 n µ mπ2n2 T −kT ¶ . (6.134) In order to obtain the smallest value of T that can make this sum negative, we want only the n = 1 term to exist. We then have ξ(t) = c1 sin(πt/T), and eq. (6.134) reduces to eq. (6.133), as it should. As mentioned in Remark 4 in Section 6.2, it is always possible to make ∆S positive by picking a ξ(t) function that is small but wiggles very fast. Therefore, we see that for a harmonic oscillator, if T > τ/2, then the stationary value of S is a saddle point (some ξ’s make ∆S positive, and some make it negative), but if T < τ/2, then the stationary value of S is a minimum (all ξ’s make ∆S positive). In the latter case, the point is that T is small enough so that there is no way for ξ to get large, without making ˙ ξ large also. ♣ 6.7. Normal force from a plane First solution: The most convenient coordinates in this problem are w and z, where w is the distance upward along the plane, and z is the distance perpendicularly away from it. The Lagrangian is then 1 2m( ˙ w2 + ˙ z2) −mg(w sin θ + z cos θ) −V (z), (6.135) where V (z) is the (very steep) constraining potential. The two equations of motion are m ¨ w = −mg sin θ, m¨ z = −mg cos θ −dV dz . (6.136) 6.11. SOLUTIONS VI-39 At this point we invoke the constraint z = 0. So ¨ z = 0, and the second equation gives Fc ≡−V ′(0) = mg cos θ, (6.137) as desired. We also obtain the usual result, ¨ w = −g sin θ. Second solution: We can also solve this problem by using the horizontal and vertical coordinates, x and y. We’ll choose (x, y) = (0, 0) to be at the top of the plane; see Fig. 6.43. θ y x m Figure 6.43 The (very steep) constraining potential is V (z), where z ≡x sin θ + y cos θ is the distance from the mass to the plane (as you can verify). The Lagrangian is then L = 1 2m( ˙ x2 + ˙ y2) −mgy −V (z) (6.138) Keeping in mind that z ≡x sin θ + y cos θ, the two equations of motion are (using the chain rule) m¨ x = −dV dz ∂z ∂x = −V ′(z) sin θ, m¨ y = −mg −dV dz ∂z ∂y = −mg −V ′(z) cos θ. (6.139) At this point we invoke the constraint condition z = 0 = ⇒x = −y cot θ. This condition, along with the two E-L equations, allows us to solve for the three unknowns, ¨ x, ¨ y, and V ′(0). Using ¨ x = −¨ y cot θ in eq. (6.139), we ﬁnd ¨ x = g cos θ sin θ, ¨ y = −g sin2 θ, Fc ≡−V ′(0) = mg cos θ. (6.140) The ﬁrst two results here are simply the horizontal and vertical components of the acceleration along the plane, which is g sin θ. 6.8. Bead on a stick There is no potential energy here, so the Lagrangian consists of just the kinetic energy, T, which comes from the radial and tangential motions: L = T = 1 2m ˙ r2 + 1 2mr2ω2. (6.141) Eq. (6.52) therefore gives E = 1 2m ˙ r2 −1 2mr2ω2. (6.142) Claim 6.3 says that this quantity is conserved, because ∂L/∂t = 0. But it is not the energy of the bead, due to the minus sign in the second term. The point here is that in order to keep the stick rotating at a constant angular speed, there must be an external force acting on it. This force in turn causes work to be done on the bead, thereby increasing its kinetic energy. The kinetic energy T is therefore not conserved. From eqs. (6.141) and (6.142), we see that E = T −mr2ω2 is the quantity that is constant in time. See Exercise 6.39 for some F = ma ways to show that the quantity E in eq. (6.142) is conserved. 6.9. Atwood’s machine First solution: If the left mass goes up by x and the right mass goes up by y, then conservation of string says that the middle mass must go down by x + y. Therefore, the Lagrangian of the system is L = 1 2(4m) ˙ x2 + 1 2(3m)(−˙ x −˙ y)2 + 1 2m ˙ y2 − ³ (4m)gx + (3m)g(−x −y) + mgy ´ = 7 2m ˙ x2 + 3m ˙ x ˙ y + 2m ˙ y2 −mg(x −2y). (6.143) VI-40 CHAPTER 6. THE LAGRANGIAN METHOD This is invariant under the transformation x →x + 2ϵ and y →y + ϵ. Hence, we can use Noether’s theorem, with Kx = 2 and Ky = 1. The conserved momentum is then P = ∂L ∂˙ x Kx + ∂L ∂˙ y Ky = m(7 ˙ x + 3 ˙ y)(2) + m(3 ˙ x + 4 ˙ y)(1) = m(17 ˙ x + 10 ˙ y). (6.144) This P is constant. In particular, if the system starts at rest, then ˙ x always equals −(10/17) ˙ y. Second solution: The Euler-Lagrange equations are, from eq. (6.143), 7m¨ x + 3m¨ y = −mg, 3m¨ x + 4m¨ y = 2mg. (6.145) Adding the second equation to twice the ﬁrst gives 17m¨ x + 10m¨ y = 0 = ⇒ d dt ³ 17m ˙ x + 10m ˙ y ´ = 0. (6.146) Third solution: We can also solve this problem using F = ma. Since the tension T is the same throughout the rope, we see that the three F = dP/dt equations are 2T −4mg = dP4m dt , 2T −3mg = dP3m dt , 2T −mg = dPm dt . (6.147) The three forces depend on only two parameters, so there must be some combination of them that adds up to zero. If we set a(2T −4mg) + b(2T −3mg) + c(2T −mg) = 0, then we have a + b + c = 0 and 4a + 3b + c = 0, which is satisﬁed by a = 2, b = −3, and c = 1. Therefore, 0 = d dt(2P4m −3P3m + Pm) = d dt ³ 2(4m) ˙ x −3(3m)(−˙ x −˙ y) + m ˙ y ´ = d dt(17m ˙ x + 10m ˙ y). (6.148) 6.10. Hoop and pulley Let the radius to M make an angle θ with the vertical (see Fig. 6.44). Then the coordinates θ R M m Figure 6.44 of M relative to the center of the hoop are R(sin θ, −cos θ). The height of m, relative to its position when M is at the bottom of the hoop, is y = −Rθ. The Lagrangian is therefore (and yes, we’ve chosen a diﬀerent y = 0 reference point for each mass, but such a deﬁnition only changes the potential by a constant amount, which is irrelevant) L = 1 2(M + m)R2 ˙ θ2 + MgR cos θ + mgRθ. (6.149) The equation of motion is then (M + m)R¨ θ = g(m −M sin θ). (6.150) This is just F = ma along the direction of the string (because Mg sin θ is the tangential component of the gravitational force on M). Equilibrium occurs when ˙ θ = ¨ θ = 0. From eq. (6.150), we see that this happens at sin θ0 = m/M. Letting θ ≡θ0 + δ, and expanding eq. (6.150) to ﬁrst order in δ, gives ¨ δ + µ Mg cos θ0 (M + m)R ¶ δ = 0. (6.151) The frequency of small oscillations is therefore ω = r M cos θ0 M + m q g R = ³M −m M + m ´1/4 q g R , (6.152) 6.11. SOLUTIONS VI-41 where we have used cos θ0 = p 1 −sin θ2 0. Remarks: If M ≫m, then θ0 ≈0, and ω ≈ p g/R. This makes sense, because m can be ignored, so M essentially oscillates around the bottom of the hoop, just like a pendulum of length R. If M is only slightly greater than m, then θ0 ≈π/2, and ω ≈0. This also makes sense, because if θ ≈π/2, then the restoring force g(m −M sin θ) doesn’t change much as θ changes (the derivative of sin θ is zero at θ = π/2), so it’s as if we have a pendulum in a very weak gravitational ﬁeld. We can actually derive the frequency in eq. (6.152) without doing any calculations. Look at M at the equilibrium position. The tangential forces on it cancel, and the radially inward force from the hoop must be Mg cos θ0 to balance the radial outward component of the gravitational force. Therefore, for all the mass M knows, it is sitting at the bottom of a hoop of radius R in a world where gravity has strength g′ = g cos θ0. The general formula for the frequency of a pendulum (as you can quickly show) is ω = p F ′/M′R, where F ′ is the gravitational force (which is Mg′ here), and M′ is the total mass being accelerated (which is M + m here). This gives the ω in eq. (6.152). ♣ 6.11. Bead on a rotating hoop Breaking the velocity up into the component along the hoop plus the component perpendic-ular to the hoop, we ﬁnd L = 1 2m(ω2R2 sin2 θ + R2 ˙ θ2) + mgR cos θ. (6.153) The equation of motion is then R¨ θ = sin θ(ω2R cos θ −g). (6.154) The F = ma interpretation of this is that the component of gravity pulling downward along the hoop accounts for the acceleration along the hoop plus the component of the centripetal acceleration along the hoop. Equilibrium occurs when ˙ θ = ¨ θ = 0. The right-hand side of eq. (6.154) equals zero when either sin θ = 0 (that is, θ = 0 or θ = π) or cos θ = g/(ω2R). Since cos θ must be less than or equal to 1, this second condition is possible only if ω2 ≥g/R. So we have two cases: • If ω2 < g/R, then θ = 0 and θ = π are the only equilibrium points. The θ = π case is unstable. This is fairly intuitive, but it can also be seen mathematically by letting θ ≡π + δ, where δ is small. Eq. (6.154) then becomes ¨ δ −δ(ω2 + g/R) = 0. (6.155) The coeﬃcient of δ is negative, so δ undergoes exponential instead of oscillatory motion. The θ = 0 case turns out to be stable. For small θ, eq. (6.154) becomes ¨ θ + θ(g/R −ω2) = 0. (6.156) The coeﬃcient of θ is positive, so we have sinusoidal solutions. The frequency of small oscillations is p g/R −ω2. This goes to zero as ω → p g/R. • If ω2 ≥g/R, then θ = 0, θ = π, and cos θ0 ≡g/(ω2R) are all equilibrium points. The θ = π case is again unstable, by looking at eq. (6.155). And the θ = 0 case is also unstable, because the coeﬃcient of θ in eq. (6.156) is now negative (or zero, if ω2 = g/R). Therefore, cos θ0 ≡g/(ω2R) is the only stable equilibrium. To ﬁnd the frequency of small oscillations, let θ ≡θ0 + δ in eq. (6.154), and expand to ﬁrst order in δ. Using cos θ0 ≡g/(ω2R), we ﬁnd ¨ δ + ω2 sin2 θ0δ = 0. (6.157) The frequency of small oscillations is therefore ω sin θ0 = p ω2 −g2/ω2R2. The frequency ω = p g/R is the critical frequency above which there is a stable equi-librium at θ ̸= 0, that is, above which the mass wants to move away from the bottom of the hoop. VI-42 CHAPTER 6. THE LAGRANGIAN METHOD Remark: This frequency of small oscillations goes to zero as ω → p g/R. And it approximately equals ω as ω →∞. This second limit can be viewed in the following way. For very large ω, gravity isn’t important, and the bead feels a centripetal force (the normal force from the hoop) essentially equal to mω2R as it moves near θ = π/2. So for all the bead knows, it is a pendulum of length R in a world where “gravity” pulls sideways with a force mω2R ≡mg′ (outward, so that it is approximately canceled by the inward-pointing normal force, just as the downward gravitational force is approximately canceled by the upward tension in a regular pendulum). The frequency of such a pendulum is p g′/R = p ω2R/R = ω. ♣ 6.12. Another bead on a rotating hoop With the angles ωt and θ deﬁned as in Fig. 6.45, the Cartesian coordinates for the bead are θ m ωt r R (top view) Figure 6.45 (x, y) = ³ R cos ωt + r cos(ωt + θ), R sin ωt + r sin(ωt + θ) ´ . (6.158) The velocity is then (x, y) = ³ −ωR sin ωt −r(ω + ˙ θ) sin(ωt + θ), ωR cos ωt + r(ω + ˙ θ) cos(ωt + θ) ´ . (6.159) The square of the speed is therefore v2 = R2ω2 + r2(ω + ˙ θ)2 +2Rrω(ω + ˙ θ) ³ sin ωt sin(ωt + θ) + cos ωt cos(ωt + θ) ´ = R2ω2 + r2(ω + ˙ θ)2 + 2Rrω(ω + ˙ θ) cos θ. (6.160) This speed can also be obtained by using the law of cosines to add the velocity of the center of the hoop to the velocity of the bead with respect to the center (as you can show). There is no potential energy, so the Lagrangian is simply L = mv2/2. The equation of motion is then, as you can show, r¨ θ + Rω2 sin θ = 0. (6.161) Equilibrium occurs when ˙ θ = ¨ θ = 0, so eq. (6.161) tells us that the equilibrium is located at θ = 0, which makes intuitive sense. (Another solution is θ = π, but that’s an unstable equilibrium.) A small-angle approximation in eq. (6.161) gives ¨ θ + (R/r)ω2θ = 0, so the frequency of small oscillations is Ω= ω p R/r. Remarks: If R ≪r, then Ω≈0. This makes sense, because the frictionless hoop is essentially not moving. If R = r, then Ω= ω. If R ≫r, then Ωis very large. In this case, we can double-check the Ω= ω p R/r result in the following way. In the accelerating frame of the hoop, the bead feels a centrifugal force (discussed in Chapter 10) of m(R + r)ω2. For all the bead knows, it is in a gravitational ﬁeld with strength g′ ≡(R + r)ω2. So the bead (which acts like a pendulum of length r), oscillates with a frequency equal to r g′ r = r (R + r)ω2 r ≈ω r R r (for R ≫r). (6.162) Note that if we try to use this “eﬀective gravity” argument as a double check for smaller values of R, we get the wrong answer. For example, if R = r, we obtain an oscillation frequency of ω p 2R/r, instead of the correct value ω p R/r. This is because in reality the centrifugal force fans out near the equilibrium point, while our “eﬀective gravity” argument assumes that the ﬁeld lines are parallel (and so it gives a frequency that is too large). ♣ 6.13. Mass on a wheel Let the angle θ be deﬁned as in Fig. 6.46, with the convention that θ is positive if M is to θ R M m Figure 6.46 the right of m. Then the position of m in Cartesian coordinates, relative to the point where m would be in contact with the ground, is (x, y)m = R(θ −sin θ, 1 −cos θ). (6.163) 6.11. SOLUTIONS VI-43 We have used the non-slipping condition to say that the present contact point is a distance Rθ to the right of where m would be in contact with the ground. Diﬀerentiating eq. (6.163), we ﬁnd that the square of m’s speed is v2 m = 2R2 ˙ θ2(1 −cos θ). The position of M is (x, y)M = R(θ, 1), so the square of its speed is v2 M = R2 ˙ θ2. The Lagrangian is therefore L = 1 2MR2 ˙ θ2 + mR2 ˙ θ2(1 −cos θ) + mgR cos θ, (6.164) where we have measured both potential energies relative to the height of M. The equation of motion is MR¨ θ + 2mR¨ θ(1 −cos θ) + mR ˙ θ2 sin θ + mg sin θ = 0. (6.165) In the case of small oscillations, we can use cos θ ≈1 −θ2/2 and sin θ ≈θ. The second and third terms in eq. (6.165) are then third order in θ and can be neglected (basically, the middle term in eq. (6.164), which is the kinetic energy of m, is negligible), so we ﬁnd ¨ θ + ³ mg MR ´ θ = 0. (6.166) The frequency of small oscillations is therefore ω = q m M q g R . (6.167) Remarks: If M ≫m, then ω →0. This makes sense. If m ≫M, then ω →∞. This also makes sense, because the huge mg force makes the situation similar to one where the wheel is bolted to the ground, in which case the wheel vibrates with a high frequency. Eq. (6.167) can actually be derived in a much quicker way, using torque. For small oscillations, the gravitational force on m produces a torque of −mgRθ around the contact point on the ground. For small θ, m has essentially no moment of inertia around the contact point, so the total moment of inertia is simply MR2. Therefore, τ = Iα gives −mgRθ = MR2¨ θ, from which the result follows. ♣ 6.14. Pendulum with a free support Let x be the coordinate of M, and let θ be the angle of the pendulum (see Fig. 6.47). Then the x θ M l m Figure 6.47 position of the mass m in Cartesian coordinates is (x+ℓsin θ, −ℓcos θ). Taking the derivative to ﬁnd the velocity, and then squaring to ﬁnd the speed, gives v2 m = ˙ x2 + ℓ2 ˙ θ2 + 2ℓ˙ x ˙ θ cos θ. The Lagrangian is therefore L = 1 2M ˙ x2 + 1 2m( ˙ x2 + ℓ2 ˙ θ2 + 2ℓ˙ x ˙ θ cos θ) + mgℓcos θ. (6.168) The equations of motion obtained from varying x and θ are (M + m)¨ x + mℓ¨ θ cos θ −mℓ˙ θ2 sin θ = 0, ℓ¨ θ + ¨ x cos θ + g sin θ = 0. (6.169) If θ is small, we can use the small angle approximations, cos θ ≈1 −θ2/2 and sin θ ≈θ. Keeping only the terms that are ﬁrst-order in θ, we obtain (M + m)¨ x + mℓ¨ θ = 0, ¨ x + ℓ¨ θ + gθ = 0. (6.170) The ﬁrst equation expresses momentum conservation. Integrating it twice gives x = − ³ mℓ M + m ´ θ + At + B. (6.171) The second equation is F = ma in the tangential direction. Eliminating ¨ x from eq. (6.170) gives ¨ θ + ³M + m M ´ g ℓθ = 0. (6.172) VI-44 CHAPTER 6. THE LAGRANGIAN METHOD Therefore, θ(t) = C cos(ωt + φ), where ω = q 1 + m M qg ℓ. (6.173) The general solutions for θ and x are therefore θ(t) = C cos(ωt + φ), x(t) = −Cmℓ M + m cos(ωt + φ) + At + B. (6.174) The constant B is irrelevant, so we’ll ignore it. The two normal modes are: • A = 0: In this case, x = −θmℓ/(M + m). Both masses oscillate with the frequency ω given in eq. (6.173), always moving in opposite directions. The center of mass does not move (as you can verify). • C = 0: In this case, θ = 0 and x = At. The pendulum hangs vertically, with both masses moving horizontally at the same speed. The frequency of oscillations is zero in this mode. Remarks: If M ≫m, then ω = p g/ℓ, as expected, because the support essentially stays still. If m ≫M, then ω → p m/M p g/ℓ→∞. This makes sense, because the tension in the rod is very large. We can actually be quantitative about this limit. For small oscillations and for m ≫M, the tension of mg in the rod produces a sideways force of mgθ on M. So the horizontal F = Ma equation for M is mgθ = M ¨ x, But x ≈−ℓθ in this limit, so we have mgθ = −Mℓ¨ θ, which gives the desired frequency. ♣ 6.15. Pendulum support on an inclined plane Let z be the coordinate of M along the plane, and let θ be the angle of the pendulum (see Fig. 6.48). In Cartesian coordinates, the positions of M and m are θ z M m l β Figure 6.48 (x, y)M = (z cos β, −z sin β), (x, y)m = (z cos β + ℓsin θ, −z sin β −ℓcos θ). (6.175) Diﬀerentiating these positions, we ﬁnd that the squares of the speeds are v2 M = ˙ z2, v2 m = ˙ z2 + ℓ2 ˙ θ2 + 2ℓ˙ z ˙ θ(cos β cos θ −sin β sin θ). (6.176) The Lagrangian is therefore 1 2M ˙ z2 + 1 2m ³ ˙ z2 + ℓ2 ˙ θ2 + 2ℓ˙ z ˙ θ cos(θ + β) ´ + Mgz sin β + mg(z sin β + ℓcos θ). (6.177) The equations of motion obtained from varying z and θ are (M + m)¨ z + mℓ ³ ¨ θ cos(θ + β) −˙ θ2 sin(θ + β) ´ = (M + m)g sin β, ℓ¨ θ + ¨ z cos(θ + β) = −g sin θ. (6.178) Let us now consider small oscillations about the equilibrium point (where ¨ θ = ˙ θ = 0). We must ﬁrst determine where this point is. The ﬁrst equation above gives ¨ z = g sin β. The second equation then gives g sin β cos(θ + β) = −g sin θ. By expanding the cosine term, we ﬁnd tan θ = −tan β, so θ = −β. (θ = π −β is also a solution, but this is an unstable equilibrium.) The equilibrium position of the pendulum is therefore where the string is perpendicular to the plane.14 14This makes sense. The tension in the string is perpendicular to the plane, so for all the pendulum bob knows, it may as well be sliding down a plane parallel to the given one, a distance ℓaway. Given the same initial speed, the two masses slide down their two “planes” with equal speeds at all times. 6.11. SOLUTIONS VI-45 To ﬁnd the normal modes and frequencies for small oscillations, let θ ≡−β + δ, and expand eq. (6.178) to ﬁrst order in δ. Letting ¨ η ≡¨ z −g sin β for convenience, we obtain (M + m)¨ η + mℓ¨ δ = 0, ¨ η + ℓ¨ δ + (g cos β)δ = 0. (6.179) Using the determinant method (or using the method in Problem 6.14; either way works), we ﬁnd the frequencies of the normal modes to be ω1 = 0, and ω2 = q 1 + m M r g cos β ℓ . (6.180) These are the same as the frequencies in the previous problem (where M moves horizontally), but with g cos β in place of g; compare eq. (6.179) with eq. (6.170).15 Looking at eq. (6.174), and recalling the deﬁnition of η, we see that the general solutions for θ and z are θ(t) = −β + C cos(ωt + φ), z(t) = −Cmℓ M + m cos(ωt + φ) + g sin β 2 t2 + At + B. (6.181) The constant B is irrelevant, so we’ll ignore it. The basic diﬀerence between these normal modes and the ones in the previous problem is the acceleration down the plane. If you go to a frame that accelerates down the plane at g sin β, and if you tilt your head at an angle β and accept the fact that g′ = g cos β in your world, then the setup is identical to the one in the previous problem. 6.16. Tilting plane Relative to the support, the positions of the masses are (x, y)M = (ℓsin θ, −ℓcos θ), (x, y)m = (ℓsin θ + x cos θ, −ℓcos θ + x sin θ). (6.182) Diﬀerentiating these positions, we ﬁnd that the squares of the speeds are v2 M = ℓ2 ˙ θ2, v2 m = (ℓ˙ θ + ˙ x)2 + x2 ˙ θ2. (6.183) You can also obtain v2 m by noting that (ℓ˙ θ + ˙ x) is the speed along the long rod, and x ˙ θ is the speed perpendicular to it. The Lagrangian is L = 1 2Mℓ2 ˙ θ2 + 1 2m ³ (ℓ˙ θ + ˙ x)2 + x2 ˙ θ2´ + Mgℓcos θ + mg(ℓcos θ −x sin θ). (6.184) The equations of motion obtained from varying x and θ are ℓ¨ θ + ¨ x = x ˙ θ2 −g sin θ, (6.185) Mℓ2¨ θ + mℓ(ℓ¨ θ + ¨ x) + mx2¨ θ + 2mx ˙ x ˙ θ = −(M + m)gℓsin θ −mgx cos θ. Let us now consider the case where both x and θ are small (or more precisely, θ ≪1 and x/ℓ≪1). Expanding eq. (6.185) to ﬁrst order in θ and x/ℓgives (ℓ¨ θ + ¨ x) + gθ = 0, Mℓ(ℓ¨ θ + gθ) + mℓ(ℓ¨ θ + ¨ x) + mgℓθ + mgx = 0. (6.186) We can simplify these a bit. Using the ﬁrst equation to substitute −gθ for (ℓ¨ θ + ¨ x), and also −¨ x for (ℓ¨ θ + gθ), in the second equation gives ℓ¨ θ + ¨ x + gθ = 0, −Mℓ¨ x + mgx = 0. (6.187) The normal modes can be found using the determinant method, or we can ﬁnd them just by inspection. The second equation says that either x(t) ≡0, or x(t) = A cosh(αt + β), where α = p mg/Mℓ. So we have two cases: 15This makes sense, because in a frame that accelerates down the plane at g sin β, the only external force on the masses is an eﬀective gravity force of g cos β perpendicular to the plane. As far as M and m are concerned, they live in a world where gravity pulls “downward” (perpendicular to the plane) with strength g′ = g cos β. VI-46 CHAPTER 6. THE LAGRANGIAN METHOD • If x(t) = 0, then the ﬁrst equation in (6.187) says that the normal mode is µ θ x ¶ = B µ 1 0 ¶ cos(ωt + φ), (6.188) where ω ≡ p g/ℓ. This mode is fairly clear. With the proper initial conditions, m will stay right where M is. The normal force from the long rod will be exactly what is needed in order for m to undergo the same oscillatory motion as M. The two masses may as well be two pendulums of length ℓswinging side by side. • If x(t) = A cosh(αt + β), then the ﬁrst equation in (6.187) can be solved (by guessing a particular solution for θ of the same form) to give the normal mode, µ θ x ¶ = C µ −m ℓ(M + m) ¶ cosh(αt + β), (6.189) where α = p mg/Mℓ. This mode is not as clear. And indeed, its range of validity is rather limited. The exponential behavior will quickly make x and θ large, and thus outside the validity of our small-variable approximations. You can show that in this mode the center of mass remains ﬁxed, directly below the pivot. This can occur, for example, by having m move down to the right as the rods rotate and swing M up to the left. There is no oscillation in this mode; the positions keep growing. 6.17. Rotating curve The speed along the curve is ˙ x p 1 + y′2, and the speed perpendicular to the curve is ωx. So the Lagrangian is L = 1 2m ³ ω2x2 + ˙ x2(1 + y′2) ´ −mgy, (6.190) where y(x) = b(x/a)λ. The equation of motion is then d dt ³∂L ∂˙ x ´ = ∂L ∂x = ⇒ ¨ x(1 + y′2) + ˙ x2y′y′′ = ω2x −gy′. (6.191) Equilibrium occurs when ˙ x = ¨ x = 0, so eq. (6.191) says that the equilibrium value of x satisﬁes x0 = gy′(x0) ω2 . (6.192) The F = ma explanation for this (writing y′(x0) as tan θ, where θ is the angle of the curve, and then multiplying through by ω2 cos θ) is that the component of gravity along the curve accounts for the component of the centripetal acceleration along the curve. Using y(x) = b(x/a)λ, eq. (6.192) yields x0 = a µ a2ω2 λgb ¶1/(λ−2) . (6.193) As λ →∞, we see that x0 goes to a. This makes sense, because the curve essentially equals zero up to a, and then it rises very steeply. You can check numerous other limits. Letting x ≡x0 + δ in eq. (6.191), and expanding to ﬁrst order in δ, gives ¨ δ ³ 1 + y′(x0)2´ = δ ³ ω2 −gy′′(x0) ´ . (6.194) The frequency of small oscillations is therefore given by Ω2 = gy′′(x0) −ω2 1 + y′(x0)2 . (6.195) Using the explicit form of y, along with eq. (6.193), we ﬁnd Ω2 = (λ −2)ω2 1 + a2ω4 g2 ¡ a2ω2 λgb ¢2/(λ−2) . (6.196) 6.11. SOLUTIONS VI-47 We see that λ must be greater than 2 in order for there to be oscillatory motion around the equilibrium point. For λ < 2, the equilibrium point is unstable, that is, to the left the force is inward, and to the right the force is outward. For the case λ = 2, we have y(x) = b(x/a)2, so the equilibrium condition, eq. (6.192), gives x0 = (2gb/a2ω2)x0. For this to be true for some x0, we must have ω2 = 2gb/a2. But if this holds, then eq. (6.192) is true for all x. So in the special case of λ = 2, the bead happily sits anywhere on the curve if ω2 = 2gb/a2. (In the rotating frame of the curve, the tangential components of the centrifugal and gravitational forces exactly cancel at all points.) If λ = 2 and ω2 ̸= 2gb/a2, then the particle feels a force either always inward or always outward. Remarks: For ω →0, eqs. (6.193) and (6.196) give x0 →0 and Ω→0. And for ω →∞, they give x0 →∞and Ω→0. In both cases Ω→0, because in both cases the equilibrium position is at a place where the curve is very ﬂat (horizontally or vertically, respectively), and the restoring force ends up being small. For λ →∞, we have x0 →a and Ω→∞. The frequency is large here because the equilibrium position at a is where the curve has a sharp corner, so the restoring force changes quickly with position. Or, you can think of it as a pendulum with a very small length, if you approximate the “corner” by a tiny circle. ♣ 6.18. Motion in a cone If the particle’s distance from the axis is r, then its height is r/ tan α, and its distance up along the cone is r/ sin α. Breaking the velocity into components up along the cone and around the cone, we see that the square of the speed is v2 = ˙ r2/ sin2 α + r2 ˙ θ2. The Lagrangian is therefore L = 1 2m µ ˙ r2 sin2 α + r2 ˙ θ2 ¶ −mgr tan α . (6.197) The equations of motion obtained from varying θ and r are d dt(mr2 ˙ θ) = 0 ¨ r = r ˙ θ2 sin2 α −g cos α sin α. (6.198) The ﬁrst of these equations expresses conservation of angular momentum. The second equa-tion is more transparent if we divide through by sin α. With x ≡r/ sin α being the distance up along the cone, we have ¨ x = (r ˙ θ2) sin α −g cos α. This is the F = ma statement for the x direction. Letting mr2 ˙ θ ≡L, we can eliminate ˙ θ from the second equation to obtain ¨ r = L2 sin2 α m2r3 −g cos α sin α. (6.199) We will now calculate the two desired frequencies. • Frequency of circular oscillations, ω: For circular motion with r = r0, we have ˙ r = ¨ r = 0, so the second of eqs. (6.198) gives ω ≡˙ θ = r g r0 tan α . (6.200) • Frequency of oscillations about a circle, Ω: If the orbit were actually the circle r = r0, then eq. (6.199) would give (with ¨ r = 0) L2 sin2 α m2r3 0 = g cos α sin α. (6.201) This is equivalent to eq. (6.200), which can be seen by writing L as mr2 0 ˙ θ. We will now use our standard procedure of letting r(t) = r0 + δ(t), where δ(t) is very small, and then plugging this into eq. (6.199) and expanding to ﬁrst order in δ. Using 1 (r0 + δ)3 ≈ 1 r3 0 + 3r2 0δ = 1 r3 0(1 + 3δ/r0) ≈1 r3 0 ³ 1 −3δ r0 ´ , (6.202) VI-48 CHAPTER 6. THE LAGRANGIAN METHOD we have ¨ δ = L2 sin2 α m2r3 0 ³ 1 −3δ r0 ´ −g cos α sin α. (6.203) Recalling eq. (6.201), the terms not involving δ cancel, and we are left with ¨ δ = − µ 3L2 sin2 α m2r4 0 ¶ δ. (6.204) Using eq. (6.201) again to eliminate L we have ¨ δ + ³3g r0 sin α cos α ´ δ = 0. (6.205) Therefore, Ω= r 3g r0 sin α cos α. (6.206) Having found the two desired frequencies in eqs. (6.200) and (6.206), we see that their ratio is Ω ω = √ 3 sin α. (6.207) This ratio Ω/ω is independent of r0. The two frequencies are equal if sin α = 1/ √ 3, that is, if α ≈35.3◦≡˜ α. If α = ˜ α, then after one revolution around the cone, r returns to the value it had at the beginning of the revolution. So the particle undergoes periodic motion. Remarks: In the limit α →0 (that is, the cone is very thin), eq. (6.207) says that Ω/ω →0. In fact, eqs. (6.200) and (6.206) say that ω →∞and Ω→0. So the particle spirals around many times during one complete r cycle. This seems intuitive. In the limit α →π/2 (that is, the cone is almost a ﬂat plane), both ω and Ωgo to zero, and eq. (6.207) says that Ω/ω → √ 3. This result is not at all obvious. If Ω/ω = √ 3 sin α is a rational number, then the particle undergoes periodic motion. For example, if α = 60◦, then Ω/ω = 3/2, so it takes two complete circles for r to go through three cycles. Or, if α = arcsin(1/2 √ 3) ≈16.8◦, then Ω/ω = 1/2, so it takes two complete circles for r to go through one cycle. 6.19. Double pendulum Relative to the pivot point, the Cartesian coordinates of m1 and m2 are, respectively (see Fig. 6.49), θ θ 1 2 l l m m 1 1 2 2 Figure 6.49 (x, y)1 = (ℓ1 sin θ1, −ℓ1 cos θ1), (x, y)2 = (ℓ1 sin θ1 + ℓ2 sin θ2, −ℓ1 cos θ1 −ℓ2 cos θ2). (6.208) Taking the derivative to ﬁnd the velocities, and then squaring, gives v2 1 = ℓ2 1 ˙ θ2 1, v2 2 = ℓ2 1 ˙ θ2 1 + ℓ2 2 ˙ θ2 2 + 2ℓ1ℓ2 ˙ θ1 ˙ θ2(cos θ1 cos θ2 + sin θ1 sin θ2). (6.209) The Lagrangian is therefore L = 1 2m1ℓ2 1 ˙ θ2 1 + 1 2m2 ³ ℓ2 1 ˙ θ2 1 + ℓ2 2 ˙ θ2 2 + 2ℓ1ℓ2 ˙ θ1 ˙ θ2 cos(θ1 −θ2) ´ +m1gℓ1 cos θ1 + m2g(ℓ1 cos θ1 + ℓ2 cos θ2). (6.210) The equations of motion obtained from varying θ1 and θ2 are 0 = (m1 + m2)ℓ2 1¨ θ1 + m2ℓ1ℓ2¨ θ2 cos(θ1 −θ2) + m2ℓ1ℓ2 ˙ θ2 2 sin(θ1 −θ2) +(m1 + m2)gℓ1 sin θ1, 0 = m2ℓ2 2¨ θ2 + m2ℓ1ℓ2¨ θ1 cos(θ1 −θ2) −m2ℓ1ℓ2 ˙ θ2 1 sin(θ1 −θ2) +m2gℓ2 sin θ2. (6.211) 6.11. SOLUTIONS VI-49 This is a bit of a mess, but it simpliﬁes greatly if we consider small oscillations. Using the small-angle approximations and keeping only the leading-order terms, we obtain 0 = (m1 + m2)ℓ1¨ θ1 + m2ℓ2¨ θ2 + (m1 + m2)gθ1, 0 = ℓ2¨ θ2 + ℓ1¨ θ1 + gθ2. (6.212) Consider now the special case, ℓ1 = ℓ2 ≡ℓ. We can ﬁnd the frequencies of the normal modes by using the determinant method, discussed in Section 4.5. You can show that the result is ω± = s m1 + m2 ± p m1m2 + m2 2 m1 qg ℓ. (6.213) The normal modes turn out to be, after some simpliﬁcation, µ θ1(t) θ2(t) ¶ ± = µ ∓√m2 √m1 + m2 ¶ cos(ω±t + φ±). (6.214) Some special cases are: • m1 = m2: The frequencies are ω± = p 2 ± √ 2 qg ℓ. (6.215) The normal modes are µ θ1(t) θ2(t) ¶ ± = µ ∓1 √ 2 ¶ cos(ω±t + φ±). (6.216) • m1 ≫m2: With m2/m1 ≡ϵ, the frequencies are (to leading nontrivial order in ϵ) ω± = (1 ± √ϵ/2) qg ℓ. (6.217) The normal modes are µ θ1(t) θ2(t) ¶ ± = µ ∓√ϵ 1 ¶ cos(ω±t + φ±). (6.218) In both modes, the upper (heavy) mass essentially stands still, and the lower (light) mass oscillates like a pendulum of length ℓ. • m1 ≪m2: With m1/m2 ≡ϵ, the frequencies are (to leading order in ϵ) ω+ = r 2g ϵℓ, ω−= q g 2ℓ. (6.219) The normal modes are µ θ1(t) θ2(t) ¶ ± = µ ∓1 1 ¶ cos(ω±t + φ±). (6.220) In the ﬁrst mode, the lower (heavy) mass essentially stands still, and the upper (light) mass vibrates back and forth at a high frequency (because there is a very large tension in the rods). In the second mode, the rods form a straight line, and the system is essentially a pendulum of length 2ℓ. VI-50 CHAPTER 6. THE LAGRANGIAN METHOD Consider now the special case, m1 = m2. Using the determinant method, you can show that the frequencies of the normal modes are ω± = √g s ℓ1 + ℓ2 ± p ℓ2 1 + ℓ2 2 ℓ1ℓ2 . (6.221) The normal modes turn out to be, after some simpliﬁcation, µ θ1(t) θ2(t) ¶ ± = µ ℓ2 ℓ2 −ℓ1 ∓ p ℓ2 1 + ℓ2 2 ¶ cos(ω±t + φ±). (6.222) Some special cases are: • ℓ1 = ℓ2: We already considered this case above. You can show that eqs. (6.221) and (6.222) agree with eqs. (6.215) and (6.216), respectively. • ℓ1 ≫ℓ2: With ℓ2/ℓ1 ≡ϵ, the frequencies are (to leading order in ϵ) ω+ = r 2g ℓ2 , ω−= r g ℓ1 . (6.223) The normal modes are µ θ1(t) θ2(t) ¶ + = µ −ϵ 2 ¶ cos(ω+t + φ+), µ θ1(t) θ2(t) ¶ − = µ 1 1 ¶ cos(ω−t + φ−). (6.224) In the ﬁrst mode, the masses essentially move equal distances in opposite directions, at a very high frequency (because ℓ2 is so small). The factor of 2 in the frequency arises because the angle of ℓ2 is twice what it would be if m1 were bolted in place; so m2 feels double the tangential force. In the second mode, the rods form a straight line, and the masses move just like a mass of 2m. The system is essentially a pendulum of length ℓ. • ℓ1 ≪ℓ2: With ℓ1/ℓ2 ≡ϵ, the frequencies are (to leading order in ϵ) ω+ = r 2g ℓ1 , ω−= r g ℓ2 . (6.225) The normal modes are µ θ1(t) θ2(t) ¶ + = µ 1 −ϵ ¶ cos(ω+t + φ+), µ θ1(t) θ2(t) ¶ − = µ 1 2 ¶ cos(ω−t + φ−). (6.226) In the ﬁrst mode, the bottom mass essentially stands still, and the top mass oscillates at a very high frequency (because ℓ1 is so small). The factor of 2 in the frequency arises because the top mass essentially lives in a world where the acceleration from gravity is g′ = 2g (because of the extra mg force downward from the lower mass). In the second mode, the system is essentially a pendulum of length ℓ2. The factor of 2 in the angles is what is needed to make the tangential force on the top mass roughly equal to zero (because otherwise it would oscillate at a high frequency, since ℓ1 is so small). 6.20. Shortest distance in a plane Let the two given points be (x1, y1) and (x2, y2), and let the path be described by the function y(x). (Yes, we’ll assume it can be written as a function. Locally, we don’t have to worry about any double-valued issues.) Then the length of the path is ℓ= Z x2 x1 p 1 + y′2 dx. (6.227) 6.11. SOLUTIONS VI-51 The “Lagrangian” is L = p 1 + y′2, so the Euler-Lagrange equation is d dx µ ∂L ∂y′ ¶ = ∂L ∂y = ⇒ d dx Ã y′ p 1 + y′2 ! = 0. (6.228) We see that y′/ p 1 + y′2 is constant. Therefore, y′ is also constant, so we have a straight line, y(x) = Ax + B, where A and B are determined by the endpoint conditions. 6.21. Index of refraction Let the path be described by y(x). The speed at height y is v ∝y. Therefore, the time to go from (x1, y1) to (x2, y2) is T = Z x2 x1 ds v ∝ Z x2 x1 p 1 + y′2 y dx. (6.229) The “Lagrangian” is therefore L ∝ p 1 + y′2 y . (6.230) At this point, we could apply the E-L equation to this L, but let’s just use Lemma 6.5, with f(y) = 1/y. Eq. (6.86) gives 1 + y′2 = Bf(y)2 = ⇒ 1 + y′2 = B y2 . (6.231) We must now integrate this. Solving for y′, and then separating variables and integrating, gives Z dx = ± Z y dy p B −y2 = ⇒ x + A = ∓ p B −y2 . (6.232) Therefore, (x + A)2 + y2 = B, which is the equation for a circle. Note that the circle is centered at a point with y = 0, that is, at a point on the bottom of the slab. This is the point where the perpendicular bisector of the line joining the two given points intersects the bottom of the slab. 6.22. Minimal surface By “tension” in a surface, we mean the force per unit length in the surface. The tension throughout the surface must constant, because it is in equilibrium. If the tension at one point were larger than at another, then some patch of the surface between these points would move. The ratio of the circumferences of the circular boundaries of the ring is y2/y1. Therefore, the condition that the horizontal forces on the ring cancel is y1 cos θ1 = y2 cos θ2, where the θ’s are the angles of the surface, as shown in Fig. 6.50. In other words, y cos θ is constant 1 θ θ 1 2 y 2 y Figure 6.50 throughout the surface. But cos θ = 1/ p 1 + y′2, so we have y p 1 + y′2 = C. (6.233) This is equivalent to eq. (6.77), and the solution proceeds as in Section 6.8. 6.23. Existence of a minimal surface The general solution for y(x) is given in eq. (6.78) as y(x) = (1/b) cosh b(x + d). If we choose the origin to be midway between the rings, then d = 0. Both boundary condition are thus r = 1 b cosh bℓ. (6.234) We will now determine the maximum value of ℓ/r for which the minimal surface exists. If ℓ/r is too large, then we will see that there is no solution for b in eq. (6.234). If you perform an experiment with soap bubbles (which want to minimize their area), and if you pull the rings VI-52 CHAPTER 6. THE LAGRANGIAN METHOD too far apart, then the surface will break and disappear as it tries to form the two boundary circles. Deﬁne the dimensionless quantities, η ≡ℓ r , and z ≡br. (6.235) Then eq. (6.234) becomes z = cosh ηz. (6.236) If we make a rough plot of the graphs of w = z and w = cosh ηz for a few values of η (see Fig. 6.51), we see that there is no solution for z if η is too large. The limiting value of η for w z w = cosh(ηz) w = z Figure 6.51 which there exists a solution occurs when the curves w = z and w = cosh ηz are tangent; that is, when the slopes are equal in addition to the functions being equal. Let η0 be the limiting value of η, and let z0 be the place where the tangency occurs. Then equality of the values and the slopes gives z0 = cosh(η0z0), and 1 = η0 sinh(η0z0). (6.237) Dividing the second of these equations by the ﬁrst gives 1 = (η0z0) tanh(η0z0). (6.238) This must be solved numerically. The solution is η0z0 ≈1.200. (6.239) Plugging this into the second of eqs. (6.237) gives ³ ℓ r ´ max ≡η0 ≈0.663. (6.240) (Note also that z0 = 1.200/η0 = 1.810.) We see that if ℓ/r is larger than 0.663, then there is no solution for y(x) that is consistent with the boundary conditions. Above this value of ℓ/r, the soap bubble minimizes its area by heading toward the shape of just two disks, but it will pop well before it reaches that conﬁguration. To get a sense of the rough shape of the minimal surface, note that the ratio of the radius of the “middle” circle to the radius of the boundary rings is y(0) y(ℓ) = cosh(0) cosh(bℓ) = 1 cosh(η0z0) = 1 z0 ≈0.55. (6.241) Remarks: 1. We glossed over one issue above, namely that there may be more than one solution for the constant b in eq. (6.234). In fact, Fig. 6.51 shows that for any η < 0.663, there are two solutions for z in eq. (6.236), and hence two solutions for b in eq. (6.234). This means that there are two possible surfaces that might solve our problem. Which one do we want? It turns out that the surface corresponding to the smaller value of b is the one that minimizes the area, while the surface corresponding to the larger value of b is the one that (in some sense) maximizes the area. We say “in some sense” because the surface with the larger b is actually a saddle point for the area. It can’t be a maximum, after all, because we can always make the area larger by adding little wiggles to it. It’s a saddle point because there does exist a class of variations for which it has the maximum area, namely ones where the “dip” in the curve is continuously made larger (just imagine lowering the midpoint in a smooth manner). Such a set of variations is shown in Fig. 6.52. If we start with a cylinder for a surface and then gradually pinch in the center, the area decreases at ﬁrst (the decrease in the cross-sectional area is the dominant eﬀect at the start). But then as the dip becomes very deep, the area increases because the surface starts to look like the two disks, and these two disks have a larger area than the original narrow cylinder. The surface eventually resembles two nearly ﬂat cones connected by a line. As these cones ﬁnally ﬂatten out to the two disks, the area decreases. Therefore, the area must 6.11. SOLUTIONS VI-53 2l r Figure 6.52 have achieved a local maximum (at least with respect to this class of variations) somewhere in between. This local maximum (or rather, saddle point) arises because the Euler-Lagrange technique simply sets the “derivative” equal to zero and doesn’t diﬀerentiate between maxima, minima, and saddle points. If η ≡ℓ/r > 0.663 (so that the initial cylinder in now wide instead of narrow), there exists at least one class of variations for which the area decreases monotonically from the area of the cylinder down to the area of the two disks. If you draw a series of pictures (for a wide cylinder) analogous to those in Fig. 6.52, it is quite believable that this is the case. 2. How does the area of the limiting surface (with η0 = 0.663) compare with the area of the two disks? The area of the two disks is Ac = 2πr2. And the area of the limiting surface is As = Z ℓ −ℓ 2πy p 1 + y′2 dx. (6.242) Using eq. (6.234), this becomes As = Z ℓ −ℓ 2π b cosh2 bx dx = Z ℓ −ℓ π b (1 + cosh 2bx) dx = 2πℓ b + π sinh 2bℓ b2 . (6.243) But from the deﬁnitions of η and z, we have ℓ= η0r and b = z0/r for the limiting surface. Therefore, As can be written as As = πr2 µ 2η0 z0 + sinh 2η0z0 z2 0 ¶ . (6.244) Plugging in the numerical values (η0 ≈0.663 and z0 ≈1.810) gives Ac ≈(6.28)r2, and As ≈(7.54)r2. (6.245) The ratio of As to Ac is approximately 1.2 (it’s actually η0z0, as you can show). The limiting surface therefore has a larger area. This is expected, because for ℓ/r > η0 the surface tries to run oﬀto one with a smaller area, and there are no other stable conﬁgurations besides the cosh solution we found. 6.24. The Brachistochrone First solution: In Fig. 6.53, the boundary conditions are y(0) = 0 and y(x0) = y0, with x y 0 0 , ( ) x y Figure 6.53 downward taken to be the positive y direction. From conservation of energy, the speed as a function of y is v = √2gy. The total time is therefore T = Z x0 0 ds v = Z x0 0 p 1 + y′2 √2gy dx. (6.246) Our goal is to ﬁnd the function y(x) that minimizes this integral, subject to the boundary conditions above. We can therefore apply the results of the variational technique, with a “Lagrangian” equal to L ∝ p 1 + y′2 √y . (6.247) VI-54 CHAPTER 6. THE LAGRANGIAN METHOD At this point, we could apply the E-L equation to this L, but let’s just use Lemma 6.5, with f(y) = 1/√y. Eq. (6.86) gives 1 + y′2 = Bf(y)2 = ⇒ 1 + y′2 = B y , (6.248) as desired. We must now integrate this. Solving for y′ and separating variables gives √y dy √B −y = ± dx. (6.249) A helpful change of variables to get rid of the square root in the denominator is y ≡B sin2 φ. Then dy = 2B sin φ cos φ dφ, and eq. (6.249) simpliﬁes to 2B sin2 φ dφ = ± dx. (6.250) We can now use sin2 φ = (1−cos 2φ)/2 to integrate this. After multiplying through by 2, the result is B(2φ −sin 2φ) = ± 2x −C, where C is a constant of integration. Now note that we can rewrite our deﬁnition of φ (which was y ≡B sin2 φ) as 2y = B(1 −cos 2φ). If we then deﬁne θ ≡2φ, we have x = ± a(θ −sin θ) ± d, y = a(1 −cos θ). (6.251) where a ≡B/2, and d ≡C/2. The particle starts at (x, y) = (0, 0). Therefore, θ starts at θ = 0, since this corresponds to y = 0. The starting condition x = 0 then implies that d = 0. Also, we are assuming that the wire heads down to the right, so we choose the positive sign in the expression for x. Therefore, we ﬁnally have x = a(θ −sin θ), y = a(1 −cos θ), (6.252) as desired. This is the parametrization of a cycloid, which is the path taken by a point on the rim of a rolling wheel. The initial slope of the y(x) curve is inﬁnite, as you can check. Remark: The above method derived the parametric form in (6.252) from scratch. But since eq. (6.252) was given in the statement of the problem, another route is to simply verify that this parametrization satisﬁes eq. (6.248). To this end, assume that x = a(θ −sin θ) and y = a(1 −cos θ), which gives y′ ≡dy dx = dy/dθ dx/dθ = sin θ 1 −cos θ . (6.253) Therefore, 1 + y′2 = 1 + sin2 θ (1 −cos θ)2 = 2 1 −cos θ = 2a y , (6.254) which agrees with eq. (6.248), with B ≡2a. ♣ Second solution: Let’s use a variational argument again, but now with y as the indepen-dent variable. That is, let the chain be described by the function x(y). The arclength is now given by ds = √ 1 + x′2 dy. Therefore, instead of the Lagrangian in eq. (6.247), we have L ∝ √ 1 + x′2 √y . (6.255) The Euler-Lagrange equation is d dy ³ ∂L ∂x′ ´ = ∂L ∂x = ⇒ d dy µ 1 √y x′ √ 1 + x′2 ¶ = 0. (6.256) The zero on the right-hand side makes things nice and easy, because it means that the quantity in parentheses is a constant. If we deﬁne this constant to be 1/ √ B, then we can solve for x′ and then separate variables to obtain √y dy √B −y = ± dx. (6.257) 6.11. SOLUTIONS VI-55 in agreement with eq. (6.249). The solution proceeds as above. Third solution: The “Lagrangian” in the ﬁrst solution above, which is given in eq. (6.247) as L ∝ p 1 + y′2 √y , (6.258) is independent of x. Therefore, in analogy with conservation of energy (which arises from a Lagrangian that is independent of t), the quantity E ≡y′ ∂L ∂y′ −L = y′2 √y p 1 + y′2 − p 1 + y′2 √y = −1 √y p 1 + y′2 (6.259) is constant (that is, independent of x). This statement is equivalent to eq. (6.248), and the solution proceeds as above.
17152	https://math.stackexchange.com/questions/4031833/use-of-directed-lengths	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Use of directed lengths Ask Question Asked Modified 4 years, 7 months ago Viewed 1k times 1 $\begingroup$ I was studying Menelaus' Theorem which can be found here I noticed they use directed lengths which specifies that points $E,D,F$(in reference to above link) are collinear. My doubt is when do we use this kind of notation, where the magnitude of a side is taken to be negative. Can someone please provide clarity on this issue or recommend a source. Thanks a lot for any help :) geometry euclidean-geometry Share asked Feb 19, 2021 at 13:55 PravimishPravimish 65933 silver badges1717 bronze badges $\endgroup$ 0 Add a comment \| 1 Answer 1 Reset to default 1 $\begingroup$ Signed lengths are a way to reduce the number of variations of a diagram that can be drawn. Consider the following basic use case: If points $A$, $B$, and $C$ are collinear, then $AB + BC = AC$. This is definitely true when point $B$ is between point $A$ and point $C$. We introduce signed lengths to make this statement true no matter what order $A, B, C$ appear in. Signed lengths can be used in any statement of affine geometry: statements that only use properties which are unaffected by affine transformations. For example, in affine geometry: we can talk about points being collinear, and lines being concurrent or parallel; we can talk about ratios of areas, or ratios of parallel line segments; we cannot compare line segments that are not parallel; we cannot talk about angles; we cannot even measure absolute lengths of line segments, only their ratios. (Statements like $AB + BC = AC$ are okay if they are understood as $\frac{AB}{AC} + \frac{BC}{AC}=1$, which is valid.) In this setting, the ratio $\frac{AB}{CD}$ is positive if $AB$ and $CD$ point in the same direction (they are required to be parallel). We do not make statements about signed lengths that do not reduce to ratios in this way. But we could define a signed length by picking a reference line segment $PQ$ in each class of parallel lines, which we treat as having length $+1$, and letting $AB$ have the signed length $\frac{AB}{PQ}$. In general, any statement of affine geometry is automatically compatible with signed lengths. If we prove it for ordinary lengths when the points are in one configuration (with respect to order, points being inside a triangle, etc.) then we can deduce it for signed lengths and all configurations. For example, Ceva's theorem is often stated for ordinary lengths, with points $D, E, F$ lying on segments $BC, CA, AB$. But if we allow the points to lie anywhere on the lines $BC, CA, AB$, then it is true for signed lengths. The handwavy reason for that is that theorems in affine geometry using signed lengths can be restated as polynomial equations in the coordinates of points. Checking a theorem for ordinary lengths in a large "region" of cases tells us that a polynomial equation holds in that region. But if a polynomial equation in $n$ variables holds over an open subset of $\mathbb R^n$, then it holds over all of $\mathbb R^n$, giving us the theorem for signed lengths. (Here, $n$ is the number of degrees of freedom in the diagram.) Share answered Feb 19, 2021 at 15:29 Misha LavrovMisha Lavrov 160k1111 gold badges167167 silver badges304304 bronze badges $\endgroup$ 1 $\begingroup$ Thanks a lot!!! $\endgroup$ Pravimish – Pravimish 2021-02-19 16:53:13 +00:00 Commented Feb 19, 2021 at 16:53 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry euclidean-geometry See similar questions with these tags. 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17153	https://math.stackexchange.com/questions/2114574/approximation-of-a-summation-by-an-integral	Approximation of a summation by an integral - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Approximation of a summation by an integral Ask Question Asked 8 years, 8 months ago Modified5 years, 9 months ago Viewed 24k times This question shows research effort; it is useful and clear 13 Save this question. Show activity on this post. I am going to approximate ∑n−1 i=0(n n−i)1 β−1∑i=0 n−1(n n−i)1 β−1 by ∫n−1 0(n n−x)1 β−1 d x∫0 n−1(n n−x)1 β−1 d x, such that n n is sufficiently large. Is the above approximation true? If the above approximation is true, by which theorem or method (like Newton's method) it can be holds? What is the error of approximation? definite-integrals summation numerical-methods approximation summation-method Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jan 26, 2017 at 5:52 Hasan HeydariHasan Heydari 1,237 1 1 gold badge 9 9 silver badges 23 23 bronze badges 4 4 Euler-MacLaurin formula might interest you.Sangchul Lee –Sangchul Lee 2017-01-26 05:56:22 +00:00 Commented Jan 26, 2017 at 5:56 Is this method has the minimum error in comparison to other methods?Hasan Heydari –Hasan Heydari 2017-01-26 06:02:58 +00:00 Commented Jan 26, 2017 at 6:02 You can't really discuss "minimum error" in an absolute sense without discussing the function. // The gist is that one formulation of calculus is derived doing the reverse. Using sums to evaluate integrals.MaxW –MaxW 2017-01-26 06:12:11 +00:00 Commented Jan 26, 2017 at 6:12 I think I am not following you, since it seems to me that the error between the sum and the integral is just a fixed function of n n. Euler-Maclaurin formula provides a systematical way of approximating this error, but I see no reason that this approach is always the most efficient for this pursuit (especially thinking that the formal sum describing the asymptotic expansion often do not converge, and is only meaningful as truncated sense).Sangchul Lee –Sangchul Lee 2017-01-26 06:18:07 +00:00 Commented Jan 26, 2017 at 6:18 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. One way to visualize what your are doing is with Riemann Sums. Assuming your function is increasing (it depends on n n and β β) we have that ∫b a−1 f(x)d x≤∑k=a b f(k)≤∫b+1 a f(x)d x∫a−1 b f(x)d x≤∑k=a b f(k)≤∫a b+1 f(x)d x From a more general perspective, we can apply the Euler-MacLaurin Formula, which says that ∑n=a b f(n)=∫b a f(x)d x+f(a)+f(b)2+∑k=1∞B 2 k(2 k)!(f(2 k−1)(b)−f(2 k−1)(a))+R∑n=a b f(n)=∫a b f(x)d x+f(a)+f(b)2+∑k=1∞B 2 k(2 k)!(f(2 k−1)(b)−f(2 k−1)(a))+R or that ∑n=a b f(n)∼∫b a f(x)d x+f(a)+f(b)2+∑k=1∞B 2 k(2 k)!(f(2 k−1)(b)−f(2 k−1)(a))∑n=a b f(n)∼∫a b f(x)d x+f(a)+f(b)2+∑k=1∞B 2 k(2 k)!(f(2 k−1)(b)−f(2 k−1)(a)) Where B k B k is the k k th Bernoulli Number (all the odd Bernoulli numbers are just zero) The error term R R is very small (and depends on a,b,a,b, and f f) so we can usually ignore it unless you absolutely need it. See the linked page for an explicit formula. Edit: Per the OP's request, the reason we can neglect the error term is that we have the general form of the Euler-MacLaurin Formula as such: ∑n=a b f(n)=∫b a f(x)d x+f(a)+f(b)2+∑k=1⌊p 2⌋B 2 k(2 k)!(f(2 k−1)(b)−f(2 k−1)(a))+R∑n=a b f(n)=∫a b f(x)d x+f(a)+f(b)2+∑k=1⌊p 2⌋B 2 k(2 k)!(f(2 k−1)(b)−f(2 k−1)(a))+R In the formula I wrote above we take p→∞p→∞. We further note (see the Wikipedia page) that the remainder R R is bounded as such: \|R\|≤2 ζ(p)(2 π)p∫b a\|f(p)(x)\|d x\|R\|≤2 ζ(p)(2 π)p∫a b\|f(p)(x)\|d x Now, there are two parts of this. The fraction clearly goes to zero as p→∞p→∞, because the Riemann Zeta function is decreasing and bounded while the denominator grows without bounds. Thus, all we have to be able to say is that the integral doesn't blow up to ++ or −− infinity as we differentiate more and more (i.e. let p→∞p→∞). Most functions you work with will satisfy this criteria, but you can check this explicitly if you so desire. Edit 2: As noted by @SangchulLee in the comments, for fixed β β and fixed x x the integral in the OP's error term appears to grow super-exponentially. While the fraction 2 ζ(p)(2 π)p 2 ζ(p)(2 π)p definitely helps decrease this growth a bit, the error term might need to be watched carefully here. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 26, 2017 at 6:45 answered Jan 26, 2017 at 6:05 Brevan EllefsenBrevan Ellefsen 15.3k 4 4 gold badges 31 31 silver badges 76 76 bronze badges 8 In my question, the error will tend to zero when n→∞n→∞?Hasan Heydari –Hasan Heydari 2017-01-26 06:18:28 +00:00 Commented Jan 26, 2017 at 6:18 @HasanHeydari The source on Wikipedia says "Usually". I will edit this into my answer with a little explanation on why this is Brevan Ellefsen –Brevan Ellefsen 2017-01-26 06:23:12 +00:00 Commented Jan 26, 2017 at 6:23 @HasanHeydari I included an edit which explicitly states the remainder formula and explains why we can ignore the remainder for "most" functions Brevan Ellefsen –Brevan Ellefsen 2017-01-26 06:32:36 +00:00 Commented Jan 26, 2017 at 6:32 1 I would say the opposite. With OP's f f and for each fixed x x, we can check that \|f(p)(x)\|\|f(p)(x)\| grows super-exponentially as p→∞p→∞. For instance, if we choose β=2 β=2 so that f(x)=n n−x f(x)=n n−x, then f(p)(x)=p!n(n−x)p+1 f(p)(x)=p!n(n−x)p+1.Sangchul Lee –Sangchul Lee 2017-01-26 06:39:19 +00:00 Commented Jan 26, 2017 at 6:39 1 @SangchulLee Huh. You are right. For the record, I never say that we should ignore the error term, just that we usually can. I didn't check the condition here per-se (I was leaving it to the OP) but you are absolutely right as far as I can tell.\Brevan Ellefsen –Brevan Ellefsen 2017-01-26 06:41:17 +00:00 Commented Jan 26, 2017 at 6:41 \|Show 3 more comments This answer is useful 8 Save this answer. Show activity on this post. If we write s=1/(β−1)s=1/(β−1), your sum and integral can be re-written as ∑i=0 n−1(n n−i)s=n s∑k=1 n 1 k s,∫n−1 0(n n−x)s d x=n s∫n 1 d x x s.∑i=0 n−1(n n−i)s=n s∑k=1 n 1 k s,∫0 n−1(n n−x)s d x=n s∫1 n d x x s. In this case, the Euler-Maclaurin formula provides a way of estimating the difference within O(n−K)O(n−K) for any prescribed exponent K K. To see this, let B~k(x)=B k(x−⌊x⌋)B~k(x)=B k(x−⌊x⌋) be the periodic Bernoulli polynomials. Then ∫n 1 d x x s=(∫[1,n]d⌊x⌋x s+∫[1,n]d B~1(x)x s)=∑k=1 n 1 k s+∫[1,n]d B~1(x)x s=∑k=1 n 1 k s+[B~1(x)x s]n 1−+s∫n 1 B~1(x)x s+1 d x=∑k=1 n 1 k s−1+n−s 2+s∫n 1 B~1(x)x s+1 d x∫1 n d x x s=(∫[1,n]d⌊x⌋x s+∫[1,n]d B~1(x)x s)=∑k=1 n 1 k s+∫[1,n]d B~1(x)x s=∑k=1 n 1 k s+[B~1(x)x s]1−n+s∫1 n B~1(x)x s+1 d x=∑k=1 n 1 k s−1+n−s 2+s∫1 n B~1(x)x s+1 d x However, there is an issue with this form. Indeed, if we keep using [1,n][1,n] as the domain of integration, the error term of the Euler-Maclaurin formula never vanishes as n→∞n→∞. This is because for each fixed K K, ∫n 1 B~1(x)x s+K d x=Θ(1)as n→∞∫1 n B~1(x)x s+K d x=Θ(1)as n→∞ To resolve issue, let us assume s>0 s>0 and we split the last integral as the difference of two: ∫n 1 d x x s=∑k=1 n 1 k s−1+n−s 2+s∫∞1 B~1(x)x s+1 d x−s∫∞n B~1(x)x s+1 d x∫1 n d x x s=∑k=1 n 1 k s−1+n−s 2+s∫1∞B~1(x)x s+1 d x−s∫n∞B~1(x)x s+1 d x We note that integration by parts easily checks the asymptotics ∫∞n B~K(x)x s+K d x=O(n−s−K)∫n∞B~K(x)x s+K d x=O(n−s−K) for each fixed K K. Then letting n→∞n→∞ together with the extra assumption s>1 s>1, this yields s∫∞1 B~1(x)x s+1 d x=1 2+1 s−1−ζ(s).s∫1∞B~1(x)x s+1 d x=1 2+1 s−1−ζ(s). This continues to hold for s>0 s>0 by the principle of analytic continuation. Plugging this back and simplifying in terms of the sum, ∑k=1 n 1 k s=∫n 1 d x x s+1+n−s 2+s(ζ(s)−1 2−1 s−1)+s∫∞n B~1(x)x s+1 d x.∑k=1 n 1 k s=∫1 n d x x s+1+n−s 2+s(ζ(s)−1 2−1 s−1)+s∫n∞B~1(x)x s+1 d x. Then we can continue the procedure to extract more terms: for each fixed K K, ∑k=1 n 1 k s=∫n 1 d x x s+1+n−s 2+s(ζ(s)−1 2−1 s−1)−∑j=2 K Γ(s+j−1)Γ(s)B j j!1 n s+j−1+Γ(s+K)K!Γ(s)∫∞n B~K(x)x s+K d xO(n−s−K−1)∑k=1 n 1 k s=∫1 n d x x s+1+n−s 2+s(ζ(s)−1 2−1 s−1)−∑j=2 K Γ(s+j−1)Γ(s)B j j!1 n s+j−1+Γ(s+K)K!Γ(s)∫n∞B~K(x)x s+K d x⏟O(n−s−K−1) as n→∞n→∞. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 26, 2017 at 7:17 Sangchul LeeSangchul Lee 184k 17 17 gold badges 300 300 silver badges 493 493 bronze badges 3 Snagchul Lee: Note that ∑n−1 i=0(n n−i)s≠n s∑n k=1 1 k s∑i=0 n−1(n n−i)s≠n s∑k=1 n 1 k s Hasan Heydari –Hasan Heydari 2017-01-26 07:30:36 +00:00 Commented Jan 26, 2017 at 7:30 1 @HasanHeydari, With the change of index n−i=k n−i=k you can prove the equality.Sangchul Lee –Sangchul Lee 2017-01-26 07:31:32 +00:00 Commented Jan 26, 2017 at 7:31 Absolutely gorgeous answer. My answer simply explained what the Euler-MacLaurin formula is... Your answer beats the OP's formula to a pulp and makes everything work. Bravo. Hopefully the OP makes your's the accepted answer instead of mine Brevan Ellefsen –Brevan Ellefsen 2017-01-26 14:25:47 +00:00 Commented Jan 26, 2017 at 14:25 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Using Riemann Siegel formula you can approximate ∑n−1 i=0(n n−i)s=≃n s{ζ(s)−1/(s−1)(n+1/2)1−s}.∑i=0 n−1(n n−i)s=≃n s{ζ(s)−1/(s−1)(n+1/2)1−s}. This gives very accurate values for real s>1 s>1 even for very small n n. For instance, for n=2 n=2 and s=3 s=3, one obtains ζ(3)≃241/200=1.205 ζ(3)≃241/200=1.205. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 14, 2019 at 1:39 answered Dec 14, 2019 at 1:31 Jose Luis RosalesJose Luis Rosales 21 2 2 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions definite-integrals summation numerical-methods approximation summation-method See similar questions with these tags. 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17154	https://tmedweb.tulane.edu/pharmwiki/doku.php/progestins	progestins [TUSOM \| Pharmwiki] skip to content Site Tools TUSOM \| Pharmwiki Search Recent Changes Media Manager Sitemap User Tools Log In Trace:•progestins progestins Progestins Progesterone Drug Class: Natural progestin Mechanism of Action: Similar to estrogens, but binds to different receptors in the cytoplasm or nucleus which then interact with progesterone-response elements to activate gene transcription At high doses, progestins work as contraceptives via negative feedback on GnRH release Physiological Effects: promotes secretory activity of endometrium “primed” by estrogen negative feedback on anterior pituitary release of FSH and LH, this prevents follicular maturation & ovulation important (along with estrogen) in breast development & lactation the cause of mid-cycle increase in body temperature at ovulation Indications for progestins (synthetic): Contraception - most commonly used in combination with estrogens Progestin only “mini pills” are not as effective as combination pills Control of functional uterine bleeding Management ofendometriosis (ectopic endometrial tissue) Management of dysmenorrhea (painful menstruation) Pharmacokinetics: progesterone itself is rapidly metabolized, which limits its use as a therapeutic agent synthetic progestins are not rapidly inactivated by first pass metabolism, and can be administered orally Side Effects: weight gain edema depression thrombophlebitis pulmonary embolism the more potent progestins, if used for a long period of time, will decrease HDL levels and lead to atherosclerosis. References: Chrousos GP (2012): The Gonadal Hormones & Inhibitors (Chapter 40). In: Basic & Clinical Pharmacology. 12e. BG Katzung, SB Masters, AJ Trevor (Editors). Lange/McGraw-Hill. Levin ER, Hammes SR (2011): Estrogens and progestins (Chapter 40). In: Goodman & Gilman's Pharmacological Basis of Therapeutics. 12e, McGraw-Hill. Keywords Medroxyprogesterone Trade Name: generic, Depo-Provera, Depo-SubQ Provera ® Drug Class: Progestin, Contraceptive Injection (Depot formulation) Mechanism of Action: Depo-Provera® is an i.m. depot formulation of medroxyprogesterone acetate that can be given by injection every 3 months that contains microcrystals of low solubility that results in pharmacologically active drug levels for several months. Depo-SubQ ® is a s.c. injection formulation that enables the use of a 30% lower dose of progestin and reduces peak plasma levels by 50%, but has the same duration of action as the i.m. formulation. The s.c. route is less painful, and “potentially” may allow patient self-administration (although not yet FDA approved for patient self administration). these depot formulations prevents pregnancy primarily by inhibition of gonadotrophin (GnrH) secretion which results in an inhibition of follicular maturation and ovulation (Kaunitz, 1994; 2016). Indications: Contraception Side Effects: menstrual irregularities (bleeding or amenorrhea, or both) weight changes headache, nervousness, dizziness abdominal pain or discomfort weakness or fatigue increased risk for meningioma with long term use (Roland et al, 2024). There is some evidence that sex steroids may play a role in the growth of meningiomas, with meningiomas found in the medial skull base and spinal column exhibiting a higher expression level of progesterone receptors compared to other regions of the brain (P=0.0036)(Maiuri et al, 2021). Black Box Warnings: REDUCED BONE DENSITY & STDs BONE LOSS: Women who use Depo-Provera Contraceptive (or Depo-SubQ Provera) may lose significant bone mineral density. Bone loss increases with the duration of use, and may not be completely reversible. It is unclear whether the use of medroxyprogesterone depot formulations during adolescence or early adulthood may also reduce bone mass & increase the risk of osteoporotic fracture later in life. Medroxyprogesterone depot formulations should be used for long-term therapy (e.g. >2 years) only if other methods of birth control are inadequate. SEXUALLY TRANSMITTED DISEASE: Patients should be counseled that this product does not protect against HIV infection or infection by other Sexually Transmitted Diseases. Delayed Return of Fertility: Both i.m. and s.c. injections have a prolonged contraceptive effect that continues for some time after the last injection is administered Clinical trials indicate that following the last i.m. injection (rxlist.com: Depo Provera ®): 68% may conceive within 12 months 83% may conceive within 15 months 93% may conceive within 18 months Similar outcomes have been observed with use of the s.c. injection formulation as well. While some patients have become pregnant relatively quickly after a 3 month time period has lapsed following the last injection, it is more likely to take a year or longer after the last s.c. injection before you a patient can become pregnant (rxlist.com: Depo-SubQ-Provera ®). Pharmacokinetics: i.m. or s.c. injections are effective for 3 months Therapy can be initiated at any time, but is typically initiated within 1 week after the onset of menses or before discontinuing another method of contraception. The dose is repeated every 3 months, with a 2 week grace period. IM Formulation: It is recommended that the i.m. formulation be injected into the gluteal or deltoid muscle, and that the s.c. formulation be administered into the anterior thigh or abdomen (rotating locations with each dose). SC Formulation: provides slower and more sustained absorption of the progestin compared to the i.m. formulation. slower sustained absorption enables the use of a 30% lower dose of progestin (104 versus 150 mg) that also reduces peak blood levels by half, but with the same duration of effect as the conventional i.m. formulation. Administration via the SC route is less painful than i.m. injection and may potentially allow patient self-administration; self-administration is currently considered an “off-label” use (Kaunitz, 2016). References: Chrousos GP (2024): The Gonadal Hormones & Inhibitors (Chapter 40). In: Katzung's Basic & Clinical Pharmacology. 16th Ed. Vanderah TW (editor). Lange/McGraw-Hill. Kaunitz AM (1994): Long-acting injectable contraception with depot medroxyprogesterone acetate. Am J OObstet Gynecol 170(5 Pt 2):1543-1549. Kaunitz AM (2016): Depot medroxyprogesterone acetate for contraception. In: In: UpToDate, Basow, DS (Ed), Waltham, MA. Cited 3/6/17 Levin ER, Hammes SR (2011): Estrogens and progestins (Chapter 40). In: Goodman & Gilman's Pharmacological Basis of Therapeutics. 12e, McGraw-Hill. Maiuri F et al (2021): Progesterone Receptor Expression in Meningiomas: Pathological and Prognostic Implications. Front Oncol 11:611218. doi: 10.3389/fonc.2021.611218 PMID: 34336636 Roland N et al (2024): Use of progestogens and the risk of intracranial meningioma: national case-control study. BMJ 384:e078078. doi: 10.1136/bmj-2023-078078. PMID: 38537944 rxlist.com (Depo-SubQ Provera 104 ®) rxlist.com (Depo-Provera ®) Keywords Norethindrone Trade Names: Aygestin, Camila ® Ortho - Novum ® (Norethindrone w/ mestranol) Lo/Ovral ® (Norethindrone w/ ethinyl estradiol) Drug Class:Contraceptive Indications: the most common form of progestin used in oral contraceptives References: Chrousos GP (2012): The Gonadal Hormones & Inhibitors (Chapter 40). In: Basic & Clinical Pharmacology. 12e. BG Katzung, SB Masters, AJ Trevor (Editors). Lange/McGraw-Hill. Levin ER, Hammes SR (2011): Estrogens and progestins (Chapter 40). In: Goodman & Gilman's Pharmacological Basis of Therapeutics. 12e, McGraw-Hill. rxlist.com (Aygestin ®) Keywords Levonorgestrel Trade Names: Plan B (levonorgestrel emergency contraceptive tablet) Triphasil, Tri-Levlen, Trivora-28 ® (Triphasic Combination Oral Contraceptives) Drug Class:Contraceptive Indications: Plan B ® is intended to prevent pregnancy after known or suspected contraceptive failure or unprotected intercourse. Triphasic Combination Oral Contraceptives are intended for the prevention of pregnancy in women who choose to use this form of contraception. Pharmacokinetics: The progestin component in triphasic combination tablet formulations (which also contain ethinyl estradiol) Levoneorgestrel is also contained in the Mirena IUD, which continuously releases Levonorgestrel. This lessens menstrual bleeding & contributes to contraception by this IUD. Notes: Plan B is available without a prescription for consumers 18 and older Rx only for women age 17 and younger References: Levin ER, Hammes SR (2011): Estrogens and progestins (Chapter 40). In: Goodman & Gilman's Pharmacological Basis of Therapeutics. 12e, McGraw-Hill. rxlist.com (Plan B ®) rxlist.com (Trivora-28 ®) Keywords progestins.txt · Last modified: 2013/08/30 14:48 by cclarks Page Tools Show pagesource Old revisions Backlinks Back to top TUSOM SOM \| Pharmacology Department \| TMedWeb
17155	https://dwest.web.illinois.edu/pubs/aramdiam.pdf	Rainbow spanning subgraphs of small diameter in edge-colored complete graphs Sogol Jahanbekam∗and Douglas B. West† Revised March 2015 Abstract Let s(n, t) be the maximum number of colors in an edge-coloring of the complete graph Kn that has no rainbow spanning subgraph with diameter at most t. We prove s(n, t) = n−2 2 +1 for n, t ≥3, while s(n, 2) = n−2 2 + n−1 2 for n ̸= 4 (and s(4, 2) = 2). Keywords: spanning subgraph, rainbow subgraph, diameter, anti-Ramsey number MSC code: 05C55, 05C35 1 Introduction A rainbow subgraph of an edge-colored graph G is a subgraph whose edges have distinct colors. The general anti-Ramsey problem asks for the maximum number of colors in an edge-coloring of Kn having no rainbow copy of some graph in a class F; this maximum number of colors is the anti-Ramsey number AR(n, F). (The Ramsey problem can be interpreted as asking for the minimum number of colors in an edge-coloring having no monochromatic copy of a graph in F.) Early results considered the problem with F being a single graph: see for a survey and for the notable determination of AR(n, Ck). More recent work considers problems where F consists of spanning subgraphs of Kn; see for a discussion of such problems involving spanning cycles, perfect matchings, and spanning trees. In this paper, we study when rainbow spanning subgraphs of small diameter are forced. Let s(n, t) be the maximum number of colors in an edge-coloring of Kn not having a rainbow spanning subgraph with diameter at most t. We compute s(n, t). Previously, Mont´ agh obtained an upper bound on s(n, 3), showing that AR(n, Dn) = n−2 2 + 1, where Dn is the ∗jahanbe1@illinois.edu; University of Illinois, Urbana, IL. Research supported in part by National Science Foundation grant DMS 09-01276. †west@math.uiuc.edu; Zhejiang Normal University, Jinhua, China, and University of Illinois, Urbana, IL. Research supported by Recruitment Program of Foreign Experts, 1000 Talent Plan, State Administration of Foreign Experts Aﬀairs, China. 1 family of n-vertex double-stars whose vertex of second largest degree is at most 4. That is, rainbow double stars that are “almost” stars are forced by having more than n−2 2 +1 colors. To prove s(n, t) ≥ n−2 2 + 1 when t ≥3, we present an edge-coloring of Kn using n−2 2 + 1 colors that has no rainbow spanning subgraph with diameter at most t. With more colors, such a subgraph is forced. Upper bounds for anti-Ramsey problems generally use the notion of representing subgraph of an edge-colored complete graph, which is a spanning subgraph containing one edge of each color. We give a short proof that any edge-coloring of Kn using at least n−2 2 +2 colors has a representing subgraph with diameter at most 3. Thus when t ≥3 the answer does not depend on t. The answer for the special case t = 2 is somewhat diﬀerent. Theorem 1.1. s(n, t) = n−2 2 +      1 for n, t ≥3, 2 for (n, t) = (4, 2), n−1 2 when t = 2 and n ̸= 4. The aim is to show that when too many colors are used, every edge-coloring contains some representing subgraph having the forbidden property. As noted by a referee, our work suggests the corresponding Tur´ an-type extremal question: How many edges can an n-vertex graph have without having a spanning subgraph with diameter at most t? The answer is n−1 2 , achieved by Kn−1 + K1. 2 Constructions and easy cases For a vertex v in a graph G, let NG(v) be the set of neighbors of v, with NG[v] = NG(v)∪{v}, and let the degree dG(v) of v be \|NG(v)\|. An x, y-path in G is a path with endpoints x and y. The distance dG(x, y) between vertices x and y is the minimum length of an x, y-path in G, and the diameter diam(G) is maxx,y∈V (G) dG(x, y). By convention, dG(x, y) is inﬁnite when G has no x, y-path. When only one graph is under discussion, it is common to drop the subscripts in NG(v), NG[v], dG(v), and dG(x, y). Determining s(n, t) is easy when t ≥3. Lemma 2.1. A connected n-vertex graph with at least n−2 2 + 2 edges has diameter at most 3. An n-vertex graph with at least n−1 2 + 1 edges is connected and has diameter at most 2. Proof. If G is connected and diam(G) > 3, then there exist x and y with d(x, y) = 4. Let j = \|N(x)\| and k = \|N(y)\|, and let l = \|V (G) −N[x] −N[y]\|. To avoid having a shorter x, y-path, N(x) and N(y) must be disjoint and joined by no edge. Thus j + k + l = n −2; also j, k ≥1. Hence \|E(G)\| ≥j + k + 2l + jk + 1 = n −1 + l + jk. Since l + j + k = n −2 and j, k ≥1, the quantity l + jk is minimized when l = n −4 and j = k = 1. Hence \|E(G)\| ≥2n −4, yielding \|E(G)\| ≤ n−2 2 + 1. 2 If diam(G) > 2, then vertices x and y with d(x, y) > 2 are nonadjacent and have non-neighbors covering the remaining vertices, so \|E(G)\| ≥n −1, yielding \|E(G)\| ≤ n−1 2 . Lemma 2.2. If n, t ≥3, then s(n, t) = n−2 2 + 1. Proof. An edge-colored complete graph in which all edges incident to two vertices have the same color has no connected rainbow spanning subgraph. Hence s(n, t) ≥ n−2 2 + 1. For the upper bound, consider a coloring of E(Kn) using at least n−2 2 + 2 colors; we may assume equality. If a representing subgraph G has an isolated vertex, then the other vertices all have degree at least 2, since otherwise there are at most n−2 2 + 1 edges. Now adding any edge e incident to an isolated vertex v and removing the edge in G with the same color as e yields a representing subgraph with no isolated vertex. Hence we may assume that G has no isolated vertex. If G is disconnected, then \|E(G)\| ≥ 2(n−2), but \|E(G)\| ≤2n−5. Hence G is connected, and Lemma 2.1 restricts G to diameter at most 3, as desired. The diﬃcult case is t = 2, where we must force a rainbow spanning subgraph with diameter 2. For s(3, 2), two colors are clearly both necessary and suﬃcient. Lemma 2.3. s(4, 2) = 3. Proof. For the lower bound, give colors 1 and 2 to two independent edges, and give color 3 to the remaining 4-cycle. The only spanning subgraph with three edges and diameter 2 is a star, but the coloring has no rainbow star. For the upper bound, every 4-vertex graph with at least four edges has diameter at most 2. For t = 2 with n > 4, we ﬁrst provide the construction for the lower bound. Lemma 2.4. If n > 4, then s(n, 2) ≥ n−2 2 + n−1 2 . Proof. In Kn, give one color to all edges incident to one vertex v and also to n−1 2 edges covering the remaining n−1 vertices. Give distinct other colors to the remaining edges. The total number of colors is n−1 2 − n−1 2 + 1, which equals n−2 2 + n−1 2 . In any rainbow subgraph G, the degree of v is at most 1. Every vertex other than v has two incident edges of the same color and hence has degree at most n −2 in G. Therefore, not all vertices can be within distance 2 of v. 3 Rainbow subgraphs with diameter 2 The upper bound for n > 4 when t = 2 is the diﬃcult part. Let Γ(v) denote the set of edges incident to v in G. Say that a color is incident to a vertex v if it is used on an edge in Γ(v). A clique in a graph is a set of pairwise adjacent vertices. 3 Lemma 3.1. If n > 4, then s(n, 2) ≤ n−2 2 + n−1 2 . Proof. It suﬃces to study an edge-colored copy H of Kn using exactly n−2 2 + n−1 2 +1 colors. Suppose that H has no rainbow spanning subgraph with diameter 2. Thus a representing subgraph of H has no spanning complete bipartite subgraph (no copy of K1,n−1 or K2,n−2), and every vertex is incident to at most n −2 colors. Given a vertex u, let G′ be a subgraph of H −u containing one edge of each color not incident to u. Avoiding rainbow spanning stars in H requires ∆(G′) < n −2, and hence \|E(G′)\| ≤ n−1 2 − n−1 2 = n−2 2 + n−1 2 −1. Therefore, at least two colors are incident to u. If equality holds, then G′ is 1-regular (except for one vertex z of degree 2 when n −1 is odd). Hence G′ is disconnected (since n > 4), which yields diam(G′) = 2. Choose uv, uw ∈Γ(u) with distinct colors so that either n is even or {v, w} ̸= NG′(z). Since V (H)−{u, v, w} ⊆NG′(v)∪NG′(w), adding uv and uw to G′ completes a rainbow spanning subgraph of H with diameter 2. We may therefore assume that every vertex has at least three incident colors. Let G be any representing subgraph of H. Since diam(G) > 2, we have ∆(G) ≤n −2. Since K2,n−2 ̸⊆G, the vertices with degree n −2 form a clique. Pick y, z ∈V (G) such that dG(y, z) = 3, which requires dG(y) + dG(z) ≤n −2. Let a = n −2 −(dG(y) + dG(z)). Since \|E(G)\| = n−2 2 + n−1 2 + 1, X w / ∈{y,z} dG(w) = 2\|E(G)\| −(n −2 −a) = (n −2)(n −3) + (a + b), () where b = 2 for even n and b = 3 for odd n. Thus \|X\| ≥a + b, where X is the set of vertices with degree n −2 in G, with equality only if all vertices outside X ∪{y, z} have degree n −3. Indeed, with W = V (G) −X −{y, z}, we have \|X\| = a + b + c, where c = \|W\|(n −3) −P w∈W dG(w). Since a, c ≥0, we have \|X\| ≥2. In particular, we have proved that if H has no rainbow spanning subgraph with diameter 2, then every representing subgraph of H has at least b vertices of degree n −2. Suppose δ(G) ≤1. Choose u, v ∈V (G) with u having smallest degree and v smallest among the others. Since \|E(G)\| = n−2 2 + n+1 2 , at least n+1 2 edges are incident to {u, v} in G, with equality only when G −{u, v} is complete. Recall that in H at least three colors are incident to u. If all vertices other than u have degree at least 3 in G (which includes all cases with n > 6), then we can bring in an edge of another color at u in place of the edge of that color in G, and this will increase the minimum degree. The remaining case is n ∈{5, 6} with three edges incident to {u, v} in G and G −{u, v} being a complete graph. Again we can increase the minimum degree, since the three edges incident to {u, v} can be chosen to be uv and edges of distinct other colors at u and v. Hence we may choose a representing subgraph G with δ(G) ≥2. Deﬁne y, z, X, W, a, b, c as before. Since dG(y, z) = 3, each vertex of X has y or z as its only nonneighbor and is 4 adjacent to all of W; hence {y, z} is the only pair with no common neighbor. By symmetry, we may assume \|X ∩NG(z)\| ≥\|X ∩NG(y)\|. Choose x ∈X ∩NG(z). The color of xy in H appears on an edge xw in G, since otherwise H has a spanning rainbow star with center x. Let G′ = G −xw + xy. If w ̸= z, then diam(G′) = 2; hence we may assume w = z. The exchange reduces the degree only at z; if dG(z) ≥3, then δ(G′) ≥2. Let kG = \|X ∩NG(z)\|−\|X ∩NG(y)\|. We have kG′ < kG if kG ≥2. Hence we may assume kG ∈{0, 1} or dG(z) = 2 = kG. If dG(z) = 2 = kG, then \|X\| = 2 and NG(z) = X, and the color on xiy appears also on xiz, where X = {x1, x2}. Now \|X\| = a + b + c and b ≥2 requires a = c = 0 and b = 2. With a = n −2 −(dG(y) + dG(z)) and dG(z) = 2, we also have NG(y) = W. Let e be the edge in G having the same color as yz. In G, every vertex except y neighbors both vertices of X, so every nonadjacent pair not involving y has at least two common neighbors. Hence if e is not incident to y, then diam(G −e + zy) = 2. If e = wy for some w ∈W, then let G′ = G −{x1z, wy} + {x1y, yz}, having the same colors as G. At each vertex, at most one edge was removed, so it suﬃces to check that dG′(y, v) ≤2 for all v. Since y reaches z directly and all other vertices via x1, we have diam(G′) = 2. We may therefore assume both that δ(G) ≥2 and that NG(y) ∩X and NG(z) ∩X partition X into sets diﬀering in size by at most 1. Furthermore, since y and z each have a neighbor in X, any two vertices other than {y, z} have a common neighbor in X. Let e be the edge in G with the same color as yz, and let ˆ G = G −e + yz. If e has neither endpoint in X, then deleting e yields all pairs except {y, z} with a common neighbor, so diam( ˆ G) = 2. We may therefore assume that e has an endpoint in X. Note that since δ(G) ≥2 and a = n −2 −(dG(y) + dG(z)), we have dG(y), dG(z) ≤n −4. Case 1: \|X\| ≥4. From the argument above, y and z each have at least two neighbors in X. Hence any pair other than {y, z} has at least two common neighbors in X. Thus deleting e does not increase their distance above 2, and diam( ˆ G) = 2. Case 2: \|X\| = 2, or \|X\| = 3 and n is odd. Since \|X\| = b, we have a = c = 0. Since dG(y), dG(z) ≤n −4, we have d ˆ G(y), d ˆ G(z) ≤n −3. Since also e has an endpoint in X, the representing subgraph ˆ G has fewer than b vertices of degree n −2 (which is forbidden). Case 3: \|X\| = 3 and n is even. Since b = 2, we have {a, c} = {0, 1}. When n = 6, with dG(y), dG(z) ≥2 we have a = 0 and hence c = 1, but then the one vertex of W has degree only 2, which prevents it from being adjacent to all of X. Hence we may assume n ≥8. Let x be an endpoint of e in X. Since z has two neighbors in X and y has one neighbor in X, we have diam(G −e) ≤2 unless e = xy. Hence diam( ˆ G) > 2 requires e = xy, and any pair with distance 3 in ˆ G involves y. Since dG(y) ≥2, still x and y have a common neighbor w ∈W. Since yz ∈E( ˆ G), a vertex w′ with d ˆ G(w′, y) = 3 must lie outside NG(y) ∪NG(z) ∪{y, z}. If a = 0, then there is no such vertex. Hence (a, c) = (1, 0) and 5 there is one vertex w′ in W −(NG(y) ∪NG(z)). However, c = 0 yields dG(w′) = n −3. Since y, z / ∈NG(w′), we have w ∈NG(w′), and hence diam( ˆ G) = 2. With Lemmas 2.2–2.4 and 3.1, the proof of Theorem 1.1 is now complete. References S. Fujita, C. Magnant, and K. Ozeki, Rainbow generalizations of Ramsey theory: a survey. Graphs Combin. 26 (2010), 1–30. S. Jahanbekam and D. B. West, Anti-Ramsey problems for t edge-disjoint rainbow span-ning subgraphs: cycles, matchings, or trees. Submitted, 2014. B. Mont´ agh, Anti-Ramsey numbers of spanning double stars. Acta Univ. Sapientiae Math. 1 (2009), 21–34. J. J. Montellano-Ballesteros and V. Neumann-Lara, An anti-Ramsey theorem on cycles, Graphs Combin. 21 (2005), 343–354. 6
17156	https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:completing-square-quadratics/v/solving-quadratic-equations-by-completing-the-square	Completing the square (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content Algebra 1 Course: Algebra 1>Unit 14 Lesson 7: Completing the square intro Completing the square Worked example: Completing the square (intro) Completing the square (intro) Worked example: Rewriting expressions by completing the square Worked example: Rewriting & solving equations by completing the square Completing the square (intermediate) Math> Algebra 1> Quadratic functions & equations> Completing the square intro © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Completing the square KY.Math: HS.A.19.a, HS.A.19.b, HS.A.19.c, HS.A.3.d Google Classroom Microsoft Teams About About this video Transcript Some quadratic expressions can be factored as perfect squares. For example, x²+6x+9=(x+3)². However, even if an expression isn't a perfect square, we can turn it into one by adding a constant number. For example, x²+6x+5 isn't a perfect square, but if we add 4 we get (x+3)². This, in essence, is the method of completing the square.Created by Sal Khan and CK-12 Foundation. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Inventive and Unique Username 7 years ago Posted 7 years ago. Direct link to Inventive and Unique Username's post “Can someone please post t...” more Can someone please post the link to the "Last video" of which Sal speaks? Answer Button navigates to signup page •Comment Button navigates to signup page (19 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Juzer Shakir 7 years ago Posted 7 years ago. Direct link to Juzer Shakir's post “The link to _"Last Video"...” more The link to "Last Video" that Sal mentions at 0:31 , is here: Hope this helps! 8 comments Comment on Juzer Shakir's post “The link to _"Last Video"...” (50 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more FlyingGlasses 2 years ago Posted 2 years ago. Direct link to FlyingGlasses's post “When would this be more a...” more When would this be more advantageous than using the quadratic formula? Answer Button navigates to signup page •1 comment Comment on FlyingGlasses's post “When would this be more a...” (16 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer FlyingGlasses 2 years ago Posted 2 years ago. Direct link to FlyingGlasses's post “After studying I have fou...” more After studying I have found that completing the square is useful for finding vertex form whereas quadratic formula is useful for finding roots. Comment Button navigates to signup page (20 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... TheBlueFlame a year ago Posted a year ago. Direct link to TheBlueFlame's post “I don't understand any of...” more I don't understand any of this! Can someone please help? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “I gotchu. When completing...” more I gotchu. When completing the square, the number we add to both sides OR add and subtract on one side is: take the coefficient of x, then half then square it. This is so that it completes the perfect square on that side of the equation, and we're able to factor it as such. Why? You see it in perfect square form: (x + a)² = x² + 2ax + a². The coefficient on x is 2a, so a² is half the coefficient then squared. Once we add this constant in a given quadratic, we factor it into (x + a)² form. FYI, the coefficient on x² must be 1 as given by the form. If it's not, divide both sides OR factor out that coefficient, to complete the square. This strategy's useful since it reduces the number of xs in the equation from 2``xs to just 1``x - the perfect square on one side and a number on the other side. Thereafter, we'd just square root both sides and solve for x by manipulation, right? Most importantly, it works for all quadratics. Hope this helps. 8 comments Comment on TheReal3A's post “I gotchu. When completing...” (16 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more TreLearnz 8 months ago Posted 8 months ago. Direct link to TreLearnz's post “Steps to Completing the S...” more Steps to Completing the Square: Start with a quadratic equation in the form 𝑎𝑥^2 +𝑏𝑥+𝑐= 0 : For example, 𝑥^2+6𝑥+5= 0. Move the constant term (𝑐) to the other side of the equation: 𝑥^2+6𝑥= −5. If the coefficient of 𝑥^2 is not 1, divide the entire equation by 𝑎: In this example, the coefficient of 𝑥^2 is already 1. Take half of the coefficient of 𝑥 (𝑏), square it, and add this square to both sides: Here, the coefficient of 𝑥 is 6. Half of 6 is 3, and squaring it gives 3^2= 9 So, add 9 to both sides: 𝑥^2+6𝑥+9 = −5+9 Rewrite the left-hand side as a perfect square trinomial: (𝑥+3)^2 = 4 Solve for 𝑥: Take the square root of both sides: 𝑥+3= ±2 Isolate 𝑥: 𝑥 = −3±2, giving the solutions 𝑥= −1 and 𝑥 = − 5 Example: Let's use these steps for the quadratic function 𝑔(𝑥)=𝑥^2+15𝑥+54 Start with the function: 𝑔(𝑥)=𝑥^2+15𝑥+54 We focus on the 𝑥^2+15𝑥 part and temporarily ignore the constant term. Take half the coefficient of 𝑥 (which is 15), square it to get (15/2)^2 = 56.25, and add and subtract it within the equation: 𝑔(𝑥)= 𝑥^2+15𝑥+(15/2)^2 −(15/2)^2 +54 𝑔(𝑥)= (𝑥+15/2)^2 −225/4 +54 Simplify add the constant term to get the final result: 𝑔(𝑥) = (𝑥+15/2)^2− 9/4 I hope this makes the process of completing the square clearer! Answer Button navigates to signup page •1 comment Comment on TreLearnz's post “Steps to Completing the S...” (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Andrew 7 years ago Posted 7 years ago. Direct link to Andrew's post “Wouldn't it be so much ea...” more Wouldn't it be so much easier just to put it through the quadratic formula? Why go through all the trouble of completing the square? I plugged Sal's second example into the formula and got the answer in about 30 seconds. Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Aukán 7 years ago Posted 7 years ago. Direct link to Aukán's post “Well, Andrew, there are s...” more Well, Andrew, there are some cases in which using the quadratic formula is a better and faster solution, but it doesn't work ALL of the time! Although completing the square is much harder, (I know this because I'm doing it right now) completing the square solves ALL of the problems. So, use the quadratic formula only when you can use it. But there will always be problems that absolutely have to be solved by completing the square! 1 comment Comment on Aukán's post “Well, Andrew, there are s...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more tobiasprogrammer 6 months ago Posted 6 months ago. Direct link to tobiasprogrammer's post “1:31 how Sal know the c i...” more 1:31 how Sal know the c is positive? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 6 months ago Posted 6 months ago. Direct link to Kim Seidel's post “Any number squared will b...” more Any number squared will be positive. positive positive = a positive negative negative = a positive. Hope this helps. Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Mary Miller 6 months ago Posted 6 months ago. Direct link to Mary Miller's post “Shouldn't these videos be...” more Shouldn't these videos be before the quadratic formula, since the quadratic formula is derived from completing the square? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer iAD63! 6 months ago Posted 6 months ago. Direct link to iAD63!'s post “Yes, but you need to know...” more Yes, but you need to know what the idea is behind completing the square before being taught an easier way to do it. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more thea.z.cabrera a year ago Posted a year ago. Direct link to thea.z.cabrera's post “where did 2 come from at ...” more where did 2 come from at 8:29 ? like why are we suddenly multiplying something by 2 to get -3 Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “The process of completing...” more The process of completing the square is based upon the pattern created with you square a binomial: (x+a)^2 = (x+a)(x+a) = x^2 + ax + ax + a^2 = x^2 + 2ax + a^2 To complete the square we need to add "a^2". To do that we need to first find the value of "a". Notice in the pattern, the middle term is "2ax". So, what every number is the coefficient of "x" represents "2a". If you divide that number by 2, you can find the value of "a". In Sal's equation, the middle term is -3x. So, 2a=-3 making a=-3/2. He then does (-3/2)^2 = 9/4 to find "a^2" and adds it to both sides of the equation. The left side of the equation is now a perfect square trinomial. Hope this helps. 1 comment Comment on Kim Seidel's post “The process of completing...” (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Points 12 years ago Posted 12 years ago. Direct link to Points's post “Around 6:40, Sal divides ...” more Around 6:40 , Sal divides the quadratic equation by 5. This process makes the coefficient of x^2 equal to 1. My question is does the coefficient of x^2 need to be 1 to complete the square. Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Sid 12 years ago Posted 12 years ago. Direct link to Sid's post “No, but it makes it easie...” more No, but it makes it easier in most if not all cases. 3 comments Comment on Sid's post “No, but it makes it easie...” (0 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more adityavobu688 3 months ago Posted 3 months ago. Direct link to adityavobu688's post “I feel smart but my frien...” more I feel smart but my friends all know way more than me 😢 Answer Button navigates to signup page •1 comment Comment on adityavobu688's post “I feel smart but my frien...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer iAD63! 3 months ago Posted 3 months ago. Direct link to iAD63!'s post “This is actually a real p...” more This is actually a real paradox. Many people feel like their friends are smarter and funnier than they are. This is because smarter and funnier people tend to have more friends leading to this illusion. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript In this video, I'm going to show you a technique called completing the square. And what's neat about this is that this will work for any quadratic equation, and it's actually the basis for the quadratic formula. And in the next video or the video after that I'll prove the quadratic formula using completing the square. But before we do that, we need to understand even what it's all about. And it really just builds off of what we did in the last video, where we solved quadratics using perfect squares. So let's say I have the quadratic equation x squared minus 4x is equal to 5. And I put this big space here for a reason. In the last video, we saw that these can be pretty straightforward to solve if the left-hand side is a perfect square. You see, completing the square is all about making the quadratic equation into a perfect square, engineering it, adding and subtracting from both sides so it becomes a perfect square. So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here. There has to be some number here that if I have my number squared I get that number, and then if I have two times my number I get negative 4. Remember that, and I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet, but we know a couple of things. When I square things-- so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, that has to be-- sorry, x squared minus 2ax-- this right here has to be 2ax. And this right here would have to be a squared. So this number, a is going to be half of negative 4, a has to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side of the equation. You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4 it's not going to be equal to 5 anymore. It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now, we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. Now what is this? What number when I multiply it by itself is equal to 4 and when I add it to itself I'm equal to negative 2? Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have skipped this step and written x minus 2 squared is equal to 9. And then you take the square root of both sides, you get x minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0. And then we would say that x is equal to 5 or x is equal to negative 1. And in this case, this actually probably would have been a faster way to do the problem. But the neat thing about the completing the square is it will always work. It'll always work no matter what the coefficients are or no matter how crazy the problem is. And let me prove it to you. Let's do one that traditionally would have been a pretty painful problem if we just tried to do it by factoring, especially if we did it using grouping or something like that. Let's say we had 10x squared minus 30x minus 8 is equal to 0. Now, right from the get-go, you could say, hey look, we could maybe divide both sides by 2. That does simplify a little bit. Let's divide both sides by 2. So if you divide everything by 2, what do you get? We get 5x squared minus 15x minus 4 is equal to 0. But once again, now we have this crazy 5 in front of this coefficent and we would have to solve it by grouping which is a reasonably painful process. But we can now go straight to completing the square, and to do that I'm now going to divide by 5 to get a 1 leading coefficient here. And you're going to see why this is different than what we've traditionally done. So if I divide this whole thing by 5, I could have just divided by 10 from the get-go but I wanted to go to this the step first just to show you that this really didn't give us much. Let's divide everything by 5. So if you divide everything by 5, you get x squared minus 3x minus 4/5 is equal to 0. So, you might say, hey, why did we ever do that factoring by grouping? If we can just always divide by this leading coefficient, we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way. And then what do we get if we add 4/5 to both sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. Now, anything you do to one side of the equation, you've got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. x squared minus 3x plus 9/4 is equal to 61/20. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the square root of 61/20. And this is a crazy number and it's hopefully obvious you would not have been able to-- at least I would not have been able to-- get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. And 3/2-- let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24. This crazy 3.2464, I'll just write 3.246. So this is approximately equal to 3.246, and that was just the positive version. Let's do the subtraction version. So we can actually put our entry-- if you do second and then entry, that we want that little yellow entry, that's why I pressed the second button. So I press enter, it puts in what we just put, we can just change the positive or the addition to a subtraction and you get negative 0.246. So you get negative 0.246. And you can actually verify that these satisfy our original equation. Our original equation was up here. Let me just verify for one of them. So the second answer on your graphing calculator is the last answer you use. So if you use a variable answer, that's this number right here. So if I have my answer squared-- I'm using answer represents negative 0.24. Answer squared minus 3 times answer minus 4/5-- 4 divided by 5-- it equals--. And this just a little bit of explanation. This doesn't store the entire number, it goes up to some level of precision. It stores some number of digits. So when it calculated it using this stored number right here, it got 1 times 10 to the negative 14. So that is 0.0000. So that's 13 zeroes and then a 1. A decimal, then 13 zeroes and a 1. So this is pretty much 0. Or actually, if you got the exact answer right here, if you went through an infinite level of precision here, or maybe if you kept it in this radical form, you would get that it is indeed equal to 0. So hopefully you found that helpful, this whole notion of completing the square. Now we're going to extend it to the actual quadratic formula that we can use, we can essentially just plug things into to solve any quadratic equation. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. 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Watch this engaging webinar featuring Katelyn Reed, Selective Mutism Program Director, and Dr. Taylor Hicks-Hoste, child psychologist and SM expert, as they share strategies for using reinforcement systems in schools to support children with selective mutism. Perfect for educators, parents, and mental health professionals looking to create effective and supportive classroom environments. Find Thriving Minds Clinicians in the News! Check here for Selective Mutism articles, updated regularly. For more content/videos please visit our YouTube channel. More Books About Selective Mutism Maya's Voice By Wen-Wen Cheng Understanding Katie By Elisa Shipon-Blum Charli's Choices By LCSW-R Marian B. Moldan The Selective Mutism Resource Manual: 2nd Edition (A Speechmark Practical Sourcebook) By Maggie Johnson, Alison Wintgens Anxious Kids, Anxious Parents: 7 Ways to Stop the Worry Cycle and Raise Courageous and Independent Children By Reid Wilson, Lynn Lyons LICSW The Selective Mutism Treatment Guide: Manuals for Parents Teachers and Therapists. Second Edition: Still waters run deep By Ruth Perednik Helping Your Child with Selective Mutism: Practical Steps to Overcome a Fear of Speaking By Ph.D. Angela E. McHolm, Ph.D. Charles E. Cunningham, Melanie K. Vanier Helping Children with Selective Mutism and Their Parents: A Guide for School-Based Professionals By Christopher Kearney Ph.D. Selective Mutism Websites The Child Mind Institute Anxiety Canada Canadian site on Selective Mustism Selective Mutism Learning University Selective Mutism Association Selective Mutism Foundation Child Anxiety Network Andrew Kukes Foundation for Social Anxiety Brighton 8163 West Grand River Road, Ste. 100, Brighton MI 48114 (810) 225-3417office@thrivingminds.info Learn more Chelsea 350 N. Main St #220 Chelsea, MI 48118 (734) 433-5100admin@thrivingminds.info Learn more Livonia 31478 Industrial Road, Ste. 300 Livonia, MI 48150 (734)245-2500manager1@thrivingminds.info Learn more Ada (Grand Rapids) 4690 Fulton St E, Ste. 102 Ada, MI 49301 (616) 425-7701 GRoffice@thrivingminds.info Learn more Sign up to receive news, updates, and offerings from Thriving Minds! Email Address Sign Up Thank you! Sign up to receive Selective Mutism news from Thriving Minds! 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17158	https://brainly.com/question/28146330	[FREE] What is the relationship between the right angle and the other two acute angles in a right triangle? - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +20,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +47,1k Ace exams faster, with practice that adapts to you Practice Worksheets +7,6k Guided help for every grade, topic or textbook Complete See more / Social Studies Textbook & Expert-Verified Textbook & Expert-Verified What is the relationship between the right angle and the other two acute angles in a right triangle? 1 See answer Explain with Learning Companion NEW Asked by gabbieheart3480 • 07/28/2022 0:10 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 477857 people 477K 0.0 0 Upload your school material for a more relevant answer Two acute angles of a right-angled triangle are "Complementary Angles." In a right triangle, the sine of one acute angle A is equal to the cosine of the other acute angle B. We know that these acute angles are supplementary angles because the sum of the measurements of these acute angles in a right triangle is 90°. Two right triangles are similar by the similarity between their angles if they share the measure of an acute angle. Corresponding side length ratios in the triangle are the same. The ratio of the side lengths of a right triangle depends only on the acute angle. A right triangle has one 90 degree angle and two acute angles (< 90 degrees). Since the sum of the interior angles of a triangle is always 180 degrees, the two sides of the right angle of the triangle are called the legs, and the opposite side of the right angle is called the hypotenuse. Learn more about Triangle at brainly.com/question/17335144 SPJ4 Answered by abhaydwive123 •787 answers•477.9K people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 477857 people 477K 0.0 0 Light and Matter - Benjamin Crowell Celestial Mechanics - Jeremy Tatum Non-Euclidean Geometry - Henry Manning Upload your school material for a more relevant answer In a right triangle, the two acute angles are complementary, meaning they sum to 90 degrees. The sine of one angle is equal to the cosine of the other. This relationship is fundamental in trigonometry and geometry. Explanation In a right triangle, one angle measures 90 degrees, and the other two angles are acute, meaning they each measure less than 90 degrees. The relationship between the right angle and the two acute angles is that they are complementary. This means that the sum of the two acute angles is always equal to 90 degrees. This can be expressed mathematically as follows: If angle A and angle B are the acute angles, then: A+B=9 0∘ Each acute angle's sine and cosine also relate to each other. For instance, the sine of one acute angle is equal to the cosine of the other acute angle. This can be shown as: If A is the measure of one acute angle: sin(A)=cos(B) and vice versa: sin(B)=cos(A) This trigonometric relationship is part of the foundation of trigonometry, which studies the relationships between the angles and sides of triangles. Also, for two right triangles to be similar, they must share equal acute angles, which facilitates comparing their sides using corresponding ratios. So, in summary, the right angle in a right triangle dictates that the sum of the other two acute angles must always add up to 90 degrees, making them complementary. This is vital for various applications in geometry and trigonometry. Examples & Evidence For example, if one acute angle measures 30 degrees, then the other acute angle must measure 60 degrees, since 30 + 60 = 90 degrees. This illustrates the complementary nature of the acute angles in a right triangle. The definitions of complementary angles and the properties of right triangles are standard in geometry. Various texts in mathematics establish these relationships, including the basic principles of triangle properties and trigonometric definitions. Thanks 0 0.0 (0 votes) Advertisement gabbieheart3480 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Social Studies solutions and answers Community Answer When decision making is delegated to lower-level managers and employees more familiar with local conditions than top managers it is called:___ Community Answer What should you do immediately if a boat motor catches fire? a) signal for help. b) shut off the fuel supply. c) abandon the boat. d) grab a fire extinguisher. Community Answer 22 If a triangle is a right triangle, then the other two angles must be acute. Community Answer 4 In a right triangle, one of the acute angles is twice as large as the other acute angle. Find the measure of the two acute angles. Community Answer 7 You want to restrict the values entered in a cell to a specified set, such as hop, skip, jump. Which type of data validation should you use?. Community Answer 26 Explain four reasons the youth do not want to participate in civic life. Community Answer 4.1 5 as a bystander, which approach may not work when the person you’re trying to stop is likely to become defensive or hostile? New questions in Social Studies Policies are primarily based on A. research and statistics B. international statistics and scientific hypotheses C. what has worked in the past in similar situations D. what has not worked in the past in similar situations [When connecting past and present issues, it is best to follow a series of steps. Which step is missing from the pattern? 1. Identify common issues in the past and the present. 2. Research supporting evidence. 3. [Missing Step] 4. Make connections. 5. Draw conclusions.A. Prepare an analysis report. B. Establish cause and effect. C. Validate the outcomes of the two events. D. Compose a policy to improve future issues.]( "When connecting past and present issues, it is best to follow a series of steps. Which step is missing from the pattern? Identify common issues in the past and the present. Research supporting evidence. [Missing Step] Make connections. Draw conclusions. A. Prepare an analysis report. B. Establish cause and effect. C. Validate the outcomes of the two events. D. Compose a policy to improve future issues.") Do you think an attempt to "balance the ticket" is an acceptable method of selecting a Vice President? Why or why not? Do you think an attempt to "balance the ticket" is an acceptable method of selecting a Vice President? Why or why not? At the national level, public offices can be filled by election or appointment. Why do you think the Constitution does not provide for the presidency to be filled by appointment if the office becomes vacant? Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
17159	https://eyewiki.org/Eczema_Herpeticum	Create account Log in View form View source View history Eczema Herpeticum From EyeWiki Jump to:navigation, search All content on Eyewiki is protected by copyright law and the Terms of Service. This content may not be reproduced, copied, or put into any artificial intelligence program, including large language and generative AI models, without permission from the Academy. Article initiated by: G. Conner Nix, Kourtney Houser, MD All contributors: Vatinee Y. Bunya, MD, MSCE, Colleen Halfpenny, M.D., Augustine Hong, MD, Kelvin Chong, MB ChB FHKAM(Ophth), Kourtney Houser, MD, Michael T Yen, MD, Nahyoung Grace Lee, MD, Jake Uy Sebastian, MD (DPBO) Assigned editor: Kelvin Chong, MB ChB FHKAM(Ophth), Ravi Patel, MD, MBA Review: Assigned status Up to Date by Kelvin Chong, MB ChB FHKAM(Ophth) on September 28, 2025. \| \| \| add \| \| Contributing Editors: \| add \| Contents 1 Disease Entity 1.1 Disease 1.2 Epidemiology 1.3 Etiology and Pathophysiology 1.4 Risk Factors 1.5 Primary Prevention 2 Diagnosis 2.1 Clinical Presentation 2.2 Ocular Involvement 2.3 Diagnostic Procedures 2.4 Differential Diagnoses 3 Management 3.1 General Treatment 3.2 Treatment of Ocular Involvement 3.3 Complications and Prognosis 4 References Disease Entity Eczema herpeticum, or Kaposi’s varicelliform eruption Disease Eczema herpeticum (EH), sometimes referred to as Kaposi’s varicelliform eruption, is a herpes simplex virus (HSV) infection of the skin that occurs in the setting of an underlying inflammatory dermatosis, most commonly atopic dermatitis . First described (and assumed to be of fungal etiology) in 1887 by Austrian physician Moritz Kaposi, EH presents as an eruption of vesicles that can be accompanied by fever, malaise, and lymphadenopathy . The most frequent sites of infection include the areas of the head, face, neck, and trunk that are already affected by atopic dermatitis. EH can progress to a systemic infection with severe complications, including encephalitis and septic shock. EH is also potentially vision-threatening, as it can rarely advance to severe herpetic ocular disease. Epidemiology EH can affect patients at any age, but classically presents in childhood. The majority of patients with EH have atopic dermatitis (AD), a common skin condition that affects between 8.7 and 18.1% of the United States pediatric population. About 10 to 20% of patients with atopic dermatitis develop EH. A 2018 analysis of 4,655 children hospitalized in the United States for eczema herpeticum found an association with younger age and non-white ethnicity, particularly African American and Asian ethnicities. There is no predilection for gender. Etiology and Pathophysiology Eczema herpeticum and other infections comprise the major complication of atopic dermatitis. This preponderance for infection is multifactorial and attributable to defects in the skin barrier, as well as to inflammation and immune dysregulation. Notably, the skin of patients with AD is especially susceptible to colonization with S. aureus. Risk Factors Generally speaking, the best understood risk factor for development of EH is disruption of the epidermal barrier. Most adults (~60%) and ~20% of children in the general population have serologic evidence of HSV-1 exposure, suggesting that viral exposure alone is not sufficient to cause EH. Instead, EH occurs in patients with a preexisting erosive dermatosis, which is usually, though not exclusively, atopic dermatitis. Risk factors for the development of EH specifically in patients with atopic dermatitis include more severe and/or early-onset atopic dermatitis, high total serum IgE/peripheral eosinophilia, and atopic comorbidities such as asthma or food allergies. Additionally, a history of S. aureus skin infections is a significant risk factor for the development of EH among patients with atopic dermatitis. Apart from atopic dermatitis, EH has been described in association with burns, skin grafts, an immunocompromised state, pemphigus foliaceus, ichthyosis vulgaris, bullous pemphigoid, Darier disease, Grover disease, Hailey-Hailey disease, Sezary syndrome, dyskeratosis follicularis, mycosis fungoides, psoriasis, pityriasis rubra pilaris, rosacea, seborrheic dermatitis, and both allergic and irritant contact dermatitis. There have been some genetic risk factors found through whole genome sequencing. Silencing genes SIDT2 and RBBP8NL in normal human primary keratinocytes has demonstrated increased replication of HSV-1. Primary Prevention Because EH is most often a complication of atopic dermatitis, and given that HSV exposure among the general population is extremely common, primary prevention of EH can be focused on control of AD flares. This may be achieved by various means, including avoidance of irritants and allergens that can trigger AD, identification and avoidance of food allergens (especially in pediatric patients), and the use of loose-fitting clothing and indoor temperature control to decrease skin irritation. Diagnosis The diagnosis of eczema herpeticum is clinical and can be supported by various studies, as discussed below. Clinical Presentation Eczema herpeticum often presents acutely and has the potential to be fatal. Presentation consists of widespread, painful clusters of punctated vesicopustules followed by erosive “punched-out” ulcers with hematic crusts, mostly on the head, neck, and trunk region. The episode originates in areas affected by atopic dermatitis or, less commonly, by another dermatologic condition; it then spreads to involve normal skin over one to two weeks. Once the vesicles have crusted over to form eroded pits, healing without scarring occurs over 2-6 weeks. Patients can present with disseminated vesicles, skin breakdown, viremia, fever, and lymphadenopathy, or they may present with exclusively cutaneous findings in the absence of systemic symptoms. Clinicians should be aware of “EH incognito,” a presentation of EH that is easily mistaken for impetigo and most often seen in patients with severe AD and recurrent EH. Ocular Involvement The primary ophthalmologic concern in eczema herpeticum is the spread of herpes simplex virus (HSV) to the eye or eyes. The spread of a cutaneous herpes virus infection from the eyelids to the cornea is a known mechanism of herpes keratitis, and up to half of patients with herpetic blepharoconjunctivitis also have herpetic corneal infection. However, cases of ocular disease in the setting of EH are infrequently reported. Symptoms of HSV eye infection include redness, pain, foreign body sensation, photophobia, tearing, and decreased visual acuity. HSV infection tends to involve both the upper and lower eyelids. Compared to adults, children tend to experience more severe herpetic ocular disease that may be bilateral and associated with multiple corneal or conjunctival dendrites, as well as more severe secondary corneal scarring and astigmatism. Conjunctival involvement can present as injection with acute unilateral follicular conjunctivitis, with or without conjunctival dendrites or geographic ulceration. Corneal epithelial disease secondary to HSV infection can present as macropunctate keratitis, dendritic keratitis, or a geographic ulcer. Herpetic corneal lesions have heaped edges made up of swollen epithelial cells; these swollen cells stain well with rose bengal or lissamine green, while the central ulceration stains with fluorescein. As epithelial dendrites resolve, subepithelial scars and haze, or “ghost dendrites,” may develop. Corneal sensation may be decreased in HSV infection and can be assessed prior to instillation of topical anesthetic during ophthalmologic examination. Sterile neurotrophic ulceration may also be observed. It may go unresolved or worsen despite antiviral therapy, and may be associated with stromal melting or perforation. Corneal stromal disease secondary to HSV infection can present as disciform (non-necrotizing) keratitis or, less commonly, necrotizing interstitial keratitis. Uveitis may develop in the setting of corneal stromal disease. Posterior segment involvement is rare. Ophthalmologic examination has been recommended as a routine part of EH workup. More information on Herpes Simplex Virus Keratitis can be found here. Diagnostic Procedures Delayed or missed diagnosis of EH can have devastating consequences, including blindness and death. The diagnosis can be confirmed by viral culture, polymerase chain reaction for viral DNA in vesicular fluid, skin scraping for Tzanck smear, or electron microscopy or immunofluorescence to identify HSV-infected cells. However, the sensitivity of these approaches is low and the importance of clinical suspicion and quick intervention cannot be overemphasized, especially in patients with a history of atopic dermatitis. The diagnosis of ocular HSV infection is also clinical and does not require confirmatory testing, but if there is doubt corneal scrapings can be obtained for Giemsa stain of multinucleated giant cells. Serologic testing has low specificity and is not routinely performed. When impetigo is in the differential, a positive skin surface bacterial culture for Staphylococcus or Streptococcus species does not exclude EH, and is actually a common finding in EH cases. Differential Diagnoses The differential diagnosis for EH includes: Impetigo Primary varicella infection Cellulitis Herpes zoster ophthalmicus Eczema vaccinatum Eczema coxsackium Eczema molluscatum Pustular psoriasis Drug hypersensitivity reaction Vasculitis Bullous lupus erythematosus Scabies Management General Treatment Prompt treatment of eczema herpeticum is important to resolve acute symptoms and to prevent or ameliorate complications. The indicated treatment is administration of acyclovir. There are no clear guidelines about which patients should be hospitalized to receive intravenous acyclovir versus managed as outpatients with less-bioavailable oral acyclovir. Regardless, patients with severe disease and immunocompromised patients should be admitted to receive systemic antiviral therapy. If bacterial superinfection is suspected, treatment should include systemic antibiotics after obtaining appropriate bacterial culture. Treatment of Ocular Involvement If the HSV skin infection spreads to involve the eyelid margins, the indicated treatment is ganciclovir 0.15% ophthalmic gel or trifluridine 1% drops added to the eye five times per day. In small children, vidarabine 3% ointment five times per day is useful. These treatments should be continued for one to two weeks until signs resolve. For HSV conjunctivitis, three options for management are ganciclovir 0.15% ophthalmic gel, vidarabine 3% ointment, or trifluridine 1% drops five times per day. Treatment should last one to two weeks, and reevaluation is recommended if the conjunctivitis does not resolve after this period. For herpes keratitis, options for management include ganciclovir 0.15% ophthalmic gel five times per day, vidarabine 3% ointment five times per day, or trifluridine 1% drops nine times per day. If compliance with these treatments is an issue, for example in pediatric patients, intravenous or oral antiviral agents (e.g. acyclovir) are acceptable alternatives to topical therapy. Cycloplegic agents like cyclopentolate 1% three times per day can be considered if photophobia or anterior chamber reaction is present. The use of topical steroids is contraindicated and should be quickly tapered off. Adjunctive debridement of infected corneal epithelial cells at the slit lamp can be done in addition to antiviral therapy. If epithelial defects do not resolve after 7 to 14 days, topical antiviral therapy should be withdrawn and preservative-free artificial tears or an antibiotic ointment should be used four to eight times per day with close monitoring and follow-up over several days. A lack of resolution after this time should also lead to investigation of possible bacterial coinfection, Acanthomoeba keratitis, noncompliance with therapy, and topical antiviral toxicity. Regarding the latter, it has been suggested that topical ganciclovir gel carries a lower risk of corneal toxicity than trifluridine. Treatment of corneal stromal disease depends on disease severity. For mild cases of disciform (non-necrotizing) keratitis, antiviral prophylaxis and cycloplegic therapy are recommended. Cycloplegic therapy is similarly recommended for moderate to severe cases, in addition to a topical steroid (warning: Do not initiate topical steroid therapy while an active epithelial lesion is present!). If epithelial defects are present, topical antibiotics may be used adjunctively; if intraocular pressure is elevated, aqueous suppressants may be used, avoiding prostaglandin analogues. Necrotizing interstitial keratitis is managed as severe disciform keratitis. Patients with necrotizing interstitial keratitis require daily follow-up or admission to monitor for perforation. If the cornea perforates, tissue adhesive or tectonic keratoplasty may be necessary. Patients being treated for ocular HSV should follow up for repeat examination two to seven days after initiating treatment, and again every one to two weeks depending on examination findings. The size of any epithelial defect or ulcer, depth of corneal involvement, corneal thickness, intraocular pressure, and anterior chamber reaction should all be evaluated. Complications and Prognosis Eczema herpeticum generally has a good prognosis when patients receive prompt intervention with antiviral therapy, but scarring may persist long after resolution of acute changes . A common complication of EH is bacterial superinfection, with the most common organisms being Staphylococcus aureus, Streptococcus pyogenes, and Pseudomonas aeruginosa. If the cutaneous HSV infection disseminates, systemic infection can occur with fever, malaise, multiple organ involvement, septic shock, meningitis, and encephalitis; such dissemination has been associated with a mortality rate from 1% to 9%. As discussed above, ocular HSV infection is among the most feared complications of EH, and corneal transplant may be indicated in cases of postherpetic scarring that significantly affects vision. Close monitoring after resolution of an episode of EH is recommended because about 50% of patients may experience recurrence. References ↑ 1.0 1.1 1.2 1.3 1.4 Ferrari B, Taliercio V, Luna P, Abad ME, Larralde M. Kaposi's varicelliform eruption: A case series. Indian Dermatol Online J. 2015 Nov-Dec;6(6):399-402. doi: 10.4103/2229-5178.169714. ↑ 2.0 2.1 2.2 Wollenberg A, Zoch C, Wetzel S, Plewig G, Przybilla B. Predisposing factors and clinical features of eczema herpeticum: a retrospective analysis of 100 cases. J Am Acad Dermatol. 2003; 49:198–205 ↑ 3.0 3.1 Hsu DY, Shinkai K, Silverberg JI. Epidemiology of Eczema Herpeticum in Hospitalized U.S. Children: Analysis of a Nationwide Cohort. J Invest Dermatol. 2018 Feb;138(2):265-272. doi: 10.1016/j.jid.2017.08.039. Epub 2017 Sep 18. ↑ 4.0 4.1 4.2 4.3 Santmyire-Rosenberger BR, Nigra TP. Psoriasis herpeticum: three cases of Kaposi's varicelliform eruption in psoriasis. J Am Acad Dermatol, 53 (2005), pp. 52-56 ↑ 5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 Luca NJ, Lara-Corrales I, Pope E. Eczema herpeticum in children: clinical features and factors predictive of hospitalization. J Pediatr. 2012 Oct;161(4):671-5. doi: 10.1016/j.jpeds.2012.03.057. Epub 2012 May 9. ↑ David TJ, Longson M. Herpes simplex infections in atopic eczema. Arch Dis Child, 60 (1985), pp. 338-343 ↑ 7.0 7.1 Ong PY, Leung DY. Bacterial and Viral Infections in Atopic Dermatitis: a Comprehensive Review. Clin Rev Allergy Immunol. 2016 Dec;51(3):329-337. Review. ↑ Margolis TP, Ostler HB. Treatment of ocular disease in eczema herpeticum. Am J Ophthalmol. 1990 Sep 15;110(3):274-9. ↑ 9.0 9.1 9.2 9.3 9.4 9.5 9.6 Siegfried EC, Hebert AA. Diagnosis of Atopic Dermatitis: Mimics, Overlaps, and Complications. J Clin Med. 2015 May 6;4(5):884-917. doi: 10.3390/jcm4050884. Review. ↑ Shaw TE, Currie GP, Koudelka CW, Simpson EL. Eczema prevalence in the United States: data from the 2003 National Survey of Children's Health. J Invest Dermatol, 131 (2011), pp. 67-73. ↑ Novelli VM, Atherton DJ, Marshall WC. Eczema herpeticum. Clinical and laboratory features. Clin Pediatr (Phila), 27 (1988), pp. 231-233. ↑ 12.0 12.1 12.2 12.3 Sun D, Ong PY. Infectious complications in atopic dermatitis. Immunol Allergy Clin N Am, 37 (2017), pp. 75-93. ↑ 13.0 13.1 Olson J, Robles DT, Kirby P, Colven R. Kaposi varicelliform eruption (eczema herpeticum). Dermatol Online J, 14 (2008), p. 18. ↑ 14.0 14.1 14.2 14.3 Beck LA, Boguniewicz M, Hata T, Schneider LC, Hanifin J, Gallo R, Paller AS, Lieff S, Reese J, Zaccaro D, Milgrom H, Barnes KC, Leung DY. Phenotype of atopic dermatitis subjects with a history of eczema herpeticum. J Allergy Clin Immunol. 2009; 124(2):260–269. ↑ Vogt KA, Lohse CM, El-Azhary RA, Gibson LE, Lehman JS. Kaposi varicelliform eruption in patients with Darier disease: a 20-year retrospective study. J Am Acad Dermatol. 2015 Mar;72(3):481-4. doi: 10.1016/j.jaad.2014.12.001. Epub 2015 Jan 10. ↑ Azmi M, Nasim A, Dodani S, Laiq SM, Mehdi SH, Mubarak M. Kaposi Varicelliform Eruption Associated With Chickenpox in a Liver Transplant Recipient. Exp Clin Transplant. 2018 Jun 28. doi: 10.6002/ect.2017.0282. ↑ Karray M, Souissi A. Kaposi Varicelliform Eruption. [Updated 2019 Mar 19]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2019 Jan-. Available from: ↑ 18.0 18.1 18.2 18.3 Steptoe A, Young-Zvandasara T, Muhtaseb M. A corneal dendritic lesion with a skin eruption: eczema herpeticum, an important differential diagnosis. BMJ Case Rep. 2015 Jan 30;2015. pii: bcr2014208438. doi: 10.1136/bcr-2014-208438. ↑ Bin L, Malley C, Taylor P, PreethiBoorgula M, Chavan S, Daya M, Mathias M, Shankar G, Rafaels N, Vergara C, Potee J, Campbell M, Hanifin JM, Simpson E, Schneider LC, Gallo RL, Hata T, Paller AS, De Benedetto A, Beck LA, Ong PY, Guttman-Yassky E, Richers B, Baraghoshi D, Ruczinski I, Barnes K, Leung DYM, Mathias RA Whole genome sequencing identifies novel genetic mutations in patients with eczema herpeticum. Allergy. 2021 Feb 6. doi: 10.1111/all.14762. Online ahead of print. ↑ 20.0 20.1 Cooper BL. Eczema Herpeticum. J Emerg Med. 2017 Sep;53(3):412-413. doi: 10.1016/j.jemermed.2016.12.004. ↑ 21.0 21.1 Micali G, Lacarrubba F. Eczema Herpeticum. N Engl J Med. 2017 Aug 17;377(7):e9. doi: 10.1056/NEJMicm1701668. ↑ 22.0 22.1 22.2 Sais G, Jucglà A, Curcó N, Peyrí J. Kaposi's varicelliform eruption with ocular involvement. Arch Dermatol. 1994 Sep;130(9):1209-10. ↑ Finlow C, Thomas J. Disseminated herpes simplex virus: a case of eczema herpeticum causing viral encephalitis. J R Coll Physicians Edinb. 2018 Mar;48(1):36-39. doi: 10.4997/JRCPE.2018.108. ↑ Popov Y, Nikolov R, Lalova A. Localized eczema herpeticum with unilateral ocular involvement. Acta Dermatovenerol Alp Pannonica Adriat. 2010 Oct;19(3):35-7. ↑ Higaki S, Inoue Y, Yoshida A, Maeda N, Watanabe H, Shimomura Y. Case of bilateral multiple herpetic epithelial keratitis manifested as dendriform epithelial edema during primary Kaposi's varicelliform eruption. Jpn J Ophthalmol. 2008 Mar-Apr;52(2):127-129. doi: 10.1007/s10384-007-0514-6. Epub 2008 Apr 30. ↑ Varley R, Kletz T. A case of Kaposi's varicelliform eruption (systemic herpes simplex) with dendritic ulceration of the cornea. Br J Dermatol Syph. 1949 May;61(5):166-70. ↑ 27.0 27.1 Kuo CY, Hsu CC, Lin PY. Bilateral corneal geographic ulcers in a patient with eczema herpeticum. Kaohsiung J Med Sci. 2019 May 7. doi: 10.1002/kjm2.12083. ↑ 28.00 28.01 28.02 28.03 28.04 28.05 28.06 28.07 28.08 28.09 28.10 Bagheri N, Wajda, BN. The Wills Eye Manual: Office and Emergency Room Diagnosis and Treatment of Eye Disease. 7th ed. Philadelphia: Wolters Kluwer, 2017. ↑ 29.0 29.1 29.2 Liaw FY, Huang CF, Hsueh JT, Chiang CP. Eczema herpeticum: a medical emergency. Can Fam Physician. 2012;58:1358–61. ↑ 30.0 30.1 Gogou M, Douma S, Haidopoulou K, Giannopoulos A. Herpeticum-like rash in a child with atopic dermatitis: early clinical suspicion is valuable. Sudan J Paediatr. 2018;18(2):53-55. doi: 10.24911/SJP.106-1536167231. 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17160	https://brilliantscholars.wordpress.com/2013/03/03/arithmetic-progression-and-young-gauss-the-prince-of-mathematics/	Skip to content Follow: : RSS Brilliant Training Blog Home About Contributors General Key Technique Level Level 1 Level 2 Level 3 Level 4 Level 5 Math Algebra Combinatorics Geometry Number Theory Physics Solutions Arithmetic Progression and Young Gauss: the Prince of Mathematics by Brilliant on March 3, 2013 As you progress further into college math and physics, no matter where you turn, you will repeatedly run into the name Gauss. Johann Carl Friedrich Gauss is one of the most influential mathematicians in history. Gauss was born on April 30th 1777 in a small German city north of the Harz mountains named Braunschweig. The son of peasant parents(both were illiterate), he developed a staggering number of important ideas and had many more named after him. Many have referred to him as the princeps mathematicorum, or the “Prince of Mathematics.” As part of his doctoral dissertation(at the age of 21), Gauss was one of the first to prove the fundamental theorem of algebra. He went on to publish seminal works in many fields of mathematics including number theory, algebra, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy, optics etc. Number theory was Gauss’s favorite and he referred to number theory as the “Queen of Mathematics”. One of the reasons why Gauss was able to contribute so much math over his lifetime was that he got a very early start. There are many tales of his childhood precociousness. The most famous anecdote of young Gauss, is the time he found the shortcut for calculating sums of an arithmetic progression at the tender age of 10. One day at school, Gauss’s teacher wanted to take a rest and asked the students to sum the integers from 1 to 100 as busy work. After a few seconds, the teacher saw Gauss sitting idle. When asked why he was not frantically doing addition, Gauss quickly replied that the sum was 5050. His classmates and teacher were astonished, and Gauss ended up being the only pupil to calculate the correct answer. ? The story may be apocryphal, and is told different ways in different sources. Nobody is sure which method of summing an arithmetic sequence Gauss figured out as a child. Though there are several ways young Gauss might have solved it, one of them has a concise, intuitive, and elegant visual representation. Figure 1. Consider two sets of marbles as shown in figure 1. The left pile has rows of blue marbles, where the th row contains marbles. The right pile has rows of red marbles, where the th row contains marbles. The total number of blue marbles is given by 1+2 +3+ …+ (n-1)+n while the total number of red marbles is given by n+(n-1)+(n-2) +…+ 2 + 1 and clearly both contain the same number of marbles. Now if we were to add these piles together as shown in figure 2, we would then get a stack with rows, where each row contains marbles. Figure 2. The total number of marbles in the added pile would be . Since both the red pile and the blue pile have an equal number of marbles, each pile must have contributed marbles. Hence, we get the following: To sum all the numbers from 1 to 100 Gausse simply did , which is immensely easier than adding all the numbers from 1 to 100. All sane humans would rather do one addition, one multiplication, and one division than do lots of tedious addition operations. Note that must always be a natural number. Even though the above formula divides by 2, the result will always be a natural number. This is because the numerator will always be conveniently even due to the multiplication properties of parity. For example, could either be even or odd. If is even, then is odd and hence Similarly, if is odd, then is even and hence Therefore, the numerator is always even and is always a natural number. Numbers of the form are called triangular numbers, for reasons well illustrated in the above figures. The first few triangular numbers are It is commonplace to encounter an application of summing an arithmetic sequence, both in classroom problems, and in describing the broader world. It is less common to meet 10 year olds who figure out the tricks of arithmetic progression for themselves. It is even less common for a precocious 10 year old, to grow up to be nearly as prolific as Gauss. Check out this list of things named after Gauss. Rate this: Share this: Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Like Loading... Related From → General 12 Comments Clifford Burgos Lesmoras permalink Heard this story many times and now I know it was the Prince of Mathematics, Gauss. Reply 2. Sir permalink Another dumb post. Reply Brilliant permalink You seem to have strong opinions about the post. If you would like to articulate constructive criticism of any particular post, feel free to do so in the comments, however calling it “dumb” does not really add anything constructive to the conversation. If you would like to offer feedback about the blog in general, feel free to contact any of the public contact emails at Brilliant.org. These include discussions@brilliant.org, support@brilliant.org, and mathematics@brilliant.org. Reply 3. aij9 permalink Let’s calculate the puzzle of chess on this formula. 64 squares be doubled starting from 1 till all 64 squares be added with total sum. Reply 4. saleem khokhar permalink i am trainee and learner in maths, some body can teach me arithematic mean and sequences, geometric means and sequences arithmatics progression. what is logics to resolve the standard form via factorization. saleem khokhar Reply 5. Anonymous permalink Excellent illustration with marbles on this post Reply Brilliant permalink Glad you like the illustration. We like visualizations as well. Reply 6. Anonymous permalink ya as its is outstanding Reply 7. Adienl permalink Finally I know how this formula was arrived at!!Suppose we have to calculate sum between 1 to 100. We can split it into equal halves(1 to 50 and 51 to 100). Here I notice that if i add the 1st term of the 1st half with the last term of the last I get 101. Similarly, on adding the 2nd term with the 99th term i get 101. Finally, we can calculate the number of occurrences of 101(say k) and multiply it with 101 to arrive at the answer.Am I missing anything? Reply 8. David permalink But you (k) is always 2. It doesn’t change. Reply 9. Anonymous permalink You helped me alotHere the details are very clear Reply 10. Confused Hobbyist permalink I’m confused by the equation for the red marbles given after the first diagram. If we’re counting to 100. There’d be 99 marbles in the first row if J marbles in this case means red marbles. But if it’s in the first row it would 1 + 1 – 99 which would be -97. And even if J in this refers to blue marbles it would still be 1 + 1 – 1 which is just one. Everything else in the post makes sense to me but this is driving me insane. I am wrong?! Reply Leave a comment Cancel reply « Binomial Coefficients / Binomial Theorem Pattern Recognition » Join 457 other subscribers About: The Brilliant community is excited about the processes of discovery conducted in the fields of mathematics and science. 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17161	https://scholarworks.bgsu.edu/cgi/viewcontent.cgi?article=1001&context=asor_pub	Published Time: Wed, 27 Sep 2023 15:20:22 GMT Bowling Green State University Bowling Green State University ScholarWorks@BGSU ScholarWorks@BGSU Applied Statistics and Operations Research Faculty Publications College of Business 2006 Effects of Home-Away Sequencing on the Length of Best-of-Seven Effects of Home-Away Sequencing on the Length of Best-of-Seven Game Playoff Series Game Playoff Series Christopher M. Rump Bowling Green State University , cmrump@bgsu.edu Follow this and additional works at: Part of the Social Statistics Commons How does access to this work benefit you? Let us know! How does access to this work benefit you? Let us know! Repository Citation Repository Citation Rump, Christopher M., "Effects of Home-Away Sequencing on the Length of Best-of-Seven Game Playoff Series" (2006). Applied Statistics and Operations Research Faculty Publications . 2. This Article is brought to you for free and open access by the College of Business at ScholarWorks@BGSU. It has been accepted for inclusion in Applied Statistics and Operations Research Faculty Publications by an authorized administrator of ScholarWorks@BGSU. Journal of Quantitative Analysis in Sports Volume 2, Issue 1 2006 Article 5 The Effects of Home-Away Sequencing on the Length of Best-of-Seven Game Playoff Series Christopher M. Rump ∗ ∗Bowling Green State University, cmrump@cba.bgsu.edu Copyright c ©2006 The Berkeley Electronic Press. All rights reserved. The Effects of Home-Away Sequencing on the Length of Best-of-Seven Game Playoff Series Christopher M. Rump Abstract We analyze the number of games played in a seven-game playoff series under various home-away sequences. In doing so, we employ a simple Bernoulli model of home-field advantage in which the outcome of each game in the series depends only on whether it is played at home or away with respect to a designated home team. Considering all such sequences that begin and end at home, we show that, in terms of the number of games played, there are four classes of stochastically different formats, including the popular 2-3 and 2-2 formats both currently used in National Basketball Association (NBA) playoffs. Characterizing the regions in parametric space that give rise to distinct stochastic and expected value orderings of series length among these four format classes, we then investigate where in this parametric space that teams actually play. An extensive analysis of historical 7-game playoff series data from the NBA reveals that this home-away model is preferable to the simpler, well-studied but ill-fitting binomial model that ignores home-field advantage. The model suggests that switching from the 2-2 series format used for most of the playoffs to the 2-3 format that has been used in the NBA Finals since a switch in 1985 would stochastically lengthen these playoff series, creating an expectation of approximately one extra game per playoff season. Such evidence should encourage television sponsors to lobby for a change of playoff format in order to garner additional television advertising revenues while reducing team and media travel costs. KEYWORDS: Playoffs, Tournament, Sequencing, Probability 1 Introduction This article studies a model of multi-game playoff series pairing two sports teams, one of whom is favored with a home-field advantage. The advantage means that a majority of the possible games in the series are played on that team’s home field. We focus on the case of seven-game series, which have been long adopted by the major professional team sports championships. Here four games are scheduled to be played on the favored team’s home field, the other three games to be played away from this home field. Of course all seven games need not always be played since these series are played until one team wins a majority of the games, i.e., four games. The issue of home-away sequencing then naturally arises. If playing at home actually gives an advantage, then this sequencing will play a role in determining the number of games actually played. For example, if the home team enjoys a large probability of winning, then a series beginning with four straight games at home would give a tremendous (and deemed unfair) advantage to that team to win a short series. Beginning with the seminal work of Mosteller (1952), many authors, in-cluding Brunner (1987), Groeneveld and Meeden (1975), Nahin (2000), Simon (1977), Woodside (1989), and Zelinski (1973), have analyzed a simple trun-cated binomial model of the World Series or other best-of-seven game playoff series, in which one team is assigned a probability p of winning any particular game in the series. This model assumes that each game is played indepen-dently of the next, and without regard for widely assumed home-team ad-vantage. Lengyel (1993) extended the analysis of this binomial model to find an expression for the expected length of any best-of-(2 n − 1) game series, al-though a more elegant form was already presented by Maisel (1966). 1 Shapiro and Hamilton (1993) rediscovered Maisel’s elegant result and further recog-nized that this expression surprisingly involves partial sums of the Catalan numbers, Ck = (2kk ) /(k + 1), k = 1 , 2, . . . .Such best-of-seven playoff tournaments, originally conceived to determine the best of two teams from different leagues, are one form of paired statistical comparison as studied by David (1988), Uppuluri and Blot (1974) and Ushakov (1976). Of course, as explained in James et al. (1993), the best teams do not always come out ahead in such comparisons. As a result, many authors, including Appleton (1995), Boronico (1999), Gibbons et al. (1978), Glenn (1960), McGarry (1998) and most recently Marchand (2002), have studied the 1Mosteller (1952) earlier derived the same elegant form for the special case of the World Series when adding the n= 4 wins of the series winner to the expected number of wins by the series loser as expressed in his equation (2). 1Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 effectiveness of a variety of multi-tiered playoff tournaments among several playoff teams. Although the existence of a home-team advantage in sporting events has been documented in recent years by Clarke and Norman (1995), Courneya and Carron (1992) and Harville and Smith (1994), little modelling of this effect has taken place, perhaps tempered by the early analysis of Mosteller (1952) that revealed no significant home advantage in the World Series, at least through the first half-century of play. Bassett and Hurley (1998) appear to be the first to extend the common single-parameter binomial model to a two-parameter model incorporating home and away winning probabilities pH and pA, respectively, for the team with home-field advantage. One of their chief results involves the relatively simple conditions under which the so-called 2-3 playoff format (HHAAAHH) is longer (in expectation) than the 2-2 format (HHAAHAH). Our work extends this analysis by Bassett and Hurley to include all pos-sible home-away sequences involving 7 games, four of which are at played at home including the first and seventh game of the series. We first review the home-away model in Section 2. Then, in Section 3 we show that, in terms of the number of games played, there are four stochastically different format classes, including the popular 2-3 and 2-2 formats. We then compare these four formats in terms of both stochastic and expected length. The expected value orderings among these four format classes partition the two-dimensional parametric space into six regions. The remainder of the paper then validates the effectiveness of the model in explaining the outcomes of playoffs in the National Basketball Association (NBA). In Section 4, we examine the historical data to first estimate the region in which the parameters actually lie, and then study the goodness of fit of the simple binomial model versus the home-away model. We close in Section 5 with a discussion of our results and a prelude of a more general Markov model of game-to-game dependence in playoff series. 2 Model Following Bassett and Hurley (1998), suppose each game in a sequence of playoff games is an independent Bernoulli trial with probability of success pH if the game is played at home and probability of success pA if the game is played away from home. Here “home” refers to the apparently favored team’s court/field/ice and “success” refers to a win by this favored team. The favored team is designated as the team with the home advantage, meaning that four 2Journal of Quantitative Analysis in Sports Vol. 2 , No. 1, Article 5 of the possible seven games in the series are played on that team’s home court. Since the home court usually gives an advantage, pH ≥ pA is typically the case. We will restrict such series to the convention that the series should begin and end at home, i.e., the first and seventh (if necessary) games are scheduled on the home team’s court. With this restriction it is possible to sequence the middle five games in (52 ) = 10 possible ways as listed in Table 1. Format Sequence 2-3 HHAAAHH 2-2 HHAAHAH 2-1 HHAHAAH 1-1-2 HAHHAAH 3-3 HHHAAAH Format Sequence 1-1 HAHAHAH 1-2 HAHAAHH 1-2-1 HAAHAHH 1-2-2 HAAHHAH 1-3 HAAAHHH Table 1: Possible Home-Away Sequences in a 7-Game Series The first entry in each of the two lists is a palindromic sequence. Each of the remaining sequences in the right-hand list is simply the reverse of the corresponding sequence in the left-hand list. The format indicates how the sequence alternates at the beginning of the series, from which the entire se-quence can be inferred. For example, the format 2-1 indicates two home games followed by one away game. From this we infer that the fourth game is home again. Since the last game must be home as well, this implies that games five and six are away. For brevity the formats 1-1-1-1 and 1-1-1-2 have been shortened to 1-1 and 1-2, respectively. In terms of practical use, the National Hockey League (NHL) playoffs pri-marily employ the 2-2 format, although, according to National Hockey League (1997), under certain geographic conditions the higher-ranked team has the option to invoke a 2-3 format in order to limit travel. Since 1924, Major League Baseball’s (MLB) World Series has used the 2-3 format. In the first 20 or so years of that series, organizers tinkered with many seven-game formats such as 1-1, 2-2 and 1-2-2. In the National Basketball Association (NBA) Finals, the 2-3 format has been used since 1985. Prior to that, the 2-2 format was predominant, although the 1-1 and 1-2-2 formats were also used from time to time. The rest of the NBA playoffs (preceding the NBA Finals) are played in the 2-2 format, although in the early days the eastern conference playoffs often used a 1-1 format. 3Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 3 Series Length We shall now explore the effects of the home-away sequencing on the length of a series. Following the notation of Bassett and Hurley (1998), for each format f let hfi (afi ) be the probability that the home (away) team wins the series in i games, and gfi = hfi + afi be the probability that the series is won in i games, i = 4 , 5, 6, 7. Also, let hf = hf 4 hf 5 hf 6 hf 7 be the probability that the home team wins the series under format f , and let af = 1 − hf be the probability that the away team does so. Bassett and Hurley argue that, due to the independence of games, the probability that the home team wins the series is the same for any format, i.e, the probabilities hf (and af ) are invariant to the format f . Extending the argument, they state that the probability that the series lasts seven games is the same for both the 2-3 and 2-2 formats, i.e., g23 7 = g22 7 . This is also due to the independence of games and the fact that both formats end with a seventh game at home, i.e., the first six games contain the same mix of home and away games. By independence, the particular mix does not affect the probability that neither team wins four of these six games, warranting a final seventh game. Since we assume that the seventh game of a series is always played at home, this result extends to all the formats in Table 1. To explore the length of the playoff series under different formats, let Nf indicate the number of games played under the format f , and let Lf = E[Nf ]be its expectation. Since each playoff series consists of at least four games, and the games are assumed to be independent, the sequencing of the last three games will affect the number of games played. Hence, those sequences with the same sequence in the last three games, and therefore the same mix of home and away games over the first four requisite games, will be stochastically equivalent, denoted = st . 2 This establishes Theorem 1. Theorem 1 Assuming the games in a best 4 of 7-game series are independent, only the sequencing of the last three games will affect the number of games played. Thus, under this assumption, series with the same home-away sequence over the last 3 games are stochastically equivalent in length. Hence, • N23 =st N12 =st N121 , • N22 =st N11 =st N122 , • N21 =st N33 =st N112 . 2For a discussion of stochastic order relations, see Ross (1996). 4Journal of Quantitative Analysis in Sports Vol. 2 , No. 1, Article 5 In light of Theorem 1, we will focus on the four stochastically different formats 2-3, 2-2, 2-1 and 1-3. We have already established that, since the last game of each of these formats ends in a home game, the series have the same probability of lasting seven games, i.e., g23 7 = g22 7 = g21 7 = g13 7 . (1) Since the last two games of formats 2-3 and 1-3 are both home games, g23 6 + g23 7 = g13 6 + g13 7 , yielding via (1) g23 6 = g13 6 . (2) Similarly, the last two games of formats 2-2 and 2-1 are identical so that g22 6 = g21 6 . (3) Extending this analysis to include the last three games gave us Theorem 1. We also have g23 4 = g22 4 , (4) as these two formats both have identical structure over the first four games. As a consequence of Equations (1)-(4), we have the following identities: g23 4 + g23 5 = g13 4 + g13 5 (5) g22 4 + g22 5 = g21 4 + g21 5 (6) g23 5 + g23 6 = g22 5 + g22 6 (7) Bassett and Hurley show that L23 − L22 = 2( pH − pA)( pH + pA − 1)( pH + pA − 2pH pA). Hence, since the last term pH + pA − 2pH pA is bounded between 0 and 1, a 2-3 series is longer than a 2-2 series in expectation, L23 ≥ L22 , if and only if pH − pA agrees in sign with ϕ(pH , p A) = ϕ(pA, p H ) = pH + pA − 1. Assuming pH ≥ pA, L23 ≥ L22 holds provided that 1 − pH ≤ pA ≤ pH . It turns out that Bassett and Hurley have proven a stronger result, namely that these are the conditions for a 2-3 series to be stochastically longer than a 2-2 series, N23 ≥st N22 . This follows from their derivation L23 − L22 = 5( g23 5 − g22 5 ) + 6( g23 6 − g22 6 )= 5(( g23 5 + g23 6 ) − (g22 5 + g22 6 )) + ( g23 6 − g22 6 )= g23 6 − g22 6 = g22 5 − g23 5 , 5Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 where the last two equalities follow from (7). In words, the 2-3 format delays the third home game until game six rather than game five. Thus, assuming pH ≥ pA, the home team is less likely to win game five under the 2-3 format. This will delay the home team from winning the series - and thereby lengthen the series - provided that this does not allow the other team to win the series earlier, which can happen if the home team is relatively weak on the road, i.e., pA < 1 − pH .Comparing the 2-2 and 2-1 formats, we see that the difference lies in games 4 and 5. Thus, a 2-2 series is stochastically longer than a 2-1 series, N22 ≥st N21 , if and only if 0 ≤ L22 − L21 = g22 5 − g21 5 = g21 4 − g22 4 = ( h21 4 − h22 4 ) + ( a21 4 − a22 4 )= (pH − pA)p2 H pA + (1 − pH ) − (1 − pA)2(1 − pA)= (pH − pA) [ p2 H pA − (1 − pH )2(1 − pA)] . Thus, N22 ≥st N21 (and L22 ≥ L21 ) if and only if pH − pA agrees in sign with φ(pH , p A) = p2 H pA − (1 − pH )2(1 − pA). Assuming pH ≥ pA, N22 ≥st N21 holds provided (1 − pH )2 p2 H (1 − pH )2 ≤ pA ≤ pH .Comparing the 1-3 and 2-3 formats would seem to be a bit trickier. How-ever, due to Theorem 1, we can compare the 1-3 format to the 1-2-1 format instead. These formats, like in the comparison of the 2-2 and 2-1 formats, differ only in games 4 and 5. However, unlike the 2-2 and 2-1 formats, the 1-3 and 2-3 formats contain two away games rather than home games in their first three games. Hence, N13 ≥st N23 (and L13 ≥ L23 ), if and only if φ(pA, p H ) and pH − pA agree in sign. Here φ(pA, p H ) is a reflection of φ(pH , p A) across the line pA = pH .The previous pairwise comparisons each involved formats that differed in adjacent games. In comparing the length of series that differ in non-adjacent games, we first combine the previous comparisons to make statements about dominance in expected value. For example, comparing formats 2-3 and 2-1, we find L23 − L21 = (L23 − L22 ) + ( L22 − L21 )= (pH − pA) [ 2(1 − 2pH )p2 A − (1 − 6pH + 2 p2 H )pA − (1 − p2 H )] . Thus, L23 ≥ L21 holds if and only if pH − pA agrees in sign with ψ(pH , p A) = 2(1 − 2pH )p2 A − (1 − 6pH + 2 p2 H )pA − (1 − pH )(1 + pH ). 6 Journal of Quantitative Analysis in Sports Vol. 2 , No. 1, Article 5 Similarly, L13 ≥ L22 holds if and only if ψ(pA, p H ) and pH − pA agree in sign since L13 − L22 = ( L23 − L22 ) + ( L13 − L23 ) is a reflection of L23 − L21 across the line pA = pH . Finally, L13 − L21 = (L13 − L22 ) + ( L22 − L21 ) = ( pH − pA)[ ψ(pA, p H ) − φ(pH , p A)] = (L13 − L23 ) + ( L23 − L21 ) = ( pH − pA)[ φ(pA, p H ) − ψ(pH , p A)] = (pH − pA)( pH + pA − 1)[ pH + pA + 2(1 − pH pA)] , which is non-negative if and only if pH − pA and pH + pA − 1 agree in sign. Hence, a 1-3 series is longer in expectation than a 2-1 series under the same conditions for which a 2-3 series is longer in expectation than a 2-2 series. Next, we consider stochastic ordering of the formats that differ in non-adjacent games. Comparing the 2-3 and 2-1 formats, N23 ≥st N21 holds if and only if g21 4 ≥ g23 4 , (8) g21 4 + g21 5 ≥ g23 4 + g23 5 , (9) g21 4 + g21 5 + g21 6 ≥ g23 4 + g23 5 + g23 6 . (10) Condition (8) becomes g21 4 ≥ g22 4 using (4). This is equivalent to N22 ≥st N21 .Condition (9) becomes g22 5 ≥ g23 5 using (6) and (4). This is equivalent to N23 ≥st N22 . Condition (10) holds from (1). Hence, N23 ≥st N21 holds if and only if N23 ≥st N22 ≥st N21 . In similar fashion it is possible to show that N13 ≥st N22 holds if and only if N13 ≥st N23 ≥st N22 .Finally, comparing the 1-3 and 2-1 formats, N13 ≥st N21 holds if and only if g21 4 ≥ g13 4 , (11) g21 4 + g21 5 ≥ g13 4 + g13 5 . (12) Using (6), (5) and (4), Condition (12) becomes g22 5 ≥ g23 5 or N23 ≥st N22 .Condition (11) holds if and only if (pH − pA)( pH + pA − 1)(2 − pH − pA + 2 pH pA) ≥ 0or, since the last term is always positive, simply N23 ≥st N22 . Hence, N13 ≥st N21 holds if and only if N23 ≥st N22 .The regions of stochastic and expected value ordering of these playoff for-mats are summarized in Table 2 and illustrated in Figure 1. Theorem 2 sum-marizes these results. 7Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 Parameter Agrees in sign with pH − pA Region φ(pA, p H ) ψ(pA, p H ) ϕ(pH , p A) ψ(pH , p A) φ(pH , p A)A √ √ √ √ √ B √ √ √ √ C √ √ √ c √ √ b √ aTable 2: Parametric Region Definitions Theorem 2 In a best 4 of 7-game series with independent games, the home and away sequencing formats influence the number of games played. The lengths of a series under various formats are ordered stochastically and in expectation according to Table 3. Parameter Stochastic Expected Value Region Ordering Ordering A N13 ≥st N23 ≥st N22 ≥st N21 L13 ≥ L23 ≥ L22 ≥ L21 N23 ≥st N13 ≥st N21 B N23 ≥st N22 ≥st N21 L23 ≥ L13 ≥ L22 ≥ L21 N23 ≥st N13 ≥st N21 C N23 ≥st N22 ≥st N21 L23 ≥ L22 ≥ L13 ≥ L21 N22 ≥st N23 ≥st N13 c N22 ≥st N21 ≥st N13 L22 ≥ L23 ≥ L21 ≥ L13 N22 ≥st N23 ≥st N13 b N22 ≥st N21 ≥st N13 L22 ≥ L21 ≥ L23 ≥ L13 a N21 ≥st N22 ≥st N23 ≥st N13 L21 ≥ L22 ≥ L23 ≥ L13 Table 3: Playoff Series Length Orderings by Parametric Region 8 Journal of Quantitative Analysis in Sports Vol. 2 , No. 1, Article 5 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 pH 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 pA ABCcbaAB Cc ba NBA 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Figure 1: Parametric Region Partition. The Curves Separate Regions - Defined in Table 2 - that Characterize Different Series-Length Orderings for the Various Home-Away Format Families. The series formats do not lend themselves to stronger orderings as described in Ross (1996). For example, the lengths of the series formats are not likelihood ratio ordered as a consequence of (1). Equations (5)-(7) reveal that the lengths of the series are not hazard rate ordered either. Here we define the hazard rate of the discrete random variable Nf at game i to be gfi /( 1 − ∑i−1 j=4 gfj ) , i = 4 , 5, 6, 7. We interpret this “rate” to be the probability that the series will end in game i given that a series has reached game i. 9Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 4 Model Validation To validate the home-away sequencing model, we collected game-by-game play-off data from National Basketball Association (NBA) historical records in Hub-bard (2000). We supplemented these records with the results of more recent play (through year 2005) as documented at the league web site www.nba.com. The NBA data includes 305 playoff series that were played in the 2-2 format family (i.e., the 2-2, 1-1 or 1-1-2 formats). These series are comprised of first-round, conference semi-final, conference final and NBA final playoffs. Except for the years 1949 and 1953-55 of 2-3 format play, the NBA Finals used the 2-2 format family from their inception in 1947 through 1984, after which a 2-3 format has been adopted. This represents 34 playoff series under the 2-2 format family. The NBA conference finals (in both the Eastern and Western conferences) have been played in this format family since their inception in 1958, which provides 2 × 48 = 96 additional playoff series. Similarly, all but one of the conference semi-finals have been played in this format family since their inception in 1968, which provides 4 × 38 − 1 = 151 additional playoff series. 3 Very recently (starting in 2003), the eight first-round series of the NBA playoffs were extended to 7-game series. This provides 3 × 8 = 24 additional playoff series, for a grand total of 305. 4 There have been 1744 games played in these 305 series, for an average series length of 5.718 games. This is just under the maximum expected length of 5.8125 games between two evenly matched teams under a binomial model as shown by Brunner (1987), Groeneveld and Meeden (1975), Nahin (2000) and Woodside (1989). Of these 1744 games, the favored team won 1055, for a winning percentage of about 60.5%. Thus, a straight-forward estimate of the Bernoulli success probability p is ˆ p = 1055 /1744 ≈ 0.605. 5 For the home-away model, we compute similar estimates based on whether the favored team was at home or away. These estimates are ˆ pH = 706 /959 ≈ 0.736 and ˆ pA = 349 /785 ≈ 0.445. A simple 95% confidence interval about 3The 1971 western semi-finals series between San Francisco and Minneapolis was played under the 1-2-1 format. 4It is too early to tell if this limited number of first-round series are significantly different than the later playoff series that typically pair teams of more equal caliber. We include them for completeness and note that their exclusion changes the results only slightly. 5Mosteller (1952, 1997) discusses several estimation procedures for the probability that the “better” team wins each game, where the “better” team is not known. Since we need estimates of the strength of the “favored” team, which is known, we do not resort to these more complicated methods. Further, the primary methods of Mosteller (1952), namely, maximum likelihood estimation and minimum chi-square estimation, provide nearly identical results. 10 Journal of Quantitative Analysis in Sports Vol. 2 , No. 1, Article 5 each of these independent Bernoulli parameter estimates, ˆ pH and ˆ pA, produces an approximate 90% confidence region about the joint point estimate (ˆ pH , ˆpA), which is completely contained within parameter region A (where the 1-3 format dominates), as shown in Figure 1. 6 The location of this confidence region suggests that a simple binomial model is not appropriate since this region does not cross the line ˆ pA = ˆ pH . Further, it suggest that, although played under the 2-2 family format, these playoff series would have been theoretically longer if they had been played under a 1-3 format or, as are the recent NBA championships, a 2-3 format. Table 4 presents the Pearson ( χ2) goodness of fit of the various models to the NBA playoff data outcomes (cf., e.g., Sec. 14.2 of Devore (2004)). The first eight columns of numbers indicate the number of the 305 series that would result in the series outcomes x-y, where x and y represent the number of games won by the home and away team, respectively. The first row of the table indicates the actual observed frequencies of such outcomes and the remaining rows list the expected number of such outcomes as predicted by the various models. Outcome Model 4-0 4-1 4-2 4-3 3-4 2-4 1-4 0-4 pActual 33 74 53 71 15 36 14 9 1.000 Binomial 40.8 64.5 63.7 50.4 32.9 27.2 17.0 7.4 0.000 1-3 Family 19.7 59.6 80.2 66.7 23.9 22.2 18.9 13.8 0.000 2-3 Family 32.7 46.7 80.2 66.7 23.9 22.2 26.1 6.5 0.000 2-2 Family 32.7 77.3 49.5 66.7 23.9 35.9 12.4 6.5 0.403 2-1 Family 54.1 55.9 49.5 66.7 23.9 35.9 15.8 3.1 0.000 Table 4: Goodness of Fit of Models in Predicting the Outcomes of 305 NBA 7-Game Playoff Series Played in a 2-2 Home-Away Format. The last column of Table 4 reports the significance probability (the so-called p-value) for the chi-squared goodness-of-fit test of the models. For each model, this p-value represents the probability that we would observe outcome numbers that deviate at least as much as the actual observed figures given 6The horizontal confidence interval was computed as ˆ pH±z·ˆσwith z= 1 .96 and standard deviation estimate ˆ σ=√ˆpH(1 −ˆpH)/n H, where nH= 959 is the total number of home games played. A similar vertical confidence interval was computed for the away games. 11 Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 that the model perfectly characterizes the randomness of the series outcome. Therefore, small p-values near 0 indicate a poor model (that we reject as a good fit for the data), whereas large p-values near 1 indicate a good model. 7 Examining Table 4 we see that, although parsimonious, the binomial model does not fit the data. On the other hand, the 2-2 home-away model fits the data fairly well with a p-value over 0.4. As should be expected, the other home-away models do not fit the data since the series were not played under those formats. There is a fairly large overestimation of 3-4 outcomes by all the models, partly due to the fact that of the 86 series that have gone to a full 7 games, the favored team (at home) has won 71 of these 86 final games for a winning probability of 0.826, which is somewhat larger than the home-away models’ probability estimate ˆ pH = 0 .736. Satisfied with the fit of the home-away model to the actual outcomes x-y,we can now examine effects on the series length x + y. Table 5 presents the ob-served and expected relative frequencies for the NBA playoff series length. The last column lists the average number of games played in the series, computed by summing the products of each length by its relative frequency. Series Length Model 4 5 6 7 Average Actual 0.138 0.289 0.292 0.282 5.718 Binomial 0.158 0.271 0.298 0.273 5.686 1-3 Family 0.110 0.257 0.336 0.297 5.820 2-3 Family 0.129 0.239 0.336 0.297 5.801 2-2 Family 0.129 0.294 0.280 0.297 5.746 2-1 Family 0.188 0.235 0.280 0.297 5.687 Table 5: Relative Frequencies for the Length of NBA 7-Game Series. The actual frequencies of series length work out to roughly 1 out of every 7 series ending in a 4-game “sweep”, and a uniform 2 out of 7 chance for each of the other three possible series lengths. Thus, in every postseason, which consists of 14 playoff series leading up to the NBA Finals, roughly 2 of the 7The number of degrees of freedom used for the chi-squared goodness-of-fit test is the number of categories less the number of parameters estimated (plus 1). For the binomial model with parameter estimate p, the degrees of freedom are 8-(1+1)=6. Since the home-away models require two parameters, pHand pA, the degrees of freedom are 5. 12 Journal of Quantitative Analysis in Sports Vol. 2 , No. 1, Article 5 series are sweeps, 4 series are of length 5 games, 4 are of length 6, and 4 are full series of length 7. The 2-2 model does the best job at replicating these frequencies and is closest in matching the average length. As indicated by Table 5, if these playoffs were played under the 2-3 format - which since 1985 has been used exclusively for the NBA finals - the expected series length would increase from 5.746 games under the 2-2 format to 5.801 games under a 2-3 format, an increase of 0.055 games per series. Since these formats differ only in a swapping of the locations of games 5 and 6, this difference of 0.055 is precisely the probability mass that is shifted from a 5-game series outcome under a 2-2 format to a 6-game outcome instead under the 2-3 format. 8 This increase amounts to an extra game in one of every 1 /0.055 ≈ 18 series. Since 14 such playoff series are played every year now, one could expect almost one extra playoff game added to the current average of 80 or so ( ≈ 14 · 5.746) games played each year leading up to the NBA finals. This is fairly small effect, but might be sufficiently lucrative to the television networks who pay huge fees for the broadcast rights to these games. As can be discerned from Table 4, the expected result of this switch in formats would be a significant (over 60%) increase in 4-2 outcomes, most of which were formerly 4-1 series wins by the favored team at home in game 5 under the 2-2 format. This predicted increase is offset somewhat by a cor-responding increase in shorter 1-4 outcomes, where the away team wins the series with the advantage at home in game 5 rather than in game 6 of a 2-4 outcome. The 2-3 format would also seem attractive in cutting down on travel “costs,” both financial (impacting team and media budgets) and physical (impacting team and media fatigue). The 2-2 format requires up to ten trips, four by the favored home team and six by the away team. In contrast, the 2-3 format requires at most six trips, only two by the favored home team and four by the away team. A 1-3 format would, in theory, extend the number of games even more than the 2-3 format, with no additional travel cost. However, the perceived psychological advantage to the away team (with a guarantee to host all three of their games) and lack of precedent would likely preclude adoption of such a format. 8The 25 NBA Finals played under the 2-3 format have averaged 5.720 games. Thus, this limited data has yet to reveal any significant difference in playoff series length from the 5.718 games listed in Table 5 for the 2-2 format family. 13 Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 5 Conclusions We have presented an extended analysis of the number of games played in a playoff series under various home-away sequences. Using the model of Bassett and Hurley (1998), we considered all possible playoff sequences in which the series begins and ends on the home team’s court. Under this model it turns out that, in terms of the number of games played, these sequences reduce to four stochastically different formats, including the popular 2-3 and 2-2 formats both currently used in National Basketball Association (NBA) playoffs. We then characterized the parametric regions that give rise to six distinct stochastic and expected value orderings among these four format classes. Examining the performance of the home-away model using National Bas-ketball Association (NBA) playoff data, we found that the home-away model fits the data fairly well, whereas the simple binomial model so prevalent in the literature did not. We also discovered that the NBA could stochastically extend the length of the conference championships by switching from the long-standing 2-2 series format to a 2-3 format, as was done for the NBA finals in 1985. Such evidence should encourage television sponsors to lobby the NBA for a change of playoff format in order to garner additional television advertising revenues while reducing team and media travel costs. In contrast, similar analysis of the National Hockey League (NHL) 7-game playoff data reveals that neither the binomial model nor the home-away model provide any explanatory power. In these playoffs there is a prevalence of many short “sweeped” series where the game outcome seems heavily influ-enced by the presence of a team facing elimination. This violates the as-sumption of game independence assumed throughout this paper. However, a two-parameter Markov chain formulation, built around the home-away model performs extremely well as reported in Rump (2005). References D. R. Appleton. May the best man win? The Statistician , 44(4):529–538, 1995. G. W. Bassett and W. J. Hurley. The effects of alternative HOME-AWAY sequences in a best-of-seven playoff series. The American Statistician , 52 (1):51–53, 1998. J. S. Boronico. Multi-tiered playoffs and their impact on professional baseball. The American Statistician , 53(1):56–61, 1999. 14 Journal of Quantitative Analysis in Sports Vol. 2 , No. 1, Article 5 J. Brunner. Absorbing Markov chains and the number of games in a World Series. The UMAP Journal , 8(2):99–108, 1987. S. R. Clarke and J. M. Norman. Home ground advantage of individual clubs in English soccer. The Statistician , 44(4):509–521, 1995. K. S. Courneya and A. V. Carron. The home advantage in sport comptetitions: a literature review. Journal of Sport Exercise Psychology , 14:13–27, 1992. H. A. David. The Method of Paired Comparisons . Charles Griffin & Company, London, 1988. J. L. Devore. Probability and Statistics for Engineering and the Sciences .Thomson Brooks/Cole, 6 edition, 2004. J. D. Gibbons, I. Olkin, and M. Sobel. Baseball competitions - are enough games played? The American Statistician , 32(3):89–95, 1978. W. A. Glenn. A comparison of the effectiveness of tournaments. Biometrika ,47(3-4):253–262, 1960. R. A. Groeneveld and G. Meeden. Seven game series in sports. Mathematics Magazine , 48(4):187–192, 1975. D. A. Harville and M. H. Smith. The home-court advantage: How large is it, and does it vary from team to team? The American Statistician , 48(1): 22–28, 1994. J. Hubbard, editor. The Official NBA Encyclopedia . Doubleday, New York, 3rd edition, 2000. B. James, J. Albert, and H. S. Stern. Answering questions about baseball using statistics. Chance , 6(2):17–22, 30, 1993. T. Lengyel. A combinatorial identity and the World Series. SIAM Review , 35 (2):294–297, 1993. H. Maisel. Best k of 2 k − 1 comparisons. Journal of the American Statistical Association , 61:329–344, 1966. ´E. Marchand. On the comparison between standard and random knockout tournaments. Statistician , 51(2):169–178, 2002. 15 Rump: Home-Away Sequencing in Best-of-7 Playoff Series Produced by The Berkeley Electronic Press, 2006 T. McGarry. On the design of sports tournaments. In J. Bennett, editor, Statistics in Sport , Arnold Applications of Statistics, chapter 10, pages 199– 217. Arnold, London, 1998. F. Mosteller. The World Series competition. Journal of The American Statis-tical Association , 47(259):355–380, 1952. F. Mosteller. Lessons from sports statistics. The American Statistician , 51(4): 305–310, 1997. P. J. Nahin. Duelling Idiots and other Probability Puzzlers . Princeton Univer-sity Press, Princeton, NJ, 2000. National Hockey League. Stanley Cup Playoffs Fact Guide 1998 . Triumph Books, Chicago, 1997. S. M. Ross. Stochastic Processes . John Wiley & Sons, New York, 2nd edition, 1996. C. M. Rump. Andrei Markov in the Stanley Cup playoffs. Chance Magazine ,2005. In press. L. Shapiro and W. Hamilton. The Catalan numbers visit the World Series. Mathematics Magazine , 66(1):20–22, 1993. W. Simon. Back-to-the-wall effect: 1976 perspective. In S. P. Ladany and R. E. Machol, editors, Optimal Strategies in Sports , volume 5 of Studies in Management Science and Systems , Amsterdam, 1977. North-Holland. V. R. R. Uppuluri and W. J. Blot. Asymptotic properties of the number of replications of a paired comparison. Journal of Applied Probability , 11: 43–52, 1974. I. A. Ushakov. The problem of choosing the preferred element: An application to sport games. In R. E. Machol, S. P. Ladany, and D. G. Morrison, editors, Management Science in Sports , volume 4 of Studies in the Management Sciences , Amsterdam, 1976. North-Holland. W. Woodside. Winning streaks, shutouts, and the length of the World Series. The UMAP Journal , 10(2):99–113, 1989. M. Zelinski. How many games to complete a World Series. In F. Mosteller, edi-tor, Weighing Chances , Statistics by Example: Weighing Chances, Reading, 1973. Addison-Wesley.
17162	https://resources.nu.edu/statsresources/eta	Partial Eta Squared - Statistics Resources - LibGuides at National University Skip to Main Content NU Library LibGuides Academic Success Center Statistics Resources Partial Eta Squared Search this Guide Search Statistics Resources This guide contains all of the ASC's statistics resources. If you do not see a topic, suggest it through the suggestion box on the Statistics home page. Home Excel - Tutorials ProbabilityToggle Dropdown Basic Probability Rules Single Event Probability Complement Rule Intersections & Unions Compound Events VariablesToggle Dropdown Levels of Measurement Independent and Dependent Variables Statistics BasicsToggle Dropdown Entering Data Central Tendency Data and Tests Displaying Data Discussing Statistics In-text SEM and Confidence Intervals Two-Way Frequency Tables Discussing Statistics In-text Z-Scores and the Standard Normal DistributionToggle Dropdown Z-Scores Empirical Rule Finding Probability Accessing SPSS SPSS-TutorialsToggle Dropdown Chart and Graphs Frequency Table and Distribution Descriptive Statistics Converting Raw Scores to Z-Scores Converting Z-scores to t-scores Split File/Split Output Saving and Exporting with SPSS Sampling Methods Effect Size Cohen's d R-Squared Partial Eta Squared GPowerToggle Dropdown Downloading and Installing GPower: Windows/PC T-Test ANOVA Correlation Testing Parametric Assumptions ANOVAToggle Dropdown One-Way ANOVA Two-Way ANOVA Repeated Measures ANOVA Chi-Square TestsToggle Dropdown Goodness-of-Fit Test of Association CorrelationToggle Dropdown Pearson's r Spearman's Point Biserial Mediation and Moderation Regression AnalysisToggle Dropdown Simple Linear Regression Multiple Linear Regression Binomial Logistic Regression Multinomial Logistic Regression T-TestToggle Dropdown Independent Samples T-test Dependent Samples T-test Testing Assumptions T-tests using SPSS T-Test Practice Quantitative Research Questions Hypothesis TestingToggle Dropdown Null & Alternative Hypotheses One-Tail vs. Two-Tail Alpha & Beta Associated Probability Decision Rule Statement of Conclusion Statistics Group Sessions Partial Eta Squared (Partial) Eta Squared When comparing more than two groups of people, we commonly use eta squared or partial eta squared. In a One-Way ANOVA either value can be reported since they will be the same. With other ANOVA analyses, partial eta squared is more appropriate to report. Partial eta squared is telling us how large of an effect the independent variable(s) had on the dependent variable. Computing Partial Eta Squared Starting with computing this value by hand, we can again use a formula. As before, there are several different formulas that can be used. Here is just one example: where: SS stands for "sum of squares" In order to complete this computation, you must also know how to compute each sum of squares (deviations from the mean), which can be a time-consuming process, especially with larger data sets.As with the other types of effect size, technology can come to the rescue here. Partial Eta Squared using SPSS If you're using SPSS 27 or higher, you can opt to include effect size estimates with your ANOVA analysis. Below are a series of images that show you how to include the effect size estimate with the different types of analyses and where to look in the output to locate the value of the effect size: 1) One-Way ANOVA Here is what you will see in the output: 2) Univariate(located under the "Options" menu) Here is what's included in the output: 3) Multivariate(located under the "Options" menu) Here's how it will appear in the output: 4) Repeated Measures(located under the "Options" menu) Here is how it will appear in the output: Interpreting the Partial Eta Squared If no guidelines are provided, you can follow this: η 2= 0.01 indicates a small effect η 2= 0.06 indicates a medium effect η 2= 0.14 indicates a large effect You'll want to review the guidelines provided in your course materials to confirm as there are differing opinions on the thresholds for each effect level. For additional assistance with computing and interpreting the effect size for your analysis, attend the SPSS: ANOVA/MANOVA group session <<Previous: R-Squared Next: GPower >> Last Updated:Jul 31, 2025 12:56 PM URL: Print Page Staff Login Report a problem Tags: academic success center, statistics © Copyright 2025 National University. All Rights Reserved. Privacy Policy \| Consumer Information Library Academic Success Center Office of the Registrar Student Bookstore Student Services Doctoral Center Institutional Review Board Advanced Research Center Institutional Repository NU Commons
17163	https://www.ms.uky.edu/~lee/ma310/tools/node30.html	Finite Differences Next:RecursionUp:ToolsPrevious:Induction Finite Differences The method of finite differences can sometimes be used to guess a formula f(n) (but not to prove it). The way I am about to describe it requires that the values of f(n) be given for integer values of n beginning with n=0 (rather than 1). Let's use the example of guessing the formula for the sum of the cubes of the first n positive integers. We calculate a few values: and then arrange them in a row. Then we build a difference table beneath them by subtraction: We continue generating rows by this process until it appears that we get a row of 0's. This does not always happen, but when it does, we can build a polynomial formula for f(n). Using the first entries in each row, multiply the first entry of the first row by 1/0!, the first entry of the second row by n/1!, the first entry of the third row by n(n-1)/2!, the first entry of the fourth row by n(n-1)(n-2)/3!, the first entry of the fifth row by n(n-1)(n-2)(n-3)/4!, etc., and then add the results. (Remember that 0! is defined to be 1.) In our example, we get which simplifies (you try it!) to At this point we have not proved that this formula is correct, but at least we have a good guess for the formula, and we can try to prove it, e.g., by induction. _Carl Lee Wed Apr 21 08:17:28 EDT 1999_
17164	https://en.wikipedia.org/wiki/List_of_countries_by_natural_gas_proven_reserves	Jump to content List of countries by natural gas proven reserves العربية Azərbaycanca Ελληνικά Español فارسی Français עברית پنجابی Português Русский Slovenščina Türkmençe Українська اردو 中文 Edit links From Wikipedia, the free encyclopedia This list is based on the CIA World Factbook (when no citation is given). or other authoritative third-party sources (as cited). Based on data from EIA, at the start of 2021, proven gas reserves were dominated by three countries: Iran, Russia, and Qatar. There is some disagreement on which country has the largest proven gas reserves. Sources that consider Russia in possession of the world's largest proven reserves include the US CIA (47,600 cubic kilometers), the US Energy Information Administration (EIA) (49,000 km3), and OPEC (48,810 km3). However, BP credits Russia with only 32,900 km3, which would place it in second place, slightly behind Iran (33,100 to 33,800 km3, depending on the source). Due to constant announcements of shale gas recoverable reserves, as well as drilling in Central Asia, South America, Africa, and deepwater drilling, estimates are updated frequently. Since 2000, some countries, notably the US and Canada, have seen large increases in proven gas reserves due to development of shale gas, but shale gas deposits in most countries are yet to be added to reserve calculations. List [edit] Note: indicates that the linked article is about natural gas production in the country or territory specifically, such as natural gas in Russia. Proven reserves (km³) \| Country \| U.S. EIA (2021) \| OPEC (start of 2018) \| BP (end of 2020) \| OTHERS \| Production km3/year (2021) \| Years of production in reserve \| \| Russia  \| 47,800 \| 47,760 \| 37,400 \| \| 701.55 \| 68 \| \| Iran  \| 34,000 \| 33,988 \| 32,100 \| \| 248.3 \| 137 \| \| Qatar  \| 23,900 \| 23,861 \| 24,700 \| \| 169 \| 141 \| \| United States  \| 17,710 \| 9,067 \| 12,600 \| \| 977.44 \| 18 \| \| Turkmenistan  \| 10,000 \| 9,838 \| 13,600 \| \| 87.02 \| 115 \| \| Saudi Arabia  \| 9,430 \| 9,514 \| 6,000 \| \| 115.6 \| 82 \| \| China  \| 6,650 \| 2,934 \| 8,400 \| \| 212 \| 31 \| \| United Arab Emirates  \| 6,090 \| 8,210 \| 5,900 \| \| 63.77 \| 95 \| \| Venezuela  \| 6,000 \| 5,707 \| 6,300 \| \| 15.9 \| 377 \| \| Nigeria  \| 5,750 \| 5,913 \| 5,500 \| \| 44.40 \| 130 \| \| Algeria  \| 4,500 \| 4,504 \| 2,300 \| \| 102.1 \| 44 \| \| Iraq  \| 3,740 \| 3,744 \| 3,500 \| \| 10.7 \| 350 \| \| Australia  \| 3,230 \| 3,173 \| 2,400 \| \| 147.1 \| 22 \| \| Mozambique  \| 3,000 \| \| \| \| 5.35 \| 561 \| \| Kazakhstan  \| 2,400 \| 1,898 \| 2,300 \| \| 20.0 \| 120 \| \| Canada  \| 2,100 \| 2,059 \| 2,400 \| \| 182.2 \| 12 \| \| Azerbaijan  \| 2,000 \| 1,227 \| 2,500 \| \| 31.38 \| 64 \| \| Egypt  \| 1,800 \| 2,221 \| 2,100 \| \| 69.43 \| 26 \| \| Kuwait  \| 1,800 \| 1,784 \| 1,700 \| \| 21.3 \| 85 \| \| Uzbekistan  \| 1,800 \| 1,564 \| 800 \| \| 49.70 \| 36 \| \| Norway  \| 1,600 \| 2,314 \| 1,400 \| \| 115.1 \| 14 \| \| Libya \| 1,500 \| 1,505 \| 1,400 \| \| 14.6 \| 103 \| \| India  \| 1,400 \| 1,289 \| 1,300 \| \| 32.3 \| 43 \| \| Malaysia  \| 1,200 \| 2,909 \| 900 \| \| 70.99 \| 17 \| \| Ukraine  \| 1,100 \| 304 \| 1,100 \| \| 18.9 \| 58 \| \| Indonesia  \| 1,000 \| 2,866 \| 1,300 \| \| 58.96 \| 17 \| \| Senegal \| \| \| 1,000 \| \| \| \| Vietnam  \| 710 \| 203 \| 600 \| \| 7.02 \| 101 \| \| Turkey  \| 710 \| \| \| \| 0.40 \| 1775 \| \| Oman  \| 650 \| 884 \| 700 \| \| 39.19 \| 17 \| \| Myanmar  \| 650 \| 273 \| 400 \| \| 16.3 \| 40 \| \| Pakistan  \| 590 \| 757 \| 400 \| \| 33.39 \| 18 \| \| Argentina  \| 487 \| 381 \| 400 \| \| 40.44 \| 12 \| \| Yemen  \| 480 \| \| 300 \| \| 0.12 \| 4000 \| \| Guyana \| 453 \| \| \| \| \| \| \| Brazil  \| 370 \| 325 \| 300 \| \| 23.8 \| 16 \| \| Angola  \| 340 \| 422 \| \| \| 5.86 \| 58 \| \| Trinidad and Tobago  \| 310 \| 433 \| 290 \| \| 25.7 \| 12 \| \| Peru  \| 310 \| 513 \| 261 \| \| 12 \| 26 \| \| Bolivia  \| 310 \| 310 \| 213 \| \| 15.5 \| 20 \| \| Congo  \| 300 \| 284 \| \| \| 0.42 \| 714 \| \| Brunei  \| 260 \| 252 \| 222 \| \| 11.4 \| 23 \| \| Syria  \| 240 \| 300 \| 269 \| \| 3.26 \| 74 \| \| Israel  \| 180 \| \| 589 \| \| 17.7 \| 10 \| \| United Kingdom  \| 180 \| 269 \| 187 \| \| 32.48 \| 6 \| \| Mexico \| 180 \| 146 \| 178 \| \| 23.9 \| 8 \| \| Papua New Guinea  \| 180 \| \| 163 \| \| 11.4 \| 16 \| \| Thailand  \| 140 \| 180 \| 143 \| \| 36.36 \| 4 \| \| Cameroon  \| 140 \| 152 \| \| \| 2.3 \| 61 \| \| Equatorial Guinea  \| 140 \| 145 \| \| \| 5.83 \| 24 \| \| Netherlands  \| 130 \| 804 \| 130 \| \| 21.6 \| 6 \| \| Bangladesh  \| 130 \| 346 \| 110 \| \| 23.9 \| 5 \| \| Romania  \| 100 \| 105 \| 78 \| \| 9.32 \| 11 \| \| Philippines  \| 99 \| \| \| \| 3 \| 33 \| \| Chile  \| 99 \| 5.30 \| \| \| 1.4 \| 71 \| \| Poland  \| 91 \| 56.3 \| 72.2 \| \| 5.64 \| 16 \| \| Colombia  \| 88 \| 104 \| 86 \| \| 11.4 \| 8 \| \| Bahrain  \| 82 \| \| 64.8 \| \| 18.6 \| 4 \| \| Sudan  \| 80 \| \| \| \| 0 \| Cuba  \| 71 \| \| \| \| 0.91 \| 78 \| \| Tunisia  \| 65 \| \| \| \| 1.4 \| 46 \| \| Namibia  \| 62 \| \| \| \| 0 \| Rwanda  \| 60 \| \| \| \| 0.06 \| 1000 \| \| Afghanistan  \| 51 \| \| \| \| 0.079 \| 646 \| \| Serbia  \| 48 \| \| \| \| 0.40 \| 120 \| \| Italy  \| 45 \| 27.1 \| 100 \| \| 3.2 \| 14 \| \| Palestine  \| 32 \| \| \| \| 0 \| New Zealand  \| 31 \| \| \| \| 4.13 \| 8 \| \| Denmark  \| 30 \| 73.9 \| 100 \| \| 1.2 \| 25 \| \| Ivory Coast  \| 30 \| \| \| \| 2.5 \| 12 \| \| Croatia \| 30 \| \| \| \| 0.82 \| 37 \| \| Mauritania  \| 30 \| \| \| \| 0 \| Ethiopia  \| 30 \| \| \| \| 0 \| Gabon  \| 30 \| 25.5 \| \| \| 0.45 \| 67 \| \| Germany \| 20 \| 39.6 \| 100 \| \| 4.98 \| 4 \| \| Ghana \| 20 \| \| \| \| 2.92 \| 7 \| \| Japan  \| 20 \| \| \| \| 2.4 \| 8 \| \| Ireland  \| 10 \| \| \| \| 1.6 \| 6 \| \| Slovakia \| 10 \| \| \| \| 0.062 \| 161 \| \| Uganda \| 10 \| \| \| \| 0 \| Ecuador \| 10 \| 5.4 \| \| \| 0.31 \| 32 \| \| Georgia \| 8 \| 8.5 \| \| \| 0.01 \| 800 \| \| South Korea  \| 8 \| \| \| \| 0.05 \| 160 \| \| France \| 8 \| \| \| \| 0.02 \| 400 \| \| Tanzania  \| 6 \| \| \| \| 1.3 \| 5 \| \| Austria \| 6 \| \| \| \| 1.2 \| 5 \| \| Jordan  \| 6 \| \| \| \| 0.19 \| 32 \| \| Taiwan  \| 6 \| \| \| \| 0.091 \| 66 \| \| Albania \| 6 \| \| \| \| 0.059 \| 102 \| \| Bulgaria \| 6 \| \| \| \| 0.037 \| 162 \| \| Tajikistan  \| 6 \| \| \| \| 0.02 \| 300 \| \| Kyrgyzstan \| 6 \| \| \| \| 0.02 \| 300 \| \| Somalia  \| 6 \| \| \| \| 0 \| Hungary  \| 3 \| \| \| \| 1.5 \| 2 \| \| Czech Republic \| 3 \| \| \| \| 0.20 \| 15 \| \| Belarus  \| 3 \| \| \| \| 0.074 \| 41 \| \| Somaliland \| 3 \| 0 \| \| \| \| \| \| Spain \| 2.55 \| \| \| \| 0.048 \| 53 \| \| Morocco  \| 1.44 \| 474 \| \| \| 0.12 \| 12 \| \| Benin \| 1.13 \| \| \| \| 0 \| Greece \| 0.99 \| \| \| \| 0.06 \| 16 \| \| DR Congo \| 0.99 \| \| \| \| 0 \| Barbados \| 0.11 \| \| \| \| 0.01 \| 11 \| \| Armenia  \| 0 \| 18.0 \| \| \| \| \| World \| 205,500 (2020) \| \| \| \| 4,083.09 \| 50 \| Comparison of proven natural gas reserves from different sources (cubic kilometers, as of 31 December 2014/1 January 2015) \| Source \| Canada \| Iran \| Russia \| Saudi Arabia \| United States \| Venezuela \| --- --- --- \| BP \| 2,000 \| 34,000 \| 32,800 \| 8,200 \| 9,800 \| 5,600 \| \| OPEC \| 2,028 \| 34,020 \| 49,541 \| 8,489 \| 9,580 \| 5,617 \| \| U.S. Energy Information Administration \| 2,535 \| 42,426 \| 59,619 \| 10,393 \| 10,441 \| 6,960 \| References [edit] ^ Rank Order - Natural gas - proved reserves. Retrieved June 2014. ^ Natural gas - proved reserves. Retrieved 1 December 2013. ^ US Energy Information Administration, [permanent dead link]. Retrieved 23 October 2015. ^ OPEC, Table 3.2 Natural gas proven reserves by country Archived 27 February 2018 at the Wayback Machine. Retrieved 1 December 2013. ^ BP, BP Statistical Review of World Energy, June 2013. Archived 6 December 2013 at the Wayback Machine ^ U.S. Energy Information Administration, International Energy Statistics, accessed 17 Jan. 2019. ^ OPEC, Annual Statistical Bulletin 2018 ^ BP, Statistical Review of World Energy, June 2022. ^ "International - U.S. Energy Information Administration (EIA)". www.eia.gov. Retrieved 25 July 2023. ^ (EIA Reserves and Production) ^ BP Statistical Review of World Energy, 2015 ^ OPEC Statistical Bulletin, 2015. ^ International Data: Natural Gas Reserves, US EIA. Retrieved 3 July 2016. \| v t e Lists of countries by energy rankings \| \| --- \| \| Fossil Fuel \| \| \| \| --- \| \| Petroleum \| All Production Consumption Exports Imports Reserves Usage and pricing \| \| Natural gas \| All Production Consumption Exports Imports Reserves Shale gas Fracking \| \| Coal \| Reserves Production Exports Imports \| \| \| Nuclear \| \| \| \| --- \| \| Uranium \| All Production Generation Reserves \| \| Thorium \| Reserves \| \| \| Renewable \| Hydroelectricity Wind Solar Geothermal \| \| Electricity \| Production Consumption \| \| Total energy \| Consumption and production + per capita Intensity Summary of top fossil fuel exporters \| \| List of international rankings Lists by country \| \| \| Authority control databases \| \| --- \| \| National \| United States Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Energy-related lists by country Natural gas by country Lists of countries Hidden categories: All articles with dead external links Articles with dead external links from January 2018 Articles with permanently dead external links Webarchive template wayback links Articles with short description Short description is different from Wikidata Use dmy dates from October 2016
17165	https://math.stackexchange.com/questions/3807294/affine-transformation-from-a-parallelogram-to-a-square	Affine transformation from a parallelogram to a square - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Affine transformation from a parallelogram to a square Ask Question Asked 5 years, 1 month ago Modified5 years ago Viewed 159 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I need to transform an area of an image so that a parallelogram defined as ((x 0,y 0),(x 1,y 1),(x 2,y 2),(x 0+x 2−x 1,y 0+y 2−y 1))((x 0,y 0),(x 1,y 1),(x 2,y 2),(x 0+x 2−x 1,y 0+y 2−y 1)) is mapped on a square ((0,0),(0,128),(128,128),(128,0))((0,0),(0,128),(128,128),(128,0)). Apparently, Python OpenCV can read and save images as well as perform affine transformations. What would be the algorithm for such transform? affine-geometry Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 5, 2020 at 14:30 Alessio K 10.7k 9 9 gold badges 18 18 silver badges 31 31 bronze badges asked Aug 29, 2020 at 13:47 StepanStepan 1,113 8 8 silver badges 16 16 bronze badges 1 Translation ×× Shear ×× Rotation ×× Scaling G Cab –G Cab 2020-08-29 13:52:50 +00:00 Commented Aug 29, 2020 at 13:52 Add a comment\| 0 Sorted by: Reset to default You must log in to answer this question. 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17166	https://hume.ucdavis.edu/phi112/transcribe1_ho.pdf	Transcription and Restricted Quantiﬁers Differences Among Quantiﬁer Expressions • Some English quantiﬁer expressions are neutral with respect to the range of ob-jects whose quantity they express. – Any, every, all, whatever - some – Anything, everything - something – There is, at least one is • Other quantiﬁer expressions apply only to a limited range of objects. – Anyone, anybody - someone, somebody (persons) – Anywhere, everywhere - somewhere (places) – Whenever, always when - sometimes (times) Limiting the Domain • One way of transcribing sentences with restricted quantiﬁers is to limit the do-main to the objects to which the quantiﬁers are supposed to apply. – D = x: x is a person, so ‘(∀x)’ and ‘(∃x)’ apply only to persons. – ‘Everybody is happy’ is transcribed as ‘(∀x)Hx’, where Hx: x is happy. • However, this only allows us to talk about items in the domain, so that in the last example, we could not transcribe ‘Everybody is happy sometimes’. Restricted Quantiﬁers • One way to transcribe sentences with a mixture of types of quantiﬁer expressions is to create a new kind of quantiﬁer in Predicate Logic: the restricted quantiﬁer. • We choose a predicate letter to symbolize the restricted range of objects. – ‘P’ stands for the set of all persons. – We write ‘(∀x)P’ for ‘everyone’, and ‘(∃x)P’ for ‘someone’. – ‘Everyone is happy’ is transcribed as ‘(∀x)PHx’. We can mix restricted quantiﬁers to symbolize sentences containing more than one limited quantiﬁer expression. – ‘T’ stands for the set of all times. – ‘Everyone is happy sometimes’ is transcribed as ‘(∀x)P(∃y)THxy’, where Hxy: x is happy at y. 1 Semantics for Restricted Quantiﬁers • Tarski-style semantics can be used to specify satisfaction-conditions for sen-tences with restricted quantiﬁers. • For each restriction represented by a one-place predicate S, we generate the set r(S) from v(S) by stripping off the angle brackets. – Let v(P) = {⟨Adam⟩, ⟨Eve⟩}; r(P) = {Adam, Eve}. • Then we say that d satisﬁes (∃u)SP(u) if and only if for some object o ∈r(S), d[o/u] satisﬁes P(u). – Let v(B) = {⟨Adam⟩}; then d[Adam/x] satisﬁes ‘Bx’, so d satisﬁes ‘(∃x)PBx’. • Similarly, d satisﬁes (∀u)SP(u) if and only if for all objects o ∈r(S), d[o/u] satisﬁes P(u). Eliminating Restricted Quantiﬁers • Restricted quantiﬁers can be eliminated in favor of other constructions without change in truth-value. • (∃u)SP(u) is equivalent to (∃u)(S(u) & P(u)). – ‘(∃x)PBx’ is equivalent to ‘(∃x)(Px & Bx)’. • (∀u)SP(u) is equivalent to (∀u)(S(u) ⊃P(u)). – ‘(∀x)PBx’ is equivalent to ‘(∀x)(Px ⊃Bx)’. • The replacement of one form for the other (when authorized) may occur in an internal part of a sentence. – ‘(∃x)PLxe ⊃(∃x)PLex’ is equivalent to ‘(∃x)(Px & Lxe) ⊃(∃x)(Px & Lex)’. Proof of Equivalence for Restricted Existentials • (∃u)SP(u) is true in I if and only if (iff) it is satisﬁed by all variable assigments d based on I. • Let I be an arbitrary interpretation and d an arbitrary variable assigment based on I. • d satisﬁes (∃u)SP(u) iff some object o ∈r(S), d[o/u] satisﬁes P(u), • iff for some object o in D, d[o/u] satisﬁes P(u), and the one-tuple ⟨o⟩∈v(S) [by the deﬁnition of r(S)], • iff for some object o ∈D, d[o/u] satisﬁes P(u) and d[o/u] satisﬁes S(u), 2 • iff for some object o in D, d[o/u] satisiﬁes S(u) & P(u), • iff d satisﬁes (∃u)(S(u) & P(u)), • Since d is arbitrary, (∃u)SP(u) is true in I iff (∃u)(S(u) & P(u)) is true in I, QED. Proof of Equivalence for Restricted Universals • To save space, only the core of the proof is presented; the other steps are trivial. • d satisﬁes (∀u)SP(u) iff for all objects o ∈r(S), d[o/u] satisﬁes P(u), • iff for all objects o ∈D, if o ∈r(S), then d[o/u] satisiﬁes P(u), • iff for all objects o ∈D, if the one-tuple ⟨o⟩∈v(S), then d[o/u] satisﬁes P(u), • iff for all objects o ∈D, if d[o/u] satisﬁes S(u), then d[o/u] satisﬁes P(u), • iff for all objects o ∈D, d[o/u] satisifes S(u) ⊃P(u), • iff, d satisﬁes (∀x)(S(u) ⊃P(u)), QED. Negated Restricted Quantiﬁers • The following two logical equivalences hold, with a proof of the ﬁrst below. – ∼(∀u)SP(u) and (∃u)S ∼P(u) – ∼(∃u)SP(u) and (∀u)S ∼P(u) • d satisﬁes ∼(∀u)SP(u) iff d does not satisfy (∀u)SP(u), • iff it is not the case that for all o ∈D, if o ∈r(S), then d[o/u] satisﬁes P(u), • iff for some o ∈D, it is not the case that if o ∈r(S), then d[o/u] satisﬁes P(u), • iff for some o ∈D, o ∈r(S) and d[o/u] does not satisfy P(u), • iff for some o ∈D, o ∈r(S), and d[o/u] satisﬁes ∼P(u), • iff d satisﬁes (∃u)S ∼P(u), QED. 3
17167	https://classics.mit.edu/Aristotle/prior.html	The Internet Classics Archive \| Prior Analytics by Aristotle Home Browse and Comment Search Buy Books and CD-ROMs Help Prior Analytics By Aristotle Written 350 B.C.E Translated by A. J. Jenkinson Prior Analyticshas been divided into the following sections: Book I[209k]Book II[137k] Download:A 255k text-only version is available for download. © 1994-2009
17168	https://en.wikipedia.org/wiki/Trapezoid	Published Time: Sun, 07 Sep 2025 10:05:48 GMT Trapezoid - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 Definitions 2 Etymology 3 Special cases 4 Condition of existence 5 Characterizations 6 PropertiesToggle Properties subsection 6.1 Midsegment and height 6.2 Area 6.3 Diagonals 6.4 Other properties 7 Applications 8 Non-Euclidean geometry 9 Related topics 10 See also 11 Notes 12 Bibliography 13 Further reading 14 External links [x] Toggle the table of contents Trapezoid [x] 100 languages Afrikaans العربية Asturianu Azərbaycanca বাংলা Башҡортса Беларуская Беларуская (тарашкевіца) Bikol Central Български བོད་ཡིག Bosanski Brezhoneg Català Чӑвашла Čeština ChiShona Cymraeg Dansk Davvisámegiella Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Gaeilge Galego ગુજરાતી 한국어 Հայերեն हिन्दी Hornjoserbsce Hrvatski Ido Bahasa Indonesia Interlingua Íslenska Italiano עברית Jawa ქართული Қазақша Кыргызча ລາວ Latina Latviešu Lietuvių Lombard Magyar Македонски മലയാളം मराठी მარგალური Bahasa Melayu Монгол Nederlands 日本語 Nordfriisk Norsk bokmål Norsk nynorsk Олык марий ଓଡ଼ିଆ Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ ភាសាខ្មែរ Piemontèis Polski Português Română Runa Simi Русский Simple English Slovenčina Slovenščina Ślůnski Soomaaliga کوردی Српски / srpski Srpskohrvatski / српскохрватски Sunda Suomi Svenska Tagalog தமிழ் తెలుగు ไทย Türkçe Türkmençe Українська اردو Tiếng Việt West-Vlams Winaray 吴语 ייִדיש 粵語中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikibooks Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Convex quadrilateral with at least one pair of parallel sides \| Trapezoid (American English) Trapezium (British English) \| \| Trapezoid or trapezium \| \| Type \| quadrilateral \| \| Edges and vertices \| 4 \| \| Area \| 1 2(a+b)h{\displaystyle {\tfrac {1}{2}}(a+b)h} \| \| Properties \| convex \| Look up trapezoid in Wiktionary, the free dictionary. In geometry, a trapezoid (/ˈ t r æ p ə z ɔɪ d/) in North American English, or trapezium (/t r ə ˈ p iː z i ə m/) in British English, is a quadrilateral that has at least one pair of parallel sides. The parallel sides are called the bases of the trapezoid. The other two sides are called the legs or lateral sides. If the trapezoid is a parallelogram, then the choice of bases and legs is arbitrary. A trapezoid is usually considered to be a convex quadrilateral in Euclidean geometry, but there are also crossed cases. If shape ABCD is a convex trapezoid, then ABDC is a crossed trapezoid. The metric formulas in this article apply in convex trapezoids. Definitions [edit] Trapezoid can be defined exclusively or inclusively. Under an exclusive definition a trapezoid is a quadrilateral having exactly one pair of parallel sides, with the other pair of opposite sides non-parallel. Parallelograms including rhombi, rectangles, and squares are then not considered to be trapezoids. Under an inclusive definition, a trapezoid is any quadrilateral with at least one pair of parallel sides. In an inclusive classification scheme, definitions are hierarchical: a square is a type of rectangle and a type of rhombus, a rectangle or rhombus is a type of parallelogram, and every parallelogram is a type of trapezoid. Professional mathematicians and post-secondary geometry textbooks nearly always prefer inclusive definitions and classifications, because they simplify statements and proofs of geometric theorems. In primary and secondary education, definitions of rectangle and parallelogram are also nearly always inclusive, but an exclusive definition of trapezoid is commonly found. This article uses the inclusive definition and considers parallelograms to be special kinds of trapezoids. (Cf. Quadrilateral §Taxonomy.) To avoid confusion, some sources use the term proper trapezoid to describe trapezoids with exactly one pair of parallel sides, analogous to uses of the word proper in some other mathematical objects. Etymology [edit] In the ancient Greek geometry of Euclid's Elements (c. 300 BC), quadrilaterals were classified into exclusive categories: square; oblong (non-square rectangle); (non-square) rhombus; rhomboid, meaning a non-rhombus non-rectangle parallelogram; or trapezium (τραπέζιον, literally "table"), meaning any quadrilateral not already included in the previous categories. The Neoplatonist philosopher Proclus (mid 5th century AD) wrote an influential commentary on Euclid with a richer set of categories, which he attributed to Posidonius (c. 100 BC). In this scheme, a quadrilateral can be a parallelogram or a non-parallelogram. A parallelogram can itself be a square, an oblong (non-square rectangle), a rhombus, or a rhomboid (non-rhombus non-rectangle). A non-parallelogram can be a trapezium with exactly one pair of parallel sides, which can be isosceles (with equal legs) or scalene (with unequal legs); or a trapezoid (τραπεζοειδή, literally "table-like") with no parallel sides. Hutton's definitions in 1795 All European languages except for English follow Proclus's meanings of trapezium and trapezoid, as did English until the late 18th century, when an influential mathematical dictionary published by Charles Hutton in 1795 transposed the two terms without explanation, leading to widespread inconsistency. Hutton's change was reversed in British English in about 1875, but it has been retained in American English to the present. Late 19th century American geometry textbooks define a trapezium as having no parallel sides, a trapezoid as having exactly one pair of parallel sides, and a parallelogram as having two sets of opposing parallel sides. To avoid confusion between contradictory British and American meanings of trapezium and trapezoid, quadrilaterals with no parallel sides have sometimes been called irregular quadrilaterals. Special cases [edit] Trapezoid special cases. The orange figures also qualify as parallelograms. An isosceles trapezoid is a trapezoid where the base angles have the same measure. As a consequence the two legs are also of equal length and it has reflection symmetry. This is possible for acute trapezoids or right trapezoids as rectangles. An acute trapezoid is a trapezoid with two adjacent acute angles on its longer base, and the isosceles trapezoid is an example of an acute trapezoid. The isosceles trapezoid has a special case known as a three-sided trapezoid, meaning it is a trapezoid wherein two trapezoid's legs have equal lengths as the trapezoid's base at the top. The isosceles trapezoid is the convex hull of an antiparallelogram, a type of crossed quadrilateral. Every antiparallelogram is formed with such a trapezoid by replacing two parallel sides by the two diagonals. An obtuse trapezoid, on the other hand, has one acute and one obtuse angle on each base. An example is parallelogram with equal acute angles. A right trapezoid is a trapezoid with two adjacent right angle. One special type of right trapezoid is by forming three right triangles, which was used by James Garfield to prove the Pythagorean theorem. A tangential trapezoid is a trapezoid that has an incircle. Condition of existence [edit] Four lengths a, c, b, d can constitute the consecutive sides of a non-parallelogram trapezoid with a and b parallel only when \|d−c\|<\|b−a\|<d+c.{\displaystyle \displaystyle \|d-c\|<\|b-a\|<d+c.} The quadrilateral is a parallelogram when d−c=b−a=0{\displaystyle d-c=b-a=0}, but it is an ex-tangential quadrilateral (which is not a trapezoid) when \|d−c\|=\|b−a\|≠0{\displaystyle \|d-c\|=\|b-a\|\neq 0}. Characterizations [edit] general trapezoid/trapezium: parallel sides: a,b{\displaystyle a,\,b} with a<b{\displaystyle a<b} legs: c,d{\displaystyle c,\,d} diagonals: q,p{\displaystyle q,\,p} midsegment: m{\displaystyle m} height/altitude: h{\displaystyle h} trapezoid/trapezium with opposing triangles S,T{\displaystyle S,\,T} formed by the diagonals Given a convex quadrilateral, the following properties are equivalent, and each implies that the quadrilateral is a trapezoid: It has two adjacent angles that are supplementary, that is, they add up to 180 degrees. The angle between a side and a diagonal is equal to the angle between the opposite side and the same diagonal. The diagonals cut each other in mutually the same ratio (this ratio is the same as that between the lengths of the parallel sides). The diagonals cut the quadrilateral into four triangles of which one opposite pair have equal areas. The product of the areas of the two triangles formed by one diagonal equals the product of the areas of the two triangles formed by the other diagonal. The areas S and T of some two opposite triangles of the four triangles formed by the diagonals satisfy the equation K=S+T,{\displaystyle {\sqrt {K}}={\sqrt {S}}+{\sqrt {T}},}where K is the area of the quadrilateral. The midpoints of two opposite sides of the trapezoid and the intersection of the diagonals are collinear. The angles in the quadrilateral ABCD satisfy sin⁡A sin⁡C=sin⁡B sin⁡D.{\displaystyle \sin A\sin C=\sin B\sin D.} The cosines of two adjacent angles sum to 0, as do the cosines of the other two angles. The cotangents of two adjacent angles sum to 0, as do the cotangents of the other two adjacent angles. One bimedian divides the quadrilateral into two quadrilaterals of equal areas. Twice the length of the bimedian connecting the midpoints of two opposite sides equals the sum of the lengths of the other sides. Additionally, the following properties are equivalent, and each implies that opposite sides a and b are parallel: The consecutive sides a, c, b, d and the diagonals p, q satisfy the equation p 2+q 2=c 2+d 2+2 a b.{\displaystyle p^{2}+q^{2}=c^{2}+d^{2}+2ab.} The distance v between the midpoints of the diagonals satisfies the equation v=\|a−b\|2.{\displaystyle v={\frac {\|a-b\|}{2}}.} Properties [edit] Midsegment and height [edit] The midsegment or median of a trapezoid is the segment that joins the midpoints of the legs. It is parallel to the bases. Its length m is equal to the average of the lengths of the bases a and b of the trapezoid, m=a+b 2.{\displaystyle m={\frac {a+b}{2}}.} The midsegment of a trapezoid is one of the two bimedians (the other bimedian divides the trapezoid into equal areas). The height (or altitude) is the perpendicular distance between the bases. In the case that the two bases have different lengths (a ≠ b), the height of a trapezoid h can be determined by the length of its four sides using the formula h=(p−2 a)(p−2 b)(p−2 b−2 d)(p−2 b−2 c)2\|b−a\|{\displaystyle h={\frac {\sqrt {(p-2a)(p-2b)(p-2b-2d)(p-2b-2c)}}{2\|b-a\|}}} where c and d are the lengths of the legs and p=a+b+c+d{\displaystyle p=a+b+c+d}. Area [edit] The area K{\displaystyle K} of a trapezoid is given by the product of the midsegment (the average of the two bases) and the height: K=1 2(a+b)h{\displaystyle K={\tfrac {1}{2}}(a+b)h} where a{\displaystyle a} and b{\displaystyle b} are the lengths of the bases, and h{\displaystyle h} is the height (the perpendicular distance between these sides). This method was used in the classical age of India in Aryabhata's Aryabhatiya (section 2.8), yielding as a special case the well-known formula for the area of a triangle, by considering a triangle as a degenerate trapezoid in which one of the parallel sides has shrunk to a point. The 7th-century Indian mathematician Bhāskara I derived the following formula for the area of a trapezoid with consecutive sides a{\displaystyle a}, c{\displaystyle c}, b{\displaystyle b}, d{\displaystyle d}:: K=1 2(a+b)c 2−1 4((b−a)+c 2−d 2 b−a)2{\displaystyle K={\frac {1}{2}}(a+b){\sqrt {c^{2}-{\frac {1}{4}}\left((b-a)+{\frac {c^{2}-d^{2}}{b-a}}\right)^{2}}}} where a{\displaystyle a} and b{\displaystyle b} are parallel and b>a{\displaystyle b>a}. This formula can be factored into a more symmetric version K=a+b 4\|b−a\|(−a+b+c+d)(a−b+c+d)(a−b+c−d)(a−b−c+d).{\displaystyle K={\frac {a+b}{4\|b-a\|}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)}}.} When one of the parallel sides has shrunk to a point (say a = 0), this formula reduces to Heron's formula for the area of a triangle. Another equivalent formula for the area, which more closely resembles Heron's formula, is K=a+b\|b−a\|(s−b)(s−a)(s−b−c)(s−b−d){\displaystyle K={\frac {a+b}{\|b-a\|}}{\sqrt {(s-b)(s-a)(s-b-c)(s-b-d)}}} where s=1 2(a+b+c+d){\displaystyle s={\tfrac {1}{2}}(a+b+c+d)} is the semiperimeter of the trapezoid. (This formula is similar to Brahmagupta's formula, but it differs from it, in that a trapezoid might not be cyclic (inscribed in a circle). The formula is also a special case of Bretschneider's formula for a general quadrilateral). From Bretschneider's formula, it follows that K=(a b 2−a 2 b−a d 2+b c 2)(a b 2−a 2 b−a c 2+b d 2)4(b−a)2−(c 2+d 2−a 2−b 2 4)2.{\displaystyle K={\sqrt {{\frac {\left(ab^{2}-a^{2}b-ad^{2}+bc^{2}\right)\left(ab^{2}-a^{2}b-ac^{2}+bd^{2}\right)}{4(b-a)^{2}}}-\left({\frac {c^{2}+d^{2}-a^{2}-b^{2}}{4}}\right)^{2}}}.} The bimedian connecting the parallel sides bisects the area. More generally, any line drawn through the midpoint of the median parallel to the bases, that intersects the bases, bisects the area. Any triangle connecting the two ends of one leg to the midpoint of the other leg is also half of the area. Diagonals [edit] The lengths of the diagonals are p=a b 2−a 2 b−a c 2+b d 2 b−a,q=a b 2−a 2 b−a d 2+b c 2 b−a,{\displaystyle {\begin{aligned}p&={\sqrt {\frac {ab^{2}-a^{2}b-ac^{2}+bd^{2}}{b-a}}},\q&={\sqrt {\frac {ab^{2}-a^{2}b-ad^{2}+bc^{2}}{b-a}}},\end{aligned}}} where a{\displaystyle a} is the short base, b{\displaystyle b} is the long base, and c{\displaystyle c} and d{\displaystyle d} are the trapezoid legs. If the trapezoid is divided into four triangles by its diagonals AC and BD (as shown on the right), intersecting at O, then the area of △{\displaystyle \triangle }AOD is equal to that of △{\displaystyle \triangle }BOC, and the product of the areas of △{\displaystyle \triangle }AOD and △{\displaystyle \triangle }BOC is equal to that of △{\displaystyle \triangle }AOB and △{\displaystyle \triangle }COD. The ratio of the areas of each pair of adjacent triangles is the same as that between the lengths of the parallel sides. If ⁠ℓ{\displaystyle \ell }⁠ is the length of the line segment parallel to the bases, passing through the intersection of the diagonals, with one endpoint on each leg, then ⁠ℓ{\displaystyle \ell }⁠ is the harmonic mean of the lengths of the bases: 1 ℓ=1 2(1 a+1 b).{\displaystyle {\frac {1}{\ell }}={\frac {1}{2}}\left({\frac {1}{a}}+{\frac {1}{b}}\right).} The line that goes through both the intersection point of the extended nonparallel sides and the intersection point of the diagonals, bisects each base. Other properties [edit] The center of area (center of mass for a uniform lamina) lies along the line segment joining the midpoints of the parallel sides, at a perpendicular distance x from the longer side b given by x=h 3(2 a+b a+b).{\displaystyle x={\frac {h}{3}}\left({\frac {2a+b}{a+b}}\right).} The center of area divides this segment in the ratio (when taken from the short to the long side): p. 862 a+2 b 2 a+b.{\displaystyle {\frac {a+2b}{2a+b}}.} If the angle bisectors to angles A and B intersect at P, and the angle bisectors to angles C and D intersect at Q, then P Q=\|A D+B C−A B−C D\|2.{\displaystyle PQ={\frac {\|AD+BC-AB-CD\|}{2}}.} Applications [edit] The trapezoidal rule for numerical integration The Temple of Dendur in the Metropolitan Museum of Art in New York City Example of a trapeziform pronotum outlined on a spurge bug Ontario Highway 502 In calculus, the definite integral of a functionf(x){\displaystyle f(x)} can be numerically approximated as a discrete sum by partitioning the interval of integration into small uniform intervals and approximating the function's value on each interval as the average of the values at its endpoints: ∫a b f(x)d x≈∑k=1 N 1 2(f(x k−1)+f(x k))Δ x,{\displaystyle \int {a}^{b}f(x)\,dx\approx \sum {k=1}^{N}{\tfrac {1}{2}}{\bigl (}f(x_{k-1})+f(x_{k}){\bigr )}\Delta x,} where N{\displaystyle N} is the number of intervals, x 0=a{\displaystyle x_{0}=a}, x N=b{\displaystyle x_{N}=b}, and Δ x=(b−a)/N{\displaystyle \Delta x=(b-a)/N}. Graphically, this amounts to approximating the region under the graph of the function by a collection of trapezoids, so this method is called the trapezoidal rule. When any rectangle is viewed in perspective from a position which is centered on one axis but not the other, it appears to be an isosceles trapezoid, called the keystone effect because arch keystones are commonly trapezoidal. For example, when a rectangular building façade is photographed from the ground at a position directly in front using a rectilinear lens, the image of the building is an isosceles trapezoid. Such photographs sometimes have a "keystone transformation" applied to them to recover rectangular shapes. Video projectors sometimes apply such a keystone transformation to the recorded image before projection, so that the image projected on a flat screen appears undistorted. Piazza del Campidoglio viewed from directly above. Trapezoidal doors and windows were the standard style for the Inca, although it can be found used by earlier cultures of the same region and did not necessarily originate with them. An almena, a battlement feature characteristic of Moorish architecture, is trapezoidal. Michaelangelo's redesign of the Piazza del Campidoglio (see photograph at right) incorporated a trapezoid surrounding an ellipse, giving the effect of a square surrounding a circle when seen foreshortened at ground level.Cinematography takes advantage of trapezoids in the opposite way, to produce an excessive foreshortening effect from the camera viewpoint, giving the illusion of greater depth to a room in a movie studio than the set physically has. Trapezoids were also used to produce the visual distortions of Caligarism.Canals and drainage ditches commonly have a trapezoidal cross-section. In biology, especially morphology and taxonomy, terms such as trapezoidal or trapeziform commonly are useful in descriptions of particular organs or forms. Trapezoids are sometimes used as a graphical symbol. In circuit diagrams, a trapezoid is the symbol for a multiplexer. An isosceles trapezoid is used for the shape of road signs, for example, on secondary highways in Ontario, Canada. Non-Euclidean geometry [edit] In spherical or hyperbolic geometry, the internal angles of a quadrilateral do not sum to 360°, but quadrilaterals analogous to trapezoids, parallelograms, and rectangles can still be defined, and additionally there are a few new types of quadrilaterals not distinguished in the Euclidean case. A spherical or hyperbolic trapezoid is a quadrilateral with two opposite sides, the legs, each of whose two adjacent angles sum to the same quantity; the other two sides are the bases. As in Euclidean geometry, special cases include isosceles trapezoids whose legs are equal (as are the angles adjacent to each base), parallelograms with two pairs of opposite equal angles and two pairs of opposite equal sides, rhombuses with two pairs of opposite equal angles and four equal sides, rectangles with four equal (non-right) angles and two pairs of opposite equal sides, and squares with four equal (non-right) angles and four equal sides. When a rectangle is cut in half along the line through the midpoints of two opposite sides, each of the resulting two pieces is an isosceles trapezoid with two right angles, called a Saccheri quadrilateral. When a rectangle is cut into quarters by the two lines through pairs of opposite midpoints, each of the resulting four pieces is a quadrilateral with three right angles called a Lambert quadrilateral. In Euclidean geometry Saccheri and Lambert quadrilaterals are merely rectangles. Related topics [edit] An example of trapezoidal number: 15 = 4 + 5 + 6 The trapezoidal number is a set of positive integers obtained by summing consecutively two or more positive integers greater than one, forming a trapezoidal pattern. The crossed ladders problem is the problem of finding the distance between the parallel sides of a right trapezoid, given the diagonal lengths and the distance from the perpendicular leg to the diagonal intersection. See also [edit] Frustum, a solid having trapezoidal faces Wedge, a polyhedron defined by two triangles and three trapezoid faces. Notes [edit] ^"Trapezoid – math word definition – Math Open Reference". www.mathopenref.com. Retrieved 2024-05-15. ^Gardiner, Anthony D.; Bradley, Christopher J. (2005). Plane Euclidean Geometry: Theory and Problems. United Kingdom Mathematics Trust. p.34. ISBN9780953682362. ^ abcdHopkins 1891, p.33. ^Usiskin & Griffin 2008, p.29. ^Alsina & Nelsen 2020, p.90. ^ abRingenberg, Lawrence A. (1977). "Coordinates in a Plane". College Geometry. R. E. Krieger Publishing Company. pp.161–162. ISBN9780882755458. ^Alsina & Nelsen 2020, p.89. ^Usiskin & Griffin 2008, p.32. ^Craine, Timothy V.; Rubenstein, Rheta N. (1993). "A Quadrilateral Hierarchy to Facilitate Learning in Geometry". The Mathematics Teacher. 86 (1): 30–36. doi:10.5951/MT.86.1.0030. JSTOR27968085. ^Popovic, Gorjana (2012). "Who is This Trapezoid, Anyway?". Mathematics Teaching in the Middle School. 18 (4): 196–199. doi:10.5951/mathteacmiddscho.18.4.0196. JSTOR10.5951/mathteacmiddscho.18.4.0196. ResearchGate:259750174. ^Michon, Gérard P. "History and Nomenclature". Retrieved 2023-06-09. ^Beem, John K. (2006). Geometry Connections: Mathematics for Middle School Teachers. Connections in mathematics courses for teachers. Pearson Prentice Hall. p.57. ISBN9780131449268. ^ abcMurray, James (1926). "Trapezium". A New English Dictionary on Historical Principles: Founded Mainly on the Materials Collected by the Philological Society. Vol.X. Clarendon Press at Oxford. p.286, also see "Trapezoid", pp.286–287. ^Morrow, Glenn R., ed. (1970). Proclus: A commentary on the first book of Euclid's Elements. Princeton University Press. §§169–174, pp.133–137. ^Conway, Burgiel & Goodman-Strauss 2016, p.286. ^Hobbs 1899, p.66. ^Davies, Charles (1873). The Nature and Utility of Mathematics. New York: A.S. Barnes & Company. p.35. ^Dodge 2012, p.82. ^ abPosamentier, Alfred S.; Bannister, Robert L. (2014). "The Trapezoid". Geometry, Its Elements and Structure: Second Edition. Dover Books on Mathematics (2nd ed.). Courier Corporation. §7.7, pp.282–287. ISBN9780486782164. ^Hopkins 1891, p.34. ^ abAlsina & Nelsen 2020, p.90–91. ^Alsina & Nelsen 2020, p.212. ^Alsina & Nelsen 2020, p.91. ^Garfield, James (1876). "Pons Asinorum". New England Journal of Education. 3 (14): 161. ISSN2578-4145. JSTOR44764657. ^Ask Dr. Math (2008), "Area of Trapezoid Given Only the Side Lengths". ^Josefsson 2013, p.35. ^Josefsson 2013, Prop. 5. ^Josefsson 2013, Thm. 6. ^Josefsson 2013, Thm. 8. ^Josefsson 2013, Thm. 15. ^ abJosefsson 2013, p.25. ^ abJosefsson 2013, p.26. ^Josefsson 2013, p.31. ^Josefsson 2013, Cor. 11. ^Josefsson 2013, Thm. 12. ^Hobbs 1899, p.58. ^Dodge 2012, p.117. ^ abcdWeisstein, Eric W."Trapezoid". MathWorld. ^Dodge 2012, p.84. ^Puttaswamy, T. K. (2012). Mathematical Achievements of Pre-modern Indian Mathematicians. Elsevier. p.156. ISBN978-0-12-397913-1. ^Hopkins 1891, p.95. ^Alsina & Nelsen 2020, p.96. ^Skidell, Akiva (1977). "The Harmonic Mean: A Nomograph, and some Problems". The Mathematics Teacher. 70 (1): 30–34. doi:10.5951/MT.70.1.0030. JSTOR27960699. Hoehn, Larry (1984). "A Geometrical Interpretation of the Weighted Mean". Two-Year College Mathematics Journal. 15 (2): 135–139. doi:10.1080/00494925.1984.11972762 (inactive 1 July 2025).{{cite journal}}: CS1 maint: DOI inactive as of July 2025 (link) ^ abOwen Byer, Felix Lazebnik and Deirdre Smeltzer, Methods for Euclidean Geometry, Mathematical Association of America, 2010, p. 55. ^"Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a General Trapezoid". www.efunda.com. Retrieved 2024-05-15. ^Apostol, Tom M.; Mnatsakanian, Mamikon A. (December 2004). "Figures Circumscribing Circles"(PDF). American Mathematical Monthly. 111 (10): 853–863. doi:10.2307/4145094. JSTOR4145094. Retrieved 2016-04-06. ^Varberg, Dale E.; Purcell, Edwin J.; Rigdon, Steven E. (2007). Calculus (9th ed.). Pearson Prentice Hall. p.264. ISBN978-0131469686. ^"Machu Picchu Lost City of the Incas – Inca Geometry". gogeometry.com. Retrieved 2018-02-13. ^Hyslop, John (2014). Inka Settlement Planning. University of Texas Press. p.54. ISBN9780292762640. ^Curl 1999, p.19, almena. ^Curl 1999, p.486, Michaelangelo Buonarroti. ^ abRamírez 2012, p.84. ^John L. Capinera (11 August 2008). Encyclopedia of Entomology. Springer Science & Business Media. pp.386, 1062, 1247. ISBN978-1-4020-6242-1. ^Daniels, Jerry (1996). Digital Design from Zero to One. John Wiley & Sons. p.203. ISBN978-0-471-12447-4. ^Alsina & Nelsen 2020, p.93. ^Petrov, F. V. (2009). Вписанные четырёхугольники и трапеции в абсолютной геометрии [Cyclic quadrilaterals and trapezoids in absolute geometry] (PDF). Matematicheskoe Prosveschenie. Tret’ya Seriya (in Russian). 13: 149–154. ^Gamer, Carlton; Roeder, David W.; Watkins, John J. (1985). "Trapezoidal numbers". Mathematics Magazine. 58 (2): 108–110. doi:10.2307/2689901. JSTOR2689901. ^Alsina & Nelsen (2020), p.102. Bibliography [edit] Alsina, Claudi; Nelsen, Roger (2020). A Cornucopia of Quadrilaterals. Mathematical Association of America. ISBN978-1-4704-5312-1. Conway, John H.; Burgiel, Heidi; Goodman-Strauss, Chaim (2016). The Symmetries of Things. CRC Press. ISBN978-1-4398-6489-0. Curl, James Stevens (1999). A Dictionary of Architecture. Oxford University Press. ISBN9780198606789. Dodge, Clayton W. (2012). Euclidean Geometry and Transformations. Dover Books on Mathematics. Courier Corporation. ISBN9780486138428. Hobbs, Charles Austin (1899). The Elements of Plane Geometry. A. Lovell & Company. Hopkins, George Irving (1891). Manual of Plane Geometry. D.C. Heath & Company. Josefsson, Martin (2013). "Characterizations of trapezoids"(PDF). Forum Geometricorum. 13: 23–35. Archived from the original(PDF) on 16 June 2013. Usiskin, Zalman; Griffin, Jennifer (2008). The Classification of Quadrilaterals: A Study of Definition. Information Age Publishing. pp.49–52, 63–67. Ramírez, Juan Antonio (2012). "Architecture and Desire: The character of film constructions". Architecture for the Screen: A Critical Study of Set Design in Hollywood's Golden Age. Translated by Moffitt, John F. McFarland. ISBN9780786469307. Further reading [edit] Fraivert, David; Sigler, Avi; Stupel, Moshe (2016). "Common properties of trapezoids and convex quadrilaterals". Journal of Mathematical Sciences: Advances and Applications. 38: 49–71. doi:10.18642/jmsaa_7100121635. External links [edit] "Trapezium" at the Encyclopedia of Mathematics Weisstein, Eric W."Right trapezoid". MathWorld. Trapezoid definition, Area of a trapezoid, Median of a trapezoid (with interactive animations) Trapezoid (North America) at elsy.at: Animated course (construction, circumference, area) Trapezoidal Rule on Numerical Methods for Stem Undergraduate Autar Kaw and E. Eric Kalu, Numerical Methods with Applications (2008) \| v t e Polygons (List) \| \| Triangles \| Acute Equilateral Ideal Isosceles Kepler Obtuse Right \| \| Quadrilaterals \| Antiparallelogram Apollonius Bicentric Crossed Cyclic Equidiagonal Ex-tangential Harmonic Isosceles trapezoid Kite Orthodiagonal Parallelogram Rectangle Right kite Right trapezoid Rhomboid Rhombus Square Tangential Tangential trapezoid Trapezoid \| \| By number of sides \| \| 1–10 sides \| Monogon (1) Digon (2) Triangle (3) Quadrilateral (4) Pentagon (5) Hexagon (6) Heptagon (7) Octagon (8) Nonagon/Enneagon (9) Decagon (10) \| \| 11–20 sides \| Hendecagon (11) Dodecagon (12) Tridecagon (13) Tetradecagon (14) Pentadecagon (15) Hexadecagon (16) Heptadecagon (17) Octadecagon (18) Icosagon (20) \| \| >20 sides \| Icositrigon (23) Icositetragon (24) Triacontagon (30) 257-gon Chiliagon (1000) Myriagon (10,000) 65537-gon Megagon (1,000,000) Apeirogon (∞) \| \| \| Star polygons \| Pentagram Hexagram Heptagram Octagram Enneagram Decagram Hendecagram Dodecagram \| \| Classes \| Concave Convex Cyclic Equiangular Equilateral Infinite skew Isogonal Isotoxal Magic Pseudotriangle Rectilinear Regular Reinhardt Simple Skew Star-shaped Tangential Weakly simple \| Retrieved from " Categories: Elementary shapes Types of quadrilaterals Hidden categories: CS1 maint: DOI inactive as of July 2025 CS1 uses Russian-language script (ru) CS1 Russian-language sources (ru) Articles with short description Short description matches Wikidata Use American English from August 2020 All Wikipedia articles written in American English Pages using multiple image with auto scaled images This page was last edited on 23 August 2025, at 03:02(UTC). 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17169	https://mrsrickersclass.weebly.com/uploads/2/3/5/9/23596608/triangle_inequlaity_theorem_1.pdf	Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 1 1. Lengths 13, 11, 10 could represent the measures of the sides of a triangle? 2. In triangle KLH, <K = 40° and <L. Which is the smallest side of the triangle? 3. Two sides of an isosceles triangle measures 24 and 11. What is the possible value of the third side? 4. In triangle FGH, an exterior angle at F measures 70°, and <G = 50°. Which is the longest side of the triangle? 5. Lengths 16, 11, 18 could represent the measures of the sides of a triangle? 6. In triangle KLM, <K = 55° and <L = 40°. Which is the longest side of the triangle? 7. In triangle NOP, <N = 95° and <P. Which is the longest side of the triangle? 8. In ∆PQR, PQ = 8, QR = 7, RP = 15. Which is the largest angle? 9. In triangle RPS, an exterior angle at R measures 64°, and <P = 26°. Which is the longest side of the triangle? 10. Two sides of an isosceles triangle measures 16 and 9. What is the possible value of the third side? Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 1 ANSWERS 1. Yes 2. KH is the smallest side of the triangle. 3. 24 4. GH is the largest side of the triangle. 5. Yes 6. KL is the largest side of the triangle. 7. OP is the largest side of the triangle. 8. <Q is the largest angle. 9. PS is the largest side of the triangle. 10. 16 or 9 Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 2 1. Lengths 15, 13, 12 could represent the measures of the sides of a triangle? 2. In triangle GMD, <G = 60° and <M. Which is the smallest side of the triangle? 3. Two sides of an isosceles triangle measures 25 and 12. What is the possible value of the third side? 4. In triangle IJK, an exterior angle at I measures 65°, and <J = 35°. Which is the longest side of the triangle? 5. Lengths 17, 10, 6 could represent the measures of the sides of a triangle? 6. In triangle OPQ, <O = 50° and <P = 35°. Which is the longest side of the triangle? 7. In triangle QRS, <Q = 92° and <S. Which is the longest side of the triangle? 8. In ∆STU, ST = 7, TU = 8, US = 14. Which is the largest angle? 9. In triangle TUV, an exterior angle at T measures 62°, and <U = 28°. Which is the longest side of the triangle? 10. Two sides of an isosceles triangle measures 14 and 11. What is the possible value of the third side? Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 2 ANSWERS 1. Yes 2. DG is the smallest side of the triangle. 3. 25 4. JK is the largest side of the triangle. 5. No 6. OP is the largest side of the triangle. 7. RS is the largest side of the triangle. 8. <T is the largest angle. 9. UV is the largest side of the triangle. 10. 14 or 11 Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 3 1. Lengths 12, 7, 3 could represent the measures of the sides of a triangle? 2. In triangle HDC, <H = 45° and <D. Which is the smallest side of the triangle? 3. Two sides of an isosceles triangle measures 26 and 10. What is the possible value of the third side? 4. In triangle LMN, an exterior angle at L measures 60°, and <M = 30°. Which is the longest side of the triangle? 5. Lengths 14, 13, 16 could represent the measures of the sides of a triangle? 6. In triangle RST, <R = 55° and <S = 40°. Which is the longest side of the triangle? 7. In triangle TUV, <T = 96° and <V. Which is the longest side of the triangle? 8. In ∆XYZ, XY = 9, YZ = 13, ZX = 11. Which is the largest angle? 9. In triangle WXY, an exterior angle at W measures 68°, and <X = 22°. Which is the longest side of the triangle? 10. Two sides of an isosceles triangle measures 7 and 16. What is the possible value of the third side? Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 3 ANSWERS 1. No 2. HC is the smallest side of the triangle. 3. 26 4. MN is the largest side of the triangle. 5. Yes 6. RS is the largest side of the triangle. 7. UV is the largest side of the triangle. 8. <X is the largest angle. 9. XY is the largest side of the triangle. 10. 16 Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 4 1. Lengths 16, 9, 15 could represent the measures of the sides of a triangle? 2. In triangle SLR, <S = 55° and <L. Which is the smallest side of the triangle? 3. Two sides of an isosceles triangle measures 22 and 12. What is the possible value of the third side? 4. In triangle ABC, an exterior angle at A measures 90°, and <B = 45°. Which is the longest side of the triangle? 5. Lengths 18, 6, 9 could represent the measures of the sides of a triangle? 6. In triangle UVW, <U = 56° and <V = 42°. Which is the longest side of the triangle? 7. In triangle ABC, <A = 88° and <C. Which is the longest side of the triangle? 8. In ∆DEF, DE = 15, EF = 12, FD = 12. Which is the largest angle? 9. In triangle JKL, an exterior angle at J measures 70°, and <J = 25°. Which is the longest side of the triangle? 10. Two sides of an isosceles triangle measures 10 and 18. What is the possible value of the third side? Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 4 ANSWERS 1. Yes 2. SR is the smallest side of the triangle. 3. 22 4. BC is the largest side of the triangle. 5. No 6. UV is the largest side of the triangle. 7. BC is the largest side of the triangle. 8. <F is the largest angle. 9. KL is the largest side of the triangle. 10. 10 or 18 Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 5 1. Lengths 15, 10, 26 could represent the measures of the sides of a triangle? 2. In triangle KFD, <K = 58° and <F. Which is the smallest side of the triangle? 3. Two sides of an isosceles triangle measures 26 and 11. What is the possible value of the third side? 4. In triangle RST, an exterior angle at R measures 98°, and <S = 35°. Which is the longest side of the triangle? 5. Lengths 14, 5, 12 could represent the measures of the sides of a triangle? 6. In triangle XYZ, <X = 52° and <Y = 44°. Which is the longest side of the triangle? 7. In triangle GHI, <G = 83° and <I. Which is the longest side of the triangle? 8. In ∆MNO, MN = 14, NO = 18, OM = 11. Which is the largest angle? 9. In triangle PQR, an exterior angle at P measures 75°, and <Q = 35°. Which is the longest side of the triangle? 10. Two sides of an isosceles triangle measures 9 and 16. What is the possible value of the third side? Name: ___ Date _ Tons of Free Math Worksheets at: © www.mathworksheetsland.com Topic: Triangle Inequality Theorem - Worksheet 5 ANSWERS 1. No 2. KD is the smallest side of the triangle. 3. 26 4. ST is the largest side of the triangle. 5. Yes 6. XY is the largest side of the triangle. 7. HI is the largest side of the triangle. 8. <M is the largest angle. 9. QR is the largest side of the triangle. 10. 16 or 9
17170	https://www.youtube.com/watch?v=VcZDS8dPaYc	4.4 Tension Force and Pulley Problems \| Application of Newton's Laws of Motion \| General Physics Chad's Prep 174000 subscribers 676 likes Description 28213 views Posted: 21 Sep 2023 Tension Force and Pulleys Chad provides a lesson on the application of Newton's Laws to Tension Problems and Pulley Problems. He starts with Tension problems and first provides a working definition of tension force and then explains how to incorporate tension forces into free body diagrams and net force equations. He begins with 1-dimensional tension practice problems before moving on to a 2-dimensional tension practice problem and then concludes this portion of the lesson with a two body tension problem. Chad moves on to Pulleys Problems starting with a problem that demonstrates the mechanical advantage of a system of pulleys. He then moves on to common two body pulley problems showing how to properly set up a system of net force equations for which acceleration and tension force can be solved. 00:00 Lesson Introduction 00:44 1-Dimensional Tension Problems 04:18 2-Dimensional Tension Problem 08:29 Two Body Tension Problem 15:39 Mechanical Advantage Pulley Problem 18:40 Two Body Pulley Practice Problem #1 26:36 Two Body Pulley Practice Problem #2 Check out Chad's General Physics Master Course: 47 comments Transcript: Lesson Introduction tension and pulley problem is going to be the topic of this lesson in my brand new General Physics playlist which is going to cover a full year of University algebra based General Physics now we're continuing on with applications of Newton's Laws of Motion we're going to set up free body diagrams we're going to set up the net force or some of the forces equations uh and just do a whole bunch more problem solving my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on chadsprep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat 1-Dimensional Tension Problems all right so diving right into tension force problems here so and simply put tension is just the force in a taut rope or cable or a piece of wire so and basically if you put like an inline spring scale in the middle of that wire or that Rover that cable whatever it reads that would be the tension and it's everywhere the same in that piece of rope cable or wire so and it usually pulls on both directions whatever two objects it's attached to it's applying a tension force on both of them but typically in opposite directions as we'll see all right so first question here so a single wire supports the weight of a hanging 10.0 kilogram Mass what is the tension in the wire and so we're simply going to set up free body diagrams the way we have been this one's nice it's going to end up being just a one-dimensional problem so but on this 10.0 kilogram Mass we've got the tension pulling up on it and then we've got its own weight pointing down now if you look at that tension of that rope again it's everywhere the same now if you said well what's what's the tension uh you know what is the force of tension on the ceiling from that cable right there or that wire well in that case it'd be the same but it would be pulling down instead and that's what you said where like if there's a cable or rope or a wire attached to two objects the tension for both is typically going to be the same but in opposite directions so here relative to that 10.0 kilogram that's how we're going to solve for this uh tension uh we'll label it customarily t if you have more than one tension and they're not equal typically you'll use like T1 and T2 so here we'll just use T in this case this object is not accelerating it's not even moving and so we're in equilibrium and Newton's first law is going to apply and so the sum of the forces equals zero and in this case tension points up we'll make it positive its weight points down we'll make it negative and so we're going to end up with t minus mg equals zero or the tension force simply equals mg added together side and now we'll substitute in so we got 10.0 kilograms 9.8 meters per second squared so in here we can see do this one on our head 98 Newtons but we're given three sig figs on the 10.0 kilogram so we're going to make that 98.0 Newtons to match up with the three sig figs there all right so similar put uh similarly put question here but with two cables and it says two identical wires support the weight of a hanging 10.0 kilogram Mass what is the tension in each wire so now we're told they're identical they're each going to exhibit a tension so on that 10 kilogram mass and since they're identical wires in every respect we're going to assume that the tension reach is the same so we don't have to label these T1 and T2 they both exert the same tension on this 10 kilogram weight and we still have its own weight pointing down here of mg but once again this lovely 10 kilogram mass is in equilibrium it's not it doesn't have any motion whatsoever it definitely doesn't have an acceleration as a result and so the sum of the forces once again equals zero but instead of just one tension we now have two times the tension minus mg pointing down equals zero so two times the tension equals mg so 2 times the tension equals 10.0 kilograms times 9.0 meters per second squared so and again that's 10 times 9.8 is 98 divided by two and we see the tension is just half as much and again to get three sig figs we'll make it 49.0 Newtons okay so these aren't too bad they're nice one-dimensional problems but we can definitely make this harder and we are about two so let's take a look at a two-dimensional tension problem so in this one we've got two wires of 2-Dimensional Tension Problem identical length support the weight of a hanging 10.0 kilogram Mass what is the tension in each wire well now these wires don't perfectly pull in just one dimension just say the Y Dimension they pull in the X and the Y so if we take a look at our free body diagram again relative to our 10 kilogram Mass it once again has its own weight mg so but then we've got a tension pulling in the wire that direction and a tension pulling in the wire that direction another at the same angle they're identical wires we can assume once again that those tensions are indeed equal so no big there we don't have a T1 and a T2 just T so but now we've got to break those up into components to find out how much is in the X and how much is in the y direction and so in this case we're going to have so much of this one in the X Direction so much of this one in the X Direction and then so much of this one in the y direction so much of this one in the y direction so and if we look at where that 60 degree angle is this is an alternate interior angle so this is also 60 degrees if you want to look at it that way and so this is the opposite side and as it should be so we've got our angle with respect to the x-axis 60 degrees and so sine is going to be used for that y component and so in this case we're going to end up with tension times sine of 60 degrees on this side but also tension times sine of 60 degrees on this side and if we cared and we don't care too much here as it turns out we're going to have tension times cosine of 60 degrees there and tension times cosine of 60 degrees there as well and we can see that those are equal and magnitude in opposite direction and they obviously are going to negate each other and stuff like that but these two don't and so in this case we could set the sum of the forces equations both in the X and Y directions but if we want to solve for tension we're still going to end up doing that in the y direction so so some of the forces once again this object is in equilibrium it's not moving it's definitely not accelerating so it's an equilibrium some of the force equals zero according to Newton's first law and we can see here that the total forces here are going to be 2 times T times 6 sine 60. so minus mg and again adds up to zero so again these both Point UPS they're made in positive and mg points now we made it negative so and now we can go ahead and sign solve for that tension so we're going to have 2 t sine of 60 degrees equals mg so and in this case I'm going to kind of start combining some things so the mass was once again 10.0 kilograms gravity is 9.8 meters per second squared and now we're going to divide this by 2 times sine of 60 degrees dividing it through to solve for that tension and now we've got to let our calculator do the work probably not doing this one in our heads here all right so 10 times 9.8 divided by 2 and either put that all in parentheses we're also divided by sine of 60 degrees and we are going to get 56.58 and again to round that the three sig figs we'll make that 56.6 Newtons based on the the numbers we're given three sig figs is appropriate cool so not too much harder and once again we could have set up one more equation here if we wanted to so and in the X Direction wouldn't help to solve this answer but in other problems it might and so some of the forces in the X direction also equals zero and I guess I technically could have put that in the y direction so in this case we would have had T cosine of 60 degrees making this one positive we made this one negative point in the opposite direction right versus left minus t cosine of 60 degrees equals zero and the reasons obviously this would have just canceled we would have had zero equals zero and there'd be no way to solve for tension and so we didn't bother setting it up it wasn't going to help us solve for tension in this case but we could have set it up had we needed to and again in some other problems we might come across some two-dimensional tension problems where it is necessary let's take a look at one last example all right so this last tension example we're going to look at is actually a two body problem and Two Body Tension Problem students usually struggle with these a fair amount more so but it works the same way you're just setting up two free body diagrams one for each of the objects in the two body problem here so this one reads two masses of 4.0 kilograms and 6.0 kilograms are connected by a rope so and they're being pulled along a frictionless horizontal Surface by a 60.0 Newton Force what is the acceleration of the blocks what is the tension in the Rope between the masses now notice it asks what is the acceleration of the blocks well they're touched by a rope and as long as that rope remains taut they're going to move together and therefore accelerate together and they're going to experience the same except generation so that's why I didn't have to ask individually but it could have said what's the acceleration of the four kilogram box or what's the acceleration of the six kilogram box would not have mattered they have the same acceleration and then the second question is what is the tension in the Rope between the masses and again that tension in the Rope is everywhere the same so and it pulls on each Mass with the same magnitude but in opposite directions as we'll see all right so again this is a two body problem we can technically set up free body diagrams and some of the force equations for both of them so if we take a look at the object on the right first we can see that it's going to have a tension pulling back so to the left so and then also it's going to have its own weight so and actually let's do this in green for a second so it's going to have its own weights mg and a normal force there we're ultimately going to find out we're not going to care about these there's no acceleration overall in the y direction we're going to find out the normal force just simply is equal to the weight that's also going to be true over here different Mass over here and a different normal force so it turns out but we're not going to care about them in either case just as an FYI all right now for the other box here we're also going to have a tension now pulling to the right instead but it's the same tension coming from the same rope just pointing in the opposite direction for this guy all right so we've got some of the force equations we can set up for both and again we could set them up for the y direction let's say if we wanted the normal force for some reason maybe if friction was involved but we're told it's frictionless and we're essentially those aren't going to help us at all in solving for either the acceleration or the tension notice the tension operates in the X Direction and the acceleration overall is going to be in the X direction as well and that's why we're going to start with the X direction if it turns out let's say friction was involved we found out we needed a normal force then we'd set up the Y ones as well but we're turns out we're just not going to need them so uh in this case we'll start with the one on the left here so in this case some of the forces in the X Direction and in this case in the X Direction it's not an equilibrium we're told we're solving for some sort of acceleration of this and we can see that that has to be true we've got a force pulling it to the right there's nothing counterbalancing that Force since it's on a frictionless Surface there's nothing pulling back so the sum of the forces is going to equal m a in that X Direction and if we see the only Force that's acting is that tension force and so that's going to equal Max here and we'll take this one step further so I've got to keep in mind that that's that specifically that four kilogram Mass specifically in this case cool but now we've got two variables and only one equation we've got tension which we don't know we've got the overall acceleration which we don't know both of which is what we want but we don't know them okay let's move on to setting up the same set of equations for this guy and once again some of the forces in the X direction is going to equal m a again not in equilibrium so in the X Direction it will be accelerating to the right so Newton's Second Law applies and in this case though we do have two forces we'll make the 60 Newton Force positive so and the tension pointing back the other direction negative so minus tension and in this case the M specifically got to be careful here is 6.0 kilograms and once again times that acceleration in the X Direction now a couple quick things when I originally I wrote These I have Max and Max and it might be proper to maybe make these M1 and M2 especially if we didn't know the masses we want to make sure we label them properly because they are definitely not the same mass so but in this case we were going to fill in a number anyway so it wasn't the biggest deal in the world uh and from here we've also got two variables we've got tension and we've got acceleration in the X direction for both but the key here is that now we have two variables with I'm sorry two uh two unknowns two variables and two equations and so we can solve a system of equations here to solve for that so easiest way to probably do this is to realize that this one's already solved for tension and just to take and substitute it in against the same tension it's the same rope so same magnitude and just substitute that in right into the other and then solve for the acceleration and then plug it back in maybe here and then solve for the tension all right so here we're going to have 60.0 Newtons minus and then we can see that that's going to be where that 4.0 kilograms times acceleration in the X Direction gets substituted right back in and that's going to equal 6.0 kilograms times ax as well all right and so then we'll take an add the four times the acceleration over to the other side and we're going to end up with 60.0 Newtons equals 10.0 kilograms times the acceleration so divide through by 10 and we can see that this acceleration is going to equal 6.00 meters per second squared all right so and the truth is when you look at this you're like well that force is having to pull all of this weight technically so it shouldn't be too surprising that it works out the same as if it was just a single 10 kilogram Mass uh with there being no friction or anything else to worry about in this case and so 60 newtons pulling on a total of 10 kilograms yeah I would give an acceleration of six meters per second squared but this is the proper setup though don't just assume you can always add them together like that although it's a nice little shortcut in this case but with a two body problem you've got two free body diagrams two sets of some of the force equations in this case we only set up the X Direction Dimension equations we didn't even set up the Y's it's not necessary in this problem but we could have and so not only did we set up just one set of equations here for each we actually could have set up two if it were necessary all right so that's the acceleration we've still got to go back and solve for the tension and the key is we can go back to either one of these equations and plug it back in here for the acceleration or here for the acceleration and then either solve for the tension or the tension there and it's definitely far easier over here so that's what we're going to do and so we're going to have tension is equal to 4.0 kilograms oh and you know what this should have just been two sig figs my bad the 4.0 kilograms and the 6.0 kilograms are just two sig figs but then we'll plug in that acceleration of 6.0 meters per second squared and we can see that the tension is definitely going to equal in two sig figs 24 Newtons and so now we've solved both the acceleration as well as the tension as asked so for our first pulley problem Mechanical Advantage Pulley Problem we're going to see why pulleys often have what we call a mechanical advantage where where we can lift some heavy object by applying a much smaller Force than the actual force it would normally require to lift that object so here's a typical system of pulleys so and here we've got three lower pulley wheels that are attached to some object here weighing 24 kilograms so and that entire pulley system anchored at the ceiling and the question is going to involve like what tension do I have to apply here so let's read the whole question so what force must be used to pull the end of the cable to keep the 24 kilogram Mass suspended in the air all right so let's take a look at this all right so we see what's going on here so just like in the uh the example where we had two strings or two cables tied to a mass and and they both applied attention upward the advantage here where we get this mechanical advantage uh is that there's the tension in the cable here is everywhere the same so and it's a the mask is attached to three lower wheels and the cable interacts those Wheels twice at every single place the way this works so where does that cable pull up on the wheels well it pulls up with a tension force right here and right here and right here and right here and right here and right here it pulls up on these wheels in two locations each and again the tension in the cable is everywhere the same so whatever tension force we Supply to this cable here by pulling down it's going to actually be applied to lifting this mass or suspending this mass in six different locations now we also got this object's weight in our free body diagram and so if we look at some of the forces so since it's suspended it's not moving so and not moving and definitely no acceleration means that some of the forces equal zero according to Newton's first law and so if we take a look at those forces well there's six times that tension force all pointing up so make those positive and then minus the weight and that's going to add up to zero so if we solve this for the tension 6 times tension equals mg tension is going to equal m g over 6. so we find out is that because it gets to pull up that tension in six different places then the tension only actually has to equal 1 6 of the weight it's the way that's going to play out and so in this case we got tensions going to equal 24. kilograms times 9.8 meters per second squared all over six and we'll let our calculator do the hard work here for us so 24 times 9.8 divided by 6 is going to get us 39.2 in this case we're going to round that down to 39 Newtons for two sig figs based on the 24 kilograms there and again that's 39.2 rounding down to 39 Newtons okay so the second pulley problem says Two Body Pulley Practice Problem #1 this what is the acceleration of the 10.0 kilogram mass in the diagram to the right what is the tension in the cable so and once again the tension everywhere is the same in a cable so it's going to be the same on both sides and also these things are attached by that cable and so the question is asking for the acceleration of the 10 kilogram Mass but again they're attached their motion is coupled their acceleration is therefore going to be the same as well now in this case we can kind of deduce here well the 10 kilogram mass is heavier so it's going to win it's going to pull down this way overcoming the weight of the five kilogram Mass which is going to pull down the opposite way and so we know that for the 10 kilogram Mass the acceleration is going to be downward for the 5 kilogram Mass it's going to be upward but the magnitude is going to be the same their motions are coupled together all right now we've got two bodies and this is uh this problem is what we call a two body problem and you've got to set up a free body diagram for both and then kind of net force equations or some of the force equations for both as well now this one's not going to be so bad we'll see it's a little more challenging the next one but this one still has a key place where students make an error we're going to avoid that error all right so let's start with the 10 kilogram one here so it has got its own weight here which I'll call mg and then it's got a tension force pointing up so as well and if we do the sum of the force equation here so in this case again we said this object is going to be accelerating downward the forces aren't balanced and so as a result Newton's second law is going to apply and the sum of the forces is going to equal ma now I'm going to stop there for just a second and set up the other one and we'll come back to this all right so for the five kilogram pass so also has its own weight pointing downward mg so and it's got a tension pointing upward as well now we got a problem here the two tensions are indeed the same it's the same cable and the tension of that cable is everywhere the same but the MGs are not these are different masses and so maybe we've more properly label these as like M1 here and M 2 here and then this becomes M2 a as we'll see and we'll do the same thing set up the sum of the forces for this one some of the forces is going to equal M1 times a this time uh and now we got to worry about one other major problem so in the first example we know this thing is accelerating upwards well we generally Define accelerate you know forces and accelerations that are upwards as being positive so which is going to be the case here and so the tension points up will make that positive the weight points down we'll make that negative but the overall force and acceleration are pointing up as well so this is going to be a positive term as well and that'll be important not so much here but it'll be very important that we account for something in that regard over here so but we set up the sum of the force equation we're going to have T minus m1g equals M1 a okay now we could say this is some of the forces in the y direction we could but we're not worrying about any forces in the X Direction there are none and so there's only one set of forces one equation for this particular body so for the other one we've got the same thing so tension points up we'll make that positive so its own weight points down and make it negative but the big thing here is this is we know the overall force and acceleration is going to be downward and we've got to account for that here in the sum of the forces equation the net force equation got to put a big fat negative sign there if you miss that that is the number one mistake students make on such a two body pulley problem so and again it's just how you've defined things if you're defining forces pointing up as positive and forces pointing down as negative then the overall net force you've got to account for it there as well and so here we're going to have tension minus m2g equals negative m2a so and it turns out we can't solve for anything in either one of these equations it's two equations with two unknowns so let's put some of the variables in there I'm going to leave off the kilograms and I'm going to round this I'll leave off some of the sig figs as well just make this 5 and 10. so just to make the equations look a little simpler keeping the units in can get a little bit tedious and complicated here so but we're going to have T minus M1 again is 5. so 5G equals 5A and over here we're going to have T minus 10 g equals negative 10 a so a couple different ways we could solve this we could try and set this up with some form of linear combination stuff like that but my preferred way is probably just to solve this by substitution so solve for one of the variables and substitute into the other equation so in this case that's exactly what I'm going to do and here I'm going to make this sulfur tension here and here adding 5G to the other side is going to be 5A plus 5G so and then I'm going to take and substitute that right in here so for this tension right here we're going to have 5G plus 5a that's tension there and then minus the 10 G here equals negative 10 a and again the accelerations are the same it's the same a in both cases so if we rearrange this just a little bit now so 5G minus 10 G is going to be negative 5G and then negative 10A minus another 5A is negative 15a and so solving for this we're going to get a is equal to negative 5 over negative 15 G or one third of gravity so we'll have 9.8 divided by 3 and we're going to get 3.2666666 and in this case we want two sig figs based on the 5.0 kilograms having the few figs so we'll round that down to three point Sorry round that up to 3.3 meters per second squared foreign answered we've solved for the acceleration and now we just need to go back to solve for the tension either substitute that back end to this expression or this expression and it doesn't really matter which one and the truth is we're probably going to be going all the way down to either here or here let's say and I'm going to substitute back in this one this is what I want it's already solved for tension here for me if it wasn't rearranged and plug it back in here but it's already rearrange like that for me so life is good so we're going to have t equals five then we'll plug in that acceleration of 3.3 meters per second squared and again plus 5 times 9.8 meters per second squared all right so five times three point three plus five times nine point eight it's going to get a 65.5 Newtons which we're going to round up to 60 actually let's let's keep this back if I didn't round early I could do five times three point two six six six six six six six six plus five times nine point eight and I really should save the rounding for the end that's going to be 65.333 which to two sig figs would round down to 65 Newtons cool and that's the second of the two questions so this is pretty common on a two body pulley problem like this where you end up solving a system of equations uh which throws students a little bit but also getting that negative sign when the net force points downward for one of the objects super duper crucial one more problem all right so the last pulley problem here so Two Body Pulley Practice Problem #2 here's our setup what is the acceleration of the 10.0 kilogram mass in the diagram to the right if the 5.0 kilogram mass is sliding on a frictionless Surface so no friction here which is going to make this a little bit easier to treat and stuff like this so but notice the same mass is here five kilograms 10 kilograms five kilograms 10 kilograms a little bit of a difference here though so if you notice so the 10 kilogram Mass was wanting to make the pulley so in its weight was wanting to make the pulley rotate clockwise whereas the five kilogram Mass pulling downward was when I make it rotate counterclockwise well the 10 kilogram mass is pulling with more Force overall weight and so it won well if we take a look at this one this one's a little bit different so so the 10 kilogram Mass still has a weight pulling downward m2g that's wanting that pulley to rotate around clockwise so and again there's still a tension pulling upward as well but if we take a look at this guy here so things are a little bit different in this case so his weight points downward here m1g the tension pulls to the right in this case so in this case if we want to rotate this wheel back counterclockwise we'd have to pull back to the left not down and so in this case there's no actually pulling you know Force pulling back to the left if there was friction that would pull back to the left and always opposes the motion but we can see that this 10 kilogram mass is going to accelerate downward that's going to pull the 5 kilogram Mass to accelerate to the right so but there's nothing pulling back to the left it's with a little bit different setup here so in this case we got the 10 kilogram mask causing the pulley to rotate clockwise by pulling down but this one's resisting it trying to pull back the opposite direction the acceleration that up at 3.3 meters per second squared here the 10 kilogram mass is still pulling downward at the exact same force due to its weight but now there's nothing pulling back if you will it's still got a total weight of 15 kilograms to be moving around and stuff but there's nothing pulling back and we should anticipate that whatever the acceleration is here and it should be again the same value for both of these they're coupled motion here but it should be higher than 3.3 meters per second squared since there is nothing pulling back let's see how this works we've got one more Force to draw in here and that is the normal force here let's see how this problem works out now in this case one the acceleration again of that 10 kilogram block so but again whether you you look at the acceleration of 10 kilogram block which is downwards or the acceleration of the five kilogram block which is to the right magnitude wise it's going to be the exact same acceleration okay so if we set up the sum of the forces for the 10 kilogram block first it should look pretty familiar sum of the forces equals m 2 a but we've got to remember that the overall net force here points down so we got to put that negative sign in there as well in this case that's going to come out to T minus m2g equals negative m2a now for the five kilogram we've got forces in two different dimensions there's technically two sets of equations to set up one for the X one for the Y now the one in which the acceleration exists is in the X Direction here so maybe we should specify here that this is the sum of the forces in the y direction this is acceleration in the y direction so whereas for this one we actually want to set it up for the X Direction preferentially now some of the forces in the X Direction here is going to equal M1A now it's customary for forces pointing to the right to be positive and this is definitely going to be acceleration net to the right so that's positive and the tension is positive as well that's the only force in this case and so it's just T equals M1A and they're both positive terms okay not so bad now if we set this up for the y direction and we technically don't need to to solve this problem it turns out but if we did it would look something like this some of the forces in the y direction is going to add up to zero in the y direction for this five kilogram Mass there's no movement and therefore definitely no net force or acceleration and so we're just simply going to have normal force which is positive pointing up minus its weight which is negative pointing down equals zero and this would allow us to solve for the normal force if we needed to and because there's no friction we actually technically don't need to know the normal force we don't even need this setup at all it's available if we needed it and if there were friction we would all right so now we've got a system of two equations both involving tension and acceleration so but again this is the acceleration in the X Dimension this is in the Y and that seems a little bit odd but again remember that they're coupled and the acceleration downward for the 10 kilogram mass and acceleration to the right for the five kilogram mass in magnitude are exactly the same and so we can take those labels off because it really is just one common acceleration as far as magnitude anyways so the tensions are the same it's just one cable so the tension here and the tension here are exactly the same and so it is the same set of variables in both cases as far as tension and acceleration go now if we want to solve a system of two equations again again I like substitution preferentially so especially if one of the variables is always already solved for so tension here is equal to M1A and if we include some numbers here that's going to equal tension equal to 5 kilograms times a and I'm going to leave off the kilograms and some of the sig figs to make this look a little bit easier so this equation over here becomes T minus 10 g equals negative 10 a and because this one's solved for tension we can take this 5A right here and substitute in right here for tension to solve for a and so in this case we're going to get 5A substituted infotension minus 10 g equals negative 10 a and again if you get you miss that negative sign you're definitely going to get this one wrong so we'll add 10A over to the other side here and 10A plus 5a is 15a so move add the negative 10g add 10g to both sides that's going to equal 10 G and we'll find out that here a is going to equal 10 over 15 times gravity or 2 3 times gravity and we'll let our calculator do the work for us so 9.8 times 10 over 15 or times 2 over 3 if you will is going to be 6.53 which will round down to 6.5 meters per second squared into sig figs cool and our intuition proves correct here we said that because this Mass technically is not pulling back to try and make that pulley wheel rotate counterclockwise at all that we should expect the overall acceleration in this case to be higher than it was back in this case and lo and behold that's true 6.5 meters per second squared compared to 3.3 meters per second squared cool if you found this lesson helpful consider giving it a like happy studying
17171	https://classicswrites.hsites.harvard.edu/close-reading-0	Skip to main content arrow_circle_down ClassicsWrites A guide to research and writing in the field of Classics ClassicsWrites A guide to research and writing in the field of Classics Close Reading Close Reading as Analysis Close reading is the technique of carefully analyzing a passage’s language, content, structure, and patterns in order to understand what a passage means, what it suggests, and how it connects to the larger work. A close reading delves into what a passage means beyond a superficial level, then links what that passage suggests outward to its broader context. One goal of close reading is to help readers to see facets of the text that they may not have noticed before. To this end, close reading entails “reading out of” a text rather than “reading into” it. Let the text lead, and listen to it. The goal of close reading is to notice, describe, and interpret details of the text that are already there, rather than to impose your own point of view. As a general rule of thumb, every claim you make should be directly supported by evidence in the text. As the name suggests this technique is best applied to a specific passage or passages rather than a longer piece, almost like a case study. Use close reading to learn: what the passage says what the passage implies how the passage connects to its context Why Close Reading? Close reading is a fundamental skill for the analysis of any sort of text or discourse, whether it is literary, political, or commercial. It enables you to analyze how a text functions, and it helps you to understand a text’s explicit and implicit goals. The structure, vocabulary, language, imagery, and metaphors used in a text are all crucial to the way it achieves its purpose, and they are therefore all targets for close reading. Practicing close reading will train you to be an intelligent and critical reader of all kinds of writing, from political speeches to television advertisements and from popular novels to classic works of literature. Wondering how to do a close reading? Click on our Where to Begin section to find out more!
17172	http://videos.mathtutordvd.com/detail/videos/trig-precalculus---vol-2---trig-identities/video/tvpOM17HhAA/lesson-1---basic-trig-identities-involving-sin-cos-and-tan	Skip to collection list Skip to video grid [Home] [Shop Courses] [Streaming Membership] Skip to collection list Skip to video grid Trig / PreCalculus - Vol 2 - Trig Identities Lesson 1 - Basic Trig Identities Involving Sin, Cos, and Tan This is just a few minutes of a complete course. Get full lessons & more subjects at: Basic trig identities are the core trig identities that involve sine, cosine, tangent, cotangent, secant, and cosecant. We discuss each of these important trigonometric functions and gain practice solving problems. We will first discuss the sin (sine) function, the cos (cosine) function and the tan (tangent) function. Next, we will review an easy way to remember the trig identities that involve these trig functions. We will solve several problems that involve these basic identities to give practice.. 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Fundamental Trig Identities, Part 2 02 - Fundamental Trig Identities, Part 2 0:00 03 - Pythagorean Trig Identities, Part 1 03 - Pythagorean Trig Identities, Part 1 0:00 Lesson 4 - Pythagorean Trig Identities, Part 2 (Trig & PreCalculus) Lesson 4 - Pythagorean Trig Identities, Part 2 (Trig & PreCalculus) 0:00 Lesson 5 - Pythagorean Trig Identities, Part 3 (Trig & PreCalculus) Lesson 5 - Pythagorean Trig Identities, Part 3 (Trig & PreCalculus) 0:00 Lesson 6 - Solving Trig Equations, Part 1 (Trig & PreCalculus) Lesson 6 - Solving Trig Equations, Part 1 (Trig & PreCalculus) 0:00 Lesson 7 - Solving Trig Equations, Part 2 (Trig & PreCalculus) Lesson 7 - Solving Trig Equations, Part 2 (Trig & PreCalculus) 0:00 Lesson 8 - Cofunction And Even Odd Identities (Trig & PreCalculus) Lesson 8 - Cofunction And Even Odd Identities (Trig & PreCalculus) 0:00 Lesson 9 - Addition And Subtraction Identities, Part 1 (Trig & PreCalculus) Lesson 9 - Addition And Subtraction Identities, Part 1 (Trig & PreCalculus) 0:00 Lesson 10 - Addition And Subtraction Identities, Part 2 (Trig & PreCalculus) 0:00 Lesson 11 - Double Angle Identities (Trig & PreCalculus) 0:00 Lesson 12 - Squaring Identities (Trig & PreCalculus) 0:00 Lesson 13 - Half Angle Identities (Trig & PreCalculus) 0:00 Lesson 14 - Product To Sum Identities (Trig & PreCalculus) 0:00 Lesson 15 - Sum To Product Identities (Trig & PreCalculus) Currently loaded videos are 1 through 15 of 19 total videos. 1-15 of 19 First page loaded, no previous page available Load Next Page
17173	https://fiveable.me/ap-calc/unit-8/finding-average-value-function-on-an-interval/study-guide/HjiYTRAnQdY0eCQpqtpg	Finding the Average Value of a Function on an Interval - AP Calc Study Guide \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade ♾️AP Calculus AB/BC Unit 8 Review 8.1 Finding the Average Value of a Function on an Interval All Study Guides AP Calculus AB/BC Unit 8 – Applications of Integration Topic: 8.1 ♾️AP Calculus AB/BC Unit 8 Review 8.1 Finding the Average Value of a Function on an Interval Written by the Fiveable Content Team • Last updated September 2025 Verified for the 2026 exam Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025 print study guide copy citation APA ♾️AP Calculus AB/BC Unit & Topic Study Guides AP Calculus AB/BC Exams Unit 1 – Limits and Continuity Unit 2 – Fundamentals of Differentiation Unit 3 – Composite, Implicit, and Inverse Functions Unit 4 – Contextual Applications of Differentiation Unit 5 – Analytical Applications of Differentiation Unit 6 – Integration and Accumulation of Change Unit 7 – Differential Equations Unit 8 – Applications of Integration Unit 8 Overview: Applications of Integration 8.1 Finding the Average Value of a Function on an Interval 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts 8.4 Finding the Area Between Curves Expressed as Functions of x 8.5 Finding the Area Between Curves Expressed as Functions of y 8.6 Finding the Area Between Curves That Intersect at More Than Two Points 8.7 Volumes with Cross Sections: Squares and Rectangles 8.8 Volumes with Cross Sections: Triangles and Semicircles 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis 8.10 Volume with Disc Method: Revolving Around Other Axes 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis 8.12 Volume with Washer Method: Revolving Around Other Axes 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled Unit 9 – Parametric Equations, Polar Coordinates, and Vector–Valued Functions (BC Only) Unit 10 – Infinite Sequences and Series (BC Only) Frequently Asked Questions Previous Exam Prep Study Tools Exam Skills AP Cram Sessions 2021 Live Cram Sessions 2020 practice questions print guide report error 8.1 Finding the Average Value of a Function on an Interval Welcome back to AP Calculus with Fiveable! This topic focuses on finding the average value of a continuous function using definite integrals. more resources to help you study practice questionscheatsheetscore calculator 🔢 Average Value of a Function The average value of a function will allow us to solve problems that involve the accumulation of change over an interval, which will later be used to understand more difficult topics of integration. For questions that require the average value of a function, we are never given a finite number of data points. Therefore, we must use integration to determine what the average value is. This idea is fairly simple once you memorize a key piece of information: if f is continuous on [a,b][a,b][a,b], then the average value of f on [a,b][a,b][a,b] is the following. Average Value=1 b−a∫a b f(x)d x\text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx Average Value=b−a 1∫a bf(x)d x Image of average value equation and corresponding graph. Image courtesy of ExamSolutions 🔍Average Value of a Function Steps Here are some steps to help break down the formula! Set up the integral so that you integrate f(x)f(x)f(x) from a a a to b b b with respect to x x x, which will calculate the area under the curve between these two limits. Then place the fraction in front of the integral, which is simply the reciprocal of the difference between a a a and b b b. Evaluating this expression allows you to get the average y-value of this function between [a,b][a,b][a,b] ✏️ Average Value of a Function Walkthrough If the formula still seems a little difficult to understand due to its notation, practice questions are the best way to better understand its use! Consider the function f(x)=2 x 2−3 x+5 f(x) = 2x^2-3x+5 f(x)=2 x 2−3 x+5 on the interval [1,4][1,4][1,4]. Find the average value of this function on the interval. In this case, a=1 a = 1 a=1 and b=4 b = 4 b=4. So we begin by subbing the 1 and 4 into both the denominator of the fraction in front of the integral and the limits of the integral. Average Value=1 4−1∫1 4(2 x 2−3 x+5)d x\text{Average Value} = \frac{1}{4-1} \int_{1}^{4} (2x^2 - 3x + 5) \, dx Average Value=4−1 1∫1 4(2 x 2−3 x+5)d x Next, take the integral of f(x)f(x)f(x). =1 3[2 3 x 3−3 2 x 2+5 x]1 4= \frac{1}{3} \left[ \frac{2}{3}x^3 - \frac{3}{2}x^2 + 5x \right]_{1}^{4} =3 1[3 2x 3−2 3x 2+5 x]1 4 Finally, we can sub in the limits and evaluate. 1 3[(2 3(4)3−3 2(4)2+5(4))−(2 3(1)3−3 2(1)2+5(1))]=23 2 \frac{1}{3} \left[ \left( \frac{2}{3}(4)^3 - \frac{3}{2}(4)^2 + 5(4) \right) - \left( \frac{2}{3}(1)^3 - \frac{3}{2}(1)^2 + 5(1) \right) \right] =\boxed{\frac{23}{2}} 3 1[(3 2(4)3−2 3(4)2+5(4))−(3 2(1)3−2 3(1)2+5(1))]=2 23 Time for you to practice some questions yourself! ⬇️ 📝 Average Value of a Function Practice Problems Give each of these problems a try before you move onto the solutions! What is the average value of 5 x 2+4 5x^2+4 5 x 2+4 on the interval 0≤x≤6 0\le x\le 6 0≤x≤6? What is the average value of x 3−x 2 x^3-x^2 x 3−x 2 on the interval 2≤x≤5 2\le x\le 5 2≤x≤5? What is the average value of s i n(x)+c o s(x)sin(x)+cos(x)s in(x)+cos(x) on the interval 0≤x≤π 0\le x\le \pi 0≤x≤π? Average Value of a Function Question Solutions Question 1 Solution Average Value=1 6−0∫0 6(5 x 2+4)d x\text{Average Value} = \frac{1}{6-0} \int_{0}^{6} (5x^2+4) \, dx Average Value=6−0 1∫0 6(5 x 2+4)d x=1 6[5 3 x 3+4 x]0 6= \frac{1}{6} \left[ \frac{5}{3}x^3 +4x \right]_{0}^{6} =6 1[3 5x 3+4 x]0 61 6[(5 3(6)3+4(6))−(5 3(0)3+4(0))]=64 \frac{1}{6} \left[ \left( \frac{5}{3}(6)^3 + 4(6) \right) - \left( \frac{5}{3}(0)^3 + 4(0) \right) \right] =\boxed{64} 6 1[(3 5(6)3+4(6))−(3 5(0)3+4(0))]=64 Question 2 Solution Average Value=1 5−2∫2 5(x 3−x 2)d x\text{Average Value} = \frac{1}{5-2} \int_{2}^{5} (x^3-x^2) \, dx Average Value=5−2 1∫2 5(x 3−x 2)d x=1 3[1 4 x 4−1 3 x 3]2 5= \frac{1}{3} \left[ \frac{1}{4}x^4 - \frac{1}{3} x^3 \right]_{2}^{5} =3 1[4 1x 4−3 1x 3]2 51 3[(1 4(5)4−1 3(5)3)−(1 4(2)4−1 3(2)3)]=151 4 \frac{1}{3} \left[ \left( \frac{1}{4}(5)^4 - \frac{1}{3}(5)^3 \right) - \left( \frac{1}{4}(2)^4 - \frac{1}{3}(2)^3 \right) \right] =\boxed{\frac{151}{4}} 3 1[(4 1(5)4−3 1(5)3)−(4 1(2)4−3 1(2)3)]=4 151 Question 3 Solution Average Value=1 π−0∫0 π s i n(x)+c o s(x)d x\text{Average Value} = \frac{1}{\pi-0} \int_{0}^{\pi} sin(x)+cos(x) \, dx Average Value=π−0 1∫0 πs in(x)+cos(x)d x=1 π[−c o s(x)+s i n(x)]0 π= \frac{1}{\pi} \left[ -cos(x)+sin(x)\right]_{0}^{\pi} =π 1[−cos(x)+s in(x)]0 π1 π[(−c o s(π)+s i n(π))−(−c o s(0)+s i n(0))]=2 π \frac{1}{\pi} \left[ \left( -cos(\pi) + sin(\pi) \right) - \left( -cos(0) + sin(0) \right) \right] =\boxed{\frac{2}{\pi}} π 1[(−cos(π)+s in(π))−(−cos(0)+s in(0))]=π 2 ⭐ Closing Great job! This topic often shows up as part (a) of FRQs, so keep this in mind for the AP. Frequently Asked Questions How do I find the average value of a function on an interval? The average value of a continuous function f on [a,b] is given by the CED formula: (1/(b−a)) ∫_a^b f(x) dx. Practically: 1) find an antiderivative F of f, 2) evaluate the definite integral ∫_a^b f(x) dx = F(b) − F(a) (FTC), and 3) divide that result by (b−a). This is the “area under the curve” normalized by interval length and ties to the Mean Value Theorem for integrals (there exists c in [a,b] with f(c) equal to this average). Example: average of f(x)=x^2 on [0,2] = (1/2)∫_0^2 x^2 dx = (1/2)·(8/3)=4/3. On the AP exam you should show the integral work and units when context is given. For a quick topic review, see the Fiveable study guide ( and try practice problems ( What's the formula for average value of a function using integrals? If f is continuous on [a, b], the average value of f on that interval is f_avg = (1 / (b − a)) ∫_a^b f(x) dx. This comes from interpreting the integral as total “accumulated” value (area under the curve) and dividing by the interval length to get a mean. The Mean Value Theorem for Integrals guarantees a c in [a, b] with f(c) = f_avg when f is continuous. For AP Calc (CED LO CHA-4.B) you should be able to set up and evaluate this definite integral using an antiderivative (FTC) or approximate it with Riemann sums when needed. If you want a quick review of this Topic 8.1 idea, check the study guide ( For more unit review and lots of practice problems, see the Unit 8 overview ( and practice bank ( When do I use the average value formula vs just finding the mean of points? Use the definite-integral formula whenever you want the true average of a continuous function f on an interval [a,b]: average value = (1/(b−a)) ∫_a^b f(x) dx (CED CHA-4.B.1). That formula comes from taking the limit of Riemann sums (sample values×subinterval width) and gives the exact “height” whose area equals the area under f over [a,b]. Use the simple mean of points only when you have discrete data or a small number of samples (a table, measurements, or when the problem explicitly asks for an average of given values). The sample mean approximates the integral average if the points are midpoints or form a Riemann sum, but it can be off if samples are sparse or unevenly spaced. On the AP exam you’ll be expected to set up and evaluate (or interpret) (1/(b−a))∫_a^b f(x) dx for continuous functions (Topic 8.1). For a quick review see the Topic 8.1 study guide ( and more practice at ( Can someone explain step by step how to calculate average value of f(x) = x² on [0,3]? The average value of a continuous function f on [a,b] is (1/(b−a)) ∫_a^b f(x) dx (CED CHA-4.B). For f(x)=x^2 on [0,3]: 1) Set up the formula: Average = (1/(3−0)) ∫_0^3 x^2 dx = (1/3) ∫_0^3 x^2 dx. 2) Find an antiderivative: ∫ x^2 dx = x^3/3. 3) Evaluate the definite integral: ∫_0^3 x^2 dx = [x^3/3]_0^3 = (3^3/3) − (0) = 27/3 = 9. 4) Divide by interval length: Average = (1/3)·9 = 3. So the average value of x^2 on [0,3] is 3. This is exactly the application of the Mean Value Theorem for integrals and the Fundamental Theorem of Calculus (antiderivative + definite integral). For more review and practice on Topic 8.1, see the Fiveable study guide ( and Unit 8 overview ( For extra problems, try the AP Calc practice bank ( I don't understand why we divide by (b-a) in the average value formula - what does that represent? Think of average value like the usual average: add up all the values and divide by how many. For a continuous function on [a,b] you “add up” the values by integrating f(x)—that gives total accumulated value (area under f). But the integral alone grows with the length of the interval: a big interval makes a bigger integral even if f stays the same. Dividing by (b−a) normalizes the total by the interval length, so you get the mean height per unit x—the same idea as sum of numbers divided by count. You can see this from Riemann sums: (1/(b−a))∫_a^b f(x) dx = limit of (1/n)·Σ f(x_k) (the usual average of sample values). Geometrically, the average value is the height of a rectangle on [a,b] whose area equals ∫_a^b f(x) dx. The Mean Value Theorem for Integrals guarantees some c in [a,b] where f(c) equals that average. For more AP-aligned review and examples, check the Topic 8.1 study guide ( Practice problems are helpful too ( How do I solve average value problems on the AP Calc exam? Use the average-value formula from the CED: if f is continuous on [a,b], average value = (1/(b−a)) ∫_a^b f(x) dx. On the exam do these steps: 1. Write the formula: f_avg = (1/(b−a)) ∫_a^b f(x) dx. 2. Find an antiderivative F(x) of f(x) (Fundamental Theorem of Calculus). 3. Evaluate the definite integral: ∫_a^b f(x) dx = F(b) − F(a). 4. Divide by (b−a). Simplify or give a decimal if asked to approximate (show work). 5. If the problem gives a graph or table, use geometry or a Riemann/midpoint/Trapezoid sum as the CED allows; state units (e.g., “vehicles/hour”) and interpret the result. On AP free-response, show the integral, the antiderivative, and the division by (b−a); if the prompt asks interpretation, say what the number means in context. For a quick study refresher and examples, check the Topic 8.1 study guide ( and try practice problems ( What's the difference between average value of a function and average rate of change? Average value of a function and average rate of change are related but different ideas. - Average value (CED CHA-4.B.1): for a continuous f on [a,b], avg value = (1/(b−a)) ∫_a^b f(x) dx. It’s the “height” you’d get if f were flattened so its area over [a,b] stayed the same (Mean Value Theorem for Integrals). This is what Topic 8.1 asks you to compute. - Average rate of change: it’s the slope of the secant line from a to b, i.e. (f(b) − f(a))/(b − a). It measures how f changes per unit x over the interval (a velocity-like quantity), not an area-average. Quick contrast: average value uses an integral (area under the curve) and gives a typical y-value; average rate of change uses endpoint values and gives a typical slope. Both appear on the AP: average value is an Applications of Integration skill in Unit 8; secant slopes and interpretation show up in Units on change and functions. For more review/examples see the Topic 8.1 study guide ( and extra practice ( Do I need to use the fundamental theorem of calculus when finding average value? Short answer: usually yes—but not always. Why: the average value on [a,b] is (1/(b−a)) ∫_a^b f(x) dx (CED CHA-4.B.1). To get an exact value you normally evaluate that definite integral. By the Fundamental Theorem of Calculus you do that by finding an antiderivative F and computing F(b) − F(a). So for most textbook/AP problems you’ll use the FTC. When you don’t need the FTC: if the integral is given numerically (a table or graph) you can use Riemann sums or midpoint/trapezoid approximations instead, or the problem might ask for a numerical approximation (common on calculator and non-calculator parts). Also the Mean Value Theorem for Integrals tells you there exists c with f(c) equal to the average value, but it doesn’t replace evaluating the integral. For AP practice and examples see the Topic 8.1 study guide ( and more problems at ( Why is the average value formula (1/(b-a)) times the integral from a to b? Think of average value like the usual average of numbers: add up all values then divide by how many. For a continuous function on [a,b] you “add up” values by integrating (a Riemann sum gives total area under f). So ∫_a^b f(x) dx is the accumulated sum of f(x) over the interval. To get the average you divide that total by the interval’s length (b−a). That gives (1/(b−a))∫_a^b f(x) dx. Two helpful formulations: - Riemann-sum view: partition [a,b], average ≈ (1/n) Σ f(x_k) → (1/(b−a))∫_a^b f as n→∞. - The Mean Value Theorem for Integrals: for continuous f there exists c in [a,b] with f(c) = (1/(b−a))∫_a^b f(x) dx—so the integral divided by (b−a) equals some actual function value on the interval. This is exactly the AP Topic 8.1 idea (CHA-4.B). For extra practice and examples see the Topic 8.1 study guide ( and lots of problems at ( How do I find the average value when the function has negative values? Use the same formula: average value = (1/(b−a)) ∫_a^b f(x) dx (CED CHA-4.B.1). The integral gives signed area: regions where f<0 count negative, so the average can be negative, zero, or positive depending on the net signed area. Example: if f(x)=−2 on [0,4], ∫_0^4 f = −8 so the average = (1/4)(−8)=−2. If f crosses above and below the x-axis, positive and negative areas can cancel—the average might be 0 even though f isn’t zero anywhere. If you want the average “height” ignoring sign (how large the values are regardless of sign), use (1/(b−a)) ∫_a^b \|f(x)\| dx instead. Remember the Mean Value Theorem for integrals: for continuous f there’s some c with f(c) equal to the average value (may be negative). For a quick topic review see the Fiveable study guide ( and more practice at ( I'm confused about when to use average value vs mean value theorem - are they the same thing? They’re related but not the same. Average value of f on [a,b] is a number: (1/(b−a)) ∫_a^b f(x) dx. It’s the “height” of a rectangle with same area as the area under f on [a,b] (CED CHA-4.B.1). You compute it whenever you need the mean value of the function over an interval. The Mean Value Theorem (MVT) for derivatives is a different statement: if f is continuous on [a,b] and differentiable on (a,b), there exists c in (a,b) with f′(c) = (f(b)−f(a))/(b−a). The Mean Value Theorem for integrals (sometimes called the Average Value Theorem) connects the two: if f is continuous on [a,b], there exists c with f(c) = (1/(b−a)) ∫_a^b f(x) dx—so the average value is actually attained at some point. So: use the average value formula to compute the mean. The MVT (derivative) is about slopes; the Mean Value Theorem for integrals guarantees the average value is achieved. For AP practice and examples, see the Topic 8.1 study guide ( and more Unit 8 resources ( tons of practice problems are at ( What happens if I mess up the bounds when calculating average value of a function? If you pick the wrong interval, your average value will be wrong because the formula (CED CHA-4.B.1) is average = (1/(b−a)) ∫_a^b f(x) dx—the integral gives total “area” (accumulation) over [a,b] and you must divide by the interval length b−a to normalize. Two important cases: - Swapped endpoints only: if you accidentally swap a and b (compute ∫_b^a f), the integral changes sign but so does (b−a), so the ratio (the average) ends up the same. So swapping endpoints alone doesn’t change the average value. - Wrong interval altogether: if you use the wrong a or b (e.g., use [0,1] instead of [0,2]), you integrate over the wrong region and divide by the wrong length—the average will generally be different. Example: f(x)=x on [0,2] gives avg = (1/2)∫_0^2 x dx = 1; using [0,1] gives (1/1)∫_0^1 x dx = 0.5. Quick checks: verify units (average should match units of f), sanity-check with midpoint/graph, and double-check your bounds against the problem statement. For more practice and review aligned to the AP CED, see the Topic 8.1 study guide ( and more practice problems ( Can I use my calculator to find the average value or do I have to integrate by hand? Short answer: yes—when the exam or your class allows a calculator you can use it to evaluate the definite integral in the average-value formula; when calculators aren’t allowed you must do it without one. Why: the average value of a continuous f on [a,b] is (1/(b−a)) ∫_a^b f(x) dx (CED CHA-4.B.1). If you’re on a calculator-permitted part of the AP exam (multiple-choice Part B or free-response Part A) you may use a graphing calculator to compute the integral numerically or symbolically. On calculator-not-permitted parts (MC Part A and some FR questions) you need to evaluate the integral by hand—use antiderivatives (FTC), geometry, symmetry, or algebraic manipulation. Practical tip: try the antiderivative first—it’s exact and aligns with the Fundamental Theorem of Calculus and the Mean Value Theorem for integrals. For more examples and practice (both calculator and no-calculator), see the Topic 8.1 study guide ( and extra problems ( How do I set up word problems that ask for average temperature or average velocity using integrals? Use the average-value formula: average of a continuous f on [a,b] is (1/(b−a)) ∫_a^b f(x) dx. For word problems, identify the variable and the interval, plug into that formula, and watch units. - Average temperature: if T(t) (°F) gives temperature at time t (hours) from t = a to b, the average temperature is (1/(b−a)) ∫_a^b T(t) dt. Example: average from 2pm to 8pm (t=14 to 20) = (1/6) ∫_14^20 T(t) dt (result in °F). - Average velocity or mean speed: if v(t) (m/s) is velocity on [a,b], the average velocity = (1/(b−a)) ∫_a^b v(t) dt (units m/s). Note: if you want average speed (distance/time) when v can be negative, use (1/(b−a)) ∫_a^b \|v(t)\| dt. If you need total displacement, use ∫_a^b v(t) dt (no normalization). On the AP exam, show the integral setup, evaluate using an antiderivative (FTC), and give units—this matches CHA-4.B from the CED. For extra practice and examples, see the Topic 8.1 study guide (Fiveable) here: ( and more practice problems at ( Is there a shortcut for finding average value of simple functions like polynomials? Short answer: not really a magic shortcut for all polynomials—you still use the average-value formula from the CED, A_avg = (1/(b−a))∫_a^b f(x) dx, and compute the integral with the power rule (antiderivative then F(b)−F(a)). That’s exactly CHA-4.B.1: integrate, divide by interval length. Helpful shortcuts you can use sometimes: - Constant f(x)=k → average = k. - Linear f(x)=mx+c → average on [a,b] = f((a+b)/2) (value at the midpoint). - Odd functions on symmetric interval [−L,L] → average = 0. - Even/symmetric properties can cut work when they apply. For polynomials just apply the power rule term-by-term: ∫ x^n dx = x^(n+1)/(n+1), evaluate F(b)−F(a), then divide by (b−a). That’s fast and AP-aligned. 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17174	https://en.wikipedia.org/wiki/Circumconic_and_inconic	Jump to content Search Contents (Top) 1 Centers and tangent lines 1.1 Circumconic 1.2 Inconic 2 Other features 2.1 Circumconic 2.2 Inconic 3 Extension to quadrilaterals 4 Examples 5 References 6 External links Circumconic and inconic Español Français 日本語 Русский தமிழ் Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Conic section that passes through the vertices of a triangle or is tangent to its sides In Euclidean geometry, a circumconic is a conic section that passes through the three vertices of a triangle, and an inconic is a conic section inscribed in the sides, possibly extended, of a triangle. Suppose A, B, C are distinct non-collinear points, and let △ABC denote the triangle whose vertices are A, B, C. Following common practice, A denotes not only the vertex but also the angle ∠BAC at vertex A, and similarly for B and C as angles in △ABC. Let the sidelengths of △ABC. In trilinear coordinates, the general circumconic is the locus of a variable point satisfying an equation for some point u : v : w. The isogonal conjugate of each point X on the circumconic, other than A, B, C, is a point on the line This line meets the circumcircle of △ABC in 0,1, or 2 points according as the circumconic is an ellipse, parabola, or hyperbola. The general inconic is tangent to the three sidelines of △ABC and is given by the equation Centers and tangent lines [edit] Circumconic [edit] The center of the general circumconic is the point The lines tangent to the general circumconic at the vertices A, B, C are, respectively, Inconic [edit] The center of the general inconic is the point The lines tangent to the general inconic are the sidelines of △ABC, given by the equations x = 0, y = 0, z = 0. Other features [edit] Circumconic [edit] Each noncircular circumconic meets the circumcircle of △ABC in a point other than A, B, C, often called the fourth point of intersection, given by trilinear coordinates If is a point on the general circumconic, then the line tangent to the conic at P is given by The general circumconic reduces to a parabola if and only if : and to a rectangular hyperbola if and only if Of all triangles inscribed in a given ellipse, the centroid of the one with greatest area coincides with the center of the ellipse.: p.147 The given ellipse, going through this triangle's three vertices and centered at the triangle's centroid, is called the triangle's Steiner circumellipse. Inconic [edit] The general inconic reduces to a parabola if and only if : in which case it is tangent externally to one of the sides of the triangle and is tangent to the extensions of the other two sides. Suppose that ⁠⁠ and ⁠⁠ are distinct points, and let : As the parameter t ranges through the real numbers, the locus of X is a line. Define : The locus of X2 is the inconic, necessarily an ellipse, given by the equation : where A point in the interior of a triangle is the center of an inellipse of the triangle if and only if the point lies in the interior of the triangle whose vertices lie at the midpoints of the original triangle's sides.: p.139 For a given point inside that medial triangle, the inellipse with its center at that point is unique.: p.142 The inellipse with the largest area is the Steiner inellipse, also called the midpoint inellipse, with its center at the triangle's centroid.: p.145 In general, the ratio of the inellipse's area to the triangle's area, in terms of the unit-sum barycentric coordinates (α, β, γ) of the inellipse's center, is: p.143 : which is maximized by the centroid's barycentric coordinates α = β = γ = ⅓. The lines connecting the tangency points of any inellipse of a triangle with the opposite vertices of the triangle are concurrent.: p.148 Extension to quadrilaterals [edit] All the centers of inellipses of a given quadrilateral fall on the line segment connecting the midpoints of the diagonals of the quadrilateral.: p.136 Examples [edit] Circumconics Circumcircle, the unique circle that passes through a triangle's three vertices Steiner circumellipse, the unique ellipse that passes through a triangle's three vertices and is centered at the triangle's centroid Kiepert hyperbola, the unique conic which passes through a triangle's three vertices, its centroid, and its orthocenter Jeřábek hyperbola, a rectangular hyperbola centered on a triangle's nine-point circle and passing through the triangle's three vertices as well as its circumcenter, orthocenter, and various other notable centers Feuerbach hyperbola, a rectangular hyperbola that passes through a triangle's orthocenter, Nagel point, and various other notable points, and has center on the nine-point circle. Inconics Incircle, the unique circle that is internally tangent to a triangle's three sides Steiner inellipse, the unique ellipse that is tangent to a triangle's three sides at their midpoints Mandart inellipse, the unique ellipse tangent to a triangle's sides at the contact points of its excircles Kiepert parabola Yff parabola References [edit] ^ Weisstein, Eric W. "Circumconic." From MathWorld--A Wolfram Web Resource. ^ Weisstein, Eric W. "Inconic." From MathWorld--A Wolfram Web Resource. ^ a b c d e f g Chakerian, G. D. "A Distorted View of Geometry." Ch. 7 in Mathematical Plums (R. Honsberger, editor). Washington, DC: Mathematical Association of America, 1979. External links [edit] Circumconic at MathWorld Inconic at MathWorld Retrieved from " Category: Conic sections Hidden categories: Articles with short description Short description with empty Wikidata description Circumconic and inconic Add topic
17175	https://www.youtube.com/watch?v=ap1Ix0w2I6s	Write and Solve DIFFERENTIAL EQUATION - Radioactive Decay - Nuclear Physics PCCL - Physics - Chemistry 29600 subscribers 22 likes Description 2480 views Posted: 26 Sep 2021 From We are going to talk about the differential equation. You have to be able to establish it. I'm going to offer you something that I find quite simple, which is to write the activity in two different ways. First by giving its definition. It is the number of disintegrations per unit of time. Don't forget the minus sign here. And you can also write that this activity is proportional to the number of nuclides. Indeed, if there are twice as many nuclides, the activity will be twice as great. Lambda is the decay constant. It is also the constant of proportionality in this equation. Which leads us to write delta N over delta t equals minus lambda times N. Well, there it is. We are almost there. Almost. The time intervals must be of infinitesimal magnitude. At this point, we replace the delta with lowercase d. And we get the derivative of the function. And that's why we call it a differential equation, because we have the function and its derivative in the same equation. You must remember that a solution of this equation is N sub 0 times e to the minus lambda t. We will demonstrate this at the end of the video. If you are given this solution in the instructions, you may be asked to verify that this solution works well for the differential equation we have established. Therefore, you will differentiate here both sides. You obtain, d N over d t equals minus lambda times N naught, times e to the minus lambda t And we find the differential equation. So here the secret is to differentiate. So that's the differential equation. Here is his solution, and I wanted to discuss with you a relation between the constant lambda and the half-life. So here is the expression of our function N as a function of t which was the solution of the differential equation. I replace N by N sub 0 over 2, if I replace t by the half-life. There is this correspondence still posted here. N sub 0 over 2 corresponds to t sub one-half. At this point, I simplify by N naught, that gets rid of, and I have, one half equals e to the minus lambda t sub one-half. I have displayed a slider at the bottom to modify lambda, so that you understand that lambda and half-life do not vary the same. They vary inversely. If I take a smaller decay constant, I will have a half-life which will be greater. They are inversely proportional. If I instead take a larger constant, I get a smaller half-life. I haven't said anything about the lambda unit yet. It can be seconds to minus one, that is to say, per second. Since here we have seconds and the logarithm of 2 is a number, 0.693, then lambda is in, per second. But we can very well express lambda in, per day, per week, per year. What is important is to be consistent. That is, if lambda is expressed in, per year, then the time must be expressed in years. It has to be consistent. How to solve the differential equation. If I pass N to the other side and d t to the right, I have d N over N and there you are thinking, but why is he doing that? Well, I take the integral of both sides of this equation. We know the antiderivative of d N over N as well as that of d t. I stop at the terminals here. If you do the integral from N naught to N, that is to say from the starting point to a given situation, it has to be consistent. So it has to match here, for the durations. This is the date chosen as the origin 0 and a date t. The antiderivative of one over N is natural log of N. That of dt is t. and it comes that natural log of N over N sub zero equals minus lambda, times t Here I am doing the opposite of what I did on time. To get rid of the logarithm, I take the exponential of both sides and it comes: N as a function of t equals N naught times e to the minus lamba t This is how we show that this, is the solution to, that. 🎧 Thanks to Gilles FOURNAT for his contribution 😲🎸 Subscribe for more licks ;-) FIND HIM: ► INSTAGRAM : ► FACEBOOK : ► YOUTUBE : Transcript: Intro [Music] hello eccl channel greets you [Music] we are going to talk about the differential equation you have to be able to establish it i'm going to offer you something that i find quite simple which is to write the activity in two different ways first by giving its definition it is the number of disintegrations per unit of time don't forget the minus sign here and you can also write that this activity is proportional to the number of nuclides indeed if there are twice as many nuclides the activity will be twice as great lambda is the decay constant it is also the constant of proportionality in this equation which leads us to write delta n over delta t equals minus lambda times n well there it is we are almost there almost the time intervals must be of infinitesimal magnitude at this point we replace the delta with lowercase t and we get the derivative of the function and that's why we call it a differential equation because we have the function and its derivative in the same equation you must remember that a solution of this equation is n sub 0 times e to the minus lambda t we will demonstrate this at the end of the video if you are given this solution in the instructions you may be asked to verify that this solution works well for the differential equation we have established therefore you will differentiate here both sides you obtain dn over dt equals minus lambda times n naught times e to the minus lambda t and we find the differential equation so here the secret is to differentiate so that's the differential equation here is his solution and i wanted to discuss with you a lambda and halflife relation between the constant lambda and the half-life so here is the expression of our function n as a function of t which was the solution of the differential equation i replace n by n sub 0 over 2 if i replace d by the half-life there is this correspondence still posted here n sub zero over two corresponds to t sub one half at this point i simplify by n naught that gets rid of and i have one half equals e to the minus lambda t sub one half if i take the natural logarithm in both sides of the equation it comes minus natural logarithm of two equals minus lambda times t sub one half and i get the formula i was looking for so here's your little card i have displayed a slider at the bottom to modify lambda so that you understand that lambda and half-life do not vary the same they vary inversely if i take a smaller decay constant i will have a half-life which will be greater they are inversely proportional if i instead take a larger constant i get a smaller half-life i haven't said anything about the lambda unit yet it can be seconds to minus 1 that is to say per second since here we have seconds and the logarithm of 2 is a number 0.693 then lambda is in per second but we can very well express lambda in per day per week per year what is important is to be consistent that is if lambda is expressed in per year then the time must be expressed in years it has to be consistent exercise i offer you an exercise here is the problem statement you stop and try to do it otherwise i will give you the solution now it is about a wooden boat which one discovered it's a dracar a viking ship so it has not been manufactured recently and we measure the activity of a sample of wood taken from the hull there are 12 disintegrations per minute and per gram whereas in the atmosphere and in living matter such as wood it is 13.6 so between the time the wood was cut and the time the activity measurement was made the activity so i warned you earlier we can use nra i purposely chose an example with a it is not to destabilize you on the contrary it is so that you are comfortable we treat subjects in the same way so justify the variation in activity of a wood sample over time then the activity is proportional to the number n of nuclei we wrote it earlier so write a as a function of t equals lambda times n is a function of t and therefore when one decreases the other also decreases and moreover in the same proportion this is why the activity of a sample decreases knowing that the law of the activity decay is written see we give it to you in this problem statement you are given the decay law express the time as a function of the other quantities i'm going to try a carbon dating this is the title we're trying to date this famous longship i therefore suggest that you pass to the left the part which contains what i am looking for namely the time and taking the logarithm to get rid of the exponential here taking on both sides the logarithm of each member it comes that minus lambda lambda t equals the natural logarithm of a as a function of t on a naught therefore d equals negative one on lambda times natural log of a as a function of t on a sub zero so i answered the question correctly since i am asked to express the duration as a function of a as a function of t a naught and the radioactive constant here lambda this is what to do the calculation of lambda i cut in half i am asked to calculate t i will first calculate lambda i'll calculate d after lambda as we saw earlier it's natural log of 2 over half-life that is to say 0.696 by 5730 i allow myself to leave lambda in per year why not if there i have years my result will be in reciprocal unit in years to the minus 1. calculating t now since i have lambda a as a function of t is here it's 12 disintegrations per minute a naught equals 13.6 i don't need to change anything in the units here why 12 over 13.6 so whatever the units the ratio will always be the same as long as i am consistent and my calculator gives me something like 1035 years the wood that was used was cut around 948 the exercise is over i suggest you see how to solve the solution differential equation this is the differential equation if i pass n to the other side and dt to the right i have d n over n and there you are thinking but why is he doing that well i take the integral of both sides of this equation we know the antiderivative of dn over n as well as that of dt i stop at the terminals here dot if you do the integral from n out to n that is to say from the starting point to a given situation it has to be consistent so it has to match here for the durations this is the date chosen as the origin zero and a day t the antiderivative of one over n is natural log of n that of dt is t and it comes that natural log of n over n sub zero equals minus lambda times t here i am doing the opposite of what i did on time to get rid of the logarithm i take the exponential of both sides and it comes n is a function of t equals n naught times e to the minus lambda t this is how we show that this is the solution to that that's it for this video thank you [Music] you
17176	https://pmc.ncbi.nlm.nih.gov/articles/PMC9986645/	Community-based management of acute malnutrition: Implementation quality, and staff and user satisfaction with services - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice J Taibah Univ Med Sci . 2023 Feb 17;18(5):988–996. doi: 10.1016/j.jtumed.2023.02.002 Search in PMC Search in PubMed View in NLM Catalog Add to search Show available content in English Arabic Community-based management of acute malnutrition: Implementation quality, and staff and user satisfaction with services Joshua A Akuu Joshua A Akuu, MSc a School of Public Health, Faculty of Health, University of Technology Sydney, Sydney, Australia Find articles by Joshua A Akuu a,∗, Moses A Amagnya Moses A Amagnya, PhD b Institute of Policing, School of Justice, Security and Sustainability, Staffordshire University, Stoke-on-Trent, United Kingdom Find articles by Moses A Amagnya b Author information Article notes Copyright and License information a School of Public Health, Faculty of Health, University of Technology Sydney, Sydney, Australia b Institute of Policing, School of Justice, Security and Sustainability, Staffordshire University, Stoke-on-Trent, United Kingdom ∗ Corresponding address: School of Public Health, Faculty of Health, University of Technology Sydney, Sydney, Australia joshua.a.akuu@student.uts.edu.au Received 2022 Sep 14; Revised 2022 Nov 22; Accepted 2023 Feb 8; Collection date 2023 Oct. © 2023 [The Author/The Authors] This is an open access article under the CC BY-NC-ND license ( PMC Copyright notice PMCID: PMC9986645 PMID: 36890797 Abstract Background Malnutrition is a problem that affects many children and therefore is the focus of multiple interventions worldwide. One intervention is community-based management of acute malnutrition (CMAM). Objective This study assessed CMAM implementation quality in the Builsa North District of Ghana, and the satisfaction among both users and CMAM staff. Design The study used a convergent mixed-method design involving in-depth interviews with CMAM staff and users, document reviews, and observations of the CMAM implementation. The data were collected across eight health care facilities in eight sub-districts. The data were qualitatively and thematically analysed in Nvivo software. Results Several factors were found to adversely affect the quality of CMAM implementation. Significant factors included inadequate training of CMAM workers; religious belief systems; and a lack of implementation materials, such as ready-to-use therapeutic food (RUTF), CMAM registration forms/cards, and computers. These factors adversely affected programme quality, thus resulting in dissatisfaction among CMAM users and staff. Conclusion This study established that the CMAM programme in the Builsa North District of Ghana is hindered by a lack of primary resources and logistics necessary for successful programme implementation. Most health facilities in the district lack such resources and are not delivering the intended results. Keywords: CMAM, Community-based management, Health professionals, Malnutrition, Quality, RUTF الملخص أهداف البحث سوء التغذية مشكلة عالمية تؤثر على العديد من الأطفال. نتيجة لذلك، تم اعتماد العديد من التدخلات في جميع أنحاء العالم لمعالجة المشكلة. أحد التدخلات هو الإدارة المجتمعية لسوء التغذية الحاد. تقيم هذه الدراسة الإدارة المجتمعية لجودة تنفيذ سوء التغذية الحاد في منطقة شمال بويلسا في غانا ورضا المستخدمين والإدارة المجتمعية لموظفي سوء التغذية الحاد. طريقة البحث تستخدم الدراسة تصميما بحثيا متقاربا متعدد الأساليب يتضمن مقابلات متعمقة مع الإدارة المجتمعية لموظفي ومستخدمي سوء التغذية الحاد، ومراجعات الوثائق، وملاحظات الإدارة المجتمعية لتنفيذ سوء التغذية الحاد. تم جمع البيانات عبر ثمانية مرافق رعاية صحية في ثماني مناطق فرعية. تم تحليل البيانات نوعيا وموضوعيا باستخدام برنامج "إن فيفو". النتائج وجدنا العديد من العوامل التي تؤثر سلبًا على جودة الإدارة المجتمعية لتنفيذ سوء التغذية الحاد. تشمل العوامل المهمة عدم كفاية التدريب على الإدارة المجتمعية للعاملين في مجال سوء التغذية الحاد ، ونظم المعتقدات الدينية ، ونقص الأغذية العلاجية الجاهزة للاستخدام، والإدارة المجتمعية لنماذج/بطاقات تسجيل سوء التغذية الحاد، وأجهزة الكمبيوتر (أي ، مواد التنفيذ). تؤثر هذه العوامل سلبا على جودة البرنامج، مما يؤدي إلى عدم رضى الإدارة المجتمعية لمستخدمي سوء التغذية الحاد والموظفين. الاستنتاجات: أثبتت هذه الدراسة أن الإدارة المجتمعية لسوء التغذية الحاد في منطقة شمال بويلسا في غانا يعوقها نقص الموارد الأولية واللوجستيات اللازمة لتنفيذ البرنامج بنجاح. تفتقر معظم المرافق الصحية في المنطقة إلى الموارد اللازمة لتنفيذ البرنامج ولا تحقق النتائج المرجوة. الكلمات المفتاحية: الإدارة المجتمعية لسوء التغذية الحاد, الإدارة المجتمعية, المهنيين الصحيين, سوء التغذية؛ الجودة, الأغذية العلاجية الجاهزة للاستخدام Introduction Malnutrition is a global public health problem. According to the 2016 Global Nutrition Report, approximately 2 billion and 800 million people have micronutrient and caloric deficiencies, respectively.1 The report further indicates that 159 million children under the age of 5 years are too short for their age (stunted growth), 50 million are underweight for their height (wasted), and 41 million are overweight.1 Malnutrition generally refers to deficiencies, excesses, or imbalances in a person's energy and nutrient intake.2 Malnutrition and its complications account for approximately half of all child deaths worldwide.3 The seriousness of malnutrition has led to several global efforts and interventions to address this health problem.4 One intervention is the community-based management of acute malnutrition (CMAM) programme recommended by the World Health Organization (WHO) to manage acute malnutrition globally.5, 6, 7 The CMAM was developed by Valid International, an organisation established in 1999 to promote evidence-based reform of humanitarian practice. The concept has been used by WHO and UNICEF as an effective intervention model to replace the management of severe acute malnutrition through rehabilitation centres.8 The CMAM is a decentralised approach in which the management of acutely malnourished children is largely shifted from facility-based to community-level treatment provided by primary health care centres and mobile teams.9 This framework has been used by many governments and non-governmental organisations (NGOs) as a long-term programme for managing severe acute malnutrition.7 The CMAM programme has been implemented in Ghana since 2010.10 However, cases of malnourishment are rising in Ghana, particularly in northern Ghana, where this study was conducted. For instance, a recent study in Ghana by Martin et al. (2019) has indicated that, whereas the national stunting rate in children under 5 years of age was 19%, that in Northern Ghana was 33%. Another study in Ghana has indicated that approximately 40% of deaths among children under 5 years of age are associated with malnutrition.11 In addition, little research has assessed implementation quality, which directly affects the outcomes and recovery of the managed cases. Consequently, this study was aimed at assessing CMAM implementation through document reviews, observations, and interviews with users and CMAM staff. To this end, the study explored two research questions: (1) What is the CMAM programme's implementation quality? (2) How satisfied are CMAM users and CMAM staff with the programme implementation and quality? In considering the effects of malnutrition in children under 5 years of age globally,11 assessing intervention implementation quality and satisfaction is useful to identify measures for improving programme service delivery to achieve better recovery. Through this study, the Ministry of Health, Ghana Health Service, policymakers, and NGOs can compare the quality of the CMAM in the district against nationally and internationally acceptable standards. In addition, assessment of satisfaction among users and CMAM staff can suggest ways to improve the CMAM programme. The next part of the article discusses the CMAM programme and its history, as well as health intervention quality and satisfaction. The theoretical or conceptual framework underpinning the study are subsequently discussed. The methods used to conduct the study are then described, including the research design, setting, sampling, data collection, and sample characteristics. Subsequently, the results of the study, including implementation quality and satisfaction among both users and health professionals, are reported. The final part discusses key results of the study. Community-based management of acute malnutrition WHO and UNICEF used the CMAM developed by Valid International as an alternative intervention model for the management of severe acute malnutrition via rehabilitation centres.8 CMAM was first introduced as an emergency intervention through in-patient care treatment and feeding centres. Because of its success and effectiveness, this framework has been used by many governments and NGOs as a long-term measure for managing severe acute malnutrition.7 The CMAM programme funded by WHO, NGOs, and governments12,13 encourages the holistic participation of community health volunteers, opinion leaders, and CMAM staff. It emphasises the management of malnutrition cases at the community level instead of in hospital settings. CMAM is grouped into four components: in-patient management of children with severe acute malnutrition and medical complications; outpatient management of children with severe acute malnutrition without medical complications; services addressing moderate acute malnutrition to prevent undernutrition; and community outreach.14 The in-patient care for acutely malnourished children with medical complications provides intensive in-patient medical and nutritional care to children with medical complications, such as a loss of appetite, high temperature, or severe oedema.15 This component connects with the outpatient therapeutic programme to allow for early discharge and continued treatment in communities.8 Outpatient management for severe acute malnutrition without complications delivers home-based treatment and rehabilitation by using ready-to-use therapeutic food (RUTF) for children with severe acute malnutrition without medical complications.7 The recovery progress in malnourished children is monitored through regular outpatient clinics. This component is critical because many children with severe acute malnourishment do not show medical complications.8 Supplementary feeding for moderate acute malnutrition ensures that primary treatment and take-home food supplies are provided to families and children with moderate acute malnutrition without medical complications. Under this component, services are extended to other people with special nutritional requirements, including pregnant women and lactating mothers.7 The community outreach work ensures strong relationships and active community participation through traditional and opinion leaders. Community volunteers for CMAM are recruited and trained in measuring the mid-upper arm circumference (MUAC) of all children under 5 years of age to identify those with acute malnutrition.8,16 Community volunteers screen children with a MUAC tape measure, and those whose MUAC is less than 11.5 cm are administered RUTF treatments.5 Children with medical complications, such as bilateral pitting oedema, generalised body oedema, anaemia, or hypothermia, are managed as in-patients in health facilities and are later transferred to community level care for continuing treatment when their condition becomes stable. Children without medical complications are scheduled to visit to collect the RUTF at specific time intervals. While community CMAM volunteers serve as a link between CMAM staff and community members, opinion leaders encourage the participation and utilisation of CMAM by community members. Quality and satisfaction with health interventions The quality of health interventions and the satisfaction of users and CMAM staff are important because they substantially influence future utilisation and uptake of services.17 According to the Institute of Medicine, health service quality refers to healthcare services for individuals that increase the likelihood of desired health outcomes and are consistent with current professional knowledge.18Oyugi et al.19 have argued that the quality of health services is based on how healthcare workers conduct themselves in delivering health services. Health intervention or service quality is inherently associated with the satisfaction of users and CMAM staff and is often difficult to differentiate.20,21 Indeed, health service quality is a force driving user satisfaction, and vice versa.22 For instance, satisfied users usually adhere to treatment directions and follow-up that help achieve desirable outcomes and influence or improve the quality of healthcare programmes.23, 24, 25 Satisfaction often contains three common elements: emotional or cognitive response to satisfaction; focus (expectation); and response time.26 We argue that satisfaction is happy feelings demonstrated when individual expectations are met during healthcare service seeking. According to prior studies, a major driving force of user satisfaction in health services is providers’ attitudes, particularly their respect and empathy towards users.22,27 Many users consider this aspect to be more important than even the technical competence of providers. In addition, users are often dissatisfied with other aspects, such as time-wasting in health facilities—an aspect that consistently ranks high in importance in user satisfaction with health services.22 Some clients also base their satisfaction on risk reduction and the value received for the price.17 Like that of users, healthcare providers' satisfaction is key to the delivery of quality healthcare services. Although variations exist in reporting of job satisfaction across several studies, most determinants of satisfaction have been similar. Satisfaction among healthcare providers is determined by the availability of training for staff, supervision, promotion, job definition, and supplies and logistics28,29; other factors are the working conditions and the organisational environment, job stress, role conflicts and ambiguity, role perception and content, and organisational and professional commitment.30 These findings suggest that good working conditions, as well as environments devoid of stress, role conflict, and ambiguity, are important for healthcare service providers' satisfaction. Beyond policies, rules, logistics, and supplies, job training is an aspect that increases providers’ satisfaction, thus ensuring the quality of services and user satisfaction.31 Theoretical/conceptual framework for the study Various theoretical or conceptual frameworks exist for measuring the quality of health interventions. One example is the SERQUAL model, which uses the SERVQUAL questionnaire measuring five dimensions in service delivery: tangibles, reliability, responsiveness, assurance and empathy. Kashf et al.32 have used this model to evaluate the quality of health services provided to 384 people referred to the health centres of Ahvaz city.33 Another example is theory-based and consumer-based evaluations, which Gross (2004) has used to identify quality indicators of ambulatory health services, such as accessibility, availability, patient satisfaction, the performance of preventive medicine, and the use of private medical services. In addition, in the conceptual framework of quality of hospital services,34,35 conceptual models have been based on patients' perspectives, as proposed by Sofaer and Firminger36; clinicians’ perspectives37; and the Donabedian38 model. In this study, the Donabedian model was used to assess the quality of the CMAM programme. This model is widely used for measuring the quality of health interventions in Ghana and other parts of the world.39, 40, 41 According to the model, health programmes' quality depends on input, process, output, outcome, and impact (Figure 1). Input focuses on resources or supplies, and the logistics needed for programme implementation, including RUTF and medicines. Process refers to the technical expertise of staff or provider–client interactions during health-seeking.38 According to the model, output relates to the number of communities and/or children covered and treated, and outcome refers to the results achieved after treatment, which focuses on the recovery, mortality and defaulting rates after the treatment (noncomplience to treatment), and user satisfaction regarding CMAM.38 Finally, impact refers to long-term contributions of recovered clients to the country's or organisation's growth.38 This study measured the quality of the CMAM intervention by focusing on the first four components of the Donabedian model. Figure 1. Open in a new tab Adapted Donabedian Model. Materials and Methods Data were collected in the Builsa North District of the Upper East Region of Ghana. We used a convergent mixed-method design involving a cross-section of CMAM users and health officials implementing the programme. Eight healthcare facilities across six sub-districts were included in this study. Six smaller facilities were randomly selected from each sub-district, whereas two larger facilities were purposively selected to ensure that data-rich facilities were included. In addition, ten users and eight CMAM staff were included according to their availability and willingness to participate in the study. We developed an interview guide based on the National CMAM Implementation Checklist to collect the data. Interviews with CMAM staff were conducted in English, whereas user interviews were conducted in Buli, the native language of the Builsa people. Interviews in English and Buli allowed respondents to express themselves comfortably and adequately. The interviews also allowed participants to provide detailed explanations of the programme's implementation by discussing what was or was not working, and to offer suggestions for improvement. In addition, secondary data in the form of CMAM treatment records from 2018 and the first quarter of 2019 were obtained from the health facilities. The variables examined in this study were the four constructs under the Donabedian model: input (qualified staff, drugs, medical equipment and consumables, and protocol/guidelines), process (supervision, documentation, and participation and follow-up), output (coverage of the programme, number of cases treated, and satisfaction), and outcome (recovery, mortality, and defaulting rates). These methods were appropriate because the quality of health services has been measured with cross-sectional surveys, observations, interviews, and literature reviews.22,42,43 Nvivo 12 data management software and Microsoft Excel were used for qualitative and quantitative data analysis. Because the quantitative data were scant, they are presented in tables. Five steps of thematic analysis were used to analyse the data through a modified version of the Donabedian model. First, the audio-recorded interviews were transcribed, and those in Buli were translated to English. Second, the transcripts were examined to ensure that no information was missing before import into Nvivo 12 data management and analysis software for coding. Third, each interview was read, and word frequencies and text queries were used to identify and create codes according to the study objectives. Finally, the codes were reflected upon and clustered to develop overarching themes, such as user satisfaction, quality of implementation, and CMAM staff satisfaction, and provide suggestions for improvement44,45 for similar processes). Results This study's results are presented according to two themes that emerged: implementation quality, and satisfaction among users and CMAM staff. Implementation quality According to the Ghana Health Service (2010) and Donabedian,38 inputs such as trained staff, protocol/guidelines for implementation, routine drugs, and supplies are the main resources and logistics needed to implement CMAM successfully. However, most health facilities were found to lack the required inputs. Table 1 presents the availability of various variables in the eight health facilities covered by the study: 0 denotes the absence of an input in a facility, and ≥1 denotes the presence of an input. All included facilities had one or two staff members. As shown in Table 1, although the eight facilities had 12 staff members, only 25% of them had received specialised training in the CMAM programme. In addition, only six facilities had a severe acute malnutrition (SAM) action protocol and access to CMAM guidelines for reference (Table 1). Two facilities each had and displayed job aids, the SAM classification algorithm, and MUAC classification tables. Only three facilities had RUTF lookup tables and key messages, and none had RUTF—the primary food for malnutrition treatment (Table 1). In addition, only five, seven, three, and two facilities had malaria test kits, albendazole, amoxicillin, and cotrimoxazole, respectively. Still under input, the CMAM officials unanimously indicated that the CMAM programme was good and was working well when it was first introduced, owing to the availability of the required resources, logistics, and supplies. Supporting this claim, a health professional from facility 2 stated: “The CMAM implementation was good, and the programme came to help the district.” Table 1. Availability of Items/Activities for CMAM Implementation in the Studied Facilities. \| Activities/Items \| Health Facilities (1–8) \| Total \| --- \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| \| \| Human Resources \| \| Staff working on CMAM \| 1 \| 2 \| 2 \| 1 \| 2 \| 1 \| 2 \| 1 \| 12 \| \| Staff trained in CMAM \| 1 \| 0 \| 1 \| 0 \| 0 \| 0 \| 1 \| 0 \| 3 \| \| Availability of Protocols/Guidelines \| \| Guidelines accessible for reference \| 1 \| 0 \| 1 \| 1 \| 0 \| 1 \| 1 \| 1 \| 6 \| \| Job aids displayed and used \| 0 \| 0 \| 1 \| 0 \| 0 \| 0 \| 0 \| 1 \| 2 \| \| SAM action protocol \| 0 \| 1 \| 1 \| 1 \| 1 \| 0 \| 1 \| 1 \| 6 \| \| RUTF lookup table and key messages \| 0 \| 0 \| 1 \| 1 \| 0 \| 0 \| 0 \| 1 \| 3 \| \| SAM classification algorithm \| 0 \| 0 \| 1 \| 0 \| 0 \| 1 \| 0 \| 0 \| 2 \| \| MUAC classification table \| 1 \| 0 \| 0 \| 1 \| 0 \| 0 \| 0 \| 0 \| 2 \| \| Availability of Supplies: Section Two \| \| Ready-to-use therapeutic food \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| \| Number of MUAC tape measures \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 8 \| \| Number of weighing scales \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 8 \| \| Availability of Routine Medications \| \| Amoxicillin \| 0 \| 0 \| 1 \| 1 \| 0 \| 0 \| 0 \| 1 \| 3 \| \| Cotrimoxazole \| 0 \| 0 \| 1 \| 0 \| 0 \| 0 \| 0 \| 1 \| 2 \| \| Albendazole \| 1 \| 1 \| 0 \| 1 \| 1 \| 1 \| 1 \| 1 \| 7 \| \| Malaria rapid test kits (para-check) \| 1 \| 1 \| 0 \| 1 \| 0 \| 1 \| 0 \| 1 \| 5 \| \| Supervision \| \| Monthly supervision of staff \| 1 \| 0 \| 0 \| 1 \| 0 \| 0 \| 1 \| 0 \| 3 \| Open in a new tab This quotation indicated that the CMAM is a good programme that was initially implemented well. However, the CMAM staff agreed that the lack of essential resources had more recently affected the programme's implementation. For instance, one CMAM official at facility 5 stated: “The main problem with the implementation now is the non-availability of the RUTFs, F-100, F-75, and ReSoMal to manage malnutrition cases.” This quotation, which reflects several concerns raised by other CMAM officials, shows that the CMAM programme's current implementation is substantially affected by the non-availability of resources and logistics. Regarding the process in the Donabedian model, this study indicated an acute lack of supervision of the CMAM programme. For instance, only three (37.5%) of the eight facilities reported monthly supervision of the programme (Table 1). In addition, most facilities had problems with documentation and follow-up, which are part of the process category in the Donabedian model. For instance, an official at facility 1 explained: “Lack of means of transport like motorbikes to go for follow-ups and sensitisation is a big challenge and is not allowing us to do our work well.” Regarding documentation, an official at facility 6 said: “The cards for admission of children … to the CMAM programme are not available for close to 1 year now.” Another official at the same facility stated: “Some of the challenges are documentation challenges: thus, no tools for us to document our actions.” The study also found challenges in measuring the programme's output (coverage and cases treated) and outcomes because of inconsistent data. The combined number of malnutrition cases obtained from the covered health facilities was higher than the total number of malnutrition cases captured by the district health directorate (Table 2). This finding was unexpected, given that the district figures were an aggregation of cases from all 22 health facilities in the district. Double counting can sometimes occur when data are collected from different health facilities within a sub-district that may report to one another instead of the district health directorate. However, double counting did not occur in this study, because the data were collected from facilities reporting directly to the district health directorate. Moreover, some health facilities did not have records for some outcomes. For example, no data were recorded for children who did not recover in 2018 and the first quarter of 2019, or for children who died within the first quarter of 2019 (Table 2). Table 2. CMAM Cases in the Selected Health Facilities and Districts for 2018 and 2019. \| Items \| Number of Cases \| --- \| \| Period \| 2018 \| First Quarter of 2019 \| \| Facilities \| Facilities Studied \| District \| Facilities Studied \| District \| \| Total malnourished cases \| 43 \| 32 \| 10 \| 20 \| \| Cured cases \| 30 \| 4 \| 7 \| 7 \| \| Fatal cases \| 1 \| – \| – \| – \| \| Defaulted cases \| 3 \| – \| 2 \| – \| \| Non-recovered cases \| – \| – \| – \| 1 \| \| Discharged cases \| 30 \| 4 \| 7 \| 8 \| Open in a new tab Likewise, the 2018 Builsa North District Health Information Management System report contained 32 CMAM cases, four (12.5%) of which were treated successfully, but no further information was available. Similarly, the 2019 first-quarter report included 20 cases admitted into the CMAM programme, seven of which were treated successfully. However, no further information on the number of deaths or defaulters was available (Table 2). When the CMAM officials at the facilities were asked about the inconsistent data, they indicated that the non-availability of documentation and registration materials is a major problem that makes keeping accurate programme records at facilities challenging. Satisfaction among stakeholders Stakeholders’ satisfaction with a health intervention is an important element of the output construct of the Donabedian model. We therefore measured the satisfaction of CMAM users and CMAM staff as the implementers of the programme. Community-based management of acute malnutrition staff The health officials were not satisfied with the CMAM programme for several reasons. The lack of specialised CMAM training (input) was a major reason for officials' dissatisfaction. The findings indicated that most CMAM officials had received no specialised training before being assigned CMAM tasks or even during their careers with the CMAM programme. For example, a health official at facility 7 said: “Since 2013, they have not trained any nurse again.” In addition, most of the staff trained in 2013 had left the service through retirement or moving on to further studies, thus leaving new and inexperienced staff without the required training and skills to manage the programme. The non-availability of logistics and supplies (input) was another reason for CMAM staff dissatisfaction with the CMAM programme. For instance, one official at health facility 3 explained the reasons for his dissatisfaction as follows: “There are new cases, but due to the absence of RUTFs, we have not been able to enrol them into the programme.” CMAM staff also noted that clients do not come to the health facilities to seek their services, because of a lack of RUTFs. These aspects together affected officials’ work and satisfaction with the programme. Community-based management of users Although users were dissatisfied with the CMAM programme because of a lack of necessary logistics and resources, they were satisfied with the treatment by CMAM staff. Most users were pleased with how nurses and CMAM officials were nice and welcoming during their visits. “They always respect me for whom I am,” “I was happy because the nurses were very nice to me,” and “I would say that the health workers have done very well, and I am pleased with their services” are some examples of sentiments among users. These findings suggest that users focus on respect from CMAM staff during visits to health facilities rather than on their children getting the needed treatments. This result is consistent with those from studies indicating that people value respect and fair treatment more than outcomes of interactions.46, 47, 48 Intriguingly, some users who were pleased with the warm reception of CMAM officials still expressed discontent with the absence of necessary logistics and resources. Notably, the shortage of RUTFs was a cause of substantial dissatisfaction among users, because their children did not receive the needed treatment for better recovery. For instance, a user from health facility 1 stated: “The shortage of the food (i.e., RUTF) negatively affected my child's health, because it is not easy to come by money to purchase it in the open market.” In addition, other users complained about the lack of follow-up by CMAM staff, particularly when users missed their reviews. For instance, a user from health facility 4 angrily noted: “I stopped going to the health centre with my child, but no nurse followed up to ask me why I do not bring my child to the health centre again. Why is it like that?” These results and quotations clearly show that CMAM users were not pleased with the programme's overall implementation. Discussion Inputs (protocol/guidelines for implementation, routine drugs supplies, and logistics supplies), process (technical competence, supervision, documentation, participation, and follow-up), and output (coverage of the programme, number of cases treated, satisfaction) are important determinants of health interventions’ implementation quality.38,49, 50, 51, 52 However, essential logistics and supplies, such as routine drugs, RUTFs, equipment, guidelines, and materials needed to implement the CMAM programme successfully, were insufficient and in some cases non-existent. These insufficient resources suggest that the CMAM programme in the Builsa North District did not meet the Donabedian38 quality standard. In addition, owing to inconsistent and incomplete data, the number of cases treated, recovered, defaulted and resulting in death, which are crucial determinants of programme quality, were impossible to measure.53 The results imply that malnourished children may not be getting the needed treatment to enable them to recover faster, and those who are treated may not be treated well, because of the staff's inadequate skills. Similarly, early identification of malnourished cases via screening at community outreach programmes, home visits, and follow-up may not occur because of a lack of logistics and means of transport. Most CMAM officials requested provision of job-specific training to improve their technical competence and working knowledge to manage malnutrition cases properly. Training on the CMAM programme could provide clear criteria for admission and discharging of malnutrition cases, and improve the overall performance of the programme.54,55 In agreement with findings from previous studies,56,57 we found inconsistency between the data from the studied facilities and the district health directorate. However, complete and consistent data are essential for assessing programme success and determining areas of improvement. The inconsistent information, as found in this study, could arguably pose a major challenge to the CMAM programme by making its success and/or determination of areas needing improvement either difficult or impossible to assess. Given the importance of data in programme assessments, governments must provide adequate registration, documentation materials, and computers to health facilities and district health directorates to help address challenges associated with inconsistent data. This aspect is very important, because research has shown that poor record-keeping; under-reporting; lack of time to complete documentation; and shortages in registration forms, computers, recording materials, tools, and equipment are substantial contributors to inconsistent records.56,57 Both CMAM users and officials were dissatisfied with the programme's implementation quality. Users based their satisfaction levels on the availability of supplies and the client–provider relationship. Users were generally unhappy with the programme's implementation, because of a lack of resources and logistics, particularly regarding the RUTF and basic medications for treating malnourished children. The user satisfaction level with health interventions is a major determinant of patients' future adherence to treatment and uptake of services.17,23, 24, 25 Arguably, most clients of the CMAM programme in the district are likely to stop attending treatment centres or to not seek healthcare for their children. In addition, some parents may resort to local means of treating their children, which could negatively affect their recovery and overall growth and development. The lack of professional healthcare can lead to malnourished children becoming severely ill or even dying. Therefore, the government and donor agencies must provide logistics and resources to health facilities, particularly RUTFs, to treat malnutrition cases effectively and improve user satisfaction. Similarly to user dissatisfaction, CMAM staff dissatisfaction was influenced by the lack of essential resources, such as RUTFs and medication, to treat malnourished children. In addition, the lack of appropriate and adequate staff training emerged as an important determinant of satisfaction. The determinants of CMAM staff satisfaction levels found in this study are consistent with those from prior studies, such as Walker,29Bilal et al.,28 and Lu et al.30 Job satisfaction is an important determinant of quality service delivery.58 Therefore, an important measure to restore CMAM service providers' satisfaction is adequate provision of logistics and supplies, and CMAM job training for staff as an ingredient to increase providers’ satisfaction,31 and also ensure the quality of services, and the satisfaction of both service providers and users. Another notable result was that some users who were dissatisfied with the programme's implementation quality were satisfied with the client–provider relationship. Thus, users were pleased with the good relationship and respect received from CMAM staff during hospital visits. This finding suggests that the CMAM providers relate well with their clients. This aspect is essential, because the client–provider relationship is a major determinant of user satisfaction that can lead to future uptake of health services or visits to health facilities.22,27 Therefore, health facilities should focus on improving their client–provider relationships to enhance user satisfaction and attract more patients for treatment. Conclusion This study has several limitations. A key limitation is that not all health facilities in the district were included, because the study was conducted during the rainy season, when some facilities were not accessible. However, a key strength of the study is the combination of data collection methods, which allowed for data triangulation. This study assessed the CMAM programme in an African context, focusing on implementation quality and stakeholders' satisfaction. The results revealed that the CMAM programme appears not to be delivering the intended results, because of a lack of supplies and logistics for managing malnutrition cases. Both CMAM staff and users are dissatisfied with the programme implementation, owing to inadequate logistics and supplies. Staff dissatisfaction with the CMAM's performance was also due to the lack of staff training and professional development. Providing adequate logistics, resources, staff numbers, and staff training is essential for improving the programme's quality. Likewise, health facilities can improve uptake and adherence to treatment by enhancing the quality of client–provider relationships and user satisfaction. Source of funding This work is drawn from a master's degree study conducted under a DAAD scholarship at Heidelberg University, Germany. However, the research did not receive any specific grant from the DAAD and other funding agencies in the public, commercial, or not-for-profit sectors for publication purposes. Conflict of interest There are no conflicts of interest to declare. Ethical approval This study was conducted according to the guidelines of the Declaration of Helsinki. The Navrongo Health Research Centre Institutional Review Board (NHRCIRB) granted ethics approval for this study on 19th July 2019 (approval ID NHRCIRB250). Verbal informed consent was obtained from all participants/patients, and was witnessed and formally recorded. Author contributions JAA (corresponding author) performed study design and data collection. He fully participated in writing the methods, results, and discussion sections. MAA participated in designing the research, software, and editing, and conducted some of the analysis of the data. He also wrote some of the results and discussion. All authors have critically reviewed and approved the final draft, and are responsible for the content and similarity index of the manuscript. Acknowledgement We acknowledge the contributions of Professor Albrecht Jahn, Dr Faith Agbozo, and Dr Abubakari Abdulai of Heidelberg University and the University for Development Studies, respectively, for their advice and supervision of the original research described herein. We also acknowledge DAAD and Heidelberg University for the scholarship under which the research was conducted and for supporting the travel to Ghana to collect the data for the study. Footnotes Peer review under responsibility of Taibah University. References 1.IFPR . 2016. Global nutrition report: from promise to impact: ending malnutrition by 2030. Retrieved from Washington DC: [Google Scholar] 2.WHO . 2017. What is malnutrition? Retrieved from. Retrieved 1st April 2019, from World Health Organisation. [Google Scholar] 3.Requejo J.H., Bhutta Z.A. 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17177	https://physics.stackexchange.com/questions/153186/how-to-theoretically-calculate-the-value-of-gravitational-acceleration-of-my-tow	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How to theoretically calculate the value of gravitational acceleration of my town? Ask Question Asked Modified 8 years, 2 months ago Viewed 3k times 3 $\begingroup$ We normally consider the value of gravitational acceleration $g = 9.8 m/s^2$ while solving the problem. But that is the value of $g$ at poles (if I am not wrong). My teacher have given as homework to find the value of $g$ of my town. I know it should be approximately 9.8, but I have no idea how to find it. homework-and-exercises newtonian-gravity acceleration centrifugal-force Share Improve this question edited Jun 28, 2017 at 9:48 Qmechanic♦ 222k5252 gold badges635635 silver badges2.5k2.5k bronze badges asked Dec 14, 2014 at 6:47 FreddyFreddy 45222 gold badges99 silver badges2222 bronze badges $\endgroup$ 6 2 $\begingroup$ I'm guessing this has to do with the centrifugal force at the coordinates of your town on account of the Earth's rotation. There are quite a few causes other than this for regional variations in $g$, like presence of mountains, other geological variations, but I don't think a high-school level class will involve that. $\endgroup$ pho – pho 2014-12-14 07:20:35 +00:00 Commented Dec 14, 2014 at 7:20 2 $\begingroup$ Did your teacher hand out an absolute gravimeter? If he didn't, then you need to look it up or calculate it from the ellipsoid shape of the planet to first order. $\endgroup$ CuriousOne – CuriousOne 2014-12-14 07:20:46 +00:00 Commented Dec 14, 2014 at 7:20 1 $\begingroup$ With the help of experiment or theoretically? $\endgroup$ Paul – Paul 2014-12-14 07:56:21 +00:00 Commented Dec 14, 2014 at 7:56 $\begingroup$ @Paul theoretically $\endgroup$ Freddy – Freddy 2014-12-14 08:03:20 +00:00 Commented Dec 14, 2014 at 8:03 $\begingroup$ Tip: Most likely your teacher wants you to assume that Earth is spherically symmetric (even though this assumption is actually inconsistent with the fact that Earth is spinning). $\endgroup$ Qmechanic – Qmechanic ♦ 2014-12-14 10:28:42 +00:00 Commented Dec 14, 2014 at 10:28 \| Show 1 more comment 3 Answers 3 Reset to default 3 $\begingroup$ Gravitational acceleration at the surface of earth varies with latitude (North-South position). This is due to 1) the outward centrifugal force produced by Earth's rotation, and 2) the equatorial bulge (itself caused by Earth's rotation). Both effects cause the gravitational acceleration to decrease away from the poles. The net effect is a gravitational acceleration of about $9.832$ $m/s^2$ at the poles, and about $9.780$ $m/s^2$ (0.5 % lower) at the equator. For any sea level position on Earth (your city, Jamnagar, at 17 m elevation can indeed be considered to be roughly at sea level), we can estimate the gravitational acceleration $g$ at any latitude $\phi$ using Helmert's equation: $$g(\phi ) = g_0 (1 + 0.0053024 sin^2 \phi - 0.0000058 sin^2 2\phi) $$ where $g_0 = 9.780327 m/s^2$ denotes the gravitational acceleration at the equator. Local variations in Earth's topography and geology cause local deviations from the above formula. Such local variations are known as gravitational anomalies. To include these would go way beyond the intentions of the exercise. Share Improve this answer edited Dec 14, 2014 at 12:33 answered Dec 14, 2014 at 11:50 JohannesJohannes 19.4k22 gold badges5151 silver badges8080 bronze badges $\endgroup$ 2 $\begingroup$ Thanks for answer. I am thinking why my teacher asked to find the stuff which requires something that i haven't learned! $\endgroup$ Freddy – Freddy 2014-12-14 13:55:07 +00:00 Commented Dec 14, 2014 at 13:55 $\begingroup$ @Freddy the problem your teacher gave you was probably to make you search for information about the problem yourself. This would make you learn both about finding information and the subject itself. Smart teacher. $\endgroup$ Communisty – Communisty 2017-06-28 13:20:14 +00:00 Commented Jun 28, 2017 at 13:20 Add a comment \| 0 $\begingroup$ You can calculate the value of acceleration due to gravity in your town by finding the latitude of your town. Once you find the latitude use this equation $g'=g-R\omega^2\cos^2\lambda$. Here $R$ is the radius of the earth, $\omega$ the angular velocity of the earth and $\lambda$ the latitude. This equation comes from the fact that the earth is rotating and a particle thus will experience a centrifugal force. Source Share Improve this answer edited Jan 16, 2015 at 4:24 answered Dec 14, 2014 at 8:04 PaulPaul 3,48522 gold badges2525 silver badges3737 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ The value of acceleration due varies with altitude as well. It actually decreases with increase in altitude. The following formula approximates the Earth's gravity variation with altitude: $g_h = g_0\left(\frac{r_\mathrm{e}}{r_\mathrm{e}+h}\right)^2$ where $g_h$ is the gravitational acceleration at height $h$ above sea level. $r_e$ is the Earth's mean radius. $g_0$ is the standard gravitational acceleration. So all you need to do is find the elevation of your town. But remember that this formula treats the Earth as a perfect sphere with a radially symmetric distribution of mass and you will only get an approximate value. However, if you take into account the fact that the earth is spinning about it's axis then it will depend on the latitude (this is because of the centrifugal force and also because of the "equatorial bulge") but this is usually very small. Moreover, it also depends on a host of other factors like the density of the ground, air density etc. You can see the Wikipedia article for more details. Share Improve this answer edited Jun 28, 2017 at 9:16 answered Dec 14, 2014 at 8:26 noir1993noir1993 2,23644 gold badges2323 silver badges3737 bronze badges $\endgroup$ 3 $\begingroup$ You say that the centrifugal contribution is really small, but what about that height variation? From see level to $2km$ height, it gives less than $0.1\%$ deviation. You should give more number to support your statement that the answer by @Paul is not relevant. $\endgroup$ Bernhard – Bernhard 2014-12-14 09:05:56 +00:00 Commented Dec 14, 2014 at 9:05 $\begingroup$ Well, I plugged in the numbers into both formulas. And the 'altitude formula' gives a value of $9.79m/s^2$ and the 'latitude formula' gives $9.77m/s^2$. (I stand corrected, both of them give nearly equal values) $\endgroup$ noir1993 – noir1993 2014-12-14 10:30:25 +00:00 Commented Dec 14, 2014 at 10:30 $\begingroup$ Both the formulas hold for different situations. I am thinking out loud here but I think what if we applied the latitude formula first to find the effective $g$ at that point and them used that value of $g$ as $g_0$ in the altitude formula, we might get a more precise value of g at the town which takes both factors into account? I am having doubts because as Qmechanic pointed out in the comment that the assumption that the earth is spherically symmetric (as we did in deriving the altitude formula) is inconsistent with the fact that earth is spinning. $\endgroup$ noir1993 – noir1993 2014-12-14 10:44:06 +00:00 Commented Dec 14, 2014 at 10:44 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions homework-and-exercises newtonian-gravity acceleration centrifugal-force See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked Why is the Earth so fat? 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17179	https://brainly.com/question/37073423	[FREE] A small 2.00 g plastic ball is suspended by a 16.7 cm long string in a uniform electric field. If the ball - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +30,9k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +11,4k Ace exams faster, with practice that adapts to you Practice Worksheets +7k Guided help for every grade, topic or textbook Complete See more / Physics Textbook & Expert-Verified Textbook & Expert-Verified A small 2.00 g plastic ball is suspended by a 16.7 cm long string in a uniform electric field. If the ball is in equilibrium when the string makes a 15° angle with the vertical, what is the net charge on the ball? Given: Electric field, E=1.00×1 0 3 N/C 1 See answer Explain with Learning Companion NEW Asked by Germanclavijo4950 • 09/07/2023 0:03 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 291790 people 291K 1.0 0 Upload your school material for a more relevant answer The net charge on the plastic ball can be figured out using the principle of equilibrium. The vertical and horizontal forces act on the ball in equilibrium and balance out hence solving the equation will provide the net charge. Substituting the given values into the derived equation reveals the value of the net charge. Explanation The net charge on the plastic ball can be found by applying the concept of equilibrium. The forces acting on the ball when it is in equilibrium are the gravitational force (mg) acting downward, the tension force (T) acting along the string, and the electric force (qE) acting horizontally, where q is the net charge on the plastic ball, m is the mass of the ball, g is acceleration due to gravity and E is the electric field. Since the ball is in equilibrium, the vertical and horizontal components of the tension must balance the electric force and the gravitational force, respectively. From the horizontal component of the tension (Tsinθ) = qE, q (net charge) can be found. Solving for q, we get q = (Tsinθ)/E. Next, calculate T using T = mg/cosθ, then substitute it into the equation for q. So, q = [mgsinθ]/(Ecosθ). When you input the given values, including g as 9.8 m/s², you will get the net charge on the ball. Learn more about Net Charge in Equilibrium here: brainly.com/question/33267245 SPJ11 Answered by kumarvijaynew •3.7K answers•291.8K people helped Thanks 0 1.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 291790 people 291K 1.0 0 The Peoples Physics Book - James H. Dann Fields and Circuits - Benjamin Crowell University Physics Volume 2 - Samuel J. Ling, William Moebs, Jeff Sanny Upload your school material for a more relevant answer The net charge on the plastic ball is approximately 5.23 µC, derived from balancing the forces acting on the ball when it is in equilibrium in an electric field. The calculations involve the gravitational force, the tension in the string, and the electric force due to the electric field. By using trigonometric relationships and substituting values accordingly, the charge can be determined accurately. Explanation To find the net charge on the plastic ball, we first analyze the forces acting on the ball when it is in equilibrium. The forces consist of: Gravitational Force (Weight): This acts downwards and can be calculated using the formula: F g=m g where m=0.002 kg (the mass of the ball) and g=9.8 m/s 2. So, F g=0.002 kg×9.8 m/s 2=0.0196 N. Tension in the String (T): This acts along the string at an angle of 15° with the vertical. The tension can be broken down into two components: the vertical component T y (which balances the weight) and the horizontal component T x (which balances the electric force). These components can be expressed as: T y=T cos(15°) T x=T sin(15°) Electric Force (F_e): This acts horizontally due to the electric field, given by: F e=qE where q is the charge on the ball and E=1.00×1 0 3 N/C. Since the ball is in equilibrium, the vertical forces must balance, giving us: T cos(15°)=m g (1) And the horizontal forces must also balance, leading to: T sin(15°)=qE (2) From equation (1), we can express tension (T) in terms of mass and gravity: T=cos(15°)m g Now substituting this expression for T into equation (2): cos(15°)m g sin(15°)=qE Rearranging gives: q=E cos(15°)m g sin(15°) Substituting in the values: m=0.002 kg g=9.8 m/s 2 θ=15° E=1.00×1 0 3 N/C Calculating: s in(15°)≈0.2588 cos(15°)≈0.9659 Therefore, plugging in: q=(1.00×1 0 3)(0.9659)(0.002)(9.8)(0.2588) Evaluating this gives: q≈965.9 0.00506≈5.23×1 0−6 C Thus, the net charge on the ball is approximately 5.23 μ C. Examples & Evidence For instance, if you consider a similar scenario with a different angle or mass, you would follow the same steps: calculate gravitational force, decompose the tension into components, and solve for charge. This method can apply to various physics problems involving charged objects in an electric field. This approach to calculating the force on charged objects in an electric field is based on established physics principles, particularly Newton's laws and the definitions of electric forces in fields. Thanks 0 1.0 (1 vote) Advertisement Germanclavijo4950 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Physics solutions and answers Community Answer A small plastic ball of mass m - 1.00 g is suspended by a string of length L = 22.5 cm in a uniform electric field, as shown in the ligure below. If the ball is in equilibrium when the string makes a Community Answer 4.7 5 A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium when the string makes a 30 degrees angle with the vertical, what is the net charge on the ball? Q = _ C Community Answer A small 15.0 g plastic ball is suspended by a 30.0 cm string in a uniform electric field of 3000 N/C, as shown in the picture. If the ball is in equilibrium when the string makes a 45° angle with the vertical as indicated, what is the net charge on the ball? Community Answer A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If the ball is in equilibrium when the string makes a 30 o angle with the vertical, what is the net charge on the ball? Q =____ Community Answer 6. A small 2 g plasic ball is suspended by a 20 cm long string in a uniform electric field E=10 3N/C directed to the right. What is the charge that should be placed on the ball to displace the string 15∘with the vertical. Community Answer 47 Whats the usefulness or inconvenience of frictional force by turning a door knob? Community Answer 5 A cart is pushed and undergoes a certain acceleration. Consider how the acceleration would compare if it were pushed with twice the net force while its mass increased by four. Then its acceleration would be? Community Answer 4.8 2 define density and give its SI unit Community Answer 9 To prevent collisions and violations at intersections that have traffic signals, use the _____ to ensure the intersection is clear before you enter it. Community Answer 4.0 29 Activity: Lab safety and Equipment Puzzle New questions in Physics The table below includes data for a ball rolling down a hill. Fill in the missing data values in the table and determine the acceleration of the rolling ball. \| Time (seconds) \| Speed (km/h) \| :--- \| \| 0 (start) \| 0 (start) \| \| 2 \| 3 \| \| \| 6 \| \| \| 9 \| \| 8 \| \| \| 10 \| 15 \| Acceleration = ____ How do wheel size and axle friction affect the speed and distance a rubber band car can travel? The distance between the two nearest molecules that are in phase with each other is called (a) In what year will we next be able to see Halley's Comet? (b) In 1997, people could see Comet Hale-Bopp. In what year will this comet return? Why do comets appear to have large tails flowing away? Microwaves were originally used in which equipment during the world wars? A. Sonar B. Radars C. Thermostats D. Telescopes Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
17180	https://www.khanacademy.org/math/cc-2nd-grade-math/cc-2nd-add-subtract-1000	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
17181	https://pubmed.ncbi.nlm.nih.gov/3046354/	Malignant external otitis: insights into pathogenesis, clinical manifestations, diagnosis, and therapy - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Malignant external otitis: insights into pathogenesis, clinical manifestations, diagnosis, and therapy J Rubin1,V L Yu Affiliations Expand Affiliation 1 Department of Otolaryngology, University of Pittsburgh, Pennsylvania. PMID: 3046354 DOI: 10.1016/0002-9343(88)90592-x Item in Clipboard Review Malignant external otitis: insights into pathogenesis, clinical manifestations, diagnosis, and therapy J Rubin et al. Am J Med.1988 Sep. Show details Display options Display options Format Am J Med Actions Search in PubMed Search in NLM Catalog Add to Search . 1988 Sep;85(3):391-8. doi: 10.1016/0002-9343(88)90592-x. Authors J Rubin1,V L Yu Affiliation 1 Department of Otolaryngology, University of Pittsburgh, Pennsylvania. PMID: 3046354 DOI: 10.1016/0002-9343(88)90592-x Item in Clipboard Full text links Cite Display options Display options Format Abstract Malignant external otitis is an infection of the external ear canal, mastoid, and base of the skull caused by Pseudomonas aeruginosa. The condition occurs primarily in elderly patients with diabetes mellitus. Current theories on pathogenesis and anatomic correlations are reviewed. Severe, unrelenting otalgia and persistent otorrhea are the symptomatic hallmarks of the disease, whereas an elevated erythrocyte sedimentation rate is the only distinctive laboratory abnormality. Iatrogenic causes such as administration of broad-spectrum antibiotics and aural irrigation may play a predisposing role in high-risk populations. The disease can result in cranial polyneuropathies (with facial nerve [VII] paralysis being the most common) and death. The mainstay of treatment is administration of antipseudomonal antibiotics for four to eight weeks. Recurrence is common, and mortality remains at about 20 percent despite antibiotic therapy. Given the increasing longevity of diabetic patients, the frequency of this disease is increasing. Internists, family practitioners, and ambulatory care physicians must now be cognizant of the presenting symptoms, while infectious disease specialists and otolaryngologists need to be appraised of strides in diagnosis and therapy. The role of surgery should be minimized. Use of new diagnostic radiologic modalities and new antipseudomonal antibiotics discussed in this review should lead to improved outcome. PubMed Disclaimer Similar articles Malignant otitis externa.Karaman E, Yilmaz M, Ibrahimov M, Haciyev Y, Enver O.Karaman E, et al.J Craniofac Surg. 2012 Nov;23(6):1748-51. doi: 10.1097/SCS.0b013e31825e4d9a.J Craniofac Surg. 2012.PMID: 23147298 Necrotizing external otitis: a case series.Soheilipour S, Meidani M, Derakhshandi H, Etemadifar M.Soheilipour S, et al.B-ENT. 2013;9(1):61-6.B-ENT. 2013.PMID: 23641593 The changing face of malignant (necrotising) external otitis: clinical, radiological, and anatomic correlations.Rubin Grandis J, Branstetter BF 4th, Yu VL.Rubin Grandis J, et al.Lancet Infect Dis. 2004 Jan;4(1):34-9. doi: 10.1016/s1473-3099(03)00858-2.Lancet Infect Dis. 2004.PMID: 14720566 Review. Malignant otitis externa: An updated review.Treviño González JL, Reyes Suárez LL, Hernández de León JE.Treviño González JL, et al.Am J Otolaryngol. 2021 Mar-Apr;42(2):102894. doi: 10.1016/j.amjoto.2020.102894. Epub 2021 Jan 5.Am J Otolaryngol. 2021.PMID: 33429178 Review. Acute pseudomonas infection of the external ear (malignant external otitis).Scherbenske JM, Winton GB, James WD.Scherbenske JM, et al.J Dermatol Surg Oncol. 1988 Feb;14(2):165-9. doi: 10.1111/j.1524-4725.1988.tb03359.x.J Dermatol Surg Oncol. 1988.PMID: 3343425 See all similar articles Cited by Diffusion MR imaging features of skull base osteomyelitis compared with skull base malignancy.Ozgen B, Oguz KK, Cila A.Ozgen B, et al.AJNR Am J Neuroradiol. 2011 Jan;32(1):179-84. doi: 10.3174/ajnr.A2237. Epub 2010 Oct 14.AJNR Am J Neuroradiol. 2011.PMID: 20947640 Free PMC article. Skull base osteomyelitis: factors implicating clinical outcome.Sokołowski J, Lachowska M, Karchier E, Bartoszewicz R, Niemczyk K.Sokołowski J, et al.Acta Neurol Belg. 2019 Sep;119(3):431-437. doi: 10.1007/s13760-019-01110-w. Epub 2019 Mar 6.Acta Neurol Belg. 2019.PMID: 30840222 Free PMC article. Epidemiology and risk factors for extension of necrotizing otitis externa.Krawiec E, Brenet E, Truong F, Nguyen Y, Papthanassiou D, Labrousse M, Dubernard X.Krawiec E, et al.Eur Arch Otorhinolaryngol. 2024 May;281(5):2383-2394. doi: 10.1007/s00405-024-08549-5. Epub 2024 Mar 18.Eur Arch Otorhinolaryngol. 2024.PMID: 38499694 The challenging diagnosis and follow-up of skull base osteomyelitis in clinical practice.Auinger AB, Dahm V, Stanisz I, Schwarz-Nemec U, Arnoldner C.Auinger AB, et al.Eur Arch Otorhinolaryngol. 2021 Dec;278(12):4681-4688. doi: 10.1007/s00405-020-06576-6. Epub 2021 Jan 28.Eur Arch Otorhinolaryngol. 2021.PMID: 33511482 Free PMC article. Malignant otitis external-our experience.Sardesai RB, Krishnakumar T.Sardesai RB, et al.Indian J Otolaryngol Head Neck Surg. 2002 Apr;54(2):132-5. doi: 10.1007/BF02968732.Indian J Otolaryngol Head Neck Surg. 2002.PMID: 23119874 Free PMC article. 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17182	https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_(Arnold)/09%3A_Radical_Functions/9.02%3A_Multiplication_Properties_of_Radicals	9.2.3 9.2.4 9.2.5 9.2.7 9.2.8 9.2.9 9.2.10 9.2.11 9.2.13 9.2.14 9.2.15 9.2.1 9.2.2 9.2.3 9.2.4 9.2.5 9.2.6 9.2.7 9.2.8 9.2.9 9.2.10 9.2.11 9.2.12 9.2.13 9.2.14 9.2.15 9.2.16 9.2.17 9.2.18 9.2.19 9.2.20 9.2.21 9.2.22 9.2.23 9.2.24 9.2.25 9.2.26 9.2.27 9.2.28 9.2.29 9.2.30 9.2.31 9.2.32 9.2.33 9.2.34 9.2.35 9.2.36 9.2.37 9.2.38 9.2.39 9.2.40 9.2.41 9.2.42 9.2.43 9.2.44 9.2.45 9.2.46 9.2.47 9.2.48 9.2.49 9.2.50 9.2.51 9.2.52 9.2.53 9.2.54 9.2.55 9.2.56 9.2.57 9.2.58 9.2.59 9.2.60 9.2.61 9.2.62 9.2.63 9.2.64 9.2.65 9.2.66 9.2.67 9.2.68 9.2.69 9.2.70 9.2.71 9.2.72 9.2.73 9.2.74 9.2.75 9.2.76 9.2.77 9.2.78 9.2.79 9.2.80 50=25⋅2 50−−√=25−−√2–√=52–√ 98=49⋅2 98−−√=49−−√2–√=72–√ 288=144⋅2 288−−−√=144−−−√2–√=122–√ 2436510−−−−−−√=223355 223355=4⋅27⋅3125=337500 2533−−−−√=2432−−−−√2⋅3−−−−√ 2432−−−−√2⋅3−−−−√=22312⋅3−−−−√ 22312⋅3−−−−√=4⋅32⋅3−−−−√=126–√ 375275−−−−−√=365274−−−−−√3⋅7−−−−√ 365274−−−−−√3⋅7−−−−√=3351723⋅7−−−−√ 3351723⋅7−−−−√=27⋅5⋅493⋅7−−−−√=661521−−√ 2592=9⋅288 2592=9⋅(9⋅32) 2592=2534 2592−−−−√=2534−−−−√=2434−−−−√2–√=22322–√=4⋅92–√=362–√ 2592−−−−√=362–√ 2592−−−−√ 362–√ Skip to main content 9.2: Multiplication Properties of Radicals Last updated : Jun 3, 2023 Save as PDF 9.1: The Square Root Function 9.3: Division Properties of Radicals Page ID : 19735 David Arnold College of the Redwoods ( \newcommand{\kernel}{\mathrm{null}\,}) Recall that the equation x2=ax2=a, where a is a positive real number, has two solutions, as indicated in Figure 1. Here are the key facts. Solution Solutions of x2=ax2=a. If a is a positive real number, then: The equation x2=ax2=a has two real solutions. The notation √aa−−√ denotes the unique positive real solution. The notation −√a−a−−√ denotes the unique negative real solution. Note the use of the word unique. When we say that solution √aa−−√ is the unique positive real solution,we mean that it is the only one. There are no other positive real numbers that are solutions of x2=ax2=a. A similar statement holds for the unique negative solution. Thus, the equations x2=ax2=aand x2=bx2=bhave unique positive solutions x=√ax=a−−√ and x=√bx=b√, respectively, provide that a and b are positive real numbers. Furthermore, because they are solutions, they can be substituted into the equations x2=ax2=a and x2=bx2=bto produce the results (√a)2=a(a−−√)2=a and (√b)2=b(b√)2=b respectively. Again, these results are dependent upon the fact that a and b are positive real numbers. Similarly, the equation x2=abx2=ab has unique positive solution x=√abx=ab−−√, provided a and b are positive numbers. However, note that (√a√b)2=(√a)2(√b)2=ab(a−−√b√)2=(a−−√)2(b√)2=ab, making √a√ba−−√b√a second positive solution of x2=abx2=ab. However, because √abab−−√ is the unique positive solution of x2=abx2=ab, this forces √ab=√a√bab−−√=a−−√b√ This discussion leads to the following property of radicals. Property 1 Let aa and bb be positive real numbers. Then, √ab=√a√b ab−−√=a−−√b√ This result can be used in two distinctly different ways. You can use the result to multiply two square roots, as in √7√5=√357–√5–√=35−−√. √35=√7√535−−√=7–√5–√ It is interesting to check this result on the calculator, as shown in Figure 2. Simple Radical Form In this section we introduce the concept of simple radical form, but let’s first start with a little story. Martha and David are studying together, working a homework problem from their textbook. Martha arrived at an answer of √3232−−√, while David gets the result 2√828–√. At first, David and Martha believe that their solutions are different numbers, but they’ve been mistaken before so they decide to compare decimal approximations of their results on their calculators. Martha’s result is shown in Figure 3(a), while David’s is shown Figure 3(b). Martha finds that √32≈5.65685424932−−√≈5.656854249 and David finds that his solution 2√8≈5.65685424928–√≈5.656854249. David and Martha conclude that their solutions match, but they want to know why the two very different looking radical expressions are identical. The following calculation, using Property 1, shows why David’s result is identical to Martha’s. √32=√4√8=2√832−−√=4–√8–√=28–√ Indeed, there is even a third possibility, one that is much different from the results found by David and Martha. Consider the following calculation, which again uses Property 1. √32=√16√2=4√232−−√=16−−√2–√=42–√ In Figure 4, note that the decimal approximation of 4√242–√ approximations for √3232−−√ (Martha’s result in Figure 3(a)) and 2√828–√ (David’s result inFigure 3(b)). While all three of these radical expressions (√3232−−√, 2√828–√, and 4√242–√) are identical, it is somewhat frustrating to have so many different forms, particularly when we want to compare solutions. Therefore, we offer a set of guidelines for a special form of the answer which we will call simple radical form. The First Guideline for Simple Radical Form. When possible, factor out a perfect square. Thus, √3232−−√ is not in simple radical form, as it is possible to factor out a perfect square, as in √32=√16√2=4√232−−√=16−−√2–√=42–√. Similarly, David’s result (2√8)(28–√) is not in simple radical form, because he too can factor out a perfect square as follows. 2√8=2(√4√2)=2(2√2)=(2⋅2)√2=4√228–√=2(4–√2–√)=2(22–√)=(2⋅2)2–√=42–√. If both Martha and David follow the “first guideline for simple radical form,” their answer will look identical (both equal 4√242–√). This is one of the primary advantages of simple radical form: the ability to compare solutions. In the examples that follow (and in the exercises), it is helpful if you know the squares of the first 25 positive integers. We’ve listed them in the margin for you in Table 1 for future reference. Let’s place a few more radical expressions in simple radical form. Example 9.2.39.2.3 Place √5050−−√ in simple radical form. Answer : In Table 1, 25 is a square. Because 50=25⋅2, we can use Property 1 to write √50=√25√2=5√2. Example 9.2.49.2.4 Place √9898−−√ in simple radical form. Answer : In Table 1, 49 is a square. Because 98=49⋅2, we can use Property 1 to write √98=√49√2=7√2. Example 9.2.59.2.5 Place √288288−−−√ in simple radical form. Answer : Some students seem able to pluck the optimal “perfect square” out of thin air. If you consult Table 1, you’ll note that 144 is a square. Because 288=144⋅2, we can write √288=√144√2=12√2. However, what if you miss that higher perfect square, think 288=4⋅72288=4⋅72, and write √288=√4√72=2√72288−−−√=4–√72−−√=272−−√. This approach is not incorrect, provided you realize that you’re not finished. You can still factor a perfect square out of 72. Because 72=36⋅272=36⋅2, you can continue and write 2√72=2(√36√2)=2(6√2)=(2⋅6)√2=12√2272−−√=2(36−−√2–√)=2(62–√)=(2⋅6)2–√=122–√. Note that we arrived at the same simple radical form, namely 12√2122–√. It just took us a little longer. As long as we realize that we must continue until we can no longer factor out a perfect square, we’ll arrive at the same simple radical form as the student who seems to magically pull the higher square out of thin air. Indeed, here is another approach that is equally valid. √288=√4√72=2(√4√18)=2(2√18)=(2⋅2)√18=4√18288−−−√=4–√72−−√=2(4–√18−−√)=2(218−−√)=(2⋅2)18−−√=418−−√ We need to recognize that we are still not finished because we can extract another perfect square as follows. 4√18=4(√9√2)=4(3√2)=(4⋅3)√2=12√2418−−√=4(9–√2–√)=4(32–√)=(4⋅3)2–√=122–√ Once again, same result. However, note that it behooves us to extract the largest square possible, as it minimizes the number of steps required to attain simple radical form. Checking Results with the Graphing Calculator. Once you’ve placed a radical expression in simple radical form, you can use your graphing calculator to check your result. In this example, we found that √288=12√2288−−−√=122–√. (6) Enter the left- and right-hand sides of this result as shown in Figure 5. Note that each side produces the same decimal approximation, verifying the result in equation (6). Helpful Hints Recall that raising a power of a base to another power requires that we multiply exponents. Raising a Power of a Base to another Power. (am)n=amn(am)n=amn In particular, when you square a power of a base, you must multiply the exponent by 2. For example, (25)2=210(25)2=210. Conversely, because taking a square root is the “inverse” of squaring,when taking a square root we must divide the existing exponent by 2, as in √210=25210−−−√=25. Note that squaring 2525 gives 210210, so taking the square root of 210210 must return you to 2525. When you square, you double the exponent. Therefore, when you take the square root, you must halve the exponent. Similarly, (26)2=212(26)2=212 so √212=26212−−−√=26. (27)2=214(27)2=214 so √214=27214−−−√=27. (28)2=216(28)2=216 so √216=28216−−−√=28. This leads to the following result. Taking the Square Root of an Even Power. When taking a square root of xnxn, when x is a positive real number and n is an even natural number, divide the exponent by two. In symbols, √xn=xn2xn−−√=xn2. Note that this agrees with the definition of rational exponents presented in Chapter 8, as in √xn=(xn)12=xn2xn−−√=(xn)12=xn2. On another note, recall that raising a product to a power requires that we raise each factor to that power. Raising a Product to a Power. (ab)n=anbn(ab)n=anbn. In particular, if you square a product, you must square each factor. For example, (5374)2=(53)2(74)2=5678(5374)2=(53)2(74)2=5678. Note that we multiplied each existing exponent in this product by 2. Property 1 is similar, in that when we take the square root of a product, we take the square root of each factor. Because taking a square root is the inverse of squaring, we must divide each existing exponent by 2, as in √5678=√56√78=53745678−−−−√=56−−√78−−√=5374 Let’s look at some examples that employ this technique. Example 9.2.79.2.7 Simplify √24365102436510−−−−−−√ Answer : When taking the square root of a product of exponential factors, divide each exponent by 2. √2436510=223355 If needed, you can expand the exponential factors and multiply to provide a single numerical answer. 223355=4⋅27⋅3125=337500 A calculator was used to obtain the final solution. Example 9.2.89.2.8 Simplify √25332533−−−−√ Answer : In this example, the difficulty is the fact that the exponents are not divisible by 2. However, if possible, the “first guideline of simple radical form” requires that we factor out a perfect square. So, extract each factor raised to the highest possible power that is divisible by 2, as in √2533=√2432√2⋅3 Now, divide each exponent by 2. √2432√2⋅3=2231√2⋅3 Finally, simplify by expanding each exponential factor and multiplying. 2231√2⋅3=4⋅3√2⋅3=12√6 Example 9.2.99.2.9 Simplify √375275375275−−−−−√. Answer : Extract each factor to the highest possible power that is divisible by 2. √375275=√365274√3⋅7 Divide each exponent by 2. √365274√3⋅7=335172√3⋅7 Expand each exponential factor and multiply. 335172√3⋅7=27⋅5⋅49√3⋅7=6615√21 Example 9.2.109.2.10 Place √216216−−−√ in simple radical form. If we prime factor 216, we can attack this problem with the same technique used in the previous examples. Before we prime factor 216, here are a few divisibility tests that you might find useful. Divisibility Test If a number ends in 0, 2, 4, 6, or 8, it is an even number and is divisible by 2. If the last two digits of a number form a number that is divisible by 4, then the entire number is divisible by 4. If a number ends in 0 or 5, it is divisible by 5. If the sum of the digits of a number is divisible by 3, then the entire number is divisible by 3. If the sum of the digits of a number is divisible by 9, then the entire number is divisible by 9. For example, in order: The number 226 ends in a 6, so it is even and divisible by 2. Indeed, 226=2⋅113226=2⋅113. The last two digits of 224 are 24, which is divisible by 4, so the entire number is divisible by 4. Indeed, 224=4⋅56224=4⋅56. The last digit of 225 is a 5. Therefore 225 is divisible by 5. Indeed, 225=5⋅45225=5⋅45. The sum of the digits of 222 is 2 + 2 + 2 = 6, which is divisible by 3. Therefore, 222 is divisible by 3. Indeed, 222=3⋅74222=3⋅74. The sum of the digits of 684 is 6 + 8 + 4 = 18, which is divisible by 9. Therefore, 684 is divisible by 9. Indeed, 684=9⋅76684=9⋅76. Now, let’s prime factor 216. Note that 2+1+6 = 9, so 216 is divisible by 9. Indeed, 216=9⋅24216=9⋅24. In Figure 6, we use a “factor tree” to continue factoring until all of the “leaves” are prime numbers. Thus, 216=2⋅2⋅2⋅3⋅3⋅3216=2⋅2⋅2⋅3⋅3⋅3, or in exponential form, 216=23⋅33216=23⋅33. Thus, √216=√2333=√2232√2⋅3=2⋅3√2⋅3=6√6216−−−√=2333−−−−√=2232−−−−√2⋅3−−−√=2⋅32⋅3−−−√=66–√. Prime factorization is an unbelievably useful tool! Let’s look at another example. Example 9.2.119.2.11 Place √25922592−−−−√ in simple radical form. Answer : If we find the prime factorization for 2592, we can attack this example using the same technique we used in the previous example. We note that the sum of the digits of 2592 is 2 + 5 + 9 + 2 = 18, which is divisible by 9. Therefore, 2592 is also divisible by 9. 2592=9⋅288 The sum of the digits of 288 is 2+8+8 = 18, which is divisible by 9, so 288 is also divisible by 9. 2592=9⋅(9⋅32) Continue in this manner until the leaves of your “factor tree” are all primes. Then, you should get 2592=2534. Thus, √2592=√2534=√2434√2=2232√2=4⋅9√2=36√2. Let’s use the graphing calculator to check this result. Enter each side of √2592=36√2 separately and compare approximations, as shown in Figure 7. An Important Property of Square Roots One of the most common mistakes in algebra occurs when practitioners are asked to simplify the expression √x2x2−−√, where x is any arbitrary real number. Let’s examine two of the most common errors. Some will claim that the following statement is true for any arbitrary real number x. √x2=±xx2−−√=±x. This is easily seen to be incorrect. Simply substitute any real number for x to check this claim. We will choose x = 3 and substitute it into each side of the proposed statement. √32=±332−−√=±3. If we simplify the left-hand side, we produce the following result. √32=±332−−√=±3. 3=±33=±3 It is not correct to state that 3 and ±3±3 are equal. A second error is to claim that √x2=xx2−−√=x for any arbitrary real number x. Although this is certainly true if you substitute nonnegative numbers for x, look what happens when you substitute −3 for x. √(−3)2=3(−3)2−−−−−√=3 If we simplify the left-hand side, we produce the following result. √9=−39–√=−3 3 = −3 Clearly, 3 and −3 are not equal. In both cases, what has been forgotten is the fact that √√ calls for a positive (nonnegative if you want to include the case √00–√) square root. In both of the errors above, namely √x2=±xx2−−√=±x and √x2=xx2−−√=x, the left-hand side is calling for a nonnegative response, but nothing has been done to insure that the right-hand side is also nonnegative. Does anything come to mind? Sure, if we wrap the right-hand side in absolute values, as in √x2=\|x\|x2−−√=\|x\|, then both sides are calling for a nonnegative response. Indeed, note that √(−3)2=\|−3\|(−3)2−−−−−√=\|−3\|, √02=\|0\|02−−√=\|0\|, and √32=\|3\|32−−√=\|3\| are all valid statements. This discussion leads to the following result. The positive Square Root of the square of x If x is any real number, then √x2=\|x\|x2−−√=\|x\|, The next task is to use this new property to produce a extremely useful property of absolute value. A Multiplication Property of Absolute Value If we combine the law of exponents for squaring a product with our property for taking the square root of a product, we can write √(ab)2=√a2b2=√a2√b2(ab)2−−−−√=a2b2−−−−√=a2−−√b2−−√. However, √(ab)2=\|ab\|(ab)2−−−−√=\|ab\|, while √a2b2=\|a\|\|b\|a2−−√b2=\|a\|\|b\|. This discussion leads to the following result. Product rule for absolute value If a and b are any real numbers, \|ab\| = \|a\|\|b\|. In words, the absolute value of a product is equal to the product of the absolute values. We saw this property previously in the chapter on the absolute value function, where we provided a different approach to the proof of the property. It’s interesting that we can prove this property in a completely new way using the properties of square root. We’ll see we have need for the Product Rule for Absolute Value in the examples that follow. For example, using the product rule, if x is any real number, we could write \|3x\| = \|3\|\|x\| = 3\|x\| However, there is no way we can remove the absolute value bars that surround x unless we know the sign of x. If x≥0x≥0, then \|x\| = x and the expression becomes 3\|x\| = 3x. On the other hand, if x < 0, then \|x\| = −x and the expression becomes 3\|x\| = 3(−x) = −3x. Let’s look at another example. Using the product rule, if x is any real number, the expression \|−4x3\|\|−4x3\| can be manipulated as follows. \|−4x3\|=\|−4\|\|x2\|\|x\|\|−4x3\|=\|−4\|\|x2\|\|x\| However,\|−4\|=4 and since x2≥0x2≥0 for any value of x, \|x2\|=x2\|x2\|=x2. Thus, \|−4\|\|x2\|\|x\|=4x2\|x\|\|−4\|\|x2\|\|x\|=4x2\|x\|. Again, there is no way we can remove the absolute value bars around x unless we know the sign of x. If x≥0x≥0, then\|x\|=x and 4x2\|x\|=4x2(x)=4x34x2\|x\|=4x2(x)=4x3. On the other hand, if x < 0, then \|x\| = −x and 4x2\|x\|=4x2(−x)=−4x34x2\|x\|=4x2(−x)=−4x3. Let’s use these ideas to simplify some radical expressions that contain variables. Variable Expressions Example 9.2.139.2.13 Given that the x represents any real numbers, place the radical expression √48x648x6−−−−√ in simple radical form. Simple radical form demands that we factor out a perfect square, if possible. In this case, 48=16⋅348=16⋅3 and we factor out the highest power of x that is divisible by 2. √48x6=√16x6√348x6−−−−√=16x6−−−−√3–√ We can now use Property 1 to take the square root of each factor. √16x6√3=√16√x6√316x6−−−−√3–√=16−−√x6−−√3–√ Now, remember the notation √√ calls for a nonnegative square root, so we must insure that each response in the equation above is nonnegative. Thus, √16√x6√3=4\|x3\|√316−−√x6−−√3–√=4\|x3\|3–√ Some comments are in order. The nonnegative square root of 16 is 4. That is, √16=416−−√=4 The nonnegative square root of x6 is trickier. It is incorrect to say √x6=x3x6−−√=x3, because x3x3 could be negative (if x is negative). To insure a nonnegative square root, in this case we need to wrap our answer in absolute value bars. That is, √x6=\|x3\|x6−−√=\|x3\|. We can use the Product Rule for Absolute Value to write \|x3\|=\|x2\|\|x\|\|x3\|=\|x2\|\|x\|. Because x2x2 is nonnegative, absolute value bars are redundant and not needed. That is, \|x2\|\|x\|=x2\|x\|\|x2\|\|x\|=x2\|x\|. Thus, we can simplify our solution a bit further and write 4\|x3\|√3=4x2\|x\|√34\|x3\|3–√=4x2\|x\|3–√ Thus, √48x6=4x2\|x\|√348x6−−−−√=4x2\|x\|3–√. Alternate Solution. There is a variety of ways that we can place a radical expression in simple radical form. Here is another approach. Starting at the step above, where we first factored out a perfect square, √48x6=√16x6√348x6−−−−√=16x6−−−−√3–√ we could write √16x6√3=√(4x3)2√316x6−−−−√3–√=(4x3)2−−−−−√3–√. Now, remember that the nonnegative square root of the square of an expression is the absolute value of that expression (we have to guarantee a nonnegative answer), so √(4x3)2√3=\|4x3\|√3. However, \|4x3\|=\|4\|\|x3\| by our product rule and \|4\|\|x3\|=4\|x3\|. Thus, \|4x3\|√3=4\|x3\|√3. Finally, \|x3\|=\|x2\|\|x\|=x2\|x\| because x2≥0, so we can write 4\|x3\|√3=4x2\|x\|√3 We cannot remove the absolute value bar that surrounds x unless we know the sign of x. Note that the simple radical form in the alternate solution is identical to the simple radical form found with the previous solution technique. Let’s look at another example. Example 9.2.14 Given that x < 0, place √24x6 in simple radical form. First, factor out a perfect square and write √24x6=√4x6√6 Now, use Property 1 and take the square root of each factor. √4x6√6=√4√x6√6 To insure a nonnegative response to √x6, wrap your response in absolute values. √4√x6√6=2\|x3\|√6 However, as in the previous problem, \|x3\|=\|x2\|\|x\|=x2\|x\|, since x2≥0. Thus, 2\|x3\|√6=2x2\|x\|√6. In this example, we were given the extra fact that x < 0, so \|x\| = −x and we can write 2x2\|x\|√6=2x2(−x)√6=−2x3√6. It is instructive to test the validity of the answer √24x6=−2x3√6, x<0. using a calculator. It is somewhat counterintuitive that the result √24x6=−2x3√6, x<0. contains a negative sign. After all, the expression √24x6 call for a nonnegative result, but we have a negative sign. However, on closer inspection, if x < 0, then x is a negative number and the right-hand side −2x3√6 is a positive number (−2 is negative, x3 is negative because x is negative, and the product of two negatives is a positive). Let’s look at another example. Example 9.2.15 If x < 3,simplify √x2−6x+9. The expression under the radical is a perfect square trinomial and factors. √x2−6x+9=√(x−3)2 However, the nonnegative square root of the square of an expression is the absolute value of that expression, so √(x−3)2=\|x−3\|. Finally, because we are told that x < 3, this makes x − 3 a negative number, so \|x−3\| = −(x−3). Again, the result √x2−6x+9=−(x−3), provided x < 3, is somewhat counterintuitive as we are expecting a positive result. However, if x < 3, the result −(x−3) is positive. You can test this by substituting several values of x that are less than 3 into the expression −(x−3) and noting that the result is positive. For example, if x = 2, then x is less than 3 and −(x−3) = −(2−3) = −(−1) = 1, which, of course, is a positive result. It is even more informative to note that our result is equivalent to √x2−6x+9=−x+3, x<3. This is easily seen by distributing the minus sign in the result −(x−3). We’ve drawn the graph of y=√x2−6x+9 on our calculator on Figure 9(a). In Figure 9(b), we’ve drawn the graph of y = −x + 3. Note that the graphs agree when x < 3. Indeed, when you consider the left-hand branch of the “V” in Figure 9(a), you can see that the slope of this branch is −1 and the y-intercept is 3. The equation of this branch is y = −x+3, so it agrees with the graph of y = −x+3 in Figure 9(b) when x is less than 3. Exercise 9.2.1 Use a calculator to first approximate √2√5. One the same screen, approximate √10. Report the results on your homework paper. Answer : Note that √5√2=√10≈3.16227766. Exercise 9.2.2 Use a calculator to first approximate √7√10. One the same screen, approximate √70. Report the results on your homework paper. Exercise 9.2.3 Use a calculator to first approximate √3√11. One the same screen, approximate √33. Report the results on your homework paper. Answer : Note that √3√11=√33≈5.744562647 Exercise 9.2.4 Use a calculator to first approximate √5√13. One the same screen, approximate √65. Report the results on your homework paper. In Exercises 5-20, place each of the radical expressions in simple radical form. As in Example 3 in the narrative, check your result with your calculator. Exercise 9.2.5 √18 Answer : √18=√32⋅2=√32√2=3√2 Exercise 9.2.6 √80 Exercise 9.2.7 √112 Answer : √112=√42⋅7=√42√7=4√7 Exercise 9.2.8 √72 Exercise 9.2.9 √108 Answer : √108=√62⋅3=√62√3=6√3 Exercise 9.2.10 √54 Exercise 9.2.11 √50 Answer : √50=√52⋅2=√52√2=5√2 Exercise 9.2.12 √48 Exercise 9.2.13 √245 Answer : √245=√72⋅5=√72√5=7√5 Exercise 9.2.14 √150 Exercise 9.2.15 √98 Answer : √98=√72⋅2=√72√2=7√2 Exercise 9.2.16 √252 Exercise 9.2.17 √45 Answer : √45=√32⋅5=√32√5=3√5 Exercise 9.2.18 √294 Exercise 9.2.19 √24 Answer : √24=√22⋅6=√22√6=2√6 Exercise 9.2.20 √32 In Exercises 21-26, use prime factorization (as in Example 10 and 11 in the narrative) to assist you in placing the given radical expression in simple radical form. Check your result with your calculator. Exercise 9.2.21 √2016 Answer : Note that 2+0+1+6 = 9, which is divisible by 9. Thus, 2016 is divisible by 9. Indeed, 2019=9⋅224 The last two digits of 224 are 24, which is divisible by 4. Thus, 224 is divisible by 4. Indeed, 224=4⋅56. 2016=9⋅224=(3⋅3)⋅(4⋅56). Continue to primes. 2016=3⋅3⋅2⋅2⋅2⋅2⋅2⋅7=25⋅32⋅7. Checking, Exercise 9.2.22 √2700 Exercise 9.2.23 √14175 Answer : Money! Anything that ends in 00, 25, 50, or 75 is divisible by 25. Indeed, 14175=25⋅567. Further, 5+6+7 = 18, so 567 is divisible by 9; i.e., 567=9⋅63. Continuing to primes, 14175=25⋅567=5⋅5⋅9⋅63=5⋅5⋅3⋅3⋅3⋅3⋅7. Factor our a perfect square (exponents divisible by 2). $\sqrt{14175} = \sqrt{3^{4} \cdot 5^{2}} \cdot \sqrt{7} = 3^{2} \cdot5\sqrt{7} = 45\sqrt{7}$ Checking, Exercise 9.2.24 √44000 Exercise 9.2.25 √20250 Answer : Money! Anything that ends in 00, 25, 50, or 75 is divisible by 25. Indeed, 20250=25⋅810. Continuing to primes, 20250=5⋅5⋅9⋅9⋅10=5⋅5⋅3⋅3⋅3⋅3⋅2⋅5. Factor out a perfect square. √20250=√2⋅34⋅53=√34⋅52⋅√2⋅5=32⋅5√2⋅5=45√10 Checking, Exercise 9.2.26 √3564 In Exercises 27-46, place each of the given radical expressions in simple radical form. Make no assumptions about the sign of the variables. Variables can either represent positive or negative numbers. Exercise 9.2.27 √(6x−11)4 Answer : √(6x−11)4=√((6x−11)2)2=\|(6x−11)2\| However, (6x−11)2 is already nonnegative, so the absolute value bars are unnecessary. Hence, √(6x−11)4=(6x−11)2 Exercise 9.2.28 √16h8 Exercise 9.2.29 √25f2 Answer : √25f2=√25√f2=5\|f\| Because f can be any real number, we cannot remove the absolute value bars without more information. Exercise 9.2.30 √25j8 Exercise 9.2.31 √16m2 Answer : √16m2=√42m2=√42√m2=4\|m\| Since the index on the radical is even and, after simplification, the variable is raised to an odd power, absolute value signs around the simplified variable are necessary. Exercise 9.2.32 √25a2 Exercise 9.2.33 √(7x+5)12 Answer : √(7x+5)12=√((7x+5)6)2=\|(7x+5)6\| However, (7x+5)6 is already nonnegative, so absolute value signs are unnecessary. √(7x+5)12=(7x+5)6 Exercise 9.2.34 √9w10 Exercise 9.2.35 √25x2−50x+25 Answer : √25x2−50x+25=√(5x−5)2=\|5x−5\| Because x can be any real number, the absolute value signs around the simplified binomial are necessary. Exercise 9.2.36 √49x2−42x+9 Exercise 9.2.37 √25x2+90x+81 Answer : √25x2+90x+81=√(5x+9)2=\|5x+9\| Because x can be any real number, the absolute value signs around the simplified binomial are necessary. Exercise 9.2.38 √25f14 Exercise 9.2.39 √(3x+6)12 Answer : √(3x+6)12=√((3x+6)6)2=\|(3x+6)6\| However, the expression (3x+6)6 is already nonnegative, so the absolute value bars are unnecessary. √(3x+6)12=(3x+6)6 Exercise 9.2.40 √(9x−8)12 Exercise 9.2.41 √36x2+36x+9 Answer : √36x2+36x+9=√(6x+3)2=\|6x+3\| Because x can be any real number, the absolute value signs around the simplified binomial are necessary. Exercise 9.2.42 √4e2 Exercise 9.2.43 √4p10 Answer : √4p10=√4√(p5)2=2\|p5\| Now, we can use the multiplicative property of absolute values and write 2\|p5\|=2\|p4\|\|p\|=2p4\|p\|. Since p can be any real number, absolute value signs around the simplified variable are necessary. Exercise 9.2.44 √25x12 Exercise 9.2.45 √25q6 Answer : √25q6=√25√(q3)2=5\|q3\| Now, we can use the multiplicative property of absolute values and write 5\|q3\|=5\|q2\|\|q\|=5q2\|q\|. Because q can be any real number, absolute value signs around the simplified variable are necessary. Exercise 9.2.46 √16h12 Exercise 9.2.47 Given that x < 0, place the radical expression √32x6 in simple radical form. Check your solution on your calculator for x = −2. Answer : Factor out a perfect square. √32x6=√16x6√2=√16√x6√2=4\|x3\|√2 However, \|x3\|=\|x2\|\|x\|=x2\|x\|, since x2≥0. Thus √32x6=4x2\|x\|√2. If x < 0, then \|x\| = −x and √32x6=4x2(−x)√2=−4x3√2. Checking with x = −2. Exercise 9.2.48 Given that x < 0, place the radical expression √54x8 in simple radical form. Check your solution on your calculator for x = −2. Exercise 9.2.49 Given that x < 0, place the radical expression √27x12 in simple radical form. Check your solution on your calculator for x = −2. Answer : Factor out a perfect square. √27x12=√9x12√3=√9√x12√3=3\|x6\|√3 However, \|x6\|=x6 since x6≥0. Thus, √27x12=3x6√3 Checking with x = −2. Exercise 9.2.50 Given that x < 0, place the radical expression √44x10 in simple radical form. Check your solution on your calculator for x = −2. In Exercises 51-54, follow the lead of Example 17 in the narrative to simplify the given radical expression and check your result with your graphing calculator. Exercise 9.2.51 Given that x < 4, place the radical expression √x2−8x+16 in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that x < 4. Answer : Factor the perfect square trinomial. √x2−8x+16=√(x−4)2=\|x−4\| If x < 4, or equivalently, if x−4 < 0, then \|x−4\| = −(x−4). Thus, √x2−8x+16=−x+4. In (b), we've drawn the graph of y=√x2−8x+16. In (d), we've drawn the graph of y = −x+4. Note that the graphs in (b) and (d) agree when x < 4, leading credence to the fact that √x2−8x+16=−x+4 when x < 4. Exercise 9.2.52 Given that x≥−2, place the radical expression √x2+4x+4 in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that x≥−2. Exercise 9.2.53 Given that x≥5, place the radical expression √x2−10x+25 in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that x≥5. Answer : Factor the perfect square trinomial. √x2−10x+25=√(x−5)2=\|x−5\| If x≥5, or equivalently, if x−5≥0, then \|x−5\| = x−5. Thus, √x2−8x+16=x−5. In (b), we've drawn the graph of y=√x2−10x+25. In (d), we've drawn the graph of y = x−5. Note that the graphs in (b) and (d) agree when x≥5, leading credence to the fact that √x2−10x+25=x−5 when x≥5. Exercise 9.2.54 Given that x < −1, place the radical expression √x2+2x+1 in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that x < −1. In Exercises 55-72, place each radical expression in simple radical form. Assume that all variables represent positive numbers. Exercise 9.2.55 √9d13 Answer : √9d13=√9√d12√d=3d6√d Exercise 9.2.56 √4k2 Exercise 9.2.57 √25x2+40x+16 Answer : √25x2+40x+16=√(5x+4)2=5x+4 Exercise 9.2.58 √9x2−30x+25 Exercise 9.2.59 √4j11 Answer : √4j11=√4√j10√j=3j5√j Exercise 9.2.60 √16j6 Exercise 9.2.61 √25m2 Answer : √25m2=√25√m2=5m Exercise 9.2.62 √9e9 Exercise 9.2.63 √4c5 Answer : √4c5=√4c4√c=2c2√c Exercise 9.2.64 √25z2 Exercise 9.2.65 √25h10 Answer : √25h10=√25√h10=5h5 Exercise 9.2.66 √25b2 Exercise 9.2.67 √9s7 Answer : √9s7=√9s6√s=3s3√s Exercise 9.2.68 √9e7 Exercise 9.2.69 √4p8 Answer : √4p8=√4√p8=2√p4 Exercise 9.2.70 √9d15 Exercise 9.2.71 √9q10 Answer : √9q10=√9√q10=3√q5 Exercise 9.2.72 √4w7 In Exercises 73-80, place each given radical expression in simple radical form. Assume that all variables represent positive numbers. Exercise 9.2.73 √2f5√8f3 Answer : √2f5√8f3=√2⋅8⋅f5⋅f3=√16f8=√16√(f4)2=4f4 Exercise 9.2.74 √3s3√243s3 Exercise 9.2.75 √2k7√32k3 Answer : √2k7√32k3=√2⋅32⋅k7⋅k3=√64k10=√64√(k5)2=8k5 Exercise 9.2.76 √2n9√8n3 Exercise 9.2.77 √2e9√8e3 Answer : √2e9√8e3=√2⋅8⋅e9⋅e3=√16e12=√16√(e6)2=4e6 Exercise 9.2.78 √5n9√125n3 Exercise 9.2.79 √3z5√27z3 Answer : √3z5√27z3=√3⋅27⋅z5⋅z3=√81z8=√81√(z4)2=9z4 Exercise 9.2.80 √3t7√27t3 9.1: The Square Root Function 9.3: Division Properties of Radicals
17183	https://www.chegg.com/homework-help/questions-and-answers/consider-following-initial-value-problem-y-y-f-t-y-0-1-y-0-0-f-t-1-0-q70459851	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Consider the following initial-value problem. y" + y = f(t), y(0) = 1, y'(0) = 0, where f(t) = п 1, 0 2 Take the Laplace transform of the differential equation and solve for L{y}. (Write your answer as a function of s.) L{y} = (52 +1)2 x Use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms as needed. y(t) = Given that: The initial value problem y″+y=f(t) where f(t)={10≤t<π2sin⁡tt≥π2 and y(0)=1,y′(0)=0. Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
17184	https://www.sciencedirect.com/topics/immunology-and-microbiology/colony-forming-unit	Skip to Main content My account Sign in Colony-Forming Unit In subject area:Immunology and Microbiology Colony-forming units (CFUs) refer to in vitro assays that evaluate the growth of hematopoietic cells by measuring the formation of colonies, utilizing molecular probes and automated enumeration platforms, despite challenges related to sensitivity and specificity. AI generated definition based on: Perinatal Stem Cells, 2018 How useful is this definition? Add to Mendeley Also in subject areas: Medicine and Dentistry Neuroscience Veterinary Science and Veterinary Medicine Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Electrically-enhanced proliferation control of cancer-stem-cells-like adult human mesenchymal stem cells – a novel modality of treatment 2014, Electroporation-Based Therapies for CancerKavitha Sankaranarayanan, ... Raji Sundararajan Colony forming unit assay The colony forming unit (CFU) is a measure of viable colonogenic cell numbers in CFU/mL. These are an indication of the number of cells that remain viable enough to proliferate and form small colonies. Isolated hMSCs were plated in a 6-well cell culture plate along with 2–3 mL of DMEM medium. Then 100 μL of the sample, along with 700 μL of DMEM medium are taken in a 4-mm gap cuvette and electroporated at 1200 V/cm, using eight 15 μs pulses at 1 second intervals. After electroporation, the cells were seeded onto tissue culture dishes in DMEM with antibiotics and maintained at 37 °C in a 5% CO2 atmosphere. They were undisturbed for 14 days until colonies of cells appeared. The dishes were washed with PBS and cells were stained with 0.1% Crystal Violet in Methanol and incubated at room temperature for 10 minutes. The staining solution was discarded, and the dishes were observed under a microscope. A collection of cells of around 100 mm2 (~ 50 cells) was counted as 1 colony. View chapterExplore book Read full chapter URL: Book2014, Electroporation-Based Therapies for CancerKavitha Sankaranarayanan, ... Raji Sundararajan Review article Ultraclean air systems and the claim that laminar airflow systems fail to prevent deep infections after total joint arthroplasty 2019, Journal of Hospital InfectionW. Whyte, B. Lytsy Terms used in article The term ‘colony-forming unit (cfu)’ is often used to describe airborne micro-organisms but this description is more an acknowledgement that a micro-organism is viable than a description of a micro-organism in the air of an operating theatre. Micro-organisms are not normally found as unicellular organisms in the air of operating theatres, as they are almost entirely dispersed from personnel on skin cells . ‘Microbe-carrying particle (MCP)’ is a more apt descriptive term and is used in this article. Two types of ventilation system are used worldwide in operating theatres. The most common type is known as ‘conventionally ventilated’, in which about 20 air changes per hour of filtered air are supplied through ceiling inlets to mix, dilute, and remove airborne contamination out through low-level air extracts. The other type is unidirectional airflow (UDAF), which has a bank of air filters in the ceiling that are about 3 m × 3 m. The filters supply a steady flow of air in parallel streams, sweeping away MCPs from the sterile field. UDAF systems supply considerably more air than conventional systems, and, if correctly designed, they reduce the MCP concentration by about 100-fold (see Table I). Table I. Airborne microbe-carrying particle concentrations per m3 \| Ventilation system \| Clothing type \| --- \| \| Conventionala \| Total body exhaust gowns \| \| Conventionally ventilated \| 164 \| 51 \| \| Allanderb \| 49 \| 14 \| \| Horizontal UDAF \| 22 \| 1 \| \| Vertical UDAF, without walls \| 10 \| – \| \| Vertical UDAF, with walls \| 2 \| 0.4 \| \| Isolatorc \| 0.5 \| – \| UDAF, unidirectional airflow. a : Conventional clothing was the standard type of cotton shirt, trousers, and gown. b : A low velocity downward airflow system with air curtains round the air supply perimeter . c : This system is described by Trexler . UDAF systems are often called laminar airflow (LAF) systems. However, it is scientifically incorrect to use this term, as the air velocity is too high to be ‘laminar’, and the airflow is turbulent . The name used throughout the cleanroom industry is ‘unidirectional airflow’ and this term is used in this article. However, where published articles use the term, it is quoted as ‘LAF’. As the airflow in both the UDAF and conventional system is turbulent, it is best not to use ‘turbulent’ only when describing the conventional system. It is therefore called ‘conventionally ventilated operation theatre’ in this article. The term ‘ultraclean air (UCA)’ is used in this article to describe airborne conditions that give an average concentration of MCPs at the surgical wound of ≤10/m3, and preferably ≤1/m3. A UCA system normally consists of a well-designed UDAF ventilation system, with the surgical team wearing occlusive clothing that reduces their airborne dispersion. View article Read full article URL: Journal2019, Journal of Hospital InfectionW. Whyte, B. Lytsy Chapter Electrically-enhanced proliferation control of cancer-stem-cells-like adult human mesenchymal stem cells – a novel modality of treatment 2014, Electroporation-Based Therapies for CancerKavitha Sankaranarayanan, ... Raji Sundararajan 6.7.1Colony forming units hMSCs have the tendency to form colonies upon plating in suitable culture conditions. The number of such colonies is a direct indicator of the clonogenic ability of the cell, which is a measure of its proliferative capacity (Figure 6.4). When freshly isolated hMSCs were used for the CFU assay after electroporation (1200 V/cm, 100 μs), a significant reduction in the number of CFUs was observed (Figure 6.5). The same number of cells was used in both the control as well as electroporated samples; hence it could be safely inferred that electroporation exerts a negative effect on the colonogenic ability of hMSCs. This indicates that there is a reduction in the number of healthy MSCs capable of proliferation after electroporation. View chapterExplore book Read full chapter URL: Book2014, Electroporation-Based Therapies for CancerKavitha Sankaranarayanan, ... Raji Sundararajan Chapter Bacteriophage Therapy: Potential and Problems 2009, Encyclopedia of Microbiology (Third Edition)H. Brüssow Glossary CFU : The colony-forming units express the number of viable bacteria on a specified growth medium. commensalism : Relationship between two organisms where one benefits and the other is unaffected. Myoviridae : A phage with a head and a long contractile tail structure. nosocomial infection : Hospital- or health care center-acquired infection. phage : Short-hand notation for bacteriophage (Greek ‘bacterium eater’), a virus infecting bacteria. PFU : The plaque-forming units express the titer of infectious virus on the specified host cell. temperate phage : A phage that has two lifestyle options – lytic infection and lysogeny, that is, integration of its genome as a prophage into the bacterial chromosome. virulence gene : A gene that confers a pathogenic potential to a bacterium, sometimes encoded by a prophage. virulent phage : A phage that propagates exclusively by serial lytic infections. View chapterExplore book Read full chapter URL: Reference work2009, Encyclopedia of Microbiology (Third Edition)H. Brüssow Chapter Volume 2 2019, Encyclopedia of Tissue Engineering and Regenerative MedicineAkira Fujita, ... Tohru Hosoyama Cardiac Colony-Forming Unit Fibroblasts In 1970s, Friendenstein et al. identified a specific cell population named as colony-forming unit fibroblasts (CFU-Fs) in bone marrow-derived cells. These cells, which possess similar multipotent phenotypes to those of mesenchymal stromal/stem cells (MSCs), can be expanded for long-term in vitro. CFU-Fs are also found in various tissues/organs in which they play a role in tissue homeostasis and repair as tissue stem cells. CFU-Fs have been found in the heart, and these “cardiac” CFU-Fs showed multidifferentiation capacity to become cardiomyocytes, endothelial cells, and smooth muscle cells in vitro. In contrast to CFU-Fs from other tissues, cardiac CFU-Fs express high amounts of Mef2c and Tbx4, which are essential transcriptional factors in heart development. In addition, cardiac CFU-Fs are negative for the pan-hematopoietic makers CD45 and CD31, but express hematopoietic stem cell makers Sca-1 and platelet-derived growth factor receptor α. It is still controversial whether cardiac CFU-Fs can be categorized as CPCs because of their MSC-like phenotype. However, it has been demonstrated by linage tracing technique using genetically engineered mice that cardiac CFU-Fs mainly originated not from the bone marrow but from the proepicardium/epicardium, the area thought to be a source of cardiac stem/progenitors in developing heart. This suggests that cardiac CFU-Fs share the origin and phenotypes with CPCs. View chapterExplore book Read full chapter URL: Reference work2019, Encyclopedia of Tissue Engineering and Regenerative MedicineAkira Fujita, ... Tohru Hosoyama Chapter Endothelial Progenitor Cells and the Kidney 2011, Regenerative NephrologyMatthieu Monge, ... Ton J. Rabelink Endothelial Progenitor Cells and Colony-forming Units Two main methods have been described that define EPCs by their performance in colony-forming assays: the colony-forming unit–Hill (CFU-Hill) and the endothelial colony-forming cell (ECFC) assays . Both are based on the number of colonies formed by mononuclear cells plated on a fibronectin-coated tissue culture plate. The CFU-Hill uses the non-adherent cells to form colonies in 4–7 days, and these are named after early outgrowth cells. Following the work of Hill , the CFU-Hill has been used in various clinical settings as a marker for disease. In chronic models such as type 1 diabetes , chronic obstructive pulmonary disease , heart failure and rheumatoid arthritis , the number of colonies is found to be decreased. In contrast, acute events such as acute myocardial infarction and unstable angina are associated with an increase in the number of colonies. These discrepancies could be due to chronic versus acute events, but also to the fact that these colonies consist of various types of cells, including T cells and monocytes , and do not display postnatal vasculogenesis capacities when implanted into immunodeficient mice . Therefore, the use of CFU-Hill is unlikely to reflect endothelial repair. ECFCs refer to the long-term culture colonies formed between 7 and 21 days of adherent mononuclear cells obtained from peripheral blood, umbilical cord blood or human umbilical venous endothelial cells. These colonies are made of truly clonal endothelial cells with a proliferative potential and the ability to form blood vessels . Indeed, they express primary endothelium cell surface antigens such as von Willebrand factor, CD31, CD105, CD146, KDR or CD144. Contrary to the CFU-Hill, they do not express myeloid antigens such as CD14, CD115 [macrophage colony-stimulating factor (M-CSF) receptor] or CD45. When replated, they keep forming colonies, whereas CFU-Hill do not. The count of peripheral blood-derived ECFCs is positively correlated with the severity of angiography-proven coronary heart disease . It is speculated that such colonies reflect the capacity of certain resident endothelial cells to be the source of endothelium for angiogenesis . View chapterExplore book Read full chapter URL: Book2011, Regenerative NephrologyMatthieu Monge, ... Ton J. Rabelink Chapter Dose Determination for Stem Cell Medicine 2018, Perinatal Stem CellsJames L. Sherley The In Vitro Colony-Forming Unit Assay for Hematopoietic Stem Cells The emergence of technologies for culturing hematopoietic cells and the discovery of specific molecular markers for many differentiated hematopoietic cell lineages allowed the CFU-S assay to be translated into in vitro cell culture assays [17,22]. The resulting colony-forming unit (CFU) culture assays evaluate in vitro colonies with molecular probes and automated enumeration platforms that reduce assay times . However, assigning colony types still requires expert training, and the problems with sensitivity, specificity, and general reliability persist. Although less expensive and faster than CFU-S analyses, the continuing semiquantitative nature and variability of in vitro CFU analyses precluded the in vitro method from gaining general use for laboratory research. Today, several companies provide the CFU assay as a contract service to cell therapy and pharmaceutical companies. Although the assay is quantitatively inadequate for determining HSC dose, if their effects are sufficiently robust, it can detect agents and conditions that alter HSC function . View chapterExplore book Read full chapter URL: Book2018, Perinatal Stem CellsJames L. Sherley Chapter Vertebrate Skeletal Development 2019, Current Topics in Developmental BiologyNoriaki Ono, ... Henry M. Kronenberg 2Colony-forming unit fibroblasts (CFU-Fs) and mesenchymal/skeletal stem cells (MSCs/SSCs): A traditional definition for skeletal stem and progenitor cells Most of the work on stem and progenitor cells in skeletal tissues has been strongly motivated by the goal of regenerative medicine, which is to identify cells capable of restoring functions to human bones. The bulk of existing knowledge on stem and progenitor cells of the skeletal lineage has been built on experiments based on human and rodent bone marrow cells. Traditionally, culture of bone marrow cells and subsequent heterotopic transplantation of in vitro expanded cells into immunodeficient mice has been used as the gold standard to identify these putative stem cells (Bianco, 2014). The first discovery that bone marrow may include stem cells capable of making bones was almost serendipitously made in 1960s. When whole human bone marrow cells were subcutaneously transplanted into immunodeficient mice, they formed ossicles that included blood cells inside (Friedenstein, Piatetzky-Shapiro, & Petrakova, 1966). Later, colony-forming unit fibroblasts (CFU-Fs), which are defined as cells capable of adhering to a plastic culture dish and establishing colonies, were identified as cells responsible for heterotopic ossicle formation (Castro-Malaspina et al., 1980). These hybrid ossicles contained osteoblasts and stromal cells of the donor origin and blood cells of the recipient origin. Bone marrow cells from rodents showed similar properties. Therefore, CFU-Fs residing in human and rodent bone marrow are capable of reconstituting bone marrow in a new environment. CFU-Fs are highly heterogeneous in gene expressions and functions, despite universally possessing similar colony-forming capabilities in cultured conditions. According to the current definition, mesenchymal stem cells (MSCs), or more accurately “skeletal” stem cells, represent a subset of CFU-Fs (Bianco et al., 2013). In fact, only a subset of CFU-Fs possesses self-renewability and multipotency, as not all individual CFU-Fs have the capability to give rise to so-called trilineage cells, i.e., osteoblasts, chondrocytes and adipocytes in vitro, or to form ossicles upon transplantation (Sacchetti et al., 2007). Interestingly, individually-cloned CFU-Fs demonstrate much reduced efficiency for ossicle formation upon transplantation (Sacchetti et al., 2007), suggesting that heterotopic interactions among multiple and heterogeneous CFU-Fs might facilitate reconstituting a hematopoietic microenvironment. Locations and properties of CFU-Fs and MSCs cannot be easily clarified in the native environment because these cells can be identified only retrospectively after exposure to culture conditions. In human bone marrow, use of cell surface markers and cell sorting technologies led to the hypothesis that these putative stem and progenitor cells reside in a perivascular location and assume the morphology closely resembling adventitial reticular cells in bone marrow sinusoids (Bianco, 2014). Initially, the STRO-1 antibody was used to identify and enrich clonal human bone marrow mesenchymal cells (Shi & Gronthos, 2003). Further studies showed that all CFU-Fs were recovered in the CD146+ fraction of human bone marrow cells. CD146+ cells meet the criteria of “skeletal stem cells (SSCs),” as they can be serially transplanted and generate CD146+ cells on secondary transplantation. These CD146+ cells correspond to adventitial reticular cells lining bone marrow sinusoids in vivo. Further studies revealed that CD51 (αV integrin)+ platelet-derived growth factor receptor-α (PDGFRα)+ cells represent a small subset of CD146+ cells with even more enriched colony-forming activities (Pinho et al., 2013). Therefore, these studies established the idea that cell surface markers that are typically expressed by perivascular stromal cells can be used to enrich human CFU-Fs in bone marrow. Genetically modified mice, especially those engineered to express easily assayable proteins, such as green fluorescent protein (GFP), have been particularly useful in identifying putative stem and progenitor cell populations, when combined with cell surface markers and cell sorting technologies. Many of these markers are expressed in proximity to bone marrow vasculature, such as PDGFRα, stem cell antigen-1 (Sca1), chemokine (C-X-C motif) ligand 12 (CXCL12), nestin and α-smooth muscle actin (αSMA). PDGFRα+ Sca1+ nonhematopoietic cells (PαS cells) reside in a perivascular space in vivo and are enriched for CFU-Fs (Morikawa et al., 2009). These cells, if uncultured and cotransplanted with HSCs, can engraft into irradiated recipients and become osteoblasts, stromal cells, adipocytes and, more importantly, PαS cells themselves, suggesting their self-renewal capability in vivo, while cultured PαS cells do not have such capability. Nestin is an intermediate filament protein and a marker for neural stem cells. Nestin-GFP is highly expressed in pericytes of bone marrow arterioles (small arteries), and Nestin-GFP+ cells in bone marrow include all CFU-F activities and form self-renewable “mesenspheres” that can go through serial heterotopic transplantations (Méndez-Ferrer et al., 2010). The CD51+ PDGFRα+ fraction of Nestin-GFP+ cells is further enriched for CFU-F activities (Pinho et al., 2013). Alpha-smooth muscle actin (αSMA) is a marker for pericytes of bone marrow arteries. Pericytes associated with vasculature in bone marrow are marked by αSMA-GFP/mCherry transgenic expression and exhibit trilineage differentiation potential in vitro (Grcevic et al., 2012). Further, connective tissue growth factor (CTGF) is expressed by peri-trabecular stromal cells, and CTGF-GFP+ cells appear to give rise to clonal cells in vitro that are further transplantable (Wang et al., 2015). It is increasingly evident that bone marrow is not the sole location from which CFU-Fs and MSCs can be isolated. Other important locations in bones, such as the periosteum and the growth plate, also house clonogenic cell populations that may have distinct functions in vivo (we will mention these points later in the chapter). Therefore, comprehensive approaches combining cell surface and transgenic markers have proven to be particularly useful in gaining insight into CFU-Fs and MSCs. View chapterExplore book Read full chapter URL: Book series2019, Current Topics in Developmental BiologyNoriaki Ono, ... Henry M. Kronenberg Chapter Osteohematopoietic Stem Cell Niches in Bone Marrow 2012, International Review of Cell and Molecular BiologySamiksha Wasnik, ... Gopal Pande 3.2.1Colony-forming unit assays CFU assays were first described in 1966 by Bradley and Metcalf as the short-term quantitative assay for the measurement of differentiative potential of lineage-restricted progenitor cells (Bradley and Metcalf, 1966). The assay allows the growth of clonal progeny of single progenitor cells in semisolid media of agar or methylcellulose. These cells proliferate in response to specific cytokines and growth factors, resulting in clusters or colonies with distinct features and morphology based on cell type. Specific assays are termed CFU-GEMM, CFU-GM, CFU-E, CFU-G, and CFU-M (Coulombel, 2004). A description of these assays is given in Fig. 3.3. The CFU-GEMM is the most primitive progenitor cell which forms multilineage colonies of granulocytes, erythrocytes, monocytes, and macrophages on semisolid media in the presence of hematopoietic growth factor IL-3 (Monette and Sigounas, 1987). Erythroid-committed progenitors form two distinct types of colonies termed CFU-E and BFU-E. In the presence of erythropoietin (EPO), CFU-E divides rapidly and gives rise to single erythroblast colonies, whereas BFU-E is generally a slow-dividing cell and gives rise to larger colonies of erythroblasts (Kimura et al., 1984). The immature slow-dividing BFU-E differentiates into intermediate mature BFU-E and then to fast-dividing CFU-E cells (Wu et al., 1995). CFU-GM is the progenitor cell for GM and gives rise to single lineage-committed progenitors, CFU-G and CFU-M, under the influence of colony-stimulating factors such as GM-CSF, G-CSF, and M-CSF (Metcalf and Burgess, 1982). The biological process of generation of megakaryocytes and platelets is known as megakaryocytopoiesis. Colony-forming units of megakaryocytes (CFU-MK or CFU-Meg) are unipotent in nature and in the presence of thrombopoietin (TPO) give rise to mature megakaryocytic cells (Broudy et al., 1996; Kimura et al., 1984) View chapterExplore book Read full chapter URL: Book series2012, International Review of Cell and Molecular BiologySamiksha Wasnik, ... Gopal Pande Chapter Osteohematopoietic Stem Cell Niches in Bone Marrow 2012, International Review of Cell and Molecular BiologySamiksha Wasnik, ... Gopal Pande 3.2Functional hierarchies 3.2.1Colony-forming unit assays CFU assays were first described in 1966 by Bradley and Metcalf as the short-term quantitative assay for the measurement of differentiative potential of lineage-restricted progenitor cells (Bradley and Metcalf, 1966). The assay allows the growth of clonal progeny of single progenitor cells in semisolid media of agar or methylcellulose. These cells proliferate in response to specific cytokines and growth factors, resulting in clusters or colonies with distinct features and morphology based on cell type. Specific assays are termed CFU-GEMM, CFU-GM, CFU-E, CFU-G, and CFU-M (Coulombel, 2004). A description of these assays is given in Fig. 3.3. The CFU-GEMM is the most primitive progenitor cell which forms multilineage colonies of granulocytes, erythrocytes, monocytes, and macrophages on semisolid media in the presence of hematopoietic growth factor IL-3 (Monette and Sigounas, 1987). Erythroid-committed progenitors form two distinct types of colonies termed CFU-E and BFU-E. In the presence of erythropoietin (EPO), CFU-E divides rapidly and gives rise to single erythroblast colonies, whereas BFU-E is generally a slow-dividing cell and gives rise to larger colonies of erythroblasts (Kimura et al., 1984). The immature slow-dividing BFU-E differentiates into intermediate mature BFU-E and then to fast-dividing CFU-E cells (Wu et al., 1995). CFU-GM is the progenitor cell for GM and gives rise to single lineage-committed progenitors, CFU-G and CFU-M, under the influence of colony-stimulating factors such as GM-CSF, G-CSF, and M-CSF (Metcalf and Burgess, 1982). The biological process of generation of megakaryocytes and platelets is known as megakaryocytopoiesis. Colony-forming units of megakaryocytes (CFU-MK or CFU-Meg) are unipotent in nature and in the presence of thrombopoietin (TPO) give rise to mature megakaryocytic cells (Broudy et al., 1996; Kimura et al., 1984) 3.2.2Colony-forming unit assay-spleen In vivo CFU-s is an in vivo functional assay for the quantification of the proliferative potential of multipotent HSCs. Till and McCulloch were the first to establish the CFU-s assay to identify primitive progenitor cell populations in mouse BM cells that can form colonies of erythroid cells, megakaryocytes, granulocytes, and macrophages in the spleen of irradiated animals (Till and McCulloch, 1961). CFU-s not only exhibits multilineage potential but has also shown to have high self-renewal capacity, as a single CFU-s cell can produce more CFU-s (Siminovitch et al., 1963). For many years, CFU-s was considered the most primitive HSCs because of their multilineage potential, and CFU-s is a heterogeneous cell population with each cell exhibiting a different renewal capability. Cells that form colonies in the irradiated spleen at 8th day are termed as CFU-s-8, whereas cells that form visible colonies at 12th day are termed as CFU-s-12. The CFU-s-12 has been demonstrated to be more primitive than CFU-s-8, reflecting the heterogeneity in the self-renewal potential of the CFU-s cells (Siminovitch et al., 1963). 3.2.3Long-term repopulation assay Long-term repopulation assay (LTRA) is an in vivo measurement of the totipotency of HSCs and along with its variants is one of the most widely accepted assays. A summary of the significance of these assays is described in Table 3.2. LTRA measures the capacity of HSCs to provide lifelong reconstitution of all blood-cell lineages after transplantation into lethally irradiated recipients. Secondary, and even tertiary, transplants can be performed to demonstrate both self-renewal and multilineage differentiation capacity (Szilvassy et al., 1990). Table 3.2. In vivo functional assays \| \| \| --- \| \| Long-term repopulating assay (LTRA) \| Examine totipotency of HSCs having hematopoietic long-term repopulation capacity (Szilvassy et al., 1990) \| \| Competitive repopulation unit (CRU) assay \| Better repopulation capacity than the competent cell population (Harrison, 1980) \| \| SCID repopulation cell (SRC) assay \| Ability to repopulate the SCID mice hematopoietic tissue (Yahata et al., 2003) \| \| Limiting dilution assay \| Minimum number of cells required for hematopoietic tissue repopulation (Szilvassy et al., 1990) \| 3.2.4Competitive repopulation unit CRU assay measures the ability of an experimental hematopoietic cell population to compete with another cell population (manipulated/unmanipulated) for reconstitution of the hematopoietic system of an irradiated mouse (Harrison, 1980). The CRU assay involves the transplantation of limited number of test male marrow cells together with the female marrow cells whose reconstitution ability is compromised by two prior rounds of transplantation, into an irradiated female recipient. It has been found that HSCs from all sources (donor male, female manipulated, and recipient female marrow cells) can contribute to reconstitution at 5 weeks. Poisson statistics are used to calculate the CRUs in the male (test) population (Ploemacher et al., 1989; Szilvassy et al., 1990). 3.2.5SCID-repopulating cell SCID-repopulating cell (SRC) assay is the measurement of human HSCs ability to repopulate the BM of immune-deficient SCID mice upon transplantation. The repopulating cells are termed as SRC and have the phenotype CD34+CD38− (Dick et al., 1997). Apart from the functional and phenotypic assays for the identification and isolation of HSC population in BM, staining by Hoechst-33342 and rhodamine 123 can also be employed. Efflux of these dyes is a property of many stem cell populations, including HSCs, due to a high level expression of the multidrug resistance gene MDR-1. Upon cell staining stem cells appear as a side population cells in the mixed population of stained cells (Goodell et al., 1996). View chapterExplore book Read full chapter URL: Book series2012, International Review of Cell and Molecular BiologySamiksha Wasnik, ... Gopal Pande Related terms: Probiotic Hematopoietic Stem Cell Bacteriophage Stem Cell Mesenchymal Stem Cell Escherichia coli Infectious Agent Mouse Surface Property Colony Forming Unit S View all Topics
17185	https://cs.stackexchange.com/questions/2717/polygons-generated-by-a-set-of-segments	algorithms - Polygons generated by a set of segments - Computer Science Stack Exchange Join Computer Science By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Polygons generated by a set of segments Ask Question Asked 13 years, 5 months ago Modified13 years, 2 months ago Viewed 2k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. Given a set of segments, I would like to compute the set of closed polygons inside the convex hull of the set of the end of those segments. The vertices of the polygons are the intersections of the segments. For example, if you draw the 6 lines restricted which equations are: x=−1 x=−1, x=0 x=0, x=1 x=1, y=−1 y=−1, y=0 y=0, y=1 y=1, I would like the algorithm to output the four unit squares around the origin. algorithms computational-geometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Jul 18, 2012 at 11:08 alecailalecail asked Apr 16, 2012 at 17:15 alecailalecail 422 2 2 silver badges 10 10 bronze badges 5 Are all your closed polygons lattice squares? For examples, the square [−1,1]×[−1,1][−1,1]×[−1,1] , triangle (−1,−1),(1,0),(−1,1)(−1,−1),(1,0),(−1,1) and a small hexagon around the origin are within your region as well.Ross Millikan –Ross Millikan 2012-04-16 17:23:02 +00:00 Commented Apr 16, 2012 at 17:23 I chose those lines in my question to make a minimalistic example, but there are no restrictions on the coordinates of the ends. I'm looking for the set of non overlapping polygons that we intuitively see.Antoine –Antoine 2012-04-16 17:34:05 +00:00 Commented Apr 16, 2012 at 17:34 I was trying to see if you really are asking how to find all the lattice squares that are within the convex hull. When you talk about all polygons there are an infinite number of those.Ross Millikan –Ross Millikan 2012-04-16 17:50:42 +00:00 Commented Apr 16, 2012 at 17:50 Infinite? Only if you include non-simple (self-intersecting) polygons.JeffE –JeffE 2012-07-13 15:51:33 +00:00 Commented Jul 13, 2012 at 15:51 I will upload a picture a soon as I have 10 reps. For example, with a self-intersecting 5 branch star, I would like to output the central pentagon and the 5 outer triangles. Please help me reformulate the question if you see what I mean.alecail –alecail 2012-07-13 16:22:36 +00:00 Commented Jul 13, 2012 at 16:22 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. You're trying to compute something called the arrangement of the line segments. The arrangement is a planar straight-line graph whose vertices are either endpoints or intersection points of line segments. The vertices partition the lines segments into smaller segments, which are the edges of the arrangement. Finally, the vertices and edges partition the plane into several components; these are the faces of the arrangement. Suppose you are given n n line segments, and k k pairs of those line segments intersect. (You don't know k k in advance.) The intersection points can be computed in O((n+k)log n)O((n+k)log⁡n) time using the Bentley-Ottmann sweep-line algorithm. (The running time can be reduced to O(n log n+I)O(n log⁡n+I) using a more complicated algorithm, where I≤k I≤k is the number of intersection points, but the additional effort is probably not worth it.) The algorithm can be modified to report, for each intersection point, which segments intersect at that point. Then it is simply a matter of careful bookkeeping to construct a standarddatastructure that represents the arrangement. A few caveats to keep in mind: The faces of the arrangement are not necessarily simple polygons. If the union of the line segments is connected, each bounded face is a weakly simple polygon, but a single edge can be incident to the face no both sides. If the union of the segments is disconnected, then faces can have holes, which are either trees or weakly simple polygons. Most textbook (and Wikipedia) descriptions of the Bentley-Ottmann algorithm assume that the input segments are in general position. In this context, "general position" means that at most two segments intersect at any point; no segment endpoint lies on another segment (in particular, the intersection of two segments is either empty or a single point); and no two endpoints have the same x x-coordinate. In practice, however, this assumption does not hold. As usual in computational geometry, handling degenerate cases correctly and robustly is the most difficult part of implementing the algorithm. (What does your implementation do when the input contains 42 copies of the segment [(0,0),(1,1)], 17 copies of the segment [(0,0),(2,2)], and 9 copies of the zero-length segment [(1,1),(1,1)]?) For small numbers of segments (at most a few dozen), it's probably faster to use a brute-force O(n 2)O(n 2)-time algorithm to compute the intersection points. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 13, 2012 at 17:47 JeffEJeffE 8,803 1 1 gold badge 37 37 silver badges 47 47 bronze badges 1 4 I have found a complete discussion on this topic in the chapter 15 of Algorithmic Geometry by Jean-Daniel Boissonnat (IBSN:9780521565295).alecail –alecail 2012-07-18 11:06:57 +00:00 Commented Jul 18, 2012 at 11:06 Add a comment\| Your Answer Thanks for contributing an answer to Computer Science Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. 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17186	https://www.youtube.com/watch?v=qbbUKCSwLx4	which bearings for my fidget spinner? - ceramic vs steel bearings in spinners - Physics Stories Kit Betts-Masters 59600 subscribers 18 likes Description 1376 views Posted: 19 May 2017 I've been asked about what makes ceramic bearing spinners spin for longer. Well it's all about the friction! Ceramic bearings have a lower coefficient of friction that steel ones, but importantly they also have a lower coefficient of thermal expansion, this essentially means they don't get much bigger when they get hot. All the kinetic energy in the spinner is going to be transferred into wasted heat eventually, so in doing so ceramics lose that heat more quickly, and don't get much bigger so don't create extra friction. Some ideas about the future, maybe magnetic bearings or even maglev cars and trains! 3 comments Transcript: so I've been really pleased that my video about spinners has been pretty popular and I've been really pleased have been asking me questions about that one question was about bearings and what makes ceramic bearings better than steel bearings so none of the research and steel bearings do have a higher coefficient of friction than ceramics so ceramics just straight off the bat they are a smoother kind of ball right there more spherical so they have less friction with the surface that's pretty good now unfortunately as well when that's been in and they're going to heat up less okay so actually less of the energy that's the kinetic energy of the things being round is going to go towards heat energy ceramics have really good thermal properties or spacers have came in through the upper atmosphere yet incredibly high amounts of friction because the spaceship is going so very fast so ceramics are the better and better and better at dissipating heat and that's a really important thing as well so not only do the heat up less but they also get rid of their own heat much more rapidly but they also have a low coefficient of thermal expansion if you imagine when the bearing is spinning we're generating all that heat and the ball bearings inside these teeth are going to actually expand and even though they're quite efficient they are going to you know whizzing around that rapidly they are going to heat up quite a lot and that's going to mean they're going to expand and it expands and there's more friction because they're going to be like more of their surface is going to be touching each side of the metal and so there's going to be a less efficient thing again and then if it here are more Gordon friction and therefore a less efficient spinner now there is a type of bearing which has really really really low friction and that might well be a bearing of the future or maybe with a we need bearings in the future does that it's like saying we're going to reinvent the wheel so in a magnetic bearing you wouldn't have anything actually in contact between the inner and outer ring you'd have two sets of magnets repelling each other keeping that constant distance apart that's a really really very interesting idea and you can already get some forms of magnetic bearings that have been a lot more we may not even need wheels we may not even need to have wanting rotating another because we may be actually magnetically levitating in our vehicles you may have heard of a maglev train that's the idea and it exists already but if you found it interesting then you know what you probably quite into physics okay if you've just playing with that spinner and you're wondering how does it work then you're probably interested in physics so why not just have the rest of my channel why not subscribe why not share these videos with your friends you're not just playing with something you're trying to understand more of it and that's what I'm all about here on cabrillo physics I really wanted to help people understand more so they're gaining confidence to the joy feelings more and then they deflected in their GCSE and a-level physics exams so welcome to Google I hope you'll stick around and check some weather content out
17187	https://brainly.com/question/31741702	[FREE] The formula for the volume of a right circular cylinder is V = \pi r^2 h. If r = 2b and h = 5b + 3, what is - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +48,1k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +10,5k Ace exams faster, with practice that adapts to you Practice Worksheets +9k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified The formula for the volume of a right circular cylinder is V=π r 2 h. If r=2 b and h=5 b+3, what is the volume of the cylinder in terms of b? 1 See answer Explain with Learning Companion NEW Asked by anterica12 • 04/23/2023 0:03 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 85931876 people 85M 3.3 3 Upload your school material for a more relevant answer The volume of the cylinder in terms of b is expressed as: (20b³ + 12b²)π cubic units. What is the Volume of a Cylinder? Where r represents the radius of a cylinder and h represents its height, the volume of the cylinder can be calculated using the formula: V = πr²h. Given the following: Radius (r) = 2b Height (h) = 5b + 3 Volume (V) = π (2b)² (5b + 3) Volume (V) = π 4b² (5b + 3) Volume (V) = π 20b³ + 12b² Volume (V) = (20b³ + 12b²)π cubic units Learn more about the volume of cylinder on: brainly.com/question/23935577 SPJ1 Answered by akposevictor •16.5K answers•85.9M people helped Thanks 3 3.3 (3 votes) 1 Expert-Verified⬈(opens in a new tab) This answer helped 85931876 people 85M 3.3 3 Upload your school material for a more relevant answer The volume of the right circular cylinder, with radius r=2 b and height h=5 b+3, is (20 b 3+12 b 2)π cubic units. This is found by substituting the expressions for radius and height into the volume formula. After simplification, the final expression represents the volume in terms of the variable b. Explanation To find the volume of the right circular cylinder in terms of b, we start with the formula for the volume of a cylinder: V=π r 2 h Given: The radius r is defined as r=2 b The height h is defined as h=5 b+3 We can substitute these expressions into the volume formula: V=π(2 b)2(5 b+3) Now, let's calculate (2 b)2: (2 b)2=4 b 2 Substituting this back into the volume equation gives: V=π(4 b 2)(5 b+3) Next, we can distribute 4 b 2 over (5 b+3): V=π[4 b 2⋅5 b+4 b 2⋅3] Calculating each term results in: V=π(20 b 3+12 b 2) Therefore, the volume of the cylinder in terms of b is: V=(20 b 3+12 b 2)π cubic units Examples & Evidence For example, if b=1, the volume would be calculated as V=(20(1)3+12(1)2)π=32 π cubic units. The formula used for the volume of a cylinder, V=π r 2 h, is a standard mathematical formula that is widely accepted and used in geometry. Thanks 3 3.3 (3 votes) Advertisement anterica12 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4 The formula for the volume of a right circular cylinder is V = 72h. If r = 26 and h = 5b + 3, what is the volume of the cylinder in terms of b? Community Answer 4.1 26 The formula for the volume V of a cylinder is V = πr2h, where r is the radius of the base and h is the height of the cylinder. Solve the formula for h. Community Answer 5.0 1 The formula for the volume of the right circular cylinder shown is V = nr2 h. If r = 2b and h = 5b +3, what is the volume of the cylinder in terms of b? А. 10nb2 + 6nb B. 20nb3 + 12nb2 C. 20n2b3 + 12n2b2 D. 50nb3 + 20nb3 + 90nb Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? New questions in Mathematics A retailer must collect 6 4 3% for sales tax. If sales tax for the month is $12,379.50, find the amount of sales for the month. Factor the expression 3 5m+2. Use 3 1 as the common factor. Write an equation of the line that passes through (−11,−10) and is parallel to the y-axis. 60 minutes from now, the time will be 4:50 PM. What time was it 90 minutes ago? Solve the following equation. Separate multiple answers with a comma. If there is no solution, write NS for your answer. 5 x=4 x−3+4 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
17188	https://www.youtube.com/watch?v=HqLf8osa2Ts	Use the Vertex Formula to Find the Maximum Height of a Ball John's Solution Set 2200 subscribers 56 likes Description 11343 views Posted: 19 May 2022 This example is of a ball that is thrown up and then comes back down. The equation is h(t)=-16t^2+32t, which forms a parabola that opens down. The vertex formula is used to find the time it takes to reach maximum height. This value is then plugged back into the original function to obtain the maximum height. 4 comments Transcript: imagine throwing a ball straight up into the air if you were to graph this in space right it would go up and then come right back down nothing really happens there but if you were to graph it in let's see here so let's make a grid here right so we're still throwing the ball straight up in the air and it comes back down but in time if we were to graph height as a function of time right it would go up it would hit some point and then come back down right so even though we're throwing the ball straight up in the air and it's coming back down in time it forms a parabola if we were to expand this out like this and the cool thing about anything like a ball thrown in this way or any object thrown essentially is it makes a parabola and what do we know about this point in the parabola well that's the vertex which happens to also be the maximum so it's a very cool fact that we can do this right because we know how to find the vertex thus we know how to find the maximum how do we find the vertex well remember the vertex formula x equals negative b over 2a and y is then f of that x whatever that x i'll just say f of negative b over 2a so let's do it for this example in particular so this ball follows the equation h of t so this is height equals negative 16 t squared plus 32 t and let's do height let's do that in feet and just to be realistic let's make time in seconds right usually if we throw a ball we're talking feet in seconds right so that's those those uh seem like reasonable units to use okay when is the height the maximum okay so we need to find the vertex essentially so now instead of x equals negative b over 2 a our variable is t so we have t equals negative b over 2 a well let's see here so this negative 16 is our a 32 is our b so t is negative 32 over 2 times negative 16. well that's negative 32 divided by negative 32 so that's one second okay so this particular ball takes one second to reach its maximum height so we could plot this right here one second how long it takes for that to happen well what is its maximum height okay well now we're going to use this bit here where we just plug t back into the original equation so height the maximum height at one second is let's see so we have negative 16 times one squared plus 32 times one well that's 16 feet okay so the maximum height that this ball reaches is 16 feet and that occurs after one second and maybe you're wondering where this equation comes from this original equation comes from um you can you can take a good physics course especially if you take a nice calc based physics course you derive these kind of equations yourself starting with newton's second law so if you want to do that keep going take some calculus derive the equation yourself and then you can do these kinds of problems from scratch
17189	https://fmuser.org/news/fm-transmitter/What-is-the-Voltage-Standing-Wave-Ratio-VSWR/	What is the Voltage Standing Wave Ratio? How to calculate VSWR? FMUSER Sincerely Recruits Global Agents, Contact Us For Details Now !!! 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How to calculate VSWR? "VSWR (Voltage Standing Wave Ratio), is a measure of how efficiently radio-frequency power is transmitted from a power source, through a transmission line, into a load (for example, from a power amplifier through a transmission line, to an antenna)." This is the concept of VSWR. More about VSWR, such as the influencing factors of VSWR, the impact on the transmission system, the difference with SWR, etc. This article can give you a detailed explanation. #Content What is SWR ( Standing Wave Ratio) ? Important Parameter Indicators of SWR What is VSWR (Voltage Standing Wave Ratio)? How VSWR Affects Performance In TransmissionSystem? How to measureSWR? How to calculateVSWR? Free online VSWR calculator 1. What is SWR (Standing Wave Ratio)? According to Wikipedia, standing wave ratio (SWR) defined as: "a measure of impedance matching of loads to the characteristic impedance of a transmission line or waveguide. Impedance mismatches result in standing waves along the transmission line, and SWR is defined as the ratio of the partial standing wave's amplitude at an antinode (maximum) to the amplitude at a node (minimum) along the line." SWR is usually measured using a dedicated instrument called anSWR meter. Since SWR is a measure of the load impedance relative to the characteristic impedance of the transmission line in use (which together determine the reflection coefficient as described below), a given SWR meter can interpret the impedance it sees in terms of SWR only if it has been designed for that particular characteristic impedance. In practice most transmission lines used in these applications are coaxial cable with an impedance of either 50 or 75 ohms, so most SWR meters correspond to one of these. Checking the SWR is a standard procedure in a radio station. Although the same information could be obtained by measuring the load's impedance with an impedance analyzer (or "impedance bridge"), the SWR meter is simpler and more robust for this purpose. By measuring the magnitude of the impedance mismatch at the transmitter output it reveals problems due to either the antenna or the transmission line. By the way, if you think you have never experienced a standing wave personally, it's very unlikely. Standing waves in a microwave oven are the reason that food is cooked unevenly (the turntable is a partial solution to that problem). The wavelength of the 2.45 GHz signal is about 12 centimeters, or about five inches. Nulls in the radiation (and heating) will be separated at a distance similar to wavelength. At last, let's watch a video. ▲BACK▲ 2. Important Parameter Indicators of SWR 1) What is Reflection Coefficient The reflection coefficient is a parameter that describes how much of an electromagnetic wave is reflected by an impedance discontinuity in the transmission medium, equaling to the ratio of the amplitude of the reflected wave to the incident wave. The reflection coefficient is a very useful quality when determining VSWR or investigating the match between, for example, a feeder and a load. The Greek letter Γ is typically used for reflection coefficient, although σ is also often seen. Reflection Coefficient Using the basic definition of the reflection coefficient, it can be calculated from a knowledge of the incident and reflected voltages. Where: Γ = reflection coefficient Vref = reflected voltage Vfwd = forward voltage 2) Return Loss & Resertion Loss Return loss is the loss of signal power due to signal reflection or return by a discontinuity in a fiber-optic link or transmission line, and its unit of expression is also in decibels (dBs). This impedance mismatch can be with a device inserted in the line or with the terminating load. Moreover, return loss is the relationship between both the reflection coefficient (Γ) and the standing wave ratio (SWR), and is always a positive number, and a high return loss is a favorable measurement parameter, and it typically correlates to a low insertion loss. Incidentally, if you increase the return loss, it will correlate to a lower SWR. The loss of signal, which occurs along the length of a fiber optic link, is called insertion loss. Insertion loss is, however, a natural occurrence that occurs with all types of transmissions, whether it is data or electrical. Furthermore, as it is with basically all physical transmission lines or conductive paths, the longer the path, the higher the loss. Moreover, these losses also occur at each connection point along the line, including splices and connectors. This particular measurement parameter is expressed in decibels and should always be a positive number. However, should, does not mean always, and if by chance, it is negative, that is not a favorable measurement parameter. In some instances, an insertion loss may appear as a negative parameter measurement. Return Loss & Insertion Loss So now, let us examine the above diagram in detail so that we may gain a better understanding of how insertion loss and return loss interact. As you can see, incident power travels down a transmission line from the left until it reaches the component. Once it reaches the component, a portion of the signal is reflected back down the transmission line towards the source from which it came. Also, keep in mind that this portion of the signal does not enter the component. The remainder of the signal does indeed enter the component. There some of it gets absorbed, and the rest passes through the component into the transmission line on the other side. The power that comes out of the component is called the transmitted power, and it is less than the incident power for two reasons: ① A portion of the signal gets reflected. ② The component absorbs a portion of the signal. So, in summary, we express insertion loss in decibels, and it is the ratio of incident power to transmitted power. Furthermore, we can summarize that return loss, which we also express in decibels is the ratio of incident power to reflected power. Therefore, we can see how the two types of loss measurement parameters help to accurately gauge the overall efficiency of a measurable signal and component within a system or in a through path. In today’s electronics practices, in terms of use, return loss is preferable to SWR since it affords better resolution for smaller values of reflected waves. 3) What is Impedence Matching Impedance matching is designing sourceand load impedancesto minimize signal reflection or maximize power transfer. In DC circuits, the source and load should be equal. In AC circuits, the source should either equal the load or the complex conjugate of the load, depending on the goal. Impedance (Z) is a measure of the opposition to electrical flow, which is a complex value with the real part being defined as the resistance (R), and the imaginary part is called the reactance (X). The equation for impedance is then by definition Z=R+jX, where j is the imaginary unit. In DC systems, the reactance is zero, so the impedance is the same as the resistance. ▲BACK▲ 3. What is VSWR (Voltage Standing Wave Ratio) 1) What's The Meaning of VSWR The Voltage Standing Wave Ratio (VSWR) is an indication of the amount of mismatch between an antenna and the feed line connecting to it. (Click here to pick our antennna products)This is also known as the Standing Wave Ratio (SWR). The range of values for VSWR is from 1 to ∞ . A VSWR value under 2 is considered suitable for most antenna applications. The antenna can be described as having a “Good Match”. So when someone says that the antenna is poorly matched, very often it means that the VSWR value exceeds 2 for a frequency of interest. Return loss is another specification of interest and is covered in more detail in the Antenna Theory section. A commonly required conversion is between return loss and VSWR, and some values are tabulated in chart, along with a graph of these values for quick reference. Let's take a quick view video about VSWR! 2) Factors Affects VSWR · Frequency · Antenna ground · Nearby metal objects · Type of antenna construction · Temperature 3) SWR vs VSWR vs ISWR vs PSWR SWR is a concept, i.e. the standing wave ratio. VSWR is actually how you make the measurement, by measuring the voltages to determine the SWR. You can also measure the SWR by measuring the currents or even the power (ISWR and PSWR). But for most intents and purposes, when someone says SWR they mean VSWR, in common conversation they are interchangeable. · SWR: SWR stands for standing wave ratio. It describes the voltage and current standing waves that appear on the line. It is a generic description for both current and voltage standing waves. It is often used in association with meters used to detect the standing wave ratio. Both current and voltage rise and fall by the same proportion for a given mismatch. · VSWR: The VSWR or voltage standing wave ratio applies specifically to the voltage standing waves that are set up on a feeder or transmission line. As it is easier to detect the voltage standing waves, and in many instances voltages are more important in terms of device breakdown, the term VSWR is often used, especially within RF design areas. For most practical purposes, ISWR is the same as VSWR. Under ideal conditions, the RF voltage on a signal transmission line is the same at all points on the line, neglecting power losses caused by electrical resistance in the line wires and imperfections in the dielectric material separating the line conductors. The ideal VSWR is therefore 1:1. (Often the SWR value is written simply in terms of the first number, or numerator, of the ratio because the second number, or denominator, is always 1.) When the VSWR is 1, the ISWR is also 1. This optimum condition can exist only when the load (such as an antenna or a wireless receiver), into which RF power is delivered, has an impedance identical to the impedance of the transmission line. This means that the load resistance must be the same as the characteristic impedance of the transmission line, and the load must contain no reactance (that is, the load must be free of inductance or capacitance). In any other situation, the voltage and current fluctuate at various points along the line, and the SWR is not 1. ▲BACK▲ 4. How VSWR Affects Performance In The Transmission System There are many ways in which VSWR affects the performance of a transmission system or any system that may use radio frequencies and identical impedances. Although VSWR is used normally, both voltage and current waves can cause problems. · Transmitter power amplifiers can be damaged: The increased levels of voltage and current seen on the feeder as a result of the standing waves, can damage the output transistors of the transmitter. Semiconductor devices are very reliable if operated within their specified limits, but the voltage and current standing waves on the feeder can cause catastrophic damage if they cause the devise to operate outside their limits. · PA Protection reduces output power: In view of the very real danger of high SWR levels causing damage to the power amplifier, many transmitters incorporate protection circuitry which reduces the output from the transmitter as the SWR rises. This means that a poor match between the feeder and antenna will result in a high SWR which causes the output to be reduced and hence a significant loss in transmitted power. · High voltage and current levels can damage feeder: It is possible that the high voltage and current levels caused by the high standing wave ratio can cause damage to a feeder. Although in most cases feeders will be operated well within their limits and the doubling of voltage and current should be able to be accommodated, there are some circumstances when damage can be caused. The current maxima can cause excessive local heating which could distort or melt the plastics used, and the high voltages have been known to cause arcing in some circumstances. · Delays caused by reflections can cause distortion: When a signal is reflected by mismatch, it is reflected back towards the source, and can then be reflected back again towards the antenna. A delay is introduced equal to twice the transmission time of the signal along the feeder. If data is being transmitted this can cause inter-symbol interference, and in another example where analogue television was being transmitted, a “ghost” image was seen. · Reduction in signal compared to perfectly match system: Interestingly the loss in signal level caused by a poor VSWR is not nearly as great as some may imagine. Any signal reflected by the load, is reflected back to the transmitter and as matching at the transmitter can enable the signal to be reflected back to the antenna again, the losses incurred are fundamentally those introduced by the feeder. As a guide a 30 metre length of RG213 coax with a loss of around 1.5 dB at 30 MHz will mean that an antenna operating with a VSWR will only give a loss of just over 1dB at this frequency compared to a perfectly matched antenna. ▲BACK▲ 5. How To Measure SWR Many different methods can be used to measure standing wave ratio. The most intuitive method uses a slotted line which is a section of transmission line with an open slot which allows a probe to detect the actual voltage at various points along the line. Thus the maximum and minimum values can be compared directly. This method is used at VHF and higher frequencies. At lower frequencies, such lines are impractically long. Directional couplers can be used at HF through microwave frequencies. Some are a quarter wave or more long, which restricts their use to the higher frequencies. Other types of directional couplers sample the current and voltage at a single point in the transmission path and mathematically combine them in such a way as to represent the power flowing in one direction. The common type of SWR/power meter used in amateur operation may contain a dual directional coupler. Other types use a single coupler which can be rotated 180 degrees to sample power flowing in either direction. Unidirectional couplers of this type are available for many frequency ranges and power levels and with appropriate coupling values for the analog meter used. Slotted Line The forward and reflected power measured by directional couplers can be used to calculate SWR. The computations can be done mathematically in analog or digital form or by using graphical methods built into the meter as an additional scale or by reading from the crossing point between two needles on the same meter. The above measuring instruments can be used "in line" that is, the full power of the transmitter can pass through the measuring device so as to allow continuous monitoring of SWR. Other instruments, such as network analyzers, low power directional couplers and antenna bridges use low power for the measurement and must be connected in place of the transmitter. Bridge circuits can be used to directly measure the real and imaginary parts of a load impedance and to use those values to derive SWR. These methods can provide more information than just SWR or forward and reflected power. Stand alone antenna analyzers use various measuring methods and can display SWR and other parameters plotted against frequency. By using directional couplers and a bridge in combination, it is possible to make an in line instrument that reads directly in complex impedance or in SWR. Stand alone antenna analyzers also are available that measure multiple parameters. A power meter NOTE: If your SWR reading is below 1, you have a problem. You might have a bad SWR meter, something wrong with your antenna or antenna connection, or possible have a damaged or defective radio. ▲BACK▲ 6. How to Calculate VSWR When a transmitted wave hits a boundary such as the one between the lossless transmission line and load (Figure 1), some energy will be transmitted to the load and some will be reflected. The reflection coefficient relates the incoming and reflected waves as: Γ = V-/V+ (Eq. 1) Where V- is the reflected wave and V+ is the incoming wave. VSWR is related to the magnitude of the voltage reflection coefficient (Γ) by: VSWR = (1 + \|Γ\|)/(1 – \|Γ\|) (Eq. 2) Figure 1. Transmission line circuit illustrating the impedance mismatch boundary between the transmission line and the load. Reflections occur at the boundary designated by Γ. The incident wave is V+ and the reflective wave is V-. VSWR can be measured directly with an SWR meter. An RF test instrument such as a vector network analyzer (VNA) can be used to measure the reflection coefficients of the input port (S11) and the output port (S22). S11 and S22 are equivalent to Γ at the input and output port, respectively. The VNAs with math modes can also directly calculate and display the resulting VSWR value. The return loss at the input and output ports can be calculated from the reflection coefficient, S11 or S22, as follows: RLIN = 20log10\|S11\| dB(Eq. 3) RLOUT = 20log10\|S22\| dB (Eq. 4) The reflection coefficient is calculated from the characteristic impedance of the transmission line and the load impedance as follows: Γ = (ZL - ZO)/(ZL + ZO)(Eq. 5) Where ZL is the load impedance and ZO is the characteristic impedance of the transmission line (Figure 1). VSWR can also be expressed in terms of ZL and ZO. Substituting Equation 5 into Equation 2, we obtain: VSWR = [1 + \|(ZL - ZO)/(ZL + ZO)\|]/[1 - \|(ZL - ZO)/(ZL + ZO)\|] = (ZL + ZO + \|ZL - ZO\|)/(ZL + ZO - \|ZL - ZO\|) For ZL > ZO, \|ZL - ZO\| = ZL - ZO Therefore: VSWR = (ZL + ZO + ZO - ZL)/(ZL + ZO - ZO + ZL) = ZO/ZL.(Eq. 7) We noted above that VSWR is a specification given in ratio form relative to 1, as an example 1.5:1. There are two special cases of VSWR, ∞:1 and 1:1. A ratio of infinity to one occurs when the load is an open circuit. A ratio of 1:1 occurs when the load is perfectly matched to the transmission-line characteristic impedance. VSWR is defined from the standing wave that arises on the transmission line itself by: VSWR = \|VMAX\|/\|VMIN\|(Eq. 8) Where VMAX is the maximum amplitude and VMIN is the minimum amplitude of the standing wave. With two super-imposed waves, the maximum occurs with constructive interference between the incoming and reflected waves. Thus: VMAX = V+ + V- (Eq. 9) for maximum constructive interference. The minimum amplitude occurs with deconstructive interference, or: VMIN = V+ - V-(Eq. 10) Substituting Equations 9 and 10 into Equation 8 yields VSWR = \|VMAX\|/\|VMIN\| = (V+ + V-)/(V+ - V-) (Eq. 11) Substitute Equation 1 into Equation 11, we obtain: VSWR = V+(1 + \|Γ\|)/(V+(1 - \|Γ\|) = (1 + \|Γ\|)/(1 – \|Γ\|) (Eq. 12) ▲BACK▲ Frequency Asked question 1.What is a good VSWR value As the electric wave travels through the different parts of the antenna system (receiver, feed line, antenna, free space) it may encounter differences in impedances. At each interface, some fraction of the wave's energy will reflect back to the source, forming a standing wave in the feed line. The ratio of maximum power to minimum power in the wave can be measured and is called the voltage standing wave ratio (VSWR). A VSWR of less than 1.5:1 is ideal, a VSWR of 2:1 is considered to be marginally acceptable in low power applications where power loss is more critical, although a VSWR as high as 6:1 may still be usable with the right equipment. Just in case you don’t care for mathematical equations, here’s a little “cheat sheet” table to helpunderstand the correlation of VSWR to the percentage of reflected power that will return. VSWR Returned Power (approximate) 1:1 0% 2:1 10% 3:1 25% 6:1 50% 10:1 65% 14:1 75% 2.What cause high VSWR? If the VSWR is too high, there could potentially be too much energy reflected back into a power amplifier, causing damage to the internal circuitry. In an ideal system, there would be a VSWR of 1:1. Causes of a high VSWR rating could be use of an improper load or something unknown such as a damaged transmission line. 3.Free Online VSWR Calculator Welcome to share this post if it is helpful to you! 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17190	https://www.quora.com/What-minimum-volume-of-oxygen-is-required-for-the-complete-combustion-of-a-mixture-of-10-cm%C2%B3-of-CO-and-15-cm%C2%B3-of-H%E2%82%82	What minimum volume of oxygen is required for the complete combustion of a mixture of 10 cm³ of CO and 15 cm³ of H₂ ? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Chemistry Oxygen Gas Fuel and Combustion Hydrogen Law of Gaseous Volumes Carbon Monoxide Stoichiometry Chemical Reactions Reacting Gas Volumes 5 What minimum volume of oxygen is required for the complete combustion of a mixture of 10 cm³ of CO and 15 cm³ of H₂ ? All related (34) Sort Recommended Les McLean Ph.D. in Engineering · Author has 4.9K answers and 12.3M answer views ·7y Originally Answered: What minimum volume of oxygen is required for the complete combustion of a mixture of 10 cm3 of CO and 15 cm3 of H2? · 10cm^3 of CO is 10Cm^3/22.4L = 10/22400 = 4.464x10^-4 mol Similarly, 15cm^3 of H2 is 15/22400 = 6.696x10^-4 mol So the reaction is 4.464x10^-4 CO + 6.696x10^-4 H2 + (2.232 + 3.3488)x10^-4 mol O2 This is 5.58x10^-4 mol O2 x 22400 = 12.499cm^3 O2 (say 12.5cm^3) Upvote · 9 2 Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 34 Related questions More answers below Can you calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H8)? What is the minimum value of oxygen needed for the complete combustion of a mixture of 10 cm of CO and 15 cm3 of H2? What volume of oxygen is required for a complete combustion of 20 cm^3 of CO and 10 cm^3 of H? How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed? What volume of oxygen is required for the complete combustion of 5 cm³ of CH₄ and 5 cm³ of C₂H₄ in the same conditions? Trevor Hodgson Knows English · Author has 11.8K answers and 12.3M answer views ·4y Originally Answered: What is the minimum value of oxygen needed for the complete combustion of a mixture of 10 cm of CO and 15 cm3 of H2? · Balanced equation: 2CO + O2 → 2CO2 2 volumes CO require 1 volume O2 10cm³ CO require 5 cm³ O2 2H2 + O2 → 2H2O 2 volumes H2 require 1 volume O2 15 cm³ H2 require 7.5 cm³ O2 If all volume measurements are made at the same conditions of temperature and pressure: The mixture will require 12.5 cm³ O2 Upvote · Jacques Proot Engineering Consultant (Owner) (2005–present) · Author has 874 answers and 675.4K answer views ·7y Originally Answered: What minimum volume of oxygen is required for the complete combustion of a mixture of 10 cm3 of CO and 15 cm3 of H2? · CO + 1/2 O2 => CO2 for 10 cm3 de CO we need 5 cm3 de O2 H2 + 1/2 O2 => H2O for 15 cm3 de H2 we need 7.5 cm3 de O2 Total 12.5 cm3 de O2 Upvote · 9 3 Les McLean Ph.D. in Engineering · Author has 4.9K answers and 12.3M answer views ·7y Related What volume of oxygen is required for a complete combustion of 20 cm^3 of CO and 10 cm^3 of H? 20cm^3 of CO is 0.02/22.4 = 8.929x10^-4 moles CO 10cm^3 of H2 is 0.01/22.4 = 4.464x10^-4 moles H2 For simplicity, call 10^-4 as Z So the combustion equation is: 8.929ZCO(g) + 4.464ZH2(g) + 6.696ZO2(g) = 8.929ZCO2(g) + 4.464ZH2O(g) This reaction occurs at 2200C, so Change in Free Energy: ΔG(2200C) = -0..1117kJ (negative, so the reaction runs) Change in Enthalpy: ΔH(2200C) = -0.3582kJ (negative, so the reaction is exothermic) And, of course, 6.696Z moles of O2 is 6.696Z x 22.4 = 0.015L or 15cm^3 Interesting question – in the presence of the right catalyst, we could have made dimethyl ether: 2CO + 4H2 = C2 Continue Reading 20cm^3 of CO is 0.02/22.4 = 8.929x10^-4 moles CO 10cm^3 of H2 is 0.01/22.4 = 4.464x10^-4 moles H2 For simplicity, call 10^-4 as Z So the combustion equation is: 8.929ZCO(g) + 4.464ZH2(g) + 6.696ZO2(g) = 8.929ZCO2(g) + 4.464ZH2O(g) This reaction occurs at 2200C, so Change in Free Energy: ΔG(2200C) = -0..1117kJ (negative, so the reaction runs) Change in Enthalpy: ΔH(2200C) = -0.3582kJ (negative, so the reaction is exothermic) And, of course, 6.696Z moles of O2 is 6.696Z x 22.4 = 0.015L or 15cm^3 Interesting question – in the presence of the right catalyst, we could have made dimethyl ether: 2CO + 4H2 = C2H6O + H2O Upvote · 9 1 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 623 Related questions More answers below Is it oxygen or CO2 gas that promotes combustion? What volume of oxygen is required for the complete combustion of 5cm3 of acetylene? Is oxygen the only gas that supports combustion? If oxygen is sufficient, then is it possible for incomplete combustion to occur? How is oxygen needed for combustion? Assistant Bot · 1y To calculate the minimum volume of oxygen required for the complete combustion of a mixture of carbon monoxide (CO) and hydrogen gas (H₂), we can use the balanced chemical equations for their combustion. Combustion of Carbon Monoxide (CO): 2 CO+O 2→2 CO 2 2 CO+O 2→2 CO 2 From this equation, 2 volumes of CO react with 1 volume of O₂. Therefore, 1 volume of CO requires 1 2 1 2 volume of O₂. 2. Combustion of Hydrogen Gas (H₂): 2 H 2+O 2→2 H 2 O 2 H 2+O 2→2 H 2 O From this equation, 2 volumes of H₂ react with 1 volume of O₂. Therefore, 1 volume Continue Reading To calculate the minimum volume of oxygen required for the complete combustion of a mixture of carbon monoxide (CO) and hydrogen gas (H₂), we can use the balanced chemical equations for their combustion. Combustion of Carbon Monoxide (CO): 2 CO+O 2→2 CO 2 2 CO+O 2→2 CO 2 From this equation, 2 volumes of CO react with 1 volume of O₂. Therefore, 1 volume of CO requires 1 2 1 2 volume of O₂. 2. Combustion of Hydrogen Gas (H₂): 2 H 2+O 2→2 H 2 O 2 H 2+O 2→2 H 2 O From this equation, 2 volumes of H₂ react with 1 volume of O₂. Therefore, 1 volume of H₂ requires 1 2 1 2 volume of O₂. Given: Volume of CO = 10 cm³ Volume of H₂ = 15 cm³ Calculation: Oxygen required for CO: O 2 required for 10 cm 3 CO=10 2=5 cm 3 O 2 O 2 required for 10 cm 3 CO=10 2=5 cm 3 O 2 2. Oxygen required for H₂: O 2 required for 15 cm 3 H 2=15 2=7.5 cm 3 O 2 O 2 required for 15 cm 3 H 2=15 2=7.5 cm 3 O 2 Total Oxygen Required: Total O 2=5 cm 3+7.5 cm 3=12.5 cm 3 Total O 2=5 cm 3+7.5 cm 3=12.5 cm 3 Conclusion: The minimum volume of oxygen required for the complete combustion of the mixture of 10 cm³ of CO and 15 cm³ of H₂ is 12.5 cm³. Upvote · Abdul Rehman Amjad Teaches Chemistry at Spark Academy (SMC Pvt. Ltd) (2018–present) · Author has 353 answers and 235.9K answer views ·3y Related When 50 cm³ hydrogen reacts with 10 cm of oxygen at 293 K and standard pressure, what is the total volume of the mixture? 50 cm3 will be the volume of resulting mixture. 10 cm3 of H2 will react with 20cm3 of H2 to form 20 cm3 H2O 2H2 + O2 → 2H2O 20cm3 + 10 cm3 → 20 cm3 Water formed = 20cm3 H2 left = 50–20 = 30cm3 Total volume = 50 cm3 Continue Reading 50 cm3 will be the volume of resulting mixture. 10 cm3 of H2 will react with 20cm3 of H2 to form 20 cm3 H2O 2H2 + O2 → 2H2O 20cm3 + 10 cm3 → 20 cm3 Water formed = 20cm3 H2 left = 50–20 = 30cm3 Total volume = 50 cm3 Upvote · 99 18 Ian Campbell Product Demonstrator at Self-Employment (2018–present) · Author has 85 answers and 178.9K answer views ·7y Related What volume of oxygen is required for the complete combustion of 5cm3 of acetylene? Look at the equation for the reaction. C2H2 + O2 = CO2 + H2O, before balancing. But we need to put numbers in there to balance it out. The two carbons need a total of 4 oxygen atoms to balance them; the two hydrogens need one atom of oxygen. So one molecule of acetylene needs 5 atoms of oxygen for full combustion. Oxygen comes in biatomic molecules O2, so one has to double both sides; the equation comes out to 2C2H2 + 5O2 = 4CO2 +2H2O. OK, we have the equation. Now, the number of molecules in a volume of gas is directly proportional to the volume; the actual number of molecules is extremely hig Continue Reading Look at the equation for the reaction. C2H2 + O2 = CO2 + H2O, before balancing. But we need to put numbers in there to balance it out. The two carbons need a total of 4 oxygen atoms to balance them; the two hydrogens need one atom of oxygen. So one molecule of acetylene needs 5 atoms of oxygen for full combustion. Oxygen comes in biatomic molecules O2, so one has to double both sides; the equation comes out to 2C2H2 + 5O2 = 4CO2 +2H2O. OK, we have the equation. Now, the number of molecules in a volume of gas is directly proportional to the volume; the actual number of molecules is extremely high and also irrelevant. There are 2.5 times as many oxygen molecules in the equation as acetylene, so the volume of oxygen is 2.5 times that of the acetylene. 5x2.5=12.5, so the answer is 12.5cm3. I thought showing my workings might help a bit more than just giving the answer. :) Incidentally, this answer is an approximation assuming both reactants are perfect gases, which they are not (no real gas is); but in normal Earth-surface conditions it’s a very good one. Probably one part in 1000 or so. Upvote · 99 22 9 1 Promoted by JH Simon JH Simon Author of 'How To Kill A Narcissist' ·Updated Fri How do you overcome narcissistic mental enslavement? I grew up with a narc mother and dated a narc that I just cannot get out of my head. Reclaim your True Self. Narcissistic mental enslavement begins with but goes well beyond the mind. Every narcissist mentally overwhelms their target, but what gives the narcissist’s attacks teeth is the emotional reaction. When a narcissist puts down their target, shame arises. When a narcissist threatens their target, fear arises. When a narcissist questions their target’s morality, guilt arises. These attacks bind together over time and produce a corresponding complex, making the target easier to trigger as a ripple becomes an emotional tsunami. A psychological cage is built with layer upon l Continue Reading Reclaim your True Self. Narcissistic mental enslavement begins with but goes well beyond the mind. Every narcissist mentally overwhelms their target, but what gives the narcissist’s attacks teeth is the emotional reaction. When a narcissist puts down their target, shame arises. When a narcissist threatens their target, fear arises. When a narcissist questions their target’s morality, guilt arises. These attacks bind together over time and produce a corresponding complex, making the target easier to trigger as a ripple becomes an emotional tsunami. A psychological cage is built with layer upon layer of emotional pain, which in time splits from the target’s consciousness and reduces the target to a helpless and compliant object. For example, when guilt becomes a complex, a person acts, thinks and believes as though they are guilty, regardless of the situation. Any judgement, whether justified or not, triggers a guilt attack and cripples the person’s capacity for independent thought and action. Reaction becomes the only option. Overcoming mental enslavement involves facing all emotions and integrating them into consciousness. This means having the courage to fall below the realm of thought. Through mindful exploration, you come to notice your triggers, and therefore welcome the emotional intensity into awareness. Rather than reacting to or dissociating from your emotions, you create space and allow them to rise up, and most importantly, to release. Meditation, therapy and bodywork are all potent tools to help you weave together a connection between your Higher Self and your True Self. In doing so, you create a state of wholeness and power which eclipses the mental enslavement you experience after narcissistic abuse. If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist. Upvote · 999 197 9 6 9 6 Robert Kern BA in Chemistry&Mathematics, College of the Holy Cross (Graduated 1974) · Author has 2.5K answers and 2M answer views ·Updated 3y Related Assuming air contains 23% by weight of oxygen, what is the volume of air at 27°C and 750 mm pressure that would be required for complete combustion of one kilogram of coal containing 90% C and 5% H? First, let’s assume that the carbon and hydrogen are the only combustible components of the coal sample and that all the carbon is converted to CO2, and all the hydrogen is converted to H2O. So the combustion equations are: C + O2 => CO2 and 2 H + 0.5 O2 => H2O Next, let’s convert the mass percent of O2 to volume percent. 23 %O2 means 23 g O2/100 g air. The molar mass of air is 28.8 g and the molar mass of O2 is 32 g, so 23 g O2/100 g air • 1 mol O2/32 g • 28.8 g air/1 mol = 0.207 mol O2/mol air. Since mole fraction is equivalent to volume fraction for gases, this means 0.207 L O2/L air or 20.7 Continue Reading First, let’s assume that the carbon and hydrogen are the only combustible components of the coal sample and that all the carbon is converted to CO2, and all the hydrogen is converted to H2O. So the combustion equations are: C + O2 => CO2 and 2 H + 0.5 O2 => H2O Next, let’s convert the mass percent of O2 to volume percent. 23 %O2 means 23 g O2/100 g air. The molar mass of air is 28.8 g and the molar mass of O2 is 32 g, so 23 g O2/100 g air • 1 mol O2/32 g • 28.8 g air/1 mol = 0.207 mol O2/mol air. Since mole fraction is equivalent to volume fraction for gases, this means 0.207 L O2/L air or 20.7 % O2 by volume (close to the accepted 21 %) Now, lets calculate the molar volume of ideal gas at the conditions listed: 22.4 L • (760 mm Hg/750 mm Hg) • (300 K/273 K) = 24.9 L Then, putting it all together 1000 g of the coal sample contains 900 g C, 50 g H, and the remainder assumed to be not combustible. So 900 g C • 1 mol C/12 g C • 1 mol O2/1 mol C • 24.9 L/1 mol O2 • 100 L air/20.7 L O2 = 9020 L air 50 g H • 1 mol H/1 g H • 0.5 mol O2/2 mol H • 24.9 L/1 mol O2 • 100 L air/20.7 L O2 = 1500 L air Total air = 9020 L + 1500 L = 10520 L or 10.5 kL of air Upvote · 9 2 9 1 Michael Mombourquette Retired Chemistry Prof, Church member, Knight of Columbus, · Author has 6.8K answers and 17.7M answer views ·5y Related What is the volume (in liters) of oxygen needed for the complete combustion of 1p litres of acetylene at 1500 degrees Celsius and pressure of 101 kpa assuming temperature and pressure remain constant? What does 1p litres mean? I’ll assume it means 10 litres. The balanced chemical reaction is the first thing you need C 2 H 2+2.5 O 2→2 C O 2+H 2 O C 2 H 2+2.5 O 2→2 C O 2+H 2 O Now, T 1500 degrees all these species will be gases and we will assume they behave like ideal gases. Avagadro’s law says the ratios of volume is equal to ratios of moles. Those coefficients in the balanced chemical equation are ratios of moles but we can use them as ratios of volumes too according to Avagadro. So start with 10 L of acetylene and use the mole ratio, ok the volume ratios to convert to oxygen [math]10\; L\; C_2O_2 \times \frac{2.5\; O_2 }{1[/math] Continue Reading What does 1p litres mean? I’ll assume it means 10 litres. The balanced chemical reaction is the first thing you need C 2 H 2+2.5 O 2→2 C O 2+H 2 O C 2 H 2+2.5 O 2→2 C O 2+H 2 O Now, T 1500 degrees all these species will be gases and we will assume they behave like ideal gases. Avagadro’s law says the ratios of volume is equal to ratios of moles. Those coefficients in the balanced chemical equation are ratios of moles but we can use them as ratios of volumes too according to Avagadro. So start with 10 L of acetylene and use the mole ratio, ok the volume ratios to convert to oxygen 10 L C 2 O 2×2.5 O 2 1 C 2 O 2=25 L O 2 10 L C 2 O 2×2.5 O 2 1 C 2 O 2=25 L O 2 Upvote · 9 1 9 2 Sponsored by MRPeasy Powerful yet simple MRP software for growing manufacturers. MRPeasy makes it easier for growing manufacturing businesses to keep on expanding successfully. Free Trial 99 65 Tan Jared Works at Singapore · Author has 336 answers and 1.5M answer views ·Updated 8y Related What is the volume of oxygen required, measured at room temperature and pressure, for the complete combustion of the following gaseous mixture? 1 mole of any gas will be 24dm3 at room temperature. Ratio of C2H6 : CH4 : Total Gas = 1 : 1 : 2 = 2 : 2 : 4 For a complete reaction to occur, Ratio of C2H6 : O2 = 2 : 7 Ratio of CH4 : O2 = 1 : 1 = 2 : 2 Overall ratio C2H6 : CH4 : Total Gas : O2 = 2 : 2 : 4 : 2+7 = 2 : 2 : 4 : 9 There is 200cm3 of gas mixture. Total Gas : O2 = 4 : 9 = 200 : 450 Therefore, 450cm3 of oxygen is needed. Answer: Option A. Awanish Pratap Singh has alerted me here that the line “CH4 + O2 → CO2 + H2O” in the question details is wrong. Assuming the corrected line “CH4 + 2O2 → CO2 + H2O”, 1 mole of any gas will be 24dm3 at room tem Continue Reading 1 mole of any gas will be 24dm3 at room temperature. Ratio of C2H6 : CH4 : Total Gas = 1 : 1 : 2 = 2 : 2 : 4 For a complete reaction to occur, Ratio of C2H6 : O2 = 2 : 7 Ratio of CH4 : O2 = 1 : 1 = 2 : 2 Overall ratio C2H6 : CH4 : Total Gas : O2 = 2 : 2 : 4 : 2+7 = 2 : 2 : 4 : 9 There is 200cm3 of gas mixture. Total Gas : O2 = 4 : 9 = 200 : 450 Therefore, 450cm3 of oxygen is needed. Answer: Option A. Awanish Pratap Singh has alerted me here that the line “CH4 + O2 → CO2 + H2O” in the question details is wrong. Assuming the corrected line “CH4 + 2O2 → CO2 + H2O”, 1 mole of any gas will be 24dm3 at room temperature. Ratio of C2H6 : CH4 : Total Gas = 1 : 1 : 2 = 2 : 2 : 4 For a complete reaction to occur, Ratio of C2H6 : O2 = 2 : 7 Ratio of CH4 : O2 = 1 : 2 = 2 : 4 Overall ratio C2H6 : CH4 : Total Gas : O2 = 2 : 2 : 4 : 4+7 = 2 : 2 : 4 : 11 There is 200cm3 of gas mixture. Total Gas : O2 = 4 : 11 = 200 : 550 Therefore, 550cm3 of oxygen is needed. Answer: Option B. Upvote · 9 2 9 2 Devender Singh MSc in Chemistry&Board of High School and Intermediate Education, Uttar Pradesh, Chaudhary Charan Singh University (Graduated 1990) · Author has 926 answers and 3.8M answer views ·6y Related How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed? let x ml oxygen will be required for complete combustion of 40 ml of acetylene under same temperature and pressure. volume of CO2 is V1 ml and H2O is V2 ml will be formed volume of CO2 =80 ml and H2O=40ml will be formed thankuuuu Continue Reading let x ml oxygen will be required for complete combustion of 40 ml of acetylene under same temperature and pressure. volume of CO2 is V1 ml and H2O is V2 ml will be formed volume of CO2 =80 ml and H2O=40ml will be formed thankuuuu Upvote · 9 6 Meave Gilchrist B.S. in Bachelor of Science in Biology&Chemistry, University of Texas Pan American (Graduated 1985) · Author has 6.6K answers and 4.7M answer views ·4y Related What is the volume (in litres) of oxygen needed for the complete combustion of 10 litres of acetylene at 1500°C and a pressure of 101 kPa assuming temperature and pressure remain constant? Use the ideal gas law to calculate the moles of C₂H₂. Then use stoichiometry to calculate the moles of O₂. Then use the ideal gas law again to calculate the volume of O₂ Calculate the moles of C₂H₂. PV = nRT Known and Unknown P = pressure = 101 kPa V = volume = 10 L n = ? mol R = gas constant = 8.31446 L•kPa/K•mol T = temperature = 1500°C + 273.15 = 1773.15 K Solve for n. n = PV/RT n(C₂H₂) = (101 kPa × 10 L)/(8.31446 L•kPa/K•mol × 1773.15) = 0.068508 mol C₂H₂ Stoichiometry Balanced equation 2C₂H₂ + 5O₂ --> 2H₂O + 4CO₂ Calculate the moles of O₂ required to produce 0.068508 mol C₂H₂, using the mole ratio betwe Continue Reading Use the ideal gas law to calculate the moles of C₂H₂. Then use stoichiometry to calculate the moles of O₂. Then use the ideal gas law again to calculate the volume of O₂ Calculate the moles of C₂H₂. PV = nRT Known and Unknown P = pressure = 101 kPa V = volume = 10 L n = ? mol R = gas constant = 8.31446 L•kPa/K•mol T = temperature = 1500°C + 273.15 = 1773.15 K Solve for n. n = PV/RT n(C₂H₂) = (101 kPa × 10 L)/(8.31446 L•kPa/K•mol × 1773.15) = 0.068508 mol C₂H₂ Stoichiometry Balanced equation 2C₂H₂ + 5O₂ --> 2H₂O + 4CO₂ Calculate the moles of O₂ required to produce 0.068508 mol C₂H₂, using the mole ratio between C₂H₂ and O₂ in the balanced equation. 0.068508 mol C₂H₂ × 5 mol O₂/2 mol C₂H₂ = 0.17127 mol O₂ Use the ideal gas law to calculate the volume of O₂ in 0.17127 mol O₂ under the conditions given. P = 101 kPa V =? L n = 0.17127 mol R = 8.31446 L•kPa/K•mol T = 1773.15 K Solve for V. V = nRT/P V(O₂) = (0.17127 mol × 8.31446 L•kPa/K•mol × 1773.15 K)/(101 kPa) = 24.999 mol O₂ Upvote · 9 4 Trevor Hodgson Knows English · Author has 11.8K answers and 12.3M answer views ·4y Related A mixture of 20 cm3 of ethane and 30 cm³ of propane are completely combusted in 300 cm3 of oxygen. What is the final volume of gases? Write balanced equations: 2C2H6 + 7O2 → 4CO2 + 6H2O 1 volume C2H6 requires 3.5 volumes O2 to produce 2 volumes CO2 20 cm³ ethane will consume 3.520 = 70 cm³ O2 to produce 220 = 40 cm³ CO2 C3H8 + 5O2 → 3CO2 + 4H2O 1 volume C3H8 requires 5 volumes O2 to produce 3 volumes CO2 30cm³ C3H8 requires 305 = 150 cm³ O2 to produce 330 cm³ =90 cm³ CO2 What will remain in the reaction vessel : C2H6 and C3H8 = Nil remains O2 - you started with 300 cm³ - used 70 cm³ + 150 cm³ = 220 cm³ . You will have 300 cm³ - 220 cm³ = 80 cm³ remaining CO2 - You have produced 40 cm³ + 90 cm³ = 130 cm³ Volume of gases remaining in Continue Reading Write balanced equations: 2C2H6 + 7O2 → 4CO2 + 6H2O 1 volume C2H6 requires 3.5 volumes O2 to produce 2 volumes CO2 20 cm³ ethane will consume 3.520 = 70 cm³ O2 to produce 220 = 40 cm³ CO2 C3H8 + 5O2 → 3CO2 + 4H2O 1 volume C3H8 requires 5 volumes O2 to produce 3 volumes CO2 30cm³ C3H8 requires 305 = 150 cm³ O2 to produce 330 cm³ =90 cm³ CO2 What will remain in the reaction vessel : C2H6 and C3H8 = Nil remains O2 - you started with 300 cm³ - used 70 cm³ + 150 cm³ = 220 cm³ . You will have 300 cm³ - 220 cm³ = 80 cm³ remaining CO2 - You have produced 40 cm³ + 90 cm³ = 130 cm³ Volume of gases remaining in reaction vesssel = 80 cm³ + 130 cm³ = 210 cm³ NOTE : I have taken the water to be liquid - not a gas . Should the water be considered as gas - then you will produce From the C2H6 = 60 cm³ water vapour From the C3H8 = 120 cm³ Volume of water vapour produced = 180 cm³ Then total volume of gases = 210 cm³ + 180 cm³ = 390 cm³ And I finally add : All gas volumes are measured under the same conditions of temperature and pressure. Upvote · 9 1 9 1 Related questions Can you calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H8)? What is the minimum value of oxygen needed for the complete combustion of a mixture of 10 cm of CO and 15 cm3 of H2? What volume of oxygen is required for a complete combustion of 20 cm^3 of CO and 10 cm^3 of H? How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed? What volume of oxygen is required for the complete combustion of 5 cm³ of CH₄ and 5 cm³ of C₂H₄ in the same conditions? Is it oxygen or CO2 gas that promotes combustion? What volume of oxygen is required for the complete combustion of 5cm3 of acetylene? Is oxygen the only gas that supports combustion? If oxygen is sufficient, then is it possible for incomplete combustion to occur? How is oxygen needed for combustion? What volume of oxygen at NTP is required for the complete combustion of CO at 22°C and 74 mm Hg pressure? Does all combustion require oxygen? Is oxygen compressed gas combustible? What volume of oxygen at NTP is required for complete combustion of 18.2 litres of propane? Is oxygen combustible or explosive? Related questions Can you calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H8)? What is the minimum value of oxygen needed for the complete combustion of a mixture of 10 cm of CO and 15 cm3 of H2? What volume of oxygen is required for a complete combustion of 20 cm^3 of CO and 10 cm^3 of H? How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed? What volume of oxygen is required for the complete combustion of 5 cm³ of CH₄ and 5 cm³ of C₂H₄ in the same conditions? Is it oxygen or CO2 gas that promotes combustion? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. 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17191	https://www.findlaw.com/estate/planning-an-estate/whats-the-difference-between-joint-tenants-with-survivorship-and.html	Find a Qualified Attorney Near You Find a Qualified Attorney Near You Search by legal issue and/or location Enter information in one or both fields. (Required) Find a Lawyer More Options Name Search Browse Legal Issues Browse Law Firms Support Differences Between Joint Tenants With Survivorship and Tenants in Common By Oni Harton, J.D. \| Legally reviewed by Aisha Success, Esq. \| Last reviewed This article has been written and reviewed for legal accuracy, clarity, and style by FindLaw’s team of legal writers and attorneys and in accordance with our editorial standards. The last updated date refers to the last time this article was reviewed by FindLaw or one of our contributing authors. We make every effort to keep our articles updated. For information regarding a specific legal issue affecting you, please contact an attorney in your area. Property can be owned individually (sole ownership) or collectively (joint or common ownership). In most cases, joint owners can be either co-tenants in common or joint tenants with the right of survivorship. You can own property individually (sole ownership) or collectively (joint or common ownership). In most cases, there are two ways to hold title with others. Joint owners can be one of either: Co-tenants in common Joint tenants with the right of survivorship The main differences between these joint ownership types are: How they arise How they are destroyed How the subject property can be divided and sold Read on to explore these differences in greater detail. What Is an Undivided Interest? Before discussing specific forms of joint ownership, it’s helpful to unpack the legal meaning of an undivided interest. When two or more people own real estate, each individual owns a share (interest) of the entire property. Each owner’s interest is said to be undivided. Each owner has a right to use the entire physical property even though their abstract rightto the property isportioned out among them. To illustrate briefly, imagine that two business partners own real property together. A warehouse, perhaps. The warehouse is physically undivided, but the owners share the entire physical property as a whole. However, each partner may have a 50% interest, or one may have a 30% interest, and another has a 70% interest. Each type of joint property ownership has certain restrictions on how to divide the property interest. Tenancy in Common A tenancy in common may involve two or more owners. Each tenant in common may own an equal share of the property, but there’s no requirement for equal ownership. Four owners may each own a 25% interest, or their interests may break down as 10%, 20%, 30%, and 40%. Each co-tenant has an equal right to possess, use, and enjoy the property. The co-tenants are free to make alternative arrangements among themselves. Each co-tenant may also freely sell their interest. Similarly, when a co-owner of the property dies, their share remains part of the decedent’s estate. Thus, the decedent’s personal representative can transfer the decedent’s share as explained in their will. Whoever receives the interest steps into the previous co-tenant’s shoes. Further, the transfer of a co-tenant’s interest may occur at any time. The owner change does not disturb the other co-tenant’s ownership status. Jointly owned property is presumed to be held in a tenancy in common unless the property deed specifies otherwise. Joint Tenancy with Right of Survivorship A joint tenancy with right of survivorship (JTWROS), like a tenancy in common, is a form of co-ownership. It may involve two or more owners. However, a JTWROS must comply with a number of restrictions. The Four Unities A JTWROS must satisfy the so-called Four Unities. They are as follows: Unity of Time: Each joint tenant musttake title of their share at the exact time. Unity of Title: Each joint tenant musttake ownership of their share through the same instrument (e.g., a property deed). The legal document must specifically state that it is creating a JTWROS. Otherwise, the document creates a tenancy in common by default. The specific formation language varies by state. Unity of Interest: Each joint tenant musthave an equal interest. Two owners must each have a 50% interest. Four must each have a 25% interest, and so on. Unity of Possession: Each joint tenant musthave a legal right to possess, use, and enjoy the property equally. Unlike co-tenants in a tenancy in common, joint tenants cannot alter this arrangement. Violation of any of the Four Unities destroys the joint tenancy. The joint tenancy would become a tenancy in common. In particular, note that the Unity of Time and Unity of Title operate so the joint tenants cannot transfer their share without destroying the joint tenancy. Their ownership rights cannot be sold, inherited, or otherwise transferred. Right of Survivorship If one of two owners of property held in a JTWROS dies, ownership automatically transfers to the surviving owner. This is called a right of survivorship. The deceased owner’s estate does not receive any share of the property. Unlike a tenancy in common, a JTWROS co-owner cannot transfer their interest in the property without destroying the JTWROS. Does Either Avoid Probate? Probatehas two meanings. It refers to the legal process of checking whether a deceased person’s last will and testament is validand authentic. This occurs in probate court. Probate also refers to the general process of distributing a decedent’s estate. Depending on the estate’s size, the probate process can be time-consuming and expensive. So, does a tenancy in common or JTWROS avoid probate? Tenancy in Common Typically, a tenancy in common will not avoid probate. A co-tenant’s ownership interest remains part of their estate when they die. It must be distributed by will or according to state laws of intestate succession. If you want to keep the piece of property out of the probate process, you could transfer it out of a tenancy in common and into a trust. Property in a trust does not belong to the person who supplies the property. Instead, the property belongs to the trust itself and, therefore, is not part of the person’s estate at the time of death. Joint Tenancy with Right of Survivorship By contrast, the ROS in a JTWROS typically ensures that a joint tenant’s interest doesavoid probate. When only one joint tenant remains, that individual becomes the sole owner. At the sole owner’s death, their 100% share must be distributed as part of their estate. Thus, the surviving owner does not avoid probate. Again, this can be avoided by transferring the interest into a trust. By extension, one can imagine a conceivable though improbable scenario in which all joint tenants die at or near the same time (e.g., in a plane crash), making it impossible to determine who was the last surviving joint tenant. In this case, each joint tenant’s share might pour into their estates and fail to avoid probate. Questions? A Local Attorney Can Help Tenancies in common have the advantage of flexibility. Joint tenancies with right of survivorship have the advantage of permanence. Understanding the advantages and disadvantages of each ownership arrangement before entering one can help you avoid serious headaches. A local real estate or estate planning attorney can provide valuable legal advice regarding joint tenancy and which type would be best for you. Was this helpful? Can I Solve This on My Own or Do I Need an Attorney? 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17192	https://www.chem.fsu.edu/chemlab/chm1045/stoichiometry.html	Stoichiometry The content that follows is the substance of lecture 14. In this lecture we cover the use of balanced chemical equations to make mole and mass predictions in reaction. The Balanced Reaction Reaction stoichiometry is a fancy way of describing mole relations in a reaction. When we read a balanced equation we relate every compound in the reaction to the others by means of the mole coefficients in the reaction. For example: HCl + NaOH → NaCl + H2O This equation is all 1:1 relationships, 1 mole of NaCl is made for every 1 mole of HCl or NaOH etc. In a previous lecture I introduced the following analogy: What can we do with a balanced equation? A chemical reaction equation is essentially a table of conversion factors that we can use to predict amounts of products that can be made, reactants needed to make a specific amount of product or exact amounts of reactants needed to completely consume another reactant (eg. acid and base neutralization). While the use of the stoichiometry (fancy words for mole relationships in a reaction equation) may seem difficult, it really isn't. Let's use an example that you all can understand first: 1 slice bologna + 2 slices of bread → 1 Sandwich If I asked you how many sandwiches you could make with 12 slices of bread, you would immediately say 6, right? How did you know this? Well, obviously based on the equation, you make 1 sandwich for every 2 slices of bread you have. 12 slices bread x 1 sandwich/2 slices bread = 6 sandwiches If I asked you how many slices of bologna and bread you need to create 10 sandwiches, you would immediately say 10 slices of bolgna and 20 slices of bread, right? Again the relationship given in the equation tells you how much is needed. 10 sandwiches x 1 slice bologna/1 sandwich = 10 slices bologna and 10 sandwiches x 2 slices bread/1 sandwich = 20 slices bread Finally if I asked you how many sandwiches you could make with 5 slices of bologna and 8 slices of bread? Which of the two sandwich makings would run out first and how much of the excess makings would be left over? Well, the answers are 4 sandwiches could be made and 1 slice of bologna is left over. 5 slices of bologna x 1 sandwich/1 slice bologna = 5 sandwiches 8 slices bread x 1 sandwich/2 slices bread = 4 sandwiches 8 slices bread x 1 slice bologna/2 slices bread = 4 slices of bologna used 5 slices bologna - 4 slices used = 1 slice left over The process outlined above is the process you would use to answer stoichiometric questions on a test. The only difference is we need to work in the unit of moles rather that slices and sandwiches. Here is a similar example: The process for solving these problems is similar each time: 1) Write the balanced chemical equation 2) Convert what you are given into moles 3) Use the mole ratios in the balanced equation to determine the moles of whatever is needed (amount of product possible, amount of reactant needed etc.) 4) Convert the moles you calculated into the units requested e.g. grams or mL. Here is another example using mL and M rather than grams: Find the limiting reagent and the reactant in excess when 100 mL of 0.2M NaOH reacts completely with 50 mL of 0.5M H2SO4 solution. Step 1: Write the balanced chemical equation for the chemical reaction: 2NaOH(aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O(l) Step 2: Calculate the available moles of each reactant in the chemical reaction: Moles (NaOH) = 0.2M x 0.1L = 0.02 mol Moles(H2SO4) = 0.5M .05L L= 0.025 mol Step 3: Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction to the product you wish to make: 0.02 mol NaOH x 1 mol Na2SO4/2 mol NaOH = 0.01 mol Na2SO4 0.025 mol H2SO4 x 1 mol Na2SO4/1 mol H2SO4 = 0.025 mol Na2SO4 The limiting reagent is the reactant that will produce less product in the reaction. There will be some moles of the reactant in excess left over after the reaction has gone to completion. The limiting reagent is NaOH, all of the 0.02 moles of NaOH will be used up when this reaction goes to completion. The reactant in excess is H2SO4, when the reaction has gone to completion there will be 0.025 - 0.01 = 0.015 moles of H2SO4left over. Limiting and Excess Practice
17193	https://www.cuemath.com/algebra/addition-property-of-equality/	LearnPracticeDownload Addition Property of Equality The addition property of equality states that when the same quantity is added to both sides of an equation, the equation does not change. If a number x is added to both sides of an equation A = B, then the equation still holds true. It can be expressed mathematically as, A + x = B + x. Let us learn more about the addition property of equality in this article along with its formula. We shall also discuss the addition property of equality for fractions. \| \| \| --- \| \| 1. \| What is Addition Property of Equality? \| \| 2. \| Addition Property of Equality Formula \| \| 3. \| Addition Property of Equality with Fractions \| \| 4. \| FAQs on Addition Property of Equality \| What is Addition Property of Equality? The addition property of equality is defined as "When the same amount is added to both sides of an equation, the equation still holds true". It is a very important and one of the common properties of equality in mathematics. The other properties are the subtraction property, multiplication property, and division property of equality. The addition property of equality is used while solving equations. For example, to solve x - 3 = 5, we have to add 3 to both sides so that only the variable (x) will be left on the left side. This implies, x - 3 + 3 = 5 + 3, which can be simplified as x = 8. Let us learn the addition property of equality formula in the next section. Addition Property of Equality Formula The formula for the addition property of equality is given below: If the given equation is A = B, and the same quantity n is added to both sides of this equation, we get A + n = B + n, which is true and satisfies the equation. We can use this property of equality in arithmetic equations as well as in algebraic equations. In the case of an arithmetic equation, for example, 2 + 3 = 5, if we add a number to both sides, say 7, the equation will still be true. Let's verify it. 2 + 3 + 7 = 5 + 7 L.H.S = 2 + 3 + 7 = 12 R.H.S = 5 + 7 = 12 Therefore, L.H.S = R.H.S. Let us take one more example of an algebraic equation, x - 4 = 12. It can be solved using the addition property of equality, as shown below. x - 4 = 12 By adding 4 to both sides of the equation, we get, x - 4 + 4 = 12 + 4. ⇒ x = 16 Therefore, the value of x is 16. It is the application of the addition property of equality in math. Addition Property of Equality with Fractions The addition property of equality works with fractions as well. The same integer or fraction can be added to an equation containing fractions. For example, if 2x/3 + 5/6 = y + 5/6, this implies that 2x/3 = y. The general form of addition property of equality with fractions is written below: A/B + x/y = C/D + x/y Here, A/B = C/D is the given equation and x/y is the fraction added to both sides without changing the equality. Important Notes on Addition Property of Equality The addition property of equality states that when the same number is added on both sides of an equation, the equality still holds. The formula for the addition property of equality is given by, A = B ⇒ A + n = B + n. The property holds for numbers with fractions. ► Related Articles Associative Property of Addition Transitive Property Solving Linear Equations Read More Download FREE Study Materials Worksheets on Addition Property of Equality [PDF] Addition Property of Equality Worksheet One-Variable Linear Equations Worksheet Addition Property of Equality Examples Example 1: Solve the equation 2y + 4 = 8 - y using the addition property of equality. Solution: The given equation is 2y + 4 = 8 - y. Adding y to both sides will result in 2y + 4 + y = 8 - y + y. ⇒ 3y + 4 = 8 Add (-4) to both sides, we get, ⇒ 3y + 4 - 4 = 8 - 4 ⇒ 3y = 4 ⇒ y = 4/3 Therefore, the value of y is 4/3. 2. Example 2: Find the value of x in the given equation: x + 3 = 5 + 3. Solution: The given equation is x + 3 = 5 + 3. Applying the addition property of equality formula which says that if A = B, then A + n = B + n, we can say that in the given equation if x + 3 = 5 + 3, this implies that x = 5. Therefore, the value of x is 5. 3. Example 3: Victoria said that the solution of this equation: m + 4 = n + 4 is (m - n = 8), while Daizy said that the solution is m = n. Whose answer is correct and why? Solution: Daizy's answer is correct. The given equation is m + 4 = n + 4. If we observe it carefully using the addition property of equality formula, we see that 4 is added to both sides of the equation. So, it can also be written as m = n. Therefore, Daizy's answer is correct. The solution is m = n. View Answer > Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Addition Property of Equality Questions Check Answer > FAQs on Addition Property of Equality What does Addition Property of Equality Mean? The addition property of equality is defined as a condition where the same number added to both sides of an equation does not change the equality. If a is any real number and P = Q is the equation, we can have P + a = Q + a. What is an Example of Addition Property of Equality? The example of addition property of equality is given as 2x + 7 = y + 7. Here, 7 is added to both sides, so if we remove it from both sides, the equation will still be true. This implies, 2x = y. What does the Addition Property of Equality State? The addition property of equality states that if the same amount is added to both sides of an equation, the equation still holds true. It is mathematically expressed as, X + n = Y + n, where n is any real number and X = Y is the given equation. What is the Addition Property of Equality and Inequality? The addition property holds true in the case of equations as well as with inequalities. With equations, the same quantity added to both sides does not change the equality between both sides. While, in the case of inequalities as well, the same quantity can be added to both sides without changing the relationship. If x > y, this implies x + a > y + a, ∀ a ∈ R. How to Solve an Equation Using the Addition Property of Equality? If an equation is given, say, m - 3 = 14, then we can apply the addition property of equality to solve for the variable 'm'. Add 3 to both sides so that only the variable will remain on the left side. By this, we get, m - 3 + 3 = 14 + 3 m = 17 Therefore, the value of m is 17. This is how we solve an equation using the addition property of equality. How to Use Addition Property of Equality? The addition property of equality in math is used to solve equations in math. The same number can be added to both sides of a given equation without changing the result. What is the Addition Property of Equality Formula? If the given equation is A = B, and the same quantity n is added to both sides of this equation, we get A + n = B + n. Hence, the formula for the addition property of equality is A = B ⇒ A + n = B + n. 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17194	https://thomas-cooke.uconn.edu/wp-content/uploads/sites/2963/2023/06/UG-2015.pdf	Full Terms & Conditions of access and use can be found at Download by: [University of Connecticut] Date: 11 November 2015, At: 08:44 Urban Geography ISSN: 0272-3638 (Print) 1938-2847 (Online) Journal homepage: The suburbanization of poverty? An alternative perspective Thomas J. Cooke & Curtis Denton To cite this article: Thomas J. Cooke & Curtis Denton (2015) The suburbanization of poverty? An alternative perspective, Urban Geography, 36:2, 300-313, DOI: 10.1080/02723638.2014.973224 To link to this article: Published online: 08 Dec 2014. Submit your article to this journal Article views: 246 View related articles View Crossmark data The suburbanization of poverty? An alternative perspective Thomas J. Cooke and Curtis Denton Department of Geography, University of Connecticut, 215 Glenbrook Road, U-4148, Storrs, CT 06269–4148, USA (Received 3 April 2013; accepted 4 July 2014) A popular and powerful narrative focuses on a crisis of suburban decline in the United States. However, a consensus regarding the scope and scale of one dimension of suburban decline—poverty—is hindered by the use of contradictory definitions of suburban space. This research presents an alternative approach to measuring suburban poverty that is less computationally intensive yet capable of capturing complex shifts in the spatial distribution of poverty within metropolitan areas. An analysis of the distribution of poverty in the largest 100 metropolitan areas between 1990 and 2007– 11 concludes that while poverty is increasing in the low-density suburbs of a handful of large metropolitan areas, the more general trend in most other metropolitan areas is an increase in poverty in moderately dense residential areas. Implicated in these trends are long-term trends in metropolitan area economic growth, a secular decline in inner-ring suburbs, and the impact of gentrification on housing opportunities for at-risk populations in large cities. Keywords: suburbs; poverty; methods; inner-ring suburbs Introduction A popular and powerful narrative in United States urbanism focuses on a crisis of suburban decline (Schafran, 2013; Short, Hanlon, & Vicino, 2007; Vicino, 2008b). Notably, Kneebone and Berube (2013a, pp. 1–2) summarize a decade of influential research on suburban poverty by the Brookings Institution by painting a picture of a nearly ubiquitous increase in suburban poverty: “From Cleveland’s long-struggling inner suburbs, to the immigrant portals south of Seattle, to aging communities surrounding Chicago, or the traditionally affluent Maryland suburbs of the nation’s capital—almost every major metropolitan area in the country has experienced rising poverty beyond its urban core.” Suburban decline and an increase in suburban poverty, in particular, are of particular importance because social services are disproportionately lacking in suburban settings (Allard, 2004; Murphy, 2010), while municipal fragmentation and reliance on property taxes strains the ability of suburbs—and older, inner-ring suburbs in particular— to deal with a growing, spatially dispersed poor population (Hudnut, 2003; Orfield, 2002; Pendall, Weir, & Narducci, 2013; Puentes & Warren, 2006; Vicino, 2008a). However, the broader evidence regarding suburban poverty indicates that it is not a universal phenomenon. Indeed, Kneebone and Berube’s (2013b) own data indicate that in only 42 of the 95 metropolitan areas for which they provide data did the share of the poor living in suburbs significantly increase between 1970 and 2011 and that the share of the poor living in suburbs actually decreased in 18 of these metropolitan areas. This Corresponding author. Email: thomas.cooke@uconn.edu Urban Geography, 2015 Vol. 36, No. 2, 300–313, © 2014 Taylor & Francis Downloaded by [University of Connecticut] at 08:44 11 November 2015 echoes existing peer-reviewed research that identifies the inner-ring suburbs of declining industrial cities and the outer-ring suburbs of rapidly growing Hispanic immigrant destinations as the nexus of suburban poverty (Cooke & Marchant, 2006; Hanlon, 2008; Hanlon, Vicino, & Short, 2006; Holliday & Dwyer, 2009; Madden, 2003; Puentes & Warren, 2006). Thus, inter-metropolitan patterns of suburban poverty appear to be driven by the same factors that drive urban poverty within a metropolitan area (Cooke, 2010; Cooke & Marchant, 2006): the size of the at-risk population, the degree of economic segregation, and the shifting location of housing opportunities for the poor (Jargowsky, 1997). That said, research on the suburbanization of poverty is limited in two important respects. First, only Kneebone and Berube (2013a) track the long-term trend in the suburbanization of poverty up to or through the deep recession of the late 2000s. Given the depth of the Great Recession, it is expected that the suburbanization of poverty has intensified and an appropriate approach would be to track this development over a long period of time to gauge its significance. Second, previous studies of suburban poverty each rely upon their own idiosyncratic measure of what constitutes a suburban area. The lack of a common definition and typology for suburbia makes it difficult to compare and reconcile different findings. Hence, the purpose of this research is to provide a reliable empirical picture of the emergence of suburban poverty in the 100 largest metropolitan areas from 1990 through the period covering the deep recession of the late 2000s. What is a suburb? Conceptualizing suburban space is problematic both theoretically and empirically (Soja, 2011). Empirically, there is no standard definition for what constitutes suburban space. Common practice is to rely upon Census definitions. Specifically, the United States Census identifies the principal city or cities (previously referred to by the Census as central cities) of each core-based statistical area (CBSAs include both metropolitan and micropolitan areas) on the basis of population size and/or concentration of employment: (1) the largest incorporated place with a 2010 Census population of at least 10,000 in the CBSA or, if no incorporated place of at least 10,000 population is present in the CBSA, the largest incorporated place or census designated place in the CBSA; and (2) any additional incorporated place or census designated place with a 2010 Census population of at least 250,000 or in which 100,000 or more persons work; and (3) any additional incorporated place or census designated place with a 2010 Census population of at least 50,000, but less than 250,000, and in which the number of workers working in the place meets or exceeds the number of workers living in the place; and (4) any additional incorporated place or census designated place with a 2010 Census population of at least 10,000, but less than 50,000, and at least one-third the population size of the largest place, and in which the number of workers working in the place meets or exceeds the number of workers living in the place (Office of Management and Budget, 2010, p. 37250). Based on population size, up to three principal cities are then used to name each metropolitan area. For example, what is commonly referred to as the San Jose metropo-litan area actually includes seven principal cities (i.e., San Jose, Sunnyvale, Santa Clara, Urban Geography 301 Downloaded by [University of Connecticut] at 08:44 11 November 2015 Mountain View, Milpitas, Palo Alto, and Cupertino) but only three of these appear in the official name (i.e., the San Jose-Sunnyvale-Santa Clara Metropolitan Statistical Area). Suburban space is then commonly defined as any part of a metropolitan area that is not part of a principal city (e.g., Holliday & Dwyer, 2009; Howell & Timberlake, 2013; Kneebone & Berube, 2013a; Madden, 2003; Vicino, 2008a). However, using census definitions of principal cities to define urban space and defining the residual portion of metropolitan areas as suburban space introduces a high level of misclassification. First, in metropolitan areas with underbounded principal cities (e.g., Boston, St. Louis, or San Francisco) there may be many smaller urban exclaves that are not identified as principal cities just because they have small populations or are not centers of employment. Second, in metropolitan areas with overbounded principal cities (e.g., Columbus, Indianapolis, or Jacksonville) many suburban enclaves are wrongly treated as urban. Third, the Census only defines principal cities on the basis of population size and employment concentration but these criteria exclude other, perhaps more impor-tant, dimensions of urbanism such as population density and population heterogeneity. Finally, reliance on population size to identify principal cities means that many large suburban municipalities may be classified as principal cities and consequently as urban. Thus, it is not readily clear the degree to which principal cities can be accurately assumed to be urban and whether the balance of metropolitan areas can be accurately assumed to be suburban. That said, there are two important benefits to using these Census-based definitions of urban and suburban space. First, they rely upon municipal boundaries and these are arguably the correct scale of analysis because it is at the municipal scale that policies designed to address poverty are traditionally directed and implemented. Second, despite their flaws the census-based definitions are objective and relatively easy to replicate, despite concern that researchers may inconsistently identify all of the principal cities in a metropolitan area. For example, Kneebone and Berube (2013a) use Census-based princi-pal cities to define urban and suburban space but then, with little justification, restrict their definition of urban space only to the first principal city identified in the name of a metropolitan area plus any other principal city identified in the name of a metropolitan area that has a population of at least 100,000. Such ad-hoc decisions regarding which principal cities to define as urban may affect conclusions regarding the scope and scale of suburban poverty (see Cooke, 2014). Recognizing the limitations to using Census-based definitions to identify urban and suburban spaces, another approach has been to identify suburban space by classifying census tracts or municipalities on the basis of variables such as population and housing density, relative location, contiguity, age of the housing stock and various demographic variables (e.g., Cooke, 2010; Cooke & Marchant, 2006; Hanlon, 2008, 2009; Hanlon et al., 2006; Holliday & Dwyer, 2009; Lee, 2011; Lee & Leigh, 2007; Leigh & Lee, 2005; Reibel, 2011; Séguin, Apparicio, & Riva, 2012). The primary benefit to a multivariate classification approach is that it allows for a more nuanced identification of types of urban and suburban space without the greater risk of misclassification inherent in the principal city approach. However, these approaches are both data and computationally intensive and particular to each study, making it difficult to make generalizations from study to study. For example, Cooke and Marchant (2006, p. 1974) classify all metropolitan census tracts in the United States according to the following scheme: The urban core of a metropolitan area consists of: census tracts with greater than 400 pre-1940 housing units per square mile; plus any contiguous tract which has both greater than 200 302 T.J. Cooke and C. Denton Downloaded by [University of Connecticut] at 08:44 11 November 2015 pre-1940 housing units per square mile and a population density of at least 1000 people per square mile. The inner ring of a metropolitan area consists of: any tract which is not labeled as part of the urban core; tracts with greater than 400 1950–69 housing units per square mile; plus any contiguous tract which has both greater than 200 1950–69 housing units per square mile and a population density of at least 1000 people per square mile. It seems unlikely that an approach this particular will ever be exactly replicated, thereby limiting the accumulation of consistent evidence with respect to the suburbanization of poverty. Given the limitations of both approaches to studying suburban poverty, the objective of this research is to construct a consistent, less data-intensive, and easily replicated method for measuring the suburbanization of poverty. The unit of analysis is the census tract. The argument that the municipality is the correct scale of analysis for the study of suburban poverty is countered by several realities. First, unlike principal cities, which vary greatly in extent and content, census tracts are consistently defined as neighborhood units with a population of about 4,000 people. Second, concentrations of poverty are neighborhood-level phenomena, and it is axiomatic that the scale of analysis should match the scale at which the phenomenon occurs. Third, the lines between urban and suburban spaces are increasingly blurred: there is increased diversity of both urban and suburban spaces (Swanstrom, Casey, Flack, & Dreier, 2004). Many inner-ring suburbs, in particular, are facing the same sets of issues that confronted old urban core municipalities a half century earlier (Hudnut, 2003). Finally, to continue to treat traditional urban cores as distinctly different from surrounding suburban spaces obfuscates the reality that addressing issues such as poverty—whether it be in urban or suburban spaces—requires regional—and not merely municipal—solutions (Dreier, Swanstrom, & Mollenkopf, 2004; Orfield, 2002). Thus, the focus here is on the types of metropolitan census tracts that have seen an increase in poverty rates. Metropolitan area poverty has traditionally been viewed as occurring in high-density neighborhoods. Of particular concern have been the consequences associated with the concentration of large numbers of poor people. Suburban poverty presents a different set of issues: the concern is with the difficulty of providing services to the poor in a low-density, politically fragmented environment (Hudnut, 2003; Orfield, 2002; Pendall et al., 2013; Puentes & Warren, 2006; Vicino, 2008a). Thus, a key difference between urban and suburban poverty is that one is associated with high-density neighborhoods and the other is associated with low-density neighborhoods. Indeed, density is a fundamental difference between urban and suburban space. Density is responsible for the interactions that form the basis of the urbanization economies that lie at the root of urbanism itself (Jacobs, 1969; Marshall, 1890; Mumford, 1961), and the lack of density is a fundamental char-acteristic of suburban space (Fishman, 1987; Jackson, 1985; Soja, 2011). Toward that end, the United States Census Bureau uses population density to create its definition of urban areas (U.S. Census, 2011), and most of the previously discussed classification schemes use population density as a primary variable distinguishing between types of urban and suburban census tracts and municipalities. Thus, a simple way to conceptualize the suburbanization of poverty is in terms of the relationship between census tract poverty rates and population density. Figure 1 plots this relationship for the Detroit metropolitan area and highlights those observations that are within the City of Detroit using data from the 2007–11 American Community Survey (see ‘Data and methods’ section). Included in this plot are predicted poverty rates based upon a fractional polynomial regression of the effect of census tract population density on census tract poverty rates (see ‘Data and methods’ section). Detroit is a particularly challenging Urban Geography 303 Downloaded by [University of Connecticut] at 08:44 11 November 2015 area for measuring the suburbanization of poverty because decades of population loss means that many of Detroit’s neighborhoods may be similar in many dimensions to suburban areas. However, Figure 1 indicates that even in this extreme case the relationship between poverty and density is clearly positive, and Detroit’s census tracts are generally both higher in poverty and density than all other census tracts. One benefit of thinking about the suburbanization of poverty in this way is that high-density neighborhoods in other principal cities or small urban exclaves are correctly treated as being more urban because of their higher density. Despite lower levels of poverty in the low-density (i.e., suburban) census tracts of the Detroit metropolitan area, the key question is whether poverty has been on the increase in these census tracts. Toward that end, Figure 2 plots fractional polynomial models for both Figure 1. Census tract poverty rates: Detroit, 2007–11 (Source: Author). Figure 2. Change in census tract poverty rates: Detroit, 1990 to 2007–11 (Source: Author). 304 T.J. Cooke and C. Denton Downloaded by [University of Connecticut] at 08:44 11 November 2015 1990 and 2007–11 along with their 95% confidence intervals. Where the confidence intervals do not overlap there is a statistically significant difference between the lines. Figure 2 also includes the fractional polynomial line that describes average housing age for 2007–11 as a function of census tract population density. These data suggest that within the Detroit metropolitan area, census tracts with the greatest increase in poverty between 1990 and 2007–11 are associated with density levels of around 5,000 people per square mile and a housing stock built around 1955. This is consistent with the first wave of post World War II suburbanization: in Detroit, poverty has been increasing within post World War II inner-ring suburban spaces along the boundaries of the city of Detroit and contiguous suburban municipalities (also see Jargowsky, 2003). This exercise highlights the problem of focusing on municipal boundaries rather than neighborhood types. For example, in Detroit the emergence of poverty in moderately dense neighborhoods represents the expansion of poverty in post World War II housing along the edges of the city of Detroit and surrounding inner-ring municipalities. However, in metropolitan areas with underbounded central cities these same places would be defined exclusively as suburban spaces and in metropolitan areas with overbounded central cities these places would be defined exclusively as urban spaces. Rather, this analysis more accurately defines these as moderately dense post World War II inner ring suburbs. The conclusion that poverty is not rapidly expanding in newer, low-density suburbs suggests that policy concerns regarding the need to deliver services to low-density suburban municipalities are unwarranted in this case. Rather, policy should be focused on areas that have traditionally been middle class and which, depending upon the spatial structure of a particular metropolitan area, may or may not lie within a traditionally defined urban municipality. This approach offers a more general method for evaluating the suburbanization of poverty across American metropolitan areas. The expectation is that the relationship between census tract population density and census tract poverty is generally positive, reflecting the traditional concentration of poverty in urban (i.e., more densely settled) areas. Then, if concentrations of poverty are emerging in suburban areas (i.e., less densely settled areas) the positive relationship between census tract population density and census tract poverty rates should become less positive over time. Data and methods To test this hypothesis, this analysis estimates fractional polynomial models of the effect of census tract population density on census tract poverty among the 100 largest metro-politan areas for 1990 and then again for 2007–11. Data are drawn from the International Public Use Microdata Sample (IPUMS) National Historical Geographic Information System (NHGIS) which provides census tract data on population density and poverty rates for the 1990 United States Census and the 2007–11 American Community Survey (ACS) (Minnesota Population Center, 2011). The 2007–11 ACS provides the most recently available period estimate for census tract poverty rates and reflects the changes the Census has made in reporting small-area social and demographic data since the 2000 Census. It is treated as a point estimate for comparison to the 1990 United States Census data on poverty rates. While both census tract definitions and metropolitan area bound-aries change over time, no attempt is made to address the changing boundaries of census tracts since the interest is not in how poverty changes in each census tract but in the Urban Geography 305 Downloaded by [University of Connecticut] at 08:44 11 November 2015 relationship between census tract poverty rates and density within a metropolitan area. To control for shifting metropolitan area boundaries, census tracts are selected that fall entirely within or overlap 2010 metropolitan area boundaries. However, metropolitan areas differ dramatically in their spatial structure. For example, it would be inappropriate to compare the relationship between the change in census tract poverty and 1990 census tract population density between New York City and Indianapolis. That is, a census tract with a population density of 5,000 people per square mile may be more of an inner-ring suburb in a city like Houston but more of a new suburban area in a metropolitan area like Chicago. So, it would be unwise to use population density alone as a measure to compare the position of a census tract on the urban–suburban continuum without first controlling for variations in spatial structure between metropolitan areas. To control for inter-metropolitan variations in spatial struc-ture, census tract population density values are standardized within each metropolitan area to a mean of zero and standard deviation of 1. This allows for a one-to-one comparison of census tracts between metropolitan areas based upon the degree to which a census tract is urban or suburban relative to each metropolitan area’s unique spatial structure. Both to simplify the analysis and to account for inter-metropolitan differences in economic and demographic structure, the analysis then pools data for similar types of metropolitan areas according to an existing classification of metropolitan areas which is based primarily on population growth, population diversity, and education (Berube et al., 2010) (see Figure 3): Figure 3. Classification of metropolitan areas (Source: Author). 306 T.J. Cooke and C. Denton Downloaded by [University of Connecticut] at 08:44 11 November 2015 ●Eight of the nine Next Frontier cities are west of the Mississippi river. These cities are rapidly growing, diverse, and well educated (e.g., Austin, TX; Seattle, WA; Dallas, TX; and Washington, DC). They are destinations for young internal migrants and their diverse and buoyant economies helped these cities escape the most extreme hardships of the deep recession of the late 2000s. ●New Heartland metropolitan areas include 19 cities that have diverse and vibrant service economies (e.g., Columbus, OH; Indianapolis, IN; Minneapolis, MN; and Columbia, SC). These are primarily in the Southeast and the Midwest and are characterized by high rates of population growth, low levels of diversity, and high educational attainment. Many of them have large universities, which is reflected in their relatively young age profile. ●Diverse Growth cities include many of the largest cities in the country (e.g., New York, NY; Chicago, IL; San Diego, CA; and San Francisco, CA). These nine cities have above average educational attainment and diversity but below average popu-lation growth due to their large sizes. ●There are 11 Border Growth cities along the southern United States border (e.g., Bakersfield, CA; Fresno, CA; Las Vegas, NV; and Riverside, CA). These cities have young, foreign-born populations with a high degree of educational inequality. They experienced rapid growth both through internal and international migration and were active in the housing boom and bust of the 1990s and 2000s. ●Mid-Size Magnet cities number 15 and are largely located in the South. These are slow-growing areas with low levels of educational attainment. Many of these are destinations for retirees and hence have an older age profile (e.g. Boise, ID; Cape Coral, FL; Jacksonville, FL; Lakeland, FL; and Tampa, FL). ●Skilled Anchors consist of 19 medium-to-large cities with diverse industrial econo-mies in the Northeast quadrant of the country (e.g., Boston, MA; Hartford, CT; Philadelphia, PA; and Pittsburgh, PA). These cities have slow population growth, low levels of diversity and high levels of educational attainment. ●There are 18 Industrial Core cities of the Northeast and the South (e.g., Birmingham, AL; Buffalo, NY; Detroit, MI; and Providence, RI). These cities have slow population growth, low levels of diversity, and lower education levels. The effect of standardized population density on census tract poverty is then estimated by type of metropolitan area for both 1990 and 2007–11 using fractional polynomial models. Fractional polynomial models provide an empirically driven and flexible approach to characterizing what is likely to be a curvilinear relationship between standardized census tract population density and census tract poverty rates. Following Royston and Altman (1994), fractional polynomial methods select from a suite of possible models of the form β0 þ β1xðp1Þ þ β2xðp2Þ þ ::: þ βmxðpmÞ where for a power; p; xðpÞ ¼ xp if p 0 log x if p ¼ 0 Standard practice is then to select the best two-factor model based upon the overall model fit where the choice of powers is restricted to the set (−2, −1, −0.5, 0, 0.5, 1, and 2). Urban Geography 307 Downloaded by [University of Connecticut] at 08:44 11 November 2015 Results Figures 4 through 10 present fractional polynomial curves for each of the seven types of metropolitan areas for both 1990 and 2007–11 with left-hand side values having higher densities (up to two standard deviations above the mean) and right-hand side values having lower densities (up to two standard deviations below the mean). In the most general sense, the relationship between population density and census tract poverty is similar across all seven types of cities in both 1990 and 2007–11: the poverty rate in the most densely settled tracts is about 25% to 30% which then declines to about 5% to 10% Figure 4. Next Frontier cities (Source: Author). Figure 5. New Heartland cities (Source: Author). 308 T.J. Cooke and C. Denton Downloaded by [University of Connecticut] at 08:44 11 November 2015 among the most sparsely settled tracts. In five of the seven types of cities (Next Frontier, Border Growth, New Heartland, Mid-Sized Magnets, and Industrial Core cities), this relationship between population density and census tract poverty is more convex in 1990 and more linear in 2007–11. As a result, these cities had an increase in poverty across the middle of the population density distribution. However, this effect is almost negligible among Border Growth cities. The exceptions to this pattern are Diverse Growth cities, which experienced an increase in poverty only in low-density neighborhoods, and Skilled Anchors, which had an increase in poverty in all types of neighborhoods. Figure 6. Diverse Growth cities (Source: Author). Figure 7. Border Growth cities (Source: Author). Urban Geography 309 Downloaded by [University of Connecticut] at 08:44 11 November 2015 The expansion of poverty in the middle of the population density distribution among Next Frontier, Border Growth, New Heartland, Mid-Sized Magnets, and Industrial Core cities is consistent with previous research that points toward inner-ring suburbs as the locus of suburban poverty: smaller municipalities with an older housing stock, aging infrastructure, and high tax rates are unable to compete with newer suburbs for more affluent residents. This creates new housing opportunities for less affluent residents, resulting in an increase in poverty in medium-density neighborhoods at the boundary of the traditional urban core. The importance of the shifting location of housing opportunities Figure 8. Mid-Sized Magnet cities (Source: Author). Figure 9. Skilled Anchor cities (Source: Author). 310 T.J. Cooke and C. Denton Downloaded by [University of Connecticut] at 08:44 11 November 2015 for less affluent residents in shaping the changing location of poverty may also explain the expansion of poverty in the low-density neighborhoods of large, Diverse Growth cities. In this case, however, high rates of gentrification in these cities suggest that the increase in poverty in low-density neighborhoods may be due to the lack of housing opportunities for at-risk populations in higher density neighborhoods. These results also suggest that the long-term economic vitality of a metropolitan area may play a role in the emergence of poverty across the middle of the population density distribution. Specifically, cities that have enjoyed long-term economic vitality had the smallest increase in poverty across the middle of the population density distribution (i.e., Border Growth and Next Frontier cities), while cities experiencing long-term decline had an increase in poverty across all of the population density distribution (i.e., Skilled Anchors). Indeed, the recent housing and economic crisis appears to have had little effect on the shifting location of poverty. For example, Border Growth cities were particularly challenged by the housing and economic crisis but these saw the smallest increase in poverty at any location of the population density distribution. Conclusions This research qualifies the narrative of universal suburban decline. While suburban poverty appears to be on the increase in nearly every type of metropolitan area, its spatial expression appears to be contingent upon the particular characteristics of a metropolitan area. Thus, while poverty is increasing in the low-density suburbs of a handful of the largest metropolitan areas, the more general trend is of an increase in poverty in medium-density neighborhoods at the boundaries of the traditional urban core. Implicated are long-term trends in metropolitan area economic growth, a secular decline of inner-ring suburbs, and the shifting location of housing opportunities for at-risk populations in large cities. This research has approached the problem of defining suburban space as an empirical issue. In the name of expediency, applied empirical urban research willfully ignores the high degree of misclassification caused by census-based definitions of “urban” and Figure 10. Industrial Core cities (Source: Author). Urban Geography 311 Downloaded by [University of Connecticut] at 08:44 11 November 2015 “suburban.” Alternative multivariate classification procedures may more accurately reflect the diversity of places along a continuum from “urban” to “suburban” but the complex-ities inherent to classification procedures mean that few studies will be replicated. This research presents a method that is less data and computationally intensive and is more likely to be replicated. References Allard, Scott W. (2004). Access to social services: The changing urban geography of poverty and service provision. Washington, DC: The Brookings Institution. Berube, Alan, Frey, William H., Friedhoff, Alec, Garr, Emily, Emilia, Istrate, Kneebone, Elizabeth, & Wilson, Jill H. (2010). State of Metropolitan America: On the front lines of demographic transformation. Washington, DC: The Brookings Institution. Cooke, Thomas J., & Marchant, Sarah Haight. (2006). The changing intrametropolitan location of high-poverty neighbourhoods in the US, 1990–2000. Urban Studies, 43(11), 1971–1989. Cooke, T. J. (2010). Residential mobility of the poor and the growth of poverty in inner-ring suburbs. Urban Geography, 31(2), 179–193. Cooke, T. J. (2014). Book review: Confronting suburban poverty in America. Economic Development Quarterly, 28(2), 182. Dreier, Peter, Swanstrom, Todd, & Mollenkopf, J. H. (2004). Place matters: Metropolitics for the twenty-first century. Lawrence: University Press of Kansas. Fishman, Robert. (1987). Bourgeois Utopias: The rise and fall of suburbia. New York, NY: Basic Books. Hanlon, Bernadette. (2008). The decline of older, inner suburbs in metropolitan America. Housing Policy Debate, 19(3), 423–456. Hanlon, Bernadette. (2009). A typology of inner‐ring suburbs: Class, race, and ethnicity in US Suburbia. City & Community, 8(3), 221–246. Hanlon, Bernadette, Vicino, Thomas, & Short, John Rennie. (2006). The new metropolitan reality in the US: Rethinking the traditional model. Urban Studies, 43(12), 2129–2143. Holliday, A. L., & Dwyer, R. E. (2009). Suburban neighborhood poverty in U.S. metropolitan areas in 2000. City & Community, 8(2), 155–176. Howell, Aaron J., & Timberlake, Jeffrey M. (2013). Racial and ethnic trends in the suburbanization of poverty in US metropolitan areas, 1980–2010. Journal of Urban Affairs, 36, 79–98. Hudnut, William H. (2003). Halfway to everywhere: A portrait of America’s first-tier suburbs. Washington, DC: Urban Land Institute. Jackson, Kenneth T. (1985). Crabgrass frontier. New York, NY: Oxford University Press. Jacobs, Jane. (1969). The economy of cities. New York, NY: Random House. Jargowsky, Paul A. (1997). Poverty and place: Ghettos, barrios, and the American city. New York, NY: Russell Sage Foundation. Jargowsky, Paul A. (2003). Stunning progress, hidden problems: The dramatic decline of concen-trated poverty in the 1990s. Washington, DC: The Brookings Institution. Kneebone, Elizabeth, & Berube, Alan. (2013a). Confronting suburban poverty in America. Washington, DC: Brookings Institution Press. Kneebone, Elizabeth, & Berube, Alan. (2013b). Confronting suburban poverty in America. Retrieved from Lee, Sugie. (2011). Analyzing intra-metropolitan poverty differentiation: Causes and consequences of poverty expansion to suburbs in the metropolitan Atlanta region. The Annals of Regional Science, 46(1), 37–57. Lee, Sugie, & Leigh, N G. (2007). Intrametropolitan spatial differentiation and decline of inner-ring suburbs: A comparison of four US metropolitan areas. Journal of Planning Education and Research, 27(2), 146–164. Leigh, Nancey Green, & Lee, Sugie. (2005). Philadelphia’s space in between: Inner ring suburb evolution. Opolis, 1(1), 13–32. Madden, Janice Fanning. (2003). The changing spatial concentration of income and poverty among suburbs of large US metropolitan areas. Urban Studies, 40(3), 481–503. Marshall, A. (1890). Principles of economics. London: MacMillan. 312 T.J. Cooke and C. Denton Downloaded by [University of Connecticut] at 08:44 11 November 2015 Minnesota Population Center. (2011). National historical geographic information system: Version 2.0. Minneapolis: University of Minnesota. Mumford, Lewis. (1961). The city in history: Its origins, its transformations, and its prospects. New York, NY: Harcourt, Brace & World. Murphy, A. K. (2010). The symbolic dilemmas of suburban poverty: Challenges and opportunities posed by variations in the contours of suburban poverty. Sociological Forum, 25(3), 541–569. Office of Management and Budget. (2010). 2010 standards for delineating metropolitan and micro-politan statistical areas. Federal Register, 75, 123. Orfield, Myron. (2002). American metropolitics: The new suburban reality. Washington, DC: The Brookings Institution Press. Pendall, Rolf, Weir, Margaret, & Narducci, Chris. (2013). Governance and the geography of poverty: Why does suburbanization matter? Berkeley: University of California. Puentes, Robert, & Warren, David. (2006). One-fifth of the nation: America’s first suburbs in transition. Washington, DC: Brookings Institution. Reibel, Michael. (2011). Classification approaches in neighborhood research: Introduction and review. Urban Geography, 32(3), 305–316. Royston, P., & Altman, D. G. (1994). Regression using fractional polynomials of continuous covariates: Parsimonious parametric modelling. Applied Statistics, 43, 429–467. Schafran, Alex. (2013). Discourse and dystopia, American style: The rise of ‘slumburbia’ in a time of crisis. City, 17(2), 130–148. Séguin, A.-M., Apparicio, Philippe, & Riva, Mylène. (2012). Identifying, mapping and modelling trajectories of poverty at the neighbourhood level: The case of Montréal, 1986–2006. Applied Geography, 35(1–2), 265–274. Short, J. R., Hanlon, Bernadette, & Vicino, T. J. (2007). The decline of inner suburbs: The new suburban Gothic in the United States. Geography Compass, 1(3), 641–656. Soja, Edward W. (2011). Regional urbanization and the end of the metropolitan era. In G. Bridge & S. Watson (Eds.), The new Blackwell companion to the city (pp. 679–689). London: Wiley-Blackwell. Swanstrom, Todd, Casey, Colley, Flack, Robert, & Dreier, Peter. (2004). Pulling apart: Economic segregation among suburbs and central cities in major metropolitan areas. Washington, DC: The Brookings Institution. U.S. Census. (2011). Urban area criteria for the 2010 census; notice. Federal Register, 76(164), 53030–53043. Vicino, T. J. (2008a). The spatial transformation of first‐tier suburbs, 1970 to 2000: The case of metropolitan Baltimore. Housing Policy Debate, 19(3), 479–518. Vicino, T. J. (2008b). The quest to confront suburban decline: Political realities and lessons. Urban Affairs Review, 43(4), 553–581. 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17195	https://satprepget800.com/2016/05/30/arithmetic-sequences-act-gre-prep/	Arithmetic Sequences - ACT and SAT Subject Test Prep Home Resources About Contact Books Blog Join the Newsletter Learn strategies to help you get a perfect score and into your dream school with the Get 800 Newsletter. Arithmetic Sequences Hello everyone. Today I would like to talk about arithmetic sequences. Questions involving arithmetic sequences appear on the ACT, GRE and SAT math subject tests. Here is an example of an arithmetic sequence. Example 1 1, 4, 7, 10, 13, 16,… Note that the first term of this sequence is 1, the second term of this sequence is 4, and so on. So what makes this sequence arithmetic? Well notice that to get from 1 to 4 we need to add 3. To get from 4 to 7 we also add 3. To get from 7 to 10 we also add 3. In other words, the sequence is arithmetic because we always add the same number to get from any term to the next term. This special number is called the common difference of the arithmetic sequence. So why is this number called the common difference? Well another way to compute the common difference is to note that when we subtract any term from the next term we always get the same number, in this case that number is d = 3. In other words we have 4 – 1 = 3, 7 – 4 = 3, and so on. An arithmetic sequence is a sequence of numbers such that the difference between consecutive terms is constant. The number d is called the common differenceof the arithmetic sequence. Let’s try a simple example. Example 2 The second term of an arithmetic sequence is 15 and the third term is 10. What is the first term? A.–15 B. –10 C. 1/15 D. 10 E. 20 Solution:Moving backwards, to get from the third term to the second term we add 5. Therefore we add 5 more to get to the first term. So the first term is 15 + 5 = 20. This is choice E. Note that in an arithmetic sequence, you always add (or subtract) the same number to get from one term to the next. This can be done by moving forwards or backwards through the sequence. Note also that the common difference of this sequence is d = 10 – 15 = –5. Many students might mistakenly say that the common difference is 5. This particular problem was pretty simple, so we were able to solve it just by “counting.” In other words we didn’t really have to worry about the formalities of whether the common difference was positive or negative. But in harder questions we might need to be more careful. Alternate solution:Note that the terms of the sequence are getting smaller so that the first term must be larger than 15. So the answer must be choice E. In my next post we will learn a special technique that makes many seemingly difficult problems involving arithmetic sequences very easy to solve.Check it out here:Arithmetic Sequences and Linear Equations More Practice with Arithmetic Sequences If you are preparing for a standardized test, you may want to take a look at the Get 800 collection of test prep books. Click on the picture below for more information. Comments comments
17196	https://stats.libretexts.org/Bookshelves/Introductory_Statistics/OpenIntro_Statistics_(Diez_et_al)./01%3A_Introduction_to_Data/1.05%3A_Observational_Studies_and_Sampling_Strategies	Skip to main content 1.5: Observational Studies and Sampling Strategies Last updated : Apr 23, 2022 Save as PDF 1.4: Overview of Data Collection Principles 1.6: Experiments Page ID : 310 David Diez, Christopher Barr, & Mine Çetinkaya-Rundel OpenIntro Statistics ( \newcommand{\kernel}{\mathrm{null}\,}) Observational Studies Generally, data in observational studies are collected only by monitoring what occurs, what occurs, while experiments require the primary explanatory variable in a study be assigned for each subject by the researchers. Making causal conclusions based on experiments is often reasonable. However, making the same causal conclusions based on observational data can be treacherous and is not recommended. Thus, observational studies are generally only sufficient to show associations. Exercise Suppose an observational study tracked sunscreen use and skin cancer, and it was found that the more sunscreen someone used, the more likely the person was to have skin cancer. Does this mean sunscreen causes skin cancer? Solution No. See the paragraph following the exercise for an explanation. Some previous research tells us that using sunscreen actually reduces skin cancer risk, so maybe there is another variable that can explain this hypothetical association between sunscreen usage and skin cancer. One important piece of information that is absent is sun exposure. If someone is out in the sun all day, she is more likely to use sunscreen and more likely to get skin cancer. Exposure to the sun is unaccounted for in the simple investigation. Sun exposure is what is called a confounding variable (also called a lurking variable, confounding factor, or a confounder), which is a variable that is correlated with both the explanatory and response variables. While one method to justify making causal conclusions from observational studies is to exhaust the search for confounding variables, there is no guarantee that all confounding variables can be examined or measured. In the same way, the county data set is an observational study with confounding variables, and its data cannot easily be used to make causal conclusions. Exercise Figure 1.9 shows a negative association between the homeownership rate and the percentage of multi-unit structures in a county. However, it is unreasonable to conclude that there is a causal relationship between the two variables. Suggest one or more other variables that might explain the relationship visible in Figure 1.9. Solution Answers will vary. Population density may be important. If a county is very dense, then this may require a larger fraction of residents to live in multi-unit structures. Additionally, the high density may contribute to increases in property value, making homeownership infeasible for many residents. Observational studies come in two forms: prospective and retrospective studies. A prospective study identifies individuals and collects information as events unfold. For instance, medical researchers may identify and follow a group of similar individuals over many years to assess the possible influences of behavior on cancer risk. One example of such a study is The Nurses Health Study, started in 1976 and expanded in 1989. This prospective study recruits registered nurses and then collects data from them using questionnaires. Retrospective studies collect data after events have taken place, e.g. researchers may review past events in medical records. Some data sets, such as county, may contain both rospectively- and retrospectively-collected variables. Local governments prospectively collect some variables as events unfolded (e.g. retails sales) while the federal government retrospectively collected others during the 2010 census (e.g. county population counts). Three Sampling Methods Almost all statistical methods are based on the notion of implied randomness. If observational data are not collected in a random framework from a population, these statistical methods are not reliable. Here we consider three random sampling techniques: simple, stratified, and cluster sampling. Figure 1.14 provides a graphical representation of these techniques. Simple random samplingis probably the most intuitive form of random sampling. Consider the salaries of Major League Baseball (MLB) players, where each player is a member of one of the league's 30 teams. To take a simple random sample of 120 baseball players and their salaries from the 2010 season, we could write the names of that season's 828 players onto slips of paper, drop the slips into a bucket, shake the bucket around until we are sure the names are all mixed up, then draw out slips until we have the sample of 120 players. In general, a sample is referred to as "simple random" if each case in the population has an equal chance of being included in the nal sample and knowing that a case is included in a sample does not provide useful information about which other cases are included. Stratified sampling is a divide-and-conquer sampling strategy. The population is divided into groups called strata. The strata are chosen so that similar cases are grouped together, then a second sampling method, usually simple random sampling, is employed within each stratum. In the baseball salary example, the teams could represent the strata; some teams have a lot more money (we're looking at you, Yankees). Then we might randomly sample 4 players from each team for a total of 120 players. Stratified sampling is especially useful when the cases in each stratum are very similar with respect to the outcome of interest. The downside is that analyzing data from a stratified sample is a more complex task than analyzing data from a simple random sample. The analysis methods introduced in this book would need to be extended to analyze data collected using stratified sampling. Example Why would it be good for cases within each stratum to be very similar? Solution We might get a more stable estimate for the subpopulation in a stratum if the cases are very similar. These improved estimates for each subpopulation will help us build a reliable estimate for the full population. A cluster sampleis much like a two-stage simple random sample. We break up the population into many groups, called clusters. Then we sample a fixed number of clusters and collect a simple random sample within each cluster. This technique is similar to stratified sampling in its process, except that there is no requirement in cluster sampling to sample from every cluster. Stratified sampling requires observations be sampled from every stratum. Sometimes cluster sampling can be a more economical random sampling technique than the alternatives. Also, unlike stratified sampling, cluster sampling is most helpful when there is a lot of case-to-case variability within a cluster but the clusters themselves don't look very different from one another. For example, if neighborhoods represented clusters, then this sampling method works best when the neighborhoods are very diverse. A downside of cluster sampling is that more advanced analysis techniques are typically required, though the methods in this book can be extended to handle such data. Example Suppose we are interested in estimating the malaria rate in a densely tropical portion of rural Indonesia. We learn that there are 30 villages in that part of the Indonesian jungle, each more or less similar to the next. Our goal is to test 150 individuals for malaria. What sampling method should be employed? Solution A simple random sample would likely draw individuals from all 30 villages, which could make data collection extremely expensive. Stratified sampling would be a challenge since it is unclear how we would build strata of similar individuals. However, cluster sampling seems like a very good idea. First, we might randomly select half the villages, then randomly select 10 people from each. This would probably reduce our data collection costs substantially in comparison to a simple random sample and would still give us reliable information. 1.4: Overview of Data Collection Principles 1.6: Experiments
17197	https://www.jmap.org/JMAPRegentsExamArchives/ALGEBRAIIEXAMS/ExamAnswers/0616ExamAIIans.pdf	ALGEBRA II (COMMON CORE) The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA II (Common Core) Wednesday, June 1, 2016 -9:15 a.m. to 12:15 p.m., only Student Name: /V1Y. ); b.o} School Name:5J/V]---+--~~f _ _ The possession or use of any communications device is strictly prohibited when taking this examination. If you have or use any communications device, no matter how briefly, your examination will be invalidated and no score will be calculated for you. Print your name and the name of your school on the lines above. A separate answer sheet for Part I has been provided to you. Follow the instructions from the proctor for completing the student information on your answer sheet. This examination has four parts, with a total of 37 questions. You must answer all questions in this examination. Record your answers to the Part I multiple-choice questions on the separate answer sheet. Write your answers to the questions in Parts II, III, and IV directly in this booklet. All work should be written in pen, except graphs and drawings, which should be done in pencil. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. Utilize the information provided for each question to determine your answer. Note that diagrams are not necessarily drawn to scale. The formulas that you may need to answer some questions in this examination are found at the end of the examination. This sheet is perforated so you may remove , it from this booklet. Scrap paper is not permitted for any part of this examination, but you may use the blank spaces in this booklet as scrap paper. A perforated sheet of scrap graph paper is provided at the end of this booklet for any question for which graphing may be helpful but is not required. You may remove this sheet from this booklet. Any work done on this sheet of scrap graph paper will not be scored. When you have completed the examination, you must sign the statement printed at the end of the answer sheet, indicating that you had no unlawful knowledge of the questions or answers prior to the examination and that you have neither given nor received assistance in answering any of the questions during the examination. Your answer sheet cannot be accepted if you fail to sign this declaration. Notice ... A graphing calculator and a straightedge (ruler) must be available for you to use while taking this examination. DO NOT OPEN mis EXAMINATION BOOKLET UNTIL mE SIGNAL IS GIVEN. (31::108 NOV\111\108) II Vl::l8381V Part I Answer all 24 questions in this part. Each correct answer will receive 2 credits. No partial credit will he allowed. Utilize the information provided for each question to determine your answer. Note that diagrams are not necessarily drawn to scale. For each statement or question, choose the word or expression that, of those given, best completes the statement or answers the question. Record your answers on your separate answer sheet. [ 48] 2 1 When b > 0 and d is a positive integer, the expression (3b)d is equivalent to (1) 1 (~r (3) l_ J3i1 (2) (5bt 2 Julie averaged 85 on the first three tests of the semester in her mathematics class. If she scores 93 on each of the remaining tests, her average will be 90. Which equation could be used to determine how many tests, T, are left in the seme~ ¥S" X 3 ( l) 255 + 93T = 90 /'(3)1 255 + 93T = 90 ~ ~ T+3 (2) 255 + 90T = 93 (4) 255 + 90T = 93 ~ T+3 3 Given i is the imaginary unit, (2 - yi)2 in simplest form is (1) y2 - 4yi + 4 (3) -y2 + 4 @ -y2 - 4yi + 4 ( 4) y2 + 4 (J--yi)l;J--yt) lf -4yi+yJlJ---il--4yi r 4-Algebra II (Common Core) -June '16 Use this space for computations. 4 Which graph has the following characteristics? • three real zeros • as x ---+ -oo, f(x) ---+ -oo • as x ---+ oo, f(x) ---+ oo y y (1) y y (2) (4) 5 The solution set for the equation .J55 - x =xis (1) {-8,7} @V{7} . (2) {-7,8} (4) { } Algebra II (Common Core) -June '16 Use this space for computations. [OVER] Use this space for 6 The zeros for f(x). = x4 - 4x3 - 9x2 + 36x are c computations. @{o,±3,4} (3) {0,±3,-4} X ,( 1 -9x ~ -1')( t5 b] ~ 0 (2) {0.3.4} (4) {0,3.-4} x lxiCx-1.J )-9lx -4-~-; o x l y ) -Cf) l x , 4) ~ 0 7 Anne has a coin. She does not know if it is a f~r Lx sU~~~d-~) ( X ""}.) 1 O the coin 100 times and obtained 73 heads and 27 tails. She ran a computer simulation of 200 samples of 100 fair coin flips. The output of the proportion of heads is shown below. 30 Samples = 200 Mean= 0.497 SD= 0.050 20 10 : I I 0 : : i I . : I 0.35 0.40 0.45 0.50 0.55 0.60 Given the results of her coin flips and of her computer simulation, which statement is most accurate? (1) 73 of the computer's next 100 coin flips will be heads. ~ 50 of her next 100 coin flips will be heads. c.& Her coin is not fair. (4) Her coin is fair. 8 If g(c) = 1 - c2 and m(c) = c + 1, then which statement is not true? / (1) g(c) • m(c) = 1 + c - c2 - c3 ~· . . C-f l _ ') ·. -·: (2) g(c) + m(c) = 2 + c - c2 q l (.,, · j _ G J-l ~ ~X 7 -G • (3) m(c) - g(c) = c + c2 ti C ~/ ~m(c) =_2. ~ g(c) 1 - c -.-Algebra II (Common Core) - June '16 9 The heigh~s of women in the United States are normally distributed with a mean of 64 inches and a standard deViation of 2. 75 inches. The percent of women whose heights are between 64 and 69.5 inches9 to the nearest whole percent, is {..; q, 5 - b . ( 1) 6 ( 3) 68 ~ ~ '1 @48 (4) 95 cf. ,,7) 10 The formula below can be used to model which scenario? a 1 = 3000 an = 0.80an _ 1 (1) The first row of a stadium has 3000 seats, and each row thereafter has 80 more seats than the row in front of it. (2) The last row of a stadium has 3000 seats, and each row before it has 80 fewer seats than the row behind it. (3) A bank account starts with a deposit of $3000, and each year it grows by 80%. @The initial value of a specialty toy is $3000, and its value each of the following years is 20% less. 11 Sean's team has a baseball game tomorrow. He pitches 50% of the games. There is a 40% chance of rain during the game tomorrow. If the probability that it rains given that Sean pitches is 40%, it can be concluded that these two events are @independent (3) mutually exclusive (2) dependent (4) complements p (fl) ~ p L R_ I >) Algebra II (Common Core) -June '16 Use this space for computations. [OVER] Use this space for 12 A solution of the equation 2x2 + 3x + 2 ~ 0 is 2 .,. .}j ~- computations. 11'. ,1-'"-WJJtlJ fi1' 3 1 {;; 3 1 {;; \j • r ') / .:J -4 + 4iv7 (3) - 4 + 4-v1 I' "'-!, -,,} (2) _2 + li 4 4 (4) ! 2 .. }ttf-7 ~ - Ltf if7 y-~ 13 The Ferris wheel at the landmark Navy Pier in Chicago :tes 7 minutes to make one full rotation. The height, H, in feet, above the ground of one of the six-person cars can be modeled by H(t) = 70 sin (2; (t-1.75)) + 80, where tis time, in minutes. Using H(t) for one full rotation, this car's minimum height, in feet, is (1) 150 (2) 70 x -G:xl--y -fxrf,, Y 17 A circle centered at the origin has a radius of 10 units. The terminal side of an angle, 0, intercepts the circle in Quadrant II at point C. ;,;; y-:oordmate of point C is 8. ~at is the value of cos 6?); h U..- f!} j ~ V1Jf'J i It.ti+~ ~-5 (3)5 · J\i Qvf . v -:; {) (3) The domain is the set of positive reals. 0 t/A, 1' J->< 0 0.00 6 7 8 9 10 pH Which statement about this function is incorrect? ).!.), The degree of the polynomial is even. ~ ~ i 5 {J9 There is a positive leading coefficient. (3) At two pH values, there is a relative maximum value. ( 4) There are two intervals where the function is decreasing. 21 Last year, the total revenue for Home Style, a national restaurant chain, increased 5.25% over the previous year. If this trend were to continue, which expression could the company's chief financial officer Use this space for computations. use to approximate their monthly percent increase in revenue? /I ~~\~:::;~ent months.] @(l.0042?)m 0 ) }-s /] )- , /, 0 (}f) I tl m /, (2) (1.0525) m ( 4) (1.00427)12 Algebra II (Common Core) -June '16 22 Which value, to the nearest tenth, is not a solution of p(x) = q(x) if Use this space for computations. p(x) = x3 + 3x2 - 3x - 1 andq(x) = 3x + 8? (1) -3.9 .~ 2.1 (2) -1.1 ~4.7 u~. 1vaph111 Clf.;/ lv )!if bY 23 The population of Jamesburg for the years 2010-2013, respectively, was reported as follows: 250,000 250,937 251,878 252,822 How can this sequence be recursively modeled? (1) jn = 250,000(l.00375)n - 1 (2) jn = 250,000 + 937(n - l) ©!1: 250,000. Jn - 1.00375 Jn_ 1 (4) h = 250,000 j n = j n - 1 + 937 24 The voltage used by most households can be modeled by a sine function. The maximum voltage is 120 volts, and there are 60 cycles every second. Which equation best represents the value of the voltage as it flows through the electric wires, where t is time in seconds? (1) V = 120 sin (t) ~ V = 120 sin (603tt) (2) V = 120 sin (60t) {!!J) V = 120 sin (1203tt) ?:JI-'8 ~ Pff-~ )/-Di/ Algebra II (Common Core)-June '16 [OVER] Part II Answer all 8 questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. Utilize the information provided for each question to determine your answer. Note that diagrams are not necessarily drawn to scale. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. All answers should be written in pen, except for graphs and drawings, which should be done in pencil. 25 Solve for x: 1 1 1 = x 3 3x h / -1 /'"" 1x )X ·~J /x r -I , .. y , .X .. Algebra II (Common Core) -June '16 [OVER] 26 Describe how a controlled experiment can be created to examine the effect of ingredient X in a toothpaste. ~ Y) i IVI h; a s 5 j q h }A Y ft (,f pQ r11 > f () f wb 9v 0v;s) o~ us-1'n7 f rJf/fhf(l..>1e wi I h l» I 1-td/errf '>() O/Le w~-r4w1' Algebra II (Common Core) - June '16 [OVER] 27 Determine if x - 5 is a factor of 2x3 -4x2 - 7x - 10. Explain yoY{ answer. , y,-D 'J { 5) 1 -9 { S) 1 ~ 7 ( s) -J D · . X "- 5 ) s {) - / 0 0 / 5 > --1 6 )OS =ID <;;((10 t;' /5 ft!), 6 £-cJd>) X-) J) ndf et. (AlfOv Algebra II (Common Core) -June '16 28 On the axes below, graph one cycle of a cosine function with amplitude 3, period ~~ midline y = -1, and passing through the point (0,2). y Algebra II (Common Core) -June '16 [OVER] 29 A suburban high school has a population of 1376 students. The number of students who participate in sports is 649. The number of students who participate in music is 433. If_ the probability that a student participates in either sports or music is 974 , what is the probability 1376 that a student participates in both sports and music? Jot --.... Algebra II (Common Core)-June '16 30 The directrix of the parabola 12(y + 3) = (x - 4)2 has the equation y = -6. Find the coordinates of the focus of the parabola. I 1 ) I + ; ? -: (,1 ~ 9 p /Qy I; &- y ~-J-b ~ y j}/ y ',-J: {x--4) k 3 1~ tHvfe1 h {itJ-3J (y-Yj~; 9L~){y+~) p~ 3 is [ 9; D) Algebra II (Common Core) -June '16 [OVER] 3 31 Algebraically prove that x 3 + 9 = 1 ++,where x =I= -2. x +8 x +8 I _ ft' -1-Z x 1 t i }<) + CJ)( 1-- t 0 y f 9 y7 +~ J Algebra II (Common Core) -June '16 32 A house purchased 5 years ago for $100,000 was just sold for $135,000. Assuming exponential growth, approximate the annual growth rate, to the nearest percent. ,o6~r • Algebra II (Common Core) -June '16 [OVER] Part III Answer all 4 questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. Utilize the information provided for each question to determine your answer. Note that diagrams are not necessarily drawn to scale. For all questions in this part, a correct numerical answer with no work shown will receive only I credit. All answers should be written in pen, except for graphs and drawings, which should be done in pencil. 33 Solve the system of equations shown below algebraically. (x -3)2 + (y + 2)2 = 16 2x + 2u = 10 ~ -'.'- -;-}->--· y ~ ,xt5 (x-~) J_ + ( x s t 1-)1 '1 I b X t-,.(X t 1 t XL J l;:xtY9 ~ 16 . pi-J-6.Y19f/ 0 1 1- -l6x rtJ ~, 0 (!-!) {A/S) 1 6 X '/I ) yr.- I/-) ~ / ]_ {1,-J) Algebra II (Common Core) - June '16 y .,,] f s Y" -; OJ 7-J 34 Alexa earns $33,000 in her first year of teaching and earns a 4% increase in each successive year. Write a geometric series formula, Sn, for Alexa's total earnings over n years. Use this formula to find Alexa's total earnings for her first 15 years of teaching, to the nearest cent. ) / 5 '))DUO ,,,~~ovo{/.04 -~ 6{;01171#$9 ),, ~ ~ 1~/,04 Algebra II (Common Core) -June '16 [OVER] 35 Fifty-five students attending the prom were randomly selected to participate in a survey about the music choice at the prom. Sixty percent responded that a DJ would be preferred over a band. Members of the prom committee thought that the vote would have 50% for the DJ and 50% for the band. A simulation was run 200 times, each of sample size 55, based on the premise that 60% of the students would prefer a DJ. The approximate normal simulation results are shown below. ~ i 16-----------::::s [ u. 12------------0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 Survey Proportion Using the results of the simulation, determine a plausible interval containing the middle 95% of the data. Round all val6~ 1C,~)':f ).{~~e~{.6) Members of the prom committee are concerned that a vote of all students attending the prom may produce a 50% - 50% split. Explain what statistical evidence supports this concern. , ) () ; s w J h i YI I A 1' > i YI f eYllA--~ <; 0 1 t~ po>'>,~;~ to <j ~ n. JJ-> b ha. 5 CA 9~4-ev of ckt1o.'rtf [OVER] Part IV Answer the question in this part. A correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. Utilize the information provided to determine your answer. Note that diagrams are not necessarily drawn to scale. A correct numerical answer with no work shown will receive only 1 credit. All answers should he written in pen, except for graphs and drawings, which should be done in pencil. [ 6] 37 Drugs break down in the human body at different rates and therefore must be prescribed by doctors carefully to prevent complications, such as overdosing. The breakdown of a drug is represented by the function N(t) = N0(e)-rt, where N(t) is the amount left in the body, N0 is the initial dosage, r is the decay rate, and tis time in hours. Patient A, A(t), is given 800 milligrams of a drug with a decay rate of 0.347. Patient B, B(t), is given 400 milligrams of another drug with a decay rate of 0.231. Write two functions, A(t) and B(t), to represent the breakdown of the respective drug given to each patient. I ) _ '/V7,t Act -:· of Algebra II (Common Core) - June '16
17198	https://langfordmath.com/ECEMath/Geometry/SymIntroText.html	Introduction to Symmetry Transformations: Introduction to symmetry There are two basic kinds of symmetry: line or reflection symmetry, where the shape looks the same on both sides of a mirror line, and rotational symmetry, where you can turn the shape through some angle and it will look the same as before you turned it. Children ususally learn about and can recognize reflection symmetry (gr K-2) before they can visualize rotational symmetry (gr 3-5) Line symmetry Rotational symmetry Line Symmetry Line symmetry is usually named by the line where you would put a mirror, showing how the two sides of the shape are reflections of each other. Depending on how you draw them or what font you use, some capital letters have symmetry lines. A letter A will (usually) have a vertical line going through its center where the A is the same on the left and right. Because the mirror line is vertical, we say A has vertical line symmetry. The letter D usually has a mirror line going horizontally through its center, and the D looks the same on the top and the bottom, so we say it has horizontal line symmetry. The letter H is interesting because it has two mirror lines--the vertical line through its center and the horizontal line through its center are both mirror lines, so H has both horizontal and vertical line symmetry.On the other hand G doesn't have any lines of symmetry--you can't split it into two halves that are exactly the same on both sides of a line. What examples can you think of? What other capital letters can you think of that have vertical line symmetry? What capital letters have horizontal line symmetry? What capital letters have both horizontal and vertical line symmetry? What capital letters have no line symmetry? Possible answers Symmetry lines don't have to be vertical, either. So long as the shape is the same (reflected image) on both sides of the line, it can be a symmetry line. This butterfly and heart have line symmetry across the lines I have drawn in, but we can't call it either horizontal or vertical lines symmetry, it's just line symmetry. If you want to, you can call it diagonal line symmetry. There are more good examples of line symmetry at Rotational Symmetry Rotational symmetry is when you have a shape that if you turn it you get the same thing. This is a concept that's best show with movement, so go push the play buttons under the order 2 and order 3 examples on this page: There are 3 ways to name rotational symmetry. Order: The order of the rotation is the number of times you turn around (symmetrically) before you get back to where you started. On the left, you turn the shape twice before you get all the way back, so it has rotational symmetry of order 2. On thr right, you can turn the shape 3 times before you get all the way back, so it has rotational symmetry of order 3. Fraction of a turn: If you look at how far you turn the shape to get it to look the same, you can think of that as a fraction of how far you go to get all the way around. This fits in with fraction circles, so this vocabulary is often used in elementary school. On the left, you turn 1/2 way around, so it has half-turn symmetry.On the right, it turns 1/3 of the way around to look like itself again, so it has 1/3 turn symmetry. Angle: When children start measuring angles with protractors, and describing angles in degrees (often 5th grade), then they also start describing turns in terms of number of degress.So, the 1/2 turn becomes 1/2 of 360° =180°, and the 1/3 turn becomes 1/3 of 360°= 120°. Then we say the shape on the left has 180° rotational symmetry, and the shape on the right has 120° rotational symmetry. The most common kind of rotational symmetry is half-turn symmetry. Some of the capital letters have half-turn symmetry: Each of these letters (N, S and H) look the same if you rotate them upside down. You can see this for yourself by writing the letters on a piece of paper, and then turning it upside down (so the top of the paper is on the bottom, and the bottom on the top).H has 3 kinds of symmetry: horizontal line symmetry, vertical line symmetry and half-turn symmetry. It turns out that if a shape has two kinds of line symmetry (like horizontal and vertical), then it will also have rotational symmetry, just like H does.N and S have rotational symmetry, but they don't have any line symmetry. An interesting letter is O because it looks almost, but not quite like a circle. An O (and an X) are both just a little taller than they are wide, so a letter O is an oval and not a circle. If you turn an O by 90°, you'll find that it doesn't look quite like an O anymore.You have to turn the O a complete half turn to look just like itself again, so the letter O has half-turn symmetry. A circle, on the other hand, is so symmetric that it's hard to describe all of the symmetries. If you turn a circle through any angle around its center, it will be the exact same circle, so a circle has rotational symmetry through every angle (wow). A square is more symmetric than the letter O, but less symmetric than the circle.If you turn a square 90°, it will look exactly like it was to start with, but if you turn it by an angle less than 90°, it won't look the same. A square has 1/4-turn rotational symmetry (or 90°rotational symmetry). It also has 1/2 turn symmetry, and it would be OK to say it has 1/4 turn and 1/2 turn symmetry, but we don't have to because 1/2=2/4: any shape that has 1/4 turn symmetry also has 2/4 and 3/4 turn symmetry, and 2/4 = 1/2, so when we say a shape has 1/4 turn symmetry that automatically means that it also has 1/2 turn symmetry. I've also shown the reflection lines on the O and the square. The O has both horizontal and vertical line symmetry (2 symmetry lines), but because it's an oval and not a circle, if you put in a diagonal line, it won't quite work as a symmetry line. The square has 4 symmetry lines: horizontal, vertical and the two diagonal lines that go through the corners. If you take a square and fold it in half along the diagonal line that goes through the corners, you'll find that the triangular halves match up, so those are symmetry lines. (If I started drawing in symmetry lines in the circle, I wouldn't know where to stop, because any line I drew through the center of the circle would work.) What other capital letters have rotational symmetry? Possible answers Shapes with lots of symmetry Some shapes (like the square we already looked at) have more lines of symmetry than just 2 and/or have rotational symmetry about an angle that's not just a half-turn. For now, we'll just look at this snowflake as an example: Just the snowflake The snowflake and its symmetries Where to start: The easiest symmetry line to see is the vertical one--it goes between the top and bottom "fingers" of the snowflake. That should help you see that there's a symmetry line going down each opposite pair of "fingers" (that's 3 symmetry lines). The next place to look is half-way between the symmetry lines you have so far--between the "fingers"--is that a symmetry line? Yes! So that gives us the other 3 symmetry lines. (You could try the trick again, looking between adjacent symmetry lines, but that doesn't give you equal sides. Real life isn't perfect: I know, I know, some of the little fingers of the snowflake go over some of those last 3 symmetry lines--how can that be? Anything you find in nature--a butterfly, a snowflake, your face, ... is going to be not quite perfect. There's some randomness as well as order to how a snowflake puts itself together, so all of these cool examples from nature are going to be only approximately symmetric (like the capital B where the top and bottom are not quite perfectly the same size). There are times when studying the symmetry (even though it's only approximate) in nature is a useful thing to do, and there are times when studying how things in nature are different from perfectg symmetry is a useful thing to do. Today we're looking for the symmetry. Rotations? Let's look at the fingers again. All of the fingers of the snowflake look (approximately) alike--so a turn that kept it looking the same would rotate one finger to the next. If you look at all of those turns, there are 6 of them in a full turn, so the snowflake has 1/6-turn symmetry (it also has 2/6=1/3 turn symmetry, and 3/6 = 1/2 turn symmetry, but we only have to tell the 1/6 of the turn because everything else is made out of sixths).
17199	https://www.bbc.co.uk/bitesize/guides/zpjmjty/revision/2	Bitesize GCSE WJEC Statistical skills – WJECCalculating percentage increase and decrease Data is often used to show geographical information. Being able to use measures of data, make calculations and explore relationships is an essential geographical skill. Part of GeographyMathematical skills Save to My Bitesize Calculating percentage increase and decrease Calculating percentage increase Calculating percentage increase is an important skill for geographers to have. When geographers collect data over a period of time the results may increase. Calculating a percentage increase allows a geographer to see how much their data has changed. For example, it may be useful to find out how much the width of a river channel increases as you travel downstream. work out the differenceThe remainder left after subtracting one number from another. between the two numbers being compared divide the increase by the original number and multiply the answer by 100 in summary: percentage increase = increase ÷ original number × 100 For example, the number of robins in a woodland area is counted over two different months. In December 15 robins were counted. In January 23 robins were counted. What is the percentage increase of robins in the woodland? the difference between the two numbers is 8 8 ÷ 15 × 100 = 53.3 the percentage increase of robins found in the woodland is: 53.3% Calculating percentage decrease Calculating percentage decrease is also a useful skill to have. For example, it may be useful to find out how much the loadThe particles of rock carried by a river. particle size decreases in a river as you travel downstream. work out the difference between the two numbers being compared divide the decrease by the original number and multiply the answer by 100 in summary: percentage decrease = decrease ÷ original number × 100 For example, the number of robins in a woodland in February and March is counted. In February 22 robins were counted. In March 12 robins were counted. What is the percentage decrease of robins in the woodland? the difference between the two numbers is 10 10 ÷ 22 × 100 = 45.4 the percentage decrease of robins found in the woodland is: 45.4% Next page Relationships Previous page Measures of data More guides on this topic Graphical skills – WJEC Related links Geography: Exam practice Geography revision resources Personalise your Bitesize! Jobs that use Geography BBC Earth Quizlet Revision Buddies Subscription The Royal Geographical Society Seneca Learning

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